{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-quotient-set-01",
  "mko_id": "mko-quotient-set",
  "title": {
    "zh-Hant": "模2關係在四元素載體上的商集",
    "en": "The quotient of a four-element carrier by parity"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-quotient-set-01",
    "source": {
      "mko_id": "mko-quotient-set",
      "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
      "sha256": "3741dd3a04bde75f8d82e8fd31849761617fb2410c548387ba4434f502159236",
      "english_sha256": "dae34ce1e56ae7006833ff81818110a2d41c8df3ecf627d111fe3e8120eb954f"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "calc"
    ],
    "goal": {
      "zh-Hant": "X/R={ {0,2},{1,3} }，且投影值是整個類。",
      "en": "X/R={{0,2},{1,3}}, and projection values are whole classes."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "X={0,1,2,3}；xRy 當且僅當2整除x-y。",
          "en": "X={0,1,2,3}; xRy exactly when 2 divides x-y."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "x\\in X",
        "after_tex": "2\\mid x-x=0",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "任意元素與自身相關，故關係反身。",
          "en": "Every element relates to itself, proving reflexivity."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "此式對全部四元素成立，不需另選代表元。",
            "en": "The formula covers all four elements without choosing representatives."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S02",
        "uses": [
          "G1"
        ],
        "before_tex": "2\\mid x-y",
        "after_tex": "2\\mid y-x=-(x-y)",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取負保留整除性，故關係對稱。",
          "en": "Negation preserves divisibility, so the relation is symmetric."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "反向再取負回到x-y，沒有方向遺失。",
            "en": "Negating again recovers x-y, so no direction is lost."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S01",
          "S02"
        ],
        "before_tex": "2\\mid x-y,\\quad2\\mid y-z",
        "after_tex": "2\\mid x-z=(x-y)+(y-z)",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個偶數之和仍偶，證傳遞性，與前兩步合成等價關係。",
          "en": "A sum of two even integers is even, proving transitivity and completing equivalence qualification."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "x、y、z均須屬同一載體X。",
            "en": "x,y,z must all lie in the same carrier X."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S03"
        ],
        "before_tex": "[x]=\\{y\\in X:2\\mid y-x\\}",
        "after_tex": "[0]=[2]=\\{0,2\\}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐查0、1、2、3與0及2的差，完整偶數類均為{0,2}。",
          "en": "Checking all four differences with 0 and 2 gives the complete even class {0,2}."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "不得只留代表0；2也是類中的元素而非另一商元素。",
            "en": "Do not retain only representative 0; 2 belongs to the same class, not another quotient element."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S03"
        ],
        "before_tex": "[1],\\ [3]",
        "after_tex": "[1]=[3]=\\{1,3\\}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "全部奇數類為{1,3}，四個元素的類已全部計算。",
          "en": "The complete odd class is {1,3}; classes of all four elements have now been computed."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "1與3相關但與0、2不相關，因此沒有漏類。",
            "en": "1 and 3 relate to each other but not to 0 or 2, so no class is missing."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "S04",
          "S05"
        ],
        "before_tex": "\\{[0],[1],[2],[3]\\}",
        "after_tex": "X/R=\\{\\{0,2\\},\\{1,3\\}\\}",
        "relation": "implies",
        "rule": "mko_definition",
        "conditions": [],
        "explanation": {
          "zh-Hant": "以元素排序的集合正規化類，再去除重複，得到兩個商元素。",
          "en": "Normalize each class as its sorted element set and deduplicate, leaving two quotient elements."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S04",
            "S05"
          ],
          "explanation": {
            "zh-Hant": "去重的是相同類而非類內互異元素；{0,1}只是代表元集合。",
            "en": "Deduplication removes repeated classes, not distinct elements within a class; {0,1} is only a representative set."
          }
        },
        "leaf_basis": "mko_definition"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "S06"
        ],
        "before_tex": "E=\\{0,2\\},\\quad O=\\{1,3\\}",
        "after_tex": "E\\cup O=X,\\quad E\\cap O=\\varnothing",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "明列的兩類覆蓋全部四元素且互斥，所以形成partition。",
          "en": "The two explicit classes cover all four elements and are disjoint, giving a partition."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S06"
          ],
          "explanation": {
            "zh-Hant": "兩類均非空；同時核覆蓋與互斥，而非只核類數。",
            "en": "Both classes are nonempty; coverage and disjointness are checked, not merely the number of classes."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "S06",
          "S07"
        ],
        "before_tex": "\\pi:X\\to X/R",
        "after_tex": "\\pi(0)=E,\\ \\pi(1)=O,\\ \\pi(2)=E,\\ \\pi(3)=O",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "完整投影表把每個元素送到唯一整個類，並命中全部商元素。",
          "en": "The full projection sends every element to its unique whole class and hits every quotient element."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S06",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "輸出是E或O，不是數字0或1；未在商集上宣告新運算。",
            "en": "Outputs are E or O, not numbers 0 or 1; no new quotient operation has been declared."
          }
        },
        "leaf_basis": "calc"
      }
    ],
    "result": {
      "zh-Hant": "X/R={ {0,2},{1,3} }，且投影值是整個類。",
      "en": "X/R={{0,2},{1,3}}, and projection values are whole classes."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "X/R={0,1}。",
          "en": "X/R={0,1}."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "0與1可作代表，但商元素是兩個集合E、O。",
          "en": "0 and 1 can be representatives, but the quotient elements are sets E and O."
        }
      },
      {
        "claim": {
          "zh-Hant": "[0]與[2]是兩個不同商元素。",
          "en": "[0] and [2] are distinct quotient elements."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "兩類逐元素均為{0,2}。",
          "en": "Both classes are exactly {0,2}."
        }
      },
      {
        "claim": {
          "zh-Hant": "任意X上的函數都會下降到商集。",
          "en": "Every function on X descends to the quotient."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "f(x)=x在0與2上值不同，故不能定義單值f([0])。",
          "en": "f(x)=x differs on 0 and 2, so it cannot define a single value f([0])."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "calc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "假設整數整除法則、集合外延相等與函數定義。",
          "en": "Assume integer divisibility laws, extensional set equality and the definition of a function."
        }
      },
      {
        "id": "mko_definition",
        "kind": "cited",
        "statement": {
          "zh-Hant": "X/R是所有distinct equivalence classes的集合，不是挑選代表元的集合。classes覆蓋X且兩兩互斥；canonical projection把每個元素送到其class。任何商運算仍需另證代表元無關。",
          "en": "X/R is the set of all distinct equivalence classes, not a set of selected representatives. The classes cover X and are pairwise disjoint. The canonical projection sends each element to its class. Any quotient operation still needs a separate proof of independence from representative choice."
        },
        "source": {
          "mko_id": "mko-quotient-set",
          "sha256": "3741dd3a04bde75f8d82e8fd31849761617fb2410c548387ba4434f502159236",
          "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f"
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": []
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "完整有限載體及模2等價關係；不自動繼承運算。",
      "en": "A complete finite carrier and parity equivalence; no automatically inherited operation."
    },
    "md_symbols": [
      [
        "E,O",
        "偶數類與奇數類",
        "even and odd classes"
      ],
      [
        "\\pi",
        "正規投影",
        "canonical projection"
      ]
    ],
    "md_error_controls": [
      {
        "title": "投影型別 / Projection type",
        "zh": "π的codomain是類的集合，不是所選代表元集合。",
        "en": "The codomain of pi is the set of classes, not selected representatives."
      },
      {
        "title": "良定另證 / Separate well-definedness",
        "zh": "要讓f([x])=g(x)合法，須核xRy推出g(x)=g(y)。",
        "en": "To define f([x])=g(x), verify xRy implies g(x)=g(y)."
      }
    ],
    "md_open_items": []
  },
  "revision": {
    "id": "ocme-new142-reader-adoption-20261021",
    "collection": "coverage-20261021",
    "source_commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
    "adopted_source_sha256": "4172954596b8f050df418adbfe5f100efbfc44e67fa2aa975c7d69dddbead328",
    "original_source_sha256": "0b6728555127f07d75805ac6a167e55fce6180e7883cf8748e68bfbb3b5aab46",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "reader_review": "142_of_142_source_bound",
    "wording_review": "original_Lead_decisions_resolved",
    "glossary_reader_paraphrases": [],
    "independent_AI_concurrence": "SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW",
    "issue_ids": [
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    "parent_manifest_sha256": "d29f4cae8229ebd24f60d496cb1d3cd4ad8c0a4dca9d44daab0c3245c544283a",
    "scope": "Owner-adopted reader revision. Lead and bilingual review concern the recorded text and premises; no new universal or Lean proof, canonical promotion or historical-label rewrite."
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}