Complement in Boolean Algebra is Unique

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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.


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


\(\ds b\) \(=\) \(\ds b \wedge \top\) $\top$ 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.