i have a simple question: assume i have something like ($\Gamma \cup \{\delta \} \models \lnot \gamma$) AND ($\Gamma \cup \{\delta\} \models \gamma$). How may I show that $\Gamma \vDash \lnot \delta$? now i understand that it occurs because of the principle of explosion $(\forall a,b :(a \land \lnot a) \vdash b$.
my question is how to mathmatically write that
thank you very much for your help!
On
$\Gamma\cup\{\delta\}\models\gamma$ means that statement $\gamma$ is always true in any evaluation that satisfies $\Gamma\cup\{\delta\}$, while $\Gamma\cup\{\delta\}\models\neg \gamma$ means that the statement $\gamma$ is never true in any evaluation that satisfies $\Gamma\cup\{\delta\}$.
This establishes that there can be no such evaluation that may satisfy both $\Gamma$ and $\{\delta\}$.
Hence if there is any evaluation which can satisfy the statements in $\Gamma$, that evaluation must deny $\delta$.
That is to say, $\Gamma\models\neg\delta$ is infered from $\Gamma\cup\{\delta\}\models\gamma$ and $\Gamma\cup\{\delta\}\models\neg\gamma$.
$$\begin{split}\Gamma\cup\{\delta\}&\models\gamma\\\Gamma\cup\{\delta\}&\models\neg\gamma\\\hline \Gamma&\models\neg \delta\end{split}$$
Because
$$\begin{array}\\ \gamma\\ \gamma\ \cup\{\delta\}\implies \lnot\ \gamma\\ \gamma\ \cup\{\delta\}\implies \gamma\\ \delta\\ \hline \gamma\ \land \lnot\ \gamma \end{array}$$
So $\delta$ has to be false, or you will arrive at a contradiction.