{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261014-advance-20261031-r03-convergence-in-distribution-0002",
  "pilot_number": "R0030-15702",
  "number": 15702,
  "family_id": "qf-mod31-convdist-bernoulli-cdf",
  "title": {
    "zh-Hant": "第 15702 題：依分布收斂",
    "en": "Question 15702: Compare Bernoulli CDF Values at an Explicit Threshold"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261014-advance-20261031-r03-convergence-in-distribution-0002",
    "batch_id": "batch-20261014-advance-20261031-r03-convergence-in-distribution-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-convergence-in-distribution"
    ],
    "title_zh": "依分布收斂",
    "stem_zh": "Bernoulli symbolic tail與極限CDF。操作=convdist-bernoulli-cdf；資料={\"p\":[1,2],\"c\":[2,20],\"shift\":3,\"n\":2,\"threshold\":[1,2]}。請依 [n,p_n,threshold,F_n,F,limit_CDF_continuity_point] 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "(2, 13/25, 1/2, 12/25, 1/2, 是)",
      "explanation_zh": "Wave4收束核對=(2, 13/25, 1/2, 12/25, 1/2, 是)；依完整明示資料重算；只在極限CDF連續點要求收斂且有限prefix不足；不要求coupling且不推出依機率、幾乎處處或Lp收斂。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 1,
      "skills": [
        "依分布收斂"
      ]
    },
    "variant": {
      "family_id": "qf-mod31-convdist-bernoulli-cdf",
      "seed": 202610313002,
      "parameters": {
        "p": [
          1,
          2
        ],
        "c": [
          2,
          20
        ],
        "shift": 3,
        "n": 2,
        "threshold": [
          1,
          2
        ],
        "domain": "Symbolic_Bernoulli_CDF_limit",
        "operation": "convdist-bernoulli-cdf",
        "semantic_template": "qf-mod31-convdist-bernoulli-cdf/{\"c\":[2,20],\"n\":2,\"p\":[1,2],\"shift\":3,\"threshold\":[1,2]}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_finite_model",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "候選須由主端獨立重算；不構成人工審定、一般不可數性／色數最優性／更強機率收斂／商域、optional stopping、Lean或形式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-20261031-r03-convergence-in-distribution-0002",
    "batch_id": "batch-20261014-advance-20261031-r03-convergence-in-distribution-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "0dfa80d64ed3e8ccda829ec0d8d8d41c4ed77d2941b446fd022f44173d3f6152",
      "family_id": "qf-mod31-convdist-bernoulli-cdf",
      "seed": 202610313002,
      "semantic_template": "qf-mod31-convdist-bernoulli-cdf/{\"c\":[2,20],\"n\":2,\"p\":[1,2],\"shift\":3,\"threshold\":[1,2]}"
    },
    "fields": {
      "title": "Compare Bernoulli CDF Values at an Explicit Threshold",
      "stem": "Use the explicit Bernoulli symbolic family with interior limit p=1/2, 0<p<1, signed c=2/20, positive shift=3, and p_n=p+c/(n+shift) for every n≥1. The whole parameter tail must stay in [0,1]. The Bernoulli CDF is 0 below 0, 1−p on [0,1), and 1 at or above 1. With this interior limit its only jumps are at 0 and 1; the definition requires CDF convergence only at its continuity points. A chosen coupling is not required, and convergence in probability, almost surely or in Lp is not inferred. Keep exact reduced rationals, omitting /1 for integers. n=2, threshold=1/2. Give (n,p_n,threshold,F_n,F,limit_CDF_continuity_point?).",
      "answer": {
        "kind": "expression",
        "value": "(2, 13/25, 1/2, 12/25, 1/2, Yes)",
        "explanation": "Compute p_n exactly, then evaluate both complete CDFs with the same threshold conventions. The final flag describes continuity of the limit CDF, not whether these two finite-n values happen to be equal. The exact result, retaining the requested order, is (2, 13/25, 1/2, 12/25, 1/2, Yes)."
      },
      "skills": [
        "Compare Bernoulli CDF Values at an Explicit Threshold"
      ],
      "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": "Bernoulli symbolic tail與極限CDF。操作=convdist-bernoulli-cdf；資料={\"p\":[1,2],\"c\":[2,20],\"shift\":3,\"n\":2,\"threshold\":[1,2]}。請依 [n,p_n,threshold,F_n,F,limit_CDF_continuity_point] 欄序回答。",
        "en": "Use the explicit Bernoulli symbolic family with interior limit p=1/2, 0<p<1, signed c=2/20, positive shift=3, and p_n=p+c/(n+shift) for every n≥1. The whole parameter tail must stay in [0,1]. The Bernoulli CDF is 0 below 0, 1−p on [0,1), and 1 at or above 1. With this interior limit its only jumps are at 0 and 1; the definition requires CDF convergence only at its continuity points. A chosen coupling is not required, and convergence in probability, almost surely or in Lp is not inferred. Keep exact reduced rationals, omitting /1 for integers. n=2, threshold=1/2. Give (n,p_n,threshold,F_n,F,limit_CDF_continuity_point?)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "interior Bernoulli limit mass at0 := sub(1, 1/2)",
      "after": "interior Bernoulli limit mass at0 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色及已核中間量精確計算，保留索引、符號、分母與運算所在載體。",
        "en": "Compute exactly from original roles and established intermediates,retaining indices,signs,denominators and the operation's carrier."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 interior Bernoulli limit mass at0，不以小數近似替代。",
          "en": "Recompute interior Bernoulli limit mass at0 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sub",
        "inputs": [
          "1",
          "1/2"
        ],
        "output": "1/2",
        "label": "interior Bernoulli limit mass at0"
      }
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "original limit parameter=[1, 2];reduced p=1/2;mass at0=1/2;mass at1=1/2",
      "after": "both endpoint masses positive;only limit-CDF jumps at0 and1;registered interior projection",
      "relation": "implies",
      "rule_id": "Bernoulli_interior_English_guard_not_degenerate_jump_set",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先遵守原英文0<p<1限制；因此0與1的跳躍都真有正質量。原始oracle雖允許端點p，也不把退化分佈套進這個連續點協定；它們須另行分析。",
        "en": "Honor the source English restriction0<p<1,so both jumps genuinely have positive mass.Although the raw oracle allows endpoint parameters,do not extend this continuity-point protocol to degenerate distributions;their jump sets require separate analysis."
      },
      "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": "Bernoulli first legal tail denominator := sum(1, 3)",
      "after": "Bernoulli first legal tail denominator = 4",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色及已核中間量精確計算，保留索引、符號、分母與運算所在載體。",
        "en": "Compute exactly from original roles and established intermediates,retaining indices,signs,denominators and the operation's carrier."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Bernoulli first legal tail denominator，不以小數近似替代。",
          "en": "Recompute Bernoulli first legal tail denominator with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1",
          "3"
        ],
        "output": "4",
        "label": "Bernoulli first legal tail denominator"
      }
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "Bernoulli first tail signed correction := div(1/10, 4)",
      "after": "Bernoulli first tail signed correction = 1/40",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色及已核中間量精確計算，保留索引、符號、分母與運算所在載體。",
        "en": "Compute exactly from original roles and established intermediates,retaining indices,signs,denominators and the operation's carrier."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Bernoulli first tail signed correction，不以小數近似替代。",
          "en": "Recompute Bernoulli first tail signed correction with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "1/10",
          "4"
        ],
        "output": "1/40",
        "label": "Bernoulli first tail signed correction"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "Bernoulli first legal parameter := sum(1/2, 1/40)",
      "after": "Bernoulli first legal parameter = 21/40",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色及已核中間量精確計算，保留索引、符號、分母與運算所在載體。",
        "en": "Compute exactly from original roles and established intermediates,retaining indices,signs,denominators and the operation's carrier."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Bernoulli first legal parameter，不以小數近似替代。",
          "en": "Recompute Bernoulli first legal parameter with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/2",
          "1/40"
        ],
        "output": "21/40",
        "label": "Bernoulli first legal parameter"
      }
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "declared all-n family p_n=1/2+(1/10)/(n+3);n>=1;first parameter=21/40;limit=1/2",
      "after": "for every n>=1,lambda=(1+3)/(n+3) in(0,1];p_n=(1-lambda)*1/2+lambda*(21/40) in[0,1]",
      "relation": "implies",
      "rule_id": "Bernoulli_whole_signed_parameter_tail_convex_combination",
      "conditions": [],
      "explanation": {
        "zh-Hant": "c可正、負或0。任意n的參數是首合法參數與極限的凸組合，所以首項與極限的機率域保證所有尾項有效；不是只看本題一個n便斷定全部有效。此分佈族不要求選同一coupling。",
        "en": "The signed c may be positive,negative or zero.Every parameter is a convex combination of the first legal parameter and the limit,so their probability bounds cover the whole tail,not merely one displayed n.No chosen coupling is required for this family of distributions."
      },
      "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": "Bernoulli selected tail denominator := sum(2, 3)",
      "after": "Bernoulli selected tail denominator = 5",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色及已核中間量精確計算，保留索引、符號、分母與運算所在載體。",
        "en": "Compute exactly from original roles and established intermediates,retaining indices,signs,denominators and the operation's carrier."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Bernoulli selected tail denominator，不以小數近似替代。",
          "en": "Recompute Bernoulli selected tail denominator with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "2",
          "3"
        ],
        "output": "5",
        "label": "Bernoulli selected tail denominator"
      }
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "Bernoulli selected signed correction := div(1/10, 5)",
      "after": "Bernoulli selected signed correction = 1/50",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色及已核中間量精確計算，保留索引、符號、分母與運算所在載體。",
        "en": "Compute exactly from original roles and established intermediates,retaining indices,signs,denominators and the operation's carrier."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Bernoulli selected signed correction，不以小數近似替代。",
          "en": "Recompute Bernoulli selected signed correction with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "1/10",
          "5"
        ],
        "output": "1/50",
        "label": "Bernoulli selected signed correction"
      }
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "Bernoulli selected exact parameter := sum(1/2, 1/50)",
      "after": "Bernoulli selected exact parameter = 13/25",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色及已核中間量精確計算，保留索引、符號、分母與運算所在載體。",
        "en": "Compute exactly from original roles and established intermediates,retaining indices,signs,denominators and the operation's carrier."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Bernoulli selected exact parameter，不以小數近似替代。",
          "en": "Recompute Bernoulli selected exact parameter with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/2",
          "1/50"
        ],
        "output": "13/25",
        "label": "Bernoulli selected exact parameter"
      }
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "selected distribution one minus Bernoulli parameter := sub(1, 13/25)",
      "after": "selected distribution one minus Bernoulli parameter = 12/25",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色及已核中間量精確計算，保留索引、符號、分母與運算所在載體。",
        "en": "Compute exactly from original roles and established intermediates,retaining indices,signs,denominators and the operation's carrier."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 selected distribution one minus Bernoulli parameter，不以小數近似替代。",
          "en": "Recompute selected distribution one minus Bernoulli parameter with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sub",
        "inputs": [
          "1",
          "13/25"
        ],
        "output": "12/25",
        "label": "selected distribution one minus Bernoulli parameter"
      }
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "selected distribution:Bernoulli parameter=13/25;same exact threshold=1/2;CDF branch=0<=t<1",
      "after": "selected distribution exact CDF value=12/25",
      "relation": "implies",
      "rule_id": "Bernoulli_CDF_same_threshold_including_endpoints",
      "conditions": [],
      "explanation": {
        "zh-Hant": "CDF為P(X≤t)，所以0點納入質量1−p，1點已納入全部質量。上下兩CDF用同一門檻分支，不把連續點旗標與當下兩值是否相等混同。",
        "en": "A CDF is P(X≤t):at0 include mass1−p,and at1 include all mass.Use the same threshold branch for both CDFs.The continuity flag is distinct from equality of two finite-n values."
      },
      "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": "limiting distribution one minus Bernoulli parameter := sub(1, 1/2)",
      "after": "limiting distribution one minus Bernoulli parameter = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色及已核中間量精確計算，保留索引、符號、分母與運算所在載體。",
        "en": "Compute exactly from original roles and established intermediates,retaining indices,signs,denominators and the operation's carrier."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 limiting distribution one minus Bernoulli parameter，不以小數近似替代。",
          "en": "Recompute limiting distribution one minus Bernoulli parameter with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sub",
        "inputs": [
          "1",
          "1/2"
        ],
        "output": "1/2",
        "label": "limiting distribution one minus Bernoulli parameter"
      }
    },
    {
      "id": "S013",
      "uses": [
        "G001",
        "S012"
      ],
      "before": "limiting distribution:Bernoulli parameter=1/2;same exact threshold=1/2;CDF branch=0<=t<1",
      "after": "limiting distribution exact CDF value=1/2",
      "relation": "implies",
      "rule_id": "Bernoulli_CDF_same_threshold_including_endpoints",
      "conditions": [],
      "explanation": {
        "zh-Hant": "CDF為P(X≤t)，所以0點納入質量1−p，1點已納入全部質量。上下兩CDF用同一門檻分支，不把連續點旗標與當下兩值是否相等混同。",
        "en": "A CDF is P(X≤t):at0 include mass1−p,and at1 include all mass.Use the same threshold branch for both CDFs.The continuity flag is distinct from equality of two finite-n values."
      },
      "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": "limit parameter=1/2;threshold=1/2;jump set=[0, 1];selected CDF=12/25;limit CDF=1/2",
      "after": "continuity-point flag=是;finite-n equality does not decide this flag",
      "relation": "implies",
      "rule_id": "convdist_continuity_is_limit_CDF_property",
      "conditions": [],
      "explanation": {
        "zh-Hant": "最後旗標屬極限CDF本身，非這次數值相等就叫連續。依分佈收斂也不推出依機率、幾乎處處或Lp收斂。",
        "en": "The final flag belongs to the limit CDF;equal displayed values do not establish continuity.Convergence in distribution does not imply convergence in probability,almost surely or in Lp."
      },
      "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,
      "13/25",
      "1/2",
      "12/25",
      "1/2",
      true
    ]
  },
  "conclusion": {
    "canonical_value": "(2, 13/25, 1/2, 12/25, 1/2, 是)",
    "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-20261031-r03-convergence-in-distribution-foundations",
      "batch_path": "public/data/questions/batches/batch-20261014-advance-20261031-r03-convergence-in-distribution-foundations/questions.jsonl",
      "batch_sha256": "6c12fd1b56e7d80adc9da9e235d6e688f1d7d0b58b6a833e3c17a5a31a1d76e9",
      "original_line": 2,
      "original_line_sha256": "6b4f5349e24429e4b673d4abc8cf6d56b83bb64843c82f9df22ece1240ece536",
      "question_compact_sha256": "0dfa80d64ed3e8ccda829ec0d8d8d41c4ed77d2941b446fd022f44173d3f6152",
      "english_row_compact_sha256": "648cc87ba5696d97855ca9ec98d22c127bd294e03fb697960fe66b345637654b",
      "english_manifest_path": "public/data/i18n/editions/batch-20261014-advance-20261031-r03-convergence-in-distribution-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261014-advance-20261031-r03-convergence-in-distribution-foundations.en.jsonl"
    },
    "mko_bindings": {
      "mko-convergence-in-distribution": {
        "sha256": "082a2a414602a248555b80f0fa923b8b69256bdacb1fe1fe32b38bb0924de21a",
        "english_sha256": "f58586abe9ae537f8873304988774caf2d138a9df726d7ab04aa6d482bdae7b2",
        "original_path": "public/data/mko/convergence-in-distribution.json",
        "english_path": "public/data/i18n/editions/mko-convergence-in-distribution.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "d85118d74859ec7e59736714f767732c6ec092eaa0bf5279abeb28e4a628e3c3",
    "original_candidate_sha256": "47a1c7eeb86368b4f421449036053999c80e86246a573c533dbd9b8e7cf14d4f",
    "original_candidate_line_sha256": "47a1c7eeb86368b4f421449036053999c80e86246a573c533dbd9b8e7cf14d4f"
  },
  "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-20261026-007-p31",
    "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": "73bc08066ffb1a81eacb2d4e288d03f522aeefd3949fa0d97da41b484ed737bc",
      "question_first_witness_sha256": "ac120452a208b257b8270f38351a09dcd832cdbc8d9db6b69933f83e43cc9336"
    },
    "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": "b8bef7d56a1f7188be911571802562ca8f8c8cce17dff2486608b6b4198d0a3a",
      "arithmetic_steps": 9
    },
    "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."
  }
}