generated at
¬P∨P⇒((P⇒Q)∨(Q⇒P))
排中律からDummettの法則を導ける

proof.tikz(tex)
\usepackage{fitch} \usepackage{amsmath} \begin{document} $\Large\begin{nd} \hypo {h} {\lnot P\lor P} \open \hypo {np} {\lnot P} \open \hypo {np-p} {P} \have {np-b} {\bot} \ne{np,{np-p}} \have {np-q} {Q} \be{{np-b}} \close \have {pq} {P\implies Q} \ii{{np-p}-{np-q}} \have {d1} {(P\implies Q)\lor(Q\implies P)} \oi{pq} \close \open \hypo {p} {P} \open \hypo {p-q} {Q} \have {p-p} {P} \r{p} \close \have {qp} {Q\implies P} \ii{{p-q}-{p-p}} \have {d2} {(P\implies Q)\lor(Q\implies P)} \oi{qp} \close \have {d} {(P\implies Q)\lor(Q\implies P)} \oe{h,np-d1,p-d2} \end{nd}$ \end{document}

#2021-11-29 16:51:26