I am reading Velleman's "How to Prove it", and in one example he proves
$\neg \exists x P(x) \implies \forall x \neg P(x)$
in the following way:
Suppose $\neg \exists x P(x)$.
Let x be arbitrary.
Suppose $P(x)$.
Since we have a specific $x$ for which P(x) is true, it follows that $\exists x P(x)$, which is a contradiction. Therefore $\neg P(x)$. Since $x$ was arbitrary, we can conclude that $\forall x \neg P(x)$, so $\neg \exists x P(x) \implies \forall x \neg P(x)$
What makes me confused is the line in bold text, that is "Suppose $P(x)$". The reason is that he writes that he is doing proof by contradiction, but $P(x) \not\iff \neg\forall x \neg P(x)$. He is not treating all possible cases where $\forall x \neg P(x)$ is false by assuming $P(x)$, does he? So how can this be a valid proof by contradiction?
Edit: I deleted my first example since it was clearly not representative for the problem.
Edit 2: So the thing that I am opposing I guess is that $\neg \forall x \neg P(x)$ is not $P(x)$. And from what I have undestood, a proof by contradiction should use the negation of the consequent which I do not think that Velleman does.
Here's the point, in some detail: Given $\neg\exists x P(x)$, we want to show that $\forall x \neg P(x)$. Pick an $x$, any $x$: we want to show that $\neg P(x)$. From this we can conclude that, as $x$ was arbitrary, $\forall x \neg P(x)$ holds. By way of contradiction, suppose the thing we want to prove of $x$ is false — namely, suppose $\neg\neg P(x)$, or, what's equivalent in classical logic, $P(x)$. Well, now we have a problem: this can't be, as it contradicts $\neg\exists x P(x)$. $P(x)$ implies a contradiction, thus we can infer that $\neg P(x)$.
It follows that $\neg P(x)$ after all. Here's why: It's a tautology that $(p \to \neg p) \to \neg p$. That is, if you assume $p$ (in this case, $p := P(x)$) and can deduce that $\neg p$, then you've shown that $(p \to \neg p)$, so by Modus Ponens $\neg p$ follows: $\neg p$ really is true.