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The Wikipedia article for $\omega$-consistency says "Now, assuming PA is really consistent, it follows that $\mathsf{PA}$ + ¬Con($\mathsf{PA}$) is also consistent, for if it were not, then PA would prove Con(PA) (since an inconsistent theory proves every sentence), contradicting Gödel's second incompleteness theorem."

I'm not sure how that follows; if $\mathsf{PA}$ + ¬Con($\mathsf{PA}$) is inconsistent, then it can obviously prove Con($\mathsf{PA}$), but I don't get how that shows that $\mathsf{PA}$ could prove Con($\mathsf{PA}$).

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    $\begingroup$ Because it is a meta-theorem of f-o logic that : "if $\mathcal T \cup \{ \lnot A \}$ is inconsistent, then $\mathcal T \vdash A$"; you can see this post. $\endgroup$ – Mauro ALLEGRANZA Apr 3 '14 at 6:17
  • $\begingroup$ ... of course, assuming the consistency of $\mathcal T$. $\endgroup$ – Mauro ALLEGRANZA Apr 3 '14 at 15:22
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If $\sf PA+\lnot \rm Con\sf (PA)$ is inconsistent but $\sf PA$ is consistent, then in every model of $\sf PA$ it is true that $\rm Con\sf (PA)$, now by completeness we get that Peano proves its own consistency.

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  • $\begingroup$ The underlying assumption is that $\mathsf{PA}$ is consistent. Now assume that $\mathsf{PA} + \lnot \Con(\mathsf{PA})$ is inconsistent. Therefore, since $\mathsf{PA}$ is consistent, conclude that $\mathsf{PA} + \Con(\mathsf{PA})$ is consistent. By completeness, there is a model for $\mathsf{PA} + \Con(\mathsf{PA})$. But this model is a model of $\mathsf{PA}$, therefore $\Con(\mathsf{PA})$ must be provable from $\mathsf{PA}$, which is impossible. Am I correct in this line of reasoning? $\endgroup$ – Rustyn Apr 3 '14 at 0:35
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    $\begingroup$ No, you have one skip (at least compared to my line of thought, it might be salvaged though). Every model of PA satisfies either Con(PA) or its negation. If none satisfy the negation they all must satisfy Con(PA). We don't need to conclude the consistency of PA+Con(PA); it follows. $\endgroup$ – Asaf Karagila Apr 3 '14 at 0:39
  • $\begingroup$ Thanks a lot, I appreciate it. $\endgroup$ – Rustyn Apr 3 '14 at 0:40
  • $\begingroup$ The one thing I'm a little unclear is what you mean by 'completeness'; if PA is consistent, doesn't that necessarily mean that it's incomplete? $\endgroup$ – Venge Apr 3 '14 at 18:24
  • $\begingroup$ @Venge: Yes, but there are different statement which could witness this incompleteness; not necessarily $\rm Con\sf (PA)$ itself. By completeness I mean the completeness theorem stating that if $\varphi$ is true in every model of $T$, then $T$ proves $\varphi$. $\endgroup$ – Asaf Karagila Apr 3 '14 at 18:29

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