# False Statement implies Every Statement/Formulation 2/Proof 1

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## Theorem

- $\vdash \neg p \implies \paren {p \implies q}$

## Proof

By the tableau method of natural deduction:

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

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

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

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

4 | 1, 2 | $q$ | Rule of Explosion: $\bot \EE$ | 3 | ||

5 | 1 | $p \implies q$ | Rule of Implication: $\implies \II$ | 2 – 4 | Assumption 2 has been discharged | |

6 | $\neg p \implies \left({p \implies q}\right)$ | Rule of Implication: $\implies \II$ | 1 – 5 | Assumption 1 has been discharged |

$\blacksquare$