{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r05-finite-group-0001","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3c42f6b65c8b66ad7b5064705401f9e81fa4cae4c0a1f8a662f0ae0c7d9d1f10","family_id":"qf-str22-group-table-qualification","seed":202610225001,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0002","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"84363736555949c3fc64fc138b9bcd921dd6a3299f198a313b5ba7f358d62b9b","family_id":"qf-str22-group-table-qualification","seed":202610225002,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0003","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"09ea32d28e9374bd9d03fe62b4ade9251bc085b19778c7c5368580e5ac637527","family_id":"qf-str22-group-table-qualification","seed":202610225003,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[1,3,0,4,2],[3,4,1,2,0],[0,1,2,3,4],[4,2,3,0,1],[2,0,4,1,3]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0004","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bbc0064a88342b8133e4e75c331b87cd823db4212f6397093dabc6561dc1c0c4","family_id":"qf-str22-group-table-qualification","seed":202610225004,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[1,2,4,0,3],[2,4,3,1,0],[4,3,0,2,1],[0,1,2,3,4],[3,0,1,4,2]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0005","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"161da3379f4771573275d2df4c90611930b241f9d9f2d1c24d49cb83aa1bd820","family_id":"qf-str22-group-table-qualification","seed":202610225005,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0006","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"59edafb68ade31dbcbfbe388934e44b5e93acc90a1b3e23c8ab27eb3f45d7e41","family_id":"qf-str22-group-table-qualification","seed":202610225006,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0007","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a98a83f07fedbad4b32cbf74fe18717b6767893b21e6139b05eb08a67e1e14c","family_id":"qf-str22-group-table-qualification","seed":202610225007,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[4,0,3,1,2],[0,1,2,3,4],[3,2,0,4,1],[1,3,4,2,0],[2,4,1,0,3]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0008","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"856c4e622d8e7c60ce28e5ce59b50afe9c079415665c74cbefc711add3847195","family_id":"qf-str22-group-table-qualification","seed":202610225008,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[4,3,0,2,1],[3,0,1,4,2],[0,1,2,3,4],[2,4,3,1,0],[1,2,4,0,3]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0009","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"51e3a9823da480cbadd0813bbe639b901f01f65b2ab435977c5afe8a402d8566","family_id":"qf-str22-group-table-qualification","seed":202610225009,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[4,2,3,0,1],[2,0,4,1,3],[3,4,1,2,0],[0,1,2,3,4],[1,3,0,4,2]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0010","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d54c1488d0c978b0b55073694986bcbec14de8654cff7bc4ce5bac3760d37ce","family_id":"qf-str22-group-table-qualification","seed":202610225010,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[3,2,4,1,0],[2,0,3,4,1],[4,3,1,0,2],[1,4,0,2,3],[0,1,2,3,4]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0011","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f63f9604aeb703b4fa793ab5bef44546d88ec30dfd54794a7c69a96639815994","family_id":"qf-str22-group-table-qualification","seed":202610225011,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0012","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"94b0e2109e533616babcd38d808df570cd4cbf733bd901f18d0d5de9fcc88795","family_id":"qf-str22-group-table-qualification","seed":202610225012,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0013","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c2d72cc0c3e7d5dc5ae77b4a7c2726f826661cc6b6228bf6be8ef08c465e1162","family_id":"qf-str22-group-table-qualification","seed":202610225013,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[4,2,0,1,3],[2,3,1,4,0],[0,1,2,3,4],[1,4,3,0,2],[3,0,4,2,1]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0014","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4cf27cbed6d4f969acdce1e95a96b82acfd0d9f6d6c8e1e5e4536e79dcc460d5","family_id":"qf-str22-group-table-qualification","seed":202610225014,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[4,3,1,0,2],[3,2,4,1,0],[1,4,0,2,3],[0,1,2,3,4],[2,0,3,4,1]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0015","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65cafebadab8c5dd7b22c5680005ee6d4f02651d3cfef5d46320694018bdd77e","family_id":"qf-str22-group-table-qualification","seed":202610225015,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[3,4,1,2,0],[4,2,3,0,1],[1,3,0,4,2],[2,0,4,1,3],[0,1,2,3,4]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0016","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"50db2522c11ce98efe56389fa5f9a4c960bc652639c2c752379c99ecbf0ac48c","family_id":"qf-str22-group-table-qualification","seed":202610225016,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0017","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5ebb5412970a03d3697b4bd7cfb2cd3e15f53bd8c9e92c4037a909223720a984","family_id":"qf-str22-group-table-qualification","seed":202610225017,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[3,0,1,4,2],[0,1,2,3,4],[1,2,4,0,3],[4,3,0,2,1],[2,4,3,1,0]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0018","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"85d2490217d2397b1a77bf5fd2106e2cb8ae97f5a0d6cc0c996be37b0b0fb4dd","family_id":"qf-str22-group-table-qualification","seed":202610225018,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[3,2,0,4,1],[2,4,1,0,3],[0,1,2,3,4],[4,0,3,1,2],[1,3,4,2,0]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0019","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65372870dbec67eef6617dc6e8a0e3ab7ee4737273fbbb972f0ef23dd41d803e","family_id":"qf-str22-group-table-qualification","seed":202610225019,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0020","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4c6001ba60f452e0bcd7d297296aa26a8828907e8124a5a9e80f77492630dbda","family_id":"qf-str22-group-table-qualification","seed":202610225020,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[2,4,3,1,0],[4,3,0,2,1],[3,0,1,4,2],[1,2,4,0,3],[0,1,2,3,4]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0021","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c396c3f58f71417d6c2188da792359e10c819b1be385661b0ab1450194351ff","family_id":"qf-str22-group-table-qualification","seed":202610225021,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0022","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"057d016b6b8066df30658293853d1f86fc9232028c11791ef6abb65216f7d76d","family_id":"qf-str22-group-table-qualification","seed":202610225022,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[3,0,4,2,1],[0,1,2,3,4],[4,2,0,1,3],[2,3,1,4,0],[1,4,3,0,2]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0023","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"214336a2ac4ec9a7cbe9e93d3da7cb4a98ac2a4a450eeb791a812156e2a5657e","family_id":"qf-str22-group-table-qualification","seed":202610225023,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0024","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1a96dd97d2101ddc159be0c2c5fb21132e4ca41dd05159e36c8197763c1677ec","family_id":"qf-str22-group-table-qualification","seed":202610225024,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[2,4,1,0,3],[4,0,3,1,2],[1,3,4,2,0],[0,1,2,3,4],[3,2,0,4,1]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0025","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81b84ce2dc99cf818a9e190e81af9ed230140e0058f227f55d9404eb658c7c4e","family_id":"qf-str22-group-table-qualification","seed":202610225025,"semantic_template":"qf-str22-group-table-qualification/{\"n\":5,\"table\":[[2,3,1,4,0],[3,0,4,2,1],[1,4,3,0,2],[4,2,0,1,3],[0,1,2,3,4]]}"},"fields":{"title":"Closure of Every Cell in a Complete Cayley Table","stem":"The complete finite Cayley table=[[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. This family permits out-of-carrier entries to test qualification. Give (carrier size,table,does every entry lie in the carrier?).","answer":{"kind":"expression","value":"(5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], Yes)","explanation":"Inspect every table cell. Closure alone does not imply associativity, an identity or inverses. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], Yes)."},"skills":["Closure of Every Cell in a Complete Cayley Table"],"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-20261013-advance-20261022-r05-finite-group-0026","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e31809e89cb2f28fa102cec9a04baf99b6134e07b93dfc0590142cec69d583aa","family_id":"qf-str22-group-associativity-audit","seed":202610225026,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0027","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5cbc23e928b367350d5b84b09264f29f2f484abdcbcc39ea6691fa5d79f6db1e","family_id":"qf-str22-group-associativity-audit","seed":202610225027,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0028","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b7cee801cb6104d2039d89975aa717d3d11ea2fef9dbfda75c3b38a20187d0df","family_id":"qf-str22-group-associativity-audit","seed":202610225028,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[1,3,0,4,2],[3,4,1,2,0],[0,1,2,3,4],[4,2,3,0,1],[2,0,4,1,3]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0029","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"55ac68fc0113b6dbce22d2f607fffed3228de27c4960527be04f770b409eeb5b","family_id":"qf-str22-group-associativity-audit","seed":202610225029,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[1,2,4,0,3],[2,4,3,1,0],[4,3,0,2,1],[0,1,2,3,4],[3,0,1,4,2]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0030","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1bb52d6060d25de3ac0b367d27f60284532fa8ee688c5884ace619dd4f90b8ff","family_id":"qf-str22-group-associativity-audit","seed":202610225030,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0031","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6c37affeb89b11795374759820839558442837c78b8c3ffc41be9daf94d9a6b3","family_id":"qf-str22-group-associativity-audit","seed":202610225031,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0032","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1a3fd0136b612b9ea690dedb41023a77e04ec938b9f8be9d8da879f031ff0fde","family_id":"qf-str22-group-associativity-audit","seed":202610225032,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[4,0,3,1,2],[0,1,2,3,4],[3,2,0,4,1],[1,3,4,2,0],[2,4,1,0,3]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0033","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"571355db2ff48101030109c275021984e024e03967652c2ae91007b3f9688992","family_id":"qf-str22-group-associativity-audit","seed":202610225033,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[4,3,0,2,1],[3,0,1,4,2],[0,1,2,3,4],[2,4,3,1,0],[1,2,4,0,3]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0034","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"04b15d234f4b8cdac8beaa614fafc69fd6773342dde9568c23c3d389c08cbbba","family_id":"qf-str22-group-associativity-audit","seed":202610225034,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[4,2,3,0,1],[2,0,4,1,3],[3,4,1,2,0],[0,1,2,3,4],[1,3,0,4,2]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0035","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"41cca2478e831c34963a7cc863539d88f25177ca96daddfb331738a7c6b5c339","family_id":"qf-str22-group-associativity-audit","seed":202610225035,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[3,2,4,1,0],[2,0,3,4,1],[4,3,1,0,2],[1,4,0,2,3],[0,1,2,3,4]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0036","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5bdd02a7cc11d8ed0bd50a38ec169edeec517cff1b2bcf4d5d18d334ededd667","family_id":"qf-str22-group-associativity-audit","seed":202610225036,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0037","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"39fcdf15d7d582b3798277da71f31cb7ff6359294f05e31a401307526713584d","family_id":"qf-str22-group-associativity-audit","seed":202610225037,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0038","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a48121a10be9427e846a5ba15d93cc126d761ebc2f3cb88303976292bc2854b1","family_id":"qf-str22-group-associativity-audit","seed":202610225038,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[4,2,0,1,3],[2,3,1,4,0],[0,1,2,3,4],[1,4,3,0,2],[3,0,4,2,1]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0039","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"123e5beee4fecedcdc4fc4772aa485bdd2b63765af712b818d22778010652188","family_id":"qf-str22-group-associativity-audit","seed":202610225039,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[4,3,1,0,2],[3,2,4,1,0],[1,4,0,2,3],[0,1,2,3,4],[2,0,3,4,1]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0040","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f52adf21a19d4e28e3be80afbaeef6b3001697502f793d6e1c4c0f19356e47c","family_id":"qf-str22-group-associativity-audit","seed":202610225040,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[3,4,1,2,0],[4,2,3,0,1],[1,3,0,4,2],[2,0,4,1,3],[0,1,2,3,4]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0041","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ac8b121d3cebacad350bb2a1a26bb60c7b79660a5a8aed5d84001ad4e370d37","family_id":"qf-str22-group-associativity-audit","seed":202610225041,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0042","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f0cd6ff8642faf446caba48f2b950ec33124c5404e2446d421a9c14290020e21","family_id":"qf-str22-group-associativity-audit","seed":202610225042,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[3,0,1,4,2],[0,1,2,3,4],[1,2,4,0,3],[4,3,0,2,1],[2,4,3,1,0]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0043","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb8ecd4e3a76d6e7c361aad882fe03203a023188b559a44f626e35f65af0aa0d","family_id":"qf-str22-group-associativity-audit","seed":202610225043,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[3,2,0,4,1],[2,4,1,0,3],[0,1,2,3,4],[4,0,3,1,2],[1,3,4,2,0]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0044","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0dc00a1de931529619984c7fdadf5cf2a694e1fd7095676cc0e5aa0d235d08f0","family_id":"qf-str22-group-associativity-audit","seed":202610225044,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0045","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ac281f97b108bfdfe7c9eb8f9fe842239db5e5ae1c9308355c1bdc9962598971","family_id":"qf-str22-group-associativity-audit","seed":202610225045,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[2,4,3,1,0],[4,3,0,2,1],[3,0,1,4,2],[1,2,4,0,3],[0,1,2,3,4]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0046","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"35c9192c2bde4e0dd2cedf1249108ae9da1df9b8f7311d12b7d3aab209b6c216","family_id":"qf-str22-group-associativity-audit","seed":202610225046,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0047","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d06b2b34d4fc911e6588016d8b5b8cb196a609f0ba6d0201873d1de98a059777","family_id":"qf-str22-group-associativity-audit","seed":202610225047,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[3,0,4,2,1],[0,1,2,3,4],[4,2,0,1,3],[2,3,1,4,0],[1,4,3,0,2]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0048","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a4a480e587a8341d674d05904bdb7e952ee95bbc805cf015f43fdce565ede64e","family_id":"qf-str22-group-associativity-audit","seed":202610225048,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0049","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c32cb3a0848f1a70457f7cddd643d3d272a43b73ba027573d7af0e9aa1858e22","family_id":"qf-str22-group-associativity-audit","seed":202610225049,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[2,4,1,0,3],[4,0,3,1,2],[1,3,4,2,0],[0,1,2,3,4],[3,2,0,4,1]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0050","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb59c0e02397c5b3c29bc0a9aa20e83199809896fd40a991751f27a2ec3315af","family_id":"qf-str22-group-associativity-audit","seed":202610225050,"semantic_template":"qf-str22-group-associativity-audit/{\"n\":5,\"table\":[[2,3,1,4,0],[3,0,4,2,1],[1,4,3,0,2],[4,2,0,1,3],[0,1,2,3,4]]}"},"fields":{"title":"Exhaustive Associativity over All Ordered Triples","stem":"The complete finite Cayley table=[[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require every entry in the carrier. Give (size,table,every [a,b,c] violating (a*b)*c=a*(b*c),associative?). List bad triples in nested increasing a,b,c order, including all triples with repeated elements.","answer":{"kind":"expression","value":"(5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], [], Yes)","explanation":"Compare both parenthesizations for every ordered triple. Symmetry or a few successful cases does not replace this exhaustive check. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], [], Yes)."},"skills":["Exhaustive Associativity over All Ordered Triples"],"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-20261013-advance-20261022-r05-finite-group-0051","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"725f2833bab2bccd83d0b83c9eaa2cfece603b6c7e1c0ed54bf3b3107d652920","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225051,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0052","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3b7e8f327f871bad34b8170065cb54982183d9425446aefa064354c7c36f824d","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225052,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0053","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cbf916044a85fcad9323985e76fc5d5d634ce842dfda2a83ec88137cf3220e0f","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225053,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[1,3,0,4,2],[3,4,1,2,0],[0,1,2,3,4],[4,2,3,0,1],[2,0,4,1,3]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0054","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"46203c8d9e6c32f1a3dedd8a024707895341314dbd4b48ec26155ad05f586982","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225054,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[1,2,4,0,3],[2,4,3,1,0],[4,3,0,2,1],[0,1,2,3,4],[3,0,1,4,2]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0055","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b0b14bd07aafb70917952fd9b693408058c631bdbba7842652badb594e8c1320","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225055,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0056","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"25c7332c1424a88e4d7f3da68aa479969d76630a7033086be1612159ffa8634c","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225056,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], [0], [[0, [0]], [1, [3]], [2, [4]], [3, [1]], [4, [2]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], [0], [[0, [0]], [1, [3]], [2, [4]], [3, [1]], [4, [2]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0057","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"203c925279578f9967707df8800b28166140f90a1e76e269992ca11df657acd6","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225057,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[4,0,3,1,2],[0,1,2,3,4],[3,2,0,4,1],[1,3,4,2,0],[2,4,1,0,3]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], [1], [[0, [3]], [1, [1]], [2, [4]], [3, [0]], [4, [2]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], [1], [[0, [3]], [1, [1]], [2, [4]], [3, [0]], [4, [2]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0058","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b60c0657795c8d3bac73445f1140f457bcbe64ccd6d4228cad9201fb5c582c33","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225058,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[4,3,0,2,1],[3,0,1,4,2],[0,1,2,3,4],[2,4,3,1,0],[1,2,4,0,3]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], [2], [[0, [3]], [1, [4]], [2, [2]], [3, [0]], [4, [1]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], [2], [[0, [3]], [1, [4]], [2, [2]], [3, [0]], [4, [1]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0059","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9fa195b55d6dbb9548e54c71a52049279e960b21a3dfd6e9a53d99faccaac24","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225059,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[4,2,3,0,1],[2,0,4,1,3],[3,4,1,2,0],[0,1,2,3,4],[1,3,0,4,2]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], [3], [[0, [2]], [1, [4]], [2, [0]], [3, [3]], [4, [1]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], [3], [[0, [2]], [1, [4]], [2, [0]], [3, [3]], [4, [1]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0060","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b2ecb79278b8709fbf1815bebc080fcb0b2e302f82312e43da61297aeb0732bd","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225060,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[3,2,4,1,0],[2,0,3,4,1],[4,3,1,0,2],[1,4,0,2,3],[0,1,2,3,4]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], [4], [[0, [2]], [1, [3]], [2, [0]], [3, [1]], [4, [4]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], [4], [[0, [2]], [1, [3]], [2, [0]], [3, [1]], [4, [4]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0061","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9c6f173e08a79cb26fa894b766a5c5d7a71f78cf978bb61a145f4fd8c5f9408b","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225061,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0062","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81fd15e3235c4671f1215323fcb73428623ad7e554dd0b41bd6bf69c022a975a","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225062,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0063","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"694483e5ee0394473bad0e2014758fdbe780c6448c7229a90ec889de5cd19cde","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225063,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[4,2,0,1,3],[2,3,1,4,0],[0,1,2,3,4],[1,4,3,0,2],[3,0,4,2,1]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0064","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"508d49483594b4d2a50fc694d389ba6b343db81891437821b97393171b00b952","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225064,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[4,3,1,0,2],[3,2,4,1,0],[1,4,0,2,3],[0,1,2,3,4],[2,0,3,4,1]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0065","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ba7622aa8f09b099d56526b34a2d007f872dae21962746200c31b6c80cf33229","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225065,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[3,4,1,2,0],[4,2,3,0,1],[1,3,0,4,2],[2,0,4,1,3],[0,1,2,3,4]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0066","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06d09615fc1485e6309c7de1d9fe1545a59710fd85595d72ad7a4da9bc0f26f0","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225066,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0067","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c884ba276679cde6d5de60ab99dc21becc719cdf962bdfa1778a7e196cb0af8","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225067,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[3,0,1,4,2],[0,1,2,3,4],[1,2,4,0,3],[4,3,0,2,1],[2,4,3,1,0]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0068","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"42905c4181b1d667fcfdbec64e62f114fe7fe055adbd505bc22c4595cca9f7ca","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225068,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[3,2,0,4,1],[2,4,1,0,3],[0,1,2,3,4],[4,0,3,1,2],[1,3,4,2,0]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0069","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"774749aafce1e379ae02ceec4e86706eedb260def6d36297bed2a01c04be4338","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225069,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0070","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91f23767389db57fc4e902c13d179e8b4ea2e5c1565af27a5822aa281d8c718b","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225070,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[2,4,3,1,0],[4,3,0,2,1],[3,0,1,4,2],[1,2,4,0,3],[0,1,2,3,4]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0071","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c60bedb35c309cf5d5c3073d86c9d16dcb2f7411e16132c46bf986adb465ee9","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225071,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0072","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fd982f127d0c50b7bfd2055b39bd6385751bfd9dc0deed049aae6d3fce7b2a89","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225072,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[3,0,4,2,1],[0,1,2,3,4],[4,2,0,1,3],[2,3,1,4,0],[1,4,3,0,2]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0073","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ab3c0125acaf8c5e3786b51a14be8866e15d8c3b3a9e5f7813a763b69ce43a8","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225073,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0074","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d28a0e233ca81590af8a735d15d40ed76c6a5631a9a543ea895635e8cc818608","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225074,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[2,4,1,0,3],[4,0,3,1,2],[1,3,4,2,0],[0,1,2,3,4],[3,2,0,4,1]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0075","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c5150dcae78507a126c90def183db11d66639f6bf41298fb30f97daef28b0af3","family_id":"qf-str22-group-identity-inverse-inventory","seed":202610225075,"semantic_template":"qf-str22-group-identity-inverse-inventory/{\"n\":5,\"table\":[[2,3,1,4,0],[3,0,4,2,1],[1,4,3,0,2],[4,2,0,1,3],[0,1,2,3,4]]}"},"fields":{"title":"Two-Sided Identity and Two-Sided Inverse Inventories","stem":"The complete finite Cayley table=[[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require closure. Give (size,table,all two-sided identity labels,every [x,all two-sided inverse labels] row relative to a unique identity,all elements have such an inverse relative to a unique identity?). If no unique identity is available, output an empty inverse-row list and a final No.","answer":{"kind":"expression","value":"(5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], Yes)","explanation":"Require both e*x=x and x*e=x. Inverse candidates must satisfy both x*y=e and y*x=e for the same identity. This inventory alone does not audit associativity. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], Yes)."},"skills":["Two-Sided Identity and Two-Sided Inverse Inventories"],"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-20261013-advance-20261022-r05-finite-group-0076","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1e3849cc1bc61e78ffb079824e8e424741d354ab1046a1b225da84c8746f03f3","family_id":"qf-str22-group-full-group-audit","seed":202610225076,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], Yes, Yes, [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]], Yes, Yes, [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0077","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0e60623f6e8fb95d3532e40ca79844fcfdc08f8393d5bc36632a064dd67b9553","family_id":"qf-str22-group-full-group-audit","seed":202610225077,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], Yes, Yes, [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]], Yes, Yes, [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0078","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"90927caffe9cd0d7bbf05faf7fe5f4190413a8e4bca7a9014173427e660e5205","family_id":"qf-str22-group-full-group-audit","seed":202610225078,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[1,3,0,4,2],[3,4,1,2,0],[0,1,2,3,4],[4,2,3,0,1],[2,0,4,1,3]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], Yes, Yes, [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 3, 0, 4, 2], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [4, 2, 3, 0, 1], [2, 0, 4, 1, 3]], Yes, Yes, [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0079","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"017836badd4a1b209f900688b0646d1fc36d2db356ed526afdbfc300e2456ce3","family_id":"qf-str22-group-full-group-audit","seed":202610225079,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[1,2,4,0,3],[2,4,3,1,0],[4,3,0,2,1],[0,1,2,3,4],[3,0,1,4,2]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], Yes, Yes, [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [0, 1, 2, 3, 4], [3, 0, 1, 4, 2]], Yes, Yes, [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0080","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2edcb8265ee372ea6a54bfe181108604ad0845aebe858d287bf19bf054d977ad","family_id":"qf-str22-group-full-group-audit","seed":202610225080,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], Yes, Yes, [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4]], Yes, Yes, [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0081","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3dd33bbae99f2ba7eef7adb953010a9d8acedcfe62ef417be5b28013acf165c5","family_id":"qf-str22-group-full-group-audit","seed":202610225081,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], Yes, Yes, [0], [[0, [0]], [1, [3]], [2, [4]], [3, [1]], [4, [2]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]], Yes, Yes, [0], [[0, [0]], [1, [3]], [2, [4]], [3, [1]], [4, [2]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0082","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"252752304e9b8a39e6e5cb1815b96290ce9fef78ed3d90e8f7dd0d38693449a2","family_id":"qf-str22-group-full-group-audit","seed":202610225082,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[4,0,3,1,2],[0,1,2,3,4],[3,2,0,4,1],[1,3,4,2,0],[2,4,1,0,3]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], Yes, Yes, [1], [[0, [3]], [1, [1]], [2, [4]], [3, [0]], [4, [2]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 3, 1, 2], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3]], Yes, Yes, [1], [[0, [3]], [1, [1]], [2, [4]], [3, [0]], [4, [2]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0083","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4d990f141d96cf10f4591efd9d9c239c306e27ec1ee8f3c8b59853d20ba6886b","family_id":"qf-str22-group-full-group-audit","seed":202610225083,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[4,3,0,2,1],[3,0,1,4,2],[0,1,2,3,4],[2,4,3,1,0],[1,2,4,0,3]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], Yes, Yes, [2], [[0, [3]], [1, [4]], [2, [2]], [3, [0]], [4, [1]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [2, 4, 3, 1, 0], [1, 2, 4, 0, 3]], Yes, Yes, [2], [[0, [3]], [1, [4]], [2, [2]], [3, [0]], [4, [1]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0084","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d17806638e84e77db153e321c1667e0dc55aa9d5631ae3dd2689001aaa1920b5","family_id":"qf-str22-group-full-group-audit","seed":202610225084,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[4,2,3,0,1],[2,0,4,1,3],[3,4,1,2,0],[0,1,2,3,4],[1,3,0,4,2]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], Yes, Yes, [3], [[0, [2]], [1, [4]], [2, [0]], [3, [3]], [4, [1]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 3, 0, 1], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [0, 1, 2, 3, 4], [1, 3, 0, 4, 2]], Yes, Yes, [3], [[0, [2]], [1, [4]], [2, [0]], [3, [3]], [4, [1]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0085","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf0aa12c07d5586d75ee7c38bd57cca6de0b914fba34bce95fe75bfe394ac0d3","family_id":"qf-str22-group-full-group-audit","seed":202610225085,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[3,2,4,1,0],[2,0,3,4,1],[4,3,1,0,2],[1,4,0,2,3],[0,1,2,3,4]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], Yes, Yes, [4], [[0, [2]], [1, [3]], [2, [0]], [3, [1]], [4, [4]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 4, 1, 0], [2, 0, 3, 4, 1], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4]], Yes, Yes, [4], [[0, [2]], [1, [3]], [2, [0]], [3, [1]], [4, [4]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0086","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eaf44f935af502cf7e2ee6d94039d881ed0cc84378b22eac38eff365e7d70129","family_id":"qf-str22-group-full-group-audit","seed":202610225086,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], Yes, Yes, [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]], Yes, Yes, [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0087","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"97822750209188fa66c825e82fbe1c79b3344b1961c8c492d659036724bda951","family_id":"qf-str22-group-full-group-audit","seed":202610225087,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], Yes, Yes, [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2]], Yes, Yes, [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0088","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6fdd4a44633b0f184444e05461f5ba193e9ce6d177c94644130aaf0bceb0c9d6","family_id":"qf-str22-group-full-group-audit","seed":202610225088,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[4,2,0,1,3],[2,3,1,4,0],[0,1,2,3,4],[1,4,3,0,2],[3,0,4,2,1]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], Yes, Yes, [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [3, 0, 4, 2, 1]], Yes, Yes, [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0089","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"309af9a3e1cfc44227bfe268d47a20f4b511e494bf6fb6fe08ddb8c06bb8a2dd","family_id":"qf-str22-group-full-group-audit","seed":202610225089,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[4,3,1,0,2],[3,2,4,1,0],[1,4,0,2,3],[0,1,2,3,4],[2,0,3,4,1]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], Yes, Yes, [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[4, 3, 1, 0, 2], [3, 2, 4, 1, 0], [1, 4, 0, 2, 3], [0, 1, 2, 3, 4], [2, 0, 3, 4, 1]], Yes, Yes, [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0090","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a4a8788a0398a50cb910bac42b18e27080f14125f93fc68a907dc012ee283aa7","family_id":"qf-str22-group-full-group-audit","seed":202610225090,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[3,4,1,2,0],[4,2,3,0,1],[1,3,0,4,2],[2,0,4,1,3],[0,1,2,3,4]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], Yes, Yes, [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 1, 2, 0], [4, 2, 3, 0, 1], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [0, 1, 2, 3, 4]], Yes, Yes, [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0091","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f16fea44d4f791c48243241affc12743b1e3d5cee0dec24bb17061965bef92f","family_id":"qf-str22-group-full-group-audit","seed":202610225091,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], Yes, Yes, [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]], Yes, Yes, [0], [[0, [0]], [1, [2]], [2, [1]], [3, [4]], [4, [3]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0092","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a353c1147ce5575f0e2bdb4d7540d1d7a83700ca830be9f2ba0c3697c74604f6","family_id":"qf-str22-group-full-group-audit","seed":202610225092,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[3,0,1,4,2],[0,1,2,3,4],[1,2,4,0,3],[4,3,0,2,1],[2,4,3,1,0]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], Yes, Yes, [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 1, 4, 2], [0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [4, 3, 0, 2, 1], [2, 4, 3, 1, 0]], Yes, Yes, [1], [[0, [2]], [1, [1]], [2, [0]], [3, [4]], [4, [3]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0093","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"61f66421f98066b97b71400dbe71bcdf3b47e4c85982594c6e8d65da73777c29","family_id":"qf-str22-group-full-group-audit","seed":202610225093,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[3,2,0,4,1],[2,4,1,0,3],[0,1,2,3,4],[4,0,3,1,2],[1,3,4,2,0]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], Yes, Yes, [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 2, 0, 4, 1], [2, 4, 1, 0, 3], [0, 1, 2, 3, 4], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0]], Yes, Yes, [2], [[0, [1]], [1, [0]], [2, [2]], [3, [4]], [4, [3]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0094","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0e3d8017243665bf8393223135bc135b502d6a5120ef98e64084fab3a91ddfe1","family_id":"qf-str22-group-full-group-audit","seed":202610225094,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], Yes, Yes, [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0]], Yes, Yes, [3], [[0, [1]], [1, [0]], [2, [4]], [3, [3]], [4, [2]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0095","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5523744e0ee5579f608107fe462f3a218ce31605b567a56dc684775cd6feaf99","family_id":"qf-str22-group-full-group-audit","seed":202610225095,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[2,4,3,1,0],[4,3,0,2,1],[3,0,1,4,2],[1,2,4,0,3],[0,1,2,3,4]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], Yes, Yes, [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 3, 1, 0], [4, 3, 0, 2, 1], [3, 0, 1, 4, 2], [1, 2, 4, 0, 3], [0, 1, 2, 3, 4]], Yes, Yes, [4], [[0, [1]], [1, [0]], [2, [3]], [3, [2]], [4, [4]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0096","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1d856ac9fa3f7230355694c1d1b4e6cc5a5c04e41c23ce96a63311629031b300","family_id":"qf-str22-group-full-group-audit","seed":202610225096,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], Yes, Yes, [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]], Yes, Yes, [0], [[0, [0]], [1, [4]], [2, [3]], [3, [2]], [4, [1]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0097","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1363b726bea783b9e8cffb87bde4db686d983c774ed8c99856694c380bf7e18d","family_id":"qf-str22-group-full-group-audit","seed":202610225097,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[3,0,4,2,1],[0,1,2,3,4],[4,2,0,1,3],[2,3,1,4,0],[1,4,3,0,2]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], Yes, Yes, [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 0, 4, 2, 1], [0, 1, 2, 3, 4], [4, 2, 0, 1, 3], [2, 3, 1, 4, 0], [1, 4, 3, 0, 2]], Yes, Yes, [1], [[0, [4]], [1, [1]], [2, [3]], [3, [2]], [4, [0]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0098","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"556d5ef24d5d0e630044f891df45a6fd3092b2227034e3e88c910354702cac01","family_id":"qf-str22-group-full-group-audit","seed":202610225098,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[3,4,0,1,2],[4,0,1,2,3],[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], Yes, Yes, [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[3, 4, 0, 1, 2], [4, 0, 1, 2, 3], [0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1]], Yes, Yes, [2], [[0, [4]], [1, [3]], [2, [2]], [3, [1]], [4, [0]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0099","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4b7b61ff6c45fa6e565eefa2e5be25c097107d510b0ba99fe18f14452621e5a1","family_id":"qf-str22-group-full-group-audit","seed":202610225099,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[2,4,1,0,3],[4,0,3,1,2],[1,3,4,2,0],[0,1,2,3,4],[3,2,0,4,1]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], Yes, Yes, [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 4, 1, 0, 3], [4, 0, 3, 1, 2], [1, 3, 4, 2, 0], [0, 1, 2, 3, 4], [3, 2, 0, 4, 1]], Yes, Yes, [3], [[0, [4]], [1, [2]], [2, [1]], [3, [3]], [4, [0]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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-20261013-advance-20261022-r05-finite-group-0100","batch_id":"batch-20261013-advance-20261022-r05-finite-group-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2bc911ebf26ffa4f775822bf5a9ae908aefc34878661c7ad6ba048d5cc935f4d","family_id":"qf-str22-group-full-group-audit","seed":202610225100,"semantic_template":"qf-str22-group-full-group-audit/{\"n\":5,\"table\":[[2,3,1,4,0],[3,0,4,2,1],[1,4,3,0,2],[4,2,0,1,3],[0,1,2,3,4]]}"},"fields":{"title":"Close Every Group Obligation on the Specified Finite Carrier","stem":"The complete finite Cayley table=[[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]] uses elements Fin(5)={0,…,4} and entry table[a][b]=a*b. Preserve element labels and row/column order. A sampled row or triple does not prove the group axioms. Require table entries in the carrier. Give (size,table,closed?,associative?,two-sided identities,inverse rows relative to a unique identity,is this table a group?). Audit all ordered triples and all elements, not samples.","answer":{"kind":"expression","value":"(5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], Yes, Yes, [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], Yes)","explanation":"A group requires closure, complete associativity, a two-sided identity and two-sided inverses for every element. Preserve the labels and all inverse candidates. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, [[2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [1, 4, 3, 0, 2], [4, 2, 0, 1, 3], [0, 1, 2, 3, 4]], Yes, Yes, [4], [[0, [3]], [1, [2]], [2, [1]], [3, [0]], [4, [4]]], Yes)."},"skills":["Close Every Group Obligation on the Specified Finite Carrier"],"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."}}