To answer the below: I'm not saying that provable(X or notX) implies provable (not X). I'm saying...I'll just put it in lemma form(P(x) means provable(x):
If P( if x then Q) AND P(if not x then Q)
Then P(not x or Q) and P(x or Q): by rules of if then
Then P( (X and not X) or Q): by rules of distribution
Then P(Q): Rules of or statements
So my proof structure is as follows: Prove that both Provable(P) and not Provable(P) imply provable(P). Then, by the above lemma, Provable(P). I don't need to prove Provable(not(Provable(P))), that's not required by the lemma. All I need to prove is that the logical operations that lead from Not(provable(P))) to Provable(P)) are truth and provability preserving
Breaking my no-comment commitment because I think I might know what you were thinking that I didn't realise that you were thinking (won't comment after this though): if you start with (provable(provable(P)) or provable(not(provable(P)))), then you can get your desired result, and indeed, provable(provable(P) or not(provable(P))). However, provable(Q or not(Q)) does not imply provable(Q) or provable(not(Q)), since there are undecideable questions in PA.
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