{ "schema_version": "ocme-reviewed-question-reader-v1", "question_id": "q-20260914-induction-proof-obligation-0007", "pilot_number": "R0070-35602", "number": 35602, "family_id": "qf-induction-proof-obligation", "title": { "zh-Hant": "第 35602 題：歸納證明義務辨識", "en": "Question 35602: Induction Proof Obligations" }, "original_question": { "schema_version": "ocme-question-v0.1", "id": "q-20260914-induction-proof-obligation-0007", "batch_id": "batch-20260914-domain-foundations", "language": "zh-Hant", "source_mko_ids": [ "mko-mathematical-induction" ], "title_zh": "歸納證明義務辨識", "stem_zh": "對從 0 起的命題族 P(n)，草稿已明示：證明 P(0)；令 n 為任意自然數並暫設 P(n)；在歸納假設下證明 P(n+1)。草稿另列出 P(0) 到 P(7) 的有限例子；這本身不構成普遍證明。下列哪個歸納角色缺漏？", "answer": { "kind": "choice", "value": "conclusion", "choices": [ "base_case", "induction_hypothesis", "induction_step", "conclusion" ], "explanation_zh": "草稿缺少「據此斷言所有 n∈N 的 P(n)」這個角色；列出有限例子不能補足歸納證明義務。" }, "difficulty": { "band": "upper_secondary", "level": 1, "skills": [ "數學歸納法", "證明義務" ] }, "variant": { "family_id": "qf-induction-proof-obligation", "seed": 20260980, "parameters": { "mode": "missing_role", "target_role": "conclusion", "provided_roles": [ "base_case", "induction_hypothesis", "induction_step" ], "base_index": 0, "step_relation": "P(n)_implies_P(n+1)", "finite_examples_checked": 7, "semantic_template": "qf-induction-proof-obligation/{\"base_index\":0,\"finite_examples_checked\":7,\"mode\":\"missing_role\",\"provided_roles\":[\"base_case\",\"induction_hypothesis\",\"induction_step\"],\"step_relation\":\"P(n)_implies_P(n+1)\",\"target_role\":\"conclusion\"}" } }, "verification": { "status": "candidate", "checks": [ "schema_pending", "independent_recomputation_pending" ], "warning_zh": "本題僅依明示的有限參數重算或分類；不構成普遍證明、形式 Evidence、人工審查或發布。" }, "provenance": { "generator_role": "question_producer", "generator_model": "GPT-5.6 Terra", "generated_at": "2026-09-14", "contract_version": "ocme-question-factory-v0.1" } }, "original_english_edition": { "schema_version": "ocme-question-localization-v0.1", "question_id": "q-20260914-induction-proof-obligation-0007", "batch_id": "batch-20260914-domain-foundations", "locale": "en", "translation_state": "ai_candidate", "source": { "question_sha256": "e48cd039cfad90c5aec8d3d6ce5b0842718e8fa556bf5d7e1fb087cc0e8003de", "family_id": "qf-induction-proof-obligation", "seed": 20260980, "semantic_template": "qf-induction-proof-obligation/{\"base_index\":0,\"finite_examples_checked\":7,\"mode\":\"missing_role\",\"provided_roles\":[\"base_case\",\"induction_hypothesis\",\"induction_step\"],\"step_relation\":\"P(n)_implies_P(n+1)\",\"target_role\":\"conclusion\"}" }, "fields": { "title": "Induction Proof Obligations", "stem": "For a proposition family P(n) beginning at 0, a draft supplies base case, induction hypothesis, induction step and checks 7 finite example(s). Which induction role is missing?", "answer": { "kind": "choice", "value": "conclusion", "choices": [ "base case", "induction hypothesis", "induction step", "conclusion" ], "explanation": "The missing role is the conclusion. Finite examples do not discharge that proof obligation." }, "skills": [ "mathematical induction", "proof obligations" ], "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": "對從 0 起的命題族 P(n)，草稿已明示：證明 P(0)；令 n 為任意自然數並暫設 P(n)；在歸納假設下證明 P(n+1)。草稿另列出 P(0) 到 P(7) 的有限例子；這本身不構成普遍證明。下列哪個歸納角色缺漏？", "en": "For a proposition family P(n) beginning at 0, a draft supplies base case, induction hypothesis, induction step and checks 7 finite example(s). Which induction role is missing?" } } ], "steps": [ { "id": "S001", "uses": [ "G001" ], "before": "source roles = [qf-induction-proof-obligation, [base_index, finite_examples_checked, mode, provided_roles, step_relation, target_role]]", "after": "derived = [induction, null, null]", "relation": "implies", "rule_id": "R70_SOURCE_bank_specific_interface", "conditions": [], "explanation": { "zh-Hant": "只使用此bank真正明示的欄位、operation/domain與型別；BASE容許保留額外metadata，不把LDL SAFE/集合上限或其他bank的去重規則搬來。", "en": "Use THIS bank’s actual fields,operation/domain and types.BASE retains permissible extra metadata.Do not import LDL SAFE/set limits or a different bank’s deduplication rules." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "反查本題原始操作數、角色與必要條件；重建下一步，不把單向推論倒當等價。數值/有限表核對不是任意命題的形式證明。原題、中英來源、MKO、AST、狀態與歷史均保留。", "en": "Reverse-audit THIS question’s original operands,roles and necessary premises;reconstruct the next step,without reversing a one-way implication.Numerical/finite-table checking is not formal proof of an arbitrary theorem.Original Q,bilingual sources,MKO,AST,status and history remain preserved." } }, "arithmetic": null }, { "id": "S002", "uses": [ "G001", "S001" ], "before": "source roles = [base_case]", "after": "derived = [prove P(0)]", "relation": "implies", "rule_id": "Induction_DEFINE_each_distinct_proof_obligation", "conditions": [], "explanation": { "zh-Hant": "歸納角色必須分開：P(0)是真正起點；IH對任意固定n是條件假設，不是先假設全稱結論；step在該假設下證下一項；最後才用歸納原理下結論。", "en": "Keep induction roles distinct:P(0)is the proved start;the IH for arbitrary fixed n is a CONDITIONAL assumption,not the universal conclusion;the step derives the next case under that assumption;only then apply induction." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "反查本題原始操作數、角色與必要條件；重建下一步，不把單向推論倒當等價。數值/有限表核對不是任意命題的形式證明。原題、中英來源、MKO、AST、狀態與歷史均保留。", "en": "Reverse-audit THIS question’s original operands,roles and necessary premises;reconstruct the next step,without reversing a one-way implication.Numerical/finite-table checking is not formal proof of an arbitrary theorem.Original Q,bilingual sources,MKO,AST,status and history remain preserved." } }, "arithmetic": null }, { "id": "S003", "uses": [ "G001", "S002" ], "before": "source roles = [induction_hypothesis]", "after": "derived = [assume P(n) for arbitrary fixed n>=0]", "relation": "implies", "rule_id": "Induction_DEFINE_each_distinct_proof_obligation", "conditions": [], "explanation": { "zh-Hant": "歸納角色必須分開：P(0)是真正起點；IH對任意固定n是條件假設，不是先假設全稱結論；step在該假設下證下一項；最後才用歸納原理下結論。", "en": "Keep induction roles distinct:P(0)is the proved start;the IH for arbitrary fixed n is a CONDITIONAL assumption,not the universal conclusion;the step derives the next case under that assumption;only then apply induction." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "反查本題原始操作數、角色與必要條件；重建下一步，不把單向推論倒當等價。數值/有限表核對不是任意命題的形式證明。原題、中英來源、MKO、AST、狀態與歷史均保留。", "en": "Reverse-audit THIS question’s original operands,roles and necessary premises;reconstruct the next step,without reversing a one-way implication.Numerical/finite-table checking is not formal proof of an arbitrary theorem.Original Q,bilingual sources,MKO,AST,status and history remain preserved." } }, "arithmetic": null }, { "id": "S004", "uses": [ "G001", "S003" ], "before": "source roles = [induction_step]", "after": "derived = [under that assumption derive P(n+1)]", "relation": "implies", "rule_id": "Induction_DEFINE_each_distinct_proof_obligation", "conditions": [], "explanation": { "zh-Hant": "歸納角色必須分開：P(0)是真正起點；IH對任意固定n是條件假設，不是先假設全稱結論；step在該假設下證下一項；最後才用歸納原理下結論。", "en": "Keep induction roles distinct:P(0)is the proved start;the IH for arbitrary fixed n is a CONDITIONAL assumption,not the universal conclusion;the step derives the next case under that assumption;only then apply induction." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "反查本題原始操作數、角色與必要條件；重建下一步，不把單向推論倒當等價。數值/有限表核對不是任意命題的形式證明。原題、中英來源、MKO、AST、狀態與歷史均保留。", "en": "Reverse-audit THIS question’s original operands,roles and necessary premises;reconstruct the next step,without reversing a one-way implication.Numerical/finite-table checking is not formal proof of an arbitrary theorem.Original Q,bilingual sources,MKO,AST,status and history remain preserved." } }, "arithmetic": null }, { "id": "S005", "uses": [ "G001", "S004" ], "before": "source roles = [conclusion]", "after": "derived = [infer every n>=0 by induction after base and step]", "relation": "implies", "rule_id": "Induction_DEFINE_each_distinct_proof_obligation", "conditions": [], "explanation": { "zh-Hant": "歸納角色必須分開：P(0)是真正起點；IH對任意固定n是條件假設，不是先假設全稱結論；step在該假設下證下一項；最後才用歸納原理下結論。", "en": "Keep induction roles distinct:P(0)is the proved start;the IH for arbitrary fixed n is a CONDITIONAL assumption,not the universal conclusion;the step derives the next case under that assumption;only then apply induction." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "反查本題原始操作數、角色與必要條件；重建下一步，不把單向推論倒當等價。數值/有限表核對不是任意命題的形式證明。原題、中英來源、MKO、AST、狀態與歷史均保留。", "en": "Reverse-audit THIS question’s original operands,roles and necessary premises;reconstruct the next step,without reversing a one-way implication.Numerical/finite-table checking is not formal proof of an arbitrary theorem.Original Q,bilingual sources,MKO,AST,status and history remain preserved." } }, "arithmetic": null }, { "id": "S006", "uses": [ "G001", "S005" ], "before": "source roles = [missing_role, [base_case, induction_hypothesis, induction_step]]", "after": "derived = [conclusion]", "relation": "implies", "rule_id": "Induction_DERIVE_role_from_provided_draft", "conditions": [], "explanation": { "zh-Hant": "缺項題由完整四角色減已提供三角色得到唯一缺項；辨識題按已給單一子句角色辨識，不抄target_role充作推導。", "en": "For a missing-role task,subtract the THREE provided roles from all FOUR.For identification,use the SINGLE given clause’s role.Do not copy target_role instead of deriving it." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "反查本題原始操作數、角色與必要條件；重建下一步，不把單向推論倒當等價。數值/有限表核對不是任意命題的形式證明。原題、中英來源、MKO、AST、狀態與歷史均保留。", "en": "Reverse-audit THIS question’s original operands,roles and necessary premises;reconstruct the next step,without reversing a one-way implication.Numerical/finite-table checking is not formal proof of an arbitrary theorem.Original Q,bilingual sources,MKO,AST,status and history remain preserved." } }, "arithmetic": null }, { "id": "S007", "uses": [ "G001", "S006" ], "before": "source roles = [7, conclusion]", "after": "derived = [role identified; arbitrary theorem NOT proved]", "relation": "implies", "rule_id": "Induction_FINITE_examples_do_NOT_discharge_universal_obligation", "conditions": [], "explanation": { "zh-Hant": "這n個有限案例不是對所有自然數的證明，也不能填補缺失的base/step；本題只辨識證明義務，未替一個未給定P證成定理。", "en": "These n FINITE examples neither prove all natural cases nor discharge a missing base/step.This task identifies an obligation;it does not prove an unspecified theorem P." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "反查本題原始操作數、角色與必要條件；重建下一步，不把單向推論倒當等價。數值/有限表核對不是任意命題的形式證明。原題、中英來源、MKO、AST、狀態與歷史均保留。", "en": "Reverse-audit THIS question’s original operands,roles and necessary premises;reconstruct the next step,without reversing a one-way implication.Numerical/finite-table checking is not formal proof of an arbitrary theorem.Original Q,bilingual sources,MKO,AST,status and history remain preserved." } }, "arithmetic": null } ], "math_model": { "result": "conclusion" }, "conclusion": { "canonical_value": "conclusion", "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-20260914-domain-foundations", "batch_path": "public/data/questions/batches/batch-20260914-domain-foundations/questions.jsonl", "batch_sha256": "726b3f94ccd05d8c91cc7e306b42fd71b6d9040e194db91739a250d488fd1de7", "original_line": 67, "original_line_sha256": "4d8111f9b45778db08e16edc1c5b0aef4d30507a4f69eff3e579cb10104f38e8", "question_compact_sha256": "e48cd039cfad90c5aec8d3d6ce5b0842718e8fa556bf5d7e1fb087cc0e8003de", "english_row_compact_sha256": "aa71c385bc78adf57fbd458442bb61bded0dd1b9f5f1e83d02e07699ef8e0fbd", "english_manifest_path": "public/data/i18n/editions/batch-20260914-domain-foundations.en.json", "english_rows_path": "public/data/i18n/questions/batch-20260914-domain-foundations.en.jsonl" }, "mko_bindings": { "mko-mathematical-induction": { "sha256": "53fe3a341f5fa25780e7aa742a25087665b257b3c91a3e4f718b50f368a8dd67", "english_sha256": "18d956a6519adafa325420611bdba90cada15d1a1db00caffeca5e4b4979cada", "original_path": "public/data/mko/mathematical-induction.json", "english_path": "public/data/i18n/editions/mko-mathematical-induction.en.json" } }, "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff", "selected_part_sha256": "20c37a7e64ee5969e7606c5cc3a00e575002ebebdfe82873521cc3d7c608caff", "original_candidate_sha256": "7309e9cc6268f8b80179b71d9333b7bfb01957c80a0c3da57469e50af438c743", "original_candidate_line_sha256": "7309e9cc6268f8b80179b71d9333b7bfb01957c80a0c3da57469e50af438c743" }, "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-20261026-012-p71", "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": "9820ac6529139bd53d516c8bfdfdd41184b0cd52dad69fa433344010db37af80", "question_first_witness_sha256": "c9c939762bbd4d3ea3403084c8e95541dc2c7e56eb82fc74eaba8c80a4517dab" }, "status": "second_context_reviewed", "mathematical_semantics": true, "bilingual_semantics": true, "steps_reviewed": 7, "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction", "exact_question_first_check": { "model_sha256": "0859edcbc9c5923d70f1e039e95b2187e43697efaee4d365fbada294efe29b56", "arithmetic_steps": 0 }, "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." } }