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  "body": "\n## Definition\n\nA *relation algebra* is a structure $(A, \\vee, \\wedge, -, 0, 1, ;, c, 1')$, where $;$ is a binary operation, $c$ is a unary operation and $1'$ is an element of $A$.\nWrite $\\breve{x}$ for $c(x)$.\nThe following hold for all $x, y, z \\in A$:\n\n1. $(A, \\vee, \\wedge, -, 0, 1)$ is a [[Boolean algebra|0XADJKM699]], with order $\\le$.\n2. $(A, ;, 1')$ is a [[monoid|0N6R902FG1]].\n3. $(x \\vee y) ; z = (x ; z) \\vee (y ; z)$.\n4. $\\breve{\\breve{x}} = x$ and $\\breve{(x \\vee y)} = \\breve{x} \\vee \\breve{y}$.\n5. $\\breve{(x ; y)} = \\breve{y} ; \\breve{x}$.\n6. $\\breve{x} ; -(x ; y) \\le -y$.\n\n## Remarks\n\nExample: let $U$ be a set and let $A$ be the set of all [[binary relations|05AVARPXA2]] on $U$, with the set operations.\nLet $;$ be the [[composition|0PYSR20HGF]] $R ; S = \\set{(a, c) : (a, b) \\in R \\text{ and } (b, c) \\in S \\text{ for some } b}$.\nLet $\\breve{R}$ be the [[converse|054K3HNN5Z]] relation and $1'$ the identity relation.\nCheck of clause 6: if $(a, c) \\in \\breve{R} ; -(R ; S)$, there is $b$ with $(b, a) \\in R$ and $(b, c) \\notin R ; S$.\nIf $(a, c)$ were in $S$, then $(b, c) \\in R ; S$; so $(a, c) \\notin S$.\n\nNon-example: on the relations of a set $U$ with at least two elements, keep the Boolean operations, $;$ and $1'$, but take $\\breve{R} = R$.\nClause 5 then says $R ; S = S ; R$ for all relations, which fails for $U = \\set{a, b}$, $R = \\set{(a, b)}$ and $S = \\set{(b, b)}$, since $R ; S = \\set{(a, b)}$ and $S ; R = \\emptyset$.\n",
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