{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0001","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"25d43e9acb6b8e8034b0d1c38ee296f10492d6794829c1dd19274292362355c2","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172001,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[],\"m\":0,\"n\":0}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(0)=∅→Fin(0) with images=[] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([], Yes, [], [], [])","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([], Yes, [], [], [])."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0002","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"04ea8ff9941b61c5567490387a6d89456d0e2ac1abc13649b66bfbafdd5c3313","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172002,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[],\"m\":2,\"n\":0}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(0)=∅→Fin(2) with images=[] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0003","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ac43c97a7a680c48999a818deb710fa858f561c7ed2b2b702371c00890d86b8c","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172003,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[0],\"m\":1,\"n\":1}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(1)={0,…,0}→Fin(1) with images=[0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([0], Yes, [0], [0], [0])","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0], Yes, [0], [0], [0])."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0004","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"030d45d5de383d1313e91237c594c60d961de2053ff1946b06ac9a43c54ca22b","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172004,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[0,1,2],\"m\":3,\"n\":3}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(3)={0,…,2}→Fin(3) with images=[0, 1, 2] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([0, 1, 2], Yes, [0, 1, 2], [0, 1, 2], [0, 1, 2])","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2], Yes, [0, 1, 2], [0, 1, 2], [0, 1, 2])."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0005","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d2f0bbc9e4f970e2715d0c78a2b80912cfa73e43349cf603d66a13336bcc6c40","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172005,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[0,0,1],\"m\":3,\"n\":3}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(3)={0,…,2}→Fin(3) with images=[0, 0, 1] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([0, 0, 1], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0006","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"caf47eba5fe6ed1e736909517601c03b25e1d8f92592d9e77fc11b5be688d6e6","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172006,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[0,1,1],\"m\":2,\"n\":3}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(3)={0,…,2}→Fin(2) with images=[0, 1, 1] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([0, 1, 1], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 1], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0007","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"66f037a7c3288f51de32185d60f2e52112b2484954073c604e346e46fa789997","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172007,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[0,1],\"m\":3,\"n\":2}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(2)={0,…,1}→Fin(3) with images=[0, 1] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([0, 1], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0008","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a3a76c7e8dff578e1dc708363e2a4382c2c14e707b877733a6d66382b670aed","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172008,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[0,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 0, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([0, 0, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0009","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8465ddde3d61f4ca86e8876d0fac29ead2bd76c924bbdbcb6fe9edf9af25df2d","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172009,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[1,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 0, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([1, 0, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0010","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"96313dda98a3307b0527ed931c3f4edfc14da66d47e18960415318b4411b5210","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172010,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[2,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 0, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([2, 0, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0011","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7a63e12d9fa21960a4619f0f99179c6e9cf72c0bf31a1da95b1f8917df134b4e","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172011,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[0,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 1, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([0, 1, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0012","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1a6a36e4709e4d91210b16219328beb4f1b91f74f60291d15eccc6d8d9ecc32e","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172012,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[1,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 1, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([1, 1, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0013","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4228eb8ab1856a0d5ee4c5b2a75b9ad886b5aed81c16aa8b5794f2a82f7f1c8a","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172013,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[2,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 1, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([2, 1, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0014","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f6a5979b834341d5ab1f9104e7dcf2ec867f04ca9545a261eea77bf2a771270","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172014,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[0,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 2, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([0, 2, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0015","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5bcba877d06c1d5e5fdae6fa21f809879032f77041804ab1e9dbc79385656b2","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172015,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[1,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 2, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([1, 2, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0016","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3dd88c3c50ad8db311c175d1809ef9efa818f596cfc92fb4be64d0791e9aaf67","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172016,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[2,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 2, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([2, 2, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 0, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0017","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3892b90bed3a93c72c677c2d50be4f8960761590787ea1336fe69089924af161","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172017,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[0,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 0, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([0, 0, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0018","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c56d1aa400acd07309d41915e92d1ffc45bfcc3030a0e6b6f0982dbcbba730aa","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172018,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[1,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 0, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([1, 0, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0019","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9fa0bf7e835b3545b3b1c356b8b723d179eb26714f9f3451b69ab2788e1d87ad","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172019,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[2,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 0, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([2, 0, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0020","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b838579ebdab42777628d84f4b1b6203aa6d7432f5cd7f8a6f6331b234fc77ca","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172020,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[0,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 1, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([0, 1, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0021","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca1e381dce6426115296453f830dc2bd668379a4a5cbcffc70695d99fdb31f63","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172021,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[1,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 1, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([1, 1, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0022","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bbc3470966ac4dbe53da3d9c286bbc783afb6928e592871cb40327799858ee9b","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172022,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[2,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 1, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([2, 1, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0023","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"48379b491a668ab52fc2ea9f4d56d5eeef6c6c5129761a67275bf70fa03fbbc1","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172023,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[0,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 2, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([0, 2, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0024","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c583bfbbc9decd17f33e5b1ceb1d682e3e08b7e78c62f53c6e7f3ba6b93f0d13","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172024,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[1,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 2, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([1, 2, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0025","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ece3440255dc566b5490cab080cbd0a130bec7126637e6dfbb5346b0885c4972","family_id":"qf-density17-inverse-bijection-gate-and-table","seed":202610172025,"semantic_template":"qf-density17-inverse-bijection-gate-and-table/{\"images\":[2,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Bijection Guard before Constructing Both Inverse Compositions","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 2, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Give (images, bijectivity, inverse table or not-an-inverse-function, g∘f values or not-evaluated, f∘g values or not-evaluated). Do not confuse inverse functions with reciprocal function values.","answer":{"kind":"expression","value":"([2, 2, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)","explanation":"Require a unique preimage for every declared target before constructing the inverse; compute the two ordered compositions separately. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 1, 0, 0], No, not-an-inverse-function, not-evaluated, not-evaluated)."},"skills":["Bijection Guard before Constructing Both Inverse Compositions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0026","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cb04835c25e7c41729eb3f8fa49f00201d39525ce34c1f2f5bb3ead71a2e6ac0","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172026,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[],\"images\":[],\"m\":0,\"n\":0}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(0)=∅→Fin(0) with images=[] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(0)→Fin(0) has g=[]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([], [], Yes, [], [], Yes, Yes)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([], [], Yes, [], [], Yes, Yes)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0027","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd640edcd3b06cfbb54d3d660b026f13ca3ddf5f11f5a991e0f01f3085fa1fd9","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172027,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[false,false],\"images\":[],\"m\":2,\"n\":0}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(0)=∅→Fin(2) with images=[] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(2)→Fin(0) has g=[-1, 0]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([], [No, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([], [No, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0028","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"60ba6dd6452f2736b1bfa980fc1d294203d964fd89e949ddfdc6816c42199d3b","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172028,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[0,1,0],\"images\":[0,1],\"m\":3,\"n\":2}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(2)={0,…,1}→Fin(3) with images=[0, 1] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(2) has g=[0, 1, 0]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([0, 1], [Yes, Yes, Yes], Yes, [0, 1], [0, 1, 0], Yes, No)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1], [Yes, Yes, Yes], Yes, [0, 1], [0, 1, 0], Yes, No)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0029","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f17bd2b5dc4eb9bfd147500604eae6b8816e6cc43adf11ecab50551a0f620bb","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172029,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[0,1],\"images\":[0,1,1],\"m\":2,\"n\":3}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(3)={0,…,2}→Fin(2) with images=[0, 1, 1] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(2)→Fin(3) has g=[0, 1]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([0, 1, 1], [Yes, Yes], Yes, [0, 1, 1], [0, 1], No, Yes)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 1], [Yes, Yes], Yes, [0, 1, 1], [0, 1], No, Yes)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0030","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ea09e7a670c1f667b4fb7752350b9cbdd0ae86d34d204f3a1499c0655dd0ea08","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172030,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[1,0],\"images\":[1,0],\"m\":2,\"n\":2}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(2)={0,…,1}→Fin(2) with images=[1, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(2)→Fin(2) has g=[1, 0]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([1, 0], [Yes, Yes], Yes, [0, 1], [0, 1], Yes, Yes)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0], [Yes, Yes], Yes, [0, 1], [0, 1], Yes, Yes)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0031","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"66d11fef88f096843be00f26c9b903ec51d958d6ad97838dd9bd17c77427d868","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172031,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[0,1],\"images\":[1,0],\"m\":2,\"n\":2}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(2)={0,…,1}→Fin(2) with images=[1, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(2)→Fin(2) has g=[0, 1]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([1, 0], [Yes, Yes], Yes, [1, 0], [1, 0], No, No)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0], [Yes, Yes], Yes, [1, 0], [1, 0], No, No)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0032","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b055e2e7d6c75d7fc9906f030a326c0029d3914ef53480e1868657a5fd3751ef","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172032,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[true,false],\"images\":[0,1],\"m\":2,\"n\":2}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(2)={0,…,1}→Fin(2) with images=[0, 1] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(2)→Fin(2) has g=[0, 2]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([0, 1], [Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1], [Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0033","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8cb0030277c6dc6296e3a29a49e76f2b975118527cf336ef3302058527920d06","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172033,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[true,true,false],\"images\":[0,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 0, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 5]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([0, 0, 0, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 0, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0034","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a4d832553a120e670a5ec69bc4f328fe4e90a78b4a571867cd6aa7d1a247e53c","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172034,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[0,2,4],\"images\":[1,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 0, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 4]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([1, 0, 0, 0, 0], [Yes, Yes, Yes], Yes, [2, 0, 0, 0, 0], [1, 0, 0], No, No)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 0, 0, 0], [Yes, Yes, Yes], Yes, [2, 0, 0, 0, 0], [1, 0, 0], No, No)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0035","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2acdff3d4482b9fb1fa98fc4c82908d7819d7ec0e5a0bd242c24db4e0f6025c4","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172035,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[true,true,false],\"images\":[2,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 0, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 5]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([2, 0, 0, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 0, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0036","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e43e87b05644b65e6e3a3b59f6ec665350135d3a519b6d1b15dd140dc3a2ab2","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172036,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[0,2,4],\"images\":[0,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 1, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 4]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([0, 1, 0, 0, 0], [Yes, Yes, Yes], Yes, [0, 2, 0, 0, 0], [0, 0, 0], No, No)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 0, 0, 0], [Yes, Yes, Yes], Yes, [0, 2, 0, 0, 0], [0, 0, 0], No, No)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0037","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"943ec6b872b52be2f9fdf97a289664b962cea3363d9799c70a0feb1e6c6106a5","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172037,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[true,true,false],\"images\":[1,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 1, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 5]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([1, 1, 0, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 0, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0038","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f2fd66c12216221346adae8cc3a247f27ffe3f05c9e36fb94e4d7b71e057e816","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172038,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[0,2,4],\"images\":[2,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 1, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 4]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([2, 1, 0, 0, 0], [Yes, Yes, Yes], Yes, [4, 2, 0, 0, 0], [2, 0, 0], No, No)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 0, 0, 0], [Yes, Yes, Yes], Yes, [4, 2, 0, 0, 0], [2, 0, 0], No, No)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0039","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"29a4cd0cb60d6d864099a068aa830acbf58c516ea2e1e728425e7c553b4660f4","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172039,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[true,true,false],\"images\":[0,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 2, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 5]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([0, 2, 0, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 0, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0040","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"18dd6aad3c086f5a5c6e682f5022b76f5c19eda6f8d4aeaf711a90fa2b0d4987","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172040,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[0,2,4],\"images\":[1,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 2, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 4]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([1, 2, 0, 0, 0], [Yes, Yes, Yes], Yes, [2, 4, 0, 0, 0], [1, 0, 0], No, No)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 0, 0, 0], [Yes, Yes, Yes], Yes, [2, 4, 0, 0, 0], [1, 0, 0], No, No)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0041","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa64e3c3efade35d230db08b1eb2cfee71a93feb2f0ee3f5de2c320270ab13f4","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172041,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[true,true,false],\"images\":[2,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 2, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 5]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([2, 2, 0, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 0, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0042","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8efcad508b01a693983fc89fa2b2fb666f361a3fa1d6e597a7ab81528aab4f84","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172042,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[0,2,4],\"images\":[0,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 0, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 4]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([0, 0, 1, 0, 0], [Yes, Yes, Yes], Yes, [0, 0, 2, 0, 0], [0, 1, 0], No, No)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1, 0, 0], [Yes, Yes, Yes], Yes, [0, 0, 2, 0, 0], [0, 1, 0], No, No)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0043","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b79a9c0c23c46a4e58edd6fce51212de18a32b8b575f6bebf15f7c21d0a68208","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172043,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[true,true,false],\"images\":[1,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 0, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 5]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([1, 0, 1, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 1, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0044","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c775733ceb6b4af7ba4eb4006c95561de9bf9c7664703042f8b2f2d550429c9b","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172044,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[0,2,4],\"images\":[2,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 0, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 4]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([2, 0, 1, 0, 0], [Yes, Yes, Yes], Yes, [4, 0, 2, 0, 0], [2, 1, 0], No, No)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 1, 0, 0], [Yes, Yes, Yes], Yes, [4, 0, 2, 0, 0], [2, 1, 0], No, No)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0045","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"48c0f6486fdb8e1c5f5f659bb186b68c82f19c7d3c4b0cc568cd8e10d6e87ddf","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172045,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[true,true,false],\"images\":[0,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 1, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 5]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([0, 1, 1, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 1, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0046","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"df23980d8f7ca23531a34802b3214390b6244f4e37147c1b56f69277fe01fed7","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172046,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[0,2,4],\"images\":[1,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 1, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 4]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([1, 1, 1, 0, 0], [Yes, Yes, Yes], Yes, [2, 2, 2, 0, 0], [1, 1, 0], No, No)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 1, 0, 0], [Yes, Yes, Yes], Yes, [2, 2, 2, 0, 0], [1, 1, 0], No, No)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0047","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e35d995d2e5aacf31f513c1412ceec747cef64b2ae197b529dbc3181f6d891c8","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172047,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[true,true,false],\"images\":[2,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 1, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 5]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([2, 1, 1, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 1, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0048","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cfdafb3fb0500f918e78420e9f504dfc0058893edca034869f549e8a1d74b6b9","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172048,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[0,2,4],\"images\":[0,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 2, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 4]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([0, 2, 1, 0, 0], [Yes, Yes, Yes], Yes, [0, 4, 2, 0, 0], [0, 1, 0], No, No)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 1, 0, 0], [Yes, Yes, Yes], Yes, [0, 4, 2, 0, 0], [0, 1, 0], No, No)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0049","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a885f0278163ccd2ade89add90a709abcf50d0c2ef60a71fffaaaa9c8db6f9d","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172049,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[true,true,false],\"images\":[1,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 2, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 5]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([1, 2, 1, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 1, 0, 0], [Yes, Yes, No], No, not-evaluated, not-evaluated, not-evaluated, not-evaluated)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0050","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fec57faaa0855ae7e40f85a2052c82dd24af2ca5698f980584e0f435510f3c1f","family_id":"qf-density17-inverse-candidate-two-sided","seed":202610172050,"semantic_template":"qf-density17-inverse-candidate-two-sided/{\"g\":[0,2,4],\"images\":[2,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Whole Candidate Inverse Type Check and Two Independent Identities","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 2, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Candidate g:Fin(3)→Fin(5) has g=[0, 2, 4]. Give (images, per-cell g type mask, whole-g eligibility, g∘f values, f∘g values, first identity?, second identity?). If any g cell is untyped, use not-evaluated in the last four fields.","answer":{"kind":"expression","value":"([2, 2, 1, 0, 0], [Yes, Yes, Yes], Yes, [4, 4, 2, 0, 0], [2, 1, 0], No, No)","explanation":"Check all candidate cells, including unused ones, before evaluating either composition. A one-sided identity does not establish the other. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 1, 0, 0], [Yes, Yes, Yes], Yes, [4, 4, 2, 0, 0], [2, 1, 0], No, No)."},"skills":["Whole Candidate Inverse Type Check and Two Independent Identities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0051","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a155203dd60e3f0dc54f4d511273aa9e5a9989171c0742bdc276c5238b70fc9e","family_id":"qf-density17-inverse-query-qualification","seed":202610172051,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[],\"m\":0,\"n\":0,\"queries\":[\"untyped\",\"untyped\"]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(0)=∅→Fin(0) with images=[] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, -1]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([], Yes, [No, No], [outside-codomain, outside-codomain])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([], Yes, [No, No], [outside-codomain, outside-codomain])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0052","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b9343f2ee9eab77375881eaf9d12ff4385206fd6db3e887cfc7bcc9a508b6742","family_id":"qf-density17-inverse-query-qualification","seed":202610172052,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[],\"m\":2,\"n\":0,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(0)=∅→Fin(2) with images=[] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 1]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0053","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"783f5ac0ada312a544b5c435ea9312a6bb3c6946a002b9f4aa6a4330f1254fa3","family_id":"qf-density17-inverse-query-qualification","seed":202610172053,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[0],\"m\":1,\"n\":1,\"queries\":[0,\"untyped\"]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(1)={0,…,0}→Fin(1) with images=[0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 1]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([0], Yes, [Yes, No], [0, outside-codomain])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0], Yes, [Yes, No], [0, outside-codomain])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0054","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"01f15dd278a4bde2e3579102a7d6f05bf758630a242816f923b77e66f835dc86","family_id":"qf-density17-inverse-query-qualification","seed":202610172054,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[0,1,2],\"m\":3,\"n\":3,\"queries\":[2,1]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(3)={0,…,2}→Fin(3) with images=[0, 1, 2] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[2, 1]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([0, 1, 2], Yes, [Yes, Yes], [2, 1])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2], Yes, [Yes, Yes], [2, 1])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0055","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f837865ffbaaa1a721d1b16dca40f9af7d0fae3f28a3b96659e5f0b4a44a4f99","family_id":"qf-density17-inverse-query-qualification","seed":202610172055,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[0,0,1],\"m\":3,\"n\":3,\"queries\":[true,false]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(3)={0,…,2}→Fin(3) with images=[0, 0, 1] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[2, -1]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([0, 0, 1], No, [Yes, No], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1], No, [Yes, No], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0056","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"52281cc4a247e5ffa628d1a925d0826443bf9cea8107eb5294636d12c614248a","family_id":"qf-density17-inverse-query-qualification","seed":202610172056,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[0,1,1],\"m\":2,\"n\":3,\"queries\":[true,false]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(3)={0,…,2}→Fin(2) with images=[0, 1, 1] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([0, 1, 1], No, [Yes, No], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 1], No, [Yes, No], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0057","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0533c4d70f21bcfc7a6560035e7cdc7f046d26dec0a58feeb95095bc5327122e","family_id":"qf-density17-inverse-query-qualification","seed":202610172057,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[0,1],\"m\":3,\"n\":2,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(2)={0,…,1}→Fin(3) with images=[0, 1] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([0, 1], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0058","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ede7538b2817f2f2fb0f99d5b3e85922c600ddd01fe29a0046c47e3328b2f7ac","family_id":"qf-density17-inverse-query-qualification","seed":202610172058,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[0,0,0,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 0, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([0, 0, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0059","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"11eceac0b20d30d7be810fa03a3e6227d30c1022da5d3da7e51a33fbfb2d655e","family_id":"qf-density17-inverse-query-qualification","seed":202610172059,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[1,0,0,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 0, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([1, 0, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0060","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a42d191121d38482dcc08da4f3d0c475dcb63abe0103d41336aed019e093eb52","family_id":"qf-density17-inverse-query-qualification","seed":202610172060,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[2,0,0,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 0, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([2, 0, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0061","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eb7a02b3cc416981860d5ec312b6d18e98d547bcf0d29d8ff3e3514a6fdf770b","family_id":"qf-density17-inverse-query-qualification","seed":202610172061,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[0,1,0,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 1, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([0, 1, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0062","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b70bf62e099212cc76e8048089473e51160733b9fc68bbf82d0a63889ac532ec","family_id":"qf-density17-inverse-query-qualification","seed":202610172062,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[1,1,0,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 1, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([1, 1, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0063","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a44fdcd21d28f52634f9dae13834e773bccb66c0f87a66c386a33cea941ae5e","family_id":"qf-density17-inverse-query-qualification","seed":202610172063,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[2,1,0,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 1, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([2, 1, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0064","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d154336a1b4945f809dd9188aca3686de9f238c00245b1ed4bc29e70eb80a74b","family_id":"qf-density17-inverse-query-qualification","seed":202610172064,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[0,2,0,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 2, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([0, 2, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0065","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7790ed2bd9ef48fce8089a89d21e95f7cec0814e44a86c61f9114e23dd5c92c6","family_id":"qf-density17-inverse-query-qualification","seed":202610172065,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[1,2,0,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 2, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([1, 2, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0066","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4dff1f4fab102badf45f462898f28d2888892e90d3b9ae37d98c067cbe879a7f","family_id":"qf-density17-inverse-query-qualification","seed":202610172066,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[2,2,0,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 2, 0, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([2, 2, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 0, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0067","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"451e144407d79395bd1d4b2edf367cd2b2b0084d527e0112259038650d0100a5","family_id":"qf-density17-inverse-query-qualification","seed":202610172067,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[0,0,1,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 0, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([0, 0, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0068","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fc67b123f8de3fa1f9f299ee7d70145a611372c9d83bf54811eac88e15dd8507","family_id":"qf-density17-inverse-query-qualification","seed":202610172068,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[1,0,1,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 0, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([1, 0, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0069","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ca824b5fe54687b6b861db20a55d21bbc8d970741e98f55227c61ebc6183c12","family_id":"qf-density17-inverse-query-qualification","seed":202610172069,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[2,0,1,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 0, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([2, 0, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0070","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7fecf5c1327e54b2c22a72317ee0f89e8f6675481e44332ca977893906ead20b","family_id":"qf-density17-inverse-query-qualification","seed":202610172070,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[0,1,1,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 1, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([0, 1, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0071","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b7741f9deca4c26d45949ee9ec10d6160496a308192ae9758314f7f85ff29dfe","family_id":"qf-density17-inverse-query-qualification","seed":202610172071,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[1,1,1,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 1, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([1, 1, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0072","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f8748de3b34525d7bc7e168852a227a3087f5c10df64dbe098200bc1e5f2ad6e","family_id":"qf-density17-inverse-query-qualification","seed":202610172072,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[2,1,1,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 1, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([2, 1, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0073","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2617a2ab26a6b599dedecb390717919c1efcccd30197b46cfa7559bba22205c5","family_id":"qf-density17-inverse-query-qualification","seed":202610172073,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[0,2,1,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 2, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([0, 2, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0074","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3c0855e6df0bf07eff0ca8390f5be2d699722998311e3875921118b675cedf88","family_id":"qf-density17-inverse-query-qualification","seed":202610172074,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[1,2,1,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 2, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([1, 2, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0075","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b753ccdb9205824ddc0942b1fac126f4a837aaabfd568b70dc01fbaf967dde2d","family_id":"qf-density17-inverse-query-qualification","seed":202610172075,"semantic_template":"qf-density17-inverse-query-qualification/{\"images\":[2,2,1,0,0],\"m\":3,\"n\":5,\"queries\":[true,true]}"},"fields":{"title":"Inverse Existence and Two Codomain Query Guards","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 2, 1, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Ordered queries=[0, 2]. Give (images, bijectivity, two codomain type guards, two inverse values). If f is not bijective use not-an-inverse-function for each value; otherwise an untyped query yields outside-codomain.","answer":{"kind":"expression","value":"([2, 2, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])","explanation":"Preserve the inverse-existence guard separately from the query masks; an inverse is not a reciprocal. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 1, 0, 0], No, [Yes, Yes], [not-an-inverse-function, not-an-inverse-function])."},"skills":["Inverse Existence and Two Codomain Query Guards"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0076","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65a9c1100c612d1620e7e5b03e787cba7a57ef80e3a66ab96bd95e2861845785","family_id":"qf-density17-inverse-affine-recovery","seed":202610172076,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":-2,\"b\":1,\"ys\":[\"-1/2\",\"-3/2\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=-2≠0, b=1; ordered rational inputs y=[1/-2, -3/2] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([-1/2, 1/2], [3/4, 5/4], [-1/2, -3/2], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-1/2, 1/2], [3/4, 5/4], [-1/2, -3/2], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0077","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4eb432a40923a66a869aa5830bf61b530beada69bc2fc50e8017d5403df823ae","family_id":"qf-density17-inverse-affine-recovery","seed":202610172077,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":-2,\"b\":-11,\"ys\":[\"-11/3\",\"1/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=-2≠0, b=-11; ordered rational inputs y=[-11/3, 2/2] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([-1/2, -11/2], [-11/3, -6/1], [-11/3, 1/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-1/2, -11/2], [-11/3, -6/1], [-11/3, 1/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0078","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"19888a227d20676f009cf2a026119f92025c4389269a15138dac67baffbf0291","family_id":"qf-density17-inverse-affine-recovery","seed":202610172078,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":3,\"b\":-10,\"ys\":[\"-10/3\",\"1/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=3≠0, b=-10; ordered rational inputs y=[-10/3, 3/3] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([1/3, 10/3], [20/9, 11/3], [-10/3, 1/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1/3, 10/3], [20/9, 11/3], [-10/3, 1/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0079","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e6869a442cf3a6ec0654047704a75e56926010731edad3663c021ae6c7e9dff","family_id":"qf-density17-inverse-affine-recovery","seed":202610172079,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":-4,\"b\":-9,\"ys\":[\"-3/1\",\"1/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=-4≠0, b=-9; ordered rational inputs y=[-9/3, 4/4] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([-1/4, -9/4], [-3/2, -5/2], [-3/1, 1/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-1/4, -9/4], [-3/2, -5/2], [-3/1, 1/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0080","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"beeca561a7a3427d7b94c43dc5e321dbed23a2e8729753ede2526776ebd9ddb4","family_id":"qf-density17-inverse-affine-recovery","seed":202610172080,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":5,\"b\":-8,\"ys\":[\"-8/3\",\"5/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=5≠0, b=-8; ordered rational inputs y=[-8/3, 5/1] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([1/5, 8/5], [16/15, 13/5], [-8/3, 5/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1/5, 8/5], [16/15, 13/5], [-8/3, 5/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0081","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2eede15d63084684c9ca9dbd94de6d8d35cb0130668d0f2360d288a25223c7b3","family_id":"qf-density17-inverse-affine-recovery","seed":202610172081,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":-1,\"b\":-7,\"ys\":[\"-7/3\",\"3/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=-1≠0, b=-7; ordered rational inputs y=[-7/3, 6/2] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([-1/1, -7/1], [-14/3, -10/1], [-7/3, 3/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-1/1, -7/1], [-14/3, -10/1], [-7/3, 3/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0082","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"80b8ea52618d6abfd5e97abda51d6d108b5828f96403c250521ad49cee813fe9","family_id":"qf-density17-inverse-affine-recovery","seed":202610172082,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":2,\"b\":-6,\"ys\":[\"-2/1\",\"7/3\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=2≠0, b=-6; ordered rational inputs y=[-6/3, 7/3] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([1/2, 3/1], [2/1, 25/6], [-2/1, 7/3], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1/2, 3/1], [2/1, 25/6], [-2/1, 7/3], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0083","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e87f17a4632263c7208546c305d1e10983b7a0db7b60e67880782f0bd0b1dc9","family_id":"qf-density17-inverse-affine-recovery","seed":202610172083,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":-3,\"b\":-5,\"ys\":[\"-5/3\",\"2/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=-3≠0, b=-5; ordered rational inputs y=[-5/3, 8/4] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([-1/3, -5/3], [-10/9, -7/3], [-5/3, 2/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-1/3, -5/3], [-10/9, -7/3], [-5/3, 2/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0084","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3baede4b1ef8f1c57a6db192cfd0e706c928299f6952b0572a088cb84ec7c86d","family_id":"qf-density17-inverse-affine-recovery","seed":202610172084,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":4,\"b\":-4,\"ys\":[\"-4/3\",\"9/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=4≠0, b=-4; ordered rational inputs y=[-4/3, 9/1] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([1/4, 1/1], [2/3, 13/4], [-4/3, 9/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1/4, 1/1], [2/3, 13/4], [-4/3, 9/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0085","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6d7c2b1c6a5b8f9de498b2665fd2d355837fdedd0f9f5d56c8ca60c47f6dd3da","family_id":"qf-density17-inverse-affine-recovery","seed":202610172085,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":-5,\"b\":-3,\"ys\":[\"-1/1\",\"5/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=-5≠0, b=-3; ordered rational inputs y=[-3/3, 10/2] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([-1/5, -3/5], [-2/5, -8/5], [-1/1, 5/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-1/5, -3/5], [-2/5, -8/5], [-1/1, 5/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0086","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"677f010af4f12c16269cfa69b9e14593fd00511fad903c947a561e1021b9aa86","family_id":"qf-density17-inverse-affine-recovery","seed":202610172086,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":1,\"b\":-2,\"ys\":[\"-2/3\",\"11/3\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=1≠0, b=-2; ordered rational inputs y=[-2/3, 11/3] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([1/1, 2/1], [4/3, 17/3], [-2/3, 11/3], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1/1, 2/1], [4/3, 17/3], [-2/3, 11/3], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0087","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b575abbc997be25378d3d09d1e51f9ac94dbaf612f54872ec80e3828329ce73","family_id":"qf-density17-inverse-affine-recovery","seed":202610172087,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":-2,\"b\":-1,\"ys\":[\"-1/3\",\"3/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=-2≠0, b=-1; ordered rational inputs y=[-1/3, 12/4] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([-1/2, -1/2], [-1/3, -2/1], [-1/3, 3/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-1/2, -1/2], [-1/3, -2/1], [-1/3, 3/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0088","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"26ad8162d70667506ab94e17d1276f3928242390c74439a3354937b3949e9637","family_id":"qf-density17-inverse-affine-recovery","seed":202610172088,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":3,\"b\":0,\"ys\":[\"0/1\",\"13/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=3≠0, b=0; ordered rational inputs y=[0/3, 13/1] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([1/3, 0/1], [0/1, 13/3], [0/1, 13/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1/3, 0/1], [0/1, 13/3], [0/1, 13/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0089","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"617288029795d0156a9c1a5bf5bae189035c11c7514505552fc86c94db446f0b","family_id":"qf-density17-inverse-affine-recovery","seed":202610172089,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":-4,\"b\":1,\"ys\":[\"1/3\",\"7/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=-4≠0, b=1; ordered rational inputs y=[1/3, 14/2] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([-1/4, 1/4], [1/6, -3/2], [1/3, 7/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-1/4, 1/4], [1/6, -3/2], [1/3, 7/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0090","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"743f82bd95518772334486864b8b875282839aeed0f8eb00b5242ea78860cbf1","family_id":"qf-density17-inverse-affine-recovery","seed":202610172090,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":5,\"b\":2,\"ys\":[\"2/3\",\"5/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=5≠0, b=2; ordered rational inputs y=[2/3, 15/3] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([1/5, -2/5], [-4/15, 3/5], [2/3, 5/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1/5, -2/5], [-4/15, 3/5], [2/3, 5/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0091","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8108a1983b1b55c63a419f6d372e0a6dce2ca7460ea3d41a9cf98c89f6878915","family_id":"qf-density17-inverse-affine-recovery","seed":202610172091,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":-1,\"b\":3,\"ys\":[\"1/1\",\"4/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=-1≠0, b=3; ordered rational inputs y=[3/3, 16/4] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([-1/1, 3/1], [2/1, -1/1], [1/1, 4/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-1/1, 3/1], [2/1, -1/1], [1/1, 4/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0092","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b57f3e24396a36d90722f2d720287c2bc8d76a449746daee176490faa99658a","family_id":"qf-density17-inverse-affine-recovery","seed":202610172092,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":2,\"b\":4,\"ys\":[\"4/3\",\"17/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=2≠0, b=4; ordered rational inputs y=[4/3, 17/1] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([1/2, -2/1], [-4/3, 13/2], [4/3, 17/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1/2, -2/1], [-4/3, 13/2], [4/3, 17/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0093","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f578522f4c448e8bdc03d01ca0c03908f5e12b7be6c39015535f16f9ce02479","family_id":"qf-density17-inverse-affine-recovery","seed":202610172093,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":-3,\"b\":5,\"ys\":[\"5/3\",\"9/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=-3≠0, b=5; ordered rational inputs y=[5/3, 18/2] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([-1/3, 5/3], [10/9, -4/3], [5/3, 9/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-1/3, 5/3], [10/9, -4/3], [5/3, 9/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0094","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b6f1a6b5e9ab968ec7962cffc01f7492821c8aa8bf1daf239e47a6edba9828b","family_id":"qf-density17-inverse-affine-recovery","seed":202610172094,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":4,\"b\":6,\"ys\":[\"2/1\",\"19/3\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=4≠0, b=6; ordered rational inputs y=[6/3, 19/3] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([1/4, -3/2], [-1/1, 1/12], [2/1, 19/3], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1/4, -3/2], [-1/1, 1/12], [2/1, 19/3], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0095","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8bf390f4b8a091bf11fb44682e8e3408699c015a50730e4fd126ef6b670cdeb4","family_id":"qf-density17-inverse-affine-recovery","seed":202610172095,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":-5,\"b\":7,\"ys\":[\"7/3\",\"5/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=-5≠0, b=7; ordered rational inputs y=[7/3, 20/4] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([-1/5, 7/5], [14/15, 2/5], [7/3, 5/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-1/5, 7/5], [14/15, 2/5], [7/3, 5/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0096","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"75444640361949ca88741ec88eb34f84db296978158da252b912c54a7c97b56d","family_id":"qf-density17-inverse-affine-recovery","seed":202610172096,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":1,\"b\":8,\"ys\":[\"8/3\",\"21/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=1≠0, b=8; ordered rational inputs y=[8/3, 21/1] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([1/1, -8/1], [-16/3, 13/1], [8/3, 21/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1/1, -8/1], [-16/3, 13/1], [8/3, 21/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0097","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"121cab84c5e8ad7db0e76efefc9a17a933cf9e72fad771e6e77a1008ffe4607e","family_id":"qf-density17-inverse-affine-recovery","seed":202610172097,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":-2,\"b\":9,\"ys\":[\"3/1\",\"11/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=-2≠0, b=9; ordered rational inputs y=[9/3, 22/2] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([-1/2, 9/2], [3/1, -1/1], [3/1, 11/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-1/2, 9/2], [3/1, -1/1], [3/1, 11/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0098","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"949ee47c037d91d4e5e50c485b7ce569f2ee1f1be9cf8c9616e8ca2124388372","family_id":"qf-density17-inverse-affine-recovery","seed":202610172098,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":3,\"b\":10,\"ys\":[\"10/3\",\"23/3\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=3≠0, b=10; ordered rational inputs y=[10/3, 23/3] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([1/3, -10/3], [-20/9, -7/9], [10/3, 23/3], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1/3, -10/3], [-20/9, -7/9], [10/3, 23/3], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0099","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5f2cfbae856098eaf708b7bd5d9651042c019107394d5274af676d24bc3db5ce","family_id":"qf-density17-inverse-affine-recovery","seed":202610172099,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":-4,\"b\":11,\"ys\":[\"11/3\",\"6/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=-4≠0, b=11; ordered rational inputs y=[11/3, 24/4] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([-1/4, 11/4], [11/6, 5/4], [11/3, 6/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([-1/4, 11/4], [11/6, 5/4], [11/3, 6/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r02-inverse-function-0100","batch_id":"batch-20261012-advance-20261017-r02-inverse-function-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b84ccb239c97a3685b98655b196364b6f3681d05a4a0bf63c81620e1122ae37","family_id":"qf-density17-inverse-affine-recovery","seed":202610172100,"semantic_template":"qf-density17-inverse-affine-recovery/{\"a\":5,\"b\":12,\"ys\":[\"4/1\",\"25/1\"]}"},"fields":{"title":"Nonzero-Slope Real Affine Inverse and Two Rational Round Trips","stem":"Declare f:R→R by f(x)=a*x+b with a=5≠0, b=12; ordered rational inputs y=[12/3, 25/1] have nonzero original denominators. Give ([inverse slope, inverse constant], recovered x values, f(x) values, do both return their original y?). Use reduced positive-denominator fractions, including /1.","answer":{"kind":"expression","value":"([1/5, -12/5], [-8/5, 13/5], [4/1, 25/1], Yes)","explanation":"Recover x=(y−b)/a and substitute into the original affine map. Negative original input denominators are normalized; inverse and reciprocal remain different operations. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1/5, -12/5], [-8/5, 13/5], [4/1, 25/1], Yes)."},"skills":["Nonzero-Slope Real Affine Inverse and Two Rational Round Trips"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}