# $\Gamma\models\phi$ 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.

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 $$\Gamma,\neg\phi\not\models\psi\land\neg\psi$$.

• Ah yes thats what i meant , yes Thank you my friend ! Mar 30 '19 at 11:00