{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe apply ordinary [[induction|0ASJVEFA9J]] to a stronger property.\nLet $S$ be the [[successor function|0832XVR512]], and write $1 = S(0)$.\nThe [[addition|0W5DSDFQ4G]] satisfies $a + 0 = a$ and $a + S(b) = S(a + b)$. So $n + 1 = S(n + 0) = S(n)$.\nThe relations $\\le$ and $\u003c$ are the [[order on the natural numbers|0Y9H52BT03]].\nFor each natural number $n$, let $Q(n)$ be the property \"$P(k)$ holds for every natural number $k \u003c n$\".\n\n1. Lemma A: each natural number $c \\neq 0$ equals $S(d)$ for some natural number $d$. We prove \"$c = 0$ or $c = S(d)$ for some $d$\" by induction on $c$.\n2. For $c = 0$, the first alternative holds. For $c + 1 = S(c)$, take $d = c$. This proves Lemma A.\n3. Lemma B: no natural number $k$ satisfies $k \u003c 0$. Suppose that $k + c = 0$ for a natural number $c$.\n4. If $c \\neq 0$, Lemma A gives $c = S(d)$. Then $k + c = S(k + d)$, and this is not $0$ because [[zero is not a successor|0D752BG155]].\n5. So $c = 0$ and $k = k + 0 = 0$. Thus $k \\le 0$ forces $k = 0$, and $k \u003c 0$ is impossible.\n6. Lemma C: if $k \u003c n + 1$, then $k \u003c n$ or $k = n$. Indeed, let $k + c = S(n)$ with $k \\neq S(n)$.\n7. Then $c \\neq 0$, because $c = 0$ would give $k = S(n)$. By Lemma A, $c = S(d)$ for some $d$.\n8. So $S(k + d) = k + S(d) = S(n)$. Since [[the successor function is injective|0TBYMWPWWB]], $k + d = n$.\n9. Thus $k \\le n$, so $k \u003c n$ or $k = n$. This proves Lemma C.\n10. Base case: by Lemma B, there is no $k \u003c 0$. So $Q(0)$ holds.\n11. Induction step: suppose that $Q(n)$ holds. The hypothesis of the statement then gives $P(n)$.\n12. Let $k \u003c n + 1$. By Lemma C, $k \u003c n$ or $k = n$. In the first case $P(k)$ holds by $Q(n)$, and in the second by step 11.\n13. So $Q(n + 1)$ holds. By induction, $Q(n)$ holds for every natural number $n$.\n14. Let $n$ be a natural number. By step 13, $Q(n)$ holds, and the hypothesis of the statement gives $P(n)$.\n\n## Idea\n\nOrdinary induction carries only one case to the next.\nThe trick is to carry all earlier cases at once, in the property $Q(n)$.\nThen the step from $Q(n)$ to $Q(n + 1)$ only needs one new case, $P(n)$, which the hypothesis gives.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "methods": [
      "0ASJVEFA9J"
    ],
    "msc": [
      "03F30"
    ],
    "name": "Proof of strong induction from ordinary induction",
    "proves": "0KMVFSHFGZ",
    "sources": [
      {
        "locator": "Complete (strong) induction",
        "url": "https://en.wikipedia.org/wiki/Mathematical_induction"
      }
    ],
    "tag": "C54QYDXFP3",
    "uses": [
      "0ASJVEFA9J",
      "0Y9H52BT03",
      "0W5DSDFQ4G",
      "0832XVR512",
      "0TBYMWPWWB",
      "0D752BG155"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/C54QYDXFP3/",
  "links": [
    {
      "field": "proves",
      "target": "0KMVFSHFGZ"
    },
    {
      "field": "uses",
      "target": "0ASJVEFA9J"
    },
    {
      "field": "uses",
      "target": "0Y9H52BT03"
    },
    {
      "field": "uses",
      "target": "0W5DSDFQ4G"
    },
    {
      "field": "uses",
      "target": "0832XVR512"
    },
    {
      "field": "uses",
      "target": "0TBYMWPWWB"
    },
    {
      "field": "uses",
      "target": "0D752BG155"
    },
    {
      "field": "methods",
      "target": "0ASJVEFA9J"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/C54QYDXFP3.md",
  "msc": [
    "03F30"
  ],
  "name": "Proof of strong induction from ordinary induction",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/C54/C54QYDXFP3.md",
  "status": null,
  "tag": "C54QYDXFP3",
  "uri": "m2://C54QYDXFP3",
  "url": "https://math2.tamnd.com/wiki/proof-of-strong-induction-from-ordinary-induction/",
  "writing": {
    "tag": "C54QYDXFP3",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}