Why is it true that if $\Sigma$ satisfies $\alpha$ and also not-$\alpha$ then $\Sigma$ is not satisfiable?
Is it true at all? it doesn't make any sense to me and I would like to know more about that statement.
If $\Sigma$ satisfies $\alpha$ and $\neg \alpha$ it means that $z(\Sigma) = z(\alpha)$ and $z(\Sigma) = z(\neg \alpha)$ for every $z$, s.t $z(\Sigma)=1$. But if $\Sigma$ is not even satisfiable it means that there isn't such $z$ s.t $z(\Sigma)=1$.
I'm confused. How can this be proven?
Let $\alpha$ be a well-formed formula.
Argue by contradiction.
Assume that $\Sigma$ satisfies both $\alpha$ and $\neg \alpha$ and assume that $\Sigma$ is satisfiable. Since $\Sigma$ is satisfiable, by definition, there is a valuation $z$ such that $\forall \phi \in \Sigma(z(\phi)=1)$. And from $\Sigma$ satisfying both $\alpha$ and $\neg \alpha$ you can conclude.