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Boolean Algebra cont’

The digital abstraction

��������������� ������������

����������

Theorem: Absorption Law

For every pair of elements a , b � B,

1. a + a · b = a

2. a · ( a + b ) = a

Proof:

(1)

abaaba ���� 1

� �ba �� 1

� �1�� ba

1�� aa�

Identity

Commutativity

Distributivity

Identity

Theorem: For any a � B, a + 1 = 1

(2) duality.

Theorem: Associative Law

In a Boolean algebra, each of the binary operations ( + ) and ( · ) is associative. That is, for every a , b , c � B,

1. a + ( b + c ) = ( a + b ) + c

2. a · ( b · c ) = ( a · b ) · c

� �� � � �� �cbacbaA ������

� �� � � �� �� �cbcbaacbaA �������Distributivity

� �� � � �� �cbaaacba �����

acabaa ���

aca��

a�

� � acbaa ���

Commutativity

Distributivity

Distributivity

Absorption Law

Absorption Law

acaba ���Idempotent Law

Proof:

(1) Let

� �� � � �� �� �cbcbaacbaA �������

� �� �� � � �� � � �� �ccbabcbacbcba ���������

� �� � � �� �cbabbcba �����

� � bcbab ���

bcbbba ���

bba��

bab��

b�

bcbba ���

Commutativity

Distributivity

Distributivity

Idempotent Law

Absorption Law

Commutativity

Absorption Law

� �� �� � � �� � � �� �ccbabcbacbcba ���������

c

Putting it all together:

� �� � � �� �� �cbcbaacbaA �������

� �� � � �� � � �� � ccbabcbaacba ���������

cba

� �cba ���

Same transitions

· before +

� � � �� � � �� �cbaccbabaA �������

� �� � � �� � � �� �cbaccbabcbaa ���������

� � cba ���

� � � �cbacbaA ������

(2) Duality

Also,

Theorem 11: DeMorgan’s Law

For every pair of elements a , b � B,

1. ( a + b )’ = a’ · b’

2. ( a · b )’ = a’ + b’

Proof:

(1) We first prove that (a+b) is the complement of a’ ·b’ .

Thus, (a+b)’ = a’ ·b’

By the definition of the complement and its uniqueness, it suffices to show: (i) (a+b)+(a’b’ ) = 1 and

(ii) (a+b)(a’b’ ) = 0.

(2) Duality (a·b)’ = a’+b’

� � � �� � � �� �bbaabababa ������������

� �� � � �� �bbaaab ��������

� �� � � �� �bbaaab ��������

� � � �11 ���� ab

11��

1�

Distributivity

Commutativity

Associativity

a’ and b’ are the complements of a and b respectively

Theorem: For any a � B, a + 1 = 1

Idempotent Law

� � � � � � � �babababa ���������

� � � �bbaaba ������

� � � �bbaaab ������

� � � �bbaaab ������

� � � �bbaaab ������

00 ������ ab

00��

0�

Commutativity

Distributivity

Commutativity

Associativity

Commutativity

a’ and b’ are the complements of a and b respectively

Theorem: For any a � B, a · 0 = 0

Idempotent Law

Algebra of SetsConsider a set S.

B = all the subsets of S (denoted by P(S)).

� �� ���,,SPM �

“plus” � set-union

“ times” � set-intersection

Additive identity element – empty set Ø

Multiplicative identity element – the set S.

Complement of X � B: XSX \��

Theorem: The algebra of sets is a Boolean algebra.

Proof:

By satisfying the axioms of Boolean algebra:

• B is a set of at least two elements

For every non empty set S:

|B| � 2.

� �SPS �,

• Closure of (

) and ( ) over B (functions ) . BBB ��

., SYX �� definitionby )(SPX �

definitionby )(SPY�

definitionby )( and SPYXSYX ����

definitionby )( and SPYXSYX ����

A1. Cummutativity of (

) and ( ).

YxXxxYX ���� or :

XxYxxXY ���� or :

An element lies in the union precisely when it lies in one of the two sets X and Y. Equally an element lies in the union precisely when it lies in one of the two sets X and Y. Hence,

YX �

XY�

XYYX ���

YxXxxYX ���� and :

XxYxxXY ���� and :

XYYX ���

A2. Distributivity of (

) and ( ).

� � � � � �ZXYXZYX ������

� � �.ZYXx ���

Xx� ZYx ��

Zx�Yx�

If ,Yx� Xx� Yx� YXx ��

YXx �� ZXx �� � � � �ZXYXx ����

Let

and

or

We have and . Hence,

If ,Zx� Xx� Zx� ZXx ��We have and . Hence,

or

� � � � � �ZXYXZYX ������

� This can be conducted in the same manner as

.

We present an alternative way:

Definition of intersection XYX �� XZX ��and

� � � � XZXYX ����

Also, definition of intersection YYX ��

definition of union ZYY ��ZYYX ���

Similarly, ZYZX ���

� � � � ZYZXYX �����

*

**

� � � � � �ZYXZXYX ������

Taking (* ) and (** ) we get,

�� � � � � �ZXYXZYX ������

Distributivity of union over intersection can be conducted in the same manner.

� � � � � �ZXYXZYX ������

A3. Existence of additive and multiplicative identity element.

XXXSX ������ .

XXSSXSX ������ .

identity additive -

identity tivemultiplica - S

A4. Existence of the complement.

XSXBX \. ����

� � � � SXXSXSXBX ������ \\.

� � � � ������ XXSXSXBX \\.

Algebra of sets is Boolean algebra.All axioms are satisfied

Dual transformation - Recursive definition:Dual: expressions expressionsbase: 0 1

1 0a a , a � B\{ 0,1}

recursion step: Let E1 and E2 be Boolean expressions.Then,

E1’ [dual(E1)]’( E1 + E2 ) [ dual(E1) · dual(E2) ] ( E1 · E2 ) [ dual(E1) + dual(E2) ]

Boolean expression - Recursive definition:base: 0 , 1 , a � B – expressions.recursion step: Let E1 and E2 be Boolean expressions.

Then,E1’( E1 + E2 ) ( E1 · E2 )

Proof:

Let f ( x1 , x2 , …, xn ) be a Boolean expression.

We show that applying the complement on the whole expression together with replacing each variable by it’s complement, yields the dual transformation definition.

Let fd be the dual of a function f ( x1 , x2 , …, xn )

Lemma: In switching algebra, fd = f’ ( x1’ , x2’ , …, xn’ )

Induction basis: 0 , 1 – expressions.

� � 0�xf� � �

���� ������ nxxxf ,,, 21 �

� ����� ������ nxxxf ,,, 21 �� � 1�xf

� � dfxf ������ 10�

� � dfxf ������ 01�

Induction hypothesis: Lemma holds for Boolean expressions: E1 and E2 .

That is:

� � � � � �� �nnn xxxExxxExxxE ,,,,,,,,, 21221121 ��� ��

� � � � � �� �������������� nnn xxxExxxExxxE ,,,,,,,,, 21221121 ���

Induction step: show that it is true for E1’( E1 + E2 ) ( E1 · E2 )

� �

� � dn

dn

ExxxE

ExxxE

,2212

,1211

,,,

,,,

�����

�����

����� �� hypothesis induction � � � �� �ndnd xxxExxxE ,,,,,, 21,221,1 �� ��

���� �� LawMorganDe' � � � �� �nn xxxExxxE ���������� ,,,,,, 212211 ��

If

then,

� �nd xxxE ,,, 21 ��

� � � � � �� �xExExE���

21 ��

� � � � � �� �������� xExExE���

21

����� �� hypothesis induction � � � �� �xExE dd

��,2,1 ��

���� �� LawMorganDe' � � � �� �xExE ��������

21

If

then,

� �xEd

��

� � � �xExE��

1��If

� � � �xExE ��������

1then,

� �� ����� xE�

1

� �� ��� xE d

�,1

� �xEd

��

����� �� hypothesis induction

Definition: A function f is called self-dual if f = fd

Lemma: For any function f and any two-valued variable A, the function g = Af + A’ fd is a self-dual.

Proof: (holds for any Boolean algebra)

� � � �dfAAfdualgdual ���

� � � �dfAdualAfdual ���

� � � �� � � � � �� �dfdualAdualfdualAdual �����

� � � �fAfA d �����

� � � � ffAAfA dd �����

� � � �dd fAffAA �����

Dual definition

Distributivity

Commutativity

� � � �dd fAffAA �����

dd fffAfAAA ������

dd fffAfA ����� 0

dd fffAfA ����

dd fffAAf ����

Notice that the above expression has the form:

ab + a’c +bc

where “a” =A, “b”=f, “c” = fd.

Distributivity

dd fffAfAAA ������Commutativity

A’ is the complement of A

Identity

Commutativity

We now prove a stronger claim:

caabbccaabBcba �������� .,,

111 bccaabbccaab �������

� �aabccaab ������ 11

abcbcacaab ������ 11

cbaabccaab ������ 11

11 cacbaababc ������

� � � �11 ����� bcacab

11 caab ���

caab ���

Identity

a’ is the complement of a

Distributivity

Commutativity

Commutativity

Distributivity

Theorem: For any a � B, a + 1 = 1

Identity

dd fffAAf ����

� � � �dfAAfdualgdual ���

dfAAf ���

caabbccaab ������

For example:

cvbf ��

� �vcbfd ��

� � � �� �vcbacvbag �����

self-dual

Easier proof (1) for switching algebra only: (using dual properties)

dd fffAAf ����

� � � �dfAAfdualgdual ���

Switching algebra

1 and 0 �� dff

0 and 1 �� dffOR 0�dff

0���� dfAAf

dfAAf ���Identity

A = 0dd ffffgdual ������ 00)(

dd ffff ����� 10

dd ffff ���� 0

ffff dd ���� 0

df�� 0

df�

dd fffg ������� �00

0’ = 1

Identity

Commutativity

Absorption Law

Theorem: For any a � B, a · 0 = 0

Identity

Easier proof (2) for switching algebra only: (case analysis)

dd fffAAf ����

� � � �dfAAfdualgdual ���

A = 1

dd ffffgdual ������ 11)(

f�

fffg d ������� �11

� � � �1,01,0: �f

� � decreasing monotone xf

� � � ��,0in decreasingstrictly xf �

� � � �1,in increasingstrictly �xf �

Example of a transfer function for an inverter

� � � � 0.,0 ����� xfx �

� � � � 0.1, ����� xfx �

� � � ��0, interval in the concave xf

� � � �,1 interval in theconvex is �xf

� �xf

� �xf

� � -1�� �f

� � -10 ��f

� � -11 ��f

� � continuous xf �

� � 1.! 11 ����� xfx �

� � 1.! 22 ����� xfx �

1

1

slope = -1

slope = -1

x

� �xf

1x � 2x

1

1�

outhighV ,

inhighV ,

outlowV ,

inlowV ,

slope = -1

slope = -1

x

� �xf

outlowinlowinhighouthigh VVVV ,,,, ���

true only if:

BUT, this is not always the case.

For example:

1

1� x

� �xf

outhighV ,

inhighV ,

outlowV ,

inlowV ,

slope = -1

slope = -1

outhighinhigh VV ,, �

Moreover, in this example it can be proved that no threshold values exist, which are consistent with definition 3 from lecture notes.

Using the assumption:

� � 00

2010

such that

:point a exists there

xxf

xxxx

��

f (x) = x

1

1

slope < -1

x

� �xf

1x 0x 2x

0,,,,

:start with

xVVVV outlowinlowinhighouthigh ��������

0x ��0x

� �0xf

� ���0xf

y�

x�

xy ���

:set �� ����� 0,, xVV inhighinhigh

�� ����� 0,, xVV inlowinlow

� � � ����� 0,, xfVfV inlowouthigh

� � � ����� 0,, xfVfV inhighoutlow

0x ��0x

� �0xf

� ���0xf

y�

x�

����� xy

� � ��� 0xf

��� 0x

� � ��� 0xf

��� 0x

f (x) = x

1

1

slope < -1

x

� �xf

1x 0x 2x

��� 0, xV inhigh

��� 0, xV inlow

��� 0, xV outhigh

��� 0, xV outlow

10,02min xxxx ����

true if:

10,02minset xxxx ���� �

f (x) = x

1

1

slope < -1

x

� �xf

1x 0x 2x

outhighV ,

inhighV ,

outlowV ,

inlowV ,

slope = -1

slope = -1outlowinlowinhighouthigh VVVV ,,,, ���

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