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

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You might in general be interested in reverse mathematics, which studies the amount of induction (and other axioms) needed in many theorems. – Noah Schweber Jul 1 '15 at 2:44
up vote 5 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 found the following paper which seems fairly relevant:

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

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@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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