Is failing to admit an axiom equivalent to proof when the axiom is false? Often, mathematicians wish to develop proofs without admitting certain axioms (e.g. the axiom of choice).
If a statement can be proven without admitting that axiom, does that mean the statement is also true when the axiom is considered to be false?
I have tried to construct a counter-example, but in every instance I can conceive, the counter-example depends on a definition which necessarily admits an axiom. I feel like the answer to my question is obvious, but maybe I am just out of practice.

Yes. Let the axiom be P. The proof that didn't make use of P followed all the rules of logic, so it still holds when you adjoin $\neg P$ to the list of axioms. (It could also happen that the other axioms sufficed to prove P, in which case the system that included $\neg P$ would be inconsistent. In an inconsistent theory, every proposition can be proved, so the thing you originally proved is still true, although vacuously. The case where the other axioms prove $\neg P$ is also OK, obviously.)
