# Complement in Boolean Algebra is Unique

## Theorem

Let $\left({S, \vee, \wedge}\right)$ be a Boolean algebra.

Then for all $a \in S$, there is a unique $b \in S$ such that:

$a \wedge b = \bot, a \vee b = \top$

i.e., a valid choice for $\neg a$ as in axiom $(BA \ 4)$ for Boolean algebras.

## Proof

Suppose $b, c \in S$ both satisfy the identities.

Then:

 $\ds b$ $=$ $\ds b \wedge \top$ $\top$ is the identity for $\wedge$ $\ds$ $=$ $\ds b \wedge \left({a \vee c}\right)$ by hypothesis $\ds$ $=$ $\ds \left({b \wedge a}\right) \vee \left({b \wedge c}\right)$ $\wedge$ distributes over $\vee$ $\ds$ $=$ $\ds \bot \vee \left({b \wedge c}\right)$ by hypothesis $\ds$ $=$ $\ds \left({a \wedge c}\right) \vee \left({b \wedge c}\right)$ Axiom $(BA \ 4)$ $\ds$ $=$ $\ds \left({a \vee b}\right) \wedge c$ $\wedge$ distributes over $\vee$ $\ds$ $=$ $\ds \top \wedge c$ by hypothesis $\ds$ $=$ $\ds c$ $\top$ is the identity for $\wedge$

That is, $b = c$.

The result follows.

$\blacksquare$