# Double Negation/Formulation 2/Proof by Truth Table

## Theorem

- $\vdash p \iff \neg \neg p$

## Proof

We apply the Method of Truth Tables to the proposition.

As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations.

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

$\blacksquare$

## Double Negation from Intuitionistic Perspective

The intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom.

This in turn invalidates the Law of Double Negation Elimination from the system of intuitionistic propositional logic.

Hence a difference is perceived between Double Negation Elimination and Double Negation Introduction, whereby it can be seen from the Principle of Non-Contradiction that if a statement is true, then it is not the case that it is false.

However, if all we know is that a statement is not false, we can not be certain that it *is* actually true without accepting that there are only two possible truth values.

Such distinctions may be important when considering, for example, multi-value logic.

However, when analysing logic from a purely classical standpoint, it is common and acceptable to make the simplification of taking just one Double Negation rule:

- $p \dashv \vdash \neg \neg p$

## Sources

- 1946: Alfred Tarski:
*Introduction to Logic and to the Methodology of Deductive Sciences*(2nd ed.) ... (previous) ... (next): $\S \text{II}$: Exercise $14 \ \text{(a)}$ - 1973: Irving M. Copi:
*Symbolic Logic*(4th ed.) ... (previous) ... (next): $2$ Arguments Containing Compound Statements: $2.4$: Statement Forms