Some theories can lie about their own consistency (for example, if $PA$ is consistent then the theory $PA + \lnot CON(PA)$ is consistent, although it proves its own inconsistency).

Now working with ZFC, one can avoid problems like ZFC lying about its own consistency, by assuming ZFC is $\omega$ - consistent.

that leads me to two questions:

the first one is - when we proved inside the metatheory of ZFC that the $\omega$ - consistency of ZFC avoids the problem mentioned above, how can we be so sure that this proof isn't another example of ZFC lying about itself? in different words, how can we know that this proof is expressing truth?

and the second question is - without assuming ZFC is $\omega$ - consistent, how can we rely on any conclusions of ZFC? is there a way to separate between theorems which can be examples of ZFC lying about itself and theorems which does express truth?

  • 3
    $\begingroup$ Hi D. Hershko, welcome to MO. This site is for research level mathematics and this appears to be a pretty standard question about Gödels incompleteness theorem and model theory unless I’m missing something, which would fit in better over at MSE. $\endgroup$ – Alec Rhea Jan 9 at 12:47
  • $\begingroup$ Please ask non-research-level questions on math.stackexhange.com. And by the way $PA + \lnot CON(PA)$ does not prove its own inconsistency. It proves inconsistency of $PA$. $\endgroup$ – Andrej Bauer Jan 9 at 13:11
  • 7
    $\begingroup$ @AndrejBauer: Re your side remark: surely PA+¬Con(PA) does prove its own inconsistency, since that follows directly from inconsistency of PA? But your main point, that this question ought to be on math.stackexchange.com , I of course agree with. $\endgroup$ – Peter LeFanu Lumsdaine Jan 9 at 13:30
  • 3
    $\begingroup$ Quite right, I shouldn't try to mix MO and administrative tasks. $\endgroup$ – Andrej Bauer Jan 9 at 15:46

The metatheory involved in proving "if ZFC is $\omega$-consistent then it doesn't prove its own inconsistency" is some tiny fragment of PA. I'd expect primitive recursive arithmetic (PRA) to be more than enough, since the proof is essentially just inspection of definitions. So we needn't worry that the metatheory argument is a case of ZFC lying to us. The only worry that I see is that PRA might lie to us; if you worry about that, then you should probably also worry about a lot of more basic things, like whether $2^n$ exists for all natural numbers $n$.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.