Why is a an implication from a found value to a null list concidered valid?

Question

Trying to create a counter-model to $\vdash \exists x P(x) \land \exists x( P(x) \rightarrow Q(x)) \rightarrow \exists x Q(x)$. What I have found out is that I should concider a counter-model $M$, in $M$ let $A = \{0,1\}$, $P^M = \{0\}$ and $Q^M = \emptyset$, because then $M \vDash \exists x P(x)$ and, here comes the part I do not understand, $M \vDash \exists x (P(x) \rightarrow Q(x))$ holds because $0 \in P^M$ and $1 \notin P^M$. Then also $M \nvDash \exists x Q(x)$ because $Q^M = \emptyset$. Through soundness we can then decide that it is not valid.

However, why does $\exists x (P(x) \rightarrow Q(x))$ hold but $\exists x Q(x)$ does not? If we find $P(x)$ but not $Q(x)$, is that not False? Seems like I am missing something, just do not understand what.

Answer 1

If our model has an $a$ such that $P(a)$ holds but $Q(a)$ fails, then that particular $a$ does not witness the truth of $\exists x(P(x)\rightarrow Q(x))$, and moreover it does witness the falsity of $\forall x(P(x)\rightarrow Q(x))$.

However, this does not prevent some other element from witnessing $\exists x(P(x)\rightarrow Q(x))$. In the model under consideration, the witnessing element is $1$: we have $\neg P(1)$ and $\neg Q(1)$, and so $P(1)\rightarrow Q(1)$ is true. (And meanwhile it's clear that $\exists x(Q(x))$ is false in this model.)

More broadly:

You cannot in general show that a sentence of the form "$\exists x(\varphi(x))$" fails in a model by considering a single element of that model.

(One example of when you can is $\exists x(x\not=x)$, which we know fails in every model; another example is $\exists x\forall y(P(y))$, in which the existential quantifier isn't really doing anything.)

Answer 2

$\exists x (P(x) \to Q(x))$ asserts that there exists at least one object such that if that particular object has property $P$, then that same object has property $Q$; equivalently either $P$ is false or $Q$ true of that $x$. If we find just one object for which the implication is true, then the existential statement is true.

Since the implication $P(x) \to Q(x)$ can be true of an object if $Q$ is false of it, namely then when $P$ is also false of it, the existence of an object that satisfies $P(x) \to Q(x)$ does not entail the existence of an object that satisfies $Q(x)$.

That's what is happening here: $M \vDash \exists x (P(x) \to Q(x)$ is satisfied because $M, v_{[x \mapsto 1]} \nvDash P(x)$ and $M, v_{[x \mapsto 1]} \nvDash Q(x)$. But because $M, v_{[x \mapsto 1]} \nvDash Q(x)$, and $M, v_{[x \mapsto 0]} \nvDash Q(x)$, and these are all the values $x$ can take, there is no object $a$ such that $M, v_{[x \mapsto a]} \vDash Q(x)$, and hence $\exists x Q(x)$ is false in $M$.