CS.462 Artificial Intelligence SOMCHAI THANGSATHITYANGKUL Lecture 06 : First Order Logic Resolution Prove.

CS.462Artificial Intelligence

SOMCHAI THANGSATHITYANGKUL

Lecture 06 : First Order Logic Resolution Prove

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Proving Theorem in FOL

• Proving theorem in FOL raises two additional problems:– Need to remove quantifiers– Need to be able to compare variable

(which represent unknown values)

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Substitution and Unification

• A substitution is a set of variable/ term pairs that one can apply to sentences to perform some needed unification or other manipulation

SUBST(, ) denoted the result of applying substitution to sentence

e.g., SUBST({x/A, y/ f(x)}, P (x,y)) = P(A, f(A))

• Unification is the process of taking 2 atomic statements and finding a substitution that make them equivalent

UNIFY(p,q) = where SUBST(, p) = SUBST(, q)

e.g. UNIFY(P(x), P(A) = { x/A }

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Unification ExampleKnows(John, x) Hates(John, x)Knows(John, Jane)Knows(y, Leonid)Knows(y, Mother(y))Knows(x, Elizabeth)

UNIFY(Knows(John, x), Knows(John, Jane)) = {x/John}

UNIFY(Knows(John, x), Knows(y, Leonid)) = {x/John, y/Leonid}

UNIFY(Knows(John, x), Knows(y, Mother(y))) = {y/John, x/Mother(John)}

UNIFY(Knows(John, x), Knows(x, Elizabeth) = fail

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Unification Example• Before beginning unification, all variable

names must be uniqueUNIFY(Knows(John, x1), Knows(x2, Elizabeth) = {x1/Elizabeth, x2/John}

• Unification makes as few commitments about variable bindings as possible

UNIFY(Knows(John, x), Knows(y, z) = {y/John, x/z}Or {y/John, x/z, w/Freda}Or {y/John, x/John, z/John}Or …

{y/John, x/z} is the most general unifier (MGU)

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First Order Resolution

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Convert to Clausal Form

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Convert to Clausal Form

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Convert to Clausal Form

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Example : Convert to Clausal Form

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Resolution Using Unification

Given: P(A) (x) (P(x) Q(f(x))), prove: (z) (Q(z))• Negate the theorem: (z) (Q(z))• Drop universal quantifiers and separate into disjunctive

clauses.1. P(A) 2. (P(x) Q(f(x)))3. Q(z)

• Clause 1 and 2 are resolved by substituting A for x {x/A}. Resulting clause is Q(f(A)), which is added to original set.

• Q(f(A)) is resolved against clause 3, using {z/f(A)}. Result is Q(f(A)) and Q(f(A))

• Contradiction proves original theorem.

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Resolution Proof• Question (informal): Is Marcus alive in the year 2001?• Axioms (informal)

1.Marcus was a man.2.Marcus was a Pompeian.3.Marcus was born in 40 A.D.4.All man are mortal.5.The volcano erupted in 79 A.D.6.All Pompeians died when the volcano erupted.7.No mortal lives longer than 150 years.

• Additional axioms are required.

8. Alive means not dead9. If someone dies, they remain dead.

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Resolution ProofAxioms in clause form

1. Man(Marcus)2. Pompeian(Marcus)3. Born(Marcus, 40)4. Man(x1) Mortal(x1)

from... x Man(x) Mortal(x)5. Erupted(Volcano, 79)6. Pomepeian(x2) Died(x2, 79)

from... x Pompeian(x) Died(x, 79)7. Mortal(x3) Born(x3, t1) Gt(t2-t1, 150)

Dead(x3, t2)from... x,t1,t2 (Mortal(x)Born(x,t)Gt(t2-t1, 150) Dead(x, t2))

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Resolution Proof

8a. Alive(x4, t3) Dead(x4, t3)

8b. Dead(x4, t3) Alive(x4, t3)

from... x,t Alive(x, t) Dead(x, t)9. Died(x6, t5) Gt(t6,t5) Dead(x6,t6)

from... x,t1,t2 Died(x, t1) Gt(t2,t1) Dead(x, t2)

Prove: alive(Marcus, 2001)

Negate clause to be proved and add to list:alive(Marcus, 2001)

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Alive(Marcus, 2001) Alive(x4, t3)Dead(x4, t3)

Dead(x4, 2001)

{x4/Marcus, t3/2001} Mortal(x3) Born(x3, t1) Gt(t2-t1, 150) Dead(x3, t2)

Mortal(Marcus) Born(Marcus, t1) Gt(2001-t1, 150)

Man(x1) Mortal(x1)

Man(Marcus) Born(Marcus, t1) Gt(2001-t1, 150)

Man(Marcus)

Born(Marcus, t1) Gt(2001-t1, 150) Born(Marcus, 40)

Gt(2001-40, 150)

Gt(1961, 150)

{x3/Marcus, t2/2001}

{x1/Marcus}

{t1/40}

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Try this• Use Resolution prove that West is a

criminal