Arithmetic theory $T$ such that $U = T + \operatorname{Con}_{T}$ is consistent, but $T + \operatorname{Con}_{U}$ is not.

Question

I have the following problem:

Give an example of arithmetic theory $T$ (i.e $T$ contains $\operatorname{PA}$ and has recursively enumerable set of axioms), such that theory $U = T + \operatorname{Con}_{T}$ is consistent, but theory $T + \operatorname{Con}_{U}$ is inconsistent.

It was already discussed here that we can easily have a consistent arithmetic theory $T$, such that $T + \operatorname{Con}_T$ is inconsistent by simply taking $T = \operatorname{PA} + \neg \operatorname{Con}_{PA}$. But what if we want to reach inconsistency by taking $\operatorname{Con}$ twice and not once?

My initial idea was to take $T = \operatorname{PA} + \neg \operatorname{Con}(\operatorname{PA} + \operatorname{Con_{PA}})$. Then $T + \operatorname{Con}_U$ is inconsistent, but I was not able to show that $T + \operatorname{Con}_T$ is consistent (and now I even think it is not the case).

Answer 1

Your initial idea is correct. $\DeclareMathOperator{Con}{Con}$ We can indeed prove that $T + \Con_T$ is consistent.

To do so, suppose $PA$ is consistent. Then by Gödel’s second incompleteness theorem, $PA + \neg \Con(PA)$ is also consistent. And clearly, $PA \vdash \Con(PA + \Con(PA)) \implies \Con(PA)$; thus, contrapositively, $PA + \neg \Con(PA) \vdash \neg \Con(PA + \Con(PA))$. So we see that $PA + \neg \Con(PA + \Con(PA))$ is consistent. That is, $T$ is consistent.

We can formalise the above argument in PA itself to show that $PA + \Con(PA) \vdash \Con(T)$. Now $\mathbb{N} \models PA$, so $PA$ is consistent. Thus, $\mathbb{N} \models \Con(PA)$, so moreover $PA + \Con(PA)$ is consistent. Apply Gödel’s incompleteness theorem to get that $Q = PA + \Con(PA) + \neg \Con(PA + \Con(PA))$ is consistent. Then $Q \vdash T$ and $Q \vdash \Con_T$, so $T + \Con_T$ is consistent.