TITLE: Can unprovability unprovable? Is there an $\omega$-fold unprovability?
QUESTION [16 upvotes]: I was just thinking about unprovability. I just wanted to know if it is possible to make a concrete boundary between provable problems and unprovable problems in a certain axiomatic system. 
We know that there is a statement that is true yet unprovable. Then is it possible that a statement is true and unprovable, but the unprovability is unprovable?
We can extend it to a question of "unprovability of unprovability of unprovability", or any finite number of times, or even $\omega$ times....
I have no clue. Has there been any research on this kind of topic?

REPLY [2 votes]: [This is not an answer to the question. It was supposed to be a response to a comment by Milo under his answer, but since it is too long and may be interesting to others, I'm putting it here.]
Different Provability Notions
Milo suggested that perhaps we can consider a different kind of "provable" than the usual one, in the hope that there might be one that satisfies all the properties he would like to have, namely (A) to (D) in his post. I commented that I think there cannot be any such notion, but here is an analysis of the suggested kind where "unprovable" includes some notion of consistency. I'll freely use what I have defined and proven in my other answer.
Let's reserve "provable" for the standard notion. We shall work in a fixed theory $T$ that is strong enough to interpret arithmetic. Let "$! P$" denote "$□ P \lor □ \neg P$" ($P$ is internally either provable or disprovable), and "$? P$" denote "$\neg □ \bot → \neg □ P \land \neg □ \neg P$" (if there is internal consistency then $P$ is internally independent), which is equivalent to "$□ P \lor □ \neg P → □ \bot$". Then these are actually useful. For example $PA \vdash{} ? R$ where $R$ is the Rosser sentence (over $PA$), by formalizing the entire Rosser trick in PA itself. I also think that $ZF \vdash{} ? AC$ (based on https://math.stackexchange.com/a/1596529), which would be a highly non-trivial example. Furthermore we always have $! P \lor{} ? P$.
Now let us see what (A) to (D) would, should or cannot be. It turns out that only (A) and (B) have some sort of true interpretation, but even then they aren't exactly very useful.
(A) should be "$! P →{} ! ! P$", which is true because $□ P → □ □ P$ and $□ \neg P → □ □ \neg P$ by (D3), and $□ □ P →{} □ ! P$ and $□ □ \neg P →{} □ ! P$ by (D1,D2).
(B) should be "$! P ↔{} ! \neg P$", which is obviously true.
(C) would be "$! P \land{} ! ( P → Q ) →{} ! Q$", which is invalid when $(P,Q) = (\bot,\neg □ \bot)$, since PA would prove $□ \neg P$ and $□ ( P → Q )$ but not $! Q$. (C) should have been (D2).
(D) says "$□ P → P$", which is even more invalid than "$□ □ P → □ P$". Even if we change it to (D*) "$\neg □ \bot \land □ P → P$", it is still invalid. To see why, (D*) gives $\neg □ \bot \land □ □ \bot → □ \bot$ and hence $□ □ \bot → □ \bot$, so by (L*) we get $□\bot$, which is about as false as it can get.
Some Correct Claims
Although Milo's arguments are flawed, there are a number of related claims that are in fact correct for any sentence $P$ over $T$:

(1) If $T$ is consistent, then $T \nvdash \neg □ P$.

Same as (U*).

(2) If $T$ is arithmetically sound or ω-consistent, then $T \nvdash □ \neg □ P$.

Proof: If $T \vdash □ \neg □ P$, then $T \vdash □ \bot$ by (U), and since $T$ is arithmetically sound or ω-consistent we can extract a proof of $\bot$ over $T$, which is impossible. Therefore $T \nvdash □ \neg □ P$.

(3) $T \vdash \neg □ \bot → \neg □ \neg □ P$.

Same as (U).
Internal Independence
Can any more can be said about internal independence, which we have denoted by "$?$", in similar manner as with "$□$"? I don't know; there does not seem to be any simple but non-trivial fact involving it, and I don't expect any either.