Show that if $S \cup \{ \neg A\} \vdash B$ then $S \vdash A$

Question

As mentioned in the title. I need to show that

$S \vdash A$ when I know that $S \cup \{ \neg A \} \vdash B$

And I can't use completeness theorem here.

I know I can say

$S \vdash \neg A \rightarrow B$

due to what we know. However I don't see how thats helpful. I have also tried to show that $S \cup \{ \neg A \}$ is inconsitent by trying to deduct $\neg B$ or A from it but no success thus far.

Edit.

Just to make sure this is bit more clear as I think the initial version might be bit messy.

So let S be a group of propositional formulas and A a propositional formula. Lets assume that for all propositional formulas B, $S \cup \{ \neg A \} \vdash B$. Now show without completeness theorem that $S \vdash A$.

Answer 1

A theory proves all propositional formulas if and only if it is inconsistent. As such, the assumption that $S \cup \{ \neg A \} \vdash B$ for all formulae $B$ implies that $S \cup \{ \neg A \}$ is inconsistent. Now either $S$ is consistent or it is inconsistent:

• If $S$ is inconsistent, then $S \vdash A$.
• If $S$ is consistent then so is $S \cup \{ C \}$ for any $C$ for which $S \vdash C$. Since $S \cup \{ \neg A \}$ is inconsistent, it follows that $S \nvdash \neg A$, so that $S \vdash A$ by completeness.

In either case we see that, $S \vdash A$.