Logic : unsatisfiable set

Question

It is obvious that for a set $\Phi$ of well-formed formulas, if $\Phi\cup\left\{\alpha\right\}$ is unsatisfiable and $\Phi\cup\left\{\left(\neg\alpha\right)\right\}$ is unsatisfiable, then $\Phi$ itself is unsatisfiable.

I understand that this is because it would mean that $\Phi$ would already contain both $\alpha$ and $\left(\neg\alpha\right)$, which is a contradiction, thereby making the set unsatisfiable by definition.

How can I rigorously prove that?

Answer 1

Assume for a contradiction that $\Phi$ is satisfiable. Consider an interpretation which satisfies $\Phi$. That interpretation must satisfy either $\Phi\cup\{\alpha\}$ or $\Phi\cup\{\neg\alpha\}$, contradicting the assumption that neither is satisfiable.

Answer 2

If $\Phi$ is satisfiable, then $\Phi \cup$ {$\alpha$} is satisfiable, if $\alpha$ is true and $\Phi \cup$ {$\neg \alpha$} is satisfiable if $\alpha$ is false.