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So I have a proof in which i have derived both ~P and (P v Q). My current objective is to extract the Q as I need it for another part of the proof. It seems obvious to me that if I have ~P true and (P v Q) true, then Q is necessarily true. However, to get Q I would need to use the disjunction elimination rule, and I am unsure of how to insert that into my proof to get the desired result (Q). Any help would be appreciated

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Indeed $\neg P$ and $P\vee Q$ do entail $Q$ by rule of Disjunctive Syllogism.

This derived rule may be proven using the fundamental rules of Disjunction Elimination and Explosion (aka. ex falso quodlibet).

A contradiction may be derived when assuming $P$, since $\neg P$ is a premise, and $Q$ may be derived from that contradiction, since anything may.

Also $Q$ is trivially derived when assuming $Q$; it is what was assumed.

And from $\neg P, P\vdash Q$ and $Q\vdash Q$ we may infer $\neg P,P\vee Q\vdash Q$ ...

$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{~~1.~\neg P\\~~2.~P\vee Q}{\fitch{~~3.~P}{~~4.~\bot\hspace{10ex}{\neg}\mathsf E~1,3\\~~5.~Q\hspace{10ex}\mathsf{X}~4}\\\fitch{~~6.~Q}{}\\~~7.~Q\hspace{14ex}{\vee}\mathsf E~2,3{-}5,6{-}6}$$

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