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Summary: Brachman and Levesque, Chapter 4 exercise 1. Determine whether the
following sentence is valid using resolution:
xyz((P(y) Q(z)) (P(x) Q(x)))
Answer To do this we need to check if from the negation of the sentence
we can derive an empty clause (a contradiction).
First transfer the negation into clausal form:
¬xyz((P(y) Q(z)) (P(x) Q(x)))
¬xyz(¬(¬P(y) Q(z)) (¬P(x) Q(x)))
xyz¬(¬(¬P(y) Q(z)) (¬P(x) Q(x)))
xyz(¬¬(¬P(y) Q(z)) ¬(¬P(x) Q(x)))
xyz((¬P(y) Q(z)) (¬¬P(x) ¬Q(x)))
xyz((¬P(y) Q(z)) P(x) ¬Q(x))
x((¬P(f(x)) Q(g(x))) P(x) ¬Q(x))
Clauses:
C1 [¬P(f(x)), Q(g(x))]
C2 [P(x)]
C3 [¬Q(x)]
Note that the definition of the resolution rule (p.58 of the textbook) pre-
supposes that all clauses have distinct variables. (We can do this without loss
of generality because variables are universally quantified, and xP(x) is equiv-
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