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 $M\models\neg\phi$
Best Answer
The only way a theory can be inconsistent is if it has no models. Because your observation is right: if a theory has a model $M$, then we can never have $M \models \phi$ and $M \models \neg \phi$ at the same time. So:
Now for your actual question: we have $\Gamma \not \models \neg \phi$. So there is a model $M$ of $\Gamma$, such that $M \not \models \neg \phi$. That is, $M \models \phi$. But then $M$ is a model of $\Gamma \cup \{\phi\}$, so we see that that theory is consistent.
Note that we did not need that $\Gamma \not \models \phi$. Indeed, if we would have $\Gamma \models \phi$ and $\Gamma$ is consistent, then $\phi$ is already true in every model so $\Gamma \cup \{\phi\}$ is definitely consistent (in fact, $\Gamma$ and $\Gamma \cup \{\phi\}$ are equivalent theories).