# Double Negation/Formulation 2/Proof 1

## Theorem

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

## Proof

By the tableau method of natural deduction:

Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|

1 | $p \implies \neg \neg p$ | Theorem Introduction | (None) | Double Negation Introduction: Formulation 2 | ||

2 | $\neg \neg p \implies p$ | Theorem Introduction | (None) | Double Negation Elimination: Formulation 2 | ||

3 | $p \iff \neg \neg p$ | Biconditional Introduction: $\iff \mathcal I$ | 1, 2 |

$\blacksquare$

## 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:

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