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By Gödel's incompleteness (which holds even for $Q$ as discussed here) we have that the Robinson arithmetic $Q$ is consistent iff $Q+\lnot Con(Q)$ is consistent.

As I understand it, without further assumptions, we cannot exclude the possibility that $Q\vdash \lnot Con(Q)$, which would imply the inconsistency of $PA$ and $ZFC$ since they both prove $Con(Q)$ but also prove every theorem of $Q$.

I want to picture the situation, where $Q\vdash \lnot Con(Q)$ but $Q$ is in fact consistent. Now my questions are the following.

  • I would like to claim that all models of natural numbers are therefore non-standard. Is there any mathematical way in which this can be formalized, or is this purely a philosophical statement under my assumptions? I guess the extreme weakness of $Q$ plays a role in this which I find hard to pin down.

  • Could we find some potentially consistent theories which still formalize infinitary mathematics such as real analysis and topology in satisfactory ways?

  • What are some combinatorial or other statements I would have to believe in, to also believe such a situation cannot occur?

Punga
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  • Just a nitpick: the first statement you make is not as straightforward as you make it out to be since Q can't prove the derivability conditions. Things get pretty subtle and weird with theories this weak. If I remember the folklore correctly, it's still true, just not a consequence of the usual statement/proof of the 2nd incompleteness theorem. – spaceisdarkgreen May 28 '22 at 01:48
  • Incidentally, after doing a little googling, seems the first bit of folklore I mentioned seems not as straightforward as I made it out to be... see the comments and linked MO from here https://math.stackexchange.com/questions/4648/can-robinsons-q-prove-presburger-arithmetic-consistent – spaceisdarkgreen May 28 '22 at 01:54
  • @spaceisdarkgreen Thank you for letting me know about this. It seems the statement that "the second Gödel theorem holds for Q" is true, but that does not simply follow from the original Gödel's work. I will edit the question accordingly. – Punga May 28 '22 at 09:14
  • One consequence is that we are working with a theory that is very very very weak as a foundation of mathematics. – Asaf Karagila May 28 '22 at 11:36
  • @Punga Yes, incidentally the result from the Bezboruah/Sheperdson paper goes a lot further and essentially says "Q does not prove the consistency of anything", which is delightful, though not exactly surprising. – spaceisdarkgreen May 28 '22 at 15:01

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