boolean algebra and logic gates - basavaraj talawar · outline boolean algebra basic theorems,...
TRANSCRIPT
Boolean Algebra and Logic Gates
Chapter 2 ndash Boolean Algebra amp Logic GatesMano amp Ciletti 6ed
OutlineBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
George Boole
(1815 ndash 1864)
An Investigation of the Laws of Thought (1854)
Boolean Algebrandash Operations conjunction
( ) disjunction ( ) and) disjunction (or) or) negation (not)
ndash Fundamental in digital electronics
ndash Provided in all modern programming languages
Boolean Algebra Switching Algebra
ndash Shannon 1938 a two-valued Boolean algebra
ndash Represents properties of bistable electrical switching circuits
Boolean Algebra Algebraic structure
ndash Set of elements B
ndash Two binary operators + and ndash Huntington postulates are satisfied
Huntington Postulates Closure
ndash
Identity Elements (0 for + 1 for )
ndash
ndash
Commutative
ndash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash
ndash
Commutativendash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash x + y = y + x
ndash x y = y x
Huntington Postulates Distributive
ndash is distributive over +
ndash x (y + z) = (x y) + (x z)
ndash + is distributive over ndash x + (y z) = (x + y) (x + z)
Huntington Postulates there exists an element
(complement of x) such thatndash x + xrsquo = 1 and x xrsquo = 0
forall xisinB x isinB
Huntington Postulates There exist at least two elements
such thatx yisinB xne y
Two Valued Boolean AlgebraB=01
Binary operators + and
Two Valued Boolean Algebra
AND () OR (+) and NOT operations
B=01 Verify Closure Commutative
properties on B
Two Valued Boolean Algebra
B=01 Distributivexsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
OutlineBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
George Boole
(1815 ndash 1864)
An Investigation of the Laws of Thought (1854)
Boolean Algebrandash Operations conjunction
( ) disjunction ( ) and) disjunction (or) or) negation (not)
ndash Fundamental in digital electronics
ndash Provided in all modern programming languages
Boolean Algebra Switching Algebra
ndash Shannon 1938 a two-valued Boolean algebra
ndash Represents properties of bistable electrical switching circuits
Boolean Algebra Algebraic structure
ndash Set of elements B
ndash Two binary operators + and ndash Huntington postulates are satisfied
Huntington Postulates Closure
ndash
Identity Elements (0 for + 1 for )
ndash
ndash
Commutative
ndash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash
ndash
Commutativendash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash x + y = y + x
ndash x y = y x
Huntington Postulates Distributive
ndash is distributive over +
ndash x (y + z) = (x y) + (x z)
ndash + is distributive over ndash x + (y z) = (x + y) (x + z)
Huntington Postulates there exists an element
(complement of x) such thatndash x + xrsquo = 1 and x xrsquo = 0
forall xisinB x isinB
Huntington Postulates There exist at least two elements
such thatx yisinB xne y
Two Valued Boolean AlgebraB=01
Binary operators + and
Two Valued Boolean Algebra
AND () OR (+) and NOT operations
B=01 Verify Closure Commutative
properties on B
Two Valued Boolean Algebra
B=01 Distributivexsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
George Boole
(1815 ndash 1864)
An Investigation of the Laws of Thought (1854)
Boolean Algebrandash Operations conjunction
( ) disjunction ( ) and) disjunction (or) or) negation (not)
ndash Fundamental in digital electronics
ndash Provided in all modern programming languages
Boolean Algebra Switching Algebra
ndash Shannon 1938 a two-valued Boolean algebra
ndash Represents properties of bistable electrical switching circuits
Boolean Algebra Algebraic structure
ndash Set of elements B
ndash Two binary operators + and ndash Huntington postulates are satisfied
Huntington Postulates Closure
ndash
Identity Elements (0 for + 1 for )
ndash
ndash
Commutative
ndash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash
ndash
Commutativendash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash x + y = y + x
ndash x y = y x
Huntington Postulates Distributive
ndash is distributive over +
ndash x (y + z) = (x y) + (x z)
ndash + is distributive over ndash x + (y z) = (x + y) (x + z)
Huntington Postulates there exists an element
(complement of x) such thatndash x + xrsquo = 1 and x xrsquo = 0
forall xisinB x isinB
Huntington Postulates There exist at least two elements
such thatx yisinB xne y
Two Valued Boolean AlgebraB=01
Binary operators + and
Two Valued Boolean Algebra
AND () OR (+) and NOT operations
B=01 Verify Closure Commutative
properties on B
Two Valued Boolean Algebra
B=01 Distributivexsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Algebra Switching Algebra
ndash Shannon 1938 a two-valued Boolean algebra
ndash Represents properties of bistable electrical switching circuits
Boolean Algebra Algebraic structure
ndash Set of elements B
ndash Two binary operators + and ndash Huntington postulates are satisfied
Huntington Postulates Closure
ndash
Identity Elements (0 for + 1 for )
ndash
ndash
Commutative
ndash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash
ndash
Commutativendash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash x + y = y + x
ndash x y = y x
Huntington Postulates Distributive
ndash is distributive over +
ndash x (y + z) = (x y) + (x z)
ndash + is distributive over ndash x + (y z) = (x + y) (x + z)
Huntington Postulates there exists an element
(complement of x) such thatndash x + xrsquo = 1 and x xrsquo = 0
forall xisinB x isinB
Huntington Postulates There exist at least two elements
such thatx yisinB xne y
Two Valued Boolean AlgebraB=01
Binary operators + and
Two Valued Boolean Algebra
AND () OR (+) and NOT operations
B=01 Verify Closure Commutative
properties on B
Two Valued Boolean Algebra
B=01 Distributivexsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Algebra Algebraic structure
ndash Set of elements B
ndash Two binary operators + and ndash Huntington postulates are satisfied
Huntington Postulates Closure
ndash
Identity Elements (0 for + 1 for )
ndash
ndash
Commutative
ndash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash
ndash
Commutativendash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash x + y = y + x
ndash x y = y x
Huntington Postulates Distributive
ndash is distributive over +
ndash x (y + z) = (x y) + (x z)
ndash + is distributive over ndash x + (y z) = (x + y) (x + z)
Huntington Postulates there exists an element
(complement of x) such thatndash x + xrsquo = 1 and x xrsquo = 0
forall xisinB x isinB
Huntington Postulates There exist at least two elements
such thatx yisinB xne y
Two Valued Boolean AlgebraB=01
Binary operators + and
Two Valued Boolean Algebra
AND () OR (+) and NOT operations
B=01 Verify Closure Commutative
properties on B
Two Valued Boolean Algebra
B=01 Distributivexsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Huntington Postulates Closure
ndash
Identity Elements (0 for + 1 for )
ndash
ndash
Commutative
ndash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash
ndash
Commutativendash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash x + y = y + x
ndash x y = y x
Huntington Postulates Distributive
ndash is distributive over +
ndash x (y + z) = (x y) + (x z)
ndash + is distributive over ndash x + (y z) = (x + y) (x + z)
Huntington Postulates there exists an element
(complement of x) such thatndash x + xrsquo = 1 and x xrsquo = 0
forall xisinB x isinB
Huntington Postulates There exist at least two elements
such thatx yisinB xne y
Two Valued Boolean AlgebraB=01
Binary operators + and
Two Valued Boolean Algebra
AND () OR (+) and NOT operations
B=01 Verify Closure Commutative
properties on B
Two Valued Boolean Algebra
B=01 Distributivexsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash
ndash
Commutativendash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash x + y = y + x
ndash x y = y x
Huntington Postulates Distributive
ndash is distributive over +
ndash x (y + z) = (x y) + (x z)
ndash + is distributive over ndash x + (y z) = (x + y) (x + z)
Huntington Postulates there exists an element
(complement of x) such thatndash x + xrsquo = 1 and x xrsquo = 0
forall xisinB x isinB
Huntington Postulates There exist at least two elements
such thatx yisinB xne y
Two Valued Boolean AlgebraB=01
Binary operators + and
Two Valued Boolean Algebra
AND () OR (+) and NOT operations
B=01 Verify Closure Commutative
properties on B
Two Valued Boolean Algebra
B=01 Distributivexsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash x + y = y + x
ndash x y = y x
Huntington Postulates Distributive
ndash is distributive over +
ndash x (y + z) = (x y) + (x z)
ndash + is distributive over ndash x + (y z) = (x + y) (x + z)
Huntington Postulates there exists an element
(complement of x) such thatndash x + xrsquo = 1 and x xrsquo = 0
forall xisinB x isinB
Huntington Postulates There exist at least two elements
such thatx yisinB xne y
Two Valued Boolean AlgebraB=01
Binary operators + and
Two Valued Boolean Algebra
AND () OR (+) and NOT operations
B=01 Verify Closure Commutative
properties on B
Two Valued Boolean Algebra
B=01 Distributivexsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Huntington Postulates Closure
ndash The structure is closed wrt the operator + The structure is closed wrt the operator
Identity Elements (0 for + 1 for )ndash x + 0 = 0 + x = x
ndash x 1 = 1 x = x
Commutativendash x + y = y + x
ndash x y = y x
Huntington Postulates Distributive
ndash is distributive over +
ndash x (y + z) = (x y) + (x z)
ndash + is distributive over ndash x + (y z) = (x + y) (x + z)
Huntington Postulates there exists an element
(complement of x) such thatndash x + xrsquo = 1 and x xrsquo = 0
forall xisinB x isinB
Huntington Postulates There exist at least two elements
such thatx yisinB xne y
Two Valued Boolean AlgebraB=01
Binary operators + and
Two Valued Boolean Algebra
AND () OR (+) and NOT operations
B=01 Verify Closure Commutative
properties on B
Two Valued Boolean Algebra
B=01 Distributivexsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Huntington Postulates Distributive
ndash is distributive over +
ndash x (y + z) = (x y) + (x z)
ndash + is distributive over ndash x + (y z) = (x + y) (x + z)
Huntington Postulates there exists an element
(complement of x) such thatndash x + xrsquo = 1 and x xrsquo = 0
forall xisinB x isinB
Huntington Postulates There exist at least two elements
such thatx yisinB xne y
Two Valued Boolean AlgebraB=01
Binary operators + and
Two Valued Boolean Algebra
AND () OR (+) and NOT operations
B=01 Verify Closure Commutative
properties on B
Two Valued Boolean Algebra
B=01 Distributivexsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Huntington Postulates there exists an element
(complement of x) such thatndash x + xrsquo = 1 and x xrsquo = 0
forall xisinB x isinB
Huntington Postulates There exist at least two elements
such thatx yisinB xne y
Two Valued Boolean AlgebraB=01
Binary operators + and
Two Valued Boolean Algebra
AND () OR (+) and NOT operations
B=01 Verify Closure Commutative
properties on B
Two Valued Boolean Algebra
B=01 Distributivexsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Huntington Postulates There exist at least two elements
such thatx yisinB xne y
Two Valued Boolean AlgebraB=01
Binary operators + and
Two Valued Boolean Algebra
AND () OR (+) and NOT operations
B=01 Verify Closure Commutative
properties on B
Two Valued Boolean Algebra
B=01 Distributivexsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Two Valued Boolean AlgebraB=01
Binary operators + and
Two Valued Boolean Algebra
AND () OR (+) and NOT operations
B=01 Verify Closure Commutative
properties on B
Two Valued Boolean Algebra
B=01 Distributivexsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Two Valued Boolean Algebra
AND () OR (+) and NOT operations
B=01 Verify Closure Commutative
properties on B
Two Valued Boolean Algebra
B=01 Distributivexsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Two Valued Boolean Algebra
B=01 Distributivexsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Two Valued Boolean Algebra
B=01 Distributive xsdot( y+z )=(xsdoty)+(xsdotz )xsdot( y+z )=(xsdoty)+(xsdotz )
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Basic Theorems amp Postulates
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Basic Theorems amp Postulates
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Basic Theorems amp PostulatesEach pair is a Dual Interchange OR and AND
change 0 to 1 1 to 0 to obtain the other
Each pair is a Dual Interchange OR and AND change 0 to 1 1 to 0 to obtain the other
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
x + xrsquo = 1x + xrsquo = 1 x middot xrsquo = 0x middot xrsquo = 0bull Postulate 5
bull Complement of xrsquo = x
bull Complement of xrsquo = (xrsquo)rsquo
bull Complement is unique
bull (xrsquo)rsquo = x(xrsquo)rsquo = x
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Theorems
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
DeMorganrsquos Law
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
DeMorganrsquos Law
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
DeMorganrsquos Law
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Operator Precedence
1parentheses
2NOT
3AND
4OR
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Function to Gate Implementation
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Function
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Function
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Gate Implementation
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Gate Implementation
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Function
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Function
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Function
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Gate Implementation
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Optimized Gate Implementation
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Optimized Gate Implementation
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Optimized Gate Implementation
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Simplification
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Simplification
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Simplification
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Simplification
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Find Complements
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Find Complements
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Minterms Consider 2 variables x amp y Minterms
ndash All combinations of variablesndash Operation AND
x y x y xy xy
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Maxterms Consider 2 variables x amp y Maxterms
ndash All combinations of variablesndash Operation OR
Complements of corresponding minterms
x+ y x+ y x + y x + y
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Minterms amp Maxterms
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Minterms amp Maxterms
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Minterms amp Maxterms
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Function fromMinterms
Given a truth table form minterm for combinations that produce a 1 take the OR of these terms
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz f 2=x yz+xy z+xyz +xyzf 2=x yz+xy z+xyz +xyz
f 1=m1+m 4+m7f 1=m1+m 4+m7 f 2=m3+m5+m6+m7
f 2=m3+m5+m6+m7
f 1(x y z )=sum (147)f 1(x y z )=sum (147) f 2(x y z )=sum (3567)f 2(x y z )=sum (3567)
Canonical Form of a Boolean FunctionCanonical Form of a Boolean Function
Boolean Function fromMinterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
Boolean Function fromMinterms Maxterms
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Function fromMinterms Maxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1=(x y z + x yz + x yz+xy z+xyz ) f 1=(x y z + x yz + x yz+xy z+xyz )
f 1 =x y z +x yz +x yz+xy z+xyz f 1 =x y z +x yz +x yz+xy z+xyz
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Function fromMaxterms
f 1=x y z+xy z +xyzf 1=x y z+xy z +xyz
f 1 =x y z +x yz +x yz+xy z+x y zf 1 =x y z +x yz +x yz+xy z+x y z
f 1=(x y z + x yz + x yz+xy z+x y z )f 1=(x y z + x yz + x yz+xy z+x y z )
f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )f 1=(x+ y+z )sdot(x+ y +z )sdot(x+ y + z )sdot(x + y+z )sdot( x + y +z )
f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6f 1=M 0sdotM 2sdotM 3sdotM5sdotM 6
f 1=prod (02 3 5 6)f 1=prod (02 3 5 6)
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Function in Mintems
Express as Minterms
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Function in Mintems
Express as Minterms
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Express as Maxterms
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Express as Maxterms
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Express as Maxterms
F=prod (0245)F=prod (0245)
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Convert between Canonical Forms
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Interchanging Canonical Forms
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Standard Form
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
All Logic Operations
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
All Logic Operations
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
All Logic Operations
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
All Logic Operations
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Boolean Functions
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Digital Logic Gates
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Digital Logic Gates
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Digital Logic Gates
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
NAND NOR
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
NAND NOR
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
XOR XNOR
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
SummaryBoolean AlgebraBasic Theorems Huntington PostulatesDeMorganrsquos LawBoolean Functions ImplementationComplements DualsCanonical Forms Standard formsDigital Logic Gates
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal
Complement through Duals
Arrive at the complement byndash Derive the dual complement each
literal