∀x(P(x)⇒Q)⇒(∃xP(x)⇒Q)

証明

https://scrapbox.io/files/6799cf6f0504d78cefbcc5f4.svg

code:proof.tikz(tex)

\usepackage{fitch}

\usepackage{amsmath}

\begin{document}

$\Large\begin{nd}

\hypo {1} {\forall x (P(x)\implies Q)}

\open

\hypo {2} {\exists x P(x)}

\hypo {Pu} {P(u)}

\have {PuQ} {P(u)\implies Q} \Ae{1}

\have {Q} {Q} \ie{Pu,Q}

\close

\have {Q2} {Q} \Ee{2,Pu-Q}

\close

\have {7} {(\exists x P(x))\implies Q} \ii{2-Q2}

\end{nd}$

\end{document}