Can $T$, $T+A$, and $T+\neg A$ all have different consistency strengths? Let $T$ be a consistent theory, and let $A$ be a statement in the same language. Consider the three theories


*

*$T$

*$T+A$

*$T+\neg A$


Is it possible for them to be pairwise distinct in consistency strength?
As a follow-up, is it possible for $T+A$ and $T+\neg A$ to be incomparable in consistency strength? (Clearly they are both stronger than $T$, so a fortiori this would mean that the three consistency strengths are distinct.)
I'm primarily interested in theories in classical, finitary first-order logic, but if it makes a difference to consider other logics, I'd find that interesting.
If $T+A$ is inconsistent, then $\neg A$ is provable in $T$, so $T+\neg A$ and $T$ are certainly equiconsistent; similarly if $T+\neg A$ is inconsistent, then $T$ and $T+A$ are equiconsistent. So the question is only interesting if all theories involved are consistent.
This question is motivated by the fact that typically if $T$ is ZFC and $A$ is a large cardinal hypothesis, then $T+\neg A$ is modeled by a substructure of any model of $T$ (by "chopping the universe off" at an inaccessible), so that $T$ and $T+\neg A$ are equiconsistent.
 A: You can construct such sentences via fixed point theorem; viz. there exists $\phi$ such that both $Con(ZFC + \phi)$, $Con(ZFC + \neg \phi)$ are unprovable in $ZFC + Con(ZFC)$. Such a $\phi$ is what is called a double jump sentence. A reference for this is P. Lindstrom, Aspects of Incompleteness, 2003.
A: Here is something that might be worth noting.
It is impossible that both $A$ and $\lnot A$ can prove something like $\operatorname{Con}\sf (ZFC)$. Suppose that they were, given any model of $\sf ZFC$ either $A$ holds there or $\lnot A$ holds there. In either case it must be that $\operatorname{Con}\sf (ZFC)$ must hold in that model; so by the completeness theorem $\sf ZFC$ must prove its own consistency.
So it raises the question, what does it mean to have different consistency strengths? Of course that $\sf ZFC$ must have the weakest consistency strength of the three; so we might expect both $\mathsf{ZFC}+A$ and $\mathsf{ZFC}+\lnot A$ to prove $\operatorname{Con}\sf (ZFC)$, since they should be strictly stronger.
But as the above shows, it's impossible for both of them to prove that. So while this is not exactly an argument while such $A$ cannot be found in the case of $\sf ZFC$, or some theory strong enough to be subjected to the second incompleteness theorem, it does suggest that the answer will not come from large cardinal and the likes of it. And that the consistency "jumps" must go in a non-obvious direction here.
