Show that $\Gamma \cup \{\neg \phi\}$ is satisfiable if and only if $\Gamma\not \models \phi$

Let $\Gamma$ be a set of formulas and $\phi$ be a formula. Show that $\Gamma \cup \{\neg \phi\}$ is satisfiable if and only if $\Gamma\not \models \phi$.

This seemed pretty obvious but I wanted to see if my proof made sense:

Proof: $(\Rightarrow)$

To derive for a contradiction, suppose that: $\Gamma \models \phi$. That means for all truth assignments $v$, for $\gamma \in \Gamma$, if $v(\gamma) = T$, then $v(\phi) = T$.

But this contradicts our assumption that $\Gamma \cup \{\neg \phi \}$ is satisfiable by the fact that $v(\phi) = T$ cannot happen so: $\Gamma \not \models \phi$ .

$(\Leftarrow)$

So by the definition of $\Gamma \not \models \phi$, we have that there is some truth assignment $v$ which satisfies $\Gamma$ but does not satisfy $\phi$. So that means $v(\phi) = F$ since $v(\phi) \not = T$ which implies that $v(\neg \phi) = T$ which means $v$ satisfies $\Gamma \cup \{\neg \phi\}$.

I feel like I'm missing something in the forward direction, but at the same time... It looks pretty trivial as well. Am I missing anything crucial?

Thank you!

• Your proof is correct. – Mauro ALLEGRANZA Oct 18 '14 at 9:07