(P⇒Q)⇒(¬Q⇒¬P)

対偶を導入する論理式

証明

論理否定を被せるだけ

https://scrapbox.io/files/65fb769ec343a200235a23a1.svg

code:proof.tikz(tex)

\usepackage{fitch}

\usepackage{amsmath}

\begin{document}

$\Large\begin{nd}

\hypo {pq} {P\implies Q}

\open

\hypo {nq} {\lnot Q}

\open

\hypo {p} {P}

\have {q} {Q} \ie{p,pq}

\have {b} {\bot} \ne{q,nq}

\close

\have {np} {\lnot P} \ii{p-b}

\close

\have {nqnp} {\lnot Q\implies\lnot P} \ii{nq-np}

\end{nd}$

\end{document}