{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261014-advance-20261107-r04-markov-inequality-0077",
  "pilot_number": "R0039-20227",
  "number": 20227,
  "family_id": "qf-mod38-markov-negative-variable-boundary",
  "title": {
    "zh-Hant": "第 20227 題：Markov不等式",
    "en": "Question 20227: A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261014-advance-20261107-r04-markov-inequality-0077",
    "batch_id": "batch-20261014-advance-20261107-r04-markov-inequality-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-markov-inequality"
    ],
    "title_zh": "Markov不等式",
    "stem_zh": "完整正規化有限有理law、明示隨機變數與正閾值。操作=markov-negative-variable-boundary；資料={\"masses\":[[1,2],[1,2]],\"values\":[-2,4],\"threshold\":3}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "([-2, 4], 否, 1/2, 1/3, Markov_not_applicable_to_signed_variable_without_transformation)",
      "explanation_zh": "Wave6根層核對=([-2, 4], 否, 1/2, 1/3, Markov_not_applicable_to_signed_variable_without_transformation)；依完整明示資料獨立重算；Markov要求X非負almost surely、有限期望與a大於0，結論是尾機率小於等於E[X]/a；上界不等於實際尾機率；signed variable不可直接套用且禁止零閾值。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 1,
      "skills": [
        "Markov不等式"
      ]
    },
    "variant": {
      "family_id": "qf-mod38-markov-negative-variable-boundary",
      "seed": 202611074077,
      "parameters": {
        "masses": [
          [
            1,
            2
          ],
          [
            1,
            2
          ]
        ],
        "values": [
          -2,
          4
        ],
        "threshold": 3,
        "domain": "Finite_rational_Markov_inequality",
        "operation": "markov-negative-variable-boundary",
        "semantic_template": "qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":3,\"values\":[-2,4]}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_finite_or_symbolic_model",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "候選須主端重算；不構成人工審定、有限sample群作用、sup等於max、basis等於subbasis、signed Markov、pigeonhole converse、nonunit Euler、finite-prefix subsequence、Lean或Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-14",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261014-advance-20261107-r04-markov-inequality-0077",
    "batch_id": "batch-20261014-advance-20261107-r04-markov-inequality-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "0a37e3262be69c5c3bcf7a725e3d53d17fc45b48e246ecae4c919521e2c80a8c",
      "family_id": "qf-mod38-markov-negative-variable-boundary",
      "seed": 202611074077,
      "semantic_template": "qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":3,\"values\":[-2,4]}"
    },
    "fields": {
      "title": "A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail",
      "stem": "Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-2, 4]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=3, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).",
      "answer": {
        "kind": "expression",
        "value": "([-2, 4], No, 1/2, 1/3, Markov_not_applicable_to_signed_variable_without_transformation)",
        "explanation": "Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-2, 4], No, 1/2, 1/3, Markov_not_applicable_to_signed_variable_without_transformation)."
      },
      "skills": [
        "A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"
      ],
      "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": "完整正規化有限有理law、明示隨機變數與正閾值。操作=markov-negative-variable-boundary；資料={\"masses\":[[1,2],[1,2]],\"values\":[-2,4],\"threshold\":3}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
        "en": "Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-2, 4]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=3, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "source operation=markov-negative-variable-boundary;original snapshot and mathematics retained",
      "after": "explicit ordered source output fields=[values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary]",
      "relation": "implies",
      "rule_id": "source_ordered_interface_explicit_without_javascript",
      "conditions": [],
      "explanation": {
        "zh-Hant": "原題快照保留歷史介面標記；本頁直接列來源欄位順序，不需JavaScript或外部表。欄位標籤不是額外數學證據，No保持原題的聲稱範圍。",
        "en": "Retain the historical interface marker in the original snapshot while listing ordered source fields here,without JavaScript or external lookup.Field labels are not additional mathematical evidence,and No retains the source claim scope."
      },
      "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": "Markov complete probability mass := sum(1/2, 1/2)",
      "after": "Markov complete probability mass = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色與已核中間量精確代值，保留索引、正負號、分母與證明適用條件。",
        "en": "Substitute original roles and established intermediates exactly,retaining indices,signs,denominators and the proof's applicability conditions."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Markov complete probability mass，不以小數近似替代。",
          "en": "Recompute Markov complete probability mass with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/2",
          "1/2"
        ],
        "output": "1",
        "label": "Markov complete probability mass"
      }
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "Markov original atom0 weighted X := mul(1/2, -2)",
      "after": "Markov original atom0 weighted X = -1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色與已核中間量精確代值，保留索引、正負號、分母與證明適用條件。",
        "en": "Substitute original roles and established intermediates exactly,retaining indices,signs,denominators and the proof's applicability conditions."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Markov original atom0 weighted X，不以小數近似替代。",
          "en": "Recompute Markov original atom0 weighted X with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "-2"
        ],
        "output": "-1",
        "label": "Markov original atom0 weighted X"
      }
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "Markov original atom1 weighted X := mul(1/2, 4)",
      "after": "Markov original atom1 weighted X = 2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色與已核中間量精確代值，保留索引、正負號、分母與證明適用條件。",
        "en": "Substitute original roles and established intermediates exactly,retaining indices,signs,denominators and the proof's applicability conditions."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Markov original atom1 weighted X，不以小數近似替代。",
          "en": "Recompute Markov original atom1 weighted X with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "4"
        ],
        "output": "2",
        "label": "Markov original atom1 weighted X"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "Markov complete SAME-law expectation := sum(-1, 2)",
      "after": "Markov complete SAME-law expectation = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色與已核中間量精確代值，保留索引、正負號、分母與證明適用條件。",
        "en": "Substitute original roles and established intermediates exactly,retaining indices,signs,denominators and the proof's applicability conditions."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Markov complete SAME-law expectation，不以小數近似替代。",
          "en": "Recompute Markov complete SAME-law expectation with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "-1",
          "2"
        ],
        "output": "1",
        "label": "Markov complete SAME-law expectation"
      }
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "ORIGINAL complete nonempty masses=[1/2, 1/2];ALL aligned values X=[-2, 4];full mass=1;ALL weighted X terms=[-1, 2]",
      "after": "finite exact expectation=1;prototype EVERY-ENTRY nonnegative=否;actual positive-mass nonnegativity=否;null-negative atoms=[];positive-mass negative atoms=[0]",
      "relation": "implies",
      "rule_id": "markov_pointwise_source_gate_not_almost_sure_confusion",
      "conditions": [],
      "explanation": {
        "zh-Hant": "原型flag逐整張表檢非負，a.s.只限制正質量點，兩者不混同。非tail題族依英文來源guard把僅null負值的歧義留給來源審查，不把它當定理前提失敗；tail本身仍可對signed X計算。",
        "en": "The prototype flag tests every table entry,whereas almost-sure nonnegativity concerns positive-mass atoms only.The English source-review guard excludes null-only negative ambiguities from non-tail families rather than calling them a theorem-premise failure.The tail itself can still be computed for signed X."
      },
      "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": "Markov original atom0 threshold-event mass := mul(1/2, 0)",
      "after": "Markov original atom0 threshold-event mass = 0",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色與已核中間量精確代值，保留索引、正負號、分母與證明適用條件。",
        "en": "Substitute original roles and established intermediates exactly,retaining indices,signs,denominators and the proof's applicability conditions."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Markov original atom0 threshold-event mass，不以小數近似替代。",
          "en": "Recompute Markov original atom0 threshold-event mass with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "0"
        ],
        "output": "0",
        "label": "Markov original atom0 threshold-event mass"
      }
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "ORIGINAL atom=0;value X=-2;positive threshold=3;original mass=1/2",
      "after": "event X>=threshold=否;EQUALITY INCLUDED;event mass contribution=0",
      "relation": "implies",
      "rule_id": "markov_full_tail_event_includes_equality_not_strict_tail",
      "conditions": [],
      "explanation": {
        "zh-Hant": "原事件是大於等於threshold，不可偷換嚴格大於；逐全部原子乘事件指示量，包括零質量結果，不以有限觀察頻率代替完整law。",
        "en": "The event is X≥threshold,not X>threshold.Multiply by the event indicator at every original atom,including null atoms;use the complete law rather than observed frequencies."
      },
      "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": "Markov original atom1 threshold-event mass := mul(1/2, 1)",
      "after": "Markov original atom1 threshold-event mass = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色與已核中間量精確代值，保留索引、正負號、分母與證明適用條件。",
        "en": "Substitute original roles and established intermediates exactly,retaining indices,signs,denominators and the proof's applicability conditions."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Markov original atom1 threshold-event mass，不以小數近似替代。",
          "en": "Recompute Markov original atom1 threshold-event mass with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "1"
        ],
        "output": "1/2",
        "label": "Markov original atom1 threshold-event mass"
      }
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "ORIGINAL atom=1;value X=4;positive threshold=3;original mass=1/2",
      "after": "event X>=threshold=是;EQUALITY INCLUDED;event mass contribution=1/2",
      "relation": "implies",
      "rule_id": "markov_full_tail_event_includes_equality_not_strict_tail",
      "conditions": [],
      "explanation": {
        "zh-Hant": "原事件是大於等於threshold，不可偷換嚴格大於；逐全部原子乘事件指示量，包括零質量結果，不以有限觀察頻率代替完整law。",
        "en": "The event is X≥threshold,not X>threshold.Multiply by the event indicator at every original atom,including null atoms;use the complete law rather than observed frequencies."
      },
      "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": "Markov complete exact tail probability := sum(0, 1/2)",
      "after": "Markov complete exact tail probability = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色與已核中間量精確代值，保留索引、正負號、分母與證明適用條件。",
        "en": "Substitute original roles and established intermediates exactly,retaining indices,signs,denominators and the proof's applicability conditions."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Markov complete exact tail probability，不以小數近似替代。",
          "en": "Recompute Markov complete exact tail probability with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "0",
          "1/2"
        ],
        "output": "1/2",
        "label": "Markov complete exact tail probability"
      }
    },
    {
      "id": "S012",
      "uses": [
        "G001",
        "S011"
      ],
      "before": "ALL original threshold-event flags=[否, 是];ALL law-weighted indicator terms=[0, 1/2]",
      "after": "exact complete-law tail probability=1/2;this event calculation ALONE does not assert Markov applicability",
      "relation": "implies",
      "rule_id": "markov_exact_event_probability_separate_from_theorem_bound",
      "conditions": [],
      "explanation": {
        "zh-Hant": "完整事件質量和是真尾機率，對signed X也可算。這個數值不自行證X非負a.s.，也不等於之後的E[X]/threshold上界。",
        "en": "The complete event-mass sum is the actual tail probability,even for signed X.That calculation neither proves almost-sure nonnegativity nor identifies the tail with the later expectation-over-threshold bound."
      },
      "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": "Markov RAW expectation over positive threshold := div(1, 3)",
      "after": "Markov RAW expectation over positive threshold = 1/3",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色與已核中間量精確代值，保留索引、正負號、分母與證明適用條件。",
        "en": "Substitute original roles and established intermediates exactly,retaining indices,signs,denominators and the proof's applicability conditions."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Markov RAW expectation over positive threshold，不以小數近似替代。",
          "en": "Recompute Markov RAW expectation over positive threshold with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "1",
          "3"
        ],
        "output": "1/3",
        "label": "Markov RAW expectation over positive threshold"
      }
    },
    {
      "id": "S014",
      "uses": [
        "G001",
        "S013"
      ],
      "before": "ACTUAL negative X at positive mass exists;original values=[-2, 4];tail=1/2;finite expectation=1;positive threshold=3",
      "after": "SOURCE pointwise applicability=否;raw E[X]/threshold=1/3;actual numeric tail<=raw quotient happens=否;NO general Markov guarantee or reverse-inequality claim",
      "relation": "implies",
      "rule_id": "markov_signed_inapplicability_not_numeric_failure_or_reverse",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正質量負值真的違反a.s.非負前提，故不能直接套Markov。這次數值不等式仍可能碰巧成立或失敗；No不給反向不等式，也不把未合格E[X]/threshold稱已證的機率上界。",
        "en": "A negative value at positive mass really violates the a.s.nonnegative premise,so Markov cannot be applied directly.This particular numeric inequality may coincidentally hold or fail.The No supplies no reverse inequality,and the unqualified raw quotient is not a certified probability bound."
      },
      "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": [
      [
        "-2",
        "4"
      ],
      false,
      "1/2",
      "1/3",
      "Markov_not_applicable_to_signed_variable_without_transformation"
    ]
  },
  "conclusion": {
    "canonical_value": "([-2, 4], 否, 1/2, 1/3, Markov_not_applicable_to_signed_variable_without_transformation)",
    "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-20261014-advance-20261107-r04-markov-inequality-foundations",
      "batch_path": "public/data/questions/batches/batch-20261014-advance-20261107-r04-markov-inequality-foundations/questions.jsonl",
      "batch_sha256": "b940549231c49394a8780469eef46b759de327644e902cfe22e155544da63a79",
      "original_line": 77,
      "original_line_sha256": "c5cb401137cbba5d902f9c261c75715f83b732ecf26bb4f89abd63254708c600",
      "question_compact_sha256": "0a37e3262be69c5c3bcf7a725e3d53d17fc45b48e246ecae4c919521e2c80a8c",
      "english_row_compact_sha256": "a83c47b427b4d676be353f37739f78e3ecf6f2cd8300c1a027e11c1fa4e0135a",
      "english_manifest_path": "public/data/i18n/editions/batch-20261014-advance-20261107-r04-markov-inequality-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261014-advance-20261107-r04-markov-inequality-foundations.en.jsonl"
    },
    "mko_bindings": {
      "mko-markov-inequality": {
        "sha256": "1363d255fe303067c7fcc800bd82bcecddd8331a0e2262159df06012d4c8bdbd",
        "english_sha256": "0c9c249b7d6fc18648349ae126e9311ddbd1af2b9589196a552d85bb6f181567",
        "original_path": "public/data/mko/markov-inequality.json",
        "english_path": "public/data/i18n/editions/mko-markov-inequality.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "521929dfe9b0919224c38312ad78e5fa56457fc886faaa12b137edd83962251e",
    "original_candidate_sha256": "2adb1e57ce087aeed0a404046ded77f6f9ae2497e119899f802e3d648c85de11",
    "original_candidate_line_sha256": "2adb1e57ce087aeed0a404046ded77f6f9ae2497e119899f802e3d648c85de11"
  },
  "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-20261026-008-p40",
    "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": "6e11e35fafe32f16900ce9b522d01acb7b2ee6b463ab179392076cfbe1af8034",
      "question_first_witness_sha256": "cc0fa361fc25151729e70264340d2eae94f179f6f2a2edfe80ba5704fbbd0682"
    },
    "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": "9179187a76f932c7b23bb0a55543fcea828d2af11ecd89e537bbaecdbd847c43",
      "arithmetic_steps": 8
    },
    "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."
  }
}