¬¬((P⇒Q)∨(Q⇒P))

証明

https://scrapbox.io/files/68af8b8976fb6820ac28fcad.svg

code:proof.tikz(tex)

\usepackage{fitch}

\usepackage{amsmath}

\begin{document}

$\Large\begin{nd}

\open

\hypo {ndummett} {\lnot((P\implies Q)\lor(Q\implies P))}

\open

\hypo {q} {Q}

\open

\hypo {p} {P}

\have {q2} {Q} \r{q}

\close

\have {pq} {P\implies Q} \ii{p-q2}

\have {pqqp} {(P\implies Q)\lor(Q\implies P)} \oi{pq}

\have {b} {\bot} \ie{ndummett,pqqp}

\have {p} {P} \be{b}

\close

\have {qp} {Q\implies P} \ii{q-p}

\have {pqqp} {(P\implies Q)\lor(Q\implies P)} \oi{qp}

\have {b} {\bot} \ie{ndummett,pqqp}

\close

\have {nndummett} {\lnot\lnot((P\implies Q)\lor(Q\implies P))} \ii{ndummett-b}

\end{nd}$

\end{document}