(P⇒R)∧(Q⇒R)⇒(P∨Q⇒R)

code:proof.tikz(tex)

\usepackage{fitch}

\usepackage{amsmath}

\begin{document}

$\Large\begin{nd}

\hypo {h} {(P\implies R)\land (Q\implies R)}

\open

\hypo {pq} {P\lor Q}

\open

\hypo {p1} {P}

\have {p2} {P\implies R} \ae{h}

\have {p3} {R} \ie{p1,p2}

\close

\open

\hypo {q1} {Q}

\have {q2} {Q\implies R} \ae{h}

\have {q3} {R} \ie{q1,q2}

\close

\have {r} {R} \oe{pq,p1-p3,q1-q3}

\close

\have {e} {P\lor Q\implies R}\ii{pq-r}

\end{nd}$

\end{document}