((P⇒Q)⇒R)⇒(((Q⇒P)⇒S)⇒¬¬(R∨S))
証明
https://scrapbox.io/files/65e7f68e9c4a7f00251bde35.svg
code:proof.tikz(tex)
\usepackage{fitch}
\usepackage{amsmath}
\begin{document}
$\Large\begin{nd}
\hypo {h1} {(P\implies Q)\implies R}
\open
\hypo {h2} {(Q\implies P)\implies S}
\open
\hypo {h3} {\lnot(R\lor S)}
\open
\hypo {q} {Q}
\open
\hypo {p1} {P}
\have {p2} {Q} \r{q}
\close
\have {q2} {P\implies Q} \ii{p1-p2}
\have {r} R \ie{q2,h1}
\have {3} {R\lor S} \oi{r}
\have {4} {\bot} \ne{h3,3}
\have {qe} {P} \be{4}
\close
\have {2} {Q\implies P} \ii{q-qe}
\have {3} {S} \ie{h2,2}
\have {4} {R\lor S} \oi{3}
\have {5} {\bot} \ne{h3,4}
\close
\have {6} {\lnot\lnot(R\lor S)} \ii{h3-5}
\close
\end{nd}$
\end{document}