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We call a set of formulas $\Sigma$ of a language $L$ consistent if there is no $\varphi$ in $L$ such that $\Sigma \vdash \varphi$ and $\Sigma \vdash \lnot \varphi$.

Apparently, an equivalent formulation is the following:

A set $\Sigma$ of formulas of $L$ is consistent iff $\Sigma \not\vdash \varphi$ for some sentence $\varphi$ of $L$.

The $\implies$ direction is clear: if we can prove all sentences then we can prove both $\varphi$ and $\lnot \varphi$ so that $\Sigma$ is inconsistent.

But I don't immediately see how to prove $\Longleftarrow$. Can someone explain this to me? Thanks!

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Hm ... seems to depend on the conclusion rules you are using, but I'd say something along the lines $\Sigma \vdash \varphi, \neg \varphi$ gives $\Sigma \vdash \varphi \land\neg\varphi$ and by $\Sigma \vdash(\varphi \land \neg\varphi \to \psi)$ (logical axiom) we have by MP $\Sigma \vdash \psi$. –  martini Nov 19 '12 at 10:48
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2 Answers

up vote 7 down vote accepted

Assume that $\Sigma$ is consistent then it cannot prove $\varphi\land\lnot\varphi$. Therefore there exists a sentence which it does not prove.

Assume that $\Sigma$ is inconsistent then it proves everything (using the principle of explosion). Therefore there is no sentence it does not prove.

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Thank you! I like concise answers. –  Matt N. Nov 22 '12 at 15:22
@Matt: I know you do. :-) I like writing long answers, though. ;-) –  Asaf Karagila Nov 22 '12 at 15:23
I know you do. ${}$ –  Matt N. Nov 22 '12 at 15:25
Perhaps you could consider writing a set theoretic novel, set 42 B.C. -- 42 before Cantor? –  Matt N. Nov 22 '12 at 15:26
@Matt: Ha. I might take you on that. It will take me a "non-standard integer"-many days to write such novel, though. –  Asaf Karagila Nov 22 '12 at 15:28
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The characterizations are not equivalent in general. Take the most trivial case: suppose $L$ is a negation-free language. Then vacuously, the set $\Sigma$ of all $L$-formulae is consistent in the first sense, but not in the second sense.

However, in a language with negation and in the presence of ex contradictione quodlibet (the rule that from $\varphi$ and $\neg\varphi$ you can derive any $\psi$), the two characterizations become co-extensional.

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I am fairly certain this is in the context of classical FOL. –  Asaf Karagila Nov 19 '12 at 11:05
@AsafKaragila Sure thing, they come to the same in FOL. But it was important to Post, to whom we owe the second definition, that it applies more widely. So I thought it was worth emphasizing that, in a more general setting, the definitions peel apart. –  Peter Smith Nov 19 '12 at 11:17
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