Proof by contradiction in predicate logic

Question

So we are given the following to prove, only by proof by contradiction

$\forall x(Q(x)\to P(y)) \vDash \forall xQ(x)\to P(y)$

Now the first thing that comes to mind in predicate logic when i am on a dead end is to perform Tarski's theorem about truth and see what types of structures ,the given types are true. But when it comes to analyzing the first part: $\forall x(Q(x)\to P(y))$ i am not very sure if i can say which structures(we want to find structures as sets, for example:for every x that makes Q true, there is a y that makes P truth etc) Anyway i need help finding the types for structures these types need to be true so after that i can move forward and start the proof by contradiction.Thanks

Answer 1

Hint:

1. Suppose $\forall x(Q(x)\to P(y))$

2. Suppose $\forall x(Q(x))\land \neg P(y)$

Obtain a contradiction, thus negating (2).

Answer 2

$\forall x(Q(x)\to P(y)) \;\models\; \forall x\,Q(x)\;\to\;P(y)$

To prove by contradiction show that $\neg(\forall x\,Q(x)\to P(y)) \;\models\; \neg\forall x(Q(x)\to P(y))$

$$\begin{array}{l} \neg(\forall x\,Q(x)\to P(y)) \\ \forall x\,Q(x) \wedge P(y) \\ \exists x\, Q(x)\wedge P(y) \\ \exists x\,(Q(x)\wedge P(y)) \\ \exists x\,\neg(Q(x)\to P(y)) \\ \neg \forall x\,(Q(x)\to P(y)) \\ \Box \end{array}$$

(Skeleton only: Justifications and several steps omitted).