TITLE: Statement true because not provable
QUESTION [6 upvotes]: It is my understanding that Gödel's second incompleteness theorem says roughly that

there exists a sentence $\varphi$ such that neither $\varphi$ nor $\neg\varphi$ is provable in Peano's arithmetic. However one of them is true (in the structure $\mathbb{N}$).

If $\varphi$ is a formula in the style : "for all integers $n$, then $\psi(n)$".
For me, if $\varphi$ is false, proving it false is simply exhibing a counterexemple $n_f$ such that $\neg\psi(n_f)$. Therefore if it is false, then it must provably false, doesn't it?
Therefore, it someone manage to prove that it not provable, does it mean that $\varphi$ must be true?
Is this reasoning correct?

REPLY [4 votes]: First of all, if I can prove that PA cannot prove $\varphi$, that does not mean that $\varphi$ is true ... for if $\varphi$ is false, then of course PA will not be able to prove it. To give a concrete example: I can (well, a good mathematical logician better than I am can do this) prove that PA cannot prove that $1+1=3$ ... but obviously that does not mean that $1+1=3$
So, I think what you are trying to say is: If I can prove that PA cannot prove either $\varphi$ or $\neg \varphi$, then that must mean that $\varphi$ is true.
Second, your argument for this works when $\varphi$ is of the form $\forall x \ \psi(x)$ where $\psi(x)$ is quantifier-free, for if it is false, then $\neg \psi(n)$ is true for some $n$, and since PA can prove all quantifier-free truths expressed in the language of arithmetic, PA can prove this, and hence would be able to prove $\neg \varphi$, which goes against the assumption that PA could not prove this. So yes, your reasoning is correct .... for those kinds of statements $\varphi$ ... 
So, the question is: what kinds of interesting statements are there that contain a single quantifier, and for which you can prove that PA cannot prove either it or its negation?