Questionnaire-2024/natural_deduction_predicate_logic/2002_sol.tex

\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}