# Con(PA) implies consistency of $\mathsf{PA}$ + ¬Con($\mathsf{PA}$)

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($\mathsf{PA}$).

• 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. – Mauro ALLEGRANZA Apr 3 '14 at 6:17
• ... of course, assuming the consistency of $\mathcal T$. – Mauro ALLEGRANZA Apr 3 '14 at 15:22

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.
• 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? – Rustyn Apr 3 '14 at 0:35
• @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 $\varphi$. – Asaf Karagila Apr 3 '14 at 18:29