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More precisely, let $T$ be a theory of some first-order language. My question is then the following. What does it mean for two sentences $\sigma_1$ and $\sigma_2$ in the same language to be independent of each other in the theory $T$?

My "hypothesis" is that it means $\neg \, ( \, T \vdash \sigma_1 \leftrightarrow \sigma_2 \,)$. However, I could only come up with this "hypothesis" by looking at some particular examples; also I could not immediately find an online reference where this concept is defined, hence this question. Can somebody confirm this?

Note. I do know what it means for one formula $\sigma$ to be independent from a theory $T$. Namely it means that $\neg \, ( \, T \vdash \sigma\,) \; \wedge \; \neg \, ( T \vdash \neg \sigma)$.

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It’s even stronger than that: it means that $T\cup\{\sigma_1\}\not\vdash\sigma_2$ and $T\cup\{\sigma_2\}\not\vdash\sigma_1$. In other words, given $T$, neither of $\sigma_1$ and $\sigma_2$ entails the other. For example, in the context of $\mathsf{ZF}$ set theory the continuum hypothesis and the axiom of choice are independent of each other: assuming that $\mathsf{ZF}$ is consistent, so are $\mathsf{ZF}+\mathsf{CH}+\neg\mathsf{AC}$ and $\mathsf{ZF}+\mathsf{AC}+\neg\mathsf{CH}$.

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  • $\begingroup$ Are you sure that this is the most "common" way of defining this? Because if this is true then it seems to me that, under this definition, a theorem in Jech's book The Axiom of Choice (Theorem 4.7 on page 51) becomes false: he claims in this theorem that "The Axiom of Choice is independent of the Ordering Principle in set theory with atoms." The Ordering Principle is the statement that any set can be linearly ordered; obviously the Axiom of Choice implies this, because the Axiom of Choice is equivalence with the Well-Ordering Theorem. $\endgroup$
    – user93680
    Sep 8, 2013 at 2:10
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    $\begingroup$ Brian's answer is correct. But there is abuse, as you see. In this particular case, what Jech means is that $\mathsf{ZFA}+$the ordering principle$ does not imply the axiom of choice. $\endgroup$
    – Andrés E. Caicedo
    Sep 8, 2013 at 2:14
  • $\begingroup$ Ok. It is weird that apparently there is not a fixed "convention" about this. This could cause confusion, especially with people who do not have a lot of experience in these matters. $\endgroup$
    – user93680
    Sep 8, 2013 at 2:53
  • $\begingroup$ (But now you know.) It is fairly common, actually, to talk of independence when one really means the restricted sense in which Jech uses the phrase. (Kunen's book is subtitled "Introduction to independence proofs", for example.) $\endgroup$
    – Andrés E. Caicedo
    Sep 8, 2013 at 2:56
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    $\begingroup$ @Andres Caicedo: Of course ZFA + the ordering principle does not disprove AC either, so Jech's claim that AC is independent of OP does not seem to be a "limited sense". $\endgroup$
    – Carl Mummert
    Sep 8, 2013 at 17:12

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