PA doesn't prove Con(PA) but ZFC does. That means the extra axiom of infinity is of tantamount importance in the proof. Not seen such a proof, think it would be interesting. Heard of it.
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closed as off topic by Timothy Chow, Qiaochu Yuan, Simon Thomas, Gjergji Zaimi, François G. Dorais♦ Dec 22 2010 at 20:13 |
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You can also prove the consistency of PA with second order logic. The key thing is that you need a higher order induction hypothesis. In first order logic + PA, the induction hypothesis are limited to first order expressions. The strength of a logic is often determined by what you allow in the induction hypothesis. |
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Within ZFC you can formalize Tarski's definition of truth, then prove that the axioms of PA are all true and that the rules of inference preserve truth. This gives a formal proof of Con(PA). This allows you to prove not just the consistency of PA, but the consistency of PA + Con(PA), or PA + Con(PA) + Con(PA+Con(PA)), etc. Nothing close to the full strength of ZFC is needed for any of this (though of course you need something beyond PA). |
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