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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}
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