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Which one of the following well-formed formulae in predicate calculus is NOT valid?
- (∀x p(x) ⇒ ∀x q(x)) ⇒ (∃x ¬p(x) ∨ ∀x q(x))
- (∃x p(x) ∨ ∃x q(x)) ⇒ ∃x (p(x) ∨ q(x))
- ∃x (p(x) ∧ q(x)) ⇒ (∃x p(x) ∧ ∃x q(x))
- ∀x (p(x) ∨ q(x)) ⇒ (∀x p(x) ∨ ∀x q(x))
Correct answer: ∀x (p(x) ∨ q(x)) ⇒ (∀x p(x) ∨ ∀x q(x))
Solution
The formula ∀x (p(x) ∨ q(x)) ⇒ (∀x p(x) ∨ ∀x q(x)) is not valid because it suggests that if every element satisfies either p or q, then every element must satisfy p or every element must satisfy q, which is not necessarily true; there can be elements that satisfy p and others that satisfy q.
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#include<stdio.h>
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