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Is it possible to prove Goodstein's theorem without transfinite induction? Is there such a proof?

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2 Answers 2

up vote 4 down vote accepted

I'm pretty sure the short answer is "no" (if you mean, can you prove Goodstein's theorem without invoking apparatus as strong as a transfinite induction which can't be reduced to an ordinary induction). For if I recall correctly, Goodstein's theorem is actually equivalent (over a weak base theory) to transfinite induction up to $\varepsilon_0$. My first port of call to check this would be Kirby and Paris's paper on 'Accessible independence results for Peano Arithmetic'.

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I can't tell if you are just being modest, but the phrasing of your reply suggests some uncertainty. If so, please let me assure you that I believe that your answer is exactly correct. –  MJD Dec 17 '12 at 14:41
I studied this a bit further. I guess that we need transfinite induction if we stick with first-order logic(?). If we allow quantification over sets, could we do with ordinal induction? I ask because I came across this link math.niu.edu/~rusin/known-math/00_incoming/goodstein Notice the the comment by Torkel Franzen. –  rank Dec 17 '12 at 19:23

I found the following paper which seems fairly relevant:

Miller J.T. On the independence of Goodstein’s theorem. 2001, Citeseer.

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Very interesting link, thank you! But I'm afraid it does not (directly) address the question? (I didn't read it through yet, though..) –  rank Dec 17 '12 at 19:25
@rank: Look at the section about the independence itself. This means that you cannot prove the theorem from first-order PA without some extra assumptions, which are ultimately equivalent to the transfinite induction. –  Asaf Karagila Dec 17 '12 at 19:46

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