I trying to study for my final exam and I can't figure out how to solve this:

$\forall x(\neg p(x)\rightarrow q(x)), \forall z(p(z)\rightarrow r(z))\vdash \forall z(\neg r(z) \rightarrow\exists yq(y))$

This is what I have so far:

1. $\forall x(\neg p(x)\rightarrow q(x))$
2. $\forall z(\neg p(z)\rightarrow r(z))$
3. $p(x\_0)\rightarrow q(x\_0)$ (using 1 and UE($S^x\_{x\_0}$))
4. $p(x\_0)\rightarrow r(x\_0)$ (using 2 and UE($S^z\_{x\_0}$))
5. $\neg p(x\_0)\rightarrow r(x\_0)$ (using 3, 4 and Hypothetical syllogism)
6. $r(x\_0) \vee q(x\_0)$ (using 5)
7. $\neg r(x\_0) \rightarrow q(x\_0)$ (using 6)
8. $\forall z(\neg r(z)\rightarrow q(z))$ (using Universal Introduction)

But I can't reach $\forall z(\neg r(z) \rightarrow\exists yq(y))$

I can't find any theorem that I can use to put the $\exists$ in the middle of the $\forall$.

Can anyone help me?

Thank you