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  "body": "\n## Proof\n\nWe use the description of the join and the meet as least upper and greatest lower bounds.\n\n1. If the [[lattice|X8WD0ABG07]] $L$ is given in algebraic form, the two absorption laws are among its axioms, and there is nothing to prove.\n2. So let $L$ be a [[partially ordered set|0QV2JS9RJ3]] $(L, \\le)$ in which every two elements have a [[join and a meet|F7YV441H6R]]. Let $a, b \\in L$, and put $c = a \\wedge b$.\n3. The meet $c$ is a lower bound of $\\set{a, b}$, so $c \\le a$.\n4. The element $a$ is an upper bound of $\\set{a, c}$, since $a \\le a$ by reflexivity and $c \\le a$ by Step 3.\n5. Every upper bound $z$ of $\\set{a, c}$ satisfies $a \\le z$, by the definition of an upper bound.\n6. By Steps 4 and 5, $a$ is the least upper bound of $\\set{a, c}$. The least upper bound is unique by antisymmetry, so $a \\vee (a \\wedge b) = a \\vee c = a$.\n7. Now put $d = a \\vee b$. The join $d$ is an upper bound of $\\set{a, b}$, so $a \\le d$.\n8. The element $a$ is a lower bound of $\\set{a, d}$, since $a \\le a$ and $a \\le d$.\n9. Every lower bound $z$ of $\\set{a, d}$ satisfies $z \\le a$.\n10. By Steps 8 and 9, $a$ is the greatest lower bound of $\\set{a, d}$, which is unique by antisymmetry. So $a \\wedge (a \\vee b) = a \\wedge d = a$.\n\n## Idea\n\nThe element $a \\wedge b$ lies below $a$, so joining it with $a$ adds nothing: $a$ is already the smallest element above both. Dually, $a \\vee b$ lies above $a$, so meeting it with $a$ gives back $a$.\n",
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    "sources": [
      {
        "url": "https://proofwiki.org/wiki/Join_Absorbs_Meet"
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        "url": "https://proofwiki.org/wiki/Meet_Absorbs_Join"
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    "tag": "FKDZ715NQS",
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  "markdown": "https://math2.tamnd.com/node/FKDZ715NQS.md",
  "msc": [
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  "name": "Proof of the absorption laws in a lattice",
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