Set of logical statements with contradiction. Let $\Sigma_1, \Sigma_2$ be two finite sets of statements in propositional calculus s.t  $\Sigma_1\cup\Sigma_2$ has a contradiction. Prove that there is a statement A s.t $$\Sigma_1\vdash{A}\  and\  \Sigma_2\vdash{\neg{A}}$$

is it still true with infinite  $\Sigma_1,\Sigma_2$?

My attempt:

in case one of the sets is $\phi$ or one of the sets has a contradiction its obvious...
So, we left with the case where both aren't empty and both without contradiction.

I didn't manage to prove this... Any kind of help will be appreciated Thanks!
Let $A_1,\ldots,A_n\in\Sigma_1$ and $B_1,\ldots,B_m\in\Sigma_2$ the sentences that generate the contradiction.

Let $A=A_1\wedge\ldots\wedge A_n$. It is clear that $A_1\wedge\ldots\wedge A_n\vdash A$. Since propositional calculus is sound, $A\wedge(B_1\wedge\ldots\wedge B_m)$ must be false in $\Sigma_2$, or, equivalently, $(B_1\wedge\ldots\wedge B_m)\to\neg A$ must be true. Since propositional calculus is complete, $(B_1\wedge\ldots\wedge B_m)\vdash\neg A$.

For the infinite sets case, I have been thinking, but no result so far.
There are a couple of problems with your puzzle.


the idea that $\Sigma_1, \Sigma_2$ be two finite sets of statements is in itself allready a contradiction, if they are closed onder modus ponens and so they are infinite sets , there are just an infinite number of theorems that belong to both sets.
If contradiction of a set means that  $ (\Sigma_1\cup\Sigma_2\vdash {A} $  and $ \Sigma_1\cup\Sigma_2\vdash{\neg{A}}$ you cannot tell which if $ A$ is from $\Sigma_1 $  or from $ \Sigma_2$ so how can you proof that $\Sigma_1\vdash {A} $?