I think you do misunderstand that, and that the proof of not(provable(consistency(PA))) is not in fact in PA (remember that the "provable()" function refers to provability in PA). Furthermore, regarding your comment before the one that I am responding to now, just because not(provable(C)) isn't provable in PA, doesn't mean that provable(C) is provable in PA: there are lots of statements P such that neither provable(P) nor provable(not(P)), since PA is incomplete (because it's consistent).
That doesn't actually answer my original question--I'll try writing out the full proof.
Premises:
P or not-P is true in PA
Also, because of that, if p -> q and not(p)-> q then q--use rules of distribution over and/or
So:
2: If provable(P), provable(P) by: switch if p then p to not p or p, premise 1
3: if not(provable(P)) Then provable( if provable(P) then P): since if p then q=not p or q and not(not(p))=p
4: therefore, if not(provable(P)) then provable(P): 3 and Lob's theorem
5: Therefore Prova...
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