(P⇒R)∧(Q⇒¬R)⇒¬(P∧Q)
証明
https://scrapbox.io/files/65e7f82e03949c0024a367d0.svg
code:proof.tikz(tex)
\usepackage{fitch}
\usepackage{amsmath}
\begin{document}
$\Large\begin{nd}
\hypo {h} {(P\implies R)\land(Q\implies\lnot R)}
\open
\hypo {pq} {P\land Q}
\have {p} P \ae{pq}
\have {pr} {P\implies R} \ae{h}
\have {r} {R} \ie{p,pr}
\have {q} {Q} \ae{pq}
\have {qr} {Q\implies\lnot R} \ae{h}
\have {nr} {\lnot R} \ie{q,qr}
\have {b} {\bot} \ne{r,nr}
\close
\have {e} {\lnot(P\land Q)} \ii{pq-b}
\end{nd}$
\end{document}