= 27d567bc2d48d9f40e9ccc20ea370b15)
DanielFilan comments on Stupid Questions December 2014 - Less Wrong
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (341)
You're right that (if X then Y) is just fancy notation for (not(X) or Y). However, I think you're mixing up levels of where things are being proved. For the purposes of the rest of this comment, I'll use provable(X) to mean that PA or whatever proves X, and not that we can prove X. Now, suppose provable(P). Then provable(not(not(P))) is derivable in PA. You then claim that not(provable(not(P))) follows in PA, that is to say, that provable(not(Q)) -> not(provable(Q)). However, this is precisely the statement that PA is consistent, which is not provable in PA. Therefore, even though we can go on to prove not(provable(not(P))), PA can't, so that last step doesn't work.
Wait. Not(provable(consistency)) is provable in PA? Then run that through the above.
I'm not sure that this is true. I can't find anything that says either way, but there's a section on Godel's second incompleteness theorem in the book "Set theory and the continuum hypothesis" by Paul Cohen that implies that the theorem is not provable in the theory that it applies to.
I'll rephrase it this way:
For all C: Either provable(C) or not(provable(C)) If provable(C), then provable(C) If not provable(C), then use the above logic to prove provable C. Therefore all C are provable.
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?