The task is to show that for every set Γ of propositions:
If $Γ,A \vDash B$ and $Γ,A \vDash \lnot B$, then $Γ \vDash \lnot A$.
This follows Gallier's notation and is actually taken right out of his book.
However, I have trouble showing this. I cannot see any case for which both premises on the left to be true, the right side would also be true. Can someone help me out?
By contradiction.
Assume not, i.e. $\Gamma \nvDash \lnot A$. This means that there is a valuation $v$ such that $v(\gamma_i)=$t, for every $\gamma_i \in \Gamma$ and $v(\lnot A)=$f.
Thus, for this valuation $v$ we have :
$v(\gamma_i)=$t, for every $\gamma_i \in \Gamma$, and $v(A)=$t.
We know that $\Gamma, A \vDash B$, and thus (because it reads : "if for all valuations ..."; see Def.3.3.4, page 42) we have $v(B)=$t.
We know also that $\Gamma, A \vDash \lnot B$, and thus $v(\lnot B)=$t.
Thus, we have a valuation $v$ such that :
$v(B)=v(\lnot B)=$t,
contradiction !