Proof that a statement involving quantifiers is false I believe that the sentence $\forall x (P(x)\rightarrow \exists y Q(x,y))$ false?
Is it sufficient to define:
1) The Domain of $x, y    $
2) The predicates $P(x)$ and $Q(x,y)$
so that for some $x$ and $y$, $P(x)$ is true while $Q(x,y)$ is false?
 A: It was established in the comments that the OP knows that to prove that
$\forall x(P(x)\to\exists yQ(x,y))$ isn't a logical truth, it suffices to find domains for $x$ and $y$ and interpretations for $P$ and $Q$ such that $\forall x(P(x)\to\exists yQ(x,y))$ is false.
What was intutive for me to think about was considering $x,y$ raging over the natural numbers and letting $P(x)$ mean '$x$ is a natural number'.
Think about an order relation for $Q$ solves the problem.
A: The Tree Proof Generator provided the following simple countermodel:

The domain contains one member, $0$. $P(0)$ is true.  There is no values from the domain for which $Q$ is true.  
With this countermodel, the antecedent is true and the consequent is false.
One way to find this manually is to consider what would make $\exists yQ(xy)$ false? One way to do that is to let $Q$ be false for all values of whatever domain we come up. That makes the consequent false. Then consider how to make the antecedent, $P$, true? Let the domain contain one member and let that member make $P$ true. 

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