Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
sounds sound :) – Hagen von Eitzen Apr 6 '13 at 15:47
up vote 2 down vote accepted

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 $\phi\in\Gamma$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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