generated at
((¬P⇒¬Q)⇒(Q⇒P))⇒二重否定除去
対偶の除去から二重否定除去を導出できる

証明

proof.tikz(tex)
\usepackage{fitch} \usepackage{amsmath} \begin{document} $\Large\begin{nd} \hypo {1} {\lnot\lnot P} \open \hypo {2} {\lnot P} \have {3} {\bot} \ne{1,2} \have {4} {\lnot\lnot\lnot P} \be{3} \close \have {5} {\lnot P\implies\lnot\lnot\lnot P} \ii{2-4} \have {6} {\lnot\lnot P\implies P} \by{$(\lnot P\implies\lnot Q)\implies(Q\implies P)$}{5} \end{nd}$ \end{document}

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