{ "schema_version": "ocme-reviewed-question-reader-v1", "question_id": "q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0003", "pilot_number": "R0046-23528", "number": 23528, "family_id": "qf-mod43-compactclosed-hypotheses-qualification", "title": { "zh-Hant": "第 23528 題：Hausdorff空間中的緊緻集為閉集", "en": "Question 23528: Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses" }, "original_question": { "schema_version": "ocme-question-v0.1", "id": "q-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-0003", "batch_id": "batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations", "language": "zh-Hant", "source_mko_ids": [ "mko-compact-subset-closed-in-hausdorff" ], "title_zh": "Hausdorff空間中的緊緻集為閉集", "stem_zh": "ambient Hausdorff與compact subset的分離－有限subcover責任。操作=compactclosed-hypotheses-qualification；資料={\"points\":[\"a2\",\"b2\",\"c2\"],\"topology\":[[\"a2\",\"b2\",\"c2\"],[\"a2\",\"b2\"],[\"a2\",\"c2\"],[\"a2\"],[\"b2\",\"c2\"],[\"b2\"],[\"c2\"],[]],\"compact_subset\":[\"a2\",\"b2\"]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。", "answer": { "kind": "expression", "value": "([a2, b2, c2], [a2, b2], [[a2, b2, [[a2, c2], [b2]]], [a2, c2, [[a2, b2], [c2]]], [b2, c2, [[a2, b2], [c2]]]], 是, 是, complete_finite_topology_Hausdorff_and_finite_subset_compact)", "explanation_zh": "Wave6第六層核對=([a2, b2, c2], [a2, b2], [[a2, b2, [[a2, c2], [b2]]], [a2, c2, [[a2, b2], [c2]]], [b2, c2, [[a2, b2], [c2]]]], 是, 是, complete_finite_topology_Hausdorff_and_finite_subset_compact)；依完整明示資料獨立重算；對每個x不在K，finite subcover產生與K不交的open neighborhood；empty K亦closed；non-Hausdorff compact subset可非closed，且定理不主張closed implies compact。" }, "difficulty": { "band": "undergraduate_core", "level": 1, "skills": [ "Hausdorff空間中的緊緻集為閉集" ] }, "variant": { "family_id": "qf-mod43-compactclosed-hypotheses-qualification", "seed": 202611123003, "parameters": { "points": [ "a2", "b2", "c2" ], "topology": [ [ "a2", "b2", "c2" ], [ "a2", "b2" ], [ "a2", "c2" ], [ "a2" ], [ "b2", "c2" ], [ "b2" ], [ "c2" ], [] ], "compact_subset": [ "a2", "b2" ], "domain": "Finite_topological_compact_closed_certificate", "operation": "compactclosed-hypotheses-qualification", "semantic_template": "qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a2\",\"b2\"],\"points\":[\"a2\",\"b2\",\"c2\"],\"topology\":[[\"a2\",\"b2\",\"c2\"],[\"a2\",\"b2\"],[\"a2\",\"c2\"],[\"a2\"],[\"b2\",\"c2\"],[\"b2\"],[\"c2\"],[]]}" } }, "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；不構成人工審定、一般定理的有限枚舉替代、未宣告瑕積分、未明示係數律、Euler criterion循環、有限word等價、Lean或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-20261112-r03-compact-subset-closed-in-hausdorff-0003", "batch_id": "batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations", "locale": "en", "translation_state": "ai_candidate", "source": { "question_sha256": "a0ab200f9a02d9ab4f7a1eae787813f218869e1f54924f8813edf377b0227d86", "family_id": "qf-mod43-compactclosed-hypotheses-qualification", "seed": 202611123003, "semantic_template": "qf-mod43-compactclosed-hypotheses-qualification/{\"compact_subset\":[\"a2\",\"b2\"],\"points\":[\"a2\",\"b2\",\"c2\"],\"topology\":[[\"a2\",\"b2\",\"c2\"],[\"a2\",\"b2\"],[\"a2\",\"c2\"],[\"a2\"],[\"b2\",\"c2\"],[\"b2\"],[\"c2\"],[]]}" }, "fields": { "title": "Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses", "stem": "Use complete finite labeled carrier=[a2, b2, c2], topology=[[a2, b2, c2], [a2, b2], [a2, c2], [a2], [b2, c2], [b2], [c2], []] in its original supplied open-set order, and compact_subset K=[a2, b2]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate).", "answer": { "kind": "expression", "value": "([a2, b2, c2], [a2, b2], [[a2, b2, [[a2, c2], [b2]]], [a2, c2, [[a2, b2], [c2]]], [b2, c2, [[a2, b2], [c2]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)", "explanation": "Require every distinct-point pair to have disjoint open neighborhoods, retaining first witnesses in supplied topology order. Report Hausdorffness and finite-subset compactness separately; do not drop the Hausdorff hypothesis in the general theorem. The exact result, retaining the requested order, is ([a2, b2, c2], [a2, b2], [[a2, b2, [[a2, c2], [b2]]], [a2, c2, [[a2, b2], [c2]]], [b2, c2, [[a2, b2], [c2]]]], Yes, Yes, complete_finite_topology_Hausdorff_and_finite_subset_compact)." }, "skills": [ "Ambient Hausdorffness and Compactness Are Both Theorem Hypotheses" ], "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": "ambient Hausdorff與compact subset的分離－有限subcover責任。操作=compactclosed-hypotheses-qualification；資料={\"points\":[\"a2\",\"b2\",\"c2\"],\"topology\":[[\"a2\",\"b2\",\"c2\"],[\"a2\",\"b2\"],[\"a2\",\"c2\"],[\"a2\"],[\"b2\",\"c2\"],[\"b2\"],[\"c2\"],[]],\"compact_subset\":[\"a2\",\"b2\"]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。", "en": "Use complete finite labeled carrier=[a2, b2, c2], topology=[[a2, b2, c2], [a2, b2], [a2, c2], [a2], [b2, c2], [b2], [c2], []] in its original supplied open-set order, and compact_subset K=[a2, b2]. Qualify the topology and every Hausdorff distinct-point pair. Finite K is compact; the general theorem still needs both ambient Hausdorffness and compactness. Give (points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate)." } } ], "steps": [ { "id": "S001", "uses": [ "G001" ], "before": "source operation=compactclosed-hypotheses-qualification;original snapshot and mathematics retained", "after": "explicit ordered source output fields=[points, compact_subset, pair_separation_rows, Hausdorff, compact, certificate]", "relation": "implies", "rule_id": "source_ordered_interface_explicit_without_javascript", "conditions": [], "explanation": { "zh-Hant": "保留原題介面快照，本頁直接列來源欄序，不需JavaScript。radius由原mode選6／11欄，不補欄；No與實際否定依來源責任分清。", "en": "Keep the source snapshot while listing ordered fields without JavaScript.Radius uses its original mode to select6 or11 fields without padding;distinguish nonclaims from actual negations." }, "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": "ORIGINAL carrier order=[a2, b2, c2];ORIGINAL open-list order=[[a2, b2, c2], [a2, b2], [a2, c2], [a2], [b2, c2], [b2], [c2], []];point=a2;ALL given sets containing it have indices=[0, 1, 2, 3]", "after": "candidate N(a2)=intersection of ALL those sets=[a2];no original open-list reordering", "relation": "implies", "rule_id": "compactclosed_ALL_given_neighborhood_intersection_candidates", "conditions": [], "explanation": { "zh-Hant": "先由全部給定鄰域求交集候選，不先假設族已是拓撲；source保原open清單次序，內部元素才按UTF16排序。", "en": "Intersect every declared neighborhood to form a candidate,without yet assuming a topology.Retain the original open-list order;only set elements use UTF16 sorting." }, "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": "ORIGINAL carrier order=[a2, b2, c2];ORIGINAL open-list order=[[a2, b2, c2], [a2, b2], [a2, c2], [a2], [b2, c2], [b2], [c2], []];point=b2;ALL given sets containing it have indices=[0, 1, 4, 5]", "after": "candidate N(b2)=intersection of ALL those sets=[b2];no original open-list reordering", "relation": "implies", "rule_id": "compactclosed_ALL_given_neighborhood_intersection_candidates", "conditions": [], "explanation": { "zh-Hant": "先由全部給定鄰域求交集候選，不先假設族已是拓撲；source保原open清單次序，內部元素才按UTF16排序。", "en": "Intersect every declared neighborhood to form a candidate,without yet assuming a topology.Retain the original open-list order;only set elements use UTF16 sorting." }, "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": "ORIGINAL carrier order=[a2, b2, c2];ORIGINAL open-list order=[[a2, b2, c2], [a2, b2], [a2, c2], [a2], [b2, c2], [b2], [c2], []];point=c2;ALL given sets containing it have indices=[0, 2, 4, 6]", "after": "candidate N(c2)=intersection of ALL those sets=[c2];no original open-list reordering", "relation": "implies", "rule_id": "compactclosed_ALL_given_neighborhood_intersection_candidates", "conditions": [], "explanation": { "zh-Hant": "先由全部給定鄰域求交集候選，不先假設族已是拓撲；source保原open清單次序，內部元素才按UTF16排序。", "en": "Intersect every declared neighborhood to form a candidate,without yet assuming a topology.Retain the original open-list order;only set elements use UTF16 sorting." }, "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": "ORIGINAL ordered open masks=[7, 3, 5, 1, 6, 2, 4, 0];ALL neighborhood-containment [x,y,N(y)_subset_N(x)] rows=[[a2, a2, 是], [b2, b2, 是], [c2, c2, 是]]", "after": "ALL carrier masks0..7 audited;EXACT qualifying U with every x in U having N(x) subset U=[0, 1, 2, 3, 4, 5, 6, 7];given family equals this COMPLETE upset family", "relation": "implies", "rule_id": "compactclosed_ALL_carrier_subsets_exact_neighborhood_upsets_topology", "conditions": [], "explanation": { "zh-Hant": "有限拓撲的open恰是其最小鄰域的upsets；反向逐全部carrier子集合比對後，upsets在任意union及有限intersection下封閉，含空／全，故得到真正拓撲與N(x) open。這涵蓋全部open-pair責任，不只列取樣，也不必打印數百萬pairs。", "en": "In a finite topology,opens are exactly upsets of its minimal neighborhoods.Conversely,after comparing every carrier subset,upsets are closed under arbitrary unions and finite intersections and contain empty/full sets,so this is a topology with open N(x).This covers all closure obligations,not sampled pairs,without printing millions of pairs." }, "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": "Compact-closed ORIGINAL pair index0-1:ORIGINAL distinct x=a2,y=b2;N(y)=[b2];open-list order=[[a2, b2, c2], [a2, b2], [a2, c2], [a2], [b2, c2], [b2], [c2], []]", "after": "ALL tested [U_index,U_intersect_N(y),disjoint_V_possible] rows=[[0, [b2], 否], [1, [b2], 否], [2, [], 是]];for FIRST possible U,ALL tested [V_index,intersection,disjoint] rows=[[0, [a2, c2], 否], [1, [a2], 否], [4, [c2], 否], [5, [], 是]];FIRST ORIGINAL witness=[[a2, c2], [b2]]", "relation": "implies", "rule_id": "compactclosed_FIRST_original_separator_with_complete_neighborhood_screen", "conditions": [], "explanation": { "zh-Hant": "若U交N(y)非空，所有含y的open都含N(y)，故不可能與U不交；若為空，N(y)本身是可用open，再按原V順序找首個。這與完整原U/V搜尋同一首見證，不是另選較小鄰域代答案。", "en": "If U meets N(y),every open containing y also meets U.If disjoint,N(y) itself is an available open,then search original V order for the first witness.This preserves exactly the original full U/V search result,not an alternate minimal-neighborhood answer." }, "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": "Compact-closed ORIGINAL pair index0-2:ORIGINAL distinct x=a2,y=c2;N(y)=[c2];open-list order=[[a2, b2, c2], [a2, b2], [a2, c2], [a2], [b2, c2], [b2], [c2], []]", "after": "ALL tested [U_index,U_intersect_N(y),disjoint_V_possible] rows=[[0, [c2], 否], [1, [], 是]];for FIRST possible U,ALL tested [V_index,intersection,disjoint] rows=[[0, [a2, b2], 否], [2, [a2], 否], [4, [b2], 否], [6, [], 是]];FIRST ORIGINAL witness=[[a2, b2], [c2]]", "relation": "implies", "rule_id": "compactclosed_FIRST_original_separator_with_complete_neighborhood_screen", "conditions": [], "explanation": { "zh-Hant": "若U交N(y)非空，所有含y的open都含N(y)，故不可能與U不交；若為空，N(y)本身是可用open，再按原V順序找首個。這與完整原U/V搜尋同一首見證，不是另選較小鄰域代答案。", "en": "If U meets N(y),every open containing y also meets U.If disjoint,N(y) itself is an available open,then search original V order for the first witness.This preserves exactly the original full U/V search result,not an alternate minimal-neighborhood answer." }, "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": "Compact-closed ORIGINAL pair index1-2:ORIGINAL distinct x=b2,y=c2;N(y)=[c2];open-list order=[[a2, b2, c2], [a2, b2], [a2, c2], [a2], [b2, c2], [b2], [c2], []]", "after": "ALL tested [U_index,U_intersect_N(y),disjoint_V_possible] rows=[[0, [c2], 否], [1, [], 是]];for FIRST possible U,ALL tested [V_index,intersection,disjoint] rows=[[0, [a2, b2], 否], [2, [a2], 否], [4, [b2], 否], [6, [], 是]];FIRST ORIGINAL witness=[[a2, b2], [c2]]", "relation": "implies", "rule_id": "compactclosed_FIRST_original_separator_with_complete_neighborhood_screen", "conditions": [], "explanation": { "zh-Hant": "若U交N(y)非空，所有含y的open都含N(y)，故不可能與U不交；若為空，N(y)本身是可用open，再按原V順序找首個。這與完整原U/V搜尋同一首見證，不是另選較小鄰域代答案。", "en": "If U meets N(y),every open containing y also meets U.If disjoint,N(y) itself is an available open,then search original V order for the first witness.This preserves exactly the original full U/V search result,not an alternate minimal-neighborhood answer." }, "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": "ORIGINAL points=[a2, b2, c2];ALL original-index distinct-pair witnesses=[[a2, b2, [[a2, c2], [b2]]], [a2, c2, [[a2, b2], [c2]]], [b2, c2, [[a2, b2], [c2]]]];actual K=[a2, b2]", "after": "Hausdorff=是;K compact=是 by finiteness:from ANY cover choose one member per K point;empty K needs no members;ambient complement=[c2],closed=是", "relation": "implies", "rule_id": "compactclosed_ALL_pairs_Hausdorff_and_finite_subset_compact_separate", "conditions": [], "explanation": { "zh-Hant": "有限K的compact性不替代母空間Hausdorff責任；pair空域vacuous True。Closed查同一母空間的補集open，不擅自取相對補集。", "en": "Finite K compactness does not replace ambient Hausdorffness.An empty pair domain is vacuous.Closedness tests the complement in this same ambient space,not a relative complement." }, "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": "CURRENT ambient Hausdorff=是;compact K=[a2, b2];consider ANY exterior x", "after": "for EACH y in K choose disjoint opens U_y containing x,V_y containing y;the V_y cover K;compactness picks FINITELY MANY indices;their U_y intersection is open,contains x and avoids ALL K,so X minus K is open", "relation": "implies", "rule_id": "compactclosed_GENERAL_Hausdorff_finite_subcover_intersection_proof", "conditions": [], "explanation": { "zh-Hant": "一般定理靠compact選有限subcover，不是只靠此有限模型的數值。每個K點由所選某個V覆蓋，對應U與V不交；有限U交便排除全部K。空K的空交是X。此方向不主張closed implies compact。", "en": "The general theorem uses compactness to select a finite subcover,not numbers from this finite model.Each K point lies in a selected V disjoint from its corresponding U;the finite intersection therefore avoids all K.For empty K,the empty intersection is X.No converse closed-implies-compact is claimed." }, "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": [ [ "a2", "b2", "c2" ], [ "a2", "b2" ], [ [ "a2", "b2", [ [ "a2", "c2" ], [ "b2" ] ] ], [ "a2", "c2", [ [ "a2", "b2" ], [ "c2" ] ] ], [ "b2", "c2", [ [ "a2", "b2" ], [ "c2" ] ] ] ], true, true, "complete_finite_topology_Hausdorff_and_finite_subset_compact" ] }, "conclusion": { "canonical_value": "([a2, b2, c2], [a2, b2], [[a2, b2, [[a2, c2], [b2]]], [a2, c2, [[a2, b2], [c2]]], [b2, c2, [[a2, b2], [c2]]]], 是, 是, complete_finite_topology_Hausdorff_and_finite_subset_compact)", "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-20261112-r03-compact-subset-closed-in-hausdorff-foundations", "batch_path": "public/data/questions/batches/batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations/questions.jsonl", "batch_sha256": "f646f6e2479be6c863608dfdb33483d0d5a6aa88b59c9106e7139dc32f74db36", "original_line": 3, "original_line_sha256": "7904e5a06e0db14b5d12964a51cb6eafbfbc2fe0344ecc1f7136992f74058ef7", "question_compact_sha256": "a0ab200f9a02d9ab4f7a1eae787813f218869e1f54924f8813edf377b0227d86", "english_row_compact_sha256": "ea393782161b7bf72e9a4b621f0c94fccf975935472856a6638db35bbe621de1", "english_manifest_path": "public/data/i18n/editions/batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations.en.json", "english_rows_path": "public/data/i18n/questions/batch-20261014-advance-20261112-r03-compact-subset-closed-in-hausdorff-foundations.en.jsonl" }, "mko_bindings": { "mko-compact-subset-closed-in-hausdorff": { "sha256": "13e0a8f9917f4a49a9cbe148c344b3f37d319fb16c750de2bb640ec93553d07d", "english_sha256": "b4153bb071eb67ca522480652f8411f2eb944d99e54b0bb5043b582af432b244", "original_path": "public/data/mko/compact-subset-closed-in-hausdorff.json", "english_path": "public/data/i18n/editions/mko-compact-subset-closed-in-hausdorff.en.json" } }, "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff", "selected_part_sha256": "04d5fe00084f382c9e6853d5f11c80b558d9cf857160ceb644fa20b64948d1fe", "original_candidate_sha256": "8bc0fe652e71940524f906defb8a7043da78ec7cc90954fcbe9d3c4d44aa78f5", "original_candidate_line_sha256": "8bc0fe652e71940524f906defb8a7043da78ec7cc90954fcbe9d3c4d44aa78f5" }, "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-20261026-009-p47", "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": "a7c0d888ade6d307703c4222c4137c78c0d2448babdcba2fa3446fe6fdd343c1", "question_first_witness_sha256": "c9571f97a37d503ad6a61034cc00108543a05c37ef246d1c4962f55136a3de02" }, "status": "second_context_reviewed", "mathematical_semantics": true, "bilingual_semantics": true, "steps_reviewed": 10, "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction", "exact_question_first_check": { "model_sha256": "984c48ff90608a38e41f708cc2106f4c5b4b85b91738ec39298d6efd6bebe6f4", "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." } }