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Let $\sigma$ be a consistent set of propositions such that for every set $\gamma$, either $\sigma$ is proofwise stronger than $\gamma$ that is {$\alpha : \sigma \vdash \alpha$} $\supseteq$ {$\alpha : \gamma \vdash \alpha$} or $\sigma \cup \gamma$ is not consistent. Prove that in that case $\sigma$ is maximally consistent.

My proof:

Suppose $\sigma$ is not maximally consistent then there exist a proposition $\alpha$ such that $\sigma \not\vdash \alpha$ and $\sigma \not\vdash \neg\alpha$. In this case, let $\gamma$ = {$\alpha$}. Since $\sigma \not\vdash \alpha$ that means $\sigma$ is not proofwise stronger than $\gamma$, and $\sigma \cup \gamma$ is consistent (by Lemma below) , which contradicts our assumption.

Lemma: For every set of propositions Σ and every proposition α, if Σ $\not\vdash$ α then Σ ∪ {(¬α)} is consistent.

Proof of the lemma: Assume, by way of contradiction, that Σ∪{(¬α)} is not consistent. Then it proves every proposition, in particular Σ ∪ {(¬α)} $\vdash$ α. By the deduction theorem we therefor get Σ $\vdash$ ((¬)α → α). However, $\vdash$(((¬)α → α) → α), for every proposition α. Now, applying MP we get Σ $\vdash$ α, which contradicts our initial assumption.

Is there a gap in my proof?

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up vote 2 down vote accepted

Yes, there's a gap in your proof because the assumption is not that $\sigma$ is proofwise stronger than $\gamma$, but that either $\sigma$ is proofwise stronger than $\gamma$ or $\sigma\cup\gamma$ is not consistent; so you have to address that second possibility, too.

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I have edited my proof to address this possibility could you please take a look? –  Mark Aug 7 '11 at 5:56
    
I don't know if we should call it a possibility; the statement is of the form $A\/B \rightarrow C)$, and you are using contraposition, so that one should then show ~$C \rightarrow$ ~$A^ $~$B $ , so that both non-maximality and non-consistency should follow for contraposition to hold. –  gary Aug 7 '11 at 6:02
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@Mark: I don't immediately see the argument from monotonicity -- could you spell that out? –  joriki Aug 7 '11 at 6:03
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@Mark: Intuitively speaking, if $\sigma \not\vdash \neg\alpha$, then $\sigma \cup\{\alpha\}$ can't be inconsistent. How to prove this formally may depend on what your exact definition of consistency is and what you've already proved and might be able to use. –  joriki Aug 7 '11 at 7:14
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@Mark: Your proof looks OK to me now. –  joriki Aug 8 '11 at 8:26
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