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Artificial Intelligence Tutorials

This tutorial discusses general unifiers and logical representations using first-order language. It also demonstrates how to translate sentences and proves statements using axioms and facts.

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Artificial Intelligence Tutorials

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  1. Artificial Intelligence Tutorials Tutorial # 7

  2. Q1: Give the most general unifier if it exists • P(A,B,B) and P(x,y,z): {x/A, y/B, z/B} • Q(y,G(A,B)) and Q(G(x,x),y): No unifier (x cannot bind to both A and B). • Older(Father(y),y) and older(Father(x),Ali): {y/Ali, x/Ali} • Knows(Father(y),y) and Knows(x,x): No unifier.

  3. Q2: Write down logical representations suitable for use with GMP • Horses, cows and pigs are mammals: Horse(x)  Mammal(x) Cow(x)  Mammal(x) Pig(x)  Mammal(x) • An offspring of a horse is a horse: Offspring(x,y)  Horse(y)  Horse(x) • Bluebeard is a horse: Horse (BlueBread) • Bluebeard is Charlie’s parent: Parent (BlueBread, Charlie) • Offspring and parent are inverse relations: Offspring(y, x)  Parent (x,y) Parent (x,y)  Offspring(y, x) • Every mammal has a parent: Mammal(x)  Parent(G(x), x) where G is Skolem function

  4. Q3: Two sentences in the language of first order language. • Translate: • For every natural number there is some other natural number that is smaller than or equal to it. • There is a particular natural number that is smaller than or equal to any natural number. • Yes. • Yes. • No, (A) does not logically entail (B) (A  B). • Yes, (B) logically entails (A) (B  A).

  5. Q4. Considering axioms • Translate into WFF: • All hounds howl at night:  h (Hound(h)  Howl(h)) • Anyone who has any cats will not have any mice:  x  c  z: Has(x, c)  Cat(c)  (Has(x,z)  Mice(z)) • Light sleepers do not have anything which howls at night:  x  h LS(x)  (Has(x,h)  Howl(h)) • Ali has either a cat or a hound: (Has(Ali, C)  Cat(C))  (Has(Ali, C)  Hound(C))

  6. Q4. Considering axioms, Cont... • Axiom • A:  h (Hound(h)  Howl(h)) • B:  x  c  z: Has(x, c)  Cat(c)  (Has(x,z)  Mice(z)) • C: x  h LS(x)  (Has(x,h)  Howl(h)) • D: (Has(Ali, C)  Cat(C))  (Has(Ali, C)  Hound(C)) • Fact • F1: LS(Ali) • Prove ((Have(Ali, z) ^ Mice(z))) • C, F1 gives *R1: LS(Ali)  (Has(Ali,h)  Howl(h)) • F1, R1 and GMP gives R2:  (Has(Ali, h) ^ Howl(h)) • A, R2 and resolution gives R3:  (Has(Ali, h) ^ Hound(h)) • D, R3 and resolution gives R4: Has(Ali, C) ^ Cat(C) • B, R4 and GMP gives **  (Has(Ali,z) ^ Mice(z)) (i.e. Ali does not have a mice).

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