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

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).

• From a quick glance, won't your $T + \operatorname{Con}_T$ be consistent if and only if $T+\operatorname{Con}_{\operatorname{PA}}$ is consistent? (the $⇒$ direction is trivial, and the $⇐$ direction follows from $\neg \operatorname{Con}(\operatorname{PA} + \operatorname{Con_{PA}})+\operatorname{Con_{PA}}$ implies $\neg \operatorname{Con}(\operatorname{PA} + \operatorname{Con_{PA}})+\operatorname{Con}(\operatorname{PA}+¬\operatorname{Con_{PA}})$
– ℋolo
Commented Jan 10, 2023 at 19:05

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.