# >Prove that if $\Gamma \not\models \phi$ and $\Gamma \not\models \neg \phi$ then $\Gamma\cup \{\phi\}$ is consistent.

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 $$M\in mod(\Gamma)$$, $$M\models \theta$$

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 $$M\models\neg\phi$$

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:
A theory $$\Gamma$$ is inconsistent if and only if $$\Gamma$$ has no models.
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).