# Negation as Implication of Bottom

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## Theorem

$p \implies \bot \dashv\vdash \neg p$

## Proof

By the tableau method of natural deduction:

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

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

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

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

4 | 1 | $\neg p$ | Proof by Contradiction: $\neg \II$ | 2 – 3 | Assumption 2 has been discharged |

$\Box$

By the tableau method of natural deduction:

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

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

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

3 | 1,2 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 1, 2 | ||

4 | 1 | $p \implies \bot$ | Rule of Implication: $\implies \II$ | 2 – 3 | Assumption 2 has been discharged |

$\blacksquare$