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\begin{logicproof}{2}
(p \implies q)
\land (q \implies p) & \prem \\
p \lor \lnot p & $\LEM$ \\
\begin{subproof}
p & \assum \\
p \implies q & $\ande{1} 1$ \\
q & $\impe 3,4$ \\
p \land q & $\andi 3,5$ \\
(p \land q) \lor (\lnot p \land \lnot q) & $\ori{1} 6$
\end{subproof}
\begin{subproof}
\lnot p & \assum \\
q \implies p & $\ande{2} 1$ \\
\lnot q & $\MT 8,9$ \\
\lnot p \land \lnot q & $\andi 8,10$ \\
(p \land q) \lor (\lnot p \land \lnot q) & $\ori{2} 11$
\end{subproof}
(p \land q) \lor (\lnot p \land \lnot q) & $\ore 2,3-7,8-12$
\end{logicproof}