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

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## 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:

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 |

$\blacksquare$