$\Gamma\models\phi$ if and only if $\Gamma,\neg\phi\models\psi\land\neg\psi$ Let $\Gamma\cup\{\phi,\psi\}\subseteq L \epsilon$ then $\Gamma\models\psi$ if and only if $\Gamma,(\neg\phi)\models(\psi\land(\neg\psi))$. I don't seem to understand how the reverse implication goes. Can anyone help me out ? Thanks.
 A: Written like this, it makes no sense. I assume you wanted to write
$$\Gamma\models\phi\text{ iff }\Gamma,\neg\phi\models\psi\land\neg\psi$$

To prove this, it is helpful to note that for $\Delta\cup\{\psi\}\subseteq\mathcal L_{FO}$, $\Delta\not\models\psi\land\neg\psi$ iff $\Delta$ is satisfiable, as then there is an interpretation $\mathcal I$ s.t. $\mathcal I\models\Delta$, and naturally $\mathcal I\not\models\psi\land\neg\psi$.
Now on to proving the equivalence. Let $\Gamma\cup\{\phi,\psi\}\subseteq\mathcal L_{FO}$.
From left to right, assume $\Gamma\models\phi$, i.e. for every interpretation $\mathcal I$: $\mathcal I\models\Gamma$ implies $\mathcal I\models\phi$. Thus, no interpretation $\mathcal I$ models $\Gamma,\neg\phi$ and thus for every interpretation $\mathcal I$: $\mathcal I\models\Gamma,\neg\phi$ implies $\mathcal I\models\psi\land\neg\psi$.
From right to left, assume $\Gamma\not\models\phi$, i.e. there is an interpretation $\mathcal I$ s.t. $\mathcal I\models\Gamma$ but $\mathcal I\not\models\phi$. The latter implies $\mathcal I\models\neg\phi$. Thus $\mathcal I\models\Gamma,\neg\phi$, i.e. $\Gamma,\neg\phi$ is satisfiable and thus $\Gamma,\neg\phi\not\models\psi\land\neg\psi$.
