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Nov 30 at 9:38 comment added Kotlopou @Wojowu It indeed isn't correct, as per this question: math.stackexchange.com/questions/5003237/…
Feb 20, 2018 at 16:51 comment added Simon Henry By all strategy are winning strategy I mean't all recursive strategy. But you're right I went back to Kirby & Paris paper and it seems that you are right and my memories of it were largely oversimmplified. This being said what they proof made me doubt that Goodstein theorem is not stronger than the consistency of PA.
Feb 20, 2018 at 16:30 comment added Wojowu @SimonHenry I'm afraid that can't be correct. Well-foundedness of $\varepsilon_0$ is a statement much stronger than Con(PA) (in particular, I believe it implies Con(PA+Con(PA))). Also, I don't think Goodstein's theorem (a single sentence) implies full Kirby-Paris theorem (which, formally, can't be even stated in the language of PA, but we can express it as a scheme like well-foundedness). I have recently discovered a reference to the fact that adding a single sentence can't increase the proof-theoretic ordinal, but I don't have it at hand.
Feb 18, 2018 at 21:20 vote accept Christopher King
Feb 18, 2018 at 16:05 answer added Payam Seraji timeline score: 8
Feb 18, 2018 at 9:54 answer added user103227 timeline score: 10
Feb 18, 2018 at 8:50 comment added Wojowu Do you have a reference for Goodstein being equivalent to Con(PA)?
Feb 18, 2018 at 2:43 history edited Christopher King CC BY-SA 3.0
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Feb 18, 2018 at 1:43 review Close votes
Feb 18, 2018 at 11:23
Feb 18, 2018 at 1:28 comment added Andrés E. Caicedo No, far from it. Consistency statements, in the sense you intend, are $\Pi^0_1$ statements. It would be more interesting to ask the version of your question where you add this additional syntactical requirement.
Feb 18, 2018 at 0:44 history edited YCor
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Feb 18, 2018 at 0:15 history asked Christopher King CC BY-SA 3.0