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

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 $S \vdash A$.

• It is not true... You have to check $B$: maybe it must be different ($\bot$ ?). – Mauro ALLEGRANZA Mar 4 '17 at 18:43
• Are you aware of the definition of inconsistent set $S$ of formulas ? – Mauro ALLEGRANZA Mar 4 '17 at 18:48
• What you mean by 'check B'? Also I assume I am familiar with definition of inconsistent set S of formulas but not entirely sure as English is not my native language. – E.K. Mar 4 '17 at 18:52
• Take $S=\varnothing$ and $B=\lnot A$, then easily $S\cup\{\lnot A\}\vdash B$, but unless $A$ is a tautology, $S$ cannot prove $A$. – Asaf Karagila Mar 4 '17 at 19:22
• @E.K.: What was missing was the "for all propositional formulas $B$" part. – Clive Newstead Mar 4 '17 at 19:33

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, $S \vdash A$.
• Been reading this answer and some materials and while this is obviously correct answer I think this could be done little bit more efficient by using some theorems. We have a theorem which says the following are equivalent 1) $S \vdash A$ 2) $S \cup \{\neg A\}$ is inconsistent. If we use this wouldn't the second case be unnecessary? – E.K. Mar 6 '17 at 11:23