{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe argue by [[induction|0ASJVEFA9J]] on $c$.\nLet $S$ be the [[successor function|0832XVR512]].\nThe [[addition|0W5DSDFQ4G]] satisfies $x + 0 = x$ and $x + S(y) = S(x + y)$.\nThe [[multiplication|0JRC32D6K2]] satisfies $x \\cdot 0 = 0$ and $x \\cdot S(y) = x \\cdot y + x$.\nFix natural numbers $a$ and $b$.\n\n1. For $c = 0$: since $b + 0 = b$, we get $a \\cdot (b + 0) = a \\cdot b$.\n2. Also $a \\cdot 0 = 0$, so $a \\cdot b + a \\cdot 0 = a \\cdot b + 0 = a \\cdot b$. Thus the law holds for $c = 0$.\n3. Suppose that $a \\cdot (b + c) = a \\cdot b + a \\cdot c$ for some natural number $c$.\n4. By the definition of addition, $b + S(c) = S(b + c)$. So $a \\cdot (b + S(c)) = a \\cdot S(b + c)$.\n5. By the definition of multiplication, $a \\cdot S(b + c) = a \\cdot (b + c) + a$.\n6. By the hypothesis of step 3, $a \\cdot (b + c) + a = (a \\cdot b + a \\cdot c) + a$.\n7. By the [[associativity of addition|0C32J27WQC]], $(a \\cdot b + a \\cdot c) + a = a \\cdot b + (a \\cdot c + a)$.\n8. By the definition of multiplication, $a \\cdot c + a = a \\cdot S(c)$.\n9. Steps 4 to 8 give $a \\cdot (b + S(c)) = a \\cdot b + a \\cdot S(c)$.\n10. Since $S(c) = c + 1$, induction gives the law for every natural number $c$. As $a$ and $b$ were arbitrary, it holds for all $a$, $b$ and $c$.\n\n## Idea\n\nBoth recursive definitions work on the last argument, so we induct on $c$.\nGoing from $c$ to $c + 1$ adds one more copy of $a$ on each side.\nAssociativity of addition lets us put that copy next to $a \\cdot c$.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "methods": [
      "0ASJVEFA9J"
    ],
    "msc": [
      "03F30"
    ],
    "name": "Proof of the distributive law by induction",
    "proves": "0KBEMAJAYA",
    "sources": [
      {
        "url": "https://proofwiki.org/wiki/Natural_Number_Multiplication_Distributes_over_Addition"
      }
    ],
    "tag": "2PZM6JSNN2",
    "uses": [
      "0W5DSDFQ4G",
      "0JRC32D6K2",
      "0832XVR512",
      "0C32J27WQC",
      "0ASJVEFA9J"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/2PZM6JSNN2/",
  "links": [
    {
      "field": "proves",
      "target": "0KBEMAJAYA"
    },
    {
      "field": "uses",
      "target": "0W5DSDFQ4G"
    },
    {
      "field": "uses",
      "target": "0JRC32D6K2"
    },
    {
      "field": "uses",
      "target": "0832XVR512"
    },
    {
      "field": "uses",
      "target": "0C32J27WQC"
    },
    {
      "field": "uses",
      "target": "0ASJVEFA9J"
    },
    {
      "field": "methods",
      "target": "0ASJVEFA9J"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/2PZM6JSNN2.md",
  "msc": [
    "03F30"
  ],
  "name": "Proof of the distributive law by induction",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/2PZ/2PZM6JSNN2.md",
  "status": null,
  "tag": "2PZM6JSNN2",
  "uri": "m2://2PZM6JSNN2",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-distributive-law-by-induction/",
  "writing": {
    "tag": "2PZM6JSNN2",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      },
      {
        "what": "a source with a locator",
        "have": 0,
        "want": 1
      }
    ]
  }
}