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  "body": "\n## Statement\n\nLet $X$ be a set and let $\\mathrm{FL}(X)$ be the [[free lattice|WP20S9NZ15]] on $X$, with [[join and meet|J2T75ATRV5]] $\\vee$ and $\\wedge$.\nThen $\\mathrm{FL}(X)$ satisfies *Whitman's condition* (W): for all $a, b, c, d \\in \\mathrm{FL}(X)$, if\n$$a \\wedge b \\le c \\vee d,$$\nthen $a \\le c \\vee d$, or $b \\le c \\vee d$, or $a \\wedge b \\le c$, or $a \\wedge b \\le d$.\n\n## Remarks\n\nThe converse implication is automatic in any lattice, since $a \\wedge b \\le a$, $a \\wedge b \\le b$, $c \\le c \\vee d$ and $d \\le c \\vee d$.\nSo (W) says that an inequality between a meet and a join holds in $\\mathrm{FL}(X)$ only for one of these four trivial reasons.\nProof idea: Whitman described the order of $\\mathrm{FL}(X)$ on lattice terms by recursive rules, the solution of the [[word problem for free lattices|Z86VGP5RMG]].\nHe showed that these rules define a lattice with the universal property.\nThe rule for comparing a meet with a join is exactly (W).\nEvery [[sublattice|PV90EW00XK]] of a lattice satisfying (W) satisfies (W), since joins and meets in a sublattice are computed in the big lattice.\nExample 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 $d = \\set{2}$.\nThen $a \\cap b = \\set{1, 2} = c \\cup d$, but $a \\not\\subseteq \\set{1, 2}$, $b \\not\\subseteq \\set{1, 2}$, $\\set{1, 2} \\not\\subseteq c$ and $\\set{1, 2} \\not\\subseteq d$.\nSo this lattice is not a sublattice of any free lattice.\n\n## Explanations\n\n### L1 human\n\nIn a free lattice nothing holds by accident.\nIf a meet $a \\wedge b$ lies below a join $c \\vee d$, then there is an obvious reason.\nEither one of $a$ or $b$ lies below $c \\vee d$, or $a \\wedge b$ lies below one of $c$ or $d$.\nLattices of sets do not behave this way, since two sets can overlap in exactly the union of two smaller sets.\n\n### L2 human\n\nWhitman's condition is the key structural property of free lattices.\nTogether with the rules for generators, it gives an algorithm that decides the order of $\\mathrm{FL}(X)$.\nIt is also the main test for deciding which finite lattices embed in free lattices, because it passes to sublattices.\nSince the subsets of a four-element set fail (W), a free lattice cannot contain this Boolean lattice as a sublattice.\n",
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