How to find a counter-model for a formula with implication and only one variable.

Question

How would I find a counter-model for a formula with implication and only one variable. I cant figure how to do it, for example $\exists x \ p(x) \rightarrow \forall x \ p(x)$. I cant find a countermodel becuasae its always true from what i know, its the same variable, it can only be $true \rightarrow true$ or $false \rightarrow false$. What would be the countermodel for $\exists x \ p(x) \rightarrow \forall x \ p(x)$?

Answer 1

$\forall x \ p(x)$ is a much stronger claim than $\exists x \ p(x)$, and so it is easy to come up with counterexamples to the claim that $\exists x \ p(x) \to \forall x \ p(x)$

For example, take as the domain all natural numbers, and let $p(x)$ be '$x$ is prime'

Then $\exists x \ p(x)$ is true (yes, there are some numbers that are prime), but $\forall x \ p(x)$ is false (no, it is not true that all numbers are prime. Hence, with this interpretation, $\exists x \ p(x) \to \forall x \ p(x)$ is false.