∀x¬P(x)⇒¬∃xP(x)
https://scrapbox.io/files/65f9566a7eb4cf0024042456.svg
code:proof.tikz(tex)
\usepackage{fitch}
\begin{document}
$\Large\begin{nd}
\hypo {1} {\forall x\lnot P(x)}
\open
\hypo {2} {\exists x P(x)}
\hypo {3} {P(u)}
\have {5} {\neg P(u)} \Ae{1}
\have {6} {\bot} \ne{3,5}
\close
\have {6a} {\bot} \Ee{2,3-6}
\close
\have {7} {\neg \exists x P(x)} \ii{2-6a}
\end{nd}$
\end{document}