Given a fixed language $L$ a fixed consistent theory $\Gamma$ and a fixed $L$ sentence $\phi$
Prove that if $\Gamma \not\models \phi$ and $\Gamma \not\models \neg \phi$ then $\Gamma\cup \{\phi\}$ is consistent.
Definition of consistent: $\Gamma$ is consistent if there is no sentence $\phi$ such that $\Gamma \models \phi$ and $\Gamma \models \neg\phi$
Definition of consequence: A sentence $\theta$ is a consequence of a theory $\Gamma$ if for all $M\in mod(\Gamma)$, $M\models \theta$
Proof Contradiction:
Suppose $\Gamma\not\models \phi$ and $\Gamma\not\models \neg\phi$ and that $\Gamma^\prime$ is inconsistent.
Since $\Gamma\subseteq \Gamma^\prime$, $\text{mod}(\Gamma^\prime)\subseteq \text{mod}(\Gamma)$.
And since $\Gamma^\prime$ is inconsistent there exists $\psi$, such that for all $M\in\text{mod}(\Gamma^\prime)$, $M\models\psi$ and $M\models\neg\psi$
I can't figure out where to go from here. I don't see how you can compare the sentences in theories.
I also don't understand how a theory can be inconsistent since the definition of truth for an $L$ structure $M$, is that if $M\models \phi$ $\iff$ $M\not\models \neg\phi$. So how can there be sentences where every $M\models\phi$ and $M\models\neg\phi$