# Does "every" first-order theory have a finitely axiomatizable conservative extension?

I've now asked this question on mathoverflow here.

There's a famous theorem (due to Montague) that states that if $$\sf ZFC$$ is consistent then it cannot be finitely axiomatized. However $$\sf NBG$$ set theory is a conservative extension of $$\sf ZFC$$ that can be finitely axiomatized.

Similarly, if $$\sf PA$$ is consistent then it is not finitely axiomatizable (Ryll-Nardzewski) but it has a conservative extension, $$\sf ACA_0$$, which is finitely axiomatizable. (Usually $$\sf ACA_0$$ is considered a second-order theory, but my understanding is that there isn't really a difference between first and second-order from the syntactic point of view.)

I'm wondering if this happens in general. At first I thought it might be that every theory had a finitely axiomatizable conservative extension. But then I realised that every theory with a finitely axiomatized conservative extension must be effectively axiomatizable. So we shouldn't hope for theories to have finitely axiomatizable conservative extensions unless they're already effectively axiomatizable. Similarly if we add countably many logical constants to the language and demand that they're all true then that's effectively axiomatizable but doesn't have a finitely axiomatizable conservative extension, so we should restrict our attention to finite languages.

Does every effectively axiomatizable first-order theory over a finite language have a finitely axiomatizable conservative extension?

• Good question. As an aside, does your statement that every theory with a finitely axiomatized conservative extension must be effectively axiomatizable have a more informative proof than arguing that the extended theory is r.e., hence the set of theorems in the unextended theory is r.e. and gives an r.e. axiomatization of itself, but r.e. axiomatizablility is equivalent to effectively axiomatizability? Commented Mar 20, 2019 at 21:01
• @RobArthan The proof I had in mind was exactly the one you gave. Commented Mar 20, 2019 at 21:31
• I asked this on mathoverflow here. Commented Mar 27, 2019 at 11:10
• The same question asked at MathOverflow had an affirmative answer. Commented Jul 11, 2019 at 8:57
• I tried to post a CW answer that pointed to the answer you've got on MO, so there would be something to accept ... but the system thought it looked too trivial and converted it to a comment :-/ Commented Jul 11, 2019 at 9:01

Let me stress that the results above apply to the literal definition of conservative extension, i.e., we extend the language of $$T$$ by additional predicate or function symbols, and we demand that any sentence in the original language is provable in the extension iff it is provable in $$T$$. If we loosen the definition so as to allow additional sorts, or extension by means of a relative interpretation, then every recursively axiomatizable first-order theory has a finitely axiomatized conservative extension.*