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

 * align="right" | 2 ||
 * align="right" | 1
 * $\left ({p \implies q}\right) \land \left ({q \implies p}\right)$
 * By definition
 * 1
 * 1


 * align="right" | 5 ||
 * align="right" |
 * $p \lor \neg p$
 * LEM
 * (None)
 * (None)


 * align="right" | 7 ||
 * align="right" | 1, 6
 * $q$
 * $\implies \mathcal E$
 * 3, 6
 * 3, 6


 * align="right" | 11 ||
 * align="right" | 1, 10
 * $\neg q$
 * MTT
 * 4, 10
 * 4, 10


 * align="right" | 14 ||
 * align="right" | 1
 * $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)$
 * $\lor \mathcal E$
 * 1, 6-9, 10-13
 * 1, 6-9, 10-13