{ "schema_version": "ocme-reviewed-question-reader-v1", "question_id": "q-20261014-advance-20261102-r04-topological-boundary-0036", "pilot_number": "R0032-16786", "number": 16786, "family_id": "qf-mod33-boundary-membership", "title": { "zh-Hant": "第 16786 題：拓撲邊界", "en": "Question 16786: Boundary Membership Is Not Just Membership in the Original Set" }, "original_question": { "schema_version": "ocme-question-v0.1", "id": "q-20261014-advance-20261102-r04-topological-boundary-0036", "batch_id": "batch-20261014-advance-20261102-r04-topological-boundary-foundations", "language": "zh-Hant", "source_mko_ids": [ "mko-topological-boundary" ], "title_zh": "拓撲邊界", "stem_zh": "完整有限拓撲與同一載體子集。操作=boundary-membership；資料={\"n\":5,\"topology\":[[],[0],[0,2],[0,2,4],[0,1,2,4],[0,1,2,3,4]],\"subset\":[2,4],\"point\":0}。請依 [point,boundary,in_boundary,scope] 欄序回答。", "answer": { "kind": "expression", "value": "(0, [1, 2, 3, 4], 否, topology_relative_boundary)", "explanation_zh": "Wave5第二層核對=(0, [1, 2, 3, 4], 否, topology_relative_boundary)；依完整明示資料重算；boundary等於closure扣除interior且與補集boundary相同；空集與全空間boundary為空；結果相對拓撲而非圖形外觀。" }, "difficulty": { "band": "undergraduate_core", "level": 2, "skills": [ "拓撲邊界" ] }, "variant": { "family_id": "qf-mod33-boundary-membership", "seed": 202611024036, "parameters": { "n": 5, "topology": [ [], [ 0 ], [ 0, 2 ], [ 0, 2, 4 ], [ 0, 1, 2, 4 ], [ 0, 1, 2, 3, 4 ] ], "subset": [ 2, 4 ], "point": 0, "domain": "Complete_finite_topological_boundary", "operation": "boundary-membership", "semantic_template": "qf-mod33-boundary-membership/{\"n\":5,\"point\":0,\"subset\":[2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,2,4],[0,2],[0]]}" } }, "verification": { "status": "candidate", "checks": [ "producer_exact_finite_or_symbolic_model", "producer_source_snapshot_check", "producer_substantive_template_deduplication", "producer_json_schema" ], "warning_zh": "候選須由主端獨立重算；不構成人工審定、任意維basis theorem、bounded-to-global kernel/image、一般compactness、因果推論、language equivalence、Lean或形式Evidence。" }, "provenance": { "generator_role": "question_producer", "generator_model": "model_unknown", "generated_at": "2026-10-14", "contract_version": "ocme-question-factory-v0.1" } }, "original_english_edition": { "schema_version": "ocme-question-localization-v0.1", "question_id": "q-20261014-advance-20261102-r04-topological-boundary-0036", "batch_id": "batch-20261014-advance-20261102-r04-topological-boundary-foundations", "locale": "en", "translation_state": "ai_candidate", "source": { "question_sha256": "160bb232368f6348a7d5d91e7db4dab13f3bd45da7e8f5cd8e803895ce466aaf", "family_id": "qf-mod33-boundary-membership", "seed": 202611024036, "semantic_template": "qf-mod33-boundary-membership/{\"n\":5,\"point\":0,\"subset\":[2,4],\"topology\":[[],[0,1,2,3,4],[0,1,2,4],[0,2,4],[0,2],[0]]}" }, "fields": { "title": "Boundary Membership Is Not Just Membership in the Original Set", "stem": "Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Subset=[2, 4], point=0. Give (point,boundary,in_boundary?,topology_relative_boundary).", "answer": { "kind": "expression", "value": "(0, [1, 2, 3, 4], No, topology_relative_boundary)", "explanation": "Compute the complete topology-relative boundary first, then test the point. Do not infer boundary status from an unstated metric or a drawing. The exact result, retaining the requested order, is (0, [1, 2, 3, 4], No, topology_relative_boundary)." }, "skills": [ "Boundary Membership Is Not Just Membership in the Original Set" ], "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": "完整有限拓撲與同一載體子集。操作=boundary-membership；資料={\"n\":5,\"topology\":[[],[0],[0,2],[0,2,4],[0,1,2,4],[0,1,2,3,4]],\"subset\":[2,4],\"point\":0}。請依 [point,boundary,in_boundary,scope] 欄序回答。", "en": "Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Subset=[2, 4], point=0. Give (point,boundary,in_boundary?,topology_relative_boundary)." } } ], "steps": [ { "id": "S001", "uses": [ "G001" ], "before": "same complete carrier Fin(5);original ordered complete topology=[[], [0], [0, 2], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]]", "after": "unique empty/whole opens=是;ordered open pairs checked=36;first union failure=null;first intersection failure=null", "relation": "implies", "rule_id": "topology_complete_finite_parent_all_pairs", "conditions": [], "explanation": { "zh-Hant": "個別子集須照原載體遞增且無重複，拓撲族須有唯一空集、全集並核全部二元聯／交。因不同開集有限，任意聯只涉及有限種開集，二元封閉加空集便足夠；不另供度量。保留拓撲原順序。", "en": "Each subset is increasing and duplicate-free.The topology has unique empty/whole opens and closure under every binary union/intersection.Finitely many distinct opens reduce arbitrary unions to finite ones,handled by binary closure and the empty set.No metric is supplied.Keep original topology order." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S002", "uses": [ "G001", "S001" ], "before": "boundary subset:original open index=0;open=[];same-carrier closed complement=[0, 1, 2, 3, 4];subset=[2, 4]", "after": "closed set contains whole subset=是;running intersection of qualifying closed sets=[0, 1, 2, 3, 4]", "relation": "implies", "rule_id": "closure_every_closed_superset_original_open_order", "conditions": [], "explanation": { "zh-Hant": "逐每個原開集取同載體補集成閉集，只有包含整個A才保留；反覆交所有合格者，而非挑一個方便閉超集。即使某點不在A，仍可能在最後交集中。", "en": "Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S003", "uses": [ "G001", "S002" ], "before": "boundary subset:original open index=1;open=[0];same-carrier closed complement=[1, 2, 3, 4];subset=[2, 4]", "after": "closed set contains whole subset=是;running intersection of qualifying closed sets=[1, 2, 3, 4]", "relation": "implies", "rule_id": "closure_every_closed_superset_original_open_order", "conditions": [], "explanation": { "zh-Hant": "逐每個原開集取同載體補集成閉集，只有包含整個A才保留；反覆交所有合格者，而非挑一個方便閉超集。即使某點不在A，仍可能在最後交集中。", "en": "Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S004", "uses": [ "G001", "S003" ], "before": "boundary subset:original open index=2;open=[0, 2];same-carrier closed complement=[1, 3, 4];subset=[2, 4]", "after": "closed set contains whole subset=否;running intersection of qualifying closed sets=[1, 2, 3, 4]", "relation": "implies", "rule_id": "closure_every_closed_superset_original_open_order", "conditions": [], "explanation": { "zh-Hant": "逐每個原開集取同載體補集成閉集，只有包含整個A才保留；反覆交所有合格者，而非挑一個方便閉超集。即使某點不在A，仍可能在最後交集中。", "en": "Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S005", "uses": [ "G001", "S004" ], "before": "boundary subset:original open index=3;open=[0, 2, 4];same-carrier closed complement=[1, 3];subset=[2, 4]", "after": "closed set contains whole subset=否;running intersection of qualifying closed sets=[1, 2, 3, 4]", "relation": "implies", "rule_id": "closure_every_closed_superset_original_open_order", "conditions": [], "explanation": { "zh-Hant": "逐每個原開集取同載體補集成閉集，只有包含整個A才保留；反覆交所有合格者，而非挑一個方便閉超集。即使某點不在A，仍可能在最後交集中。", "en": "Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S006", "uses": [ "G001", "S005" ], "before": "boundary subset:original open index=4;open=[0, 1, 2, 4];same-carrier closed complement=[3];subset=[2, 4]", "after": "closed set contains whole subset=否;running intersection of qualifying closed sets=[1, 2, 3, 4]", "relation": "implies", "rule_id": "closure_every_closed_superset_original_open_order", "conditions": [], "explanation": { "zh-Hant": "逐每個原開集取同載體補集成閉集，只有包含整個A才保留；反覆交所有合格者，而非挑一個方便閉超集。即使某點不在A，仍可能在最後交集中。", "en": "Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S007", "uses": [ "G001", "S006" ], "before": "boundary subset:original open index=5;open=[0, 1, 2, 3, 4];same-carrier closed complement=[];subset=[2, 4]", "after": "closed set contains whole subset=否;running intersection of qualifying closed sets=[1, 2, 3, 4]", "relation": "implies", "rule_id": "closure_every_closed_superset_original_open_order", "conditions": [], "explanation": { "zh-Hant": "逐每個原開集取同載體補集成閉集，只有包含整個A才保留；反覆交所有合格者，而非挑一個方便閉超集。即使某點不在A，仍可能在最後交集中。", "en": "Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S008", "uses": [ "G001", "S007" ], "before": "boundary subset:ALL qualifying closed supersets in original complement order=[[0, 1, 2, 3, 4], [1, 2, 3, 4]]", "after": "topology-relative closure=[1, 2, 3, 4]", "relation": "implies", "rule_id": "closure_all_candidates_intersection_not_metric_guess", "conditions": [], "explanation": { "zh-Hant": "全集是閉超集，所以交的家族不缺；所得是相對這份拓撲最小閉超集，原序只影響候選列表，不改集合值。", "en": "The whole carrier is a closed superset,so the family is present.The intersection is the least closed superset in this topology.Original order affects the candidate list,not its set value." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S009", "uses": [ "G001", "S008" ], "before": "boundary subset:original open index=0;open=[];subset=[2, 4]", "after": "whole open contained in subset=是;running union of qualifying opens=[]", "relation": "implies", "rule_id": "interior_every_contained_open_original_order", "conditions": [], "explanation": { "zh-Hant": "須整個開集包含在A才合格，不是只跟A相交。把全部合格開集聯起來，才有相對此拓撲的最大開子集。", "en": "The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S010", "uses": [ "G001", "S009" ], "before": "boundary subset:original open index=1;open=[0];subset=[2, 4]", "after": "whole open contained in subset=否;running union of qualifying opens=[]", "relation": "implies", "rule_id": "interior_every_contained_open_original_order", "conditions": [], "explanation": { "zh-Hant": "須整個開集包含在A才合格，不是只跟A相交。把全部合格開集聯起來，才有相對此拓撲的最大開子集。", "en": "The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S011", "uses": [ "G001", "S010" ], "before": "boundary subset:original open index=2;open=[0, 2];subset=[2, 4]", "after": "whole open contained in subset=否;running union of qualifying opens=[]", "relation": "implies", "rule_id": "interior_every_contained_open_original_order", "conditions": [], "explanation": { "zh-Hant": "須整個開集包含在A才合格，不是只跟A相交。把全部合格開集聯起來，才有相對此拓撲的最大開子集。", "en": "The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S012", "uses": [ "G001", "S011" ], "before": "boundary subset:original open index=3;open=[0, 2, 4];subset=[2, 4]", "after": "whole open contained in subset=否;running union of qualifying opens=[]", "relation": "implies", "rule_id": "interior_every_contained_open_original_order", "conditions": [], "explanation": { "zh-Hant": "須整個開集包含在A才合格，不是只跟A相交。把全部合格開集聯起來，才有相對此拓撲的最大開子集。", "en": "The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S013", "uses": [ "G001", "S012" ], "before": "boundary subset:original open index=4;open=[0, 1, 2, 4];subset=[2, 4]", "after": "whole open contained in subset=否;running union of qualifying opens=[]", "relation": "implies", "rule_id": "interior_every_contained_open_original_order", "conditions": [], "explanation": { "zh-Hant": "須整個開集包含在A才合格，不是只跟A相交。把全部合格開集聯起來，才有相對此拓撲的最大開子集。", "en": "The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S014", "uses": [ "G001", "S013" ], "before": "boundary subset:original open index=5;open=[0, 1, 2, 3, 4];subset=[2, 4]", "after": "whole open contained in subset=否;running union of qualifying opens=[]", "relation": "implies", "rule_id": "interior_every_contained_open_original_order", "conditions": [], "explanation": { "zh-Hant": "須整個開集包含在A才合格，不是只跟A相交。把全部合格開集聯起來，才有相對此拓撲的最大開子集。", "en": "The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S015", "uses": [ "G001", "S014" ], "before": "boundary subset:ALL contained opens in original topology order=[[]]", "after": "topology-relative interior=[]", "relation": "implies", "rule_id": "interior_all_candidates_union_not_subset_membership", "conditions": [], "explanation": { "zh-Hant": "空開集永遠合格；結果只依原拓撲，不把A內每個點都自動當內點。", "en": "The empty open always qualifies.The result depends on the specified topology;membership in A alone does not make a point interior." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S016", "uses": [ "G001", "S015" ], "before": "boundary subset:same-topology closure=[1, 2, 3, 4];same-topology interior=[]", "after": "boundary=closure minus interior=[1, 2, 3, 4]", "relation": "implies", "rule_id": "boundary_same_topology_closure_minus_interior", "conditions": [], "explanation": { "zh-Hant": "把內部從閉包扣掉，不是從原A扣掉；非A成員仍可能是邊界點，結果沒有使用歐氏圖形或度量。", "en": "Subtract the interior from the closure,not from A itself.A point outside A can be a boundary point.No Euclidean drawing or metric is being used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S017", "uses": [ "G001", "S016" ], "before": "original point=0;original subset=[2, 4];complete relative boundary=[1, 2, 3, 4]", "after": "point in boundary=否;topology_relative_boundary", "relation": "implies", "rule_id": "boundary_point_not_original_membership_or_picture", "conditions": [], "explanation": { "zh-Hant": "只對已算出的完整邊界作成員測試，不按點是否在A或圖上看似邊緣來判。", "en": "Test the computed whole boundary,not membership in A or a visual impression of an edge." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null } ], "math_model": { "result": [ 0, [ 1, 2, 3, 4 ], false, "topology_relative_boundary" ] }, "conclusion": { "canonical_value": "(0, [1, 2, 3, 4], 否, topology_relative_boundary)", "explanation": { "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。", "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged." } }, "definition_note": { "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。", "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate." }, "source": { "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db", "question_binding": { "batch_id": "batch-20261014-advance-20261102-r04-topological-boundary-foundations", "batch_path": "public/data/questions/batches/batch-20261014-advance-20261102-r04-topological-boundary-foundations/questions.jsonl", "batch_sha256": "9d7eda81e75fc8dc1c598acc75ff8f07f47d136fcf7b8177617ba4e6df1fbbba", "original_line": 36, "original_line_sha256": "7938475b600a9876c3fb7fa7077bd6188abbc6e655054bbd8d2f84fcb8d1e3f0", "question_compact_sha256": "160bb232368f6348a7d5d91e7db4dab13f3bd45da7e8f5cd8e803895ce466aaf", "english_row_compact_sha256": "b10413e1e913264feea28437bc52e711c2ec39740b588bf93e17162aa7ae859d", "english_manifest_path": "public/data/i18n/editions/batch-20261014-advance-20261102-r04-topological-boundary-foundations.en.json", "english_rows_path": "public/data/i18n/questions/batch-20261014-advance-20261102-r04-topological-boundary-foundations.en.jsonl" }, "mko_bindings": { "mko-topological-boundary": { "sha256": "31f941e14ac65e7366835ceef8264cd169c8739bdaa6e04182c4f7b160ec8a13", "english_sha256": "c6431291562b0578aba8220926ac7692cbde83c7b924a496bb5e74c445028500", "original_path": "public/data/mko/topological-boundary.json", "english_path": "public/data/i18n/editions/mko-topological-boundary.en.json" } }, "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff", "selected_part_sha256": "fe4c48735753bdd58c10db9b2ca3a2a3baeb31d0424c3fc09dea62317185ddae", "original_candidate_sha256": "5a986b178a213ce66dc8068a59f8c7969438400717d230ba474a8ed639145de6", "original_candidate_line_sha256": "5a986b178a213ce66dc8068a59f8c7969438400717d230ba474a8ed639145de6" }, "historical_source_state": { "review_state": { "derivation": "candidate_authored_and_locally_checked", "source_question_state": "candidate", "independent_AI_or_human_review": "NOT_PERFORMED", "formalization": "not_formalized", "Lean": "NOT_RUN" }, "scope_note": { "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。", "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged." }, "label": "Frozen producer metadata; current per-question review is separate below." }, "revision": { "id": "review-20261026-007-p33", "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": "d64b4e08bda1c25dd393b7c7e4140a8734610a9b84ae47e6f0adcff7f30a0b8e", "question_first_witness_sha256": "790fd6623c04ba52a6ffb0a95cea52b532a02b23bb9453926d2d9f7a825fdec8" }, "status": "second_context_reviewed", "mathematical_semantics": true, "bilingual_semantics": true, "steps_reviewed": 17, "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction", "exact_question_first_check": { "model_sha256": "437578085704ff4e9b70c7e19a12f3ec681dfff7e9ab89724b11f385847778ec", "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." } }