{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-0064",
  "pilot_number": "R0053-27239",
  "number": 27239,
  "family_id": "qf-mod49-clt-binomial-lattice",
  "title": {
    "zh-Hant": "第 27239 題：有限變異數IID中央極限定理",
    "en": "Question 27239: An Exact Binomial Lattice Is Not a Normal Approximation"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-0064",
    "batch_id": "batch-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-central-limit-theorem-finite-variance-iid"
    ],
    "title_zh": "有限變異數IID中央極限定理",
    "stem_zh": "明示無限IID、共同mu、0<sigma^2<infinity與sigma*sqrt(n)標準化責任。操作=clt-binomial-lattice；資料={\"success_probability\":[1,2],\"n\":5,\"threshold\":1}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "(5, 1/2, [[0, 1/32], [1, 5/32], [2, 5/16], [3, 5/16], [4, 5/32], [5, 1/32]], 1, 1, 3/16, 是, z_threshold=(1-n*p)/sqrt(n*p*(1-p)), exact_lattice_not_normal_approximation_or_convergence_evidence)",
      "explanation_zh": "Wave6第十二層核對=(5, 1/2, [[0, 1/32], [1, 5/32], [2, 5/16], [3, 5/16], [4, 5/32], [5, 1/32]], 1, 1, 3/16, 是, z_threshold=(1-n*p)/sqrt(n*p*(1-p)), exact_lattice_not_normal_approximation_or_convergence_evidence)；依完整明示資料獨立重算；finite moments與binomial lattice完整重算但不作normal approximation或limit evidence；non-IID為不適用，zero variance、wrong scaling與finite histogram不得冒充CLT。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 2,
      "skills": [
        "有限變異數IID中央極限定理"
      ]
    },
    "variant": {
      "family_id": "qf-mod49-clt-binomial-lattice",
      "seed": 202611181064,
      "parameters": {
        "success_probability": [
          1,
          2
        ],
        "n": 5,
        "threshold": 1,
        "domain": "Exact_finite_IID_limit_theorem_boundary",
        "operation": "clt-binomial-lattice",
        "semantic_template": "clt-binomial-lattice|{n:5,success_probability:[1,2],threshold:1}"
      }
    },
    "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；不構成一般limit theorem、coverage、number-theory或language theorem證明、人工審定、Lean或Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-15",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-0064",
    "batch_id": "batch-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "24b6f02d06ccd6e04690fd5e679d461d8c70ec5543eb166925c0733361715416",
      "family_id": "qf-mod49-clt-binomial-lattice",
      "seed": 202611181064,
      "semantic_template": "clt-binomial-lattice|{n:5,success_probability:[1,2],threshold:1}"
    },
    "fields": {
      "title": "An Exact Binomial Lattice Is Not a Normal Approximation",
      "stem": "Use Bernoulli success probability p=1/2, sample size n=5, integer threshold=1, and independent coordinates for this finite sum. Give (n, success_probability, binomial_distribution, total_probability, threshold, cdf_at_threshold, normalized, standardized_threshold_description, finite_scope).",
      "answer": {
        "kind": "expression",
        "value": "(5, 1/2, [[0, 1/32], [1, 5/32], [2, 5/16], [3, 5/16], [4, 5/32], [5, 1/32]], 1, 1, 3/16, Yes, z_threshold=(1-n*p)/sqrt(n*p*(1-p)), exact_lattice_not_normal_approximation_or_convergence_evidence)",
        "explanation": "Compute the exact probabilities for every integer count0 through n and the inclusive CDF at the threshold. Retain the symbolic standardized threshold; neither that expression nor a finite lattice table establishes an asymptotic limit or an approximation error. The exact result, retaining the requested order, is (5, 1/2, [[0, 1/32], [1, 5/32], [2, 5/16], [3, 5/16], [4, 5/32], [5, 1/32]], 1, 1, 3/16, Yes, z_threshold=(1-n*p)/sqrt(n*p*(1-p)), exact_lattice_not_normal_approximation_or_convergence_evidence)."
      },
      "skills": [
        "An Exact Binomial Lattice Is Not a Normal Approximation"
      ],
      "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": "明示無限IID、共同mu、0<sigma^2<infinity與sigma*sqrt(n)標準化責任。操作=clt-binomial-lattice；資料={\"success_probability\":[1,2],\"n\":5,\"threshold\":1}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
        "en": "Use Bernoulli success probability p=1/2, sample size n=5, integer threshold=1, and independent coordinates for this finite sum. Give (n, success_probability, binomial_distribution, total_probability, threshold, cdf_at_threshold, normalized, standardized_threshold_description, finite_scope)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "source roles = [clt-binomial-lattice]",
      "after": "derived = [[n, success_probability, binomial_distribution, total_probability, threshold, cdf_at_threshold, normalized, standardized_threshold_description, finite_scope]]",
      "relation": "implies",
      "rule_id": "source_ordered_fields_without_JS",
      "conditions": [],
      "explanation": {
        "zh-Hant": "此頁直接列完整來源欄序與mode特定角色，不靠JavaScript或由答案外形猜介面；No、undefined、infinity各保原意。",
        "en": "List the complete source field order and mode-specific roles without JavaScript or answer-shape inference;No,undefined and infinity retain their distinct meanings."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。",
          "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "CLT binomial ORIGINAL one minus probability := sub(1, 1/2)",
      "after": "CLT binomial ORIGINAL one minus probability = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL one minus probability，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL one minus probability with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sub",
        "inputs": [
          "1",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT binomial ORIGINAL one minus probability"
      }
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "source roles = [CLT binomial k0 success, 1/2, 0]",
      "after": "derived = [1]",
      "relation": "implies",
      "rule_id": "CLT_geometric_probability_power_zero_empty_product",
      "conditions": [],
      "explanation": {
        "zh-Hant": "零次冪是空乘積1，不因n0或base0改成0。",
        "en": "A zeroth power is the empty product1,not0 because n or the base is0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。",
          "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "CLT binomial k0 failure ORIGINAL power multiply1 := mul(1, 1/2)",
      "after": "CLT binomial k0 failure ORIGINAL power multiply1 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k0 failure ORIGINAL power multiply1，不以小數近似替代。",
          "en": "Recompute CLT binomial k0 failure ORIGINAL power multiply1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT binomial k0 failure ORIGINAL power multiply1"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "CLT binomial k0 failure ORIGINAL power multiply2 := mul(1/2, 1/2)",
      "after": "CLT binomial k0 failure ORIGINAL power multiply2 = 1/4",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k0 failure ORIGINAL power multiply2，不以小數近似替代。",
          "en": "Recompute CLT binomial k0 failure ORIGINAL power multiply2 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "1/2"
        ],
        "output": "1/4",
        "label": "CLT binomial k0 failure ORIGINAL power multiply2"
      }
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "CLT binomial k0 failure ORIGINAL power multiply3 := mul(1/4, 1/2)",
      "after": "CLT binomial k0 failure ORIGINAL power multiply3 = 1/8",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k0 failure ORIGINAL power multiply3，不以小數近似替代。",
          "en": "Recompute CLT binomial k0 failure ORIGINAL power multiply3 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/4",
          "1/2"
        ],
        "output": "1/8",
        "label": "CLT binomial k0 failure ORIGINAL power multiply3"
      }
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "CLT binomial k0 failure ORIGINAL power multiply4 := mul(1/8, 1/2)",
      "after": "CLT binomial k0 failure ORIGINAL power multiply4 = 1/16",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k0 failure ORIGINAL power multiply4，不以小數近似替代。",
          "en": "Recompute CLT binomial k0 failure ORIGINAL power multiply4 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/8",
          "1/2"
        ],
        "output": "1/16",
        "label": "CLT binomial k0 failure ORIGINAL power multiply4"
      }
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "CLT binomial k0 failure ORIGINAL power multiply5 := mul(1/16, 1/2)",
      "after": "CLT binomial k0 failure ORIGINAL power multiply5 = 1/32",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k0 failure ORIGINAL power multiply5，不以小數近似替代。",
          "en": "Recompute CLT binomial k0 failure ORIGINAL power multiply5 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/16",
          "1/2"
        ],
        "output": "1/32",
        "label": "CLT binomial k0 failure ORIGINAL power multiply5"
      }
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "CLT binomial ORIGINAL power product k0 := mul(1, 1/32)",
      "after": "CLT binomial ORIGINAL power product k0 = 1/32",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL power product k0，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL power product k0 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1/32"
        ],
        "output": "1/32",
        "label": "CLT binomial ORIGINAL power product k0"
      }
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "CLT binomial ORIGINAL choose times joint k0 := mul(1, 1/32)",
      "after": "CLT binomial ORIGINAL choose times joint k0 = 1/32",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL choose times joint k0，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL choose times joint k0 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1/32"
        ],
        "output": "1/32",
        "label": "CLT binomial ORIGINAL choose times joint k0"
      }
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "source roles = [5, 0, 1/2, 1/2]",
      "after": "derived = [1, 1, 1/32, 1/32]",
      "relation": "implies",
      "rule_id": "CLT_EACH_binomial_lattice_exact_count_and_probability",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每一組k個success的位置有相同p^k(1-p)^(n-k)機率，選位數C(n,k)；由完整finite independence給這個lattice質量，不是normal近似。相同純項配方後續總和可引用已算量，不改任何source profile。",
        "en": "Every choice of k success positions has probability p^k(1-p)^(n-k),with C(n,k) choices.This complete finite-independence calculation gives an exact lattice mass,not a normal approximation.Later sums reuse the SAME established pure term recipe without changing any source profile."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。",
          "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S012",
      "uses": [
        "G001",
        "S011"
      ],
      "before": "CLT binomial k1 success ORIGINAL power multiply1 := mul(1, 1/2)",
      "after": "CLT binomial k1 success ORIGINAL power multiply1 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k1 success ORIGINAL power multiply1，不以小數近似替代。",
          "en": "Recompute CLT binomial k1 success ORIGINAL power multiply1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT binomial k1 success ORIGINAL power multiply1"
      }
    },
    {
      "id": "S013",
      "uses": [
        "G001",
        "S012"
      ],
      "before": "CLT binomial k1 failure ORIGINAL power multiply1 := mul(1, 1/2)",
      "after": "CLT binomial k1 failure ORIGINAL power multiply1 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k1 failure ORIGINAL power multiply1，不以小數近似替代。",
          "en": "Recompute CLT binomial k1 failure ORIGINAL power multiply1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT binomial k1 failure ORIGINAL power multiply1"
      }
    },
    {
      "id": "S014",
      "uses": [
        "G001",
        "S013"
      ],
      "before": "CLT binomial k1 failure ORIGINAL power multiply2 := mul(1/2, 1/2)",
      "after": "CLT binomial k1 failure ORIGINAL power multiply2 = 1/4",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k1 failure ORIGINAL power multiply2，不以小數近似替代。",
          "en": "Recompute CLT binomial k1 failure ORIGINAL power multiply2 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "1/2"
        ],
        "output": "1/4",
        "label": "CLT binomial k1 failure ORIGINAL power multiply2"
      }
    },
    {
      "id": "S015",
      "uses": [
        "G001",
        "S014"
      ],
      "before": "CLT binomial k1 failure ORIGINAL power multiply3 := mul(1/4, 1/2)",
      "after": "CLT binomial k1 failure ORIGINAL power multiply3 = 1/8",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k1 failure ORIGINAL power multiply3，不以小數近似替代。",
          "en": "Recompute CLT binomial k1 failure ORIGINAL power multiply3 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/4",
          "1/2"
        ],
        "output": "1/8",
        "label": "CLT binomial k1 failure ORIGINAL power multiply3"
      }
    },
    {
      "id": "S016",
      "uses": [
        "G001",
        "S015"
      ],
      "before": "CLT binomial k1 failure ORIGINAL power multiply4 := mul(1/8, 1/2)",
      "after": "CLT binomial k1 failure ORIGINAL power multiply4 = 1/16",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k1 failure ORIGINAL power multiply4，不以小數近似替代。",
          "en": "Recompute CLT binomial k1 failure ORIGINAL power multiply4 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/8",
          "1/2"
        ],
        "output": "1/16",
        "label": "CLT binomial k1 failure ORIGINAL power multiply4"
      }
    },
    {
      "id": "S017",
      "uses": [
        "G001",
        "S016"
      ],
      "before": "CLT binomial ORIGINAL power product k1 := mul(1/2, 1/16)",
      "after": "CLT binomial ORIGINAL power product k1 = 1/32",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL power product k1，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL power product k1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "1/16"
        ],
        "output": "1/32",
        "label": "CLT binomial ORIGINAL power product k1"
      }
    },
    {
      "id": "S018",
      "uses": [
        "G001",
        "S017"
      ],
      "before": "CLT binomial ORIGINAL choose times joint k1 := mul(5, 1/32)",
      "after": "CLT binomial ORIGINAL choose times joint k1 = 5/32",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL choose times joint k1，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL choose times joint k1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "5",
          "1/32"
        ],
        "output": "5/32",
        "label": "CLT binomial ORIGINAL choose times joint k1"
      }
    },
    {
      "id": "S019",
      "uses": [
        "G001",
        "S018"
      ],
      "before": "source roles = [5, 1, 1/2, 1/2]",
      "after": "derived = [5, 1/2, 1/16, 5/32]",
      "relation": "implies",
      "rule_id": "CLT_EACH_binomial_lattice_exact_count_and_probability",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每一組k個success的位置有相同p^k(1-p)^(n-k)機率，選位數C(n,k)；由完整finite independence給這個lattice質量，不是normal近似。相同純項配方後續總和可引用已算量，不改任何source profile。",
        "en": "Every choice of k success positions has probability p^k(1-p)^(n-k),with C(n,k) choices.This complete finite-independence calculation gives an exact lattice mass,not a normal approximation.Later sums reuse the SAME established pure term recipe without changing any source profile."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。",
          "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S020",
      "uses": [
        "G001",
        "S019"
      ],
      "before": "CLT binomial k2 success ORIGINAL power multiply1 := mul(1, 1/2)",
      "after": "CLT binomial k2 success ORIGINAL power multiply1 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k2 success ORIGINAL power multiply1，不以小數近似替代。",
          "en": "Recompute CLT binomial k2 success ORIGINAL power multiply1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT binomial k2 success ORIGINAL power multiply1"
      }
    },
    {
      "id": "S021",
      "uses": [
        "G001",
        "S020"
      ],
      "before": "CLT binomial k2 success ORIGINAL power multiply2 := mul(1/2, 1/2)",
      "after": "CLT binomial k2 success ORIGINAL power multiply2 = 1/4",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k2 success ORIGINAL power multiply2，不以小數近似替代。",
          "en": "Recompute CLT binomial k2 success ORIGINAL power multiply2 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "1/2"
        ],
        "output": "1/4",
        "label": "CLT binomial k2 success ORIGINAL power multiply2"
      }
    },
    {
      "id": "S022",
      "uses": [
        "G001",
        "S021"
      ],
      "before": "CLT binomial k2 failure ORIGINAL power multiply1 := mul(1, 1/2)",
      "after": "CLT binomial k2 failure ORIGINAL power multiply1 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k2 failure ORIGINAL power multiply1，不以小數近似替代。",
          "en": "Recompute CLT binomial k2 failure ORIGINAL power multiply1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT binomial k2 failure ORIGINAL power multiply1"
      }
    },
    {
      "id": "S023",
      "uses": [
        "G001",
        "S022"
      ],
      "before": "CLT binomial k2 failure ORIGINAL power multiply2 := mul(1/2, 1/2)",
      "after": "CLT binomial k2 failure ORIGINAL power multiply2 = 1/4",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k2 failure ORIGINAL power multiply2，不以小數近似替代。",
          "en": "Recompute CLT binomial k2 failure ORIGINAL power multiply2 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "1/2"
        ],
        "output": "1/4",
        "label": "CLT binomial k2 failure ORIGINAL power multiply2"
      }
    },
    {
      "id": "S024",
      "uses": [
        "G001",
        "S023"
      ],
      "before": "CLT binomial k2 failure ORIGINAL power multiply3 := mul(1/4, 1/2)",
      "after": "CLT binomial k2 failure ORIGINAL power multiply3 = 1/8",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k2 failure ORIGINAL power multiply3，不以小數近似替代。",
          "en": "Recompute CLT binomial k2 failure ORIGINAL power multiply3 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/4",
          "1/2"
        ],
        "output": "1/8",
        "label": "CLT binomial k2 failure ORIGINAL power multiply3"
      }
    },
    {
      "id": "S025",
      "uses": [
        "G001",
        "S024"
      ],
      "before": "CLT binomial ORIGINAL power product k2 := mul(1/4, 1/8)",
      "after": "CLT binomial ORIGINAL power product k2 = 1/32",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL power product k2，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL power product k2 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/4",
          "1/8"
        ],
        "output": "1/32",
        "label": "CLT binomial ORIGINAL power product k2"
      }
    },
    {
      "id": "S026",
      "uses": [
        "G001",
        "S025"
      ],
      "before": "CLT binomial ORIGINAL choose times joint k2 := mul(10, 1/32)",
      "after": "CLT binomial ORIGINAL choose times joint k2 = 5/16",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL choose times joint k2，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL choose times joint k2 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "10",
          "1/32"
        ],
        "output": "5/16",
        "label": "CLT binomial ORIGINAL choose times joint k2"
      }
    },
    {
      "id": "S027",
      "uses": [
        "G001",
        "S026"
      ],
      "before": "source roles = [5, 2, 1/2, 1/2]",
      "after": "derived = [10, 1/4, 1/8, 5/16]",
      "relation": "implies",
      "rule_id": "CLT_EACH_binomial_lattice_exact_count_and_probability",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每一組k個success的位置有相同p^k(1-p)^(n-k)機率，選位數C(n,k)；由完整finite independence給這個lattice質量，不是normal近似。相同純項配方後續總和可引用已算量，不改任何source profile。",
        "en": "Every choice of k success positions has probability p^k(1-p)^(n-k),with C(n,k) choices.This complete finite-independence calculation gives an exact lattice mass,not a normal approximation.Later sums reuse the SAME established pure term recipe without changing any source profile."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。",
          "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S028",
      "uses": [
        "G001",
        "S027"
      ],
      "before": "CLT binomial k3 success ORIGINAL power multiply1 := mul(1, 1/2)",
      "after": "CLT binomial k3 success ORIGINAL power multiply1 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k3 success ORIGINAL power multiply1，不以小數近似替代。",
          "en": "Recompute CLT binomial k3 success ORIGINAL power multiply1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT binomial k3 success ORIGINAL power multiply1"
      }
    },
    {
      "id": "S029",
      "uses": [
        "G001",
        "S028"
      ],
      "before": "CLT binomial k3 success ORIGINAL power multiply2 := mul(1/2, 1/2)",
      "after": "CLT binomial k3 success ORIGINAL power multiply2 = 1/4",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k3 success ORIGINAL power multiply2，不以小數近似替代。",
          "en": "Recompute CLT binomial k3 success ORIGINAL power multiply2 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "1/2"
        ],
        "output": "1/4",
        "label": "CLT binomial k3 success ORIGINAL power multiply2"
      }
    },
    {
      "id": "S030",
      "uses": [
        "G001",
        "S029"
      ],
      "before": "CLT binomial k3 success ORIGINAL power multiply3 := mul(1/4, 1/2)",
      "after": "CLT binomial k3 success ORIGINAL power multiply3 = 1/8",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k3 success ORIGINAL power multiply3，不以小數近似替代。",
          "en": "Recompute CLT binomial k3 success ORIGINAL power multiply3 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/4",
          "1/2"
        ],
        "output": "1/8",
        "label": "CLT binomial k3 success ORIGINAL power multiply3"
      }
    },
    {
      "id": "S031",
      "uses": [
        "G001",
        "S030"
      ],
      "before": "CLT binomial k3 failure ORIGINAL power multiply1 := mul(1, 1/2)",
      "after": "CLT binomial k3 failure ORIGINAL power multiply1 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k3 failure ORIGINAL power multiply1，不以小數近似替代。",
          "en": "Recompute CLT binomial k3 failure ORIGINAL power multiply1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT binomial k3 failure ORIGINAL power multiply1"
      }
    },
    {
      "id": "S032",
      "uses": [
        "G001",
        "S031"
      ],
      "before": "CLT binomial k3 failure ORIGINAL power multiply2 := mul(1/2, 1/2)",
      "after": "CLT binomial k3 failure ORIGINAL power multiply2 = 1/4",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k3 failure ORIGINAL power multiply2，不以小數近似替代。",
          "en": "Recompute CLT binomial k3 failure ORIGINAL power multiply2 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "1/2"
        ],
        "output": "1/4",
        "label": "CLT binomial k3 failure ORIGINAL power multiply2"
      }
    },
    {
      "id": "S033",
      "uses": [
        "G001",
        "S032"
      ],
      "before": "CLT binomial ORIGINAL power product k3 := mul(1/8, 1/4)",
      "after": "CLT binomial ORIGINAL power product k3 = 1/32",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL power product k3，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL power product k3 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/8",
          "1/4"
        ],
        "output": "1/32",
        "label": "CLT binomial ORIGINAL power product k3"
      }
    },
    {
      "id": "S034",
      "uses": [
        "G001",
        "S033"
      ],
      "before": "CLT binomial ORIGINAL choose times joint k3 := mul(10, 1/32)",
      "after": "CLT binomial ORIGINAL choose times joint k3 = 5/16",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL choose times joint k3，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL choose times joint k3 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "10",
          "1/32"
        ],
        "output": "5/16",
        "label": "CLT binomial ORIGINAL choose times joint k3"
      }
    },
    {
      "id": "S035",
      "uses": [
        "G001",
        "S034"
      ],
      "before": "source roles = [5, 3, 1/2, 1/2]",
      "after": "derived = [10, 1/8, 1/4, 5/16]",
      "relation": "implies",
      "rule_id": "CLT_EACH_binomial_lattice_exact_count_and_probability",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每一組k個success的位置有相同p^k(1-p)^(n-k)機率，選位數C(n,k)；由完整finite independence給這個lattice質量，不是normal近似。相同純項配方後續總和可引用已算量，不改任何source profile。",
        "en": "Every choice of k success positions has probability p^k(1-p)^(n-k),with C(n,k) choices.This complete finite-independence calculation gives an exact lattice mass,not a normal approximation.Later sums reuse the SAME established pure term recipe without changing any source profile."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。",
          "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S036",
      "uses": [
        "G001",
        "S035"
      ],
      "before": "CLT binomial k4 success ORIGINAL power multiply1 := mul(1, 1/2)",
      "after": "CLT binomial k4 success ORIGINAL power multiply1 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k4 success ORIGINAL power multiply1，不以小數近似替代。",
          "en": "Recompute CLT binomial k4 success ORIGINAL power multiply1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT binomial k4 success ORIGINAL power multiply1"
      }
    },
    {
      "id": "S037",
      "uses": [
        "G001",
        "S036"
      ],
      "before": "CLT binomial k4 success ORIGINAL power multiply2 := mul(1/2, 1/2)",
      "after": "CLT binomial k4 success ORIGINAL power multiply2 = 1/4",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k4 success ORIGINAL power multiply2，不以小數近似替代。",
          "en": "Recompute CLT binomial k4 success ORIGINAL power multiply2 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "1/2"
        ],
        "output": "1/4",
        "label": "CLT binomial k4 success ORIGINAL power multiply2"
      }
    },
    {
      "id": "S038",
      "uses": [
        "G001",
        "S037"
      ],
      "before": "CLT binomial k4 success ORIGINAL power multiply3 := mul(1/4, 1/2)",
      "after": "CLT binomial k4 success ORIGINAL power multiply3 = 1/8",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k4 success ORIGINAL power multiply3，不以小數近似替代。",
          "en": "Recompute CLT binomial k4 success ORIGINAL power multiply3 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/4",
          "1/2"
        ],
        "output": "1/8",
        "label": "CLT binomial k4 success ORIGINAL power multiply3"
      }
    },
    {
      "id": "S039",
      "uses": [
        "G001",
        "S038"
      ],
      "before": "CLT binomial k4 success ORIGINAL power multiply4 := mul(1/8, 1/2)",
      "after": "CLT binomial k4 success ORIGINAL power multiply4 = 1/16",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k4 success ORIGINAL power multiply4，不以小數近似替代。",
          "en": "Recompute CLT binomial k4 success ORIGINAL power multiply4 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/8",
          "1/2"
        ],
        "output": "1/16",
        "label": "CLT binomial k4 success ORIGINAL power multiply4"
      }
    },
    {
      "id": "S040",
      "uses": [
        "G001",
        "S039"
      ],
      "before": "CLT binomial k4 failure ORIGINAL power multiply1 := mul(1, 1/2)",
      "after": "CLT binomial k4 failure ORIGINAL power multiply1 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k4 failure ORIGINAL power multiply1，不以小數近似替代。",
          "en": "Recompute CLT binomial k4 failure ORIGINAL power multiply1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT binomial k4 failure ORIGINAL power multiply1"
      }
    },
    {
      "id": "S041",
      "uses": [
        "G001",
        "S040"
      ],
      "before": "CLT binomial ORIGINAL power product k4 := mul(1/16, 1/2)",
      "after": "CLT binomial ORIGINAL power product k4 = 1/32",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL power product k4，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL power product k4 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/16",
          "1/2"
        ],
        "output": "1/32",
        "label": "CLT binomial ORIGINAL power product k4"
      }
    },
    {
      "id": "S042",
      "uses": [
        "G001",
        "S041"
      ],
      "before": "CLT binomial ORIGINAL choose times joint k4 := mul(5, 1/32)",
      "after": "CLT binomial ORIGINAL choose times joint k4 = 5/32",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL choose times joint k4，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL choose times joint k4 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "5",
          "1/32"
        ],
        "output": "5/32",
        "label": "CLT binomial ORIGINAL choose times joint k4"
      }
    },
    {
      "id": "S043",
      "uses": [
        "G001",
        "S042"
      ],
      "before": "source roles = [5, 4, 1/2, 1/2]",
      "after": "derived = [5, 1/16, 1/2, 5/32]",
      "relation": "implies",
      "rule_id": "CLT_EACH_binomial_lattice_exact_count_and_probability",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每一組k個success的位置有相同p^k(1-p)^(n-k)機率，選位數C(n,k)；由完整finite independence給這個lattice質量，不是normal近似。相同純項配方後續總和可引用已算量，不改任何source profile。",
        "en": "Every choice of k success positions has probability p^k(1-p)^(n-k),with C(n,k) choices.This complete finite-independence calculation gives an exact lattice mass,not a normal approximation.Later sums reuse the SAME established pure term recipe without changing any source profile."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。",
          "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S044",
      "uses": [
        "G001",
        "S043"
      ],
      "before": "CLT binomial k5 success ORIGINAL power multiply1 := mul(1, 1/2)",
      "after": "CLT binomial k5 success ORIGINAL power multiply1 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k5 success ORIGINAL power multiply1，不以小數近似替代。",
          "en": "Recompute CLT binomial k5 success ORIGINAL power multiply1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT binomial k5 success ORIGINAL power multiply1"
      }
    },
    {
      "id": "S045",
      "uses": [
        "G001",
        "S044"
      ],
      "before": "CLT binomial k5 success ORIGINAL power multiply2 := mul(1/2, 1/2)",
      "after": "CLT binomial k5 success ORIGINAL power multiply2 = 1/4",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k5 success ORIGINAL power multiply2，不以小數近似替代。",
          "en": "Recompute CLT binomial k5 success ORIGINAL power multiply2 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "1/2"
        ],
        "output": "1/4",
        "label": "CLT binomial k5 success ORIGINAL power multiply2"
      }
    },
    {
      "id": "S046",
      "uses": [
        "G001",
        "S045"
      ],
      "before": "CLT binomial k5 success ORIGINAL power multiply3 := mul(1/4, 1/2)",
      "after": "CLT binomial k5 success ORIGINAL power multiply3 = 1/8",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k5 success ORIGINAL power multiply3，不以小數近似替代。",
          "en": "Recompute CLT binomial k5 success ORIGINAL power multiply3 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/4",
          "1/2"
        ],
        "output": "1/8",
        "label": "CLT binomial k5 success ORIGINAL power multiply3"
      }
    },
    {
      "id": "S047",
      "uses": [
        "G001",
        "S046"
      ],
      "before": "CLT binomial k5 success ORIGINAL power multiply4 := mul(1/8, 1/2)",
      "after": "CLT binomial k5 success ORIGINAL power multiply4 = 1/16",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k5 success ORIGINAL power multiply4，不以小數近似替代。",
          "en": "Recompute CLT binomial k5 success ORIGINAL power multiply4 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/8",
          "1/2"
        ],
        "output": "1/16",
        "label": "CLT binomial k5 success ORIGINAL power multiply4"
      }
    },
    {
      "id": "S048",
      "uses": [
        "G001",
        "S047"
      ],
      "before": "CLT binomial k5 success ORIGINAL power multiply5 := mul(1/16, 1/2)",
      "after": "CLT binomial k5 success ORIGINAL power multiply5 = 1/32",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial k5 success ORIGINAL power multiply5，不以小數近似替代。",
          "en": "Recompute CLT binomial k5 success ORIGINAL power multiply5 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/16",
          "1/2"
        ],
        "output": "1/32",
        "label": "CLT binomial k5 success ORIGINAL power multiply5"
      }
    },
    {
      "id": "S049",
      "uses": [
        "G001",
        "S048"
      ],
      "before": "source roles = [CLT binomial k5 failure, 1/2, 0]",
      "after": "derived = [1]",
      "relation": "implies",
      "rule_id": "CLT_geometric_probability_power_zero_empty_product",
      "conditions": [],
      "explanation": {
        "zh-Hant": "零次冪是空乘積1，不因n0或base0改成0。",
        "en": "A zeroth power is the empty product1,not0 because n or the base is0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。",
          "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S050",
      "uses": [
        "G001",
        "S049"
      ],
      "before": "CLT binomial ORIGINAL power product k5 := mul(1/32, 1)",
      "after": "CLT binomial ORIGINAL power product k5 = 1/32",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL power product k5，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL power product k5 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/32",
          "1"
        ],
        "output": "1/32",
        "label": "CLT binomial ORIGINAL power product k5"
      }
    },
    {
      "id": "S051",
      "uses": [
        "G001",
        "S050"
      ],
      "before": "CLT binomial ORIGINAL choose times joint k5 := mul(1, 1/32)",
      "after": "CLT binomial ORIGINAL choose times joint k5 = 1/32",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL choose times joint k5，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL choose times joint k5 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1/32"
        ],
        "output": "1/32",
        "label": "CLT binomial ORIGINAL choose times joint k5"
      }
    },
    {
      "id": "S052",
      "uses": [
        "G001",
        "S051"
      ],
      "before": "source roles = [5, 5, 1/2, 1/2]",
      "after": "derived = [1, 1/32, 1, 1/32]",
      "relation": "implies",
      "rule_id": "CLT_EACH_binomial_lattice_exact_count_and_probability",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每一組k個success的位置有相同p^k(1-p)^(n-k)機率，選位數C(n,k)；由完整finite independence給這個lattice質量，不是normal近似。相同純項配方後續總和可引用已算量，不改任何source profile。",
        "en": "Every choice of k success positions has probability p^k(1-p)^(n-k),with C(n,k) choices.This complete finite-independence calculation gives an exact lattice mass,not a normal approximation.Later sums reuse the SAME established pure term recipe without changing any source profile."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。",
          "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S053",
      "uses": [
        "G001",
        "S052"
      ],
      "before": "CLT binomial ORIGINAL total mass sum0 := sum(0, 1/32)",
      "after": "CLT binomial ORIGINAL total mass sum0 = 1/32",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL total mass sum0，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL total mass sum0 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "0",
          "1/32"
        ],
        "output": "1/32",
        "label": "CLT binomial ORIGINAL total mass sum0"
      }
    },
    {
      "id": "S054",
      "uses": [
        "G001",
        "S053"
      ],
      "before": "CLT binomial ORIGINAL INCLUSIVE CDF sum0 := sum(0, 1/32)",
      "after": "CLT binomial ORIGINAL INCLUSIVE CDF sum0 = 1/32",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL INCLUSIVE CDF sum0，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL INCLUSIVE CDF sum0 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "0",
          "1/32"
        ],
        "output": "1/32",
        "label": "CLT binomial ORIGINAL INCLUSIVE CDF sum0"
      }
    },
    {
      "id": "S055",
      "uses": [
        "G001",
        "S054"
      ],
      "before": "CLT binomial ORIGINAL total mass sum1 := sum(1/32, 5/32)",
      "after": "CLT binomial ORIGINAL total mass sum1 = 3/16",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL total mass sum1，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL total mass sum1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/32",
          "5/32"
        ],
        "output": "3/16",
        "label": "CLT binomial ORIGINAL total mass sum1"
      }
    },
    {
      "id": "S056",
      "uses": [
        "G001",
        "S055"
      ],
      "before": "CLT binomial ORIGINAL INCLUSIVE CDF sum1 := sum(1/32, 5/32)",
      "after": "CLT binomial ORIGINAL INCLUSIVE CDF sum1 = 3/16",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL INCLUSIVE CDF sum1，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL INCLUSIVE CDF sum1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/32",
          "5/32"
        ],
        "output": "3/16",
        "label": "CLT binomial ORIGINAL INCLUSIVE CDF sum1"
      }
    },
    {
      "id": "S057",
      "uses": [
        "G001",
        "S056"
      ],
      "before": "CLT binomial ORIGINAL total mass sum2 := sum(3/16, 5/16)",
      "after": "CLT binomial ORIGINAL total mass sum2 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL total mass sum2，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL total mass sum2 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "3/16",
          "5/16"
        ],
        "output": "1/2",
        "label": "CLT binomial ORIGINAL total mass sum2"
      }
    },
    {
      "id": "S058",
      "uses": [
        "G001",
        "S057"
      ],
      "before": "CLT binomial ORIGINAL total mass sum3 := sum(1/2, 5/16)",
      "after": "CLT binomial ORIGINAL total mass sum3 = 13/16",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL total mass sum3，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL total mass sum3 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/2",
          "5/16"
        ],
        "output": "13/16",
        "label": "CLT binomial ORIGINAL total mass sum3"
      }
    },
    {
      "id": "S059",
      "uses": [
        "G001",
        "S058"
      ],
      "before": "CLT binomial ORIGINAL total mass sum4 := sum(13/16, 5/32)",
      "after": "CLT binomial ORIGINAL total mass sum4 = 31/32",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL total mass sum4，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL total mass sum4 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "13/16",
          "5/32"
        ],
        "output": "31/32",
        "label": "CLT binomial ORIGINAL total mass sum4"
      }
    },
    {
      "id": "S060",
      "uses": [
        "G001",
        "S059"
      ],
      "before": "CLT binomial ORIGINAL total mass sum5 := sum(31/32, 1/32)",
      "after": "CLT binomial ORIGINAL total mass sum5 = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT binomial ORIGINAL total mass sum5，不以小數近似替代。",
          "en": "Recompute CLT binomial ORIGINAL total mass sum5 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "31/32",
          "1/32"
        ],
        "output": "1",
        "label": "CLT binomial ORIGINAL total mass sum5"
      }
    },
    {
      "id": "S061",
      "uses": [
        "G001",
        "S060"
      ],
      "before": "source roles = [5, 1/2, 1, [[0, 1/32], [1, 5/32], [2, 5/16], [3, 5/16], [4, 5/32], [5, 1/32]]]",
      "after": "derived = [1, 3/16, z_threshold=(1-n*p)/sqrt(n*p*(1-p)), 否]",
      "relation": "implies",
      "rule_id": "CLT_complete_binomial_total_inclusive_CDF_NO_normal_approximation",
      "conditions": [],
      "explanation": {
        "zh-Hant": "k=0..n全列且總和=(p+(1-p))^n=1；CDF累加k<=threshold含等號。z僅保標準化threshold符號，不提供Gaussian近似值、誤差界或一般CLT證明。",
        "en": "List every k=0..n;the total is(p+(1-p))^n=1.The CDF includes k<=threshold,equality included.The z expression is a symbolic standardized threshold,not a Gaussian approximation,error bound or general CLT proof."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。",
          "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "result": [
      5,
      "1/2",
      [
        [
          0,
          "1/32"
        ],
        [
          1,
          "5/32"
        ],
        [
          2,
          "5/16"
        ],
        [
          3,
          "5/16"
        ],
        [
          4,
          "5/32"
        ],
        [
          5,
          "1/32"
        ]
      ],
      "1",
      1,
      "3/16",
      true,
      "z_threshold=(1-n*p)/sqrt(n*p*(1-p))",
      "exact_lattice_not_normal_approximation_or_convergence_evidence"
    ]
  },
  "conclusion": {
    "canonical_value": "(5, 1/2, [[0, 1/32], [1, 5/32], [2, 5/16], [3, 5/16], [4, 5/32], [5, 1/32]], 1, 1, 3/16, 是, z_threshold=(1-n*p)/sqrt(n*p*(1-p)), exact_lattice_not_normal_approximation_or_convergence_evidence)",
    "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-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-foundations",
      "batch_path": "public/data/questions/batches/batch-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-foundations/questions.jsonl",
      "batch_sha256": "6eaac14f97913d20b3a00dc8becf4943e2b1791210c8c5bbf3a89ecced69c6d1",
      "original_line": 64,
      "original_line_sha256": "c2d235fcd53a565bea889f154b7b015117659c3bffcdf8b95b851b9584374a45",
      "question_compact_sha256": "24b6f02d06ccd6e04690fd5e679d461d8c70ec5543eb166925c0733361715416",
      "english_row_compact_sha256": "fec610e73eb7fc57d49116a0de77be4fd475a731f019469eb2cb15bcc189ae7f",
      "english_manifest_path": "public/data/i18n/editions/batch-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-foundations.en.jsonl"
    },
    "mko_bindings": {
      "mko-central-limit-theorem-finite-variance-iid": {
        "sha256": "c37700292553db11f9702a86ba165e9287d7830d81ee8351eff28de1c07ea9fe",
        "english_sha256": "951e71ab789a556b03eeb44c3472b5e89e3ac252d23992bc8abf538f39a494cb",
        "original_path": "public/data/mko/central-limit-theorem-finite-variance-iid.json",
        "english_path": "public/data/i18n/editions/mko-central-limit-theorem-finite-variance-iid.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "a54409c0bc3f68730e1072c6d3d772970462c1bc7ffc203b4030a6136a610283",
    "original_candidate_sha256": "cf948c96a9576ee42e61b2fbd43252f235cd51a2764ce947abf08cdb81fec00c",
    "original_candidate_line_sha256": "cf948c96a9576ee42e61b2fbd43252f235cd51a2764ce947abf08cdb81fec00c"
  },
  "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-20261026-010-p54",
    "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": "1b7a62514347b82518305d5feb4369a52d7d44c442a0f2b131e9eb3cd7b45f62",
      "question_first_witness_sha256": "79dc2a1077081f3ea2a03f75e32778163bf05e814cdf4c29ed86e567295f9ee1"
    },
    "status": "second_context_reviewed",
    "mathematical_semantics": true,
    "bilingual_semantics": true,
    "steps_reviewed": 61,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "a53b0ad1d4ccedf3023a5ec0bf190fb42ea57ea688dcb299c50ff39de8066cf8",
      "arithmetic_steps": 51
    },
    "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."
  }
}