This is a somewhat open-ended question — I’m curious whether there are any known results which are able to prove that one theory is not recursively-axiomatizable over another, despite both having the same turing degree. In particular the situation I’m curious about is the following:

We have two first-order theories $A, B$ over a common language, and we want to show that for any recursively-enumerable subtheory $T \subseteq B$, $A + T \not\vdash B$.

Of course there is a straightforward reason this could be true, namely that $A$ is R.E. while $B$ is not, or more generally that $B$ has a sufficiently higher turing degree then $A$. In such cases we can invoke the first-order completeness theorem to argue that if $B$ were recursively-axiomatizable over $A$, then $B$ would in fact be recursively enumerable with oracle access to an enumeration of $A$, which would contradict the fact that $B$ is “harder to enumerate” then $A$.

What I’m curious about is if their are known methods for proving such results in the case where $A$ and $B$ have the same degree of undecidability, so that we would not obtain a contradiction from that existence of an enumeration of $B$ with an oracle for $A$, but nonetheless $B$ cannot be recursively axiomatized over $A$ for some more “logical”/model-theoretic reasons.