Proving the following about sets of propositions.

Question

I'm wondering how proving this question.If $\Gamma_1 \cup \Gamma_2 \vdash \bot$ then there exists a proposition such that $\Gamma_1\vdash A$ and $\Gamma_2 \vdash \lnot A$.I tried some idea.for instance we know that if $\Gamma_1 \cup \Gamma_2 \vdash \bot $ then there exists a finite set of propositions $\Delta$ that is subset of $\Gamma_1\cup \Gamma_2$ and $\Delta \vdash \bot$ and so there exists $\Delta_1$ and $\Delta_2$ and $A$ such that $\Delta_1 \cup \Delta_2 =\Delta$ and $\Delta_1\vdash A$ and $\Delta_2 \vdash \lnot A$ now it's enough to prove that for example $\Delta_1 \subset \Gamma_1$ and $\Delta_2 \subset \Gamma_2$ but I don't see this.I would appreciate any help.

Answer 1

You're almost done.

For the last short step, let $A$ be the conjunction of the finitely many elements of $\Delta_1$.