{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0001","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8aa1e0a4855ab4a3915b1b2143c640aa84e975f290213315e473d63534c74be2","family_id":"qf-density17-bij-fibers-and-class","seed":202610171001,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[],\"m\":0,\"n\":0}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([], [], [], [], Yes, Yes, Yes)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([], [], [], [], Yes, Yes, Yes)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0002","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa0d0977ef6ad852a2d90d273bbac14473343d47a171b27f3bb4401d60d2885d","family_id":"qf-density17-bij-fibers-and-class","seed":202610171002,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[],\"m\":2,\"n\":0}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([], [[], []], [], [0, 1], Yes, No, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([], [[], []], [], [0, 1], Yes, No, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0003","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f577dc70f98fac0d8530f1dbb4fab2f4f98906afd303b81b700cc717750e6cdd","family_id":"qf-density17-bij-fibers-and-class","seed":202610171003,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[0],\"m\":1,\"n\":1}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([0], [[0]], [], [], Yes, Yes, Yes)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0], [[0]], [], [], Yes, Yes, Yes)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0004","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6f8dd5a4c2f2491a922fe9b3aad4517997a8696372940a52b5c4680d7552e237","family_id":"qf-density17-bij-fibers-and-class","seed":202610171004,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[0,1,2],\"m\":3,\"n\":3}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([0, 1, 2], [[0], [1], [2]], [], [], Yes, Yes, Yes)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2], [[0], [1], [2]], [], [], Yes, Yes, Yes)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0005","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3cdd452df14a6cd49eaa8a749bad263f0ba712a15747d73807ea97883ed036d8","family_id":"qf-density17-bij-fibers-and-class","seed":202610171005,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[0,0,1],\"m\":3,\"n\":3}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([0, 0, 1], [[0, 1], [2], []], [[0, 1]], [2], No, No, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1], [[0, 1], [2], []], [[0, 1]], [2], No, No, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0006","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e8e230f6a04f50555e7c21d9dbb70b0e1c969e450774fa24ae4cc8d0fef0667f","family_id":"qf-density17-bij-fibers-and-class","seed":202610171006,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[0,1,1],\"m\":2,\"n\":3}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([0, 1, 1], [[0], [1, 2]], [[1, 2]], [], No, Yes, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 1], [[0], [1, 2]], [[1, 2]], [], No, Yes, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0007","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"78d438082c1fb78ffc0c861600a4b3f564449988c0fae2faaabc3dd67c527ed8","family_id":"qf-density17-bij-fibers-and-class","seed":202610171007,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[0,1],\"m\":3,\"n\":2}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([0, 1], [[0], [1], []], [], [2], Yes, No, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1], [[0], [1], []], [], [2], Yes, No, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0008","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"44e9d0b1a1641c86396cf2a68953819c55aa1ddd74af519d653f62fb313eeda1","family_id":"qf-density17-bij-fibers-and-class","seed":202610171008,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[0,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([0, 0, 0, 0, 0], [[0, 1, 2, 3, 4], [], []], [[0, 1], [0, 2], [0, 3], [0, 4], [1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]], [1, 2], No, No, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 0, 0, 0], [[0, 1, 2, 3, 4], [], []], [[0, 1], [0, 2], [0, 3], [0, 4], [1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]], [1, 2], No, No, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0009","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9d1f6318a585704ce0424975817a8c499224e5d12058a64581a4632f638bf76","family_id":"qf-density17-bij-fibers-and-class","seed":202610171009,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[1,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([1, 0, 0, 0, 0], [[1, 2, 3, 4], [0], []], [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]], [2], No, No, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 0, 0, 0], [[1, 2, 3, 4], [0], []], [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]], [2], No, No, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0010","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0366552161a6aa2e97a309177037bbe53bc8ba60c7eade8c6b85360d469744c3","family_id":"qf-density17-bij-fibers-and-class","seed":202610171010,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[2,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([2, 0, 0, 0, 0], [[1, 2, 3, 4], [], [0]], [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]], [1], No, No, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 0, 0, 0], [[1, 2, 3, 4], [], [0]], [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]], [1], No, No, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0011","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c0a3a5f0ef716368714899ac4117c70fdd65d96d9a2c8ff117dac6e6ca4f9e1","family_id":"qf-density17-bij-fibers-and-class","seed":202610171011,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[0,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([0, 1, 0, 0, 0], [[0, 2, 3, 4], [1], []], [[0, 2], [0, 3], [0, 4], [2, 3], [2, 4], [3, 4]], [2], No, No, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 0, 0, 0], [[0, 2, 3, 4], [1], []], [[0, 2], [0, 3], [0, 4], [2, 3], [2, 4], [3, 4]], [2], No, No, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0012","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9bb2063ff9a1988a277edc9d734cef72b5d46acde91257e2605849d2aeb6891d","family_id":"qf-density17-bij-fibers-and-class","seed":202610171012,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[1,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([1, 1, 0, 0, 0], [[2, 3, 4], [0, 1], []], [[0, 1], [2, 3], [2, 4], [3, 4]], [2], No, No, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 0, 0, 0], [[2, 3, 4], [0, 1], []], [[0, 1], [2, 3], [2, 4], [3, 4]], [2], No, No, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0013","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fdef79a4372b7816c7c2cba5148e64d773431fa3a1f7369d6d58f1da834a2f76","family_id":"qf-density17-bij-fibers-and-class","seed":202610171013,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[2,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([2, 1, 0, 0, 0], [[2, 3, 4], [1], [0]], [[2, 3], [2, 4], [3, 4]], [], No, Yes, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 0, 0, 0], [[2, 3, 4], [1], [0]], [[2, 3], [2, 4], [3, 4]], [], No, Yes, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0014","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"00378706db4ff3ba2d1db4cd9d8f977358c00f93cd647c58bdba2ad303c7b2cd","family_id":"qf-density17-bij-fibers-and-class","seed":202610171014,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[0,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([0, 2, 0, 0, 0], [[0, 2, 3, 4], [], [1]], [[0, 2], [0, 3], [0, 4], [2, 3], [2, 4], [3, 4]], [1], No, No, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 0, 0, 0], [[0, 2, 3, 4], [], [1]], [[0, 2], [0, 3], [0, 4], [2, 3], [2, 4], [3, 4]], [1], No, No, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0015","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d6ab4d6ca29f877a1671081baef8a278dd4257200c621e194fe3db4c7e948892","family_id":"qf-density17-bij-fibers-and-class","seed":202610171015,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[1,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([1, 2, 0, 0, 0], [[2, 3, 4], [0], [1]], [[2, 3], [2, 4], [3, 4]], [], No, Yes, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 0, 0, 0], [[2, 3, 4], [0], [1]], [[2, 3], [2, 4], [3, 4]], [], No, Yes, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0016","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c4ca564f28e340b88556d4980af10f0736f1fb3ce34263ada269800c0d0aede4","family_id":"qf-density17-bij-fibers-and-class","seed":202610171016,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[2,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([2, 2, 0, 0, 0], [[2, 3, 4], [], [0, 1]], [[0, 1], [2, 3], [2, 4], [3, 4]], [1], No, No, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 0, 0, 0], [[2, 3, 4], [], [0, 1]], [[0, 1], [2, 3], [2, 4], [3, 4]], [1], No, No, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0017","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"270e3651414a2c38673f45a0677de35be806a2d8358e313e957252c64a50b0e0","family_id":"qf-density17-bij-fibers-and-class","seed":202610171017,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[0,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([0, 0, 1, 0, 0], [[0, 1, 3, 4], [2], []], [[0, 1], [0, 3], [0, 4], [1, 3], [1, 4], [3, 4]], [2], No, No, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1, 0, 0], [[0, 1, 3, 4], [2], []], [[0, 1], [0, 3], [0, 4], [1, 3], [1, 4], [3, 4]], [2], No, No, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0018","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b0541365ae6c8e844c008c373831d91664e627ff4fd6d848555756febf5ed8c1","family_id":"qf-density17-bij-fibers-and-class","seed":202610171018,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[1,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([1, 0, 1, 0, 0], [[1, 3, 4], [0, 2], []], [[0, 2], [1, 3], [1, 4], [3, 4]], [2], No, No, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 1, 0, 0], [[1, 3, 4], [0, 2], []], [[0, 2], [1, 3], [1, 4], [3, 4]], [2], No, No, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0019","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a646666c9e93da020be3ae60ee51bd81d391a2e029bf181e9e4ac5c7031d988e","family_id":"qf-density17-bij-fibers-and-class","seed":202610171019,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[2,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([2, 0, 1, 0, 0], [[1, 3, 4], [2], [0]], [[1, 3], [1, 4], [3, 4]], [], No, Yes, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 1, 0, 0], [[1, 3, 4], [2], [0]], [[1, 3], [1, 4], [3, 4]], [], No, Yes, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0020","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a3b2c0b1b619cfe7aa77fd41522c8e91950e9971d1be444f9732743bb3bcd6c2","family_id":"qf-density17-bij-fibers-and-class","seed":202610171020,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[0,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([0, 1, 1, 0, 0], [[0, 3, 4], [1, 2], []], [[0, 3], [0, 4], [1, 2], [3, 4]], [2], No, No, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 1, 0, 0], [[0, 3, 4], [1, 2], []], [[0, 3], [0, 4], [1, 2], [3, 4]], [2], No, No, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0021","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a23966da718c93f9d75e5c84633327ce89f4fc13f5d22239cc706d4b54b4a376","family_id":"qf-density17-bij-fibers-and-class","seed":202610171021,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[1,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([1, 1, 1, 0, 0], [[3, 4], [0, 1, 2], []], [[0, 1], [0, 2], [1, 2], [3, 4]], [2], No, No, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 1, 0, 0], [[3, 4], [0, 1, 2], []], [[0, 1], [0, 2], [1, 2], [3, 4]], [2], No, No, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0022","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9e2af08542f7b31bfce0997a3b30f087f7b8ea8e542c5f9f01701276db1563fa","family_id":"qf-density17-bij-fibers-and-class","seed":202610171022,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[2,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([2, 1, 1, 0, 0], [[3, 4], [1, 2], [0]], [[1, 2], [3, 4]], [], No, Yes, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 1, 0, 0], [[3, 4], [1, 2], [0]], [[1, 2], [3, 4]], [], No, Yes, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0023","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5f45ccc199e896d5b31d1d7f8aa9abf8fff9db6260de6b3e373ac7772c02a716","family_id":"qf-density17-bij-fibers-and-class","seed":202610171023,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[0,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([0, 2, 1, 0, 0], [[0, 3, 4], [2], [1]], [[0, 3], [0, 4], [3, 4]], [], No, Yes, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 1, 0, 0], [[0, 3, 4], [2], [1]], [[0, 3], [0, 4], [3, 4]], [], No, Yes, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0024","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f0e318ebc6336938a5b118bec49e37fb5b27cf4ff6876c782bb41469364b290","family_id":"qf-density17-bij-fibers-and-class","seed":202610171024,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[1,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([1, 2, 1, 0, 0], [[3, 4], [0, 2], [1]], [[0, 2], [3, 4]], [], No, Yes, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 1, 0, 0], [[3, 4], [0, 2], [1]], [[0, 2], [3, 4]], [], No, Yes, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0025","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"59c760c71c52735d81af91389b93ce6210fb4c719d8d8dd47f4b57a9b7e4db93","family_id":"qf-density17-bij-fibers-and-class","seed":202610171025,"semantic_template":"qf-density17-bij-fibers-and-class/{\"images\":[2,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Complete Fibers, Collisions, Missing Targets, and Map Class","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, every codomain fiber, all lexicographically ordered distinct-input collision pairs, missing targets, injectivity, surjectivity, bijectivity).","answer":{"kind":"expression","value":"([2, 2, 1, 0, 0], [[3, 4], [2], [0, 1]], [[0, 1], [3, 4]], [], No, Yes, No)","explanation":"Enumerate all declared targets, including empty fibers, and keep the three function properties separate. The empty-to-empty map is bijective. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 1, 0, 0], [[3, 4], [2], [0, 1]], [[0, 1], [3, 4]], [], No, Yes, No)."},"skills":["Complete Fibers, Collisions, Missing Targets, and Map Class"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0026","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"31eb88f88628cb1c90f555d05b775312b0a4de1ef47ba960bdf7bb54e93cbf2b","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171026,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[],\"m\":0,\"n\":0}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([], [], Yes, [])","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([], [], Yes, [])."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0027","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5205703ed4040dbe256c540099a8890904d6e9b4bec11bb628d2cd0f0b6f469b","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171027,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[],\"m\":2,\"n\":0}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([], [0, 0], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([], [0, 0], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0028","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"218552c2f02add497b081fcae6272ae5d64abe988fcca7eef57b9f9086154fb2","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171028,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[0],\"m\":1,\"n\":1}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 0]], [1], Yes, [0])","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0]], [1], Yes, [0])."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0029","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68c3dd3c49e5b861adb4291205ed9a6e072755d58df8194453b35a266cdd8da1","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171029,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[0,1,2],\"m\":3,\"n\":3}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 0], [1, 1], [2, 2]], [1, 1, 1], Yes, [0, 1, 2])","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0], [1, 1], [2, 2]], [1, 1, 1], Yes, [0, 1, 2])."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0030","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"da79892e0f286706f75f7523026fbd865fce8eae17981f9e00450d9836fdd46d","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171030,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[0,0,1],\"m\":3,\"n\":3}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 0], [0, 1], [1, 2]], [2, 1, 0], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0], [0, 1], [1, 2]], [2, 1, 0], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0031","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0049681360dbe4f8cbaa2ca904701094d97b0015e48332a78993118bd628c222","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171031,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[0,1,1],\"m\":2,\"n\":3}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 0], [1, 1], [1, 2]], [1, 2], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0], [1, 1], [1, 2]], [1, 2], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0032","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"918fe562d647a408ca3592b888f8d81ab4eefa5fb9c8c779dc2c32e4010b2e25","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171032,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[0,1],\"m\":3,\"n\":2}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 0], [1, 1]], [1, 1, 0], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0], [1, 1]], [1, 1, 0], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0033","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ec5924748fc986ac991e980c1f95390ff7821e628db46d32684f1f05a2e80d7","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171033,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[0,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 0], [0, 1], [0, 2], [0, 3], [0, 4]], [5, 0, 0], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0], [0, 1], [0, 2], [0, 3], [0, 4]], [5, 0, 0], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0034","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa512bc741abe6d28cc4d667ffb7aed7fec8db0c3aff2f6506160b4e08ac1a6b","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171034,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[1,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 1], [0, 2], [0, 3], [0, 4], [1, 0]], [4, 1, 0], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 1], [0, 2], [0, 3], [0, 4], [1, 0]], [4, 1, 0], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0035","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"50b53d17c01e9dc53305e4ff08003c55b12034bd3dc491f7852147d1929cbd57","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171035,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[2,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 1], [0, 2], [0, 3], [0, 4], [2, 0]], [4, 0, 1], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 1], [0, 2], [0, 3], [0, 4], [2, 0]], [4, 0, 1], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0036","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6a55f413406cbb84aa26ebfd4f3f32fc6d88c0b824ba521a9cb18b19361d4b89","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171036,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[0,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 0], [0, 2], [0, 3], [0, 4], [1, 1]], [4, 1, 0], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0], [0, 2], [0, 3], [0, 4], [1, 1]], [4, 1, 0], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0037","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"19bd7eb3932a2fa91861ff40bab43cbceed8b0ce6c15e206d216c7b368794fd0","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171037,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[1,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 2], [0, 3], [0, 4], [1, 0], [1, 1]], [3, 2, 0], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 2], [0, 3], [0, 4], [1, 0], [1, 1]], [3, 2, 0], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0038","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ffebe8d020206ae6906587e8da43a3daaec526bdaa39ed0e1f3143ace30988cc","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171038,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[2,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 2], [0, 3], [0, 4], [1, 1], [2, 0]], [3, 1, 1], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 2], [0, 3], [0, 4], [1, 1], [2, 0]], [3, 1, 1], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0039","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4327a8c7cbceecb7405a0e43a6bd5e8c494c920527d68f27000a74b4f861be0f","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171039,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[0,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 0], [0, 2], [0, 3], [0, 4], [2, 1]], [4, 0, 1], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0], [0, 2], [0, 3], [0, 4], [2, 1]], [4, 0, 1], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0040","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"20d1fc51130df5c50886173945011fcbba9bc20ccfe21777cd27d9856c7dae4f","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171040,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[1,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 2], [0, 3], [0, 4], [1, 0], [2, 1]], [3, 1, 1], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 2], [0, 3], [0, 4], [1, 0], [2, 1]], [3, 1, 1], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0041","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4199bf3709da515059509d2dde7e3c211d0e16867bfb6026e3a381954b0e9867","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171041,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[2,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 2], [0, 3], [0, 4], [2, 0], [2, 1]], [3, 0, 2], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 2], [0, 3], [0, 4], [2, 0], [2, 1]], [3, 0, 2], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0042","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"20c07af3ffa1c31eaec62c6acb2e8f0417c015ef78df6154becb9d2e3455dcbc","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171042,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[0,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 0], [0, 1], [0, 3], [0, 4], [1, 2]], [4, 1, 0], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0], [0, 1], [0, 3], [0, 4], [1, 2]], [4, 1, 0], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0043","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"079717747d938d8e465d249c1bb983bcb70165c9ba15fd20f32c49ed7475e785","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171043,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[1,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 1], [0, 3], [0, 4], [1, 0], [1, 2]], [3, 2, 0], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 1], [0, 3], [0, 4], [1, 0], [1, 2]], [3, 2, 0], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0044","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e216ec8f8239161eb424768b44312ba71aa377bb4b66dcd682c457799bb8cd82","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171044,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[2,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 1], [0, 3], [0, 4], [1, 2], [2, 0]], [3, 1, 1], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 1], [0, 3], [0, 4], [1, 2], [2, 0]], [3, 1, 1], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0045","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e51bd9002e510e8ffe57fb0dca7aba601832f7f162dbe3608d743d84dce5dd90","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171045,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[0,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 0], [0, 3], [0, 4], [1, 1], [1, 2]], [3, 2, 0], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0], [0, 3], [0, 4], [1, 1], [1, 2]], [3, 2, 0], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0046","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"df76afd91e4ebe9cb01f4dc3c44c73d467538086edf726c0bfa59fbefb829a30","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171046,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[1,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 3], [0, 4], [1, 0], [1, 1], [1, 2]], [2, 3, 0], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 3], [0, 4], [1, 0], [1, 1], [1, 2]], [2, 3, 0], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0047","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"64a290e59d5e0e66bfa6f089dea4592ff659d9d890453354a8b0150de2445c41","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171047,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[2,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 3], [0, 4], [1, 1], [1, 2], [2, 0]], [2, 2, 1], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 3], [0, 4], [1, 1], [1, 2], [2, 0]], [2, 2, 1], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0048","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a1ce0bb58a1c32152aeb7869ba49e9c4f8c78979166075e839e7e55157640e9","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171048,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[0,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 0], [0, 3], [0, 4], [1, 2], [2, 1]], [3, 1, 1], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 0], [0, 3], [0, 4], [1, 2], [2, 1]], [3, 1, 1], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0049","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ad8bcc3746ec927427c6bc7d3220bb01b50599101600c6de354893637da596d","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171049,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[1,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 3], [0, 4], [1, 0], [1, 2], [2, 1]], [2, 2, 1], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 3], [0, 4], [1, 0], [1, 2], [2, 1]], [2, 2, 1], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0050","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b7ecb44ce00f71a4544592d73e2e506014f62944df31230368ffadf882f5c78","family_id":"qf-density17-bij-swapped-relation-audit","seed":202610171050,"semantic_template":"qf-density17-bij-swapped-relation-audit/{\"images\":[2,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Converse Relation before Inverse-Function Qualification","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 (sorted swapped relation pairs [f(x),x], each codomain fiber size, inverse-function eligibility, inverse table or not-an-inverse-function).","answer":{"kind":"expression","value":"([[0, 3], [0, 4], [1, 2], [2, 0], [2, 1]], [2, 1, 2], No, not-an-inverse-function)","explanation":"Swapping pairs creates a relation; it creates a total inverse function exactly when every target has one preimage. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([[0, 3], [0, 4], [1, 2], [2, 0], [2, 1]], [2, 1, 2], No, not-an-inverse-function)."},"skills":["Converse Relation before Inverse-Function Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0051","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"97f9c64d58f2ff5401832788c4821d4d5c1aa8c6d09505525648cb5669e98782","family_id":"qf-density17-bij-output-transposition","seed":202610171051,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[],\"images\":[],\"m\":2,\"n\":0}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=1 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([], [], [[], []], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([], [], [[], []], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0052","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4e1d31f37a6d2f33522a0dd3abf2cccd4d202d13faf1b965304e3a60729a1b9e","family_id":"qf-density17-bij-output-transposition","seed":202610171052,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[1,0],\"images\":[0,1],\"m\":2,\"n\":2}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=1 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([0, 1], [1, 0], [[1], [0]], Yes, Yes, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1], [1, 0], [[1], [0]], Yes, Yes, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0053","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e44db34c367c720173b0ae35a5d83554efb41f8cd7d28bcbf799e65f2cc7bd24","family_id":"qf-density17-bij-output-transposition","seed":202610171053,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[1,1,0],\"images\":[0,0,1],\"m\":3,\"n\":3}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=1 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([0, 0, 1], [1, 1, 0], [[2], [0, 1], []], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1], [1, 1, 0], [[2], [0, 1], []], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0054","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"627b04980a145931637b3606579fccbefa0e37c2174acb2b6e72c9ba2d033e89","family_id":"qf-density17-bij-output-transposition","seed":202610171054,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[2,2,2,2,2],\"images\":[0,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([0, 0, 0, 0, 0], [2, 2, 2, 2, 2], [[], [], [0, 1, 2, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 0, 0, 0], [2, 2, 2, 2, 2], [[], [], [0, 1, 2, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0055","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e1c04fc7f579454259b7a48752720403e6c0d315b1177e7b1d48325483e0c3a","family_id":"qf-density17-bij-output-transposition","seed":202610171055,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[1,2,2,2,2],\"images\":[1,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([1, 0, 0, 0, 0], [1, 2, 2, 2, 2], [[], [0], [1, 2, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 0, 0, 0], [1, 2, 2, 2, 2], [[], [0], [1, 2, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0056","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2fdebcc58cb8fda713d528721ee4eee21c1e38bdfe3645737ccb69eea59438d5","family_id":"qf-density17-bij-output-transposition","seed":202610171056,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[0,2,2,2,2],\"images\":[2,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([2, 0, 0, 0, 0], [0, 2, 2, 2, 2], [[0], [], [1, 2, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 0, 0, 0], [0, 2, 2, 2, 2], [[0], [], [1, 2, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0057","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"89c1be85ad4aaf3606e63ef3b54c4f1eb86d662c9a7ba56699bd83cb35de88c3","family_id":"qf-density17-bij-output-transposition","seed":202610171057,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[2,1,2,2,2],\"images\":[0,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([0, 1, 0, 0, 0], [2, 1, 2, 2, 2], [[], [1], [0, 2, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 0, 0, 0], [2, 1, 2, 2, 2], [[], [1], [0, 2, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0058","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"37c8f12ef8240409e7860e62d6964f5ec667e7af0a02e507eae7c0b384c85461","family_id":"qf-density17-bij-output-transposition","seed":202610171058,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[1,1,2,2,2],\"images\":[1,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([1, 1, 0, 0, 0], [1, 1, 2, 2, 2], [[], [0, 1], [2, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 0, 0, 0], [1, 1, 2, 2, 2], [[], [0, 1], [2, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0059","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8484282bcced2c0bf2ceb67422526c3bd6c94541989cc388c388ea0783222c40","family_id":"qf-density17-bij-output-transposition","seed":202610171059,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[0,1,2,2,2],\"images\":[2,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([2, 1, 0, 0, 0], [0, 1, 2, 2, 2], [[0], [1], [2, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 0, 0, 0], [0, 1, 2, 2, 2], [[0], [1], [2, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0060","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ea0cb5c65f79373d6bc31cbf8d79883d9c4f08a71b5969b7cc8d10839b79f700","family_id":"qf-density17-bij-output-transposition","seed":202610171060,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[2,0,2,2,2],\"images\":[0,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([0, 2, 0, 0, 0], [2, 0, 2, 2, 2], [[1], [], [0, 2, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 0, 0, 0], [2, 0, 2, 2, 2], [[1], [], [0, 2, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0061","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0a6804978d95e316f4ec4265aac4d929bf807c56f63a8a67e1d41d7047bc3653","family_id":"qf-density17-bij-output-transposition","seed":202610171061,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[1,0,2,2,2],\"images\":[1,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([1, 2, 0, 0, 0], [1, 0, 2, 2, 2], [[1], [0], [2, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 0, 0, 0], [1, 0, 2, 2, 2], [[1], [0], [2, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0062","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f8960804326467f7ff688854fcef0b608fb981717abac87b15e14c6584561cfa","family_id":"qf-density17-bij-output-transposition","seed":202610171062,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[0,0,2,2,2],\"images\":[2,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([2, 2, 0, 0, 0], [0, 0, 2, 2, 2], [[0, 1], [], [2, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 0, 0, 0], [0, 0, 2, 2, 2], [[0, 1], [], [2, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0063","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a334ecc6d86e9f0e5490199d7f90a288033176b69cb26e35909fa4008cf9658","family_id":"qf-density17-bij-output-transposition","seed":202610171063,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[2,2,1,2,2],\"images\":[0,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([0, 0, 1, 0, 0], [2, 2, 1, 2, 2], [[], [2], [0, 1, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1, 0, 0], [2, 2, 1, 2, 2], [[], [2], [0, 1, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0064","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"be76cba0f3cebfcdb44ec72e3f989834a823c4f4dacf8ba024f062258b5e7398","family_id":"qf-density17-bij-output-transposition","seed":202610171064,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[1,2,1,2,2],\"images\":[1,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([1, 0, 1, 0, 0], [1, 2, 1, 2, 2], [[], [0, 2], [1, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 1, 0, 0], [1, 2, 1, 2, 2], [[], [0, 2], [1, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0065","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0fd12ab6f9ecd6a2d74c6a091a1120bf68ee2eb7c3e4c4bbf3a5de9e251b1452","family_id":"qf-density17-bij-output-transposition","seed":202610171065,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[0,2,1,2,2],\"images\":[2,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([2, 0, 1, 0, 0], [0, 2, 1, 2, 2], [[0], [2], [1, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 1, 0, 0], [0, 2, 1, 2, 2], [[0], [2], [1, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0066","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"30c771df997c39de5b48c9fde9601d8d906b2196d8a4cffe3541c731dce325f0","family_id":"qf-density17-bij-output-transposition","seed":202610171066,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[2,1,1,2,2],\"images\":[0,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([0, 1, 1, 0, 0], [2, 1, 1, 2, 2], [[], [1, 2], [0, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 1, 0, 0], [2, 1, 1, 2, 2], [[], [1, 2], [0, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0067","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e6d843d7feb828f019a19c9b1c438105312dd12ad4f4731e9315dbc074fe87d","family_id":"qf-density17-bij-output-transposition","seed":202610171067,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[1,1,1,2,2],\"images\":[1,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([1, 1, 1, 0, 0], [1, 1, 1, 2, 2], [[], [0, 1, 2], [3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 1, 0, 0], [1, 1, 1, 2, 2], [[], [0, 1, 2], [3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0068","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"daf4322ceee9732ce56a2826e16a889f7144a3cd340e6a41dfaf2b41eb5e1beb","family_id":"qf-density17-bij-output-transposition","seed":202610171068,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[0,1,1,2,2],\"images\":[2,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([2, 1, 1, 0, 0], [0, 1, 1, 2, 2], [[0], [1, 2], [3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 1, 0, 0], [0, 1, 1, 2, 2], [[0], [1, 2], [3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0069","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"02a90cd35db6307621438aeea0733499bf00a2c3326f70dea05743a379f8257a","family_id":"qf-density17-bij-output-transposition","seed":202610171069,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[2,0,1,2,2],\"images\":[0,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([0, 2, 1, 0, 0], [2, 0, 1, 2, 2], [[1], [2], [0, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 1, 0, 0], [2, 0, 1, 2, 2], [[1], [2], [0, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0070","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"531f697547f03a86c61c2934a061757a38dc41c5c50022df9e5e0cea67d864c4","family_id":"qf-density17-bij-output-transposition","seed":202610171070,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[1,0,1,2,2],\"images\":[1,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([1, 2, 1, 0, 0], [1, 0, 1, 2, 2], [[1], [0, 2], [3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 1, 0, 0], [1, 0, 1, 2, 2], [[1], [0, 2], [3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0071","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"66262ac74a63bba82a922b098148725019db4c7f6bb5d23ae289e051a0b96cc8","family_id":"qf-density17-bij-output-transposition","seed":202610171071,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[0,0,1,2,2],\"images\":[2,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","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. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([2, 2, 1, 0, 0], [0, 0, 1, 2, 2], [[0, 1], [2], [3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 1, 0, 0], [0, 0, 1, 2, 2], [[0, 1], [2], [3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0072","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"61b9668ddca508399af76af919dcafef83a10e28f84a08faf10da8b15861c480","family_id":"qf-density17-bij-output-transposition","seed":202610171072,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[2,2,0,2,2],\"images\":[0,0,2,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 0, 2, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([0, 0, 2, 0, 0], [2, 2, 0, 2, 2], [[2], [], [0, 1, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 2, 0, 0], [2, 2, 0, 2, 2], [[2], [], [0, 1, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0073","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b604f2d43a7a28e19a22d9f76358bd8a23f115d48f8d06592433407a1873813","family_id":"qf-density17-bij-output-transposition","seed":202610171073,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[1,2,0,2,2],\"images\":[1,0,2,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[1, 0, 2, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([1, 0, 2, 0, 0], [1, 2, 0, 2, 2], [[2], [0], [1, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 2, 0, 0], [1, 2, 0, 2, 2], [[2], [0], [1, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0074","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c467500b603cc82d1872a1ceb1437f3fed6f7c43b4a305ef3844b2a20716a35","family_id":"qf-density17-bij-output-transposition","seed":202610171074,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[0,2,0,2,2],\"images\":[2,0,2,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[2, 0, 2, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([2, 0, 2, 0, 0], [0, 2, 0, 2, 2], [[0, 2], [], [1, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 2, 0, 0], [0, 2, 0, 2, 2], [[0, 2], [], [1, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0075","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e76b266e9199551231b8fa87548b7931977301c8d218af49a21dcc79ed4f7f72","family_id":"qf-density17-bij-output-transposition","seed":202610171075,"semantic_template":"qf-density17-bij-output-transposition/{\"after\":[2,1,0,2,2],\"images\":[0,1,2,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Swap Two Output Labels and Recheck Bijectivity","stem":"Use the complete total map Fin(5)={0,…,4}→Fin(3) with images=[0, 1, 2, 0, 0] in source-index order. Validate every image in the declared codomain first. Equal finite cardinalities do not make the given map bijective. Interchange codomain labels a=0, b=2 in every image (equal labels are allowed). Give (original images, new images, new fibers, original bijectivity, new bijectivity, do these two flags agree?).","answer":{"kind":"expression","value":"([0, 1, 2, 0, 0], [2, 1, 0, 2, 2], [[2], [1], [0, 3, 4]], No, No, Yes)","explanation":"Apply the same output-label transposition to every input and recompute the complete fibers. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 0, 0], [2, 1, 0, 2, 2], [[2], [1], [0, 3, 4]], No, No, Yes)."},"skills":["Swap Two Output Labels and Recheck Bijectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-r01-bijective-0076","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3936e03a54d6db55f5857650ae1918b18f9225901bfd98b25697eed9a2e74113","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171076,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[],\"m\":0,\"n\":0}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([0, 0], [], Yes, [], Yes)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0], [], Yes, [], Yes)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0077","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2bca4e3eb6ccd6a761d6ed61547446a075e145350212033a70bf4cc0ec861fd8","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171077,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[],\"m\":2,\"n\":0}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([0, 2], [], No, [[], []], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2], [], No, [[], []], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0078","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d1164ea8032fc12238387549f580343d5a7c8ce5b44d6b0aff33cfc99b01f5e","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171078,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[0],\"m\":1,\"n\":1}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([1, 1], [0], Yes, [[0]], Yes)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1], [0], Yes, [[0]], Yes)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0079","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"01a73e9894cc7b061be767c5ccd64d214dc00b37434acf984db1e3afd53c4817","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171079,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[0,1,2],\"m\":3,\"n\":3}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([3, 3], [0, 1, 2], Yes, [[0], [1], [2]], Yes)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([3, 3], [0, 1, 2], Yes, [[0], [1], [2]], Yes)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0080","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"97075821dba121e444c6921e193baf225be9ae14aeb7b0c268b94477d6fe674e","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171080,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[0,0,1],\"m\":3,\"n\":3}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([3, 3], [0, 0, 1], Yes, [[0, 1], [2], []], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([3, 3], [0, 0, 1], Yes, [[0, 1], [2], []], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0081","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7156230e62377ad6929f08ce445b2c77fc0f13e3a55ece6e5ad26381a87ddb70","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171081,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[0,1,1],\"m\":2,\"n\":3}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([3, 2], [0, 1, 1], No, [[0], [1, 2]], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([3, 2], [0, 1, 1], No, [[0], [1, 2]], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0082","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"da704f6fa79ea6ec10b82016e01a670a734b2d935986c005791117f34e2c741c","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171082,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[0,1],\"m\":3,\"n\":2}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([2, 3], [0, 1], No, [[0], [1], []], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 3], [0, 1], No, [[0], [1], []], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0083","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"27db2a06c311ec897a2bbbc3f108f645ea8b7697c1a9c349be596f3267f88ee3","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171083,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[0,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [0, 0, 0, 0, 0], No, [[0, 1, 2, 3, 4], [], []], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [0, 0, 0, 0, 0], No, [[0, 1, 2, 3, 4], [], []], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0084","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"11529a7a5652e8324f9dc76f59a02e9c327c2c82d4e36e1eda669ab1a1c8159c","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171084,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[1,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [1, 0, 0, 0, 0], No, [[1, 2, 3, 4], [0], []], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [1, 0, 0, 0, 0], No, [[1, 2, 3, 4], [0], []], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0085","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bdd8fd01c3f304a5d7a1445ab181ee0846b7cdbda4ffa59f3dd1f4abcb886081","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171085,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[2,0,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [2, 0, 0, 0, 0], No, [[1, 2, 3, 4], [], [0]], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [2, 0, 0, 0, 0], No, [[1, 2, 3, 4], [], [0]], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0086","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a7834f8bb011e6b40ec2ce01f8866155d5bfbc7fe57b95e49e1e367c81acf69d","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171086,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[0,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [0, 1, 0, 0, 0], No, [[0, 2, 3, 4], [1], []], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [0, 1, 0, 0, 0], No, [[0, 2, 3, 4], [1], []], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0087","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6123eb145ebf8afc5373db8775da04957d0a44f766d4d0bca2df3c476e1773c1","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171087,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[1,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [1, 1, 0, 0, 0], No, [[2, 3, 4], [0, 1], []], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [1, 1, 0, 0, 0], No, [[2, 3, 4], [0, 1], []], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0088","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c4b4c1d550958b52da33d2c0aae7063802493fc219c1f4e15369c8a297e13de7","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171088,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[2,1,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [2, 1, 0, 0, 0], No, [[2, 3, 4], [1], [0]], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [2, 1, 0, 0, 0], No, [[2, 3, 4], [1], [0]], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0089","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b43300c2a9e4c635eba6de4e6fd16af29ea2ce2f221d9cd1089c9a7fb441688c","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171089,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[0,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [0, 2, 0, 0, 0], No, [[0, 2, 3, 4], [], [1]], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [0, 2, 0, 0, 0], No, [[0, 2, 3, 4], [], [1]], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0090","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2cf61cd26946ccceded8a6921106d23ac91bc033ef9298427bdeb772549e160b","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171090,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[1,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [1, 2, 0, 0, 0], No, [[2, 3, 4], [0], [1]], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [1, 2, 0, 0, 0], No, [[2, 3, 4], [0], [1]], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0091","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"34b5843b6f7a658c0b32a9d377d3d302f8a476276c01a5964127534cdfb3055c","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171091,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[2,2,0,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [2, 2, 0, 0, 0], No, [[2, 3, 4], [], [0, 1]], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [2, 2, 0, 0, 0], No, [[2, 3, 4], [], [0, 1]], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0092","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9fb353438d1bac67e97f095863756ee51e2eb7b5025a7cad65eeef592ee051da","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171092,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[0,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [0, 0, 1, 0, 0], No, [[0, 1, 3, 4], [2], []], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [0, 0, 1, 0, 0], No, [[0, 1, 3, 4], [2], []], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0093","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2cdf3585d5882895e497c8bb76062bd66f08c4e7b019ecae9e42a88d27ae1631","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171093,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[1,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [1, 0, 1, 0, 0], No, [[1, 3, 4], [0, 2], []], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [1, 0, 1, 0, 0], No, [[1, 3, 4], [0, 2], []], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0094","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06aa360d637c8a2680a816adfdb6150f21ae64712adf9945e038eb2ed8985a92","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171094,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[2,0,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [2, 0, 1, 0, 0], No, [[1, 3, 4], [2], [0]], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [2, 0, 1, 0, 0], No, [[1, 3, 4], [2], [0]], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0095","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cf9c3333e82b3caaa7dbec4a00cd42e7ecf6d0fcfe4e308a0aa0cdc2111b859e","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171095,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[0,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [0, 1, 1, 0, 0], No, [[0, 3, 4], [1, 2], []], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [0, 1, 1, 0, 0], No, [[0, 3, 4], [1, 2], []], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0096","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f0e1704dfeee2f34d27a9799f67523b6df9539e48471d0f30a595565ea51e4de","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171096,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[1,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [1, 1, 1, 0, 0], No, [[3, 4], [0, 1, 2], []], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [1, 1, 1, 0, 0], No, [[3, 4], [0, 1, 2], []], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0097","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7ffe52c0a32c3d70dd88021dffcdd1a24365755462b36841e51281146352fce","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171097,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[2,1,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [2, 1, 1, 0, 0], No, [[3, 4], [1, 2], [0]], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [2, 1, 1, 0, 0], No, [[3, 4], [1, 2], [0]], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0098","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"24deb839c23db9f89f22f492e926b045d29231639e3280be89065416ba140fa3","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171098,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[0,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [0, 2, 1, 0, 0], No, [[0, 3, 4], [2], [1]], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [0, 2, 1, 0, 0], No, [[0, 3, 4], [2], [1]], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0099","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2cea54daa7d585309dd1094b8f0233afcec6e476ed2366c3fbb30020d5df8ff8","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171099,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[1,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [1, 2, 1, 0, 0], No, [[3, 4], [0, 2], [1]], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [1, 2, 1, 0, 0], No, [[3, 4], [0, 2], [1]], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261017-r01-bijective-0100","batch_id":"batch-20261012-advance-20261017-r01-bijective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"16cd7e4e76e471728f094c5ac5393120c4ff720539acfbba414fca77b0476ee5","family_id":"qf-density17-bij-size-obstruction-with-table","seed":202610171100,"semantic_template":"qf-density17-bij-size-obstruction-with-table/{\"images\":[2,2,1,0,0],\"m\":3,\"n\":5}"},"fields":{"title":"Equal Cardinality versus the Given Total Table","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 ([n,m], images, n=m?, every fiber, is the given map bijective?).","answer":{"kind":"expression","value":"([5, 3], [2, 2, 1, 0, 0], No, [[3, 4], [2], [0, 1]], No)","explanation":"Equal finite sizes address cardinality, not whether this particular table has repeated or missing values. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5, 3], [2, 2, 1, 0, 0], No, [[3, 4], [2], [0, 1]], No)."},"skills":["Equal Cardinality versus the Given Total Table"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}