Complement in Distributive Lattice is Unique
Jump to navigation
Jump to search
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.
$\blacksquare$