I am trying to prove the equivalence of two ways soundness of propositional logic is presented.
Soundness version 1 : If $\Gamma \vdash A$ then $\Gamma \models A$.
Soundness version 2 : If $\Gamma$ has a model (Satisfiable) then $\Gamma$ is consistent
As part of the proof from version 2 to version 1, I need to prove that if $\Gamma \cup \{\lnot\phi\}$ has no model, then $\Gamma \models \phi$.
How do I prove this? Any help appreciated.
If $$\Gamma\not \models A$$ that means that there is a model of $\Gamma\cup\{\neg A\}$ and so by $2$, $\Gamma\cup\{\neg A\}$ is consistent. Therefore $$\Gamma\not \vdash A$$ ]eg