How to prove Con(PA) in ZFC? - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-25T01:47:16Z http://mathoverflow.net/feeds/question/50173 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50173/how-to-prove-conpa-in-zfc How to prove Con(PA) in ZFC? Zirui Wang 2010-12-22T16:41:44Z 2010-12-22T18:32:47Z <p>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.</p> http://mathoverflow.net/questions/50173/how-to-prove-conpa-in-zfc/50174#50174 Answer by Steven Landsburg for How to prove Con(PA) in ZFC? Steven Landsburg 2010-12-22T17:03:20Z 2010-12-22T18:32:47Z <p>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).</p> <p>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 <em>something</em> beyond PA).</p> http://mathoverflow.net/questions/50173/how-to-prove-conpa-in-zfc/50176#50176 Answer by Lucas K. for How to prove Con(PA) in ZFC? Lucas K. 2010-12-22T18:26:28Z 2010-12-22T18:26:28Z <p>You can also prove the consistency of PA with second order logic.</p> <p>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. </p> <p>The strength of a logic is often determined by what you allow in the induction hypothesis. </p>