Is $\exists y\forall x(Pxy\land Qy)$ always false?

Question

Is $\exists y\forall x(Pxy\land Qy)$ always false? I ask because of the following:

1. Let the domain of discourse be the set where x and y are natural numbers and y is always greater than x.
2. Let P be the set of Prime Numbers
3. $\exists x\exists y\left(y>x\land y\in P\right)\rightarrow\exists y\forall x\left(y>x\rightarrow y\in P\right)$ is valid. Also, $\exists x\exists y\left(y>x\land y\in P\right)$ is true. Thus, $\exists y\forall x(y>x\rightarrow y\in P)$ is true.
4. $\forall x\forall y(y>x)\land\exists y\forall x(y>x\rightarrow y\in P)\rightarrow\exists y\forall x(y>x\land y\in P)$ is valid.
5. However, $\exists y\forall x(y>x\land y\in P)$ seems false. Yet, $\forall x\forall y(y>x)$ and $\forall x\exists y(y>x\rightarrow y\in P)$ are true.
6. Thus, it must be the case that $\exists y\forall x(y>x\land y\in P)$ is true.