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  "body": "\n## Example\n\nThe *diamond lattice* $M_3$ has five elements $0, a, b, c, 1$.\nIts order is $0 \u003c a \u003c 1$, $0 \u003c b \u003c 1$ and $0 \u003c c \u003c 1$, with $a$, $b$, $c$ pairwise incomparable.\nThe join of any two distinct elements among $a, b, c$ is $1$ and their meet is $0$, so $M_3$ is a [[lattice|WH4SFAACAS]].\n\n$M_3$ is a [[modular lattice|QX7WZ92QAK]].\nIf $u \\le v$ and $u = v$, or $u = 0$, or $v = 1$, the modular law $u \\vee (x \\wedge v) = (u \\vee x) \\wedge v$ reduces to a trivial identity; these cases cover all comparable pairs.\nFor example, with $u = 0$ both sides equal $x \\wedge v$, and with $v = 1$ both sides equal $u \\vee x$.\n\n$M_3$ is not a [[distributive lattice|0HM0HQCK6N]].\nIndeed $a \\wedge (b \\vee c) = a \\wedge 1 = a$, while $(a \\wedge b) \\vee (a \\wedge c) = 0 \\vee 0 = 0$.\n\n## Remarks\n\n$M_3$ is isomorphic to the lattice of subspaces of the plane $\\F_2^2$ over the field with two elements: the zero subspace, the three lines and the whole plane.\nBy the [[M3–N5 characterization of distributivity|C6GH4VHAMQ]], a lattice is distributive exactly when it has no sublattice isomorphic to $M_3$ or to the [[pentagon lattice N5|DHC8TZXGYS]].\n\n## Explanations\n\n### L1 human\n\nDraw a diamond with an extra middle point: a bottom, a top and three points side by side in between.\nEach middle point meets each other one at the bottom and joins it at the top.\nSo $a$ meets \"$b$ or $c$\" in $a$ itself, but $a$ meets $b$ and meets $c$ only in the bottom.\nThis is the simplest place where meet does not distribute over join, although the weaker modular law still holds.\n",
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