\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}