Disprove $\{(\exists x(P(x)\to Q(x))),(\exists y P(y))\}\vdash(\exists x Q(x))$ 
Given the statement $$\{(\exists x(P(x)\to Q(x))),(\exists y P(y))\}\vdash(\exists x Q(x))$$
Show that there is no natural deduction proof for the statement.

I know I need to use the completeness property of natural deduction in the predicate case, but I need some help exhibiting a contradiction. I know that I need to find a interpretation where the $(\exists x(P(x)\to Q(x)))$and $(\exists y P(y))$ holds for any predicate P and Q, but $(\exists x Q(x))$ is false. But I'm having trouble coming up with anything.
I can understand showing the entailment $\{(\exists x(P(x)\to Q(x))),(\exists y P(y))\}\nvDash(\exists x Q(x))$ for one choice of $P$ and $Q$, but to get the correct answer, it seems like I need to show that for any $P$ and $Q$, the entailment doesn't hold.
 A: Counterexample: consider a bag with $n$ balls inside, $n−1$ of which are Black and the last one is Red. 
Interpret $P$ as "is Black" and interpret $Q$ as "is White". 
Clearly: $∃yPy$ is satisfied by this interpretation, while $∃xQx$ is not (no White balls). 
What about $∃x(Px→Qx)$ ? Call $r$ the Red ball: it is such that both $P$ an $Q$ do not hold of it. Thus, $Pr → Qr$ is True (False $\to$ False is True) and thus the above interpretation satisfies $∃x(Px→Qx)$.
Having found an interpretation that satisfies the premises and not the conclusion, we can conclude that:

$∃x(Px→Qx), ∃yPy \nvDash ∃xQx$.

A: One way to get a countermodel is to use a tree proof generator.  Here is one countermodel:

This model has a domain with two members. Only for the member $1$ is $P$ true. There is no member that makes $Q$ true.
With that assignment, $\exists x Qx$ is false since $Q$ is false for both members of the domain. This makes the consequent false, but the antecedent is true because for $y=1$ we have $P$ true and for $x=0$ we have $P$ false and $Q$ false making the conditional true.

Tree Proof Generator. https://www.umsu.de/trees/
