I'm having trouble finding counterexamples for these or finding a model to show their validity:
$∃x∀y(Q(x, y)→∀zR(x, y, z))$
$P(x)→∀yP(y)$
I know that we have to make up an interpretation and a domain where we can dis/prove these. Would appreciate some guidance.
On
Neither of the two are valid. For the first one, you just need to find an interpretation for $Q$ and $R$ such that $Q$ holds for all $y$ and $R$ is false for all $z$ in the domain. For the second one, you need a domain containing $x$ such that $P(x)$ holds where the domain also contains some $y$ for which $P(y)$ is false.
For the first, let the set of the model $M$ be $\Bbb N$, and interpret $$ Q^M(x,y) \iff x = y $$ and $$ R^M(x,y,z) \iff x = y*z. $$ Then $M\models \exists x\forall y\,(Q(x, y)\to \forall z\,R(x, y, z))$, because $M\models \forall y\,(Q(x, y)\to \forall z\,R(x, y, z))[x/0]$ — that is, for all $y\in \Bbb N$, if $y = 0$ then for all $z\in\Bbb N, 0 = y*z$.
Thus the sentence is satisfiable. However, it's also possible to find a model $M'$ in which the sentence is false. Let the underlying set of $M'$ be $\Bbb N$ again, and interpret the predicates as $$ Q^{M'}(x,y)\iff x\le y $$ and $$ R^{M'}(x,y,z)\iff y\le z. $$ Then $M'\models \exists x\forall y\,(Q(x, y)\to \forall z\,R(x, y, z))$ means: there is $x\in\Bbb N$ such that for all $y\in\Bbb N$, if $x\le y$ then for all $z\in\Bbb N, y\le z$. Suppose there is such an $x$. Letting $y=x+1$, we have $x\le y$. But $y>0$ so it's certainly not the case that for all $z\in\Bbb N, y\le z$.
Hence, both the sentence and its negation are satisfiable, so the sentence isn't valid.
As for the formula $\phi(x) := (P(x)\to\forall y\,P(y))$, the answer depends upon what your definition is of a formula with free variables being true in a model. Presumably it means, true under all assignments of values to variables, so for $M$ any model, $M\models \phi$ iff $M\models \forall x\,\phi(x)$. Note that $$\begin{align} \forall x\,\phi(x) &\iff \forall x\,(P(x)\to\forall y\,P(y)) \\ &\iff (\exists x\,P(x)\to\forall y\,P(y)) \\ &\iff (\neg\exists x\,P(x)\lor\forall y\,P(y)). \end{align}$$ This sentence is true in a model $M = (U^M, P^M)$ iff $P^M = \emptyset$ or $P^M = U^M$ (that is, iff the interpretation of $P$ "is either everything or nothing"), and false iff $P^M$ is neither. In particular, if $U^M$ has just one element then the sentence is true in $M$.