# Biconditional with Contradiction/Proof 1

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## Theorem

- $p \iff \bot \dashv \vdash \neg p$

## Proof

By the tableau method of natural deduction:

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

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

2 | 1 | $p \implies \bot$ | Biconditional Elimination: $\iff \EE_1$ | 1 | ||

3 | 1 | $\neg p$ | Sequent Introduction | 2 | Contradictory Consequent |

$\Box$

By the tableau method of natural deduction:

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

1 | 1 | $\neg p$ | Assumption | (None) | ||

2 | 1 | $p \implies \bot$ | Sequent Introduction | 1 | Contradictory Consequent | |

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

4 | $\bot \implies p$ | Sequent Introduction | 3 | Contradictory Antecedent | ||

5 | 1 | $p \iff \bot$ | Biconditional Introduction: $\iff \II$ | 2, 4 |

$\blacksquare$