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 $M\models\neg\phi$


1 Answer 1


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).

  • $\begingroup$ 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? $\endgroup$ Nov 14, 2019 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. $\endgroup$ Nov 14, 2019 at 0:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.