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DanielFilan comments on Stupid Questions December 2014 - Less Wrong Discussion
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Comments (341)
Which "above logic" are you referring to? If you mean your OP, I don't think that the logic holds, for reasons that I've explained in my replies.
Your reasons were that not(provable(c)) isn't provable in PA, right? If so, then I will rebut thusly: the setup in my comment immediately above(I.e. either provable(c) or not provable(c)) gets rid of that.
I'm not claiming that there is no proposition C such that not(provable(C)), I'm saying that there is no proposition C such that provable(not(provable(C))) (again, where all of these 'provable's are with respect to PA, not our whole ability to prove things). I'm not seeing how you're getting from not(provable(not(provable(C)))) to provable(C), unless you're commuting 'not's and 'provable's, which I don't think you can do for reasons that I've stated in an ancestor to this comment.
Well, there is, unless i misunderstand what meta level provable(not(provable(consistency))) is on.
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: 1. provable(P) or not(provable(P)) by premise 1
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 Provable(P): By premise 2, line 2, and line 4.
Where's the flaw? Is it between lines 3 and 4?
I think step 3 is wrong. Expanding out your logic, you are saying that if not(provable(P)), then (if provable(P) then P), then provable(if provable(P) then P). The second step in this chain is wrong, because there are true facts about PA that we can prove, that PA cannot prove.
So the statement (if not(p) then (if p then q)) is not provable in PA? Doesn't it follow immediately from the definition of if-then in PA?
(if not(p) then (if p then q)) is provable. What I'm claiming isn't necessarily provable is (if not(p) then provable(if provable(p) then q)), which is a different statement.
Oh, that's what I've been failing to get across.
I'm not saying if not(p) then (if provable(p) then q). I'm saying if not provable(p) then (if provable(p) then q)