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Assume that $\Gamma$ is a maximally consistent set of formulas of $\mathcal{L}$. Show that if $\varphi$ is a validity, then $\varphi \in \Gamma$.


Can I check if what I am doing is sound, no pun intended? Please point out any mistakes! Sincere thanks.

Let $\varphi$ be a validity.

So $(\mathcal{M},\nu)\models\varphi$ for all $\mathcal{M}$ and for all $\nu$.

Suppose $\varphi\notin \Gamma$. Then, $(\neg \varphi )\in\Gamma$. $\therefore \Gamma \vdash (\neg \varphi )$.

By Completeness Theorem, $\Gamma$ is consistent implies $\Gamma$ is satisfiable.

Let $\mathcal{M}$ be a structure and $\nu$ an $\mathcal{M}$-assignment such that $(\mathcal{M},\nu ) \models \Gamma$.

By Soundness, $( \mathcal{M}, \nu )\models (\neg \varphi )$

This is a contradiction.

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    $\begingroup$ sounds sound :) $\endgroup$ – Hagen von Eitzen Apr 6 '13 at 15:47
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Assuming that all the results you're invoking (for example that a maximal consistent set must contain either $\phi$ or $\neg\phi$) are available, your argument looks correct but unnecessarily complicated. You could just say that, since $\Gamma$ is consistent, it is satisfied by some $(\mathcal M,\nu)$, and the same $(\mathcal M,\nu)$ then satisfies $\Gamma\cup\{\phi\}$ because $\phi$ is valid. So $\Gamma\cup\{\phi\}$ is consistent; by maximality of $\Gamma$, we get $\phi\in\Gamma$.

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