Zrjydtj

It is my understanding that Gödel's second incompleteness theorem says roughly that

there exists a sentence $varphi$ such that neither $varphi$ nor $negvarphi$ is provable in Peano's arithmetic. However one of them is true (in the structure $mathbbN$).

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 $negpsi(n_f)$. 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?

No, because you may not be able to prove that the counterexample is false in Peano arithmetic. In other words, $negpsi(n_f)$ itself may be another statement that is true but not provable in PA.

However, if for every $n$, you know that either $psi(n)$ or $negpsi(n)$ is provable in Peano arithmetic, then your conclusion is correct. For example, this will be the case if $psi$ is quantifier-free, because then $psi$ is a Boolean combination of purely numerical statements like 2+2=4 and you can always prove or refute such a statement in Peano Arithmetic.

You are confusing theories and models. Theories are all the statements that are true given a number of axioms. Gödels incompleteness theorem deals with theories. A model is a mathematical structure that satisfies a theory. In a model every statement is either true or false and your counterexample idea would work, but that would only prove something for the model.

Consider groups for example. If we take the theory of group then we should not be able to prove $(forall x)(forall y)(xy=yx)$, i.e. every group is abelian as there exist abelian groups and non-abelian groups.

First of all, suppose that some universal statement $forall x psi(x)$ is in fact true, i.e. there is no counterexample to it. How would you know that there is no counterexample to it? You can't go though all the numbers one by one, and see if any of them forms a counterexample, because you would never end that process: you would forever be checking ever-increasing numbers, and never find a counterexample ... but at no point would you know that there is no counterexample. That is, at no point can you say with 100% confidence that there is no counterexample, if you just go through the numbers one by one.

Second, even if some number is a counterexample for some claim, you might still run into the same problem. For example, what if the original universal statement is of the form $forall x exists y varphi(x,y)$? Now, to demonstrate that by setting $x=n$ you would obtain a counterexample, you would have to show that $exists y varphi(n,y)$ is false, i.e. you need to show that $forall y neg varphi(n,y)$ is true, ... and hence you're back at the problem of proving some universal to be true.