{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r01-injective-0001","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"302577b553b2b54ff931833a3e53dcaa465e1ae7d1251f266548541204d22719","family_id":"qf-injective-table-property","seed":202610051001,"semantic_template":"qf-injective-table-property/{\"A\":[],\"operation\":\"injective\",\"pairs\":[]}"},"fields":{"title":"Injectivity of a Finite Total Function","stem":"The declared total function f:A→B has A=∅, B=∅, and table ∅. Is f injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"Yes","explanation":"Check whether each output fiber has at most one permitted input, giving Yes. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0002","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6be562f0e89d716adc84f6b7b05bb42a371c581aac742e23b83785a1728b200b","family_id":"qf-injective-table-property","seed":202610051002,"semantic_template":"qf-injective-table-property/{\"A\":[-1,0],\"operation\":\"injective\",\"pairs\":[[-1,1],[0,0]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"Yes","explanation":"Check whether each output fiber has at most one permitted input, giving Yes. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0003","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa626c971d01bdcda2ef902f7cc6eb45715f5e5448eeb9b2e1a58d6026fa6845","family_id":"qf-injective-table-property","seed":202610051003,"semantic_template":"qf-injective-table-property/{\"A\":[-1,0],\"operation\":\"injective\",\"pairs\":[[-1,1],[0,3]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"Yes","explanation":"Check whether each output fiber has at most one permitted input, giving Yes. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0004","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"072ff51226035749bc4cf77bdbad6096c7710f6cec40d1dd585304d0cb2345f7","family_id":"qf-injective-table-property","seed":202610051004,"semantic_template":"qf-injective-table-property/{\"A\":[-1,0],\"operation\":\"injective\",\"pairs\":[[-1,0],[0,0]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"No","explanation":"Check whether each output fiber has at most one permitted input, giving No. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0005","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65824e86b193b408391f2419c510c65529a6f31a71a272a4fcc1f98fcdf053a4","family_id":"qf-injective-table-property","seed":202610051005,"semantic_template":"qf-injective-table-property/{\"A\":[-1,0],\"operation\":\"injective\",\"pairs\":[[-1,-1],[0,-1]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"No","explanation":"Check whether each output fiber has at most one permitted input, giving No. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0006","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1cbc28b9aa29d2fc890df12d2b4aade67c8a7ae81f82a3e49fb9071ebbf99fa0","family_id":"qf-injective-table-property","seed":202610051006,"semantic_template":"qf-injective-table-property/{\"A\":[-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-2,1],[-1,2],[0,0]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"Yes","explanation":"Check whether each output fiber has at most one permitted input, giving Yes. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0007","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5f3a187c005993b0e00a8fedfff235b6964aa242d694ddcbe1344192db15f568","family_id":"qf-injective-table-property","seed":202610051007,"semantic_template":"qf-injective-table-property/{\"A\":[-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-2,1],[-1,3],[0,5]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"Yes","explanation":"Check whether each output fiber has at most one permitted input, giving Yes. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0008","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5aa782f80b3928f6390ed2e9242e6b3dd69fa71aca6076fb8e9d5eb0293d3472","family_id":"qf-injective-table-property","seed":202610051008,"semantic_template":"qf-injective-table-property/{\"A\":[-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-2,0],[-1,1],[0,0]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"No","explanation":"Check whether each output fiber has at most one permitted input, giving No. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0009","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc26a483e922033846c3bdc7430eda75b6eeee150f3381881a5d471f3784ad39","family_id":"qf-injective-table-property","seed":202610051009,"semantic_template":"qf-injective-table-property/{\"A\":[-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-2,-1],[-1,-1],[0,-2]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"No","explanation":"Check whether each output fiber has at most one permitted input, giving No. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0010","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c5fe17ab400446e430e088d2e3aab5dcfd8fe32e1fdb31ed9243b6e7fd8f8f40","family_id":"qf-injective-table-property","seed":202610051010,"semantic_template":"qf-injective-table-property/{\"A\":[-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-3,1],[-2,2],[-1,3],[0,0]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"Yes","explanation":"Check whether each output fiber has at most one permitted input, giving Yes. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0011","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e8ed1810d75da31f27bce08cd5b86f7c2a4a082f409ba20034b325439250520c","family_id":"qf-injective-table-property","seed":202610051011,"semantic_template":"qf-injective-table-property/{\"A\":[-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-3,1],[-2,3],[-1,5],[0,7]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"Yes","explanation":"Check whether each output fiber has at most one permitted input, giving Yes. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0012","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9679e23f8fa8179ccaba9f44686161aec0613fb1b8f20e2b5d33cc9b92642d6e","family_id":"qf-injective-table-property","seed":202610051012,"semantic_template":"qf-injective-table-property/{\"A\":[-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-3,0],[-2,1],[-1,2],[0,0]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"No","explanation":"Check whether each output fiber has at most one permitted input, giving No. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0013","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f90e942d69372c8959abe5ba9a1ca7cb71b11faea485f05c665a3625e9e2dbe4","family_id":"qf-injective-table-property","seed":202610051013,"semantic_template":"qf-injective-table-property/{\"A\":[-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-3,-1],[-2,-1],[-1,-2],[0,-2]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"No","explanation":"Check whether each output fiber has at most one permitted input, giving No. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0014","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ea6739aab3abc5434b8959466a3fec024ebc57289daca8386fa7dd9093d403a0","family_id":"qf-injective-table-property","seed":202610051014,"semantic_template":"qf-injective-table-property/{\"A\":[-4,-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"Yes","explanation":"Check whether each output fiber has at most one permitted input, giving Yes. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0015","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0a05c2a64d3d35418bee19d9b88d0cdb51fc054c9c295c32f3050c4d4cb008e7","family_id":"qf-injective-table-property","seed":202610051015,"semantic_template":"qf-injective-table-property/{\"A\":[-4,-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-4,1],[-3,3],[-2,5],[-1,7],[0,9]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"Yes","explanation":"Check whether each output fiber has at most one permitted input, giving Yes. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0016","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6ddc22cec26fb016a984f5b1cf889c757a95b3c68937d0e698ea3e1f639d0594","family_id":"qf-injective-table-property","seed":202610051016,"semantic_template":"qf-injective-table-property/{\"A\":[-4,-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-4,0],[-3,1],[-2,2],[-1,3],[0,0]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"No","explanation":"Check whether each output fiber has at most one permitted input, giving No. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0017","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa9fea2de99d3544eae9190d9b25985562a9bef7fb456aafc59027ce8f1cc134","family_id":"qf-injective-table-property","seed":202610051017,"semantic_template":"qf-injective-table-property/{\"A\":[-4,-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-4,-1],[-3,-1],[-2,-2],[-1,-2],[0,-3]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"No","explanation":"Check whether each output fiber has at most one permitted input, giving No. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0018","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb4b029caee30d9cf089e09c40fa9f9500422cabc0114315c6914a2e077a092d","family_id":"qf-injective-table-property","seed":202610051018,"semantic_template":"qf-injective-table-property/{\"A\":[-5,-4,-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"Yes","explanation":"Check whether each output fiber has at most one permitted input, giving Yes. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0019","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b3fcd8eb3b906f02b0369113bf833c6603337bdb5019c8b4edb21bc5a03c18c","family_id":"qf-injective-table-property","seed":202610051019,"semantic_template":"qf-injective-table-property/{\"A\":[-5,-4,-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-5,1],[-4,3],[-3,5],[-2,7],[-1,9],[0,11]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"Yes","explanation":"Check whether each output fiber has at most one permitted input, giving Yes. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0020","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"83652b0a583a184ce932e752ad275353e259ed98a0bfd7cdc65366469217beb3","family_id":"qf-injective-table-property","seed":202610051020,"semantic_template":"qf-injective-table-property/{\"A\":[-5,-4,-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-5,0],[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"No","explanation":"Check whether each output fiber has at most one permitted input, giving No. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0021","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"794aefc2444ad285d885a226d9e8be91510d065b8d6d56a72e7d1453da1165ec","family_id":"qf-injective-table-property","seed":202610051021,"semantic_template":"qf-injective-table-property/{\"A\":[-5,-4,-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-5,-1],[-4,-1],[-3,-2],[-2,-2],[-1,-3],[0,-3]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"No","explanation":"Check whether each output fiber has at most one permitted input, giving No. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0022","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2af91e76b6b8fb5bd04d5c235d06d5431099a8ada40e82c602e656e362600d86","family_id":"qf-injective-table-property","seed":202610051022,"semantic_template":"qf-injective-table-property/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-6,1],[-5,2],[-4,3],[-3,4],[-2,5],[-1,6],[0,0]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"Yes","explanation":"Check whether each output fiber has at most one permitted input, giving Yes. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0023","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"21806a8fa2aa768a8357e880d7e049b80f92b2e7fbfc921d2a589150099fdcda","family_id":"qf-injective-table-property","seed":202610051023,"semantic_template":"qf-injective-table-property/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-6,1],[-5,3],[-4,5],[-3,7],[-2,9],[-1,11],[0,13]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"Yes","explanation":"Check whether each output fiber has at most one permitted input, giving Yes. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0024","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e6fa90499ca7092b39ef4dce5aafa769e8c1cddaa53e2df28771dcea4b2b813e","family_id":"qf-injective-table-property","seed":202610051024,"semantic_template":"qf-injective-table-property/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-6,0],[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"No","explanation":"Check whether each output fiber has at most one permitted input, giving No. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0025","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"be80721ed8155be21b9b20b0082b7e2488259677e9f2595ed1e544be3dcb1134","family_id":"qf-injective-table-property","seed":202610051025,"semantic_template":"qf-injective-table-property/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"operation\":\"injective\",\"pairs\":[[-6,-1],[-5,-1],[-4,-2],[-3,-2],[-2,-3],[-1,-3],[0,-4]]}"},"fields":{"title":"Injectivity 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 injective? Compare distinct permitted inputs rather than guessing from the codomain size.","answer":{"kind":"short_text","value":"No","explanation":"Check whether each output fiber has at most one permitted input, giving No. Uncovered codomain elements do not defeat injectivity."},"skills":["Injectivity 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-r01-injective-0026","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a984c7b013f06d5b08a9a91b944ded8d4af99c25f8dd25f3acdd6337ff3424be","family_id":"qf-injective-first-collision","seed":202610051026,"semantic_template":"qf-injective-first-collision/{\"A\":[],\"operation\":\"first_collision\",\"pairs\":[]}"},"fields":{"title":"Lexicographically First Collision","stem":"The declared total function f:A→B has A=∅, B=∅, and table ∅. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"none","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is none; repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0027","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9965ec671d498c23db62300886f7b7b5d89d9b567be31f0a38022d3df1473b44","family_id":"qf-injective-first-collision","seed":202610051027,"semantic_template":"qf-injective-first-collision/{\"A\":[-1,0],\"operation\":\"first_collision\",\"pairs\":[[-1,1],[0,0]]}"},"fields":{"title":"Lexicographically First Collision","stem":"The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"none","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is none; repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0028","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b79754b34b8590c8051da577882ba10bc3cd9f5a2ba0458ce3cc50c63d14becd","family_id":"qf-injective-first-collision","seed":202610051028,"semantic_template":"qf-injective-first-collision/{\"A\":[-1,0],\"operation\":\"first_collision\",\"pairs\":[[-1,1],[0,3]]}"},"fields":{"title":"Lexicographically First Collision","stem":"The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"none","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is none; repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0029","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"804f69dd4292ab975c787dcf129f9a8231da25de71f392a751c7d375cb2db379","family_id":"qf-injective-first-collision","seed":202610051029,"semantic_template":"qf-injective-first-collision/{\"A\":[-1,0],\"operation\":\"first_collision\",\"pairs\":[[-1,0],[0,0]]}"},"fields":{"title":"Lexicographically First Collision","stem":"The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"(-1, 0)","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is (-1, 0); repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0030","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9bb6efa7913be2010df2847b79c81343c94633eca542448c97d0c0d2a1a4f28b","family_id":"qf-injective-first-collision","seed":202610051030,"semantic_template":"qf-injective-first-collision/{\"A\":[-1,0],\"operation\":\"first_collision\",\"pairs\":[[-1,-1],[0,-1]]}"},"fields":{"title":"Lexicographically First Collision","stem":"The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"(-1, 0)","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is (-1, 0); repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0031","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7664b0c71a44a90f54d104919d72a4ab36851bfa5c5934a35489fab446c6dafd","family_id":"qf-injective-first-collision","seed":202610051031,"semantic_template":"qf-injective-first-collision/{\"A\":[-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-2,1],[-1,2],[0,0]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"none","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is none; repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0032","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c4cc1803505b9471beae1b86c22f23a977f5580b5526490deace92524cdd6703","family_id":"qf-injective-first-collision","seed":202610051032,"semantic_template":"qf-injective-first-collision/{\"A\":[-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-2,1],[-1,3],[0,5]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"none","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is none; repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0033","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a4818ee7ff1f1047c562b22b8c14619428d78b372a98a8dda39f106834fe35c7","family_id":"qf-injective-first-collision","seed":202610051033,"semantic_template":"qf-injective-first-collision/{\"A\":[-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-2,0],[-1,1],[0,0]]}"},"fields":{"title":"Lexicographically First Collision","stem":"The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"(-2, 0)","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is (-2, 0); repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0034","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"470314b653822b2422d403a9ff25927672a4eac97a37a9e78b7b8d1239af9c04","family_id":"qf-injective-first-collision","seed":202610051034,"semantic_template":"qf-injective-first-collision/{\"A\":[-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-2,-1],[-1,-1],[0,-2]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"(-2, -1)","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is (-2, -1); repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0035","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"92fd4dee3e29d212dc16902a9251d523f04831e1951aab86b34855eb8d20c34f","family_id":"qf-injective-first-collision","seed":202610051035,"semantic_template":"qf-injective-first-collision/{\"A\":[-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-3,1],[-2,2],[-1,3],[0,0]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"none","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is none; repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0036","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e98b98d159caf00d958501ee220cab3b78ec27b1c21519f910d24317c03dde02","family_id":"qf-injective-first-collision","seed":202610051036,"semantic_template":"qf-injective-first-collision/{\"A\":[-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-3,1],[-2,3],[-1,5],[0,7]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"none","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is none; repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0037","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"086acabb5813e8ac94c1f9ba7a4fe7878c0cddcf13fbcbda0f6eb36963728f7b","family_id":"qf-injective-first-collision","seed":202610051037,"semantic_template":"qf-injective-first-collision/{\"A\":[-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-3,0],[-2,1],[-1,2],[0,0]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"(-3, 0)","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is (-3, 0); repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0038","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"28a04ed05b523170ef5c42464c27f5f7cbb87658d613391cccb400087ce327b7","family_id":"qf-injective-first-collision","seed":202610051038,"semantic_template":"qf-injective-first-collision/{\"A\":[-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-3,-1],[-2,-1],[-1,-2],[0,-2]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"(-3, -2)","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is (-3, -2); repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0039","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ba9f18666a11da06a34c2da3697832bf194850bcfad4734bf4dd3118038f979f","family_id":"qf-injective-first-collision","seed":202610051039,"semantic_template":"qf-injective-first-collision/{\"A\":[-4,-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"none","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is none; repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0040","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7fe8c5fcde3509b8c606301c35b12f805d73a147844b2c7533cf5ba2cb790a15","family_id":"qf-injective-first-collision","seed":202610051040,"semantic_template":"qf-injective-first-collision/{\"A\":[-4,-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-4,1],[-3,3],[-2,5],[-1,7],[0,9]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"none","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is none; repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0041","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a7a2ef1ef0f8776e3c0dee68873d50b803b33dec1dcb3103c6c2ef7b0faf6b2","family_id":"qf-injective-first-collision","seed":202610051041,"semantic_template":"qf-injective-first-collision/{\"A\":[-4,-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-4,0],[-3,1],[-2,2],[-1,3],[0,0]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"(-4, 0)","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is (-4, 0); repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0042","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2222d48ee07d5a7e0f50172a992d331d0dc8cce9657c40e2edbdf48fc848a72e","family_id":"qf-injective-first-collision","seed":202610051042,"semantic_template":"qf-injective-first-collision/{\"A\":[-4,-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-4,-1],[-3,-1],[-2,-2],[-1,-2],[0,-3]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"(-4, -3)","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is (-4, -3); repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0043","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e55965e5e30b1b24272717d1290136d040b3831df92bdfcd9a85fb22e61d8b22","family_id":"qf-injective-first-collision","seed":202610051043,"semantic_template":"qf-injective-first-collision/{\"A\":[-5,-4,-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"none","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is none; repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0044","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d07fefaadebc2a6d4565744337daaccc764dc346207303337dedf8105330a775","family_id":"qf-injective-first-collision","seed":202610051044,"semantic_template":"qf-injective-first-collision/{\"A\":[-5,-4,-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-5,1],[-4,3],[-3,5],[-2,7],[-1,9],[0,11]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"none","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is none; repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0045","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ad33701d7509bd82aadef217165aa256a50a49c14b75c1943defe88b89a429a1","family_id":"qf-injective-first-collision","seed":202610051045,"semantic_template":"qf-injective-first-collision/{\"A\":[-5,-4,-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-5,0],[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"(-5, 0)","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is (-5, 0); repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0046","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c672d0dd9de506df4b59d9aef108510a1dfb19f3a4f20bf000c3b9da604c3f0f","family_id":"qf-injective-first-collision","seed":202610051046,"semantic_template":"qf-injective-first-collision/{\"A\":[-5,-4,-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-5,-1],[-4,-1],[-3,-2],[-2,-2],[-1,-3],[0,-3]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"(-5, -4)","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is (-5, -4); repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0047","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fc65edf8a21dae7f1eac723e6fa65ec4199dac61e648ffbc6e8710391163daa5","family_id":"qf-injective-first-collision","seed":202610051047,"semantic_template":"qf-injective-first-collision/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-6,1],[-5,2],[-4,3],[-3,4],[-2,5],[-1,6],[0,0]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"none","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is none; repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0048","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3dc76cf33849ddd6801879785724f92863efa2f4085699d05230195130f83be2","family_id":"qf-injective-first-collision","seed":202610051048,"semantic_template":"qf-injective-first-collision/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-6,1],[-5,3],[-4,5],[-3,7],[-2,9],[-1,11],[0,13]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"none","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is none; repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0049","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1e46a56b8dce49a1700a5f650f8c3fc17a7ed7a03d7ef2a7b83d18e1b15e46f8","family_id":"qf-injective-first-collision","seed":202610051049,"semantic_template":"qf-injective-first-collision/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-6,0],[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"(-6, 0)","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is (-6, 0); repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0050","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d7f87b0e8aa8afa7eca972bb843fce32f14186d87569c6fc57f4c6b07d4e8fa0","family_id":"qf-injective-first-collision","seed":202610051050,"semantic_template":"qf-injective-first-collision/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"operation\":\"first_collision\",\"pairs\":[[-6,-1],[-5,-1],[-4,-2],[-3,-2],[-2,-3],[-1,-3],[0,-4]]}"},"fields":{"title":"Lexicographically First Collision","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)}. Find the first pair x<x′ with f(x)=f(x′), ordered first by x and then x′. Answer none if no collision exists.","answer":{"kind":"short_text","value":"(-6, -5)","explanation":"Group permitted inputs by equal output and order the distinct-input pairs numerically and lexicographically. The first collision is (-6, -5); repeated copies of the same pair are not distinct inputs."},"skills":["Lexicographically First Collision"],"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-r01-injective-0051","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fc64d0f602143a7d9970cec9a566b6b79fa9ad3b3321754da91f8a6c1255cff4","family_id":"qf-injective-fiber-cardinality","seed":202610051051,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[],\"operation\":\"fiber_cardinality\",\"pairs\":[],\"t\":0}"},"fields":{"title":"Cardinality of a Specified Fiber","stem":"The declared total function f:A→B has A=∅, B={0}, and table ∅. For target t=0∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"0","explanation":"Enumerate inputs in A whose output is exactly 0. The fiber cardinality is 0; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0052","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"423e54a9e9bd79944003a5f007cdc320c7877528e02faff0ee6833fdfab2ebd2","family_id":"qf-injective-fiber-cardinality","seed":202610051052,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-1,1],[0,0]],\"t\":1}"},"fields":{"title":"Cardinality of a Specified Fiber","stem":"The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. For target t=1∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"1","explanation":"Enumerate inputs in A whose output is exactly 1. The fiber cardinality is 1; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0053","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1ed1e72cf064f906d69b19c844a16b0e02f579320327d8298d6d8f59d84db934","family_id":"qf-injective-fiber-cardinality","seed":202610051053,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-1,1],[0,3]],\"t\":5}"},"fields":{"title":"Cardinality of a Specified Fiber","stem":"The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. For target t=5∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"0","explanation":"Enumerate inputs in A whose output is exactly 5. The fiber cardinality is 0; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0054","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f6831d5059346d4f6a4d0bd6c2a400ac92150619948276f2bd90e7615ced3fbc","family_id":"qf-injective-fiber-cardinality","seed":202610051054,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-1,0],[0,0]],\"t\":0}"},"fields":{"title":"Cardinality of a Specified Fiber","stem":"The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. For target t=0∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"2","explanation":"Enumerate inputs in A whose output is exactly 0. The fiber cardinality is 2; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0055","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"32aec59f30c3cad59a08cd881fff614239e239b3562151c38d72198ac69aa1a3","family_id":"qf-injective-fiber-cardinality","seed":202610051055,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-1,-1],[0,-1]],\"t\":-2}"},"fields":{"title":"Cardinality of a Specified Fiber","stem":"The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. For target t=-2∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"0","explanation":"Enumerate inputs in A whose output is exactly -2. The fiber cardinality is 0; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0056","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c1da45447fcc188214730f174cae1486eeb19808eacf80ae10c6334fd1f6de05","family_id":"qf-injective-fiber-cardinality","seed":202610051056,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-2,1],[-1,2],[0,0]],\"t\":2}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=2∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"1","explanation":"Enumerate inputs in A whose output is exactly 2. The fiber cardinality is 1; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0057","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36b9aabef5aa71420019a58221e3dbc158099b3b2dd771fcfea2df6d128a1df0","family_id":"qf-injective-fiber-cardinality","seed":202610051057,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-2,1],[-1,3],[0,5]],\"t\":7}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=7∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"0","explanation":"Enumerate inputs in A whose output is exactly 7. The fiber cardinality is 0; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0058","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3063b8b8ff49f61cd00054ab0f06b9954bd667cdb5ea0d752ece93588816db83","family_id":"qf-injective-fiber-cardinality","seed":202610051058,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-2,0],[-1,1],[0,0]],\"t\":1}"},"fields":{"title":"Cardinality of a Specified Fiber","stem":"The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. For target t=1∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"1","explanation":"Enumerate inputs in A whose output is exactly 1. The fiber cardinality is 1; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0059","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"515cf1472ca61804f5dfc52c45ca1e970e9e4a44d35552f9395d0cb7eb342859","family_id":"qf-injective-fiber-cardinality","seed":202610051059,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-2,-1],[-1,-1],[0,-2]],\"t\":-4}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=-4∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"0","explanation":"Enumerate inputs in A whose output is exactly -4. The fiber cardinality is 0; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0060","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"37f3bada3676a4a5f181e0ac501097da7477c08f92cc6be922a2071eba0047f0","family_id":"qf-injective-fiber-cardinality","seed":202610051060,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-3,1],[-2,2],[-1,3],[0,0]],\"t\":1}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=1∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"1","explanation":"Enumerate inputs in A whose output is exactly 1. The fiber cardinality is 1; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0061","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7f0f8d5bd451db6b541feb9207108967409c819e3ee40fd52f61ec78649c7dbc","family_id":"qf-injective-fiber-cardinality","seed":202610051061,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-3,1],[-2,3],[-1,5],[0,7]],\"t\":9}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=9∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"0","explanation":"Enumerate inputs in A whose output is exactly 9. The fiber cardinality is 0; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0062","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0de4512dd6e2b0286fbb7f8c497f0ba7dfa6f5e8da705de5521b9f9c9360a62a","family_id":"qf-injective-fiber-cardinality","seed":202610051062,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-3,0],[-2,1],[-1,2],[0,0]],\"t\":2}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=2∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"1","explanation":"Enumerate inputs in A whose output is exactly 2. The fiber cardinality is 1; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0063","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7c111fd64ef379ba7c2d15fe4f4b22b4ff4bfaab86ecc345c9fa87ed3f2f274b","family_id":"qf-injective-fiber-cardinality","seed":202610051063,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-3,-1],[-2,-1],[-1,-2],[0,-2]],\"t\":-3}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=-3∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"0","explanation":"Enumerate inputs in A whose output is exactly -3. The fiber cardinality is 0; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0064","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4d9b3f993580ee905acc18ec63568c5a346126cbd0aa775a69d756d7a3938db7","family_id":"qf-injective-fiber-cardinality","seed":202610051064,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-4,-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-4,1],[-3,2],[-2,3],[-1,4],[0,0]],\"t\":3}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=3∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"1","explanation":"Enumerate inputs in A whose output is exactly 3. The fiber cardinality is 1; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0065","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0953fd54e4639ca73a7d3d4baabd25d2c7861166e55a07bff7c68dd4e26160d7","family_id":"qf-injective-fiber-cardinality","seed":202610051065,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-4,-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-4,1],[-3,3],[-2,5],[-1,7],[0,9]],\"t\":11}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=11∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"0","explanation":"Enumerate inputs in A whose output is exactly 11. The fiber cardinality is 0; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0066","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b5ecd4fd928c983ed0de838d93de8591841c674a0519570dd1d3e5d5652bd8cc","family_id":"qf-injective-fiber-cardinality","seed":202610051066,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-4,-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-4,0],[-3,1],[-2,2],[-1,3],[0,0]],\"t\":3}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=3∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"1","explanation":"Enumerate inputs in A whose output is exactly 3. The fiber cardinality is 1; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0067","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"df840bb2422f9a471a8ea5412f16cb76cbf1fbdc292c30d2ccd5084839009f67","family_id":"qf-injective-fiber-cardinality","seed":202610051067,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-4,-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-4,-1],[-3,-1],[-2,-2],[-1,-2],[0,-3]],\"t\":-2}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=-2∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"2","explanation":"Enumerate inputs in A whose output is exactly -2. The fiber cardinality is 2; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0068","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a3ed842dcbb5ba15c1355c3ec59eae8e39f16fd3b976d4b759585a14ddd1dbd","family_id":"qf-injective-fiber-cardinality","seed":202610051068,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-5,-4,-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]],\"t\":5}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=5∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"1","explanation":"Enumerate inputs in A whose output is exactly 5. The fiber cardinality is 1; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0069","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"96068c5f50e2bbb5b52e3272066489762cb952788b17da30c95ecc3b4bc75af0","family_id":"qf-injective-fiber-cardinality","seed":202610051069,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-5,-4,-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-5,1],[-4,3],[-3,5],[-2,7],[-1,9],[0,11]],\"t\":13}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=13∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"0","explanation":"Enumerate inputs in A whose output is exactly 13. The fiber cardinality is 0; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0070","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d56a0fc330efa5c68ca15a76261c3883e8e82b87f870b8ea9dfd4ef54f9c81d2","family_id":"qf-injective-fiber-cardinality","seed":202610051070,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-5,-4,-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-5,0],[-4,1],[-3,2],[-2,3],[-1,4],[0,0]],\"t\":4}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=4∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"1","explanation":"Enumerate inputs in A whose output is exactly 4. The fiber cardinality is 1; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0071","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db13219cd8d10a21e944271ded8715cc2fb4d14e2e565e4bf71860c7a213444f","family_id":"qf-injective-fiber-cardinality","seed":202610051071,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-5,-4,-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-5,-1],[-4,-1],[-3,-2],[-2,-2],[-1,-3],[0,-3]],\"t\":-1}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=-1∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"2","explanation":"Enumerate inputs in A whose output is exactly -1. The fiber cardinality is 2; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0072","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5bff91a2f4905928df10e767b2f5e77d945613801cfe2c44c1d0b68c31a63b88","family_id":"qf-injective-fiber-cardinality","seed":202610051072,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-6,1],[-5,2],[-4,3],[-3,4],[-2,5],[-1,6],[0,0]],\"t\":0}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=0∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"1","explanation":"Enumerate inputs in A whose output is exactly 0. The fiber cardinality is 1; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0073","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1b9311e011372ec58f1fe2f89ae21b89be41d9b391e6f927f70d1eafa0977a24","family_id":"qf-injective-fiber-cardinality","seed":202610051073,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-6,1],[-5,3],[-4,5],[-3,7],[-2,9],[-1,11],[0,13]],\"t\":15}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=15∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"0","explanation":"Enumerate inputs in A whose output is exactly 15. The fiber cardinality is 0; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0074","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c143c4b9311aa5bc5cee97b48c57047835742e6b62dc75aac66daa38276a189","family_id":"qf-injective-fiber-cardinality","seed":202610051074,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-6,0],[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]],\"t\":5}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=5∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"1","explanation":"Enumerate inputs in A whose output is exactly 5. The fiber cardinality is 1; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0075","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"41a2051963df91fb629d55fc8dfa2d3b2857d3dd5a0dbb6cda7425c02f8200d7","family_id":"qf-injective-fiber-cardinality","seed":202610051075,"semantic_template":"qf-injective-fiber-cardinality/{\"A\":[-6,-5,-4,-3,-2,-1,0],\"operation\":\"fiber_cardinality\",\"pairs\":[[-6,-1],[-5,-1],[-4,-2],[-3,-2],[-2,-3],[-1,-3],[0,-4]],\"t\":-8}"},"fields":{"title":"Cardinality of a Specified Fiber","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)}. For target t=-8∈B, find the cardinality of its preimage fiber, including a zero-size fiber. The target is not an input.","answer":{"kind":"short_text","value":"0","explanation":"Enumerate inputs in A whose output is exactly -8. The fiber cardinality is 0; it is neither an output value nor a probability."},"skills":["Cardinality of a Specified Fiber"],"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-r01-injective-0076","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"640dec7a262b5b09f0efda1d3c94cb7c317c315699717d61dc6cb85547b42492","family_id":"qf-injective-restriction-property","seed":202610051076,"semantic_template":"qf-injective-restriction-property/{\"C\":[],\"operation\":\"restricted_injective\",\"pairs\":[]}"},"fields":{"title":"Injectivity after Domain Restriction","stem":"The declared total function f:A→B has A={0, 1}, B={0}, and table {(0, 0), (1, 0)}. For C=∅⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0077","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b1fb52281869a93753a2de8951501c600201d5af8300020882bd01bf4f63255","family_id":"qf-injective-restriction-property","seed":202610051077,"semantic_template":"qf-injective-restriction-property/{\"C\":[0],\"operation\":\"restricted_injective\",\"pairs\":[[0,0]]}"},"fields":{"title":"Injectivity after Domain Restriction","stem":"The declared total function f:A→B has A={-1, 0}, B={0, 1}, and table {(-1, 1), (0, 0)}. For C={0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0078","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0fa569feeed80a965af0f27d1bb6b110fe01815b45543ee39659ea469cdda372","family_id":"qf-injective-restriction-property","seed":202610051078,"semantic_template":"qf-injective-restriction-property/{\"C\":[0],\"operation\":\"restricted_injective\",\"pairs\":[[0,3]]}"},"fields":{"title":"Injectivity after Domain Restriction","stem":"The declared total function f:A→B has A={-1, 0}, B={1, 3, 5}, and table {(-1, 1), (0, 3)}. For C={0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0079","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb482ddd4aab9d9ea962650dba5d160f9eb80e5796ddab7d0e4a55bd3f7cda19","family_id":"qf-injective-restriction-property","seed":202610051079,"semantic_template":"qf-injective-restriction-property/{\"C\":[-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-1,0],[0,0]]}"},"fields":{"title":"Injectivity after Domain Restriction","stem":"The declared total function f:A→B has A={-1, 0}, B={0}, and table {(-1, 0), (0, 0)}. For C={-1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"No","explanation":"Compare outputs of distinct inputs in C only, obtaining No. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0080","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0dec7beee80c75c6d40589a996ff7831b1c6fcca662dc1802a41ccaad3cb1a75","family_id":"qf-injective-restriction-property","seed":202610051080,"semantic_template":"qf-injective-restriction-property/{\"C\":[0],\"operation\":\"restricted_injective\",\"pairs\":[[0,-1]]}"},"fields":{"title":"Injectivity after Domain Restriction","stem":"The declared total function f:A→B has A={-1, 0}, B={-3, -2, -1}, and table {(-1, -1), (0, -1)}. For C={0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0081","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"54beb3bfec06b303146c1ec4aab33f072499fa742fe1a4f8dfb41372693c90de","family_id":"qf-injective-restriction-property","seed":202610051081,"semantic_template":"qf-injective-restriction-property/{\"C\":[-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-1,2],[0,0]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0082","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca476598694d5b5fe487f44b4753b36f593d47c9c6ddc05569e4f539c182ac59","family_id":"qf-injective-restriction-property","seed":202610051082,"semantic_template":"qf-injective-restriction-property/{\"C\":[-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-2,1],[-1,3],[0,5]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0083","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3d5a65d155b4c8fbcf12e858e6ed6d049523ee7822f815e0a69b9e1c66f2326e","family_id":"qf-injective-restriction-property","seed":202610051083,"semantic_template":"qf-injective-restriction-property/{\"C\":[-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-1,1],[0,0]]}"},"fields":{"title":"Injectivity after Domain Restriction","stem":"The declared total function f:A→B has A={-2, -1, 0}, B={0, 1}, and table {(-2, 0), (-1, 1), (0, 0)}. For C={-1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0084","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa23572df0a7b81e5b5efc17951b4c006a620c27968a8962cebd96b1eca39547","family_id":"qf-injective-restriction-property","seed":202610051084,"semantic_template":"qf-injective-restriction-property/{\"C\":[-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-1,-1],[0,-2]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0085","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa99dc7f9ee775c44db71d1f8849100c8d5842238326c0c96fa7360c95e13ce5","family_id":"qf-injective-restriction-property","seed":202610051085,"semantic_template":"qf-injective-restriction-property/{\"C\":[-3,-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-3,1],[-2,2],[-1,3],[0,0]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-3, -2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0086","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"60f11b1c31b2cdeca42a86a16a1e06c807b4accba11c3260925b5718c3911539","family_id":"qf-injective-restriction-property","seed":202610051086,"semantic_template":"qf-injective-restriction-property/{\"C\":[-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-2,3],[-1,5],[0,7]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0087","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb8d3c26aa5b49eca898977faa5acb55db8f7f7a28518175e62e0366cc6c398a","family_id":"qf-injective-restriction-property","seed":202610051087,"semantic_template":"qf-injective-restriction-property/{\"C\":[-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-2,1],[-1,2],[0,0]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0088","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"593909a73c4bd5a449a58c9f70891778387308963d4b2ec483d5fcd1bb9bfb3e","family_id":"qf-injective-restriction-property","seed":202610051088,"semantic_template":"qf-injective-restriction-property/{\"C\":[-3,-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-3,-1],[-2,-1],[-1,-2],[0,-2]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-3, -2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"No","explanation":"Compare outputs of distinct inputs in C only, obtaining No. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0089","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d612d7c093863d5fe1773ec25e3ab0853b249d72ea459255ae4ec2b9c8ac9ee7","family_id":"qf-injective-restriction-property","seed":202610051089,"semantic_template":"qf-injective-restriction-property/{\"C\":[-3,-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-3,2],[-2,3],[-1,4],[0,0]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-3, -2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0090","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4e47cb43345bb875b4f4774a97ef28d8bfd762aa9006e3f151856f0dce9ec695","family_id":"qf-injective-restriction-property","seed":202610051090,"semantic_template":"qf-injective-restriction-property/{\"C\":[-3,-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-3,3],[-2,5],[-1,7],[0,9]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-3, -2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0091","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7339cf1a1bc8e2eb5491ca028ee9eea59e7f60f7344b92db96373a535c36c027","family_id":"qf-injective-restriction-property","seed":202610051091,"semantic_template":"qf-injective-restriction-property/{\"C\":[-4,-3,-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-4,0],[-3,1],[-2,2],[-1,3],[0,0]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-4, -3, -2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"No","explanation":"Compare outputs of distinct inputs in C only, obtaining No. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0092","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"00b5c0523e2f861de1d080ace5e8e64fd267f0547059335d27937e100b623644","family_id":"qf-injective-restriction-property","seed":202610051092,"semantic_template":"qf-injective-restriction-property/{\"C\":[-3,-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-3,-1],[-2,-2],[-1,-2],[0,-3]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-3, -2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"No","explanation":"Compare outputs of distinct inputs in C only, obtaining No. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0093","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"46dfaf71e821f7d7cde3bd1ec91a57c14eaef40af5d05e8dd52b1430eeb5f7ef","family_id":"qf-injective-restriction-property","seed":202610051093,"semantic_template":"qf-injective-restriction-property/{\"C\":[-4,-3,-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-4, -3, -2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0094","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a7738fcdb1f1f5ab65689528dc3c2d71ba453828fec3c136cdaf221b71427ee7","family_id":"qf-injective-restriction-property","seed":202610051094,"semantic_template":"qf-injective-restriction-property/{\"C\":[-5,-4,-3,-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-5,1],[-4,3],[-3,5],[-2,7],[-1,9],[0,11]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-5, -4, -3, -2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0095","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"092ec819cca7b092036e2a028270346147378d10d0f224d573c37a1f1a7e2eb7","family_id":"qf-injective-restriction-property","seed":202610051095,"semantic_template":"qf-injective-restriction-property/{\"C\":[-4,-3,-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-4,1],[-3,2],[-2,3],[-1,4],[0,0]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-4, -3, -2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0096","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ea26cf0e880527f41b2eae0694d5c176745a07d7c63d97a3dbdfd60c1b480b19","family_id":"qf-injective-restriction-property","seed":202610051096,"semantic_template":"qf-injective-restriction-property/{\"C\":[-4,-3,-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-4,-1],[-3,-2],[-2,-2],[-1,-3],[0,-3]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-4, -3, -2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"No","explanation":"Compare outputs of distinct inputs in C only, obtaining No. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0097","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f316de18a454a8dcc80d6efb905d6370f74ca35dc259a820905def3b89f28c8e","family_id":"qf-injective-restriction-property","seed":202610051097,"semantic_template":"qf-injective-restriction-property/{\"C\":[-6,-5,-4,-3,-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-6,1],[-5,2],[-4,3],[-3,4],[-2,5],[-1,6],[0,0]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-6, -5, -4, -3, -2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0098","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e4e172fb45eceb7b5431351b6949bd853c9bd88be24b3dcc46c01e2ada072ddf","family_id":"qf-injective-restriction-property","seed":202610051098,"semantic_template":"qf-injective-restriction-property/{\"C\":[-5,-4,-3,-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-5,3],[-4,5],[-3,7],[-2,9],[-1,11],[0,13]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-5, -4, -3, -2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0099","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"13a467c15a77f66139dad12f10fe92b7bba0de0328474954a75912143854c9d1","family_id":"qf-injective-restriction-property","seed":202610051099,"semantic_template":"qf-injective-restriction-property/{\"C\":[-5,-4,-3,-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-5,1],[-4,2],[-3,3],[-2,4],[-1,5],[0,0]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-5, -4, -3, -2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"Yes","explanation":"Compare outputs of distinct inputs in C only, obtaining Yes. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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-r01-injective-0100","batch_id":"batch-20261004-advance-20261005-r01-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4646fc6e7ffc1fc00c4e2e67886b2d67a1a590a96f44c1a6d354796467c307bc","family_id":"qf-injective-restriction-property","seed":202610051100,"semantic_template":"qf-injective-restriction-property/{\"C\":[-6,-5,-4,-3,-2,-1,0],\"operation\":\"restricted_injective\",\"pairs\":[[-6,-1],[-5,-1],[-4,-2],[-3,-2],[-2,-3],[-1,-3],[0,-4]]}"},"fields":{"title":"Injectivity after Domain Restriction","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)}. For C={-6, -5, -4, -3, -2, -1, 0}⊆A, is the restriction f|C:C→B injective? Consider only the restricted function.","answer":{"kind":"short_text","value":"No","explanation":"Compare outputs of distinct inputs in C only, obtaining No. Collisions outside C do not count, and empty C satisfies the condition. This does not establish injectivity of the original function."},"skills":["Injectivity after Domain Restriction"],"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."}}