Predicate calculus validity.

Question

Is this predicate calculus statement , valid?

∃x,(P(x)∨Q(x))⇒(~∀x,P(x))∨∃x,Q(x)

Please explain the proof in detail.

Answer 1

It's false. Let $P$ be "$\top$" (or any tautology) and $Q$ "$\bot$". The left-hand side is true - let $x=\emptyset$, for instance. The right-hand side is false: it is not true that "not forall $x$, $\top$" - that is, it's true that for all $x$, $\top$ - and there does not exist $x$ such that $\bot$.

Answer 2

It is not valid : to show it, it is enough to produce a counterexample.

Consider a "universe" with only one white ball.

Consider the following interpretation for the predicate letters $P$ and $Q$ :

$P(x)$ is interpreted as "$x$ is white" and $Q(x)$ is interpreted as "$x$ is black".

According to this interpetation, $\exists x (P(x)\lor Q(a))$ is true : in our interpretation it is true that "there is a ball $b$ such that : either $b$ is white or $b$ is black".

But $\exists x Q(x)$ is false : there are no balck balls, and also $\lnot \forall x P(x)$ is false, because we have only one white ball, and thus it is true that "all balls are white".

Thus, the disjunction : $\lnot \forall x P(x) \lor \exists x Q(x)$ is false, because both its disjuncts are.

Conclusion : in the said interpretation the antecedent of the conditional is true while the consequent is false, and thus the conditional is false.

Having found an interpretation where the conditional is false it is enough to conclude that it is not valid (because valid means : "true in all interpretations").