Complement in Distributive Lattice is Unique

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Theorem

Let $\struct {S, \vee, \wedge, \preceq}$ be a bounded distributive lattice.


Then every $a \in S$ admits at most one complement.


Corollary

Let $\struct {S, \vee, \wedge, \preceq}$ be a Boolean lattice.


Then every $a \in S$ has a unique complement $\neg a$.


Proof

Let $a \in S$, and suppose that $b, c \in S$ are complements for $a$.

Then:

\(\ds b\) \(=\) \(\ds \top \wedge b\) $\top$ is the identity for $\wedge$
\(\ds \) \(=\) \(\ds \paren {c \vee a} \wedge b\) $c$ is a complement for $a$
\(\ds \) \(=\) \(\ds \paren {c \wedge b} \vee \paren {a \wedge b}\) $S$ is a distributive lattice
\(\ds \) \(=\) \(\ds \paren {c \wedge b} \vee \bot\) $b$ is a complement for $a$
\(\ds \) \(=\) \(\ds c \wedge b\) $\bot$ is the identity for $\vee$

Interchanging $c$ and $b$ in the above gives that $c = c \wedge b$ as well.

Hence $b = c$, as desired.

$\blacksquare$