Counterexample for first order logic argument

Question

Find a counterexample to show that the following argument is not valid:

∃xP(x), ∃x(P(x) → Q(x)) |= ∃xQ(x)

To my understanding, I have to a select a single element $x$ such that the premises are true but conclusion is false.

I can't find a counterexample: if I go with the domain: D={a}, $Q(a)=0$ (to make the conclusion false), then $P(a)$ must be $0$ to make the implication true. In doing so $∃xP(x)$ becomes false, so this isn't a countermodel to show that the argument is invalid.

What am I doing wrong?

Answer 1

HINT 1: There's two ways an element $a$ could witness "$\exists x(P(x)\implies Q(x))$": either by having $P(a)\wedge Q(a)$ hold, or by having $\neg P(a)$ hold (so that $P(a)\implies Q(a)$ becomes vacuously true).

HINT 2: The witnesses for "$\exists x (P(x))$" and "$\exists x(P(x)\implies Q(x))$" don't need to be the same element. That is, your sentence

To my understanding, I have to a select a single element $x$ such that the premises are true but conclusion is false.

is incorrect.

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Does this help?