False Statement implies Every Statement/Formulation 1

Theorem

$\ds \neg p$

$\ds $

$\ds \vdash \ \ $

$\ds p \implies q$

$\ds $

Proof 1

By the tableau method of natural deduction:

$\neg p \vdash p \implies q$
Line	Pool	Formula	Rule	Depends upon	Notes
1	1	$\neg p$	Premise	(None)
2	1	$\neg p \lor q$	Rule of Addition: $\lor \II_1$	1
3	1	$p \implies q$	Sequent Introduction	2	Rule of Material Implication

$\blacksquare$

Proof by Truth Table

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 $1.5: \ 2 \ \text{(b)}$