Conjunction with Contradiction

Theorem

A conjunction with a contradiction:

$p \land \bot \dashv \vdash \bot$

Proof

Proof by Natural deduction

This is proved by the Tableau method:

$p \land \bot \vdash \bot$

Line		Pool	Formula	Rule	Depends upon
1		1	$p \land \bot$	P	(None)
2		1	$\bot$	$\land \mathcal E_2$	1

$\blacksquare$

$\bot \vdash p \land \bot$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\bot$	P	(None)
2		1	$p \land \bot$	$\bot \mathcal E$	1	From a bottom, we can prove what we like.

$\blacksquare$

Proof by Truth Table

We apply the Method of Truth Tables to the proposition.

As can be seen by inspection, in each case, the truth values in the appropriate columns match for all models.

$\begin{array}{|c|ccc||c|ccc|} \hline \bot & p & \land & \bot & p \\ \hline F & F & F & F & F \\ F & T & F & F & T \\ \hline \end{array}$

$\blacksquare$