Tautology Negation of Contradiction

Theorem

A tautology implies and is implied by the negation of a contradiction:

$\top \dashv \vdash \neg \bot$

That is, a truth can not be false, and a non-falsehood must be a truth.

Proof

Proof by Natural deduction

This is proved by the Tableau method:

$\top \vdash \neg \bot $

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\top$	P	(None)
2		2	$\bot$	A	(None)	If we were to assume a contradiction ...
3		2	$\neg \top$	$\bot \mathcal E$	2	... we would be able to prove anything (including non-true) ...
4		1	$\neg \bot$	$\neg \mathcal E$	1, 3	... so a contradiction must be false.

$\blacksquare$

$\neg \bot \vdash \top$

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\neg \bot$	P	(None)
2		2	$\neg \top$	A	(None)	To assume a non-truth ...
3		2	$\bot$	SI: Use the above result	2	... implies falsehood ...
4		1	$\top$	RAA	1, 3	... which is not the case, so the assumption of non-truth must be incorrect.

$\blacksquare$

Comment

Note that the proof of:

$\neg \bot \vdash \top$

relies indirectly, via Reductio Ad Absurdum, upon the Law of the Excluded Middle, and it can be seen that this is just another way of stating that truth.

The proposition:

If it's not false, it must be true

is indeed valid only in the context where there are only two truth values.

From the intuitionist perspective, this result does not hold.

Proof by Truth Table

We apply the Method of Truth Tables to the proposition.

As can be seen by inspection, the truth values in the appropriate columns match.

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

$\blacksquare$

Proof by Boolean Interpretation

Let $p$ be a logical formula.

Let $v$ be any arbitrary boolean interpretation of $p$.

Then $v \left({p}\right) = T \iff v \left({\neg p}\right) = F$ by the definition of the logical not.

Since $v$ is arbitrary, $p$ is true in all interpretations iff $\neg p$ is false in all interpretations.

Hence $\top \dashv \vdash \neg \bot$.

$\blacksquare$