Contradiction is Negation of Tautology

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


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

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

Proof

 * align="right" | 3 ||
 * align="right" | 1
 * $\neg \top$
 * $\bot \mathcal E$
 * 1
 * Any statement we want ...
 * align="right" | 4 ||
 * align="right" | 1
 * $\bot$
 * $\neg \mathcal I$
 * 2, 3
 * ... and the one we picked contradicts our assumption, which must therefore be false.
 * align="right" | 5 ||
 * align="right" | 1
 * $\neg \top$
 * $\bot \mathcal E$
 * 2-4
 * }
 * $\bot \mathcal E$
 * 2-4
 * }
 * }


 * align="right" | 2 ||
 * align="right" | -
 * $p \lor \neg p$
 * LEM
 * (None)
 * From the Law of Excluded Middle ...
 * align="right" | 3 ||
 * align="right" | -
 * $\top$
 * LEM
 * 2
 * ... we deduce that truth ...
 * align="right" | 4 ||
 * align="right" | 1
 * $\bot$
 * $\neg \mathcal E$
 * 1, 3
 * ... is contrary to the assumption of non-truth, which must therefore be false.
 * }
 * 1, 3
 * ... is contrary to the assumption of non-truth, which must therefore be false.
 * }

Comment
Note that the proof of:
 * $\neg \top \vdash \bot$

relies directly 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 true, it must be false

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.

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

Proof by Boolean Interpretation
Let $p$ be a logical formula.

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

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

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

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