Showing equivalence of two equalities I'm working through a book on logic and wanted to check one of my steps in a derivation I'm working out. Given the two quantifier negation equalities:


*

*$\lnot\exists P(x) = \forall\lnot P(x)$

*$\exists\lnot P(x) = \lnot\forall P(x)$
I'm trying to derive (2) from (1). I'm not asking for the solution if I have it wrong -- I think I've got it, but I want to find out if my last step is valid.
Negating both sides of (1):
$\lnot\lnot\exists P(x) = \lnot\forall\lnot P(x)$
Then removing the double negation, you get:
$\exists P(x) = \lnot\forall\lnot P(x)$
Now, it's not at all clear to me that algebraically, you can just slap a negation of both sides inside the quantifiers to get to (2), preserving correctness. But I notice that this equation has the same "shape" as (2). Intuitively, it seems to be saying the same thing, because you could define a predicate $R(x) = \lnot P(x)$ and obtain:
$\exists\lnot R(x) = \lnot\forall R(x)$
...but I don't like "intuitively". This step feels like a hand-wavy leap of faith -- is it valid? If so, is there a name for it?

Yes, it is valid. The $P$ refers to an arbitrary predicate. That means that it can be anything, including $\lnot P'$ for some predicate $P'$. That is exactly what you did.
