# Biconditional is Commutative/Formulation 1/Proof 1

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## Theorem

- $p \iff q \dashv \vdash q \iff p$

## Proof

By the tableau method of natural deduction:

Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|

1 | 1 | $p \iff q$ | Premise | (None) | ||

2 | 1 | $p \implies q$ | Biconditional Elimination: $\iff \mathcal E_1$ | 1 | ||

3 | 1 | $q \implies p$ | Biconditional Elimination: $\iff \mathcal E_2$ | 1 | ||

4 | 1 | $q \iff p$ | Biconditional Introduction: $\iff \mathcal I$ | 3, 2 |

$\Box$

By the tableau method of natural deduction:

Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|

1 | 1 | $q \iff p$ | Premise | (None) | ||

2 | 1 | $q \implies p$ | Biconditional Elimination: $\iff \mathcal E_1$ | 1 | ||

3 | 1 | $p \implies q$ | Biconditional Elimination: $\iff \mathcal E_2$ | 1 | ||

4 | 1 | $p \iff q$ | Biconditional Introduction: $\iff \mathcal I$ | 3, 2 |

$\blacksquare$