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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 $M\models\neg\phi$

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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).

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  • $\begingroup$ Ok. My definition of inconsistent is just not being consistent but this makes sense. Is proving that a model being enough for consistency stronger then what I'm trying to prove though? $\endgroup$ – AColoredReptile Nov 14 '19 at 0:31
  • $\begingroup$ @AColoredReptile Your definition is correct: consistent is the negation of inconsistent. As I explained, being inconsistent is equivalent to having no models. So being consistent is then equivalent to having a model. So I used no stronger notions. $\endgroup$ – Mark Kamsma Nov 14 '19 at 0:46

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