proof.tikz(tex)\usepackage{fitch}
\usepackage{amsmath}
\begin{document}
$\Large\begin{nd}
\hypo {pq} {P\implies Q}
\open
\hypo {nq} {\lnot Q}
\open
\hypo {p} {P}
\have {q} {Q} \ie{p,pq}
\have {b} {\bot} \ne{q,nq}
\close
\have {np} {\lnot P} \ii{p-b}
\close
\have {nqnp} {\lnot Q\implies\lnot P} \ii{nq-np}
\end{nd}$
\end{document}