\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}