Undo a weakened statement in sequent calculus later in the inferences

Question

I'm working on an answer to (b) of Mathematical Logic, Ebbinghaus et. al. 1984 p. 64

Consider the following inference: $$ \frac{\begin{align}\Gamma \vdash A\\ \Gamma \vdash B\end{align}}{\Gamma \vdash A\land B} $$ I show it is a valid inference, the question only allowing a restricted set of rules, using Weakening, Contraposition, and the $\lor$-Antecedent rules. But I'm stuck on the last part:

$Γ\vdash A$

$Γ\vdash B$

$Γ \, \neg C \vdash A$

$Γ \, \neg C \vdash B$

$Γ \, \neg A \vdash C$

$Γ \, \neg B \vdash C$

$Γ \, (\neg A\lor \neg B) \vdash C$

$Γ \, \neg C \vdash \neg(\neg A\lor\neg B)$

How would I get rid of the $\neg C$ I acquired from weakening? I have access to explosion, Contraposition, Modus ponens, and these (the last formula on each line is made true by the ones before it): I also have $$ \frac{\begin{align}\Gamma \vdash \phi\\ \Gamma \phi \vdash \psi \end{align}}{\Gamma \vdash \psi} $$

Answer 1

Follow the same proof using $\neg C$ in place of $C$, concluding $\Gamma, \neg \neg C \vdash A \land B$. Together with the original proof of $\Gamma, \neg C \vdash A \land B$, we can use the rule PC to eliminate $C$ entirely and conclude $\Gamma \vdash A \land B$.

$C$ seems to just be a placeholder which is needed to apply some of the rules. So there really isn't much that changes in the proof if we replace $C$ by something else.