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\setlength{\subproofhorizspace}{1.5em}
\begin{logicproof}{3}
\lnot \exists x~P(x) \lor \lnot \exists y~Q(y) &prem \\
\begin{subproof}
\begin{subproof}
\hspace*{-2.75em}
\llap{$z_0\enspace \;$} \: \hspace{2.75em} Q(z_0) \land P(z_0) &assum \\
\begin{subproof}
\lnot \exists x~P(x) & assum \\
P(z_0) &$\land \mathrm{e} 2$ \\
\exists x~P(x) &$\exists \mathrm{i} 4$ \\
\bot &$\lnot \mathrm{e} 3,5$
\end{subproof}
\begin{subproof}
\lnot \exists y~Q(y) & assum \\
Q(z_0) &$\land \mathrm{e} 2$ \\
\exists y~Q(y) &$\exists \mathrm{i} 8$ \\
\bot &$\lnot \mathrm{e} 7,9$
\end{subproof}
\bot &$\lor \mathrm{e} 1,3-6,7-10$
\end{subproof}
\lnot(Q(z_0) \land P(z_0)) &$\lnot \mathrm{i} 3-11$
\end{subproof}
\forall z \; \lnot(Q(z) \land P(z)) &$\forall \mathrm{i} 3-12$
\end{logicproof}