# Can every first order theory have a finitely axiomatizable conservative extension in this sense?

Note: This is an edit of the previous question.

By $$T_2$$ conservatively extends $$T_1$$ if and only if:

There exists a function $$F$$ such that $$T_2$$ extends $$T_1$$ through $$F$$; and for every function $$G$$ such that $$T_2$$ extends $$T_1$$ through $$G$$, we have $$T_2$$ conservatively extends $$T_1$$ through $$G$$.

For definitions of the above terminology see "About conservative extensions of First Order theories?"

Now lets assume $$T_1$$ and $$T_2$$ to have the same language, like for example with the case of $$ZFC$$ and $$Predicative \ MK$$ set theories, where the later is a weakening of Morse-Kelley set theory by restricting class comprehension scheme to formulas in which all quantifiers are bounded in $$V$$. I think that $$Predicative \ MK$$ would conservatively extends $$ZFC$$ according to the above definition. The former theory is finitely axiomatizable.

Now my question is can we apply methods present in the answer to this question as to get also finitely axiomatizable conservative extensions [in the above sense] for all first order theories [meeting qualifications in that posting] even if they were written in the same language?

• Such a theory clearly has to be effectively axiomatizable. In the light of that, how does the linked question not answer yours? – Wojowu Mar 27 at 19:41
• the answer to the linked question included a change in the language of the interpreting theory. I demand it here to be made in the SAME language – Zuhair Al-Johar Mar 27 at 19:43
• In the language with only equality, let $T$ be the theory of an infinite set. The axioms are the infinitely many sentences expressing "there are at least $n$ elements" for each natural number $n$. Since $T$ is complete, its only consistent extension in the same language is $T$ itself, and that isn't finitely axiomatized. – Andreas Blass Mar 28 at 12:24
• No, incompleteness doesn't help. Note that the only conservative extension of a theory in the same language would be the same theory, just axiomatized differently. So any incomplete theory that isn't finitely axiomatizable will be a counterexample. One example is the theory, in the language with only equality, that says "there are not exactly $n$ objects" for all natural numbers $n$ except $7$. – Andreas Blass Mar 28 at 12:42
• @ZuhairAl-Johar That’s not a conservative extension, but a faithful interpretation. Any given theory has one, and only one, conservative extension in the same language, namely itself. You are confusing yourself by using sloppy terminology. If you choose to formulate ZF and this weakening of MK in the same language (as opposed to, say, formulating MK in a two-sorted language), then MK is not an extension of ZF at all, let alone a conservative extension. For example, ZF proves $\forall x\,\exists y\,x\in y$, while MK does not (it actually proves its negation). – Emil Jeřábek Mar 28 at 14:17