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 than $A$. 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.
