Negation of Quantifiers confusion

Question

I will appreciate it if anyone can check if the following negation is correct. The question from my class practice problem doesn't seem to include parenthesis, and I'm uncertain if I did it correctly:

Question: $\sim\! \exists x \in \mathbb{H}_{\sqrt{2}}, \forall n \in N, \sim\! \exists z \in \mathbb{R}, (x^n > z) \land \sim\! (z < n)$

After applying the negation:

Initially I thought the answer should be:

A) $\forall x \in \mathbb{H}_{\sqrt{2}}, \exists n \in N, \exists z \in \mathbb{R}, (x^n \leq z) \lor (z < n)$

On the second thought, the negation in front of the third quantifier is negated by the negation in front of the first quantifier. So what's after $\lnot \exists z \in \mathbb{R}$ should stay intact:

B) $\forall x \in \mathbb{H}_{\sqrt{2}}, \exists n \in N, \exists z \in \mathbb{R}, (x^n > z) \land \lnot(z < n)$

I'm not sure if my logic is correct, but I think answer B) should be correct.

Answer 1

Yes the second (B) is correct; the double negations cancel.

Using some abreviation, we basically have : ${\quad\neg\exists~\forall~\neg\exists~P~\\\equiv~\forall~\neg\forall~\neg\exists~P\\\equiv~\forall~\exists~\neg\neg\exists~P\\\equiv~\forall~\exists~\exists~P}$

Answer 2

Answer B is correct. You can approach this two ways.

You can apply the first negation, which you found lead to $\neg\neg\exists z\in\mathbb{R} \iff \exists z \in\mathbb{R}$

Alternatively, you can apply the second negation fully and get $\forall z\in \mathbb{R}, \neg (stuff)$, and just negate that again once applying the second negation.