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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.

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    I suppose the questio$n$ is equivalent to "is failing to admit an axiom equivalent to when the axiom is true," or "is proof without admitting an axiom equivalent to independence of that axiom."2012-08-10

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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.)

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    Neither is 'correct', whatever that word means. It just means that axiom $P$ has nothing to do with your statement. I could add as an axiom that 'Reptilians are in charge of the Earth' to the axioms of Euclid and still prove Pythagoras' theorem.2012-08-10