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Are there sentences $\phi$ and $\psi$ in the language of PA and models $\mathcal{M}_\phi$ and $\mathcal{M}_\psi$ of PA such that $\mathcal{M}_\phi\models\phi$ and $\mathcal{M}_\psi\models\psi$, but $\text{PA}\vdash\neg(\phi\wedge\psi)$ and $\mathbb{N}\models(\neg\phi)\wedge(\neg\psi)$ (where $\mathbb{N}$ is the standard model of PA)?

I assume that the answer is yes, but I do not know how to construct them or even show non-constructively that they exist. This answer seems to suggest that $\phi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ and $\psi :\leftrightarrow \neg\text{Con}(\text{PA})\wedge\neg\text{Con}(\text{PA}+\neg\text{Con}(\text{PA}))$ should work, but it's not obvious to me from the form of the second incompleteness theorem that I'm familiar with that $\phi$ is consistent: $\text{PA}+\neg\text{Con}(\text{PA})$ cannot prove its own consistency, but is it clear that it can't prove its own inconsistency either?

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closed as off-topic by Eric Wofsey, David White, Ricardo Andrade, Benjamin Steinberg, Andrés E. Caicedo Oct 8 '13 at 17:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Eric Wofsey, Ricardo Andrade, Andrés E. Caicedo
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ How about $\phi=\neg Con(PA)$ and $\psi=Con(PA) \wedge \neg Con(PA+Con(PA))$? $\endgroup$ – Eric Wofsey Oct 3 '13 at 9:05
  • $\begingroup$ D'oh, why didn't I think of that? Thanks, Eric! $\endgroup$ – Benya Oct 3 '13 at 9:44
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    $\begingroup$ @Benja It seems to me that $\mathtt{PA}+\lnot\mathtt{Con}(\mathtt{PA})$ not only can, but in fact does prove its own inconsistency. $\endgroup$ – Noel Vaillant Oct 3 '13 at 9:58
  • $\begingroup$ From $PA\vdash Con(PA+\lnot Con(PA))\to Con(PA)$ you obtain $PA\vdash\lnot Con(PA)\to\lnot Con(PA+\lnot Con(PA))$ and thus $PA+\lnot Con(PA)\vdash\lnot Con(PA+\lnot Con(PA))$ $\endgroup$ – Noel Vaillant Oct 3 '13 at 10:09
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    $\begingroup$ This question appears to be off-topic because it appears to have been answered very easily in the comments. $\endgroup$ – David White Oct 3 '13 at 12:11

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