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\setlength{\subproofhorizspace}{1.5em}
\begin{logicproof}{2}
\lnot \exists x \; Q(x) &prem \\
\begin{subproof}
\hspace*{-1.5em}
\llap{$x_0\enspace \;$}
\begin{subproof}
Q(x_0) &assum \\
\exists x \; Q(x) &$\exists \mathrm{i} 2$ \\
\bot &$\lnot \mathrm{e} 1,3$
\end{subproof}
\lnot Q(x_0) &$\lnot \mathrm{i} 2-4$
\end{subproof}
\forall x \; \lnot Q(x) &$\forall \mathrm{i} 2-5$
\end{logicproof}