(∃xP(x)⇒Q)⇒∀x(P(x)⇒Q)

証明

https://scrapbox.io/files/6799d0c0ebf36b9e7f4a011a.svg

code:proof.tikz(tex)

\usepackage{fitch}

\usepackage{amsmath}

\begin{document}

$\Large\begin{nd}

\hypo {s} {(\exists x P(x))\implies Q}

\open

\hypo {Pu} {P(u)}

\have {EPx} {\exists x P(x)} \Ei{Pu}

\have {Q} {Q} \ie{s,EPx}

\close

\have {3} {P(u)\implies Q} \ii{Pu-Q}

\close

\have {e} {\forall x(P(x)\implies Q)} \Ai{3}

\end{nd}$

\end{document}