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.

| cite | improve this question | | | | |
  • 1
    $\begingroup$ sounds sound :) $\endgroup$ – Hagen von Eitzen Apr 6 '13 at 15:47

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 $\phi\in\Gamma$.

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.