{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261013-advance-20261018-r01-topological-space-0056",
  "pilot_number": "R0017-09231",
  "number": 9231,
  "family_id": "qf-ext18-topology-discrete-indiscrete",
  "title": {
    "zh-Hant": "第 9231 題：有限拓撲完整核對",
    "en": "Question 9231: Topology Qualification before Discrete or Indiscrete Classification"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261013-advance-20261018-r01-topological-space-0056",
    "batch_id": "batch-20261013-advance-20261018-r01-topological-space-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-topological-space"
    ],
    "title_zh": "有限拓撲完整核對",
    "stem_zh": "有限底集合完整資料。操作=topology-discrete-indiscrete；資料={\"n\":4,\"opens\":[[],[3],[0,3],[0,1,3],[0,1,2,3]]}。請依 [n,opens,topology,classification] 的欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "(4, [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], 是, 其他拓撲)",
      "explanation_zh": "擴展基礎核對=(4, [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], 是, 其他拓撲)；依完整明示資料逐欄精確重算；有限／合成資料核對不是一般形式證明；抽查不構成拓撲公理；開集族與底集合分離。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 1,
      "skills": [
        "有限拓撲完整核對"
      ]
    },
    "variant": {
      "family_id": "qf-ext18-topology-discrete-indiscrete",
      "seed": 202610181056,
      "parameters": {
        "n": 4,
        "opens": [
          [],
          [
            3
          ],
          [
            0,
            3
          ],
          [
            0,
            1,
            3
          ],
          [
            0,
            1,
            2,
            3
          ]
        ],
        "domain": "Fin_topology_family",
        "operation": "topology-discrete-indiscrete",
        "semantic_template": "qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[3],[0,3],[0,1,3],[0,1,2,3]]}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_finite_arithmetic",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "僅為產生端候選及自檢；主端須另做驗收，不構成人工審定、普遍證明、Lean執行或形式Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-13",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261013-advance-20261018-r01-topological-space-0056",
    "batch_id": "batch-20261013-advance-20261018-r01-topological-space-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "ac3692ab5ba36c9c5dbd28696b494d7ce158e3dd6aa1c3fca6fc2b77c2e28363",
      "family_id": "qf-ext18-topology-discrete-indiscrete",
      "seed": 202610181056,
      "semantic_template": "qf-ext18-topology-discrete-indiscrete/{\"n\":4,\"opens\":[[],[3],[0,3],[0,1,3],[0,1,2,3]]}"
    },
    "fields": {
      "title": "Topology Qualification before Discrete or Indiscrete Classification",
      "stem": "Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.",
      "answer": {
        "kind": "expression",
        "value": "(4, [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology)",
        "explanation": "Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology)."
      },
      "skills": [
        "Topology Qualification before Discrete or Indiscrete Classification"
      ],
      "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."
    }
  },
  "givens": [
    {
      "id": "G001",
      "text": {
        "zh-Hant": "有限底集合完整資料。操作=topology-discrete-indiscrete；資料={\"n\":4,\"opens\":[[],[3],[0,3],[0,1,3],[0,1,2,3]]}。請依 [n,opens,topology,classification] 的欄序回答。",
        "en": "Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "space=given;n=4;raw family=[[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]]",
      "after": "universe=[0, 1, 2, 3];canonical opens=[[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]]",
      "relation": "implies",
      "rule_id": "finite_topology_normalize_relative_universe",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每個候選先合型為同一Fin(n)的子集，去重元素及重複子集，再以二進位mask排序；開集族與底集合是不同層資料。",
        "en": "Type every candidate as a subset of the same Fin(n),deduplicate elements and members,and order by binary masks.The open family and underlying set are distinct data layers."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "space=given;canonical family=[[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]]",
      "after": "self pairs are idempotent;unordered distinct pairs cover all ordered pairs",
      "relation": "implies",
      "rule_id": "binary_set_operation_symmetry_coverage",
      "conditions": [],
      "explanation": {
        "zh-Hant": "A∪A=A、A∩A=A自動在候選族；聯集與交集可交換，因此列全部不同無序對即可完整核所有有序對，不是抽樣。",
        "en": "Self pairs are covered by A∪A=A and A∩A=A.Commutativity means all distinct unordered pairs cover every ordered pair;this is not sampling."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "space=given;A=[];B=[3]",
      "after": "union=[3];in family=是;intersection=[];in family=是",
      "relation": "implies",
      "rule_id": "finite_topology_complete_binary_pair",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依集合定義逐一求聯集與交集，分別查是否仍在原族；保留首次反例的原兩集合與實際結果。",
        "en": "Form union and intersection from the set definitions and test each against the original family.Preserve the original pair and result for the first failure."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "space=given;A=[];B=[0, 3]",
      "after": "union=[0, 3];in family=是;intersection=[];in family=是",
      "relation": "implies",
      "rule_id": "finite_topology_complete_binary_pair",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依集合定義逐一求聯集與交集，分別查是否仍在原族；保留首次反例的原兩集合與實際結果。",
        "en": "Form union and intersection from the set definitions and test each against the original family.Preserve the original pair and result for the first failure."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "space=given;A=[];B=[0, 1, 3]",
      "after": "union=[0, 1, 3];in family=是;intersection=[];in family=是",
      "relation": "implies",
      "rule_id": "finite_topology_complete_binary_pair",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依集合定義逐一求聯集與交集，分別查是否仍在原族；保留首次反例的原兩集合與實際結果。",
        "en": "Form union and intersection from the set definitions and test each against the original family.Preserve the original pair and result for the first failure."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "space=given;A=[];B=[0, 1, 2, 3]",
      "after": "union=[0, 1, 2, 3];in family=是;intersection=[];in family=是",
      "relation": "implies",
      "rule_id": "finite_topology_complete_binary_pair",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依集合定義逐一求聯集與交集，分別查是否仍在原族；保留首次反例的原兩集合與實際結果。",
        "en": "Form union and intersection from the set definitions and test each against the original family.Preserve the original pair and result for the first failure."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "space=given;A=[3];B=[0, 3]",
      "after": "union=[0, 3];in family=是;intersection=[3];in family=是",
      "relation": "implies",
      "rule_id": "finite_topology_complete_binary_pair",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依集合定義逐一求聯集與交集，分別查是否仍在原族；保留首次反例的原兩集合與實際結果。",
        "en": "Form union and intersection from the set definitions and test each against the original family.Preserve the original pair and result for the first failure."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "space=given;A=[3];B=[0, 1, 3]",
      "after": "union=[0, 1, 3];in family=是;intersection=[3];in family=是",
      "relation": "implies",
      "rule_id": "finite_topology_complete_binary_pair",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依集合定義逐一求聯集與交集，分別查是否仍在原族；保留首次反例的原兩集合與實際結果。",
        "en": "Form union and intersection from the set definitions and test each against the original family.Preserve the original pair and result for the first failure."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "space=given;A=[3];B=[0, 1, 2, 3]",
      "after": "union=[0, 1, 2, 3];in family=是;intersection=[3];in family=是",
      "relation": "implies",
      "rule_id": "finite_topology_complete_binary_pair",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依集合定義逐一求聯集與交集，分別查是否仍在原族；保留首次反例的原兩集合與實際結果。",
        "en": "Form union and intersection from the set definitions and test each against the original family.Preserve the original pair and result for the first failure."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "space=given;A=[0, 3];B=[0, 1, 3]",
      "after": "union=[0, 1, 3];in family=是;intersection=[0, 3];in family=是",
      "relation": "implies",
      "rule_id": "finite_topology_complete_binary_pair",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依集合定義逐一求聯集與交集，分別查是否仍在原族；保留首次反例的原兩集合與實際結果。",
        "en": "Form union and intersection from the set definitions and test each against the original family.Preserve the original pair and result for the first failure."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "space=given;A=[0, 3];B=[0, 1, 2, 3]",
      "after": "union=[0, 1, 2, 3];in family=是;intersection=[0, 3];in family=是",
      "relation": "implies",
      "rule_id": "finite_topology_complete_binary_pair",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依集合定義逐一求聯集與交集，分別查是否仍在原族；保留首次反例的原兩集合與實際結果。",
        "en": "Form union and intersection from the set definitions and test each against the original family.Preserve the original pair and result for the first failure."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S012",
      "uses": [
        "G001",
        "S011"
      ],
      "before": "space=given;A=[0, 1, 3];B=[0, 1, 2, 3]",
      "after": "union=[0, 1, 2, 3];in family=是;intersection=[0, 1, 3];in family=是",
      "relation": "implies",
      "rule_id": "finite_topology_complete_binary_pair",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依集合定義逐一求聯集與交集，分別查是否仍在原族；保留首次反例的原兩集合與實際結果。",
        "en": "Form union and intersection from the set definitions and test each against the original family.Preserve the original pair and result for the first failure."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S013",
      "uses": [
        "G001",
        "S012"
      ],
      "before": "space=given;empty=是;full=是;union closed=是;intersection closed=是",
      "after": "topology=是;first union failure=無;first intersection failure=無",
      "relation": "implies",
      "rule_id": "finite_topology_axioms_joint_gate",
      "conditions": [],
      "explanation": {
        "zh-Hant": "四條件同時成立才是拓撲。底集合有限，所以不同開集也只有有限多個；任意聯集可約成有限不同成員的聯集，由二元閉合與空聯集得到。空交集需全集，無成員時二元條件真也不能略過空集／全集。",
        "en": "All four conditions are required.On a finite underlying set there are finitely many distinct opens,so an arbitrary union reduces to a finite distinct-member union,covered by binary closure and the empty union.The empty intersection needs the whole set.Vacuous binary closure does not waive empty/whole-set membership."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S014",
      "uses": [
        "G001",
        "S013"
      ],
      "before": "topology qualified=是;family size=5;powerset size=16",
      "after": "classification=其他拓撲",
      "relation": "implies",
      "rule_id": "topology_classification_after_qualification",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先過拓撲資格再分。全部2^n子集為離散，否則只含空集及全集為不可分；空底集合兩者相同，協定優先標離散。",
        "en": "Qualify the topology first.All2^n subsets give the discrete topology;otherwise only empty and whole give the indiscrete topology.On the empty space both coincide,and the protocol classifies discrete first."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "result": [
      4,
      [
        [],
        [
          3
        ],
        [
          0,
          3
        ],
        [
          0,
          1,
          3
        ],
        [
          0,
          1,
          2,
          3
        ]
      ],
      true,
      "其他拓撲"
    ]
  },
  "conclusion": {
    "canonical_value": "(4, [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], 是, 其他拓撲)",
    "explanation": {
      "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。",
      "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged."
    }
  },
  "definition_note": {
    "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。",
    "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate."
  },
  "source": {
    "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db",
    "question_binding": {
      "batch_id": "batch-20261013-advance-20261018-r01-topological-space-foundations",
      "batch_path": "public/data/questions/batches/batch-20261013-advance-20261018-r01-topological-space-foundations/questions.jsonl",
      "batch_sha256": "9d63d2019d578bf3d3a3d09635666b7dd08cd8ca93894403258ded79a01fe94f",
      "original_line": 56,
      "original_line_sha256": "5185dc99a20f0605893052ce642cac6461c2ac5d0e477527d9ac263c3a0d7aa0",
      "question_compact_sha256": "ac3692ab5ba36c9c5dbd28696b494d7ce158e3dd6aa1c3fca6fc2b77c2e28363",
      "english_row_compact_sha256": "824073a1b69cf3051a7c0341f3f3dc0149252abc831995ea713ffd177bc4e24d",
      "english_manifest_path": "public/data/i18n/editions/batch-20261013-advance-20261018-r01-topological-space-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261013-advance-20261018-r01-topological-space-foundations.en.jsonl"
    },
    "mko_bindings": {
      "mko-topological-space": {
        "sha256": "95f78b4f323b3139dd05068842f7209ac33b39b3d86feb5b73e0eb8849cdb43d",
        "english_sha256": "d4be991db50b46add5676c87b921ccea446b23b0f93e496d3dac77e5f07c3d7c",
        "original_path": "public/data/mko/topological-space.json",
        "english_path": "public/data/i18n/editions/mko-topological-space.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "20e59f30c0b6218e6d9614c16170d770e56a2f1e4ae5ce39d0a8a3db206afa61",
    "original_candidate_sha256": "bca42d8f4c4d826811b47e7b4417b6a2ab561d2e232b0a5df4b2c9112832e50e",
    "original_candidate_line_sha256": "bca42d8f4c4d826811b47e7b4417b6a2ab561d2e232b0a5df4b2c9112832e50e"
  },
  "historical_source_state": {
    "review_state": {
      "derivation": "candidate_authored_and_locally_checked",
      "source_question_state": "candidate",
      "independent_AI_or_human_review": "NOT_PERFORMED",
      "formalization": "not_formalized",
      "Lean": "NOT_RUN"
    },
    "scope_note": {
      "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。",
      "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged."
    },
    "label": "Frozen producer metadata; current per-question review is separate below."
  },
  "revision": {
    "id": "review-20261026-006-p18",
    "review_date": "2026-10-26",
    "owner_authorized_reader_publication": true,
    "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.",
    "review_receipts": {
      "semantic_review_sha256": "abad28564830bb9b6ef069ca5a1aa128ce7b1a67a5cb8fe6d868a06db73bcfb8",
      "question_first_witness_sha256": "43126988479968837abe086263dac1b24c37463ad80e00d2e280a7b8e9036a04"
    },
    "status": "second_context_reviewed",
    "mathematical_semantics": true,
    "bilingual_semantics": true,
    "steps_reviewed": 14,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "35556c105ce8b3f2604a7e441f1f2cc889f28ec05bc51adfb9595f029661f969",
      "arithmetic_steps": 0
    },
    "uses_answer_as_premise": false,
    "human_review": "NOT_PERFORMED",
    "formalization": "not_formalized",
    "Lean": "NOT_RUN",
    "Canonical_truth_promotion": false,
    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
  }
}