Biconditional as Disjunction of Conjunctions/Formulation 1/Forward Implication

Theorem

 * $p \iff q \vdash \left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)$

Proof

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 * align="right" | 1
 * $\left ({p \implies q}\right) \land \left ({q \implies p}\right)$
 * By definition
 * 1
 * 1