# False Statement implies Every Statement/Formulation 1

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## Theorem

- $\neg p \vdash p \implies q$

## Proof 1

By the tableau method of natural deduction:

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

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

2 | 1 | $\neg p \lor q$ | Rule of Addition: $\lor \mathcal I_1$ | 1 | ||

3 | 1 | $p \implies q$ | Sequent Introduction | 2 | Rule of Material Implication |

$\blacksquare$

## Proof 2

We apply the Method of Truth Tables.

As can be seen by inspection, where the truth value in the relevant column on the left hand side is $T$, that under the one on the right hand side is also $T$:

$\begin{array}{|cc||ccc|} \hline \neg & p & p & \implies & q \\ \hline T & F & F & T & F \\ T & F & F & T & T \\ F & F & T & F & F \\ F & F & T & T & T \\ \hline \end{array}$

$\blacksquare$

## Sources

- 2000: Michael R.A. Huth and Mark D. Ryan:
*Logic in Computer Science: Modelling and reasoning about systems*... (previous) ... (next): $\S 1.2.1$: Rules for natural deduction: Exercises $1.5: \ 2 \ \text{(b)}$