Contradictory Consequent

Theorem

A conditional with a contradiction as consequent:

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

Proof by Natural Deduction

By the tableau method of natural deduction:

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

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:

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

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

$\blacksquare$

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 boolean interpretations.

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

$\blacksquare$

Also see

• Proof by Contradiction

Sources

• 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.3.3$