{
  "index": "1971-B-1",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "B-1. Let \\( S \\) be a set and let \\( \\circ \\) be a binary operation on \\( S \\) satisfying the two laws\n\\[\n\\begin{array}{l}\nx \\circ x=x \\text { for all } x \\text { in } S \\text {, and } \\\\\n(x \\circ y) \\circ z=(y \\circ z) \\circ x \\text { for all } x, y, z \\text { in } S .\n\\end{array}\n\\]\n\nShow that \\( \\circ \\) is associative and commutative.",
  "solution": "B-1 Using the given laws we have\n\\[\n\\begin{aligned}\nx \\circ y & =(x \\circ y) \\circ(x \\circ y)=[(x \\circ y) \\circ x] \\circ y=[(y \\circ x) \\circ x] \\circ y \\\\\n& =[(x \\circ x) \\circ y] \\circ y=(x \\circ y) \\circ y=(y \\circ y) \\circ x=y \\circ x .\n\\end{aligned}\n\\]\n\nFrom this commutative law we obtain\n\\[\n(x \\circ y) \\circ z=(y \\circ z) \\circ x=x \\circ(y \\circ z)\n\\]",
  "vars": [
    "x",
    "y",
    "z"
  ],
  "params": [
    "S"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "elementalpha",
        "y": "elementbeta",
        "z": "elementgamma",
        "S": "collection"
      },
      "question": "B-1. Let \\( collection \\) be a set and let \\( \\circ \\) be a binary operation on \\( collection \\) satisfying the two laws\n\\[\n\\begin{array}{l}\nelementalpha \\circ elementalpha = elementalpha \\text { for all } elementalpha \\text { in } collection \\text {, and } \\\\\n(elementalpha \\circ elementbeta) \\circ elementgamma = (elementbeta \\circ elementgamma) \\circ elementalpha \\text { for all } elementalpha, elementbeta, elementgamma \\text { in } collection .\n\\end{array}\n\\]\n\nShow that \\( \\circ \\) is associative and commutative.",
      "solution": "B-1 Using the given laws we have\n\\[\n\\begin{aligned}\nelementalpha \\circ elementbeta & =(elementalpha \\circ elementbeta) \\circ(elementalpha \\circ elementbeta)=[(elementalpha \\circ elementbeta) \\circ elementalpha] \\circ elementbeta=[(elementbeta \\circ elementalpha) \\circ elementalpha] \\circ elementbeta \\\\\n& =[(elementalpha \\circ elementalpha) \\circ elementbeta] \\circ elementbeta=(elementalpha \\circ elementbeta) \\circ elementbeta=(elementbeta \\circ elementbeta) \\circ elementalpha=elementbeta \\circ elementalpha .\n\\end{aligned}\n\\]\n\nFrom this commutative law we obtain\n\\[\n(elementalpha \\circ elementbeta) \\circ elementgamma=(elementbeta \\circ elementgamma) \\circ elementalpha=elementalpha \\circ(elementbeta \\circ elementgamma)\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "S": "labyrinth",
        "x": "daffodil",
        "y": "saxophone",
        "z": "pendulum"
      },
      "question": "B-1. Let \\( labyrinth \\) be a set and let \\( \\circ \\) be a binary operation on \\( labyrinth \\) satisfying the two laws\n\\[\n\\begin{array}{l}\ndaffodil \\circ daffodil=daffodil \\text { for all } daffodil \\text { in } labyrinth \\text {, and } \\\\\n(daffodil \\circ saxophone) \\circ pendulum=(saxophone \\circ pendulum) \\circ daffodil \\text { for all } daffodil, saxophone, pendulum \\text { in } labyrinth .\n\\end{array}\n\\]\n\nShow that \\( \\circ \\) is associative and commutative.",
      "solution": "B-1 Using the given laws we have\n\\[\n\\begin{aligned}\ndaffodil \\circ saxophone & =(daffodil \\circ saxophone) \\circ(daffodil \\circ saxophone)=[(daffodil \\circ saxophone) \\circ daffodil] \\circ saxophone=[(saxophone \\circ daffodil) \\circ daffodil] \\circ saxophone \\\\\n& =[(daffodil \\circ daffodil) \\circ saxophone] \\circ saxophone=(daffodil \\circ saxophone) \\circ saxophone=(saxophone \\circ saxophone) \\circ daffodil=saxophone \\circ daffodil .\n\\end{aligned}\n\\]\n\nFrom this commutative law we obtain\n\\[\n(daffodil \\circ saxophone) \\circ pendulum=(saxophone \\circ pendulum) \\circ daffodil=daffodil \\circ(saxophone \\circ pendulum)\n\\]\n"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantone",
        "y": "constanttwo",
        "z": "constantthree",
        "S": "emptiness"
      },
      "question": "B-1. Let \\( emptiness \\) be a set and let \\( \\circ \\) be a binary operation on \\( emptiness \\) satisfying the two laws\n\\[\n\\begin{array}{l}\nconstantone \\circ constantone=constantone \\text { for all } constantone \\text { in } emptiness \\text {, and } \\\\\n(constantone \\circ constanttwo) \\circ constantthree=(constanttwo \\circ constantthree) \\circ constantone \\text { for all } constantone, constanttwo, constantthree \\text { in } emptiness .\n\\end{array}\n\\]\n\nShow that \\( \\circ \\) is associative and commutative.",
      "solution": "B-1 Using the given laws we have\n\\[\n\\begin{aligned}\nconstantone \\circ constanttwo & =(constantone \\circ constanttwo) \\circ(constantone \\circ constanttwo)=[(constantone \\circ constanttwo) \\circ constantone] \\circ constanttwo=[(constanttwo \\circ constantone) \\circ constantone] \\circ constanttwo \\\\\n& =[(constantone \\circ constantone) \\circ constanttwo] \\circ constanttwo=(constantone \\circ constanttwo) \\circ constanttwo=(constanttwo \\circ constanttwo) \\circ constantone=constanttwo \\circ constantone .\n\\end{aligned}\n\\]\n\nFrom this commutative law we obtain\n\\[\n(constantone \\circ constanttwo) \\circ constantthree=(constanttwo \\circ constantthree) \\circ constantone=constantone \\circ(constanttwo \\circ constantthree)\n\\]\n"
    },
    "garbled_string": {
      "map": {
        "S": "lkjrevsn",
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "z": "mofnlcier"
      },
      "question": "B-1. Let \\( lkjrevsn \\) be a set and let \\( \\circ \\) be a binary operation on \\( lkjrevsn \\) satisfying the two laws\n\\[\n\\begin{array}{l}\nqzxwvtnp \\circ qzxwvtnp=qzxwvtnp \\text { for all } qzxwvtnp \\text { in } lkjrevsn \\text {, and } \\\\\n(qzxwvtnp \\circ hjgrksla) \\circ mofnlcier=(hjgrksla \\circ mofnlcier) \\circ qzxwvtnp \\text { for all } qzxwvtnp, hjgrksla, mofnlcier \\text { in } lkjrevsn .\n\\end{array}\n\\]\n\nShow that \\( \\circ \\) is associative and commutative.",
      "solution": "B-1 Using the given laws we have\n\\[\n\\begin{aligned}\nqzxwvtnp \\circ hjgrksla & =(qzxwvtnp \\circ hjgrksla) \\circ(qzxwvtnp \\circ hjgrksla)=[(qzxwvtnp \\circ hjgrksla) \\circ qzxwvtnp] \\circ hjgrksla=[(hjgrksla \\circ qzxwvtnp) \\circ qzxwvtnp] \\circ hjgrksla \\\\\n& =[(qzxwvtnp \\circ qzxwvtnp) \\circ hjgrksla] \\circ hjgrksla=(qzxwvtnp \\circ hjgrksla) \\circ hjgrksla=(hjgrksla \\circ hjgrksla) \\circ qzxwvtnp=hjgrksla \\circ qzxwvtnp .\n\\end{aligned}\n\\]\n\nFrom this commutative law we obtain\n\\[\n(qzxwvtnp \\circ hjgrksla) \\circ mofnlcier=(hjgrksla \\circ mofnlcier) \\circ qzxwvtnp=qzxwvtnp \\circ(hjgrksla \\circ mofnlcier)\n\\]"
    },
    "kernel_variant": {
      "question": "Let $\\Omega$ be a non-empty set equipped with a binary operation $\\star : \\Omega\\times\\Omega \\to \\Omega$ satisfying the following two identities.\n\n1.  (Interchange) \\[(u\\star v)\\star w\\;=\\;(v\\star w)\\star u\\quad\\text{for all }u,v,w\\in\\Omega.\\]\n2.  (Idempotence) \\[t\\star t\\;=\\;t\\quad\\text{for all }t\\in\\Omega.\\]\n\nProve that $\\star$ is both commutative and associative on $\\Omega$.",
      "solution": "Step 1. Commutativity.\nFix arbitrary u,v∈Ω.  Idempotence lets us duplicate their product:\n  u★v = (u★v)★(u★v).   (1)\nApply the interchange law to the right–hand side of (1), viewing (u★v), u, v as the three slots:\n  (u★v)★(u★v) = ((u★v)★u)★v.   (2)\nAnother interchange with u,v,u in the slots gives\n  (u★v)★u = (v★u)★u.   (3)\nInsert (3) into (2) and simplify, using interchange and idempotence as needed:\n  u★v\n  = ((v★u)★u)★v\n  = ((u★u)★v)★v\n  = (u★v)★v\n  = (v★v)★u\n  = v★u.\nHence u★v=v★u; ★ is commutative.\n\nStep 2. Associativity.\nFor arbitrary u,v,w∈Ω the interchange identity, followed by commutativity, yields\n  (u★v)★w = (v★w)★u = u★(v★w),\nwhich is the associative law.\n\nThus ★ is both commutative and associative on Ω.  ■",
      "_meta": {
        "core_steps": [
          "Duplicate an arbitrary product x∘y using idempotence: x∘y = (x∘y)∘(x∘y).",
          "Apply the interchange law repeatedly to that duplicate to reverse the two factors, obtaining x∘y = y∘x (commutativity).",
          "Insert commutativity into the interchange law to rewrite (x∘y)∘z as x∘(y∘z), giving associativity."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Name given to the underlying set.",
            "original": "S"
          },
          "slot2": {
            "description": "Symbol chosen for the binary operation.",
            "original": "∘"
          },
          "slot3": {
            "description": "Letters used for the three generic elements in the laws.",
            "original": "x, y, z"
          },
          "slot4": {
            "description": "Order in which the two axioms are listed in the statement.",
            "original": "1st: idempotence, 2nd: interchange"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}