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" | 14 ||
 * align="right" | 10
 * $q \implies p$
 * Sequent Introduction
 * 13
 * Rule of Transposition
 * Rule of Transposition


 * align="right" | 18 ||
 * align="right" | 10
 * $p \implies q$
 * Sequent Introduction
 * 17
 * Rule of Transposition
 * Rule of Transposition


 * align="right" | 21 ||
 * align="right" | 1
 * $p \iff q$
 * By definition
 * 20
 * 20