How do you make a universal quantifier a existential quantifier in a multiple-quantifier statement? So I'm studying for a final- and one of the study questions is "Express (as simply as you can) each of the following sentences without the use of universal quantification:"
a) (∀x)(∃y)(∀z)[P(x,y,z)]
And I'm stuck- the textbook mentions nothing here. At first I thought that maybe it was a tricky set of negations, but I think that would just leave me with (∃x)(∀y)(∃z)[¬P(x,y,z)].
The solution to the problem is apparently ¬(∃x)¬(∃y)¬(∃z) ¬P(x,y,z)
So far the closest I can think is to go from
(∀x)(∃y)(∀z)[P(x,y,z)]            (Start)
¬( (∀x)(∃y)(∀z)[P(x,y,z)] )       (Negate the whole thing)
(∃x)(∀y)(∃z)[¬P(x,y,z)]           (Thus swap all quantifiers, negate the inside)
¬( (∃x)(∀y)(∃z)[¬P(x,y,z)] )      (Negate Everything again)              
¬(∃x)(∃y)(¬∃z)[P(x,y,z)]          (Instead of swapping existential quantifiers,
                                    negate them. But we still have no negation on y,
                                    and we had to negate the negation on the inside?)

 A: Perhaps adding many more parentheses will make more sense:
$(\forall x)~(~(\exists y) ~(~(\forall z)~P(x,y,z)~)~)$.
Negate 4 times; this is equivalent to doing nothing to the proposition, so this step is justified.
$\neg\neg\neg\neg(~(\forall x)~(~(\exists y) ~(~(\forall z)~P(x,y,z)~)~)~)$.
Take the first negation inside:
$\neg\neg\neg(~(\exists x)~(~(\forall y) ~(~(\exists z)~\neg P(x,y,z)~)~)~)$.
Take the second negation inside, stop after you've negated the $\forall y$.
$\neg\neg(~(\forall x)~(~(\exists y) ~(~\neg(\exists z)~\neg P(x,y,z)~)~)~)$.
Take the third negation inside, stop after you've negated the $\forall x$.
$\neg(~(\exists x)~(~\neg(\exists y) ~(~\neg(\exists z)~\neg P(x,y,z)~)~)~)$.
This is the answer.
The point is that negating a quantified statement like $(\forall x)P(x)$ always results in a statement like $(\exists x)\neg P(x)$, even if $P(x)$ itself is a quantified statement. So there's no difference in meaning if you go down to the last level of the statement or not.
A: Especially knowing the answer, I would try going backwards to "simplify" the solution.
$$\neg(\exists x)\neg(\exists y)\neg(\exists z)\neg P(x,y,z)$$
$$(\forall x)\neg\neg(\exists y)\neg(\exists z)\neg P(x,y,z)$$
$$(\forall x)(\exists y)\neg(\exists z)\neg P(x,y,z)$$
$$(\forall x)(\exists y)(\forall z)\neg\neg P(x,y,z)$$
$$(\forall x)(\exists y)(\forall z) P(x,y,z)$$
Run these steps backwards and you have the solution. Alternatively, if you want to repeatedly negate on the outside and bring the negations in, you can do that as well:
$$(\forall x)(\exists y)(\forall z) P(x,y,z)$$
$$\neg\neg(\forall x)(\exists y)(\forall z) P(x,y,z)$$
$$\neg(\exists x)(\forall y)(\exists z)\neg P(x,y,z)$$
$$(\forall x)(\exists y)\neg(\exists z)\neg P(x,y,z)$$
$$\neg\neg(\forall x)(\exists y)\neg(\exists z)\neg P(x,y,z)$$
$$\neg(\exists x)\neg(\exists y)\neg(\exists z)\neg P(x,y,z)$$
