Am I correct for this predicate logic? Is my statement correct? If it is correct, is it possible to further simplify?
$$
\neg (\forall z )(\exists!y)
[\neg P(z, y) \land Q(z, y)]
= (\exists z)(\forall y)
\big[ (P(z, y) \lor \neg Q(z, y) ) 
\lor (\exists x)(\neg P(z, x) \land Q(z, x) \land x \neq y)\big]
$$
 A: Close!
It's clear what you're trying to do, and it's a good idea: if it is not true that for all $z$ there is a unique $y$ such that [bla bla], then there must be at least one $z$ for which there either are no $y$ such that [bla bla] or there are at least two things for which [bla bla]
You just didn't execute this very well. Note that the pattern you need is:
$$\exists z (\text{zero things such that [bla bla]} \lor \text{at least two things such that [bla bla]})$$
Now, you figured that 'zero things such that [bla bla]' is the same as 'everything not [bla bla]', i.e. you would get $\forall y \neg$ [bla bla]. And yes, with [bla bla] being $[\neg P(z, y) \land Q(z, y)]$, you get that $\neg$ [bla bla] amounts to $[P(z, y) \lor \neg Q(z, y)]$, and thus 'zero things such that [bla bla]' can be formalized as $\forall z \ [P(z, y) \lor \neg Q(z, y)]$
OK, so far so good ... but now you made a mistake. For the 'two things that [bla bla] you need two quantifiers $y$ and $x$, but you figured that since you already are using a $y$ quantifier for the 'nothing bla bla', you can just use that one as long as you extend the scope of it, i.e. you need to 'take out' the $y$ quantifier. In effect, you tried to go from:
$$\exists z (\forall y \ [P(z, y) \lor \neg Q(z, y)] \lor \text{at least two things such that [bla bla]})$$
to:
$$\exists z \forall y \ ( [P(z, y) \lor \neg Q(z, y)] \lor \text{one more thing such that [bla bla]})$$
OK, so you made 2 mistakes here in pulling out the $\forall y$:

*

*You need an existential $\exists y$ for the 'at least two things ...', rather than a universal


*Even if a universal was used in the second half, you still cannot pull a universal over a disjunction. That is, $\forall y \phi \lor \forall \psi)$ is not equivalant to $\forall y (\phi \lor \psi)$
OK, so keep the $\forall y$ inside, and introduce a new $y$, but with an existential:
$$\exists z (\forall y \  [P(z, y) \lor \neg Q(z, y)] \lor \exists y \exists x ...)$$
A final mistake you made was that you never specified that in the second half, the $y$ is something that has [bla bla]. That is, you specified that [bla bla] is true for $x$, but not for $y$. So, what you need is
$$\exists z (\forall y \  [P(z, y) \lor \neg Q(z, y)] \lor \exists y \exists x [\neg P(z, y) \land Q(z, y) \land \neg P(z, x) \land Q(z, x) \land x \neq y])$$
