Is it possible to prove Goodstein's theorem without transfinite induction? Is there such a proof?
Goodstein's theorem without transfinite induction
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logic
proof-theory
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1You might in general be interested in *reverse mathematics*, which studies the amount of induction (and other axioms) needed in many theorems. – 2015-07-01
2 Answers
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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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1I 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. – 2012-12-17
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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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0@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. – 2012-12-17