# Double negation

Type Theorem .mw-parser-output .plainlist ol,.mw-parser-output .plainlist ul{line-height:inherit;list-style:none;margin:0;padding:0}.mw-parser-output .plainlist ol li,.mw-parser-output .plainlist ul li{margin-bottom:0} If a statement is true, then it is not the case that the statement is not true." $A\equiv \sim (\sim A)$ In propositional logic, double negation is the theorem that states that "If a statement is true, then it is not the case that the statement is not true." This is expressed by saying that a proposition A is logically equivalent to not (not-A), or by the formula A ≡ ~(~A) where the sign ≡ expresses logical equivalence and the sign ~ expresses negation.

Like the law of the excluded middle, this principle is considered to be a law of thought in classical logic, but it is disallowed by intuitionistic logic. The principle was stated as a theorem of propositional logic by Russell and Whitehead in Principia Mathematica as:

$\mathbf {*4\cdot 13} .\ \ \vdash .\ p\ \equiv \ \thicksim (\thicksim p)$ "This is the principle of double negation, i.e. a proposition is equivalent of the falsehood of its negation."

## Elimination and introduction

Double negation elimination and double negation introduction are two valid rules of replacement. They are the inferences that, if not not-A is true, then A is true, and its converse, that, if A is true, then not not-A is true, respectively. The rule allows one to introduce or eliminate a negation from a formal proof. The rule is based on the equivalence of, for example, It is false that it is not raining. and It is raining.

The double negation introduction rule is:

P $\Rightarrow$ $\neg$ $\neg$ P

and the double negation elimination rule is:

$\neg$ $\neg$ P $\Rightarrow$ P

Where "$\Rightarrow$ " is a metalogical symbol representing "can be replaced in a proof with."

In logics that have both rules, negation is an involution.

### Formal notation

The double negation introduction rule may be written in sequent notation:

$P\vdash \neg \neg P$ The double negation elimination rule may be written as:

$\neg \neg P\vdash P$ In rule form:

${\frac {P}{\neg \neg P}}$ and

${\frac {\neg \neg P}{P}}$ or as a tautology (plain propositional calculus sentence):

$P\to \neg \neg P$ and

$\neg \neg P\to P$ These can be combined into a single biconditional formula:

$\neg \neg P\leftrightarrow P$ .

Since biconditionality is an equivalence relation, any instance of ¬¬A in a well-formed formula can be replaced by A, leaving unchanged the truth-value of the well-formed formula.

Double negative elimination is a theorem of classical logic, but not of weaker logics such as intuitionistic logic and minimal logic. Double negation introduction is a theorem of both intuitionistic logic and minimal logic, as is $\neg \neg \neg A\vdash \neg A$ .

Because of their constructive character, a statement such as It's not the case that it's not raining is weaker than It's raining. The latter requires a proof of rain, whereas the former merely requires a proof that rain would not be contradictory. This distinction also arises in natural language in the form of litotes.

## Proofs

### In classical propositional calculus system

In Hilbert-style deductive systems for propositional logic, double negation is not always taken as an axiom (see list of Hilbert systems), and is rather a theorem. We describe a proof of this theorem in the system of three axioms proposed by Jan Łukasiewicz:

A1. $\phi \to \left(\psi \to \phi \right)$ A2. $\left(\phi \to \left(\psi \rightarrow \xi \right)\right)\to \left(\left(\phi \to \psi \right)\to \left(\phi \to \xi \right)\right)$ A3. $\left(\lnot \phi \to \lnot \psi \right)\to \left(\psi \to \phi \right)$ We use the lemma $p\to p$ proved here, which we refer to as (L1), and use the following additional lemma, proved here:

(L2) $p\to ((p\to q)\to q)$ We first prove $\neg \neg p\to p$ . For shortness, we denote $q\to (r\to q)$ by φ0. We also use repeatedly the method of the hypothetical syllogism metatheorem as a shorthand for several proof steps.

(1) $\varphi _{0}$ (instance of (A1))
(2) $(\neg \neg \varphi _{0}\to \neg \neg p)\to (\neg p\to \neg \varphi _{0})$ (instance of (A3))
(3) $(\neg p\to \neg \varphi _{0})\to (\varphi _{0}\to p)$ (instance of (A3))
(4) $(\neg \neg \varphi _{0}\to \neg \neg p)\to (\varphi _{0}\to p)$ (from (2) and (3) by the hypothetical syllogism metatheorem)
(5) $\neg \neg p\to (\neg \neg \varphi _{0}\to \neg \neg p)$ (instance of (A1))
(6) $\neg \neg p\to (\varphi _{0}\to p)$ (from (4) and (5) by the hypothetical syllogism metatheorem)
(7) $\varphi _{0}\to ((\varphi _{0}\to p)\to p)$ (instance of (L2))
(8) $(\varphi _{0}\to p)\to p$ (from (1) and (7) by modus ponens)
(9) $\neg \neg p\to p$ (from (6) and (8) by the hypothetical syllogism metatheorem)

We now prove $p\to \neg \neg p$ .

(1) $\neg \neg \neg p\to \neg p$ (instance of the first part of the theorem we have just proven)
(2) $(\neg \neg \neg p\to \neg p)\to (p\to \neg \neg p)$ (instance of (A3))
(3) $p\to \neg \neg p$ (from (1) and (2) by modus ponens)

And the proof is complete.