proof.tikz(tex)\usepackage{fitch}
\usepackage{amsmath}
\begin{document}
$\Large\begin{nd}
\hypo {1} {\lnot\lnot P}
\open
\hypo {2} {\lnot P}
\have {3} {\bot} \ne{1,2}
\have {4} {\lnot\lnot\lnot P} \be{3}
\close
\have {5} {\lnot P\implies\lnot\lnot\lnot P} \ii{2-4}
\have {6} {\lnot\lnot P\implies P}
\by{$(\lnot P\implies\lnot Q)\implies(Q\implies P)$}{5}
\end{nd}$
\end{document}