{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0001","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fbd8459b6f6bd98977483f99f5bf2d94879905923887b0fedb080abeee262de3","family_id":"qf-bijective-table-property","seed":202610053001,"semantic_template":"qf-bijective-table-property/{\"A\":[],\"B\":[],\"operation\":\"bijective\",\"pairs\":[]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A=∅, B=∅, and table ∅. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"Yes","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is Yes."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0002","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca9f86ece6b2bb88cd97dc8a45fbc6628bfaad82a54cea03937d8df0459451f6","family_id":"qf-bijective-table-property","seed":202610053002,"semantic_template":"qf-bijective-table-property/{\"A\":[-1,0],\"B\":[0,1],\"operation\":\"bijective\",\"pairs\":[[-1,1],[0,0]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"Yes","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is Yes."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0003","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f315fe11962b1fdd566b7bc55d2997989d7389fae9c1d975fc6e6172aedb4d67","family_id":"qf-bijective-table-property","seed":202610053003,"semantic_template":"qf-bijective-table-property/{\"A\":[-1,0],\"B\":[1,3,5],\"operation\":\"bijective\",\"pairs\":[[-1,1],[0,3]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0004","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1df201c3e157d5973524a22cd63bade80131cf0b38fbda2164fc88642bf0012a","family_id":"qf-bijective-table-property","seed":202610053004,"semantic_template":"qf-bijective-table-property/{\"A\":[-1,0],\"B\":[0],\"operation\":\"bijective\",\"pairs\":[[-1,0],[0,0]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0005","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d98f7ca4b85966a5ca5432ada9f68309956ec0d8c5fc00df0f205bbc2f3e30cd","family_id":"qf-bijective-table-property","seed":202610053005,"semantic_template":"qf-bijective-table-property/{\"A\":[-1,0],\"B\":[-3,-2,-1],\"operation\":\"bijective\",\"pairs\":[[-1,-1],[0,-1]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0006","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"43990cbc66c1a79a2564a62ae52962e9a8c7edfc857a089428bf51b84929abe6","family_id":"qf-bijective-table-property","seed":202610053006,"semantic_template":"qf-bijective-table-property/{\"A\":[-2,-1,0],\"B\":[0,1,2],\"operation\":\"bijective\",\"pairs\":[[-2,1],[-1,2],[0,0]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-2, -1, 0}, B={0, 1, 2}, and table {(-2, 1), (-1, 2), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"Yes","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is Yes."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0007","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6a09b0e236b660477dd49dfb24e87bf796e18385f2b1ade57aeabf1f8af83eab","family_id":"qf-bijective-table-property","seed":202610053007,"semantic_template":"qf-bijective-table-property/{\"A\":[-2,-1,0],\"B\":[1,3,5,7],\"operation\":\"bijective\",\"pairs\":[[-2,1],[-1,3],[0,5]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-2, -1, 0}, B={1, 3, 5, 7}, and table {(-2, 1), (-1, 3), (0, 5)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0008","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"82f8fa610865d9bb7e7de127886a2c33dbc66ce7fe12d231c0615f8ba5557c33","family_id":"qf-bijective-table-property","seed":202610053008,"semantic_template":"qf-bijective-table-property/{\"A\":[-2,-1,0],\"B\":[0,1],\"operation\":\"bijective\",\"pairs\":[[-2,0],[-1,1],[0,0]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0009","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b7d6548f1222a3c27e056e1443d76d120bf3db8dc7ab58dfe67d14c9862f03af","family_id":"qf-bijective-table-property","seed":202610053009,"semantic_template":"qf-bijective-table-property/{\"A\":[-2,-1,0],\"B\":[-4,-3,-2,-1],\"operation\":\"bijective\",\"pairs\":[[-2,-1],[-1,-1],[0,-2]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-2, -1, 0}, B={-4, -3, -2, -1}, and table {(-2, -1), (-1, -1), (0, -2)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0010","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1d5ef6738c6c90c34869cf104602d2d63485dcbbc1016b9e002fc7897ef24471","family_id":"qf-bijective-table-property","seed":202610053010,"semantic_template":"qf-bijective-table-property/{\"A\":[-3,-2,-1,0],\"B\":[0,1,2,3],\"operation\":\"bijective\",\"pairs\":[[-3,1],[-2,2],[-1,3],[0,0]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-3, 1), (-2, 2), (-1, 3), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"Yes","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is Yes."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0011","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ea7f98b86f48f306e6df1fb2314f9d9b82b2b083168e3397a2698e6b51f3dfd","family_id":"qf-bijective-table-property","seed":202610053011,"semantic_template":"qf-bijective-table-property/{\"A\":[-3,-2,-1,0],\"B\":[1,3,5,7,9],\"operation\":\"bijective\",\"pairs\":[[-3,1],[-2,3],[-1,5],[0,7]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-3, -2, -1, 0}, B={1, 3, 5, 7, 9}, and table {(-3, 1), (-2, 3), (-1, 5), (0, 7)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0012","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de6df9218a99f915e8da43f9c89eaa592458ea7fbb35edea99edfcbc01df10b0","family_id":"qf-bijective-table-property","seed":202610053012,"semantic_template":"qf-bijective-table-property/{\"A\":[-3,-2,-1,0],\"B\":[0,1,2],\"operation\":\"bijective\",\"pairs\":[[-3,0],[-2,1],[-1,2],[0,0]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2}, and table {(-3, 0), (-2, 1), (-1, 2), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0013","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc9e17f436176512400693126be2a54556decb675a8e585cde3713ec42787795","family_id":"qf-bijective-table-property","seed":202610053013,"semantic_template":"qf-bijective-table-property/{\"A\":[-3,-2,-1,0],\"B\":[-5,-4,-3,-2,-1],\"operation\":\"bijective\",\"pairs\":[[-3,-1],[-2,-1],[-1,-2],[0,-2]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-3, -2, -1, 0}, B={-5, -4, -3, -2, -1}, and table {(-3, -1), (-2, -1), (-1, -2), (0, -2)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0014","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a3bbdcf211d0b3bf35ee96ae3749d275dbe44733ab40579649aa19011dfc391b","family_id":"qf-bijective-table-property","seed":202610053014,"semantic_template":"qf-bijective-table-property/{\"A\":[-4,-3,-2,-1,0],\"B\":[0,1,2,3,4],\"operation\":\"bijective\",\"pairs\":[[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"Yes","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is Yes."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0015","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8045a8c0f2dd7fbab391180a63770e0e8a4770765bb661dfa51117e37afee838","family_id":"qf-bijective-table-property","seed":202610053015,"semantic_template":"qf-bijective-table-property/{\"A\":[-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11],\"operation\":\"bijective\",\"pairs\":[[-4,1],[-3,3],[-2,5],[-1,7],[0,9]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11}, and table {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0016","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"82db00ab92f883eddba34eb22c053be96028dc41c7e210fed3c1af9c0d893419","family_id":"qf-bijective-table-property","seed":202610053016,"semantic_template":"qf-bijective-table-property/{\"A\":[-4,-3,-2,-1,0],\"B\":[0,1,2,3],\"operation\":\"bijective\",\"pairs\":[[-4,0],[-3,1],[-2,2],[-1,3],[0,0]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0017","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3ae3f37c38e94dcf5f79d52bd7241d596a3366548dd617a2315fe143ff9bf944","family_id":"qf-bijective-table-property","seed":202610053017,"semantic_template":"qf-bijective-table-property/{\"A\":[-4,-3,-2,-1,0],\"B\":[-6,-5,-4,-3,-2,-1],\"operation\":\"bijective\",\"pairs\":[[-4,-1],[-3,-1],[-2,-2],[-1,-2],[0,-3]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={-6, -5, -4, -3, -2, -1}, and table {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0018","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f2e5c09661712c49ba3e97e62ddaf0001a753471764fe99ffc975533604b5fc6","family_id":"qf-bijective-table-property","seed":202610053018,"semantic_template":"qf-bijective-table-property/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5],\"operation\":\"bijective\",\"pairs\":[[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"Yes","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is Yes."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0019","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ce6e0caa5551920703d977e4014d6aee08748444ccc1fbb53b2394f8d1e020d4","family_id":"qf-bijective-table-property","seed":202610053019,"semantic_template":"qf-bijective-table-property/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11,13],\"operation\":\"bijective\",\"pairs\":[[-5,1],[-4,3],[-3,5],[-2,7],[-1,9],[0,11]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13}, and table {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0020","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a92fbb6715069280ec8befa431549fc02b164beae6dbbefe7078dadc5880957b","family_id":"qf-bijective-table-property","seed":202610053020,"semantic_template":"qf-bijective-table-property/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4],\"operation\":\"bijective\",\"pairs\":[[-5,0],[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0021","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a12eef716e1003fc223f0ffc69fa438f4855a65e6b8c948055a8ddbf794ff55","family_id":"qf-bijective-table-property","seed":202610053021,"semantic_template":"qf-bijective-table-property/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[-7,-6,-5,-4,-3,-2,-1],\"operation\":\"bijective\",\"pairs\":[[-5,-1],[-4,-1],[-3,-2],[-2,-2],[-1,-3],[0,-3]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={-7, -6, -5, -4, -3, -2, -1}, and table {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0022","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a9d6273608ccba04c62d3da352ad47e2d0f84ea53733fffd1f817c7d3e3e96b0","family_id":"qf-bijective-table-property","seed":202610053022,"semantic_template":"qf-bijective-table-property/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5,6],\"operation\":\"bijective\",\"pairs\":[[-6,1],[-5,2],[-4,3],[-3,4],[-2,5],[-1,6],[0,0]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5, 6}, and table {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"Yes","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is Yes."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0023","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3d8f45fb93d998063f733e3ddc2ecf78a734857cca67576a584030a1eef077ea","family_id":"qf-bijective-table-property","seed":202610053023,"semantic_template":"qf-bijective-table-property/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11,13,15],\"operation\":\"bijective\",\"pairs\":[[-6,1],[-5,3],[-4,5],[-3,7],[-2,9],[-1,11],[0,13]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13, 15}, and table {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0024","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"651a6c6c2000e0fae3502ffc5c5ac6150841472ce7f97650c9c313b6580d27a7","family_id":"qf-bijective-table-property","seed":202610053024,"semantic_template":"qf-bijective-table-property/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5],\"operation\":\"bijective\",\"pairs\":[[-6,0],[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0025","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa8108a79119abc95785d90afd3e0dee1598fb83e0d6e52c9c9d2be6d765a772","family_id":"qf-bijective-table-property","seed":202610053025,"semantic_template":"qf-bijective-table-property/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[-8,-7,-6,-5,-4,-3,-2,-1],\"operation\":\"bijective\",\"pairs\":[[-6,-1],[-5,-1],[-4,-2],[-3,-2],[-2,-3],[-1,-3],[0,-4]]}"},"fields":{"title":"Bijectivity of a Finite Total Function","stem":"The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={-8, -7, -6, -5, -4, -3, -2, -1}, and table {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}. Is f both injective and surjective? Decide bijectivity from the actual table.","answer":{"kind":"short_text","value":"No","explanation":"Both conditions must hold, equivalently every codomain target must have exactly one permitted preimage. The answer is No."},"skills":["Bijectivity of a Finite Total Function"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0026","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87738bceea0998241f6e305bff949dddda90c306c62f68e5538e7282d56ac129","family_id":"qf-bijective-classification","seed":202610053026,"semantic_template":"qf-bijective-classification/{\"A\":[],\"B\":[],\"operation\":\"map_classification\",\"pairs\":[]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A=∅, B=∅, and table ∅. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"bijective","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give bijective. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0027","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"37c1e7571b75fb6fd9cc8d227995dc7e2e0da715957ebda018fc1f028f34d597","family_id":"qf-bijective-classification","seed":202610053027,"semantic_template":"qf-bijective-classification/{\"A\":[-1,0],\"B\":[0,1],\"operation\":\"map_classification\",\"pairs\":[[-1,1],[0,0]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"bijective","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give bijective. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0028","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7f59582bf68f4e4b76cefac0485ae150437b49952f4e33f4aae108eecfdada45","family_id":"qf-bijective-classification","seed":202610053028,"semantic_template":"qf-bijective-classification/{\"A\":[-1,0],\"B\":[1,3,5],\"operation\":\"map_classification\",\"pairs\":[[-1,1],[0,3]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"injective only","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give injective only. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0029","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb40e74e5f39eb058e02769f8138923c6b6867aff3cd7a3f609810246f3093d6","family_id":"qf-bijective-classification","seed":202610053029,"semantic_template":"qf-bijective-classification/{\"A\":[-1,0],\"B\":[0],\"operation\":\"map_classification\",\"pairs\":[[-1,0],[0,0]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"surjective only","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give surjective only. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0030","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a332b1653c1cd41d7054c3afeb511cbe60136768dcb27d706215892101c8da7a","family_id":"qf-bijective-classification","seed":202610053030,"semantic_template":"qf-bijective-classification/{\"A\":[-1,0],\"B\":[-3,-2,-1],\"operation\":\"map_classification\",\"pairs\":[[-1,-1],[0,-1]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"neither injective nor surjective","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give neither injective nor surjective. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0031","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"25077788e1844a9e0e76582a5b0095bebd09e6569eab78e5ca0a280f7157ac30","family_id":"qf-bijective-classification","seed":202610053031,"semantic_template":"qf-bijective-classification/{\"A\":[-2,-1,0],\"B\":[0,1,2],\"operation\":\"map_classification\",\"pairs\":[[-2,1],[-1,2],[0,0]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-2, -1, 0}, B={0, 1, 2}, and table {(-2, 1), (-1, 2), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"bijective","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give bijective. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0032","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f13806d1864f78188407e19fbfe6b1dfa89d18937df3e557ab7ea44c7945770","family_id":"qf-bijective-classification","seed":202610053032,"semantic_template":"qf-bijective-classification/{\"A\":[-2,-1,0],\"B\":[1,3,5,7],\"operation\":\"map_classification\",\"pairs\":[[-2,1],[-1,3],[0,5]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-2, -1, 0}, B={1, 3, 5, 7}, and table {(-2, 1), (-1, 3), (0, 5)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"injective only","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give injective only. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0033","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ee6aed07a7542a7025d51b62695883cebc070750d9f9398c4cdbb52df1a7e486","family_id":"qf-bijective-classification","seed":202610053033,"semantic_template":"qf-bijective-classification/{\"A\":[-2,-1,0],\"B\":[0,1],\"operation\":\"map_classification\",\"pairs\":[[-2,0],[-1,1],[0,0]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"surjective only","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give surjective only. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0034","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"982c09a837e664f4de0c80028a306cf4e0f0ff5559f508027429caf0388b40fc","family_id":"qf-bijective-classification","seed":202610053034,"semantic_template":"qf-bijective-classification/{\"A\":[-2,-1,0],\"B\":[-4,-3,-2,-1],\"operation\":\"map_classification\",\"pairs\":[[-2,-1],[-1,-1],[0,-2]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-2, -1, 0}, B={-4, -3, -2, -1}, and table {(-2, -1), (-1, -1), (0, -2)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"neither injective nor surjective","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give neither injective nor surjective. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0035","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5d52e9880c2b83c3f419c1d67d21fc06947e928c0716b23bc69ab94ec4340121","family_id":"qf-bijective-classification","seed":202610053035,"semantic_template":"qf-bijective-classification/{\"A\":[-3,-2,-1,0],\"B\":[0,1,2,3],\"operation\":\"map_classification\",\"pairs\":[[-3,1],[-2,2],[-1,3],[0,0]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-3, 1), (-2, 2), (-1, 3), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"bijective","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give bijective. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0036","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8fb1dd2220561665eec47da98258a3a6ecc7ab2fd09dc9542cde17819ea10902","family_id":"qf-bijective-classification","seed":202610053036,"semantic_template":"qf-bijective-classification/{\"A\":[-3,-2,-1,0],\"B\":[1,3,5,7,9],\"operation\":\"map_classification\",\"pairs\":[[-3,1],[-2,3],[-1,5],[0,7]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-3, -2, -1, 0}, B={1, 3, 5, 7, 9}, and table {(-3, 1), (-2, 3), (-1, 5), (0, 7)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"injective only","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give injective only. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0037","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"83ab020005c614aaebbb29579a191f9cd2b5cdf92e3f05c654330ca0ab300cbb","family_id":"qf-bijective-classification","seed":202610053037,"semantic_template":"qf-bijective-classification/{\"A\":[-3,-2,-1,0],\"B\":[0,1,2],\"operation\":\"map_classification\",\"pairs\":[[-3,0],[-2,1],[-1,2],[0,0]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2}, and table {(-3, 0), (-2, 1), (-1, 2), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"surjective only","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give surjective only. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0038","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e496b3812af8c3d5c6b82af87d407d48a51440b8a783ca6baaaa07ab1d3e9b37","family_id":"qf-bijective-classification","seed":202610053038,"semantic_template":"qf-bijective-classification/{\"A\":[-3,-2,-1,0],\"B\":[-5,-4,-3,-2,-1],\"operation\":\"map_classification\",\"pairs\":[[-3,-1],[-2,-1],[-1,-2],[0,-2]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-3, -2, -1, 0}, B={-5, -4, -3, -2, -1}, and table {(-3, -1), (-2, -1), (-1, -2), (0, -2)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"neither injective nor surjective","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give neither injective nor surjective. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0039","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a972daed69126e7cbb46f033af8a90405bef1bb60d3dd81a5a2700d82859697","family_id":"qf-bijective-classification","seed":202610053039,"semantic_template":"qf-bijective-classification/{\"A\":[-4,-3,-2,-1,0],\"B\":[0,1,2,3,4],\"operation\":\"map_classification\",\"pairs\":[[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"bijective","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give bijective. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0040","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dc769d2f08f746b429657c24b623ce9818ab2cfa99d26d30e32d2f3e55317676","family_id":"qf-bijective-classification","seed":202610053040,"semantic_template":"qf-bijective-classification/{\"A\":[-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11],\"operation\":\"map_classification\",\"pairs\":[[-4,1],[-3,3],[-2,5],[-1,7],[0,9]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11}, and table {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"injective only","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give injective only. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0041","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ac8c70f38596e6b0774e8354b54316aadbf3269902c052a721a54ad5e0327be1","family_id":"qf-bijective-classification","seed":202610053041,"semantic_template":"qf-bijective-classification/{\"A\":[-4,-3,-2,-1,0],\"B\":[0,1,2,3],\"operation\":\"map_classification\",\"pairs\":[[-4,0],[-3,1],[-2,2],[-1,3],[0,0]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"surjective only","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give surjective only. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0042","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06a154f13384231477b836197fe7834e419a47b9f862cf6cc729e5bee574e35e","family_id":"qf-bijective-classification","seed":202610053042,"semantic_template":"qf-bijective-classification/{\"A\":[-4,-3,-2,-1,0],\"B\":[-6,-5,-4,-3,-2,-1],\"operation\":\"map_classification\",\"pairs\":[[-4,-1],[-3,-1],[-2,-2],[-1,-2],[0,-3]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={-6, -5, -4, -3, -2, -1}, and table {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"neither injective nor surjective","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give neither injective nor surjective. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0043","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87293d799011e4b52524fbd87ff7a1dda746ae4f181b0088f156a157db8c9440","family_id":"qf-bijective-classification","seed":202610053043,"semantic_template":"qf-bijective-classification/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5],\"operation\":\"map_classification\",\"pairs\":[[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"bijective","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give bijective. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0044","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"41ecdb06c31448dbab1a8ebf0ff8614c99aeff1eee8b43e2730f741beead16f0","family_id":"qf-bijective-classification","seed":202610053044,"semantic_template":"qf-bijective-classification/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11,13],\"operation\":\"map_classification\",\"pairs\":[[-5,1],[-4,3],[-3,5],[-2,7],[-1,9],[0,11]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13}, and table {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"injective only","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give injective only. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0045","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f76718e30b2b8f7e169624ca56708e7626db975cc0a83e69a653fdef91b4b59","family_id":"qf-bijective-classification","seed":202610053045,"semantic_template":"qf-bijective-classification/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4],\"operation\":\"map_classification\",\"pairs\":[[-5,0],[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"surjective only","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give surjective only. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0046","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b6540b85d60a191a41e92e64dbdbf9eccab22e48ac6f8e0d070a3c4161a03e3","family_id":"qf-bijective-classification","seed":202610053046,"semantic_template":"qf-bijective-classification/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[-7,-6,-5,-4,-3,-2,-1],\"operation\":\"map_classification\",\"pairs\":[[-5,-1],[-4,-1],[-3,-2],[-2,-2],[-1,-3],[0,-3]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={-7, -6, -5, -4, -3, -2, -1}, and table {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"neither injective nor surjective","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give neither injective nor surjective. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0047","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"15af6eaf72aea4121df5714006d96e0109d2e4b37de5bb676adefbf6394f5563","family_id":"qf-bijective-classification","seed":202610053047,"semantic_template":"qf-bijective-classification/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5,6],\"operation\":\"map_classification\",\"pairs\":[[-6,1],[-5,2],[-4,3],[-3,4],[-2,5],[-1,6],[0,0]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5, 6}, and table {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"bijective","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give bijective. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0048","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a9843bc8a2e21d06a784ac844323f3474305c218b9156c41f9704aeb5a471ff","family_id":"qf-bijective-classification","seed":202610053048,"semantic_template":"qf-bijective-classification/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11,13,15],\"operation\":\"map_classification\",\"pairs\":[[-6,1],[-5,3],[-4,5],[-3,7],[-2,9],[-1,11],[0,13]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13, 15}, and table {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"injective only","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give injective only. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0049","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cf7bce2584b83f04354b985f1dbb7b36c79d3b2ed1bfc1f70a5b0fcdab50ee50","family_id":"qf-bijective-classification","seed":202610053049,"semantic_template":"qf-bijective-classification/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5],\"operation\":\"map_classification\",\"pairs\":[[-6,0],[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"surjective only","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give surjective only. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0050","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf3604c1076144b93e0812d27f4d393b38e90eb3f40f7e229ada246c627079d6","family_id":"qf-bijective-classification","seed":202610053050,"semantic_template":"qf-bijective-classification/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[-8,-7,-6,-5,-4,-3,-2,-1],\"operation\":\"map_classification\",\"pairs\":[[-6,-1],[-5,-1],[-4,-2],[-3,-2],[-2,-3],[-1,-3],[0,-4]]}"},"fields":{"title":"Four-Way Function Classification","stem":"The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={-8, -7, -6, -5, -4, -3, -2, -1}, and table {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}. Classify f as bijective, injective only, surjective only, or neither injective nor surjective.","answer":{"kind":"short_text","value":"neither injective nor surjective","explanation":"Check distinct-input collisions and full codomain coverage separately; their two truth values give neither injective nor surjective. Set sizes alone do not classify this supplied function."},"skills":["Four-Way Function Classification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0051","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"46fe8918568e17b32c0b60cea76bc3cb9b22b99d1d5e25a6e754406124e7ca78","family_id":"qf-bijective-fiber-profile","seed":202610053051,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[],\"B\":[],\"operation\":\"fiber_profile\",\"pairs\":[]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A=∅, B=∅, and table ∅. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"∅","explanation":"Count the permitted preimages of every element of B to obtain ∅. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0052","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c6ff346ea3771a4f3db8656a062c599024e554b7a236352133e9a31dacb7b418","family_id":"qf-bijective-fiber-profile","seed":202610053052,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-1,0],\"B\":[0,1],\"operation\":\"fiber_profile\",\"pairs\":[[-1,1],[0,0]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(0, 1), (1, 1)}","explanation":"Count the permitted preimages of every element of B to obtain {(0, 1), (1, 1)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0053","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"32182fdfcd9b746386cb05998c9d1a744d2611bc4941d8b66391724e94e853c5","family_id":"qf-bijective-fiber-profile","seed":202610053053,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-1,0],\"B\":[1,3,5],\"operation\":\"fiber_profile\",\"pairs\":[[-1,1],[0,3]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(1, 1), (3, 1), (5, 0)}","explanation":"Count the permitted preimages of every element of B to obtain {(1, 1), (3, 1), (5, 0)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0054","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"019e40cbaa73742e5f2f96e514a91946562c551a4c3b07617a4d41b345c2608d","family_id":"qf-bijective-fiber-profile","seed":202610053054,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-1,0],\"B\":[0],\"operation\":\"fiber_profile\",\"pairs\":[[-1,0],[0,0]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(0, 2)}","explanation":"Count the permitted preimages of every element of B to obtain {(0, 2)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0055","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b4a6e8000c8522e0dab034bb6adcdf910be5935aa823028003af8fedb66b69a","family_id":"qf-bijective-fiber-profile","seed":202610053055,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-1,0],\"B\":[-3,-2,-1],\"operation\":\"fiber_profile\",\"pairs\":[[-1,-1],[0,-1]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(-3, 0), (-2, 0), (-1, 2)}","explanation":"Count the permitted preimages of every element of B to obtain {(-3, 0), (-2, 0), (-1, 2)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0056","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"afc2cca9984aa07326e100d3903d592ddb1151564919b4ddc2214b3d7ffc3272","family_id":"qf-bijective-fiber-profile","seed":202610053056,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-2,-1,0],\"B\":[0,1,2],\"operation\":\"fiber_profile\",\"pairs\":[[-2,1],[-1,2],[0,0]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-2, -1, 0}, B={0, 1, 2}, and table {(-2, 1), (-1, 2), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(0, 1), (1, 1), (2, 1)}","explanation":"Count the permitted preimages of every element of B to obtain {(0, 1), (1, 1), (2, 1)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0057","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"945cc87dfe0ba3e6e050ebc70ab1ad95832a1b8d4b473387c26e5c5f72b4cf12","family_id":"qf-bijective-fiber-profile","seed":202610053057,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-2,-1,0],\"B\":[1,3,5,7],\"operation\":\"fiber_profile\",\"pairs\":[[-2,1],[-1,3],[0,5]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-2, -1, 0}, B={1, 3, 5, 7}, and table {(-2, 1), (-1, 3), (0, 5)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(1, 1), (3, 1), (5, 1), (7, 0)}","explanation":"Count the permitted preimages of every element of B to obtain {(1, 1), (3, 1), (5, 1), (7, 0)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0058","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b597ca45283b8bcfdcb19580558b5e6c0644dc6dca5437f0c32d86b9c4e41bea","family_id":"qf-bijective-fiber-profile","seed":202610053058,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-2,-1,0],\"B\":[0,1],\"operation\":\"fiber_profile\",\"pairs\":[[-2,0],[-1,1],[0,0]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(0, 2), (1, 1)}","explanation":"Count the permitted preimages of every element of B to obtain {(0, 2), (1, 1)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0059","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d3bf40c42dde30c9d077eb46db02c4a52293f21ee9701006d11e5af66932a4e7","family_id":"qf-bijective-fiber-profile","seed":202610053059,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-2,-1,0],\"B\":[-4,-3,-2,-1],\"operation\":\"fiber_profile\",\"pairs\":[[-2,-1],[-1,-1],[0,-2]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-2, -1, 0}, B={-4, -3, -2, -1}, and table {(-2, -1), (-1, -1), (0, -2)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(-4, 0), (-3, 0), (-2, 1), (-1, 2)}","explanation":"Count the permitted preimages of every element of B to obtain {(-4, 0), (-3, 0), (-2, 1), (-1, 2)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0060","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"054cf68035b913c34bb392e08ac107ee380555033ed8be1a949a3b8825cb5eba","family_id":"qf-bijective-fiber-profile","seed":202610053060,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-3,-2,-1,0],\"B\":[0,1,2,3],\"operation\":\"fiber_profile\",\"pairs\":[[-3,1],[-2,2],[-1,3],[0,0]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-3, 1), (-2, 2), (-1, 3), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(0, 1), (1, 1), (2, 1), (3, 1)}","explanation":"Count the permitted preimages of every element of B to obtain {(0, 1), (1, 1), (2, 1), (3, 1)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0061","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"808de77851100e2be0b1abf957d8a44898b8dc4b4932d10a576bbe1b3105e61f","family_id":"qf-bijective-fiber-profile","seed":202610053061,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-3,-2,-1,0],\"B\":[1,3,5,7,9],\"operation\":\"fiber_profile\",\"pairs\":[[-3,1],[-2,3],[-1,5],[0,7]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-3, -2, -1, 0}, B={1, 3, 5, 7, 9}, and table {(-3, 1), (-2, 3), (-1, 5), (0, 7)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(1, 1), (3, 1), (5, 1), (7, 1), (9, 0)}","explanation":"Count the permitted preimages of every element of B to obtain {(1, 1), (3, 1), (5, 1), (7, 1), (9, 0)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0062","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"61a3bcabaae806c643903cb8514e95ec3acb5acd5637b9126ce52fd143f4fd86","family_id":"qf-bijective-fiber-profile","seed":202610053062,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-3,-2,-1,0],\"B\":[0,1,2],\"operation\":\"fiber_profile\",\"pairs\":[[-3,0],[-2,1],[-1,2],[0,0]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-3, -2, -1, 0}, B={0, 1, 2}, and table {(-3, 0), (-2, 1), (-1, 2), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(0, 2), (1, 1), (2, 1)}","explanation":"Count the permitted preimages of every element of B to obtain {(0, 2), (1, 1), (2, 1)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0063","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ed5a5f261dc8945bc1c1d85a4a14a71845f89aec953e1e98d24094e151571a33","family_id":"qf-bijective-fiber-profile","seed":202610053063,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-3,-2,-1,0],\"B\":[-5,-4,-3,-2,-1],\"operation\":\"fiber_profile\",\"pairs\":[[-3,-1],[-2,-1],[-1,-2],[0,-2]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-3, -2, -1, 0}, B={-5, -4, -3, -2, -1}, and table {(-3, -1), (-2, -1), (-1, -2), (0, -2)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(-5, 0), (-4, 0), (-3, 0), (-2, 2), (-1, 2)}","explanation":"Count the permitted preimages of every element of B to obtain {(-5, 0), (-4, 0), (-3, 0), (-2, 2), (-1, 2)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0064","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"efd1d9fc8a9fe6bbf01c6b6a1397c6e89a2bdf64ecacc2da5ec3c2a4104ba610","family_id":"qf-bijective-fiber-profile","seed":202610053064,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-4,-3,-2,-1,0],\"B\":[0,1,2,3,4],\"operation\":\"fiber_profile\",\"pairs\":[[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(0, 1), (1, 1), (2, 1), (3, 1), (4, 1)}","explanation":"Count the permitted preimages of every element of B to obtain {(0, 1), (1, 1), (2, 1), (3, 1), (4, 1)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0065","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ec1bb1a5b4ff9ab55082753f2b3f05e47c085268007779c4eadf35184e49630","family_id":"qf-bijective-fiber-profile","seed":202610053065,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11],\"operation\":\"fiber_profile\",\"pairs\":[[-4,1],[-3,3],[-2,5],[-1,7],[0,9]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11}, and table {(-4, 1), (-3, 3), (-2, 5), (-1, 7), (0, 9)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(1, 1), (3, 1), (5, 1), (7, 1), (9, 1), (11, 0)}","explanation":"Count the permitted preimages of every element of B to obtain {(1, 1), (3, 1), (5, 1), (7, 1), (9, 1), (11, 0)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0066","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"78fef2b6e5d078671f158c0a8559c370349490d451c862ca1bde57f456154a79","family_id":"qf-bijective-fiber-profile","seed":202610053066,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-4,-3,-2,-1,0],\"B\":[0,1,2,3],\"operation\":\"fiber_profile\",\"pairs\":[[-4,0],[-3,1],[-2,2],[-1,3],[0,0]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={0, 1, 2, 3}, and table {(-4, 0), (-3, 1), (-2, 2), (-1, 3), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(0, 2), (1, 1), (2, 1), (3, 1)}","explanation":"Count the permitted preimages of every element of B to obtain {(0, 2), (1, 1), (2, 1), (3, 1)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0067","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"984657049e8128eb8b278aa984cb24a3cc77525d30c87ca999384dba4856dcb1","family_id":"qf-bijective-fiber-profile","seed":202610053067,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-4,-3,-2,-1,0],\"B\":[-6,-5,-4,-3,-2,-1],\"operation\":\"fiber_profile\",\"pairs\":[[-4,-1],[-3,-1],[-2,-2],[-1,-2],[0,-3]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-4, -3, -2, -1, 0}, B={-6, -5, -4, -3, -2, -1}, and table {(-4, -1), (-3, -1), (-2, -2), (-1, -2), (0, -3)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(-6, 0), (-5, 0), (-4, 0), (-3, 1), (-2, 2), (-1, 2)}","explanation":"Count the permitted preimages of every element of B to obtain {(-6, 0), (-5, 0), (-4, 0), (-3, 1), (-2, 2), (-1, 2)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0068","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"79b6d8c183dfe11a5a59064888ac540607287814678a94212a752b26bf74720f","family_id":"qf-bijective-fiber-profile","seed":202610053068,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5],\"operation\":\"fiber_profile\",\"pairs\":[[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(0, 1), (1, 1), (2, 1), (3, 1), (4, 1), (5, 1)}","explanation":"Count the permitted preimages of every element of B to obtain {(0, 1), (1, 1), (2, 1), (3, 1), (4, 1), (5, 1)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0069","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a2cf95a8186e78e9e5ffa9b4000830581a23f2670986261fed8b6dbb9057ad19","family_id":"qf-bijective-fiber-profile","seed":202610053069,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11,13],\"operation\":\"fiber_profile\",\"pairs\":[[-5,1],[-4,3],[-3,5],[-2,7],[-1,9],[0,11]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13}, and table {(-5, 1), (-4, 3), (-3, 5), (-2, 7), (-1, 9), (0, 11)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(1, 1), (3, 1), (5, 1), (7, 1), (9, 1), (11, 1), (13, 0)}","explanation":"Count the permitted preimages of every element of B to obtain {(1, 1), (3, 1), (5, 1), (7, 1), (9, 1), (11, 1), (13, 0)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0070","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5f2097599820b4815ee31a9a9eb848799cc2dbd7a63038a2d45390cb17bef40","family_id":"qf-bijective-fiber-profile","seed":202610053070,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4],\"operation\":\"fiber_profile\",\"pairs\":[[-5,0],[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4}, and table {(-5, 0), (-4, 1), (-3, 2), (-2, 3), (-1, 4), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(0, 2), (1, 1), (2, 1), (3, 1), (4, 1)}","explanation":"Count the permitted preimages of every element of B to obtain {(0, 2), (1, 1), (2, 1), (3, 1), (4, 1)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0071","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f6f368545936063be230609227b57bb75b2cd1d8d6b34b04ae6b0d0379a0f2e1","family_id":"qf-bijective-fiber-profile","seed":202610053071,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-5,-4,-3,-2,-1,0],\"B\":[-7,-6,-5,-4,-3,-2,-1],\"operation\":\"fiber_profile\",\"pairs\":[[-5,-1],[-4,-1],[-3,-2],[-2,-2],[-1,-3],[0,-3]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-5, -4, -3, -2, -1, 0}, B={-7, -6, -5, -4, -3, -2, -1}, and table {(-5, -1), (-4, -1), (-3, -2), (-2, -2), (-1, -3), (0, -3)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(-7, 0), (-6, 0), (-5, 0), (-4, 0), (-3, 2), (-2, 2), (-1, 2)}","explanation":"Count the permitted preimages of every element of B to obtain {(-7, 0), (-6, 0), (-5, 0), (-4, 0), (-3, 2), (-2, 2), (-1, 2)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0072","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6444aec172efe40f0120ae4916e363ec30f0db9b75948e2f14c5b01922f94988","family_id":"qf-bijective-fiber-profile","seed":202610053072,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5,6],\"operation\":\"fiber_profile\",\"pairs\":[[-6,1],[-5,2],[-4,3],[-3,4],[-2,5],[-1,6],[0,0]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5, 6}, and table {(-6, 1), (-5, 2), (-4, 3), (-3, 4), (-2, 5), (-1, 6), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(0, 1), (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)}","explanation":"Count the permitted preimages of every element of B to obtain {(0, 1), (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0073","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f6132f0bee930932684f0680f5f07b32ea73c15be7c6e03bbb2e0858c4cde992","family_id":"qf-bijective-fiber-profile","seed":202610053073,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[1,3,5,7,9,11,13,15],\"operation\":\"fiber_profile\",\"pairs\":[[-6,1],[-5,3],[-4,5],[-3,7],[-2,9],[-1,11],[0,13]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={1, 3, 5, 7, 9, 11, 13, 15}, and table {(-6, 1), (-5, 3), (-4, 5), (-3, 7), (-2, 9), (-1, 11), (0, 13)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(1, 1), (3, 1), (5, 1), (7, 1), (9, 1), (11, 1), (13, 1), (15, 0)}","explanation":"Count the permitted preimages of every element of B to obtain {(1, 1), (3, 1), (5, 1), (7, 1), (9, 1), (11, 1), (13, 1), (15, 0)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0074","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a8765ef9e0237e8efae8833b99b4b09055831673ae5f7f595b57a85f37710a36","family_id":"qf-bijective-fiber-profile","seed":202610053074,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[0,1,2,3,4,5],\"operation\":\"fiber_profile\",\"pairs\":[[-6,0],[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={0, 1, 2, 3, 4, 5}, and table {(-6, 0), (-5, 1), (-4, 2), (-3, 3), (-2, 4), (-1, 5), (0, 0)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(0, 2), (1, 1), (2, 1), (3, 1), (4, 1), (5, 1)}","explanation":"Count the permitted preimages of every element of B to obtain {(0, 2), (1, 1), (2, 1), (3, 1), (4, 1), (5, 1)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0075","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c8043ffa40676668427d59f58cec28003fbeb2c82ab6320d05348d06b959b945","family_id":"qf-bijective-fiber-profile","seed":202610053075,"semantic_template":"qf-bijective-fiber-profile/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"B\":[-8,-7,-6,-5,-4,-3,-2,-1],\"operation\":\"fiber_profile\",\"pairs\":[[-6,-1],[-5,-1],[-4,-2],[-3,-2],[-2,-3],[-1,-3],[0,-4]]}"},"fields":{"title":"Complete Codomain Fiber Profile","stem":"The declared total function f:A→B has A={-6, -5, -4, -3, -2, -1, 0}, B={-8, -7, -6, -5, -4, -3, -2, -1}, and table {(-6, -1), (-5, -1), (-4, -2), (-3, -2), (-2, -3), (-1, -3), (0, -4)}. In increasing target order, list every (b,|f⁻¹({b})|), including zero for targets not attained.","answer":{"kind":"short_text","value":"{(-8, 0), (-7, 0), (-6, 0), (-5, 0), (-4, 1), (-3, 2), (-2, 2), (-1, 2)}","explanation":"Count the permitted preimages of every element of B to obtain {(-8, 0), (-7, 0), (-6, 0), (-5, 0), (-4, 1), (-3, 2), (-2, 2), (-1, 2)}. The profile includes zero-size fibers, not only attained targets."},"skills":["Complete Codomain Fiber Profile"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0076","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1f193e176669ee81ef600352d088478a5d64ffa8472c4ad994cbaeefff9bb14b","family_id":"qf-bijective-cardinality-possibility","seed":202610053076,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":0,\"b_size\":0,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=0 and |B|=0, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"Yes","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0077","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6a593c04eff173c43730f51c798a1e079249475ff02f8b0ffb8d476a141e467b","family_id":"qf-bijective-cardinality-possibility","seed":202610053077,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":0,\"b_size\":1,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=0 and |B|=1, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"No","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0078","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"321e000c8b99bb2191c6c000fc3b38a390639012126123d517011a0498df0b67","family_id":"qf-bijective-cardinality-possibility","seed":202610053078,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":1,\"b_size\":1,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=1 and |B|=1, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"Yes","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0079","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0745a9e1c410a21e6cd4f189f353797e7372645102e1bd8938674fb20d659892","family_id":"qf-bijective-cardinality-possibility","seed":202610053079,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":1,\"b_size\":2,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=1 and |B|=2, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"No","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0080","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d960ebdb9e524c3393b55ec131317f3f1e56408398e2c408680aa8b636986b7d","family_id":"qf-bijective-cardinality-possibility","seed":202610053080,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":2,\"b_size\":2,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=2 and |B|=2, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"Yes","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0081","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3b93fae5acfcc50d3b9272d3b2b8635ea18a60f753e30508f02d5db840b69418","family_id":"qf-bijective-cardinality-possibility","seed":202610053081,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":2,\"b_size\":3,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=2 and |B|=3, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"No","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0082","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1bd7939ecb5f1c069a3672009ca88d6a73e1ea74d1489b05f77d37e6d3a5c31f","family_id":"qf-bijective-cardinality-possibility","seed":202610053082,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":3,\"b_size\":3,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=3 and |B|=3, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"Yes","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0083","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65cea4233cf9d07f9352ade72525c171f26fcc5c88495eeddc9680a3e0af462d","family_id":"qf-bijective-cardinality-possibility","seed":202610053083,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":3,\"b_size\":4,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=3 and |B|=4, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"No","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0084","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"26b89066cc7a45e079e7c93e190e567b56d663683f89d974d86cebfac8f958a1","family_id":"qf-bijective-cardinality-possibility","seed":202610053084,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":4,\"b_size\":4,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=4 and |B|=4, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"Yes","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0085","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"58f15249ab22011ee7e19f773f77569c37e1f787c0ccbfb98dc62d3b4fae1364","family_id":"qf-bijective-cardinality-possibility","seed":202610053085,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":4,\"b_size\":5,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=4 and |B|=5, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"No","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0086","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"01dbe9ae1c08d1a73c4fe79ce0dc1b4ea49997d1548e6e2c846aa9676d5b10ab","family_id":"qf-bijective-cardinality-possibility","seed":202610053086,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":5,\"b_size\":5,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=5 and |B|=5, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"Yes","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0087","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"692feba4d7e1f8a6d92e71ca8a8cd184e8b62c104be0830297f1c15a466bf4de","family_id":"qf-bijective-cardinality-possibility","seed":202610053087,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":5,\"b_size\":6,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=5 and |B|=6, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"No","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0088","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"24027406457a1ae560412cacc1a6019e5d2a4137f2b76d87df709d047bae8721","family_id":"qf-bijective-cardinality-possibility","seed":202610053088,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":6,\"b_size\":6,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=6 and |B|=6, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"Yes","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0089","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06475a48aa4dd09e338f43bf51e134551b8c130d86c5160f5b671a4985e26471","family_id":"qf-bijective-cardinality-possibility","seed":202610053089,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":6,\"b_size\":7,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=6 and |B|=7, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"No","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0090","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b28cf74cf98ded28cd18b1975071fd4039121411a6d35bad134c0214a0370d7","family_id":"qf-bijective-cardinality-possibility","seed":202610053090,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":7,\"b_size\":7,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=7 and |B|=7, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"Yes","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0091","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6d46d0549eca32f797f3fc410f4474b3576b5e601ebe8f278c0482a3869f56c9","family_id":"qf-bijective-cardinality-possibility","seed":202610053091,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":7,\"b_size\":8,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=7 and |B|=8, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"No","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0092","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fca836e0fab6b0e8ca355893957096b60bf880adfbe6f9c611bd332bf552a411","family_id":"qf-bijective-cardinality-possibility","seed":202610053092,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":8,\"b_size\":8,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=8 and |B|=8, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"Yes","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0093","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ce9c5fd751b2396190c8753a571e67e1b80962f9e52c6390970d251acef1a2fd","family_id":"qf-bijective-cardinality-possibility","seed":202610053093,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":8,\"b_size\":9,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=8 and |B|=9, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"No","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0094","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"44cea468c5dc5e95eee6b39b36f671245d9b86539ce784939b0cedb2647ef31b","family_id":"qf-bijective-cardinality-possibility","seed":202610053094,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":9,\"b_size\":9,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=9 and |B|=9, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"Yes","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0095","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8229f1621538331c95f0fe51337ca6e64ee8af3997cfa535175a6e0056896785","family_id":"qf-bijective-cardinality-possibility","seed":202610053095,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":9,\"b_size\":10,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=9 and |B|=10, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"No","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0096","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fdcab74567c8867249aa0bf44989b7c62e7b865ddc7fab0c5daa0ec49dc134c3","family_id":"qf-bijective-cardinality-possibility","seed":202610053096,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":10,\"b_size\":10,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=10 and |B|=10, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"Yes","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0097","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"75e0b00bdb1b9776a8bdeec88d312d2f3bb06cc1185e38b65ab0aa1e431b999b","family_id":"qf-bijective-cardinality-possibility","seed":202610053097,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":10,\"b_size\":11,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=10 and |B|=11, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"No","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0098","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe83c5460f0330bbb3055afa39316b9cc00e61a5e2897c120b8eba758824bdd5","family_id":"qf-bijective-cardinality-possibility","seed":202610053098,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":11,\"b_size\":11,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=11 and |B|=11, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"Yes","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0099","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"400ad24032ce7a5ba0d46eb9c759c1f219ee72921581813b7986c5c47e79d295","family_id":"qf-bijective-cardinality-possibility","seed":202610053099,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":11,\"b_size\":12,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=11 and |B|=12, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"No","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving No. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r03-bijective-0100","batch_id":"batch-20261004-advance-20261005-r03-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3379a5131650b4174dc903b63294734fbad2ca3367ca863cfe75ff9ea349db2","family_id":"qf-bijective-cardinality-possibility","seed":202610053100,"semantic_template":"qf-bijective-cardinality-possibility/{\"a_size\":12,\"b_size\":12,\"operation\":\"cardinality_possibility\"}"},"fields":{"title":"Existence of a Bijection from Finite Cardinalities","stem":"The explicit finite cardinalities are |A|=12 and |B|=12, with no particular function table supplied. Does some bijection A→B exist?","answer":{"kind":"short_text","value":"Yes","explanation":"Finite sets admit a bijection exactly when their cardinalities agree, giving Yes. This asks about existence rather than testing a given function; two empty sets admit the unique empty bijection."},"skills":["Existence of a Bijection from Finite Cardinalities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}