# Biconditional as Disjunction of Conjunctions/Formulation 2/Proof 1

## Theorem

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

## Proof

By the tableau method of natural deduction:

$\vdash \left({p \iff q}\right) \iff \left({\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $p \iff q$ Assumption (None)
2 1 $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)$ Sequent Introduction 1 Biconditional as Disjunction of Conjunctions: Formulation 1
3 $\left({p \iff q}\right) \implies \left({\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)}\right)$ Rule of Implication: $\implies \mathcal I$ 1 – 2 Assumption 1 has been discharged
4 4 $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)$ Assumption (None)
5 4 $p \iff q$ Sequent Introduction 4 Biconditional as Disjunction of Conjunctions: Formulation 1
6 $\left({\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)}\right) \implies \left({p \iff q}\right)$ Rule of Implication: $\implies \mathcal I$ 4 – 5 Assumption 4 has been discharged
7 $\left({p \iff q}\right) \iff \left({\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)}\right)$ Biconditional Introduction: $\iff \mathcal I$ 3, 6

$\blacksquare$