\setlength\subproofhorizspace{1.3em}
\begin{logicproof}{2}
  \exists x \;(P(x) \lor Q(x))                     & prem.\\
  \begin{subproof}
    \llap{$x_0\enspace \;$} P(x_0) \lor Q(x_0)     & ass.\\
    \begin{subproof}
      P(x_0)                                       & ass.\\
      \exists x \; P(x)                            &$ \exists\mathrm{i} 3$\\
      \exists x \; P(x) \lor \exists x \; Q(x)       &$ \lor\mathrm{i} 4$
    \end{subproof}
    \begin{subproof}
      Q(x_0)                                       & ass.\\
      \exists x \; Q(x)                            &$ \exists\mathrm{i} 6$\\
      \exists x \; P(x) \lor \exists x \; Q(x)       &$ \lor\mathrm{i} 7$
    \end{subproof}
    \exists x \;P(x) \lor \exists x \; Q(x)       &$ \lor\mathrm{e} 2,3-8$
  \end{subproof}
  \exists x \;P(x) \lor \exists x \; Q(x)         & $\exists\mathrm{e} 1,2-9$
\end{logicproof}