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Is there a consistent, complete axiom system that proves its own consistency?

I know that this question isn't exact and I haven't defined when an axiom system proves its own consistency because that's just human interpretation.

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    $\begingroup$ The question was asked by Hilbert and answered in the negative by Gödel en.m.wikipedia.org/wiki/Gödel%27s_incompleteness_theorems $\endgroup$ – Jonathan Julián Huerta Jun 7 '15 at 16:28
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    $\begingroup$ That wouldn't be a contradiction to a positive answer of my question. I read Gödel's original paper. $\endgroup$ – asdfusername Jun 7 '15 at 16:30
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    $\begingroup$ @Jonathan: It would be a contradiction if the theory is also recursively enumerable. $\endgroup$ – Asaf Karagila Jun 7 '15 at 16:35
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    $\begingroup$ No, that's wrong. $\endgroup$ – asdfusername Jun 7 '15 at 16:37
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    $\begingroup$ You're right, it needs to also interpret arithmetic. But how do you plan on formulating "$T$ is consistent" without interpreting arithmetic? $\endgroup$ – Asaf Karagila Jun 7 '15 at 16:58
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Dan Willard published several papers about this topic in the Journal of Symbolic Logic. One place to start is the short Wikipedia article "Self-verifying theories". I am not familiar with the detailed proofs about Willard's theories, but when I have heard him talk about them he indicated they do not prove that multiplication is a total function, and in that way manage to remain weak enough to avoid the incompleteness theorem.

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    $\begingroup$ Thanks. In the Wikipedia article it is written that there are consistent theories proving their own consistency. Are there also such theories which are complete and prove their own completeness? $\endgroup$ – asdfusername Jun 8 '15 at 15:04
  • $\begingroup$ Do you know whether there are such theories? $\endgroup$ – asdfusername Jun 11 '15 at 13:26

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