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  "body": "\n## Statement\n\nLet $X$ be a set and let $\\mathrm{FL}(X)$ be the [[free lattice|WP20S9NZ15]] on $X$.\nFor lattice [[terms|6D3WA7N2W4]] $s$ and $t$ over $X$, write $s \\le t$ when the value of $s$ is below the value of $t$ in $\\mathrm{FL}(X)$.\nThen $s \\le t$ holds if and only if one of the following holds:\n\n1. $s$ and $t$ are the same letter $x \\in X$;\n2. $s = s_1 \\vee s_2$, and $s_1 \\le t$ and $s_2 \\le t$;\n3. $t = t_1 \\wedge t_2$, and $s \\le t_1$ and $s \\le t_2$;\n4. $s$ is a letter $x \\in X$, $t = t_1 \\vee t_2$, and $x \\le t_1$ or $x \\le t_2$;\n5. $s = s_1 \\wedge s_2$, $t$ is a letter $x \\in X$, and $s_1 \\le x$ or $s_2 \\le x$;\n6. $s = s_1 \\wedge s_2$, $t = t_1 \\vee t_2$, and $s_1 \\le t$ or $s_2 \\le t$ or $s \\le t_1$ or $s \\le t_2$.\n\nConsequently, equality of two lattice terms in $\\mathrm{FL}(X)$ is [[decidable|PPGD5KW8JC]]: $s = t$ in $\\mathrm{FL}(X)$ exactly when $s \\le t$ and $t \\le s$.\n\n## Remarks\n\nIn each clause the comparisons on the right involve terms of smaller total length, so the clauses give a recursive algorithm to decide $s \\le t$.\nThe \"if\" direction holds in every lattice.\nThe \"only if\" direction is the content of the theorem; clause 6 is [[Whitman's condition|5TNA2HBG7Y]].\nExample: $x \\wedge (y \\vee z) \\le (x \\wedge y) \\vee (x \\wedge z)$ fails in $\\mathrm{FL}(\\set{x, y, z})$.\nBy clause 6 it would need one of $x \\le (x \\wedge y) \\vee (x \\wedge z)$, $y \\vee z \\le (x \\wedge y) \\vee (x \\wedge z)$, $x \\wedge (y \\vee z) \\le x \\wedge y$ or $x \\wedge (y \\vee z) \\le x \\wedge z$, and the clauses refute each of them in turn.\nSo the distributive law is not an identity of all lattices.\n\n## Explanations\n\n### L1 human\n\nTwo lattice formulas, such as $x \\wedge (y \\vee x)$ and $x$, may look different but always give the same value.\nThis theorem gives a mechanical test for when this happens.\nTo compare two formulas, look at their outermost operations and reduce the question to smaller comparisons, until only letters are left.\n\n### L2 human\n\nThe word problem asks whether two terms define the same element of a free algebra.\nFor finitely presented groups it can be undecidable, but for free lattices it has this explicit solution, due to Whitman.\nThe procedure also decides whether a lattice identity holds in all lattices.\nIndeed, $s = t$ holds in every lattice exactly when $s = t$ in the free lattice on the letters of $s$ and $t$.\nWhitman also used these rules to show that each element of $\\mathrm{FL}(X)$ has a canonical form: a shortest term, unique up to commutativity and associativity.\n",
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