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FOL: Extra problems to solve

FOL: Extra problems to solve. IT 422: intelligent systems. Tutorial #7: Q4. For each pair of atomic sentences, give the most general unifier if it exists : P(a,b,b ), P(x, y, z). { x /a , y /b , z /b } b. Q(y, G(a, b)), Q(G(x, x), y).

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FOL: Extra problems to solve

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  1. FOL: Extra problems to solve IT 422: intelligent systems

  2. Tutorial #7: Q4 • For each pair of atomic sentences, give the most general unifier if it exists: • P(a,b,b), P(x, y, z). { x /a , y /b , z /b } b. Q(y, G(a, b)), Q(G(x, x), y). No unifier ({y/G(A,B),G(A,B)/G(x,x)}? x cannot take both the values a and b.

  3. Tutorial #7: Q4 c. Older(Father(y), y), Older(Father(x), John). { y / John , x / John } d. Knows(Father(y), y), Knows(x, x). fail. Start with (partial) substitution x/father(y). One clause is now knows(father(y),y), and the other knows(father(y),father(y)). Unification fails here because we can’t unify y and father(y), due to the occurs check.

  4. Tutorial #7: Q7 Consider the following axioms: • All hounds howl at night. • Anyone who has any cats will not have any mice. • Light sleepers do not have anything which howls at night. • Ali has either a cat or a hound. • Using FOL translate each of the above axioms into a well formed formula (WFF). • Given the fact that Ali is a light sleeper, give the sequence of inference rules to prove that Ali does not have any mice.

  5. Q7: Translating the sentences into WFF • All hounds howl at night. R1: ∀ x (Hound(x) ⇒ Howl(x)) • Anyone who has any cats will not have any mice. R2: ∀ x ∀y ∀z (Have(x, y) ^ Cat(y) ⇒ ¬(Have(x, z) ^ Mouse(z))) • Light sleepers do not have anything which howls at night. R3: ∀ x ∀y (LS(x) ⇒ ¬(Have(x, y) ⋀Howl(y)) • Ali has either a cat or a hound. R4: Ǝx(Have(Ali, x) ^ (Cat(x) V Hound(x)))

  6. Q7: Ali does not have any mice? Cont. • Given the fact that Ali is a light sleeper, give the sequence of inference rules to prove that Ali does not have any mice. • Translate: (if Ali is a light sleeper then Ali does not have any mice)into WFF R5: LS(Ali) ⇒¬Ǝz (Have(Ali, z) ^ Mouse(z))

  7. Q7: Ali does not have any mice? Cont.Negated conclusion • To prove LS(Ali) ⇒¬Ǝz (Have(Ali, z) ^ Mouse(z)) • Negate the conclusion, and convert the premise and the negated conclusion into conjunctive normal form. • Use resolution to show that the conclusion follows from the premise ( you should find a contradiction, an empty clause).

  8. Q7: CNF Write the set of sentences (KB) as CNF in order to use the resolution rule. • ∀ x (Hound(x) ⇒ Howl(x) ¬Hound(x) V Howl(x) • ∀ x ∀y ∀z (Have(x, y) ^ Cat(y) ⇒ ¬(Have(x, z) ^ Mouse(z))) ¬Have(x, y) V ¬Cat(y) V ¬Have(x, z) V ¬Mouse(z) • ∀ x ∀y (LS(x) ⇒ ¬(Have(x, y) ⋀Howl(y)) ¬LS(x) V ¬Have(x, y) V ¬Howl(y) • Ǝx (Have(Ali, x) ^ (Cat(x) V Hound(x))) Have(Ali, A) ^(Cat(A) V Hound(A)) (Skolemize, it is CNF)

  9. Q7: CNF 5. ¬R5 (negated conclusion) ¬ (LS(Ali) ⇒¬Ǝz (Have(Ali, z) ^ Mouse(z))) ¬ (¬LS(Ali) V (¬Ǝz (Have(Ali, z) ^ Mouse(z)))) LS(Ali) ^Ǝz (Have(Ali, z) ^ Mouse(z) LS(Ali) ^Have(Ali, B) ^ Mouse(B) (Skolemize) • It is written in CFN

  10. Q7: The set of clauses obtained The set of clauses are: 1. ¬Hound(x) V Howl(x) 2. ¬Have(x, y) V ¬Cat(y) V ¬Have(x, z) V ¬Mouse(z) 3. ¬LS(x) V ¬Have(x, y) V ¬Howl(y) 4. (a) Have(Ali, A) (b) Cat(A) V Hound(A) 5. (a) LS(Ali) (b) Have(Ali, B) (c) Mouse(B)

  11. Q7: Apply the resolution rule ¬Hound(x) V Howl(x), Cat(A) V Hound(A) Subst(θ, Howl(x) V Cat(A)) where θ={x/A} The conclusion is Howl(A) V Cat(A)

  12. Q7: Apply the resolution rule ¬Have(x, y) V ¬Cat(y) V ¬Have(x, z) V ¬Mouse(z), Mouse(B) Subst(θ, ¬Have(x, y) V ¬Cat(y) V ¬Have(x, z)) Where θ={z/B} The conclusion is: ¬Have(x, y) V ¬Cat(y) V ¬Have(x, B)

  13. Q7: Apply the resolution rule ¬Have(x, y) V ¬Cat(y) V ¬Have(x, B), Have(Ali,B) Subst(θ, ¬Have(x, y) V ¬Cat(y)) Where θ={x/Ali} The conclusion ¬Have(Ali, y) V ¬Cat(y)

  14. Q7: Apply the resolution rule Howl(A) V Cat(A), ¬Have(Ali, y) V ¬Cat(y) Subst(θ, Howl(A) V ¬Have(Ali, y)) Where θ={y/A} is Howl(A) V ¬Have(Ali, A)

  15. Q7: Apply the resolution rule Have(Ali,A), Howl(A) V ¬Have(Ali, A) Howl(A) ¬LS(x) V ¬Have(x, y) V ¬Howl(y), Howl(A) Subst(θ, ¬LS(x) V ¬Have(x, y)) Where θ={y/A} is¬LS(x) V ¬Have(x, A)

  16. Q7: Apply the resolution rule ¬LS(x) V ¬Have(x, A), Have(Ali,A) Subst(θ, ¬LS(x)) Where θ={x/Ali} is ¬LS(Ali) And finally, ¬LS(Ali), LS(Ali) – emptyContradiction Ali does not have any mice

  17. Problem 1 • Consider the following knowledge base: • 1. boss(sami) = nabil • 2. boss(nabil) = fatima • 3. paycut(fatima) • 4. x (paycut(boss(x)) paycut(x)) Draw a resolution proof to show that Sami is getting a paycutie • 5: paycut(sami)?.

  18. Problem 1: solution • CNF : • paycut(boss(x)) paycut(x) paycut(boss(sami)) • 6 : paycut(boss(sami))  paycut(boss(sami)) unify {x/sami} ,5 • 7: paycut(boss(sami)) 5+ 6 • 8: paycut(nabil) 1+7 • 9: paycut(boss(nabil)) unify {x/nabil} , 5+9 • 10: paycut(fatima) 2+9 • 11: {} 10+3

  19. Problem 2 Express the following sentences in first-order logic: (a) All students love AI (b) Some students love AI (c) At least two students love AI (d) Exactly two students love AI (e) At most two students love AI

  20. Problem 2: solution • All students love AI • ∀ x Student(x) => Loves(x, AI) • Some students love AI • Ǝ x Student(x)  Loves(x, AI) • At least two students love AI  • Ǝ x,y ┐(x=y)  Student(x)  Loves(x,AI)  Student(y)  Loves(y,AI) • or • ∃x∃y (Student(x)  Loves(x,AI )∧ Student(y)  Loves(y,AI) ∧ x ≠ y • Exactly two students love AI • Ǝ x,y (x≠y Student(x)  Loves(x,AI)  Student(y)  Loves(y,AI)  ∀ z (Student(z)  Loves(z,AI) < --> (x=z V y=z)) ) • At most two students love AI • ∀ x,y ,z Student(x)  Loves(x,AI)  Student(y)  Loves(y,AI)  Student(z)  Loves(z,AI)  (x=y Vx=z V y=z)

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