Biconditional as Disjunction of Conjunctions/Formulation 1/Reverse Implication

Theorem

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

Proof

 * align="right" | 5 ||
 * align="right" | 2
 * $p \implies q$
 * $\implies \mathcal I$
 * 3, 4
 * 3, 4


 * align="right" | 8 ||
 * align="right" | 2
 * $q \implies p$
 * $\implies \mathcal I$
 * 6, 7
 * 6, 7


 * align="right" | 13 ||
 * align="right" | 10
 * $\neg p \implies \neg q$
 * $\implies \mathcal I$
 * 11, 12
 * align="right" | 14 ||
 * align="right" | 10
 * $q \implies p$
 * Sequent Introduction
 * 13
 * Rule of Transposition
 * Sequent Introduction
 * 13
 * Rule of Transposition


 * align="right" | 17 ||
 * align="right" | 10
 * $\neg q \implies \neg p$
 * $\implies \mathcal I$
 * 15, 16
 * align="right" | 18 ||
 * align="right" | 10
 * $p \implies q$
 * Sequent Introduction
 * 17
 * Rule of Transposition
 * Sequent Introduction
 * 17
 * Rule of Transposition


 * align="right" | 20 ||
 * align="right" | 1
 * $\left ({p \implies q}\right) \land \left ({q \implies p}\right)$
 * $\lor \mathcal E$
 * 1, 2-9, 10-20
 * align="right" | 20 ||
 * align="right" | 1
 * $p \iff q$
 * By definition
 * 19
 * $p \iff q$
 * By definition
 * 19