From comments by user Mauro ALLEGRANZA:
> If $\Sigma \models A$ and $\Sigma \models \
eg A$, this means that there is **no** valuation that satisfy $\Sigma$.
>
> Apply this to show $\Sigma \models B$ by contradiction.
From comments by user Mauro ALLEGRANZA:
> If $\Sigma \models A$ and $\Sigma \models \
eg A$, this means that there is **no** valuation that satisfy $\Sigma$.
>
> Apply this to show $\Sigma \models B$ by contradiction.