# Biconditional with Tautology/Proof 1

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## Theorem

- $p \iff \top \dashv \vdash p$

## Proof

By the tableau method of natural deduction:

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

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

2 | $\top$ | Rule of Top-Introduction: $\top \mathcal I$ | (None) | |||

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

4 | 1 | $p$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2, 3 |

$\Box$

By the tableau method of natural deduction:

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

1 | 1 | $\top$ | Premise | (None) | ||

2 | 2 | $p$ | Assumption | (None) | ||

3 | $\top$ | Rule of Top-Introduction: $\top \mathcal I$ | (None) | |||

4 | $p \implies \top$ | Rule of Implication: $\implies \mathcal I$ | 2 – 3 | Assumption 2 has been discharged | ||

5 | 2 | $\top \implies p$ | Rule of Implication: $\implies \mathcal I$ | 1 – 2 | Assumption 1 has been discharged | |

6 | 2 | $p \iff \top$ | Biconditional Introduction: $\iff \mathcal I$ | 4, 5 |

$\blacksquare$