Weird "proof" recursion Assume we have some statement called $\alpha$. We will define a new statement $P(\alpha):="(\alpha\text{ can be proved})\lor(\lnot\alpha \text{ can be proved})"$
I claim that $P(P(\alpha))$ is always correct (that is, we can always prove\disprove whether a statement can be proved or not).
The following is a &quot;proof&quot; for that:
Assume towards contradiction that $P(P(\alpha))$ is not correct. Then, there is no proof for $P(\alpha)$ and also there isn't a proof for $\lnot P(\alpha)$.
Lemma: If $\alpha$ can be proved, say using some proof $p_\alpha$, then this $p_\alpha$ also proves that &quot;$\alpha$ can be proved&quot;, hence $p_\alpha$ proves that $P(\alpha)$ is also true.
Use this lemma again and get that $P(P(\alpha))$ is true.
If $\alpha$ cannot be proved, then either that $\lnot\alpha$ can be proved or not. If it is possible to prove $\lnot \alpha$, then by the lemma we know that $P(\lnot\alpha)$ can be proved. But notice that by the definition of $P$ we have $P(\alpha)\equiv P(\lnot \alpha)$, so also $P(\alpha)$ can be proved, and once again, using the lemma we get that $P(P(\alpha))$ is true.
If its also impossible to prove $\lnot\alpha$, then by the definition of $P(\alpha)$ we have $P(\alpha)$ is false. However the &quot;proof&quot; up until this point proves that $P(\alpha)$ was false, since we assumed that there is no proof for $P(\alpha)$ but in every other case there was a proof. Thus, a contradiction - and hence $P(P(\alpha))$ must be true in this case as well.
My question

Is this idea &quot;correct&quot;?
If it is &quot;correct&quot;, then is it possible to formalize it? how would it look like?
I feel like there is a kind of circularity in the last part of your argument (and I think the conclusion is wrong, there are undecidable problems which we cannot prove to be undecidable).
You are basically making a case distinction:

if $\alpha$ can be proved or $\neg \alpha$ can be proved: Then $P(\alpha)$ can be proved and thus $P(P(\alpha))$ is true. (I agree here)
if neither $\alpha$ nor $\neg \alpha$ can be proved: Then $\neg P(\alpha)$ can be proved and thus $P(P(\alpha))$ is true. I disagree here. The correct assumption to make this conclusion would be: &quot;if we can prove that neither $\alpha$ nor $\neg \alpha$ can be proved&quot;.

Note that you cannot in general prove that you are in the second case. You say you can because otherwise you could prove that you are in the first case. But what you need is to be able to prove that you cannot find a proof that you are in the first case, which is a different thing.