(P∧Q)∨P⇔P

証明

https://scrapbox.io/files/68aebcc5c3e535e281afbae2.svg

https://scrapbox.io/files/68aebc37141a884d8b7159d2.svg

code:l2r.tikz(tex)

\usepackage{fitch}

\usepackage{amsmath}

\begin{document}

$\Large\begin{nd}

\hypo {l} {(P\land Q)\lor P}

\open

\hypo {pq} {P\land Q}

\have {p} {P} \ae{pq}

\close

\open

\hypo {p2} P

\have {p3} P \r{p2}

\close

\have {r} {P} \oe{l,pq-p,p2-p3}

\end{nd}$

\end{document}

code:r2l.tikz(tex)

\usepackage{fitch}

\usepackage{amsmath}

\begin{document}

$\Large\begin{nd}

\hypo {r} {P}

\have {l} {(P\land Q)\lor P}\oi{r}

\end{nd}$

\end{document}