T $\models \varphi$ iff $T \cup \{\lnot \varphi\}$ inconsistent, proof verification

Question

I would like to show that:

T $\models \varphi$ iff $T \cup \{\lnot \varphi\}$ inconsistent

My attempt:

$(\Rightarrow)$ Suppose that $T \models \varphi$. Then $\varphi$ is true in every model of $T$. Since $\varphi$ and $\lnot \varphi$ cannot be true in the same model, there doesn't exists a model where $\lnot \varphi$ is true. So we look at $T \cup \{\lnot \varphi\}$. If $M$ is a model of $T \cup \{\lnot \varphi\}$, then $M \models \psi$ for all $\psi \in T \cup \{\lnot \varphi\}$. Since this is not true for $\lnot \varphi$, there doesn't exists a model of $T \cup \{\lnot \varphi\}$, so $T \cup \{\lnot \varphi\}$ is inconsistent.

$(\Leftarrow)$ Suppose that $T \not\models \varphi$. Then there doesn't exists a model where $\varphi$ is true, so $\lnot \varphi$ is true in every model of $T$. So, $T \cup \{\lnot \varphi\}$ has a model, but that is impossible since $T \cup \{\lnot \varphi\}$ is inconsistent, so $T \models \varphi$

Can somebody check this proof, and correct me if I am wrong?

Answer 1

Your $(\Rightarrow)$ direction is correct. For the other direction you are on the right track, but there is an error. The statement $T \not \models \varphi$ does not mean that there does not exist a model of $T$ where $\varphi$ is true. It just says that $\varphi$ is not true in every model of $T$. In other words, there is some model of $T$ where $\varphi$ is false. Can you now tie this to your last sentence to arrive at the required conclusion again?

I am sure you will be able to complete the proof. Just in case you want to double check, I have put the full proof for the $(\Leftarrow)$ in a spoiler tag below.

We prove the contraposition. So suppose that $T \not \models \varphi$. Then there is a model $M$ of $T$ such that $M \not \models \varphi$. That is $M \models \neg \varphi$, and thus $M \models T \cup \{\neg \varphi\}$. We thus see that $T \cup \{ \neg \varphi \}$ is consistent, which concludes the proof.