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\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}