Rewriting predicate sentences to logically equivalent statements that doesn't use the negation operator

Question

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

b) $\exists x \in D, \sim\! \forall y \in C, \sim\! \exists n \in \{2,3,4\}, \sim\! (Height(x) > \frac{1}{n} \cdot Height(y))$

I got:

a) $ \forall x \in \mathbb{H}_{\sqrt{2}}, \forall n \in N, \sim\! \exists z \in \mathbb{R}, (x^n \le z) \lor (z < n)$

b) $\exists x \in D, \exists y \in C, \forall n \in \{2,3,4\}, (Height(x) \le \frac{1}{n} \cdot Height(y))$

Hoping for some feedback to see whether I've done it right.

Answer 1

You have to "move inside" the leading negation sign step-by-step.

Thus, regarding

a) $\lnot ∃x \ ∀n \ \lnot ∃z \ (\ldots)$,

we have that the initial $\lnot ∃x$ must be rewritten as the equivalent $∀x \lnot$.

This means that the resulting formula will be :

a') $∀x \ \lnot ∀n \ \lnot ∃z \ (\ldots)$.

Now we have to rewrite $\lnot ∀n$ as $∃n \lnot$ and we get :

a'') $∀x \ ∃n \ \lnot \lnot ∃z \ (\ldots)$, i.e. $∀x \ ∃n \ ∃z \ (\ldots)$.

Same for b).