Timeline for Are there two mutually incompatible consistent sentences in the language of PA, neither of which is true in the standard model? [closed]
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21 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Oct 8, 2013 at 17:05 | history | closed |
Eric Wofsey David White Ricardo Andrade Benjamin Steinberg Andrés E. Caicedo |
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| Oct 3, 2013 at 12:11 | comment | added | David White | This question appears to be off-topic because it appears to have been answered very easily in the comments. | |
| Oct 3, 2013 at 11:58 | comment | added | Noel Vaillant | @Benja sorry I missed your comment, thank you ! | |
| Oct 3, 2013 at 11:52 | comment | added | Noel Vaillant | @Benja Let $T=PA$, $\phi=\lnot Con(PA)$ and $T'=T+\phi$. Yes, the inconsistency of $T$ implies that of $T'$, and it does so provably, i.e. $T\vdash Con(T')\to Con(T)$ (as I claimed earlier). If this is not clear to you (it wasn't to me), with obvious notation $Axiom_{T'}(n)=Axiom_{T}(n)\lor(n=\ulcorner\phi\urcorner)$. This will mess up the Godel numbering of proofs so we can't argue that $T\vdash Prf_{T}[n,m]\to Prf_{T'}[n,m]$, but we shall have $T\vdash \exists n Prf_{T}[n,m]\to \exists n Prf_{T'}[n,m]$. | |
| Oct 3, 2013 at 11:23 | comment | added | Benya | @NoelVaillant As Joel points out, if $\neg\text{Con(PA)}$, then there is a proof of a contradiction from the axioms of $\text{PA}$, which is also a proof of a contradiction from the axioms of $\text{PA} + \neg\text{Con(PA)}$, so I think your reasoning is just fine. | |
| Oct 3, 2013 at 11:21 | comment | added | Benya | @JoelDavidHamkins Right, missed that initially, thanks! | |
| Oct 3, 2013 at 10:36 | comment | added | Joel David Hamkins | Benja, you linked to my answer in your question, but the tree there was built using the Rosser sentence to branch at each node, rather than the consistency sentences. This allows one to surmount the problem you are facing with the fact that the inconsistency of PA implies the inconsistency of any extension of PA. (And that is easy to prove, since any proof from PA is also a proof from any stronger theory.) | |
| Oct 3, 2013 at 10:34 | comment | added | Benya | @NoelVaillant Ah, I see -- the comment I linked to uses the Rosser sentence, not Con(T). By Rosser's theorem, both the Rosser sentence for T and its negation is consistent with T, resolving my confusion. Thanks! | |
| Oct 3, 2013 at 10:32 | comment | added | Noel Vaillant | @Benja, if I am honest, I am not sure my premise is true in the above argument. Going back to my notes.... | |
| Oct 3, 2013 at 10:09 | comment | added | Noel Vaillant | 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 $PA+\lnot Con(PA)\vdash\lnot Con(PA+\lnot Con(PA))$ | |
| Oct 3, 2013 at 9:58 | comment | added | Noel Vaillant | @Benja It seems to me that $\mathtt{PA}+\lnot\mathtt{Con}(\mathtt{PA})$ not only can, but in fact does prove its own inconsistency. | |
| Oct 3, 2013 at 9:48 | history | edited | Benya | CC BY-SA 3.0 |
fix body of question to be in accord with the title
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| Oct 3, 2013 at 9:44 | comment | added | Benya | D'oh, why didn't I think of that? Thanks, Eric! | |
| Oct 3, 2013 at 9:19 | review | Close votes | |||
| Oct 6, 2013 at 11:52 | |||||
| Oct 3, 2013 at 9:05 | comment | added | Eric Wofsey | How about $\phi=\neg Con(PA)$ and $\psi=Con(PA) \wedge \neg Con(PA+Con(PA))$? | |
| Oct 3, 2013 at 7:46 | history | edited | Benya | CC BY-SA 3.0 |
fix bug in previous edit
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| Oct 3, 2013 at 7:39 | review | First posts | |||
| Oct 3, 2013 at 7:59 | |||||
| Oct 3, 2013 at 7:32 | history | undeleted | Benya | ||
| Oct 3, 2013 at 7:31 | history | deleted | Benya | via Vote | |
| Oct 3, 2013 at 7:19 | history | asked | Benya | CC BY-SA 3.0 |