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

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

2 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)

3 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 }

4 Unification Example Knows(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

5 Unification Example Before beginning unification, all variable names must be unique UNIFY(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)

6 First Order Resolution

7 Convert to Clausal Form

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10 Example : Convert to Clausal Form

11 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.

12 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 dead 9. If someone dies, they remain dead.

13 Resolution Proof Axioms 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))

14 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)

15 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}

16 Try this Use Resolution prove that West is a criminal

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