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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}
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