{
  "aliases": [
    "proof-by-exhaustion",
    "case-analysis"
  ],
  "backlinks": null,
  "body": "\n## Summary\n\nA proof by cases splits the situation into a finite number of cases that cover all possibilities.\nThen it proves the statement in each case.\nThe basic form is this rule: from $p \\lor q$, $p \\to r$ and $q \\to r$, conclude $r$.\n\n## Remarks\n\nAnother name is proof by exhaustion, mainly when the cases are many.\nExample: $n^2 + n$ is even for each integer $n$. If $n$ is even, $n^2$ and $n$ are even. If $n$ is odd, $n^2$ and $n$ are odd. In both cases, the sum is even.\nA split into $p$ and $\\neg p$ uses the [[law of excluded middle|0RYYK8PG4S]].\n\n## Explanations\n\n### L1 human\n\nWhen one argument does not work for all cases, split the cases.\nIntegers are even or odd, so a proof for even integers and a proof for odd integers together cover all integers.\nMake sure that no case is missing.\n",
  "commit": "6ac761d40018146541484a9e8b9eafd6ecc303dc",
  "front": {
    "aliases": [
      "proof-by-exhaustion",
      "case-analysis"
    ],
    "kind": "concept",
    "msc": [
      "00A35"
    ],
    "name": "Proof by cases",
    "same_as": {
      "wikidata": "Q9299950",
      "wikipedia": "Proof by exhaustion"
    },
    "sources": [
      {
        "locator": "lead section",
        "url": "https://en.wikipedia.org/wiki/Proof_by_exhaustion"
      }
    ],
    "tag": "0WCP0PVET6",
    "uses": [
      "0B94TKFQJX",
      "0H5GKDTB01",
      "0ZNX6DQ3EJ"
    ]
  },
  "kind": "concept",
  "link": "https://math2.tamnd.com/node/0WCP0PVET6/",
  "links": [
    {
      "field": "uses",
      "target": "0B94TKFQJX"
    },
    {
      "field": "uses",
      "target": "0H5GKDTB01"
    },
    {
      "field": "uses",
      "target": "0ZNX6DQ3EJ"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/0WCP0PVET6.md",
  "msc": [
    "00A35"
  ],
  "name": "Proof by cases",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/0WC/0WCP0PVET6.md",
  "status": null,
  "tag": "0WCP0PVET6",
  "uri": "m2://0WCP0PVET6",
  "url": "https://math2.tamnd.com/wiki/proof-by-cases/",
  "writing": {
    "tag": "0WCP0PVET6",
    "kind": "concept",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}