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My curiosity was piqued by this discussion:

Presburger Arithmetic

I'll state this question loosely (rather than formally in terms of first-order logic or another formal framework) but I think it's clear what I'm asking.

Is there any formal system that can prove its own consistency?

PA is essentially too strong to prove its own consistency. The choice of axioms makes the theory "too complicated" a structure for it to be able to deal with itself in this way.

(From my limited understanding) Presburger arithmetic is too weak to prove its own consistency. Indeed the question of its consistency is not representable within it.

Is there something in the middle? A system powerful enough to represent a formula (or some other formal construct) stating that it itself is consistent, yet not too powerful to allow the second incompleteness theorem to take over?

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closed as no longer relevant by Tom Ellis, Andrej Bauer, Dan Petersen, Qiaochu Yuan, Felipe Voloch May 21 '11 at 11:08

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An inconsistent formal system can! – Qiaochu Yuan May 21 '11 at 7:34
@Qiaochu: your comment makes me realise that stating my question formally is going to be harder than I initially imagined ... – Tom Ellis May 21 '11 at 7:42
Poking around, it seems that Dan Willard has worked on this question, and as far as I can make out with my amateur understanding the answer is "yes": – Tom Ellis May 21 '11 at 7:47
@Tom: doesn't Neel Krishnaswamis answer in the discussion you cite provide an answer? – Michael Bächtold May 21 '11 at 8:20
Voting to close as no longer relevant, since the question is answered here:… – Tom Ellis May 21 '11 at 8:43

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