We say that $\Gamma$ is consistent if for no $\phi$, then both $\Gamma\vdash \neg\phi$ and $\Gamma\vdash\phi$. We say that $\Gamma$ is satisfiable if there is a structure $M$ s.t. $\models_M\gamma[s]$ for alle $\gamma\in\Gamma$.
I want to show that if $\Gamma$ is satisfiable, then $\Gamma$ is consistent.
I'm thinking we assume that $\Gamma$ is not consistent, and then show that $\Gamma$ is not satisfiable. So if $\Gamma$ is not consistent then both $\Gamma\vdash \neg\phi$ and $\Gamma\vdash\phi$ for some $\phi$. But by soundness theorem then $\Gamma\models\neg\phi$ and $\Gamma\models\phi$, meaning for structure $M$, we have $\models_M\neg\phi[s]$ (whereas $\not\models_M\phi[s]$) and $\models_M\phi[s]$. So would this mean that $\Gamma$ is not satisfiable? I'm a little confused, because wouldn't this instead be a contradiction?
The soundness theorem does not say that if $\Gamma\vdash\phi$ then $\models_M\phi$. It says that if $\Gamma\vdash\phi$ then $\Gamma\models\phi$, meaning that in any $M$ in which $\Gamma$ is satisfied we have $\models_M\phi$.
So if $\Gamma$ is satisfiable then there exist $M$ with $\models_M\phi$ and also $\models_M\neg\phi$; since that's impossible, $\Gamma$ is not satisfiable.