{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0001","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A=∅、B=∅，函數表為 ∅。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"單射=是；滿射=是。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052001,"parameters":{"A":[],"B":[],"pairs":[],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[],\"B\":[],\"operation\":\"surjective\",\"pairs\":[]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0002","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={0, 1}，函數表為 {(-1, 1), (0, 0)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"單射=是；滿射=是。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052002,"parameters":{"A":[-1,0],"B":[0,1],"pairs":[[-1,1],[0,0]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-1,0],\"B\":[0,1],\"operation\":\"surjective\",\"pairs\":[[-1,1],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0003","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={1, 3, 5}，函數表為 {(-1, 1), (0, 3)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"單射=是；滿射=否。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052003,"parameters":{"A":[-1,0],"B":[1,3,5],"pairs":[[-1,1],[0,3]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-1,0],\"B\":[1,3,5],\"operation\":\"surjective\",\"pairs\":[[-1,1],[0,3]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0004","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={0}，函數表為 {(-1, 0), (0, 0)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"單射=否；滿射=是。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052004,"parameters":{"A":[-1,0],"B":[0],"pairs":[[-1,0],[0,0]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-1,0],\"B\":[0],\"operation\":\"surjective\",\"pairs\":[[-1,0],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0005","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={-3, -2, -1}，函數表為 {(-1, -1), (0, -1)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"單射=否；滿射=否。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052005,"parameters":{"A":[-1,0],"B":[-3,-2,-1],"pairs":[[-1,-1],[0,-1]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-1,0],\"B\":[-3,-2,-1],\"operation\":\"surjective\",\"pairs\":[[-1,-1],[0,-1]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0006","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={0, 1, 2}，函數表為 {(-2, 1), (-1, 2), (0, 0)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"單射=是；滿射=是。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052006,"parameters":{"A":[-2,-1,0],"B":[0,1,2],"pairs":[[-2,1],[-1,2],[0,0]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-2,-1,0],\"B\":[0,1,2],\"operation\":\"surjective\",\"pairs\":[[-2,1],[-1,2],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0007","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={1, 3, 5, 7}，函數表為 {(-2, 1), (-1, 3), (0, 5)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"單射=是；滿射=否。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052007,"parameters":{"A":[-2,-1,0],"B":[1,3,5,7],"pairs":[[-2,1],[-1,3],[0,5]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-2,-1,0],\"B\":[1,3,5,7],\"operation\":\"surjective\",\"pairs\":[[-2,1],[-1,3],[0,5]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0008","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={0, 1}，函數表為 {(-2, 0), (-1, 1), (0, 0)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"單射=否；滿射=是。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052008,"parameters":{"A":[-2,-1,0],"B":[0,1],"pairs":[[-2,0],[-1,1],[0,0]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-2,-1,0],\"B\":[0,1],\"operation\":\"surjective\",\"pairs\":[[-2,0],[-1,1],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0009","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={-4, -3, -2, -1}，函數表為 {(-2, -1), (-1, -1), (0, -2)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"單射=否；滿射=否。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052009,"parameters":{"A":[-2,-1,0],"B":[-4,-3,-2,-1],"pairs":[[-2,-1],[-1,-1],[0,-2]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-2,-1,0],\"B\":[-4,-3,-2,-1],\"operation\":\"surjective\",\"pairs\":[[-2,-1],[-1,-1],[0,-2]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0010","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={0, 1, 2, 3}，函數表為 {(-3, 1), (-2, 2), (-1, 3), (0, 0)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"單射=是；滿射=是。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052010,"parameters":{"A":[-3,-2,-1,0],"B":[0,1,2,3],"pairs":[[-3,1],[-2,2],[-1,3],[0,0]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-3,-2,-1,0],\"B\":[0,1,2,3],\"operation\":\"surjective\",\"pairs\":[[-3,1],[-2,2],[-1,3],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0011","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={1, 3, 5, 7, 9}，函數表為 {(-3, 1), (-2, 3), (-1, 5), (0, 7)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"單射=是；滿射=否。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052011,"parameters":{"A":[-3,-2,-1,0],"B":[1,3,5,7,9],"pairs":[[-3,1],[-2,3],[-1,5],[0,7]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-3,-2,-1,0],\"B\":[1,3,5,7,9],\"operation\":\"surjective\",\"pairs\":[[-3,1],[-2,3],[-1,5],[0,7]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0012","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={0, 1, 2}，函數表為 {(-3, 0), (-2, 1), (-1, 2), (0, 0)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"單射=否；滿射=是。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052012,"parameters":{"A":[-3,-2,-1,0],"B":[0,1,2],"pairs":[[-3,0],[-2,1],[-1,2],[0,0]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-3,-2,-1,0],\"B\":[0,1,2],\"operation\":\"surjective\",\"pairs\":[[-3,0],[-2,1],[-1,2],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0013","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={-5, -4, -3, -2, -1}，函數表為 {(-3, -1), (-2, -1), (-1, -2), (0, -2)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"單射=否；滿射=否。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052013,"parameters":{"A":[-3,-2,-1,0],"B":[-5,-4,-3,-2,-1],"pairs":[[-3,-1],[-2,-1],[-1,-2],[0,-2]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-3,-2,-1,0],\"B\":[-5,-4,-3,-2,-1],\"operation\":\"surjective\",\"pairs\":[[-3,-1],[-2,-1],[-1,-2],[0,-2]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0014","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4}，函數表為 {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"單射=是；滿射=是。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052014,"parameters":{"A":[-4,-3,-2,-1,0],"B":[0,1,2,3,4],"pairs":[[-4,1],[-3,2],[-2,3],[-1,4],[0,0]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-4,-3,-2,-1,0],\"B\":[0,1,2,3,4],\"operation\":\"surjective\",\"pairs\":[[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0015","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={1, 3, 5, 7, 9, 11}，函數表為 {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"單射=是；滿射=否。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052015,"parameters":{"A":[-4,-3,-2,-1,0],"B":[1,3,5,7,9,11],"pairs":[[-4,1],[-3,3],[-2,5],[-1,7],[0,9]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11],\"operation\":\"surjective\",\"pairs\":[[-4,1],[-3,3],[-2,5],[-1,7],[0,9]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0016","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={0, 1, 2, 3}，函數表為 {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"單射=否；滿射=是。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052016,"parameters":{"A":[-4,-3,-2,-1,0],"B":[0,1,2,3],"pairs":[[-4,0],[-3,1],[-2,2],[-1,3],[0,0]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-4,-3,-2,-1,0],\"B\":[0,1,2,3],\"operation\":\"surjective\",\"pairs\":[[-4,0],[-3,1],[-2,2],[-1,3],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0017","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={-6, -5, -4, -3, -2, -1}，函數表為 {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"單射=否；滿射=否。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052017,"parameters":{"A":[-4,-3,-2,-1,0],"B":[-6,-5,-4,-3,-2,-1],"pairs":[[-4,-1],[-3,-1],[-2,-2],[-1,-2],[0,-3]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-4,-3,-2,-1,0],\"B\":[-6,-5,-4,-3,-2,-1],\"operation\":\"surjective\",\"pairs\":[[-4,-1],[-3,-1],[-2,-2],[-1,-2],[0,-3]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0018","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4, 5}，函數表為 {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"單射=是；滿射=是。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052018,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4,5],"pairs":[[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5],\"operation\":\"surjective\",\"pairs\":[[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0019","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={1, 3, 5, 7, 9, 11, 13}，函數表為 {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"單射=是；滿射=否。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052019,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[1,3,5,7,9,11,13],"pairs":[[-5,1],[-4,3],[-3,5],[-2,7],[-1,9],[0,11]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11,13],\"operation\":\"surjective\",\"pairs\":[[-5,1],[-4,3],[-3,5],[-2,7],[-1,9],[0,11]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0020","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4}，函數表為 {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"單射=否；滿射=是。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052020,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4],"pairs":[[-5,0],[-4,1],[-3,2],[-2,3],[-1,4],[0,0]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4],\"operation\":\"surjective\",\"pairs\":[[-5,0],[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0021","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={-7, -6, -5, -4, -3, -2, -1}，函數表為 {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"單射=否；滿射=否。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052021,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[-7,-6,-5,-4,-3,-2,-1],"pairs":[[-5,-1],[-4,-1],[-3,-2],[-2,-2],[-1,-3],[0,-3]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[-7,-6,-5,-4,-3,-2,-1],\"operation\":\"surjective\",\"pairs\":[[-5,-1],[-4,-1],[-3,-2],[-2,-2],[-1,-3],[0,-3]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0022","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4, 5, 6}，函數表為 {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"單射=是；滿射=是。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052022,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4,5,6],"pairs":[[-6,1],[-5,2],[-4,3],[-3,4],[-2,5],[-1,6],[0,0]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5,6],\"operation\":\"surjective\",\"pairs\":[[-6,1],[-5,2],[-4,3],[-3,4],[-2,5],[-1,6],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0023","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={1, 3, 5, 7, 9, 11, 13, 15}，函數表為 {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"單射=是；滿射=否。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052023,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[1,3,5,7,9,11,13,15],"pairs":[[-6,1],[-5,3],[-4,5],[-3,7],[-2,9],[-1,11],[0,13]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11,13,15],\"operation\":\"surjective\",\"pairs\":[[-6,1],[-5,3],[-4,5],[-3,7],[-2,9],[-1,11],[0,13]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0024","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4, 5}，函數表為 {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"單射=否；滿射=是。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052024,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4,5],"pairs":[[-6,0],[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5],\"operation\":\"surjective\",\"pairs\":[[-6,0],[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0025","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"有限總函數的滿射性","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={-8, -7, -6, -5, -4, -3, -2, -1}，函數表為 {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}。f 是否覆蓋宣告陪域中每個目標而為滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"單射=否；滿射=否。必須逐一查全部 B；多個原像可以存在，不把實際值域自動改名成陪域。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["有限總函數的滿射性"]},"variant":{"family_id":"qf-surjective-table-property","seed":202610052025,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[-8,-7,-6,-5,-4,-3,-2,-1],"pairs":[[-6,-1],[-5,-1],[-4,-2],[-3,-2],[-2,-3],[-1,-3],[0,-4]],"operation":"surjective","semantic_template":"qf-surjective-table-property/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[-8,-7,-6,-5,-4,-3,-2,-1],\"operation\":\"surjective\",\"pairs\":[[-6,-1],[-5,-1],[-4,-2],[-3,-2],[-2,-3],[-1,-3],[0,-4]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0026","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A=∅、B=∅，函數表為 ∅。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"∅","explanation_zh":"缺少目標=∅。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052026,"parameters":{"A":[],"B":[],"pairs":[],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[],\"B\":[],\"operation\":\"missing_targets\",\"pairs\":[]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0027","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={0, 1}，函數表為 {(-1, 1), (0, 0)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"∅","explanation_zh":"缺少目標=∅。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052027,"parameters":{"A":[-1,0],"B":[0,1],"pairs":[[-1,1],[0,0]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-1,0],\"B\":[0,1],\"operation\":\"missing_targets\",\"pairs\":[[-1,1],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0028","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={1, 3, 5}，函數表為 {(-1, 1), (0, 3)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"{5}","explanation_zh":"缺少目標={5}。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052028,"parameters":{"A":[-1,0],"B":[1,3,5],"pairs":[[-1,1],[0,3]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-1,0],\"B\":[1,3,5],\"operation\":\"missing_targets\",\"pairs\":[[-1,1],[0,3]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0029","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={0}，函數表為 {(-1, 0), (0, 0)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"∅","explanation_zh":"缺少目標=∅。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052029,"parameters":{"A":[-1,0],"B":[0],"pairs":[[-1,0],[0,0]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-1,0],\"B\":[0],\"operation\":\"missing_targets\",\"pairs\":[[-1,0],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0030","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={-3, -2, -1}，函數表為 {(-1, -1), (0, -1)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"{-3, -2}","explanation_zh":"缺少目標={-3, -2}。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052030,"parameters":{"A":[-1,0],"B":[-3,-2,-1],"pairs":[[-1,-1],[0,-1]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-1,0],\"B\":[-3,-2,-1],\"operation\":\"missing_targets\",\"pairs\":[[-1,-1],[0,-1]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0031","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={0, 1, 2}，函數表為 {(-2, 1), (-1, 2), (0, 0)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"∅","explanation_zh":"缺少目標=∅。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052031,"parameters":{"A":[-2,-1,0],"B":[0,1,2],"pairs":[[-2,1],[-1,2],[0,0]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-2,-1,0],\"B\":[0,1,2],\"operation\":\"missing_targets\",\"pairs\":[[-2,1],[-1,2],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0032","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={1, 3, 5, 7}，函數表為 {(-2, 1), (-1, 3), (0, 5)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"{7}","explanation_zh":"缺少目標={7}。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052032,"parameters":{"A":[-2,-1,0],"B":[1,3,5,7],"pairs":[[-2,1],[-1,3],[0,5]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-2,-1,0],\"B\":[1,3,5,7],\"operation\":\"missing_targets\",\"pairs\":[[-2,1],[-1,3],[0,5]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0033","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={0, 1}，函數表為 {(-2, 0), (-1, 1), (0, 0)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"∅","explanation_zh":"缺少目標=∅。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052033,"parameters":{"A":[-2,-1,0],"B":[0,1],"pairs":[[-2,0],[-1,1],[0,0]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-2,-1,0],\"B\":[0,1],\"operation\":\"missing_targets\",\"pairs\":[[-2,0],[-1,1],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0034","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={-4, -3, -2, -1}，函數表為 {(-2, -1), (-1, -1), (0, -2)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"{-4, -3}","explanation_zh":"缺少目標={-4, -3}。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052034,"parameters":{"A":[-2,-1,0],"B":[-4,-3,-2,-1],"pairs":[[-2,-1],[-1,-1],[0,-2]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-2,-1,0],\"B\":[-4,-3,-2,-1],\"operation\":\"missing_targets\",\"pairs\":[[-2,-1],[-1,-1],[0,-2]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0035","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={0, 1, 2, 3}，函數表為 {(-3, 1), (-2, 2), (-1, 3), (0, 0)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"∅","explanation_zh":"缺少目標=∅。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052035,"parameters":{"A":[-3,-2,-1,0],"B":[0,1,2,3],"pairs":[[-3,1],[-2,2],[-1,3],[0,0]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-3,-2,-1,0],\"B\":[0,1,2,3],\"operation\":\"missing_targets\",\"pairs\":[[-3,1],[-2,2],[-1,3],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0036","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={1, 3, 5, 7, 9}，函數表為 {(-3, 1), (-2, 3), (-1, 5), (0, 7)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"{9}","explanation_zh":"缺少目標={9}。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052036,"parameters":{"A":[-3,-2,-1,0],"B":[1,3,5,7,9],"pairs":[[-3,1],[-2,3],[-1,5],[0,7]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-3,-2,-1,0],\"B\":[1,3,5,7,9],\"operation\":\"missing_targets\",\"pairs\":[[-3,1],[-2,3],[-1,5],[0,7]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0037","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={0, 1, 2}，函數表為 {(-3, 0), (-2, 1), (-1, 2), (0, 0)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"∅","explanation_zh":"缺少目標=∅。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052037,"parameters":{"A":[-3,-2,-1,0],"B":[0,1,2],"pairs":[[-3,0],[-2,1],[-1,2],[0,0]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-3,-2,-1,0],\"B\":[0,1,2],\"operation\":\"missing_targets\",\"pairs\":[[-3,0],[-2,1],[-1,2],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0038","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={-5, -4, -3, -2, -1}，函數表為 {(-3, -1), (-2, -1), (-1, -2), (0, -2)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"{-5, -4, -3}","explanation_zh":"缺少目標={-5, -4, -3}。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052038,"parameters":{"A":[-3,-2,-1,0],"B":[-5,-4,-3,-2,-1],"pairs":[[-3,-1],[-2,-1],[-1,-2],[0,-2]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-3,-2,-1,0],\"B\":[-5,-4,-3,-2,-1],\"operation\":\"missing_targets\",\"pairs\":[[-3,-1],[-2,-1],[-1,-2],[0,-2]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0039","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4}，函數表為 {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"∅","explanation_zh":"缺少目標=∅。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052039,"parameters":{"A":[-4,-3,-2,-1,0],"B":[0,1,2,3,4],"pairs":[[-4,1],[-3,2],[-2,3],[-1,4],[0,0]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-4,-3,-2,-1,0],\"B\":[0,1,2,3,4],\"operation\":\"missing_targets\",\"pairs\":[[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0040","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={1, 3, 5, 7, 9, 11}，函數表為 {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"{11}","explanation_zh":"缺少目標={11}。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052040,"parameters":{"A":[-4,-3,-2,-1,0],"B":[1,3,5,7,9,11],"pairs":[[-4,1],[-3,3],[-2,5],[-1,7],[0,9]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11],\"operation\":\"missing_targets\",\"pairs\":[[-4,1],[-3,3],[-2,5],[-1,7],[0,9]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0041","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={0, 1, 2, 3}，函數表為 {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"∅","explanation_zh":"缺少目標=∅。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052041,"parameters":{"A":[-4,-3,-2,-1,0],"B":[0,1,2,3],"pairs":[[-4,0],[-3,1],[-2,2],[-1,3],[0,0]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-4,-3,-2,-1,0],\"B\":[0,1,2,3],\"operation\":\"missing_targets\",\"pairs\":[[-4,0],[-3,1],[-2,2],[-1,3],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0042","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={-6, -5, -4, -3, -2, -1}，函數表為 {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"{-6, -5, -4}","explanation_zh":"缺少目標={-6, -5, -4}。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052042,"parameters":{"A":[-4,-3,-2,-1,0],"B":[-6,-5,-4,-3,-2,-1],"pairs":[[-4,-1],[-3,-1],[-2,-2],[-1,-2],[0,-3]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-4,-3,-2,-1,0],\"B\":[-6,-5,-4,-3,-2,-1],\"operation\":\"missing_targets\",\"pairs\":[[-4,-1],[-3,-1],[-2,-2],[-1,-2],[0,-3]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0043","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4, 5}，函數表為 {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"∅","explanation_zh":"缺少目標=∅。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052043,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4,5],"pairs":[[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5],\"operation\":\"missing_targets\",\"pairs\":[[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0044","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={1, 3, 5, 7, 9, 11, 13}，函數表為 {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"{13}","explanation_zh":"缺少目標={13}。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052044,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[1,3,5,7,9,11,13],"pairs":[[-5,1],[-4,3],[-3,5],[-2,7],[-1,9],[0,11]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11,13],\"operation\":\"missing_targets\",\"pairs\":[[-5,1],[-4,3],[-3,5],[-2,7],[-1,9],[0,11]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0045","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4}，函數表為 {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"∅","explanation_zh":"缺少目標=∅。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052045,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4],"pairs":[[-5,0],[-4,1],[-3,2],[-2,3],[-1,4],[0,0]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4],\"operation\":\"missing_targets\",\"pairs\":[[-5,0],[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0046","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={-7, -6, -5, -4, -3, -2, -1}，函數表為 {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"{-7, -6, -5, -4}","explanation_zh":"缺少目標={-7, -6, -5, -4}。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052046,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[-7,-6,-5,-4,-3,-2,-1],"pairs":[[-5,-1],[-4,-1],[-3,-2],[-2,-2],[-1,-3],[0,-3]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[-7,-6,-5,-4,-3,-2,-1],\"operation\":\"missing_targets\",\"pairs\":[[-5,-1],[-4,-1],[-3,-2],[-2,-2],[-1,-3],[0,-3]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0047","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4, 5, 6}，函數表為 {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"∅","explanation_zh":"缺少目標=∅。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052047,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4,5,6],"pairs":[[-6,1],[-5,2],[-4,3],[-3,4],[-2,5],[-1,6],[0,0]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5,6],\"operation\":\"missing_targets\",\"pairs\":[[-6,1],[-5,2],[-4,3],[-3,4],[-2,5],[-1,6],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0048","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={1, 3, 5, 7, 9, 11, 13, 15}，函數表為 {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"{15}","explanation_zh":"缺少目標={15}。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052048,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[1,3,5,7,9,11,13,15],"pairs":[[-6,1],[-5,3],[-4,5],[-3,7],[-2,9],[-1,11],[0,13]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11,13,15],\"operation\":\"missing_targets\",\"pairs\":[[-6,1],[-5,3],[-4,5],[-3,7],[-2,9],[-1,11],[0,13]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0049","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4, 5}，函數表為 {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"∅","explanation_zh":"缺少目標=∅。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052049,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4,5],"pairs":[[-6,0],[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5],\"operation\":\"missing_targets\",\"pairs\":[[-6,0],[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0050","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"陪域中未覆蓋的目標","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={-8, -7, -6, -5, -4, -3, -2, -1}，函數表為 {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}。求缺少原像的目標集合 B∖f[A]，空集合記 ∅。","answer":{"kind":"short_text","value":"{-8, -7, -6, -5}","explanation_zh":"缺少目標={-8, -7, -6, -5}。先得到實際值域，再從宣告陪域移除已取得的值；缺少任一目標即不滿射。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["陪域中未覆蓋的目標"]},"variant":{"family_id":"qf-surjective-missing-targets","seed":202610052050,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[-8,-7,-6,-5,-4,-3,-2,-1],"pairs":[[-6,-1],[-5,-1],[-4,-2],[-3,-2],[-2,-3],[-1,-3],[0,-4]],"operation":"missing_targets","semantic_template":"qf-surjective-missing-targets/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[-8,-7,-6,-5,-4,-3,-2,-1],\"operation\":\"missing_targets\",\"pairs\":[[-6,-1],[-5,-1],[-4,-2],[-3,-2],[-2,-3],[-1,-3],[0,-4]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0051","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A=∅、B={0}，函數表為 ∅。對 t=0∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"不存在","explanation_zh":"纖維(0)=∅。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052051,"parameters":{"A":[],"B":[0],"pairs":[],"t":0,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[],\"B\":[0],\"operation\":\"first_preimage\",\"pairs\":[],\"t\":0}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0052","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={0, 1}，函數表為 {(-1, 1), (0, 0)}。對 t=1∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"-1","explanation_zh":"纖維(1)={-1}。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052052,"parameters":{"A":[-1,0],"B":[0,1],"pairs":[[-1,1],[0,0]],"t":1,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-1,0],\"B\":[0,1],\"operation\":\"first_preimage\",\"pairs\":[[-1,1],[0,0]],\"t\":1}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0053","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={1, 3, 5}，函數表為 {(-1, 1), (0, 3)}。對 t=5∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"不存在","explanation_zh":"纖維(5)=∅。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052053,"parameters":{"A":[-1,0],"B":[1,3,5],"pairs":[[-1,1],[0,3]],"t":5,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-1,0],\"B\":[1,3,5],\"operation\":\"first_preimage\",\"pairs\":[[-1,1],[0,3]],\"t\":5}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0054","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={0}，函數表為 {(-1, 0), (0, 0)}。對 t=0∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"-1","explanation_zh":"纖維(0)={-1, 0}。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052054,"parameters":{"A":[-1,0],"B":[0],"pairs":[[-1,0],[0,0]],"t":0,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-1,0],\"B\":[0],\"operation\":\"first_preimage\",\"pairs\":[[-1,0],[0,0]],\"t\":0}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0055","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={-3, -2, -1}，函數表為 {(-1, -1), (0, -1)}。對 t=-2∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"不存在","explanation_zh":"纖維(-2)=∅。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052055,"parameters":{"A":[-1,0],"B":[-3,-2,-1],"pairs":[[-1,-1],[0,-1]],"t":-2,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-1,0],\"B\":[-3,-2,-1],\"operation\":\"first_preimage\",\"pairs\":[[-1,-1],[0,-1]],\"t\":-2}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0056","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={0, 1, 2}，函數表為 {(-2, 1), (-1, 2), (0, 0)}。對 t=2∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"-1","explanation_zh":"纖維(2)={-1}。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052056,"parameters":{"A":[-2,-1,0],"B":[0,1,2],"pairs":[[-2,1],[-1,2],[0,0]],"t":2,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-2,-1,0],\"B\":[0,1,2],\"operation\":\"first_preimage\",\"pairs\":[[-2,1],[-1,2],[0,0]],\"t\":2}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0057","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={1, 3, 5, 7}，函數表為 {(-2, 1), (-1, 3), (0, 5)}。對 t=7∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"不存在","explanation_zh":"纖維(7)=∅。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052057,"parameters":{"A":[-2,-1,0],"B":[1,3,5,7],"pairs":[[-2,1],[-1,3],[0,5]],"t":7,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-2,-1,0],\"B\":[1,3,5,7],\"operation\":\"first_preimage\",\"pairs\":[[-2,1],[-1,3],[0,5]],\"t\":7}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0058","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={0, 1}，函數表為 {(-2, 0), (-1, 1), (0, 0)}。對 t=1∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"-1","explanation_zh":"纖維(1)={-1}。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052058,"parameters":{"A":[-2,-1,0],"B":[0,1],"pairs":[[-2,0],[-1,1],[0,0]],"t":1,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-2,-1,0],\"B\":[0,1],\"operation\":\"first_preimage\",\"pairs\":[[-2,0],[-1,1],[0,0]],\"t\":1}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0059","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={-4, -3, -2, -1}，函數表為 {(-2, -1), (-1, -1), (0, -2)}。對 t=-4∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"不存在","explanation_zh":"纖維(-4)=∅。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052059,"parameters":{"A":[-2,-1,0],"B":[-4,-3,-2,-1],"pairs":[[-2,-1],[-1,-1],[0,-2]],"t":-4,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-2,-1,0],\"B\":[-4,-3,-2,-1],\"operation\":\"first_preimage\",\"pairs\":[[-2,-1],[-1,-1],[0,-2]],\"t\":-4}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0060","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={0, 1, 2, 3}，函數表為 {(-3, 1), (-2, 2), (-1, 3), (0, 0)}。對 t=1∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"-3","explanation_zh":"纖維(1)={-3}。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052060,"parameters":{"A":[-3,-2,-1,0],"B":[0,1,2,3],"pairs":[[-3,1],[-2,2],[-1,3],[0,0]],"t":1,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-3,-2,-1,0],\"B\":[0,1,2,3],\"operation\":\"first_preimage\",\"pairs\":[[-3,1],[-2,2],[-1,3],[0,0]],\"t\":1}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0061","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={1, 3, 5, 7, 9}，函數表為 {(-3, 1), (-2, 3), (-1, 5), (0, 7)}。對 t=9∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"不存在","explanation_zh":"纖維(9)=∅。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052061,"parameters":{"A":[-3,-2,-1,0],"B":[1,3,5,7,9],"pairs":[[-3,1],[-2,3],[-1,5],[0,7]],"t":9,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-3,-2,-1,0],\"B\":[1,3,5,7,9],\"operation\":\"first_preimage\",\"pairs\":[[-3,1],[-2,3],[-1,5],[0,7]],\"t\":9}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0062","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={0, 1, 2}，函數表為 {(-3, 0), (-2, 1), (-1, 2), (0, 0)}。對 t=2∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"-1","explanation_zh":"纖維(2)={-1}。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052062,"parameters":{"A":[-3,-2,-1,0],"B":[0,1,2],"pairs":[[-3,0],[-2,1],[-1,2],[0,0]],"t":2,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-3,-2,-1,0],\"B\":[0,1,2],\"operation\":\"first_preimage\",\"pairs\":[[-3,0],[-2,1],[-1,2],[0,0]],\"t\":2}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0063","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={-5, -4, -3, -2, -1}，函數表為 {(-3, -1), (-2, -1), (-1, -2), (0, -2)}。對 t=-3∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"不存在","explanation_zh":"纖維(-3)=∅。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052063,"parameters":{"A":[-3,-2,-1,0],"B":[-5,-4,-3,-2,-1],"pairs":[[-3,-1],[-2,-1],[-1,-2],[0,-2]],"t":-3,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-3,-2,-1,0],\"B\":[-5,-4,-3,-2,-1],\"operation\":\"first_preimage\",\"pairs\":[[-3,-1],[-2,-1],[-1,-2],[0,-2]],\"t\":-3}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0064","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4}，函數表為 {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}。對 t=3∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"-2","explanation_zh":"纖維(3)={-2}。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052064,"parameters":{"A":[-4,-3,-2,-1,0],"B":[0,1,2,3,4],"pairs":[[-4,1],[-3,2],[-2,3],[-1,4],[0,0]],"t":3,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-4,-3,-2,-1,0],\"B\":[0,1,2,3,4],\"operation\":\"first_preimage\",\"pairs\":[[-4,1],[-3,2],[-2,3],[-1,4],[0,0]],\"t\":3}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0065","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={1, 3, 5, 7, 9, 11}，函數表為 {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}。對 t=11∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"不存在","explanation_zh":"纖維(11)=∅。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052065,"parameters":{"A":[-4,-3,-2,-1,0],"B":[1,3,5,7,9,11],"pairs":[[-4,1],[-3,3],[-2,5],[-1,7],[0,9]],"t":11,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11],\"operation\":\"first_preimage\",\"pairs\":[[-4,1],[-3,3],[-2,5],[-1,7],[0,9]],\"t\":11}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0066","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={0, 1, 2, 3}，函數表為 {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}。對 t=3∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"-1","explanation_zh":"纖維(3)={-1}。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052066,"parameters":{"A":[-4,-3,-2,-1,0],"B":[0,1,2,3],"pairs":[[-4,0],[-3,1],[-2,2],[-1,3],[0,0]],"t":3,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-4,-3,-2,-1,0],\"B\":[0,1,2,3],\"operation\":\"first_preimage\",\"pairs\":[[-4,0],[-3,1],[-2,2],[-1,3],[0,0]],\"t\":3}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0067","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={-6, -5, -4, -3, -2, -1}，函數表為 {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}。對 t=-2∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"-2","explanation_zh":"纖維(-2)={-2, -1}。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052067,"parameters":{"A":[-4,-3,-2,-1,0],"B":[-6,-5,-4,-3,-2,-1],"pairs":[[-4,-1],[-3,-1],[-2,-2],[-1,-2],[0,-3]],"t":-2,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-4,-3,-2,-1,0],\"B\":[-6,-5,-4,-3,-2,-1],\"operation\":\"first_preimage\",\"pairs\":[[-4,-1],[-3,-1],[-2,-2],[-1,-2],[0,-3]],\"t\":-2}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0068","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4, 5}，函數表為 {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}。對 t=5∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"-1","explanation_zh":"纖維(5)={-1}。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052068,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4,5],"pairs":[[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]],"t":5,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5],\"operation\":\"first_preimage\",\"pairs\":[[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]],\"t\":5}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0069","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={1, 3, 5, 7, 9, 11, 13}，函數表為 {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}。對 t=13∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"不存在","explanation_zh":"纖維(13)=∅。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052069,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[1,3,5,7,9,11,13],"pairs":[[-5,1],[-4,3],[-3,5],[-2,7],[-1,9],[0,11]],"t":13,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11,13],\"operation\":\"first_preimage\",\"pairs\":[[-5,1],[-4,3],[-3,5],[-2,7],[-1,9],[0,11]],\"t\":13}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0070","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4}，函數表為 {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}。對 t=4∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"-1","explanation_zh":"纖維(4)={-1}。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052070,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4],"pairs":[[-5,0],[-4,1],[-3,2],[-2,3],[-1,4],[0,0]],"t":4,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4],\"operation\":\"first_preimage\",\"pairs\":[[-5,0],[-4,1],[-3,2],[-2,3],[-1,4],[0,0]],\"t\":4}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0071","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={-7, -6, -5, -4, -3, -2, -1}，函數表為 {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}。對 t=-1∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"-5","explanation_zh":"纖維(-1)={-5, -4}。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052071,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[-7,-6,-5,-4,-3,-2,-1],"pairs":[[-5,-1],[-4,-1],[-3,-2],[-2,-2],[-1,-3],[0,-3]],"t":-1,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[-7,-6,-5,-4,-3,-2,-1],\"operation\":\"first_preimage\",\"pairs\":[[-5,-1],[-4,-1],[-3,-2],[-2,-2],[-1,-3],[0,-3]],\"t\":-1}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0072","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4, 5, 6}，函數表為 {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}。對 t=0∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"0","explanation_zh":"纖維(0)={0}。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052072,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4,5,6],"pairs":[[-6,1],[-5,2],[-4,3],[-3,4],[-2,5],[-1,6],[0,0]],"t":0,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5,6],\"operation\":\"first_preimage\",\"pairs\":[[-6,1],[-5,2],[-4,3],[-3,4],[-2,5],[-1,6],[0,0]],\"t\":0}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0073","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={1, 3, 5, 7, 9, 11, 13, 15}，函數表為 {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}。對 t=15∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"不存在","explanation_zh":"纖維(15)=∅。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052073,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[1,3,5,7,9,11,13,15],"pairs":[[-6,1],[-5,3],[-4,5],[-3,7],[-2,9],[-1,11],[0,13]],"t":15,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11,13,15],\"operation\":\"first_preimage\",\"pairs\":[[-6,1],[-5,3],[-4,5],[-3,7],[-2,9],[-1,11],[0,13]],\"t\":15}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0074","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4, 5}，函數表為 {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}。對 t=5∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"-1","explanation_zh":"纖維(5)={-1}。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052074,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4,5],"pairs":[[-6,0],[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]],"t":5,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5],\"operation\":\"first_preimage\",\"pairs\":[[-6,0],[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]],\"t\":5}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0075","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"目標的最小合法原像","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={-8, -7, -6, -5, -4, -3, -2, -1}，函數表為 {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}。對 t=-8∈B，求按數值最小的合法原像；沒有原像則答不存在。","answer":{"kind":"short_text","value":"不存在","explanation_zh":"纖維(-8)=∅。先找全部原像，再取其最小輸入；原像存在不代表唯一，無原像時不虛構函數輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["目標的最小合法原像"]},"variant":{"family_id":"qf-surjective-first-preimage","seed":202610052075,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[-8,-7,-6,-5,-4,-3,-2,-1],"pairs":[[-6,-1],[-5,-1],[-4,-2],[-3,-2],[-2,-3],[-1,-3],[0,-4]],"t":-8,"operation":"first_preimage","semantic_template":"qf-surjective-first-preimage/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[-8,-7,-6,-5,-4,-3,-2,-1],\"operation\":\"first_preimage\",\"pairs\":[[-6,-1],[-5,-1],[-4,-2],[-3,-2],[-2,-3],[-1,-3],[0,-4]],\"t\":-8}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0076","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A=∅、B=∅，函數表為 ∅。新陪域 B2=∅ 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"新陪域未覆蓋=∅。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052076,"parameters":{"A":[],"B":[],"pairs":[],"B2":[],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[],\"B2\":[],\"operation\":\"changed_codomain\",\"pairs\":[]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0077","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={0, 1}，函數表為 {(-1, 1), (0, 0)}。新陪域 B2={0, 1, 101} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"新陪域未覆蓋={101}。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052077,"parameters":{"A":[-1,0],"B":[0,1],"pairs":[[-1,1],[0,0]],"B2":[1,0,101],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-1,0],\"B2\":[0,1,101],\"operation\":\"changed_codomain\",\"pairs\":[[-1,1],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0078","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={1, 3, 5}，函數表為 {(-1, 1), (0, 3)}。新陪域 B2={1, 3} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"新陪域未覆蓋=∅。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052078,"parameters":{"A":[-1,0],"B":[1,3,5],"pairs":[[-1,1],[0,3]],"B2":[1,3],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-1,0],\"B2\":[1,3],\"operation\":\"changed_codomain\",\"pairs\":[[-1,1],[0,3]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0079","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={0}，函數表為 {(-1, 0), (0, 0)}。新陪域 B2={0, 103} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"新陪域未覆蓋={103}。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052079,"parameters":{"A":[-1,0],"B":[0],"pairs":[[-1,0],[0,0]],"B2":[0,103],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-1,0],\"B2\":[0,103],\"operation\":\"changed_codomain\",\"pairs\":[[-1,0],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0080","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-1, 0}、B={-3, -2, -1}，函數表為 {(-1, -1), (0, -1)}。新陪域 B2={-1} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"新陪域未覆蓋=∅。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052080,"parameters":{"A":[-1,0],"B":[-3,-2,-1],"pairs":[[-1,-1],[0,-1]],"B2":[-1],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-1,0],\"B2\":[-1],\"operation\":\"changed_codomain\",\"pairs\":[[-1,-1],[0,-1]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0081","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={0, 1, 2}，函數表為 {(-2, 1), (-1, 2), (0, 0)}。新陪域 B2={0, 1, 2, 105} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"新陪域未覆蓋={105}。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052081,"parameters":{"A":[-2,-1,0],"B":[0,1,2],"pairs":[[-2,1],[-1,2],[0,0]],"B2":[1,2,0,105],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-2,-1,0],\"B2\":[0,1,2,105],\"operation\":\"changed_codomain\",\"pairs\":[[-2,1],[-1,2],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0082","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={1, 3, 5, 7}，函數表為 {(-2, 1), (-1, 3), (0, 5)}。新陪域 B2={1, 3, 5} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"新陪域未覆蓋=∅。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052082,"parameters":{"A":[-2,-1,0],"B":[1,3,5,7],"pairs":[[-2,1],[-1,3],[0,5]],"B2":[1,3,5],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-2,-1,0],\"B2\":[1,3,5],\"operation\":\"changed_codomain\",\"pairs\":[[-2,1],[-1,3],[0,5]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0083","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={0, 1}，函數表為 {(-2, 0), (-1, 1), (0, 0)}。新陪域 B2={0, 1, 107} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"新陪域未覆蓋={107}。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":1,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052083,"parameters":{"A":[-2,-1,0],"B":[0,1],"pairs":[[-2,0],[-1,1],[0,0]],"B2":[0,1,107],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-2,-1,0],\"B2\":[0,1,107],\"operation\":\"changed_codomain\",\"pairs\":[[-2,0],[-1,1],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0084","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-2, -1, 0}、B={-4, -3, -2, -1}，函數表為 {(-2, -1), (-1, -1), (0, -2)}。新陪域 B2={-2, -1} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"新陪域未覆蓋=∅。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052084,"parameters":{"A":[-2,-1,0],"B":[-4,-3,-2,-1],"pairs":[[-2,-1],[-1,-1],[0,-2]],"B2":[-1,-2],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-2,-1,0],\"B2\":[-2,-1],\"operation\":\"changed_codomain\",\"pairs\":[[-2,-1],[-1,-1],[0,-2]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0085","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={0, 1, 2, 3}，函數表為 {(-3, 1), (-2, 2), (-1, 3), (0, 0)}。新陪域 B2={0, 1, 2, 3, 109} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"新陪域未覆蓋={109}。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052085,"parameters":{"A":[-3,-2,-1,0],"B":[0,1,2,3],"pairs":[[-3,1],[-2,2],[-1,3],[0,0]],"B2":[1,2,3,0,109],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-3,-2,-1,0],\"B2\":[0,1,2,3,109],\"operation\":\"changed_codomain\",\"pairs\":[[-3,1],[-2,2],[-1,3],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0086","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={1, 3, 5, 7, 9}，函數表為 {(-3, 1), (-2, 3), (-1, 5), (0, 7)}。新陪域 B2={1, 3, 5, 7} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"新陪域未覆蓋=∅。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052086,"parameters":{"A":[-3,-2,-1,0],"B":[1,3,5,7,9],"pairs":[[-3,1],[-2,3],[-1,5],[0,7]],"B2":[1,3,5,7],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-3,-2,-1,0],\"B2\":[1,3,5,7],\"operation\":\"changed_codomain\",\"pairs\":[[-3,1],[-2,3],[-1,5],[0,7]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0087","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={0, 1, 2}，函數表為 {(-3, 0), (-2, 1), (-1, 2), (0, 0)}。新陪域 B2={0, 1, 2, 111} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"新陪域未覆蓋={111}。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052087,"parameters":{"A":[-3,-2,-1,0],"B":[0,1,2],"pairs":[[-3,0],[-2,1],[-1,2],[0,0]],"B2":[0,1,2,111],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-3,-2,-1,0],\"B2\":[0,1,2,111],\"operation\":\"changed_codomain\",\"pairs\":[[-3,0],[-2,1],[-1,2],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0088","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-3, -2, -1, 0}、B={-5, -4, -3, -2, -1}，函數表為 {(-3, -1), (-2, -1), (-1, -2), (0, -2)}。新陪域 B2={-2, -1} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"新陪域未覆蓋=∅。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052088,"parameters":{"A":[-3,-2,-1,0],"B":[-5,-4,-3,-2,-1],"pairs":[[-3,-1],[-2,-1],[-1,-2],[0,-2]],"B2":[-1,-2],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-3,-2,-1,0],\"B2\":[-2,-1],\"operation\":\"changed_codomain\",\"pairs\":[[-3,-1],[-2,-1],[-1,-2],[0,-2]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0089","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4}，函數表為 {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}。新陪域 B2={0, 1, 2, 3, 4, 113} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"新陪域未覆蓋={113}。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052089,"parameters":{"A":[-4,-3,-2,-1,0],"B":[0,1,2,3,4],"pairs":[[-4,1],[-3,2],[-2,3],[-1,4],[0,0]],"B2":[1,2,3,4,0,113],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-4,-3,-2,-1,0],\"B2\":[0,1,2,3,4,113],\"operation\":\"changed_codomain\",\"pairs\":[[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0090","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={1, 3, 5, 7, 9, 11}，函數表為 {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}。新陪域 B2={1, 3, 5, 7, 9} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"新陪域未覆蓋=∅。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052090,"parameters":{"A":[-4,-3,-2,-1,0],"B":[1,3,5,7,9,11],"pairs":[[-4,1],[-3,3],[-2,5],[-1,7],[0,9]],"B2":[1,3,5,7,9],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-4,-3,-2,-1,0],\"B2\":[1,3,5,7,9],\"operation\":\"changed_codomain\",\"pairs\":[[-4,1],[-3,3],[-2,5],[-1,7],[0,9]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0091","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={0, 1, 2, 3}，函數表為 {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}。新陪域 B2={0, 1, 2, 3, 115} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"新陪域未覆蓋={115}。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052091,"parameters":{"A":[-4,-3,-2,-1,0],"B":[0,1,2,3],"pairs":[[-4,0],[-3,1],[-2,2],[-1,3],[0,0]],"B2":[0,1,2,3,115],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-4,-3,-2,-1,0],\"B2\":[0,1,2,3,115],\"operation\":\"changed_codomain\",\"pairs\":[[-4,0],[-3,1],[-2,2],[-1,3],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0092","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-4, -3, -2, -1, 0}、B={-6, -5, -4, -3, -2, -1}，函數表為 {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}。新陪域 B2={-3, -2, -1} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"新陪域未覆蓋=∅。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":2,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052092,"parameters":{"A":[-4,-3,-2,-1,0],"B":[-6,-5,-4,-3,-2,-1],"pairs":[[-4,-1],[-3,-1],[-2,-2],[-1,-2],[0,-3]],"B2":[-1,-2,-3],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-4,-3,-2,-1,0],\"B2\":[-3,-2,-1],\"operation\":\"changed_codomain\",\"pairs\":[[-4,-1],[-3,-1],[-2,-2],[-1,-2],[0,-3]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0093","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4, 5}，函數表為 {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}。新陪域 B2={0, 1, 2, 3, 4, 5, 117} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"新陪域未覆蓋={117}。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052093,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4,5],"pairs":[[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]],"B2":[1,2,3,4,5,0,117],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-5,-4,-3,-2,-1,0],\"B2\":[0,1,2,3,4,5,117],\"operation\":\"changed_codomain\",\"pairs\":[[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0094","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={1, 3, 5, 7, 9, 11, 13}，函數表為 {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}。新陪域 B2={1, 3, 5, 7, 9, 11} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"新陪域未覆蓋=∅。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052094,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[1,3,5,7,9,11,13],"pairs":[[-5,1],[-4,3],[-3,5],[-2,7],[-1,9],[0,11]],"B2":[1,3,5,7,9,11],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-5,-4,-3,-2,-1,0],\"B2\":[1,3,5,7,9,11],\"operation\":\"changed_codomain\",\"pairs\":[[-5,1],[-4,3],[-3,5],[-2,7],[-1,9],[0,11]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0095","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4}，函數表為 {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}。新陪域 B2={0, 1, 2, 3, 4, 119} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"新陪域未覆蓋={119}。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052095,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4],"pairs":[[-5,0],[-4,1],[-3,2],[-2,3],[-1,4],[0,0]],"B2":[0,1,2,3,4,119],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-5,-4,-3,-2,-1,0],\"B2\":[0,1,2,3,4,119],\"operation\":\"changed_codomain\",\"pairs\":[[-5,0],[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0096","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-5, -4, -3, -2, -1, 0}、B={-7, -6, -5, -4, -3, -2, -1}，函數表為 {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}。新陪域 B2={-3, -2, -1} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"新陪域未覆蓋=∅。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052096,"parameters":{"A":[-5,-4,-3,-2,-1,0],"B":[-7,-6,-5,-4,-3,-2,-1],"pairs":[[-5,-1],[-4,-1],[-3,-2],[-2,-2],[-1,-3],[0,-3]],"B2":[-1,-2,-3],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-5,-4,-3,-2,-1,0],\"B2\":[-3,-2,-1],\"operation\":\"changed_codomain\",\"pairs\":[[-5,-1],[-4,-1],[-3,-2],[-2,-2],[-1,-3],[0,-3]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0097","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4, 5, 6}，函數表為 {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}。新陪域 B2={0, 1, 2, 3, 4, 5, 6, 121} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"新陪域未覆蓋={121}。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052097,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4,5,6],"pairs":[[-6,1],[-5,2],[-4,3],[-3,4],[-2,5],[-1,6],[0,0]],"B2":[1,2,3,4,5,6,0,121],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B2\":[0,1,2,3,4,5,6,121],\"operation\":\"changed_codomain\",\"pairs\":[[-6,1],[-5,2],[-4,3],[-3,4],[-2,5],[-1,6],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0098","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={1, 3, 5, 7, 9, 11, 13, 15}，函數表為 {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}。新陪域 B2={1, 3, 5, 7, 9, 11, 13} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"新陪域未覆蓋=∅。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052098,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[1,3,5,7,9,11,13,15],"pairs":[[-6,1],[-5,3],[-4,5],[-3,7],[-2,9],[-1,11],[0,13]],"B2":[1,3,5,7,9,11,13],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B2\":[1,3,5,7,9,11,13],\"operation\":\"changed_codomain\",\"pairs\":[[-6,1],[-5,3],[-4,5],[-3,7],[-2,9],[-1,11],[0,13]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0099","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={0, 1, 2, 3, 4, 5}，函數表為 {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}。新陪域 B2={0, 1, 2, 3, 4, 5, 123} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"否","explanation_zh":"新陪域未覆蓋={123}。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052099,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[0,1,2,3,4,5],"pairs":[[-6,0],[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]],"B2":[0,1,2,3,4,5,123],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B2\":[0,1,2,3,4,5,123],\"operation\":\"changed_codomain\",\"pairs\":[[-6,0],[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261004-advance-20261005-r02-surjective-0100","batch_id":"batch-20261004-advance-20261005-r02-surjective-foundations","language":"zh-Hant","source_mko_ids":["mko-surjective"],"title_zh":"合法更換陪域後的覆蓋","stem_zh":"明示總函數 f:A→B，A={-6, -5, -4, -3, -2, -1, 0}、B={-8, -7, -6, -5, -4, -3, -2, -1}，函數表為 {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}。新陪域 B2={-4, -3, -2, -1} 包含全部既有輸出；同一輸出表視為 A→B2 時是否滿射？","answer":{"kind":"short_text","value":"是","explanation_zh":"新陪域未覆蓋=∅。值域包含於 B2 先保證總函數宣告合法，再比較是否等於 B2；不能為求滿射而丟掉既有輸出。"},"difficulty":{"band":"undergraduate_core","level":3,"skills":["合法更換陪域後的覆蓋"]},"variant":{"family_id":"qf-surjective-codomain-change","seed":202610052100,"parameters":{"A":[-6,-5,-4,-3,-2,-1,0],"B":[-8,-7,-6,-5,-4,-3,-2,-1],"pairs":[[-6,-1],[-5,-1],[-4,-2],[-3,-2],[-2,-3],[-1,-3],[0,-4]],"B2":[-1,-2,-3,-4],"operation":"changed_codomain","semantic_template":"qf-surjective-codomain-change/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B2\":[-4,-3,-2,-1],\"operation\":\"changed_codomain\",\"pairs\":[[-6,-1],[-5,-1],[-4,-2],[-3,-2],[-2,-3],[-1,-3],[0,-4]]}"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"Codex independent producer (inherited model; exact runtime model not exposed)","generated_at":"2026-10-04","contract_version":"ocme-question-factory-v0.1"}}