{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261015-advance-20261118-r04-wilson-theorem-0030",
  "pilot_number": "R0054-27780",
  "number": 27780,
  "family_id": "qf-mod49-wilson-prime-inverse-pairing",
  "title": {
    "zh-Hant": "第 27780 題：Wilson定理",
    "en": "Question 27780: All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261015-advance-20261118-r04-wilson-theorem-0030",
    "batch_id": "batch-20261015-advance-20261118-r04-wilson-theorem-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-wilson-theorem"
    ],
    "title_zh": "Wilson定理",
    "stem_zh": "n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing；資料={\"p\":11}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "(11, [[1, 1], [2, 6], [3, 4], [4, 3], [5, 9], [6, 2], [7, 8], [8, 7], [9, 5], [10, 10]], [1, 10], [[2, 6, 1], [3, 4, 1], [5, 9, 1], [7, 8, 1]], self_inverse_residues_1_and_p_minus_1, 10, 是)",
      "explanation_zh": "Wave6第十二層核對=(11, [[1, 1], [2, 6], [3, 4], [4, 3], [5, 9], [6, 2], [7, 8], [8, 7], [9, 5], [10, 10]], [1, 10], [[2, 6, 1], [3, 4, 1], [5, 9, 1], [7, 8, 1]], self_inverse_residues_1_and_p_minus_1, 10, 是)；依完整明示資料獨立重算；p=2只有一個self-inverse residue 1=-1；composite instances與finite testing不取代converse proof，Wilson不等於Fermat。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 1,
      "skills": [
        "Wilson定理"
      ]
    },
    "variant": {
      "family_id": "qf-mod49-wilson-prime-inverse-pairing",
      "seed": 202611184030,
      "parameters": {
        "p": 11,
        "domain": "Exact_Wilson_congruence",
        "operation": "wilson-prime-inverse-pairing",
        "semantic_template": "wilson-prime-inverse-pairing|{n:11}"
      }
    },
    "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-r04-wilson-theorem-0030",
    "batch_id": "batch-20261015-advance-20261118-r04-wilson-theorem-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "89ad139f7e9fe25e1052123c77d4ea5fc5641afb22fa6f50466986257ff7a89f",
      "family_id": "qf-mod49-wilson-prime-inverse-pairing",
      "seed": 202611184030,
      "semantic_template": "wilson-prime-inverse-pairing|{n:11}"
    },
    "fields": {
      "title": "All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case",
      "stem": "Use prime modulus p=11. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).",
      "answer": {
        "kind": "expression",
        "value": "(11, [[1, 1], [2, 6], [3, 4], [4, 3], [5, 9], [6, 2], [7, 8], [8, 7], [9, 5], [10, 10]], [1, 10], [[2, 6, 1], [3, 4, 1], [5, 9, 1], [7, 8, 1]], self_inverse_residues_1_and_p_minus_1, 10, Yes)",
        "explanation": "List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (11, [[1, 1], [2, 6], [3, 4], [4, 3], [5, 9], [6, 2], [7, 8], [8, 7], [9, 5], [10, 10]], [1, 10], [[2, 6, 1], [3, 4, 1], [5, 9, 1], [7, 8, 1]], self_inverse_residues_1_and_p_minus_1, 10, Yes)."
      },
      "skills": [
        "All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"
      ],
      "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": "n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing；資料={\"p\":11}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
        "en": "Use prime modulus p=11. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "source roles = [wilson-prime-inverse-pairing]",
      "after": "derived = [[p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence]]",
      "relation": "implies",
      "rule_id": "source_ordered_fields_without_JS",
      "conditions": [],
      "explanation": {
        "zh-Hant": "直接明列此操作與mode完整欄序，不依JavaScript、不由答案形狀猜角色；布林、字串、null與未作聲稱保持不同。",
        "en": "List the complete operation/mode field order without JavaScript or answer-shape inference;Boolean,string,null and an unclaimed result remain distinct."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "source roles = [11, 1]",
      "after": "derived = [1, 1]",
      "relation": "implies",
      "rule_id": "Wilson_EVERY_nonzero_prime_residue_unique_inverse",
      "conditions": [],
      "explanation": {
        "zh-Hant": "prime modulus的非零residue均為units，逐a找來源首inverse；若有兩個inverse，乘a的消去性使其相同。",
        "en": "Every nonzero residue at a prime modulus is a unit.For each a find the first source inverse;two inverses coincide by cancellation of a."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "source roles = [11, 2]",
      "after": "derived = [6, 1]",
      "relation": "implies",
      "rule_id": "Wilson_EVERY_nonzero_prime_residue_unique_inverse",
      "conditions": [],
      "explanation": {
        "zh-Hant": "prime modulus的非零residue均為units，逐a找來源首inverse；若有兩個inverse，乘a的消去性使其相同。",
        "en": "Every nonzero residue at a prime modulus is a unit.For each a find the first source inverse;two inverses coincide by cancellation of a."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "source roles = [11, 3]",
      "after": "derived = [4, 1]",
      "relation": "implies",
      "rule_id": "Wilson_EVERY_nonzero_prime_residue_unique_inverse",
      "conditions": [],
      "explanation": {
        "zh-Hant": "prime modulus的非零residue均為units，逐a找來源首inverse；若有兩個inverse，乘a的消去性使其相同。",
        "en": "Every nonzero residue at a prime modulus is a unit.For each a find the first source inverse;two inverses coincide by cancellation of a."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "source roles = [11, 4]",
      "after": "derived = [3, 1]",
      "relation": "implies",
      "rule_id": "Wilson_EVERY_nonzero_prime_residue_unique_inverse",
      "conditions": [],
      "explanation": {
        "zh-Hant": "prime modulus的非零residue均為units，逐a找來源首inverse；若有兩個inverse，乘a的消去性使其相同。",
        "en": "Every nonzero residue at a prime modulus is a unit.For each a find the first source inverse;two inverses coincide by cancellation of a."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "source roles = [11, 5]",
      "after": "derived = [9, 1]",
      "relation": "implies",
      "rule_id": "Wilson_EVERY_nonzero_prime_residue_unique_inverse",
      "conditions": [],
      "explanation": {
        "zh-Hant": "prime modulus的非零residue均為units，逐a找來源首inverse；若有兩個inverse，乘a的消去性使其相同。",
        "en": "Every nonzero residue at a prime modulus is a unit.For each a find the first source inverse;two inverses coincide by cancellation of a."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "source roles = [11, 6]",
      "after": "derived = [2, 1]",
      "relation": "implies",
      "rule_id": "Wilson_EVERY_nonzero_prime_residue_unique_inverse",
      "conditions": [],
      "explanation": {
        "zh-Hant": "prime modulus的非零residue均為units，逐a找來源首inverse；若有兩個inverse，乘a的消去性使其相同。",
        "en": "Every nonzero residue at a prime modulus is a unit.For each a find the first source inverse;two inverses coincide by cancellation of a."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "source roles = [11, 7]",
      "after": "derived = [8, 1]",
      "relation": "implies",
      "rule_id": "Wilson_EVERY_nonzero_prime_residue_unique_inverse",
      "conditions": [],
      "explanation": {
        "zh-Hant": "prime modulus的非零residue均為units，逐a找來源首inverse；若有兩個inverse，乘a的消去性使其相同。",
        "en": "Every nonzero residue at a prime modulus is a unit.For each a find the first source inverse;two inverses coincide by cancellation of a."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "source roles = [11, 8]",
      "after": "derived = [7, 1]",
      "relation": "implies",
      "rule_id": "Wilson_EVERY_nonzero_prime_residue_unique_inverse",
      "conditions": [],
      "explanation": {
        "zh-Hant": "prime modulus的非零residue均為units，逐a找來源首inverse；若有兩個inverse，乘a的消去性使其相同。",
        "en": "Every nonzero residue at a prime modulus is a unit.For each a find the first source inverse;two inverses coincide by cancellation of a."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "source roles = [11, 9]",
      "after": "derived = [5, 1]",
      "relation": "implies",
      "rule_id": "Wilson_EVERY_nonzero_prime_residue_unique_inverse",
      "conditions": [],
      "explanation": {
        "zh-Hant": "prime modulus的非零residue均為units，逐a找來源首inverse；若有兩個inverse，乘a的消去性使其相同。",
        "en": "Every nonzero residue at a prime modulus is a unit.For each a find the first source inverse;two inverses coincide by cancellation of a."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "source roles = [11, 10]",
      "after": "derived = [10, 1]",
      "relation": "implies",
      "rule_id": "Wilson_EVERY_nonzero_prime_residue_unique_inverse",
      "conditions": [],
      "explanation": {
        "zh-Hant": "prime modulus的非零residue均為units，逐a找來源首inverse；若有兩個inverse，乘a的消去性使其相同。",
        "en": "Every nonzero residue at a prime modulus is a unit.For each a find the first source inverse;two inverses coincide by cancellation of a."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S012",
      "uses": [
        "G001",
        "S011"
      ],
      "before": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product1 := mul(1, 1)",
      "after": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product1 = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "以原始角色與已證中間量精確代值，不用近似或自洽但換過的操作數；額外教學代數不另縮來源域。",
        "en": "Substitute original roles and established intermediates exactly,not approximations or internally consistent changed operands;extra teaching algebra adds no source-domain gate."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Wilson ORIGINAL prime factorial ORIGINAL running factorial product1，不以小數近似替代。",
          "en": "Recompute Wilson ORIGINAL prime factorial ORIGINAL running factorial product1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1"
        ],
        "output": "1",
        "label": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product1"
      }
    },
    {
      "id": "S013",
      "uses": [
        "G001",
        "S012"
      ],
      "before": "source roles = [Wilson ORIGINAL prime factorial, 11, 1, 1, 1]",
      "after": "derived = [0, 1, 是, 是]",
      "relation": "implies",
      "rule_id": "Wilson_EACH_running_factorial_exact_Euclidean_remainder",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐k保前residue與完整乘積，再以唯一Euclidean餘數縮回0..n-1；累積不變量是value≡k! mod n，不把table最後一項當未算的假設。",
        "en": "Retain the prior residue and full product at every k,then its unique Euclidean remainder in0..n-1.The invariant is value congruent to k! mod n,not an assumed last table entry."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S014",
      "uses": [
        "G001",
        "S013"
      ],
      "before": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product2 := mul(1, 2)",
      "after": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product2 = 2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "以原始角色與已證中間量精確代值，不用近似或自洽但換過的操作數；額外教學代數不另縮來源域。",
        "en": "Substitute original roles and established intermediates exactly,not approximations or internally consistent changed operands;extra teaching algebra adds no source-domain gate."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Wilson ORIGINAL prime factorial ORIGINAL running factorial product2，不以小數近似替代。",
          "en": "Recompute Wilson ORIGINAL prime factorial ORIGINAL running factorial product2 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "2"
        ],
        "output": "2",
        "label": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product2"
      }
    },
    {
      "id": "S015",
      "uses": [
        "G001",
        "S014"
      ],
      "before": "source roles = [Wilson ORIGINAL prime factorial, 11, 2, 1, 2]",
      "after": "derived = [0, 2, 是, 是]",
      "relation": "implies",
      "rule_id": "Wilson_EACH_running_factorial_exact_Euclidean_remainder",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐k保前residue與完整乘積，再以唯一Euclidean餘數縮回0..n-1；累積不變量是value≡k! mod n，不把table最後一項當未算的假設。",
        "en": "Retain the prior residue and full product at every k,then its unique Euclidean remainder in0..n-1.The invariant is value congruent to k! mod n,not an assumed last table entry."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S016",
      "uses": [
        "G001",
        "S015"
      ],
      "before": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product3 := mul(2, 3)",
      "after": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product3 = 6",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "以原始角色與已證中間量精確代值，不用近似或自洽但換過的操作數；額外教學代數不另縮來源域。",
        "en": "Substitute original roles and established intermediates exactly,not approximations or internally consistent changed operands;extra teaching algebra adds no source-domain gate."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Wilson ORIGINAL prime factorial ORIGINAL running factorial product3，不以小數近似替代。",
          "en": "Recompute Wilson ORIGINAL prime factorial ORIGINAL running factorial product3 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "2",
          "3"
        ],
        "output": "6",
        "label": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product3"
      }
    },
    {
      "id": "S017",
      "uses": [
        "G001",
        "S016"
      ],
      "before": "source roles = [Wilson ORIGINAL prime factorial, 11, 3, 2, 6]",
      "after": "derived = [0, 6, 是, 是]",
      "relation": "implies",
      "rule_id": "Wilson_EACH_running_factorial_exact_Euclidean_remainder",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐k保前residue與完整乘積，再以唯一Euclidean餘數縮回0..n-1；累積不變量是value≡k! mod n，不把table最後一項當未算的假設。",
        "en": "Retain the prior residue and full product at every k,then its unique Euclidean remainder in0..n-1.The invariant is value congruent to k! mod n,not an assumed last table entry."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S018",
      "uses": [
        "G001",
        "S017"
      ],
      "before": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product4 := mul(6, 4)",
      "after": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product4 = 24",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "以原始角色與已證中間量精確代值，不用近似或自洽但換過的操作數；額外教學代數不另縮來源域。",
        "en": "Substitute original roles and established intermediates exactly,not approximations or internally consistent changed operands;extra teaching algebra adds no source-domain gate."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Wilson ORIGINAL prime factorial ORIGINAL running factorial product4，不以小數近似替代。",
          "en": "Recompute Wilson ORIGINAL prime factorial ORIGINAL running factorial product4 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "6",
          "4"
        ],
        "output": "24",
        "label": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product4"
      }
    },
    {
      "id": "S019",
      "uses": [
        "G001",
        "S018"
      ],
      "before": "source roles = [Wilson ORIGINAL prime factorial, 11, 4, 6, 24]",
      "after": "derived = [2, 2, 是, 是]",
      "relation": "implies",
      "rule_id": "Wilson_EACH_running_factorial_exact_Euclidean_remainder",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐k保前residue與完整乘積，再以唯一Euclidean餘數縮回0..n-1；累積不變量是value≡k! mod n，不把table最後一項當未算的假設。",
        "en": "Retain the prior residue and full product at every k,then its unique Euclidean remainder in0..n-1.The invariant is value congruent to k! mod n,not an assumed last table entry."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S020",
      "uses": [
        "G001",
        "S019"
      ],
      "before": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product5 := mul(2, 5)",
      "after": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product5 = 10",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "以原始角色與已證中間量精確代值，不用近似或自洽但換過的操作數；額外教學代數不另縮來源域。",
        "en": "Substitute original roles and established intermediates exactly,not approximations or internally consistent changed operands;extra teaching algebra adds no source-domain gate."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Wilson ORIGINAL prime factorial ORIGINAL running factorial product5，不以小數近似替代。",
          "en": "Recompute Wilson ORIGINAL prime factorial ORIGINAL running factorial product5 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "2",
          "5"
        ],
        "output": "10",
        "label": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product5"
      }
    },
    {
      "id": "S021",
      "uses": [
        "G001",
        "S020"
      ],
      "before": "source roles = [Wilson ORIGINAL prime factorial, 11, 5, 2, 10]",
      "after": "derived = [0, 10, 是, 是]",
      "relation": "implies",
      "rule_id": "Wilson_EACH_running_factorial_exact_Euclidean_remainder",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐k保前residue與完整乘積，再以唯一Euclidean餘數縮回0..n-1；累積不變量是value≡k! mod n，不把table最後一項當未算的假設。",
        "en": "Retain the prior residue and full product at every k,then its unique Euclidean remainder in0..n-1.The invariant is value congruent to k! mod n,not an assumed last table entry."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S022",
      "uses": [
        "G001",
        "S021"
      ],
      "before": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product6 := mul(10, 6)",
      "after": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product6 = 60",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "以原始角色與已證中間量精確代值，不用近似或自洽但換過的操作數；額外教學代數不另縮來源域。",
        "en": "Substitute original roles and established intermediates exactly,not approximations or internally consistent changed operands;extra teaching algebra adds no source-domain gate."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Wilson ORIGINAL prime factorial ORIGINAL running factorial product6，不以小數近似替代。",
          "en": "Recompute Wilson ORIGINAL prime factorial ORIGINAL running factorial product6 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "10",
          "6"
        ],
        "output": "60",
        "label": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product6"
      }
    },
    {
      "id": "S023",
      "uses": [
        "G001",
        "S022"
      ],
      "before": "source roles = [Wilson ORIGINAL prime factorial, 11, 6, 10, 60]",
      "after": "derived = [5, 5, 是, 是]",
      "relation": "implies",
      "rule_id": "Wilson_EACH_running_factorial_exact_Euclidean_remainder",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐k保前residue與完整乘積，再以唯一Euclidean餘數縮回0..n-1；累積不變量是value≡k! mod n，不把table最後一項當未算的假設。",
        "en": "Retain the prior residue and full product at every k,then its unique Euclidean remainder in0..n-1.The invariant is value congruent to k! mod n,not an assumed last table entry."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S024",
      "uses": [
        "G001",
        "S023"
      ],
      "before": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product7 := mul(5, 7)",
      "after": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product7 = 35",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "以原始角色與已證中間量精確代值，不用近似或自洽但換過的操作數；額外教學代數不另縮來源域。",
        "en": "Substitute original roles and established intermediates exactly,not approximations or internally consistent changed operands;extra teaching algebra adds no source-domain gate."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Wilson ORIGINAL prime factorial ORIGINAL running factorial product7，不以小數近似替代。",
          "en": "Recompute Wilson ORIGINAL prime factorial ORIGINAL running factorial product7 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "5",
          "7"
        ],
        "output": "35",
        "label": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product7"
      }
    },
    {
      "id": "S025",
      "uses": [
        "G001",
        "S024"
      ],
      "before": "source roles = [Wilson ORIGINAL prime factorial, 11, 7, 5, 35]",
      "after": "derived = [3, 2, 是, 是]",
      "relation": "implies",
      "rule_id": "Wilson_EACH_running_factorial_exact_Euclidean_remainder",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐k保前residue與完整乘積，再以唯一Euclidean餘數縮回0..n-1；累積不變量是value≡k! mod n，不把table最後一項當未算的假設。",
        "en": "Retain the prior residue and full product at every k,then its unique Euclidean remainder in0..n-1.The invariant is value congruent to k! mod n,not an assumed last table entry."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S026",
      "uses": [
        "G001",
        "S025"
      ],
      "before": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product8 := mul(2, 8)",
      "after": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product8 = 16",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "以原始角色與已證中間量精確代值，不用近似或自洽但換過的操作數；額外教學代數不另縮來源域。",
        "en": "Substitute original roles and established intermediates exactly,not approximations or internally consistent changed operands;extra teaching algebra adds no source-domain gate."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Wilson ORIGINAL prime factorial ORIGINAL running factorial product8，不以小數近似替代。",
          "en": "Recompute Wilson ORIGINAL prime factorial ORIGINAL running factorial product8 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "2",
          "8"
        ],
        "output": "16",
        "label": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product8"
      }
    },
    {
      "id": "S027",
      "uses": [
        "G001",
        "S026"
      ],
      "before": "source roles = [Wilson ORIGINAL prime factorial, 11, 8, 2, 16]",
      "after": "derived = [1, 5, 是, 是]",
      "relation": "implies",
      "rule_id": "Wilson_EACH_running_factorial_exact_Euclidean_remainder",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐k保前residue與完整乘積，再以唯一Euclidean餘數縮回0..n-1；累積不變量是value≡k! mod n，不把table最後一項當未算的假設。",
        "en": "Retain the prior residue and full product at every k,then its unique Euclidean remainder in0..n-1.The invariant is value congruent to k! mod n,not an assumed last table entry."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S028",
      "uses": [
        "G001",
        "S027"
      ],
      "before": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product9 := mul(5, 9)",
      "after": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product9 = 45",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "以原始角色與已證中間量精確代值，不用近似或自洽但換過的操作數；額外教學代數不另縮來源域。",
        "en": "Substitute original roles and established intermediates exactly,not approximations or internally consistent changed operands;extra teaching algebra adds no source-domain gate."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Wilson ORIGINAL prime factorial ORIGINAL running factorial product9，不以小數近似替代。",
          "en": "Recompute Wilson ORIGINAL prime factorial ORIGINAL running factorial product9 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "5",
          "9"
        ],
        "output": "45",
        "label": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product9"
      }
    },
    {
      "id": "S029",
      "uses": [
        "G001",
        "S028"
      ],
      "before": "source roles = [Wilson ORIGINAL prime factorial, 11, 9, 5, 45]",
      "after": "derived = [4, 1, 是, 是]",
      "relation": "implies",
      "rule_id": "Wilson_EACH_running_factorial_exact_Euclidean_remainder",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐k保前residue與完整乘積，再以唯一Euclidean餘數縮回0..n-1；累積不變量是value≡k! mod n，不把table最後一項當未算的假設。",
        "en": "Retain the prior residue and full product at every k,then its unique Euclidean remainder in0..n-1.The invariant is value congruent to k! mod n,not an assumed last table entry."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S030",
      "uses": [
        "G001",
        "S029"
      ],
      "before": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product10 := mul(1, 10)",
      "after": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product10 = 10",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "以原始角色與已證中間量精確代值，不用近似或自洽但換過的操作數；額外教學代數不另縮來源域。",
        "en": "Substitute original roles and established intermediates exactly,not approximations or internally consistent changed operands;extra teaching algebra adds no source-domain gate."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 Wilson ORIGINAL prime factorial ORIGINAL running factorial product10，不以小數近似替代。",
          "en": "Recompute Wilson ORIGINAL prime factorial ORIGINAL running factorial product10 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "10"
        ],
        "output": "10",
        "label": "Wilson ORIGINAL prime factorial ORIGINAL running factorial product10"
      }
    },
    {
      "id": "S031",
      "uses": [
        "G001",
        "S030"
      ],
      "before": "source roles = [Wilson ORIGINAL prime factorial, 11, 10, 1, 10]",
      "after": "derived = [0, 10, 是, 是]",
      "relation": "implies",
      "rule_id": "Wilson_EACH_running_factorial_exact_Euclidean_remainder",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐k保前residue與完整乘積，再以唯一Euclidean餘數縮回0..n-1；累積不變量是value≡k! mod n，不把table最後一項當未算的假設。",
        "en": "Retain the prior residue and full product at every k,then its unique Euclidean remainder in0..n-1.The invariant is value congruent to k! mod n,not an assumed last table entry."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S032",
      "uses": [
        "G001",
        "S031"
      ],
      "before": "source roles = [11, [[1, 1], [2, 6], [3, 4], [4, 3], [5, 9], [6, 2], [7, 8], [8, 7], [9, 5], [10, 10]]]",
      "after": "derived = [[1, 10], [[2, 6, 1], [3, 4, 1], [5, 9, 1], [7, 8, 1]], 10, 是]",
      "relation": "implies",
      "rule_id": "Wilson_GENERAL_inverse_pairing_and_p2_exception",
      "conditions": [],
      "explanation": {
        "zh-Hant": "非self-inverse成對相乘1；self-inverse解(r-1)(r+1)=0 mod prime，故只±1。p>2相異乘積-1；p2僅1=-1一個。這是一般教學理由，有限表／本程式不冒稱Lean proof。",
        "en": "Nonself inverse pairs multiply to1.A self-inverse satisfies(r-1)(r+1)=0 modulo a prime,so is±1.For p>2 their product is-1;at p2 only the single1=-1 remains.This general teaching argument does not turn the finite table/program into a Lean proof."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。",
          "en": "Reverse-check original data,operands,complete scope and premises;finite witnesses,conditional theorem use and formal proof remain distinct."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "result": [
      11,
      [
        [
          1,
          1
        ],
        [
          2,
          6
        ],
        [
          3,
          4
        ],
        [
          4,
          3
        ],
        [
          5,
          9
        ],
        [
          6,
          2
        ],
        [
          7,
          8
        ],
        [
          8,
          7
        ],
        [
          9,
          5
        ],
        [
          10,
          10
        ]
      ],
      [
        1,
        10
      ],
      [
        [
          2,
          6,
          1
        ],
        [
          3,
          4,
          1
        ],
        [
          5,
          9,
          1
        ],
        [
          7,
          8,
          1
        ]
      ],
      "self_inverse_residues_1_and_p_minus_1",
      10,
      true
    ]
  },
  "conclusion": {
    "canonical_value": "(11, [[1, 1], [2, 6], [3, 4], [4, 3], [5, 9], [6, 2], [7, 8], [8, 7], [9, 5], [10, 10]], [1, 10], [[2, 6, 1], [3, 4, 1], [5, 9, 1], [7, 8, 1]], self_inverse_residues_1_and_p_minus_1, 10, 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-20261015-advance-20261118-r04-wilson-theorem-foundations",
      "batch_path": "public/data/questions/batches/batch-20261015-advance-20261118-r04-wilson-theorem-foundations/questions.jsonl",
      "batch_sha256": "d1aac7811ce398b2e86d71092a3480737f06aa76137b9fe7b65b1fe52487ff92",
      "original_line": 30,
      "original_line_sha256": "51fab5ee36751c316c6de1ca9927a4e8eac2a781a0271d5f3d839cc32ba1f900",
      "question_compact_sha256": "89ad139f7e9fe25e1052123c77d4ea5fc5641afb22fa6f50466986257ff7a89f",
      "english_row_compact_sha256": "c1cfcdd30a7a9436ab2bf25cc9535f36f44293289e38f302a119bf35ccb68a7e",
      "english_manifest_path": "public/data/i18n/editions/batch-20261015-advance-20261118-r04-wilson-theorem-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261015-advance-20261118-r04-wilson-theorem-foundations.en.jsonl"
    },
    "mko_bindings": {
      "mko-wilson-theorem": {
        "sha256": "32b4a0da9e832eb6c7c0c33d723d8f66608b7013dcd990ad19fa2596126d9eb8",
        "english_sha256": "0ea5bedaf95ddb7a9d900b474d7830c9ce429c691d48ecd945d3920fcd1d2b48",
        "original_path": "public/data/mko/wilson-theorem.json",
        "english_path": "public/data/i18n/editions/mko-wilson-theorem.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "ad40cc18f9292c43924a7679f3af46574a5284aa5447f5f7f78a0de98840acb8",
    "original_candidate_sha256": "7c34ef79c364ac5c9d1e0fe93eb102083310674484f9abab1159bcbd2b57fbf0",
    "original_candidate_line_sha256": "7c34ef79c364ac5c9d1e0fe93eb102083310674484f9abab1159bcbd2b57fbf0"
  },
  "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-p55",
    "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": "0a76d8e93c08bb16998a0483f3dfc0504f269fc3e1982e6cd868cab0a2856d54",
      "question_first_witness_sha256": "09127b736c225714dfa0b7ceb0e6a6e66df27438b480cad8217d3c2f99ce9437"
    },
    "status": "second_context_reviewed",
    "mathematical_semantics": true,
    "bilingual_semantics": true,
    "steps_reviewed": 32,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "29da3ead1751d336d6ab7877541c83d964f38a1ab9a1dcf7467dfa9d2a5f96c7",
      "arithmetic_steps": 10
    },
    "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."
  }
}