{ "schema_version": "ocme-reviewed-question-reader-v1", "question_id": "q-20261015-advance-20261118-r04-wilson-theorem-0073", "pilot_number": "R0054-27748", "number": 27748, "family_id": "qf-mod49-wilson-composite-instance", "title": { "zh-Hant": "第 27748 題：Wilson定理", "en": "Question 27748: A Composite Factorial Computation Is an Instance, Not the Converse Proof" }, "original_question": { "schema_version": "ocme-question-v0.1", "id": "q-20261015-advance-20261118-r04-wilson-theorem-0073", "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-composite-instance；資料={\"n\":35}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。", "answer": { "kind": "expression", "value": "(35, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 15], [6, 20], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0], [33, 0], [34, 0]], 0, 34, 是, finite_composite_instance_not_converse_proof)", "explanation_zh": "Wave6第十二層核對=(35, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 15], [6, 20], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0], [33, 0], [34, 0]], 0, 34, 是, finite_composite_instance_not_converse_proof)；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": 3, "skills": [ "Wilson定理" ] }, "variant": { "family_id": "qf-mod49-wilson-composite-instance", "seed": 202611184073, "parameters": { "n": 35, "domain": "Exact_Wilson_congruence", "operation": "wilson-composite-instance", "semantic_template": "wilson-composite-instance|{n:35}" } }, "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-0073", "batch_id": "batch-20261015-advance-20261118-r04-wilson-theorem-foundations", "locale": "en", "translation_state": "ai_candidate", "source": { "question_sha256": "3a05c5893d762179f074fbaae998dc73f5313611b5f0b8bc28c5412836fa5c6c", "family_id": "qf-mod49-wilson-composite-instance", "seed": 202611184073, "semantic_template": "wilson-composite-instance|{n:35}" }, "fields": { "title": "A Composite Factorial Computation Is an Instance, Not the Converse Proof", "stem": "Use composite integer n=35. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).", "answer": { "kind": "expression", "value": "(35, No, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 15], [6, 20], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0], [33, 0], [34, 0]], 0, 34, Yes, finite_composite_instance_not_converse_proof)", "explanation": "Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (35, No, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 15], [6, 20], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0], [33, 0], [34, 0]], 0, 34, Yes, finite_composite_instance_not_converse_proof)." }, "skills": [ "A Composite Factorial Computation Is an Instance, Not the Converse Proof" ], "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-composite-instance；資料={\"n\":35}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。", "en": "Use composite integer n=35. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope)." } } ], "steps": [ { "id": "S001", "uses": [ "G001" ], "before": "source roles = [wilson-composite-instance]", "after": "derived = [[n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope]]", "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": "Wilson ORIGINAL factorial ORIGINAL running factorial product1 := mul(1, 1)", "after": "Wilson ORIGINAL 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 factorial ORIGINAL running factorial product1，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product1 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "1", "1" ], "output": "1", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product1" } }, { "id": "S003", "uses": [ "G001", "S002" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 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": "S004", "uses": [ "G001", "S003" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product2 := mul(1, 2)", "after": "Wilson ORIGINAL 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 factorial ORIGINAL running factorial product2，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product2 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "1", "2" ], "output": "2", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product2" } }, { "id": "S005", "uses": [ "G001", "S004" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 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": "S006", "uses": [ "G001", "S005" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product3 := mul(2, 3)", "after": "Wilson ORIGINAL 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 factorial ORIGINAL running factorial product3，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product3 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "2", "3" ], "output": "6", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product3" } }, { "id": "S007", "uses": [ "G001", "S006" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 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": "S008", "uses": [ "G001", "S007" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product4 := mul(6, 4)", "after": "Wilson ORIGINAL 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 factorial ORIGINAL running factorial product4，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product4 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "6", "4" ], "output": "24", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product4" } }, { "id": "S009", "uses": [ "G001", "S008" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 4, 6, 24]", "after": "derived = [0, 24, 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": "S010", "uses": [ "G001", "S009" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product5 := mul(24, 5)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product5 = 120", "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 factorial ORIGINAL running factorial product5，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product5 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "24", "5" ], "output": "120", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product5" } }, { "id": "S011", "uses": [ "G001", "S010" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 5, 24, 120]", "after": "derived = [3, 15, 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": "S012", "uses": [ "G001", "S011" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product6 := mul(15, 6)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product6 = 90", "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 factorial ORIGINAL running factorial product6，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product6 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "15", "6" ], "output": "90", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product6" } }, { "id": "S013", "uses": [ "G001", "S012" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 6, 15, 90]", "after": "derived = [2, 20, 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 factorial ORIGINAL running factorial product7 := mul(20, 7)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product7 = 140", "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 factorial ORIGINAL running factorial product7，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product7 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "20", "7" ], "output": "140", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product7" } }, { "id": "S015", "uses": [ "G001", "S014" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 7, 20, 140]", "after": "derived = [4, 0, 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 factorial ORIGINAL running factorial product8 := mul(0, 8)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product8 = 0", "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 factorial ORIGINAL running factorial product8，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product8 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "8" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product8" } }, { "id": "S017", "uses": [ "G001", "S016" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 8, 0, 0]", "after": "derived = [0, 0, 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 factorial ORIGINAL running factorial product9 := mul(0, 9)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product9 = 0", "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 factorial ORIGINAL running factorial product9，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product9 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "9" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product9" } }, { "id": "S019", "uses": [ "G001", "S018" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 9, 0, 0]", "after": "derived = [0, 0, 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 factorial ORIGINAL running factorial product10 := mul(0, 10)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product10 = 0", "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 factorial ORIGINAL running factorial product10，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product10 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "10" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product10" } }, { "id": "S021", "uses": [ "G001", "S020" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 10, 0, 0]", "after": "derived = [0, 0, 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 factorial ORIGINAL running factorial product11 := mul(0, 11)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product11 = 0", "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 factorial ORIGINAL running factorial product11，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product11 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "11" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product11" } }, { "id": "S023", "uses": [ "G001", "S022" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 11, 0, 0]", "after": "derived = [0, 0, 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 factorial ORIGINAL running factorial product12 := mul(0, 12)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product12 = 0", "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 factorial ORIGINAL running factorial product12，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product12 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "12" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product12" } }, { "id": "S025", "uses": [ "G001", "S024" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 12, 0, 0]", "after": "derived = [0, 0, 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 factorial ORIGINAL running factorial product13 := mul(0, 13)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product13 = 0", "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 factorial ORIGINAL running factorial product13，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product13 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "13" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product13" } }, { "id": "S027", "uses": [ "G001", "S026" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 13, 0, 0]", "after": "derived = [0, 0, 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 factorial ORIGINAL running factorial product14 := mul(0, 14)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product14 = 0", "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 factorial ORIGINAL running factorial product14，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product14 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "14" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product14" } }, { "id": "S029", "uses": [ "G001", "S028" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 14, 0, 0]", "after": "derived = [0, 0, 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 factorial ORIGINAL running factorial product15 := mul(0, 15)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product15 = 0", "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 factorial ORIGINAL running factorial product15，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product15 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "15" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product15" } }, { "id": "S031", "uses": [ "G001", "S030" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 15, 0, 0]", "after": "derived = [0, 0, 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": "Wilson ORIGINAL factorial ORIGINAL running factorial product16 := mul(0, 16)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product16 = 0", "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 factorial ORIGINAL running factorial product16，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product16 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "16" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product16" } }, { "id": "S033", "uses": [ "G001", "S032" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 16, 0, 0]", "after": "derived = [0, 0, 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": "S034", "uses": [ "G001", "S033" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product17 := mul(0, 17)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product17 = 0", "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 factorial ORIGINAL running factorial product17，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product17 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "17" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product17" } }, { "id": "S035", "uses": [ "G001", "S034" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 17, 0, 0]", "after": "derived = [0, 0, 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": "S036", "uses": [ "G001", "S035" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product18 := mul(0, 18)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product18 = 0", "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 factorial ORIGINAL running factorial product18，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product18 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "18" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product18" } }, { "id": "S037", "uses": [ "G001", "S036" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 18, 0, 0]", "after": "derived = [0, 0, 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": "S038", "uses": [ "G001", "S037" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product19 := mul(0, 19)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product19 = 0", "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 factorial ORIGINAL running factorial product19，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product19 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "19" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product19" } }, { "id": "S039", "uses": [ "G001", "S038" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 19, 0, 0]", "after": "derived = [0, 0, 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": "S040", "uses": [ "G001", "S039" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product20 := mul(0, 20)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product20 = 0", "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 factorial ORIGINAL running factorial product20，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product20 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "20" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product20" } }, { "id": "S041", "uses": [ "G001", "S040" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 20, 0, 0]", "after": "derived = [0, 0, 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": "S042", "uses": [ "G001", "S041" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product21 := mul(0, 21)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product21 = 0", "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 factorial ORIGINAL running factorial product21，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product21 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "21" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product21" } }, { "id": "S043", "uses": [ "G001", "S042" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 21, 0, 0]", "after": "derived = [0, 0, 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": "S044", "uses": [ "G001", "S043" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product22 := mul(0, 22)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product22 = 0", "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 factorial ORIGINAL running factorial product22，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product22 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "22" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product22" } }, { "id": "S045", "uses": [ "G001", "S044" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 22, 0, 0]", "after": "derived = [0, 0, 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": "S046", "uses": [ "G001", "S045" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product23 := mul(0, 23)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product23 = 0", "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 factorial ORIGINAL running factorial product23，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product23 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "23" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product23" } }, { "id": "S047", "uses": [ "G001", "S046" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 23, 0, 0]", "after": "derived = [0, 0, 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": "S048", "uses": [ "G001", "S047" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product24 := mul(0, 24)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product24 = 0", "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 factorial ORIGINAL running factorial product24，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product24 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "24" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product24" } }, { "id": "S049", "uses": [ "G001", "S048" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 24, 0, 0]", "after": "derived = [0, 0, 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": "S050", "uses": [ "G001", "S049" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product25 := mul(0, 25)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product25 = 0", "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 factorial ORIGINAL running factorial product25，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product25 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "25" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product25" } }, { "id": "S051", "uses": [ "G001", "S050" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 25, 0, 0]", "after": "derived = [0, 0, 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": "S052", "uses": [ "G001", "S051" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product26 := mul(0, 26)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product26 = 0", "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 factorial ORIGINAL running factorial product26，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product26 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "26" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product26" } }, { "id": "S053", "uses": [ "G001", "S052" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 26, 0, 0]", "after": "derived = [0, 0, 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": "S054", "uses": [ "G001", "S053" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product27 := mul(0, 27)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product27 = 0", "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 factorial ORIGINAL running factorial product27，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product27 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "27" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product27" } }, { "id": "S055", "uses": [ "G001", "S054" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 27, 0, 0]", "after": "derived = [0, 0, 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": "S056", "uses": [ "G001", "S055" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product28 := mul(0, 28)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product28 = 0", "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 factorial ORIGINAL running factorial product28，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product28 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "28" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product28" } }, { "id": "S057", "uses": [ "G001", "S056" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 28, 0, 0]", "after": "derived = [0, 0, 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": "S058", "uses": [ "G001", "S057" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product29 := mul(0, 29)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product29 = 0", "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 factorial ORIGINAL running factorial product29，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product29 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "29" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product29" } }, { "id": "S059", "uses": [ "G001", "S058" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 29, 0, 0]", "after": "derived = [0, 0, 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": "S060", "uses": [ "G001", "S059" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product30 := mul(0, 30)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product30 = 0", "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 factorial ORIGINAL running factorial product30，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product30 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "30" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product30" } }, { "id": "S061", "uses": [ "G001", "S060" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 30, 0, 0]", "after": "derived = [0, 0, 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": "S062", "uses": [ "G001", "S061" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product31 := mul(0, 31)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product31 = 0", "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 factorial ORIGINAL running factorial product31，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product31 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "31" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product31" } }, { "id": "S063", "uses": [ "G001", "S062" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 31, 0, 0]", "after": "derived = [0, 0, 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": "S064", "uses": [ "G001", "S063" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product32 := mul(0, 32)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product32 = 0", "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 factorial ORIGINAL running factorial product32，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product32 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "32" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product32" } }, { "id": "S065", "uses": [ "G001", "S064" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 32, 0, 0]", "after": "derived = [0, 0, 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": "S066", "uses": [ "G001", "S065" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product33 := mul(0, 33)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product33 = 0", "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 factorial ORIGINAL running factorial product33，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product33 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "33" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product33" } }, { "id": "S067", "uses": [ "G001", "S066" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 33, 0, 0]", "after": "derived = [0, 0, 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": "S068", "uses": [ "G001", "S067" ], "before": "Wilson ORIGINAL factorial ORIGINAL running factorial product34 := mul(0, 34)", "after": "Wilson ORIGINAL factorial ORIGINAL running factorial product34 = 0", "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 factorial ORIGINAL running factorial product34，不以小數近似替代。", "en": "Recompute Wilson ORIGINAL factorial ORIGINAL running factorial product34 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "34" ], "output": "0", "label": "Wilson ORIGINAL factorial ORIGINAL running factorial product34" } }, { "id": "S069", "uses": [ "G001", "S068" ], "before": "source roles = [Wilson ORIGINAL factorial, 35, 34, 0, 0]", "after": "derived = [0, 0, 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": "S070", "uses": [ "G001", "S069" ], "before": "source roles = [35, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 15], [6, 20], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0], [33, 0], [34, 0]]]", "after": "derived = [否, 0, 34, 否, 是]", "relation": "implies", "rule_id": "Wilson_prime_classification_instance_flags_AND_separate_converse_reason", "conditions": [], "explanation": { "zh-Hant": "prime與congruence分別驗；合數時兩者都False也可equivalenceTrue。一般converse另有理由：若合數有proper divisor d>=2，d同除n與(n-1)!；假設後者≡-1 mod n便使d除1，矛盾。此推理不等於一個有限instance或finite cutoff測試。", "en": "Qualify primality and congruence separately.Both may be False at a composite while equivalence is True.The converse has a separate argument:a proper divisor d>=2 of composite n divides both n and(n-1)!;a congruence to-1 would force d to divide1,a contradiction.This reasoning is not a single instance or finite-cutoff test." }, "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": [ 35, false, [ [ 1, 1 ], [ 2, 2 ], [ 3, 6 ], [ 4, 24 ], [ 5, 15 ], [ 6, 20 ], [ 7, 0 ], [ 8, 0 ], [ 9, 0 ], [ 10, 0 ], [ 11, 0 ], [ 12, 0 ], [ 13, 0 ], [ 14, 0 ], [ 15, 0 ], [ 16, 0 ], [ 17, 0 ], [ 18, 0 ], [ 19, 0 ], [ 20, 0 ], [ 21, 0 ], [ 22, 0 ], [ 23, 0 ], [ 24, 0 ], [ 25, 0 ], [ 26, 0 ], [ 27, 0 ], [ 28, 0 ], [ 29, 0 ], [ 30, 0 ], [ 31, 0 ], [ 32, 0 ], [ 33, 0 ], [ 34, 0 ] ], 0, 34, true, "finite_composite_instance_not_converse_proof" ] }, "conclusion": { "canonical_value": "(35, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 15], [6, 20], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0], [33, 0], [34, 0]], 0, 34, 是, finite_composite_instance_not_converse_proof)", "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": 73, "original_line_sha256": "5f8972e8b40ccdbec89f5cdcd50ec47d100cf9d9f342ddbaccb6b1c7f96945f7", "question_compact_sha256": "3a05c5893d762179f074fbaae998dc73f5313611b5f0b8bc28c5412836fa5c6c", "english_row_compact_sha256": "403b1a9f8ad506978a6d8cd3991e74e25ac28e536a722522ff7844ef8e962fd8", "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": "1a71ab408e893f29c1edd304e7d2d1443fc4a2c47ab0e3927278494177a072ef", "original_candidate_line_sha256": "1a71ab408e893f29c1edd304e7d2d1443fc4a2c47ab0e3927278494177a072ef" }, "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": 70, "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction", "exact_question_first_check": { "model_sha256": "7ec5a3dac8f950bffdb820d957a09fbfdc3ba523ac3370f197b9857c4844889f", "arithmetic_steps": 34 }, "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." } }