generated at
(P⇒R)∧(Q⇒R)⇒(P∨Q⇒R)
これは∨除去含意で書き換えたのと等しい


proof.tikz(tex)
\usepackage{fitch} \usepackage{amsmath} \begin{document} $\Large\begin{nd} \hypo {h} {(P\implies R)\land (Q\implies R)} \open \hypo {pq} {P\lor Q} \open \hypo {p1} {P} \have {p2} {P\implies R} \ae{h} \have {p3} {R} \ie{p1,p2} \close \open \hypo {q1} {Q} \have {q2} {Q\implies R} \ae{h} \have {q3} {R} \ie{q1,q2} \close \have {r} {R} \oe{pq,p1-p3,q1-q3} \close \have {e} {P\lor Q\implies R}\ii{pq-r} \end{nd}$ \end{document}

#2025-02-11 17:45:50
#2024-03-10 21:24:32
#2021-12-07 16:41:56