Whitman's condition

Remarks

The converse implication is automatic in any lattice, since $a \wedge b \le a$, $a \wedge b \le b$, $c \le c \vee d$ and $d \le c \vee d$. So (W) says that an inequality between a meet and a join holds in $\mathrm{FL}(X)$ only for one of these four trivial reasons. Proof idea: Whitman described the order of $\mathrm{FL}(X)$ on lattice terms by recursive rules, the solution of the word problem for free lattices. He showed that these rules define a lattice with the universal property. The rule for comparing a meet with a join is exactly (W). Every sublattice of a lattice satisfying (W) satisfies (W), since joins and meets in a sublattice are computed in the big lattice. Example of failure: in the lattice of subsets of $\set{1, 2, 3, 4}$, take $a = \set{1, 2, 3}$, $b = \set{1, 2, 4}$, $c = \set{1}$ and $d = \set{2}$. Then $a \cap b = \set{1, 2} = c \cup d$, but $a \not\subseteq \set{1, 2}$, $b \not\subseteq \set{1, 2}$, $\set{1, 2} \not\subseteq c$ and $\set{1, 2} \not\subseteq d$. So this lattice is not a sublattice of any free lattice.

Going deeper

For a student

Whitman's condition is the key structural property of free lattices. Together with the rules for generators, it gives an algorithm that decides the order of $\mathrm{FL}(X)$. It is also the main test for deciding which finite lattices embed in free lattices, because it passes to sublattices. Since the subsets of a four-element set fail (W), a free lattice cannot contain this Boolean lattice as a sublattice.