So is there a countable order of metaprovability? i.e. are there problems that if they're impossible to prove, it's impossible to prove whether it's possible to prove whether they're unprovable, but _that_ can be proven? Or does that mean the same thing? What does it mean if that goes infinitely deep? That we'll never know anything about a problem in that set unless we prove it? I guess it's not possible to prove that a problem is in that set, right? Is it possible to prove that problems do exist in that set at all?
edit: (non-vacuously. Provable problems are obviously vacuously in all the sets).
edit2: oh I see now (I think). Above I'd need an "AND if you can't prove it undecideable ... then you can't prove its undecidability's decideability either, and that can be proved"
So basically each level needs to be conditioned on all 0..n-2 levels not holding, then you can prove level n-1 is undecideable.
level 0: provable
level 1: provably undecideable
level 2: if not provable (can't prove level 0) then can't prove undecideable. (Goldbach, Collatz)
level 3: if it's not in level 0 or 1, then can't prove level 2.
level 4: if it's not in levels 0, 1, or 2, then can't prove level 3.
...
Which means there's no level infinity, since there's no level infinity minus 1 to talk about so it's meaningless. Correct?