according to HPC:
Let S be a set of sentences and α that is not in S. Prove or disprove :
If $S\cup\{\alpha\} \vdash \beta$ and $S\cup\{\neg \alpha\} \vdash \beta$ then $S\vdash \beta$.
It seems like this is true, i tried to do some examples with a truth table and it always worked, i'm not sure how i can formaly prove this. Thanks in advance !
Edit : the Hilbert system i am refering to :
If you have proved the Deduction Theorem for your Hilbert-style calculus, you know that the premises $S\cup\{\alpha\}\vdash\beta$ and $S\cup\{\neg\alpha\}\vdash\beta$ imply, respectively, $S\vdash \alpha\to\beta$ and $S\vdash\neg\alpha\to\beta$.
Now $(\alpha\to\beta)\to((\neg\alpha\to\beta)\to\beta)$ is a classical propositional tautology. You ought to have shown that your Hilbert-style calculus can prove all classical tautologies, and from that argument you can extract a formal proof of $$\varnothing\vdash(\alpha\to\beta)\to((\neg\alpha\to\beta)\to\beta)$$ which of course also proves $$S\vdash(\alpha\to\beta)\to((\neg\alpha\to\beta)\to\beta)$$ Now $S\vdash\beta$ is only two applications of Modus Ponens away!