Complement in Boolean Algebra is Unique
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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.
$\blacksquare$