What does ∀x¬∃xQ(x,x)

Question

I got two questions. I am supposed to convert logical forms into prenex normal form, and haveing troubles with two forms.

Firstly What does ∀x¬∃xQ(x,x) ∨ ∃xS(x) means?. I believe I cannot just call one of the first two a "y", since I don't know if Q() symmetric is. But can I call the last x a y? and How to interpret the first part?

and secondly how to interpret this: ¬∀x¬∀y(Q(x,y)→¬∃yS(y))

can I think of it as

                                          ⟺¬∀x(¬∀y(Q(x,y)→¬∃yS(y))) 
                                          ⟺ ∃x¬{¬∀y(Q(x,y) → ¬∃wS(w))}
                                          ⟺ ∃x¬{¬∀y(Q(x,y) → ¬∃wS(w))}
                                          ⟺ ∃x¬{¬∀y(¬Q(x,y) ∨ ¬∃wS(w))}
                                          ⟺ ∃x{∀y(Q(x,y) ∧ ¬∃wS(w))}

or

                                          ⟺¬∀x¬∀y(Q(x,y)→¬∃yS(y))
                                          ⟺ ∃x¬∀y¬(Q(x,y)→¬∃yS(y))
                                          ⟺ ∃x∃y(Q(x,y)→¬∃yS(y))
                                          ⟺ ∃x∃y(Q(x,y)→¬∃wS(w))

Answer 1

In $\forall x \neg \exists x Q(x,x)$, the first quantifier is called a 'Null quantifier': it effectively quantifies nothing at all, since the $x$'s in $Q(x,x)$ are quantified by the $\exists x$

So, this statement is simply equivalent to $\neg \exists x Q(x,x)$ ... and its prenex normal thus $\forall x \neg Q(x,x)$

In $\neg \forall x \neg \forall y (Q(x,y) \rightarrow \neg \exists y \ S(y))$, the $y$ in $S(y)$ is quantified by the $\exists y$, and therefore not by the $\forall y$. The $\forall y$ does quantifiy the $y$ in $Q(x,y)$ though, so it is not a null quantifier. In such a case, it's best to replace the duplicate variable with a new variable,e.g:

$\neg \forall x \neg \forall y (Q(x,y) \rightarrow \neg \exists w \ S(w))$

... which is what you did in the first option.

I do note that you did not get that one all the way in prenex normal form though, so you need to take a few more steps:

$\neg \forall x \neg \forall y (Q(x,y) \rightarrow \neg \exists w \ S(w))=$

$\exists x \forall y (Q(x,y) \rightarrow \forall w \ \neg S(w))=$

$\exists x \forall y \forall w (Q(x,y) \rightarrow \ \neg S(w))$