---
tag: VH4R85Y6Y3
kind: definition
name: Huntington's axioms
msc: ["06E05", "03G05"]
defines: M3GFCVX549
uses: [0XADJKM699]
sources:
  - url: https://en.wikipedia.org/wiki/Boolean_algebra_(structure)
---

## Definition

Let $A$ be a set with a binary operation $+$ and a unary operation $n$.
The structure $(A, +, n)$ satisfies *Huntington's axioms* if for all $x, y, z \in A$:

1. $x + y = y + x$.
2. $(x + y) + z = x + (y + z)$.
3. $n(n(x) + y) + n(n(x) + n(y)) = x$.

The third axiom is called *Huntington's equation*.

## Remarks

A structure satisfies Huntington's axioms if and only if it comes from a [[Boolean algebra|0XADJKM699]] with $x + y = x \vee y$ and $n(x) = \neg x$.
The other operations are recovered as $x \wedge y = n(n(x) + n(y))$, $1 = x + n(x)$ and $0 = n(1)$, where the last two do not depend on $x$.

Example: $\set{0, 1}$ with $x + y = x \vee y$ and $n(x) = \neg x$.
By De Morgan's laws, the left side of axiom 3 is $(x \wedge \neg y) \vee (x \wedge y) = x \wedge (\neg y \vee y) = x$.

Non-example: $\set{0, 1}$ with $x + y = x \vee y$ and $n(x) = x$.
Axioms 1 and 2 hold, but axiom 3 becomes $(x \vee y) \vee (x \vee y) = x$, which fails for $x = 0$, $y = 1$.