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Recently, this post got me interested about various Incompleteness phenomena. Let me first lay out Pudlák's idea, so that misunderstandings will be easier to catch, if there are any.

By the Second Incompleteness Theorem, if $\mathsf{PA}$ is consistent, $\mathsf{PA} \nvdash \mathrm{Con}(\mathsf{PA})$. $\mathsf{PA} \nvdash \mathrm{Con}(\mathsf{PA})$ is equivalent to $\mathsf{PA} \nvDash \mathrm{Con}(\mathsf{PA})$ by the Completeness Theorem for FOL. Let's assume that $\mathsf{PA}$ is consistent. It follows by the above that there is a model $M$ such that $M \vDash \mathsf{PA}$ and $M \vDash \neg\mathrm{Con}(\mathsf{PA})$. Now Pudlák lays out his argument: If we live in this model $M$, then this model is our actual metatheoretic arithmetic; it's how our numbers, and hence our mechanical deduction procedures, actually behave. $M \vDash \neg\mathrm{Con}(\mathsf{PA})$ means that there is an actual derivation of a contradiction from $\mathsf{PA}$ in our world, since $M$ is our world.

So far so good. But now the problems start. Let's say that we do live in $M$. This implies that there is a mechanical deduction of $\mathsf{PA} \vdash \bot$ (of course, what $\bot$ is depends on the definition of $\mathrm{Con}$ and is not essential). By Completeness, this means that $\mathsf{PA} \vDash \bot$, which implies that $\mathsf{PA}$ has no model, which contradicts our assumption.

If we take this as a straight-forward case of proof by contradiction, we have proven that $\mathsf{PA}$ is inconsistent, and we could run the same proof for any theory $T$ satisfying the criteria of the Second Incompleteness Theorem. But this is clearly a wrong conclusion. So I want to know where I made a mistake in my reasoning.

God bless
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    I don't understand the last bit of your proof, "PA has no model, which contradicts our assumption." The lack of a model was established in the $M$ world, whereas the assumption that PA is consistent was in our ("real") world. How is this a contradiction? – Andreas Blass Jul 09 '22 at 16:33
  • @AndreasBlass Perhaps I'm confused. Let's say that we run the entire argument from the point of view of people from the $M$ world. Then the "real" world and $M$ are one and the same. – God bless Jul 09 '22 at 16:45
  • If we run the whole argument from the point of view of people in $M$, then those people will assume that PA is consistent, and they will use that to obtain another model $M'$ satisfying PA + "PA is inconsistent". Then people in $M'$ will think PA is inconsistent, while people in $M$ disagree. I still see no contradiction. – Andreas Blass Jul 09 '22 at 16:48
  • @AndreasBlass That could be, but my question is precisely what happens if $M = M'$. – God bless Jul 09 '22 at 16:52
  • If $M'=M$ then pigs can fly and circles are square. In other words, $M'$ cannot be equal to $M$. – Andreas Blass Jul 09 '22 at 16:54
  • @AndreasBlass Isn't this conclusion too strong? If $M = M'$ for some $T$ (in our case, the toy example was $\mathsf{PA}$), this means that $T$ is in fact inconsistent, per the above argument. But we can still work with theories too weak for Incompleteness, such as Presburger Arithmetic. – God bless Jul 09 '22 at 16:57
  • The only way $M'$ exists at all is if $M$ thinks $T$ is consistent. – Andreas Blass Jul 09 '22 at 17:00
  • @AndreasBlass What you seem to be saying is that my argument proves that $M \neq M'$ by contradiction. That makes sense.

    I think the main thing I'd like to understand here is how to formalize "thinking" inside of a model. Thus far, I've always taken syntactic deduction to exist over and above all models.

     – God bless Jul 09 '22 at 17:07
  • @AndreasBlass I've read my argument again, and it seems to me that it's a $\alpha, \beta \vdash \bot$ kind of situation. If we want to hold that $\mathsf{PA}$ is consistent, we have to part with $M = M'$ and vice versa, if we want to hold that $M = M'$, we have to say that $\mathsf{PA}$ is inconsistent. – God bless Jul 09 '22 at 17:09

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