logical representations and resolution. boolean logic conjunctive normal form resolution

Logical Representations and Resolution

Boolean Logic

•Conjunctive normal form

•Resolution

Conjunctive Normal Form

A literal is a variable or a negated variable.

A clause is either a single literal or the disjunction of two or more literals.

P, P P, and P Q R S are clauses.

(R S ) and P Q are not clauses.

A wff is in conjunctive normal form iff it is either a single clause or the conjunction of two or more clauses.

(P Q R S) (P R) is in cnf

(P Q R S) (P R) is not in cnf

Conjunctive Normal Form Theorem

Theorem: Every wff can be converted to an equivalent wff in conjunctive normal form.

Proof: By construction: conjunctiveBoolean(w: wff) = 1. Eliminate and from w, using the fact that P Q is

equivalent to P Q.2. Reduce the scope of each to a single term, using the facts:

• Double negation: (P) = P• deMorgan’s laws:

(P Q) (P Q) (P Q) (P Q)

3. Convert w to a conjunction of clauses using the fact that both and are associative and the fact that and distribute over each other.

Conversion to Conjunctive Normal Form

Let w be the wff :

P (R Q).

Then w can be converted to conjunctive normal form as follows:

Step 1 produces: P (R Q).

Step 2 produces: P (R Q).

Step 3 produces: (P R) (P Q).

Resolution

Suppose we know:

winter summerwinter summersummer nil

winter summerwinter cold summer cold

Resolution

From the premises: (P Q) and (R Q), Conclude: (P R).

Resolution is sound: If (P Q) and (R Q) are True, then:

• If Q is True, R must be True.• If Q True, P must be True.

Since either Q or Q must be True, P R must be True.

Resolution – The Algorithm

To prove ST from A:

1. Convert A to a list L of clauses in conjunctive normal form.

2. Construct ST and add it to L.3. Resolve.

Complementary LiteralsA pair of complementary literals is a pair of literals that are not

mutually satisfiable. So two literals are complementary iff one is positive, one is negative, and they contain the same propositional symbol. For example, Q and Q.

Choose two parent clauses that contain a pair of complementary literals. For example:

C1 = R1 R2 … Rj QC2 = S1 S2 … Sk Q.

Resolve C1 and C2 to derive:

R1 R2 … Rj S1 S2 … Sk

Resolution – Generating nil

Consider:

Q Q

nil

The empty clause is unsatisfiable since it contains no literals that can be made True. So if it is ever generated, the resolution procedure halts and reports that, since adding ST to A has led to a contradiction, ST is a theorem given A.

Resolution – The Algorithm

resolve-Boolean(A, ST) =1. Construct L, the list of of clauses from A. 2. Negate ST, convert the result to conjunctive normal form,

and add the resulting clauses to L.3. Until either nil is generated or no progress is being made

do:3. 1 Choose two parent clauses. 3. 2 Resolve the parent clauses together. 3. 3 If the resolvent is not nil and is not in L, add it to L.

4. If nil was generated, a contradiction has been found. Return success. ST must be true.

5. If nil was not generated and there was nothing left to do, return failure.

Resolution – An Example

Prove R given:

Given Axioms: Clauses: P P (P Q) R P Q R (S T) Q S Q

T Q T T

Add: R R

Resolution – An Example

PP Q RS QT QT R

P Q R R

P Q P

T Q Q

T T

nil

Resolution – Only Select One Pair to Resolve

P Q (1)P Q R (2)

Prove R:

P Q R R

P Q P Q

? nil

But is R entailed by the two facts we have been given?

Facts in English(1) Marcus was a man.

(2) Marcus was a Pompeian.

(3) All Pompeians were Romans.

(4) Caesar was a ruler.

(5) All Romans were either loyal to Caesar or hated him.

(6) Everyone is loyal to someone.

(7) People only try to assassinate rulers they are not loyal to.

(8) Marcus tried to assassinate Caesar.

An Example - Facts in FOL(1) Marcus was a man.

man(Marcus)(2) Marcus was a Pompeian.

Pompeian(Marcus)(3) All Pompeians were Romans.

x Pompeian(x) Roman(x)(4) Caesar was a ruler.

ruler(Caesar)(5) All Romans were either loyal to Caesar or hated him.

x Roman(x) loyalto(x, Caesar) hate(x, Caesar)(6) Everyone is loyal to someone.

x y loyalto(x, y)(7) People only try to assassinate rulers they are not loyal to.

x y person(x) ruler(y) tryassassinate(x, y) loyalto(x, y)(8) Marcus tried to assassinate Caesar.

tryassassinate(Marcus, Caesar)

Question Answering(1) Marcus was a man.


Pompeian(Marcus)(3) All Pompeians were Romans.

x Pompeian(x) Roman(x)(4) Caesar was a ruler.

ruler(Caesar)(5) All Romans were either loyal to Caesar or hated him.

x Roman(x) loyalto(x, Caesar) hate(x, Caesar)(6) Everyone is loyal to someone.

x y loyalto(x, y)(7) People only try to assassinate rulers they are not loyal to.

x y person(x) ruler(y) tryassassinate(x, y) loyalto(x, y)(8) Marcus tried to assassinate Caesar.

tryassassinate(Marcus, Caesar)

Was Marcus loyal to Caesar?

Reasoning Backward(1) man(Marcus)(2) Pompeian(Marcus)(3) x Pompeian(x) Roman(x)(4) ruler(Caesar)(5) x Roman(x) loyalto(x, Caesar) hate(x, Caesar)(6) x y loyalto(x, y)(7) x y person(x) ruler(y) tryassassinate(x, y) loyalto(x, y)(8) tryassassinate(Marcus, Caesar)

loyalto(Marcus, Caesar)

(Marcus/x) (Caesar/y)

person(Marcus) ruler(Caesar) tryassassinate(Marcus, Caesar)

Reasoning Backward(1) man(Marcus)(2) Pompeian(Marcus)(3) x Pompeian(x) Roman(x)(4) ruler(Caesar)(5) x Roman(x) loyalto(x, Caesar) hate(x, Caesar)(6) x y loyalto(x, y)(7) x y person(x) ruler(y) tryassassinate(x, y) loyalto(x, y)(8) tryassassinate(Marcus, Caesar)(9) x man(x) person(x)

loyalto(Marcus, Caesar)

(Marcus/x) (Caesar/y)

person(Marcus) ruler(Caesar) tryassassinate(Marcus, Caesar)

(Marcus/x)

man(Marcus)

Functions and Predicates(1) Marcus was a man.


Pompeian(Marcus)(3) Marcus was born in 40 A.D.

born(Marcus, 40)(4) All men are mortal.

x man(x) mortal(x)(5, 6) AllPompeians died when the volcano erupted in 79 A.D.

erupted(volcano, 79) x Pompeian(x) died(x,79) (7) No mortal lives longer than 150 years.

x t1 t2 mortal(x) born(x, t1) gt(t2-t1, 150) dead(x,t2)(8) It is now 2004.

now = 2004

Is Marcus alive?

Functions and Predicates(1) man(Marcus)(2) Pompeian(Marcus)(3) born(Marcus, 40)(4) x man(x) mortal(x)(5) erupted(volcano, 79) (6) x Pompeian(x) died(x,79) (7) x t1 t2 mortal(x) born(x, t1) gt(t2-t1, 150) dead(x,t2)(8) now = 2007

alive(Marcus, now)

Functions and Predicates - Filling in the Blanks(1) man(Marcus)(2) Pompeian(Marcus)(3) born(Marcus, 40)(4) x man(x) mortal(x)(5) erupted(volcano, 79) (6) x Pompeian(x) died(x,79) (7) x t2 t2 mortal(x) born(x, t1) gt(t2-t1, 150) dead(x,t2)(8) now = 2007(9a) x t alive(x, t) dead(x, t)(9b) x t dead(x,t) alive(x, t)(10) x t2 t2 died(x, t1) gt(t2, t1) dead(x, t2)

alive(Marcus, now)

Showing that Marcus is Not Alive

alive(Marcus, now) (9a) (Marcus/x) (now/t)

dead(Marcus, now) (10)(Marcus/x)(now/t2) (7) (Marcus/x)(now/t2) died(Marcus, t1) gt(now, t1) mortal(Marcus) born(Marcus, t1) gt(now-t1,150)

(5) (Marcus/x) (4) (Marcus/x) (3) (40/t1) subst (79/t1)Pompeian(Marcus) gt(now, 79) man(Marcus) born(Marcus, 40) gt(now-40,150) (2) subst (1) subst

T gt(2004, 79) T T gt(2007-40,150) eval eval

T T

A Harder One

Given:x [Roman(x) know(x, Marcus)]

[hate(x, Caesar) (y (z hate(y, z)) thinkcrazy(x, y))]Roman(Isaac)hate(Isaac, Caesar)hate(Paulus, Marcus)thinkcrazy(Isaac, Paulus)

Prove:know(Isaac, Marcus)

Prenex Normal Form

A sentence in first-order logic is in prenex normal form iff it is of the form:

<quantifier list> <matrix>,

where <quantifier list> is a list of quantified variables and <matrix> is quantifier-free.

x (y ((P(x) Q(y)) z (R(x, y, z))) is not in pnf.

x y z (P(x) Q(y)) R(x, y, z) is in pnf. Its matrix is:

(P(x) Q(y)) R(x, y, z).

Clause FormA sentence is in clause form iff:

• It has been converted to prenex normal form,• Its quantifier list contains only universal quantifiers, • Its quantifier list is no longer explicitly represented, • It is in conjunctive normal form, and• There are no variable names that appear in more than one

clause.

Begin with: x (P(x) Q(x)) y (S(y)) In prenex normal form: x y (P(x) Q(x)) S(y)In clause form: (P(x) Q(x)) S(y)

Clause Form Simplifies the Process

x [Roman(x) know(x, Marcus)] [hate(x, Caesar) (y (z hate(y, z)) thinkcrazy(x, y))]

becomes:

Roman(x) know(x, Marcus) hate(x, Caesar) hate(y, z) thinkcrazy(x, y)

Conversion to Clause Form - Step 1

1. Eliminate , using the fact that a b is equivalent to a b




2. Reduce the scope of each to a single term, using:(p) = p•deMorgan’s lawsx P(x) x P(x)x P(x) x P(x)


x [ Roman(x) know(x, Marcus)] [hate(x, Caesar) (y z hate(y, z) thinkcrazy(x, y))]


3. Standardize variables so that each quantifier binds a unique variable.

x P(x) x Q(x)

x P(x) y Q(y)


4. Move all quantifiers to the left without changing their relative order.

x [ Roman(x) know(x, Marcus)] [hate(x, Caesar) (y z hate(y, z) thinkcrazy(x, y))]

x y z [ Roman(x) know(x, Marcus)] [hate(x, Caesar) ( hate(y, z) thinkcrazy(x, y))]

At this point, we have prenex normal form.


5. Eliminate existential quantifiers through the use of Skolem functions and constants.

x Roman(x) Roman(S1)

x z father-of(x, z) x father-of(x, S2(x))


6. Drop the prefix since all remaining quantifiers are universal.

x y z [ Roman(x) know(x, Marcus)] [hate(x, Caesar) ( hate(y, z) thinkcrazy(x, y))]

[ Roman(x) know(x, Marcus)] [hate(x, Caesar) ( hate(y, z) thinkcrazy(x, y))]


7. Convert the matrix into a conjunction of disjuncts by using:•Associative properties of and .

[ Roman(x) know(x, Marcus)] [hate(x, Caesar) ( hate(y, z) thinkcrazy(x, y))]

Roman(x) know(x, Marcus) hate(x, Caesar) hate(y, z) thinkcrazy(x, y)

•Distributivity of and .

(P(x) Q(x)) T(x)

(P(x) T(x)) (Q(x)) T(x))


8. Create a separate clause for each conjunct.

(P(x) T(x)) (Q(x)) T(x))

(P(x) T(x)) (Q(x)) T(x))


9. Standardize apart the variables.

(P(x) T(x)) (Q(x)) T(x))

(P(x) T(x)) (Q(y)) T(y))

Resolution in FOLTo find a contradiction, we must show that the KB, augmented with

P, is unsatisfiable.

Herbrand’s theorem tells us:1. To show that a set of clauses S is unsatisfiable, it is necessary to

consider only interpretations over a particular set called the Herbrand universe of S, which is the set of all ground terms constructable from the following:1. The function symbols in S, if any.2. The constant symbols in S, if any. If none, then the constant

symbol A.

2. A set of clauses S is unsatisfiable iff a finite subset of ground instances (in which all bound variables have had a value substituted for them) of S is unsatisfiable.

Resolution is an algorithm that finds contradictions without enumerating most of the Herbrand universe.

Unification

In propositional logic, it is easy to identify complementary literals such as P and P.

But in FOL, what should we do about:x,y (hate(x, y)) hate(Marcus, Caesar)

Or: hate(Marcus, Caesar) hate(Paulus, Caesar)

Or: x,y (P(x, x)) P(Marcus, Caesar)

Before We StartBefore we consider a set of formulas and begin to reason with them, we need to standardize the variables apart:

Consider:x, y Lived-In(x, y)x, y Mother-Of(x, y)

Although we have used x and y in both of these formulas, the variables in these formulas are logically distinct since they occur inside the scope of quantifiers that are independent (i.e., neither lies inside the scope of the other). So an equivalent way to write these facts is:

x1, y1 Lived-In(x1, y1)x2, y2 Mother-Of(x2, y2)

UnificationWe need an algorithm that does two things given a sentence S and a term T:

•Determine whether T matches some term T´ in S, and•Return a substitution that can be applied to the remainder of S.

Example:

S: x,y In(x, y) Climate(y, moderate) Climate(x, moderate)T: In(Italy, Europe)

The process we will call unification will determine that T matches the first term of S and it will return the substitution Italy/x, Europe/y. When we apply that to the remainder of S, we have:

Climate(Europe, moderate) Climate(Italy, moderate)

The Unification Algorithm

unify-for-resolution(lit1, lit2) = If either lit1 or lit2 is a variable or a constant then:

Case (checking the conditions in order and executing only the first one that matches):

lit1 and lit2 are identical: return nil. /* Succeed with no substitution. lit1 is a variable that occurs in lit2: return Fail. /* These two cases implement the lit2 is a variable that occurs in lit1: return Fail. occur check. lit1 is a variable: return (lit2/lit1). lit2 is a variable: return (lit1/lit2). otherwise: return Fail. /* No match. If the initial predicate or function symbols of lit1 and lit2 are not the same, return Fail. If lit1 and lit2 do not have the same number of arguments, return Fail. substitution-list = nil. For i = 1 to the number of arguments of lit1 do

Let S be the result of unify-for-resolution on the ith argument of lit1 and of lit2. If S contains Fail, return Fail. If S is not equal to nil then:

Apply S to the remainder of both lit1 and lit2. Append S to substitution-list.

Return substitution-list.

The Occur CheckSuppose we are trying to unify: P(x, x)

P(g(x), g(x))

What happens if we skip the occur check?

x and g(x) unify and yield the substitution g(x)/x.

Now we must apply that substitution :

g(x) g(g(x)) g(g(x)) g(g(g(x)))

But if we’d standardized apart the variables before we started, this wouldn’t have been a problem.

The Occur Check is Necessary Even with Standardizing Apart

Suppose we are trying to unify: P(x, f(x), x)P(f(a), a, a)

x and f(a) unify and yield the substitution f(a)/x. Applying it:

f(f(a)), f(a) a, a

Without the occur check, f(f(a)) and a will unify, yielding the substitution f(f(a))/a. To apply it, we must replace all a’s in the remaining terms. But that process never terminates.

Most General Unifiers (MGUs)

Suppose we are trying to prove x,y (A(x, y) B(x, y))

Using the fact z (A(John, z))

Unifying A(x, y) with A(John, z) yields John/x, z/y. Applying it: z (B(John, z))

But we could also have matched with the substitution John/x, John/y, and been left trying to prove B(John, John), which would be harder.

The algorithm we have presented always returns the most general unifier. The MGU is unique up to variable name substitution.

Unification Examples

Inputs [1] Roman(x), Roman(Paulus) [2] Roman(x), Ancient(Paulus) [3] Roman(father-of(Marcus)), Roman(x) [4] Roman(father-of(Marcus)), Roman(Flavius), [5] Roman(x), Roman(y), [6] Roman(father-of(x)), Roman(x) [7] Likes(x, y), Likes(Flavius, Marcus)

Result

Succeed

Fail

Succeed

Fail

Succeed

Fail (fails occur check)Succeed

Substitution

Paulus/x

father-of(Marcus)/x

x/y

Flavius/x, Marcus/y

FOL Resolution – An Example

Prove: Mortal(Marcus) given:

x (Man(x) Mortal(x)) Man(x) Mortal(x)Man(Marcus) Man(Marcus)

Add: Mortal(Marcus)

Man(x) Mortal(x) Mortal(Marcus)

Marcus/x

Man(Marcus) Man(Marcus)

nil

FOL Resolution – the Algorithmresolve-FOL(A, ST) =Construct L, the list of clauses from A.Rename all variables in ST so that they do not conflict with any variables in L.Negate ST, convert the result to clause form, and add the resulting clauses to L. Until either nil is generated or no progress is being made do:

Choose from L two (parent) clauses that contain a pair CL of complementary literals.

Resolve the parent clauses together to produce a resolvent: Initially, let the resolvent be the disjunction of all of the literals in both

parent clauses except for the two literals in CL.

Apply to all of the literals in the resolvent the substitution that was constructed when the literals in CL were unified.

Rename all of the variables in the resolvent so that they do not conflict with any of the variables in L.

If the resolvent is not nil and is not already in L, add it to L.If nil was generated, a contradiction has been found. Return success. ST must be true.If nil was not generated and there was nothing left to do, return failure. ST may or may

not be true. But no proof of ST has been found.

Heuristics to Aid the Resolution Procedure

•Only resolve pairs of clauses that contain complementary literals. P(x) Q(x)

•Eliminate certain clauses as soon as they are generated:•Tautologies•Clauses that are subsumed by other clauses.

•Set of support strategy: whenever possible resolve with a clause that is part of the statement we are trying to prove.

•Unit preference strategy: whenever possible, resolve with clauses that have a single literal.

Subsumption

Clause A subsumes clause B iff B must be true in any interpretation in which A is true. (A is more general than B.)

True

P Q R T W

P Q R T

P Q

P Q

P Q W

Subsumption in FOL

Clause A subsumes clause B iff B must be true in any interpretation in which A is true. (A is more general than B.)

True

x (P(x))

P(Marcus)

x(P(x))

Using ResolutionPainter(Leonardo)Composer(Palistrina)Country(Italy)Lived-In(Leonardo, Italy)In(Italy,Europe)Climate(Europe, moderate)Painter(Rubens)Lived-In(Rubens, Europe)x,y (In(x, y) Climate(y, moderate) Climate(x, moderate))

Prove that Leonardo was a painter who lived in a moderate climate.

Prove: y (Painter(Leonardo) Lived-In(Leonardo,y) Climate(y, moderate))

Negate it: (y (Painter(Leonardo) Lived-In(Leonardo,y) Climate(y, moderate)))

Converting to Clause FromPainter(Leonardo)Composer(Palistrina)Country(Italy)Lived-In(Leonardo, Italy)In(Italy,Europe)Climate(Europe, moderate)Painter(Rubens)Lived-In(Rubens, Europe)

x,y (In(x, y) Climate(y, moderate) Climate(x, moderate)) In(x, y) Climate(y, moderate) Climate(x, moderate)

(y (Painter(Leonardo) Lived-In(Leonardo,y) Climate(y, moderate))) y ((Painter(Leonardo) Lived-In(Leonardo,y) Climate(y, moderate)))

Painter(Leonardo) Lived-In(Leonardo,y) Climate(y, moderate)

Using Resolution1. Painter(Leonardo) 2. Composer(Palistrina)3. Country(Italy) 4. Lived-In(Leonardo, Italy)5. In(Italy,Europe) 6. Climate(Europe, moderate)7. Painter(Rubens) 8. Lived-In(Rubens, Europe)9. Painter(Sargent)10. In(x, y) Climate(y, moderate) Climate(x, moderate)

Painter(Leonardo) Lived-In(Leonardo,y) Climate(y, moderate) Painter(Leonardo)

Leonardo/xLived-In(Leonardo, Italy) Lived-In(Leonardo,y1) Climate(y1, moderate)

Italy/y1

Climate(Italy, moderate)

Using Resolution1. Painter(Leonardo) 2. Composer(Palistrina)3. Country(Italy) 4. Lived-In(Leonardo, Italy)5. In(Italy,Europe) 6. Climate(Europe, moderate)7. Painter(Rubens) 8. Lived-In(Rubens, Europe)9. Painter(Sargent)10. In(x, y) Climate(y, moderate) Climate(x, moderate)

Painter(Leonardo) Lived-In(Leonardo,y) Climate(y, moderate) Painter(Leonardo)

Leonardo/xLived-In(Leonardo, Italy) Lived-In(Leonardo,y1) Climate(y1, moderate)

Italy/y1

In(x, y) Climate(y, moderate)Climate(x, moderate) Climate(Italy, moderate)

Italy/xIn(Italy,Europe) In(Italy, y2) Climate(y2, moderate)

Europe/y2

Climate(Europe, moderate) Climate(Europe, moderate)

nil

Answering QuestionsA Question: Name a painter who lived in a moderate climate.

?Painter(x) Lived-In(x,y) Climate(y, moderate)

Construct an existence statement to be proved:

x, y (Painter(x) Lived-In(x,y) Climate(y, moderate))

Negate it and resolve, tracking the bindings:

(x, y (Painter(x) Lived-In(x,y) Climate(y, moderate))) x,y ((Painter(x) Lived-In(x,y) Climate(y, moderate))) Painter(x) Lived-In(x,y) Climate(y, moderate)

Tracking the Bindings

Painter(x) Painter(x) Lived-In(x,y) Climate(y, moderate) Painter(Leonardo)

Leonardo/xPainter(Leonardo) Lived-In(Leonardo,y1) Climate(y1, moderate)

Lived-In(Leonardo, Italy) Italy/y1

Painter(Leonardo) Climate(Italy, moderate)

In(x, y) Climate(y, moderate)Climate(x, moderate) Italy/x

In(Italy,Europe) Painter(Leonardo) In(Italy, y2) Climate(y2, moderate)

Europe/y2

Climate(Europe, moderate) Painter(Leonardo) Climate(Europe, moderate)

Painter(Leonardo)

A Reminder About Standardizing Variables Apart

Prove father(Chris, Bill) given:father(x, y) woman(x) (1)

{father(x, y) woman(x)}mother(x, y) woman(x) (2)

{mother(x, y) woman(x)}mother(Chris, Mary) (3)

Add: father(Chris, Bill)

father(x, y) woman(x) mother(x, y) woman(x)

* father(x, y) mother(x, y) mother(Chris, Mary)

Chris/x, Mary/y

father(Chris, Mary)

Back to Marcus

(1) man(Marcus)(2) Pompeian(Marcus)(3) x Pompeian(x) Roman(x)(4) ruler(Caesar)(5) x Roman(x) loyalto(x, Caesar) hate(x, Caesar)(6) x y loyalto(x, y)(7) x y man(x) ruler(y) tryassassinate(x, y) loyalto(x, y)(8) tryassassinate(Marcus, Caesar)

(1) man(Marcus)(2) Pompeian(Marcus)(3) Pompeian(x1) Roman(x1)(4) ruler(Caesar)(5) Roman(x2) loyalto(x2, Caesar) hate(x2, Caesar)(6) loyalto(x3, S1(x3))(7) man(x4) ruler(y1) tryassassinate(x4, y1) loyalto(x4, y1)(8) tryassassinate(Marcus, Caesar)

Proving Marcus Not Loyal to Caesar

loyalto(M, C) man(x4) ruler(y1) tryassassinate(x4, y1) loyalto(x4, y1)

(M/x4)(C/y1)

man(M) ruler(C) tryassassinate(M,C) man(M)

ruler(C) tryassassinate(M,C) ruler(C)

tryassassinate(M,C) tryassassinate(M,C)

nil

Does Isaac Know Marcus?

Given:1. x [Roman(x) know(x, Marcus)]

[hate(x, Caesar) (y (z hate(y, z)) thinkcrazy(x, y))] Roman(x) know(x, Marcus) hate(x, Caesar)

hate(y, z) thinkcrazy(x, y)

2. Roman(Isaac)3. hate(Isaac, Caesar)4. hate(Paulus, Marcus)5. thinkcrazy(Isaac, Paulus)

Prove:know(Isaac, Marcus) * know(Isaac, Marcus)

Does Marcus Hate Caesar?Try to prove hate(M, C):

Hate(M, C)

Try to Prove that Marcus Does Not Hate Caesar

Try to prove hate(M, C):

Hate(M, C)

Try (Again) to Prove that Marcus Does Not Hate Caesar

Now try to prove hate(M, C):

Hate(M, C)

(1) man(Marcus)(2) Pompeian(Marcus)(3) Pompeian(x1) Roman(x1)(4) ruler(Caesar)(5) Roman(x2) loyalto(x2, Caesar) hate(x2, Caesar)(6) loyalto(x3, S1(x3))(7) man(x4) ruler(y1) tryassassinate(x4, y1) loyalto(x4, y1)(8) tryassassinate(Marcus, Caesar)(9) persecute(x5, y2) hate(y2, x5) persecute(x, y) hate(y, x)(10) hate(x6, y3) persecute(y3, x6) hate(x, y) persecute(y, x)

A Winnie Example

"And if anyone knows anything about anything," said Bear to himself,"It's Owl who knows something about something," he said, "Or my nameis not Winnie-the-Pooh," he said. "Which it is," he added."So there you are."

When Does Resolution Work?

Resolution is sound and refutation-complete.

If there is a contradiction, resolution will eventually find it if resolvents are chosen systematically.

But:

•It may take exponential time.

•Satisfiability is semi-decidable: if a set of clauses is satisfiable, resolution will never find a conflict and it may not notice that it is failing to do so.

Satisfiability is Semi Decidable

Suppose we want to know whether P(x) is true. There are three possibilities:

(1) P(x) is entailed by KB and thus:

P(x) KB is unsatisfiable.

(2) P(x) is entailed by KB and thus:

P(x) KB is unsatisfiable.

(3)

Horn Clauses

A Horn Clause is a clause in which at most one literal is positive.

A(x) B(x) C(x) becomes

A(x) B(x) C(x) a Horn clause

Deciding entailment with Horn clauses can be done in time that is linear in the size of the KB by using either backward or forward chaining.

logical representations and resolution. boolean logic conjunctive normal form resolution

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