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)]
Proof:
(1) [Q(g(x))] from C1 and C2, x/f(x)
(2) [] from (1) and C3, x/g(x)

Source: Alechina, Natasha - School of Computer Science, University of Nottingham

Collections: Computer Technologies and Information Sciences