# How to prove Con(PA) in ZFC? [closed]

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.

## closed as off topic by Timothy Chow, Qiaochu Yuan, Simon Thomas, Gjergji Zaimi, François G. Dorais♦Dec 22 '10 at 20:13

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• ZFC proves that the natural numbers (which exist by the axiom of infinity) are a model of PA, and therefore by soundness that PA is consistent? – Gabriel Ebner Dec 22 '10 at 16:52
• Is this a research-level question? – Andrej Bauer Dec 22 '10 at 18:23
• Vote to close since this is too elementary a question for MO. – Timothy Chow Dec 22 '10 at 18:36
• Wikipedia's article en.wikipedia.org/wiki/Axiom_of_infinity has a good explanation of how ZF proves that there is a set $\omega$ and an operation $S$ obeying the Peano axioms. In other words, ZF proves that there is a model of PA. (continued...) – David E Speyer Dec 23 '10 at 13:45
• This no doubt reveals my ignorance of set theory, but it seems to me to be a little tricky to finish from here. I would like a theorem of ZF saying "For any theory T, if T has a model then Con(T)". It's not clear to me that this claim can be expressed in ZF! Everytime I try, I wind up wanting a truth predicate planetmath.org/encyclopedia/… . (continued) – David E Speyer Dec 23 '10 at 13:53