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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 question 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." –  Arkamis Aug 10 '12 at 4:12
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up vote 4 down vote accepted

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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So my question is then, if a proof is complete without using $P$, then you can adjoin $\neg P$ to your list of axioms; however, you could also adjoin $P$ to your list of axioms. Which is correct? –  Arkamis Aug 10 '12 at 4:21
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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. –  Kris Aug 10 '12 at 4:45
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