# Proving P → (Q → R) ∴ (Q ∧ ¬R) → ¬P in with Fitch [closed]

I'm having a bit of trouble proving this question my prof gave to me. We won't have a lecture for next 2 weeks so I'm curious if I'm doing it right. If not, I'm going to back to drawing board. please help. Here's my idea

$$%PREAMBLE \newcommand{\fitch}[1]{\begin{array}{rlr}#1\end{array}} \newcommand{\fcol}[1]{\begin{array}{r}#1\end{array}} %FirstColumn \newcommand{\scol}[1]{\begin{array}{l}#1\end{array}} %SecondColumn \newcommand{\tcol}[1]{\begin{array}{l}#1\end{array}} %ThirdColumn \newcommand{\subcol}[1]{\begin{array}{|l}#1\end{array}} %SubProofColumn \newcommand{\startsub}{\\[-0.29em]} %adjusts line spacing slightly \newcommand{\endsub}{\startsub} %adjusts line spacing slightly \newcommand{\fendl}{\\[0.044em]} %adjusts line spacing slightly$$

$$\fitch{\fcol{\fendl 1 \fendl 2 \fendl 3 \fendl 4 \fendl 5 \fendl 6 \fendl 7 \fendl 8 \fendl 9 \fendl 10} & \scol{\startsub{\subcol{P \to (Q \to R) \\ \hline \subcol{Q \wedge \neg R \\ \hline Q \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \wedge\text{E, 2} \\ \neg R \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \wedge\text{E, 2} \\ \subcol{P \\ \hline Q \to R \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \to\text{E, 5, 1} \\ R \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \to\text{E, 6, 3} \\ \perp \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \neg\text{ E, 7, 4}}\endsub \neg P \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{RAA (reductio ad absurdum)}} \endsub (Q \wedge \neg R) \to \neg P \ \ \ \to\text{I, 2, 9}}}}}$$

• Line $9$ should be justified as Negation Introduction [$\lnot\mathrm I~5{-}8$], rather than Reductio Ad Absurdum.$$\begin{array}{|l}\quad\begin{array}{|l}~~P\\\hline~~\bot\end{array}\\\lnot P\qquad\lnot\mathsf I\end{array}\qquad\text{versus}\qquad\begin{array}{|l}\quad\begin{array}{|l}\lnot P\\\hline~~\bot\end{array}\\~~P\qquad\mathsf{RAA}\end{array}$$ May 29 at 9:10