{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261012-advance-20261016-r06-injective-0001","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e92e7a17d36284b38f147aeb4b6cef6541496432d643a1c8905f068358d5a54","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166001,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[],\"m\":0,\"n\":0}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(0)=∅→Fin(0), complete images=[]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([], [], Yes)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([], [], Yes)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0002","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a9d34b4dc56c8afb4f23583648589e300dec7996c4957b2550d8754b99c977b0","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166002,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[0],\"m\":1,\"n\":1}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(1)={0,…,0}→Fin(1), complete images=[0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([0], [], Yes)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0], [], Yes)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0003","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"62aa8e4accf8cd3a49382535760621a69eef108e0efa6b47f8ec9873d312234e","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166003,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[0,1,2],\"m\":3,\"n\":3}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(3)={0,…,2}→Fin(3), complete images=[0, 1, 2]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([0, 1, 2], [], Yes)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2], [], Yes)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0004","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e13312fa31c55a3df60675418c5a478e8958fe9d5c255cd5b6b57e005639938c","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166004,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[0,0,0,0,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[0, 0, 0, 0, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([0, 0, 0, 0, 0], [[0, 1], [0, 2], [0, 3], [0, 4], [1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. 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], [0, 2], [0, 3], [0, 4], [1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0005","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8896712cacd63c3505664cc65ea2f1db8ac633546d355fae2a8db35d184b99b7","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166005,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[1,0,0,0,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[1, 0, 0, 0, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([1, 0, 0, 0, 0], [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. 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], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0006","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef45e12f9d88fa7a8de0434bb6fa78283ee4496058682bbb3ee901bae22b62f4","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166006,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[0,1,0,0,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[0, 1, 0, 0, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([0, 1, 0, 0, 0], [[0, 2], [0, 3], [0, 4], [2, 3], [2, 4], [3, 4]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. 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], [0, 3], [0, 4], [2, 3], [2, 4], [3, 4]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0007","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"340534baabc345afd7f7b1a50937907578bb11d3b6c71d37bc1fcdbd10ff2591","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166007,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[1,1,0,0,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[1, 1, 0, 0, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([1, 1, 0, 0, 0], [[0, 1], [2, 3], [2, 4], [3, 4]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 0, 0, 0], [[0, 1], [2, 3], [2, 4], [3, 4]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0008","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"400e24974640455aa0a5f65a8d9ade220578a49ec777fc9eedb9dec48744fee6","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166008,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[0,0,1,0,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[0, 0, 1, 0, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([0, 0, 1, 0, 0], [[0, 1], [0, 3], [0, 4], [1, 3], [1, 4], [3, 4]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. 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], [0, 3], [0, 4], [1, 3], [1, 4], [3, 4]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0009","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6f2b0aeedcd269b95c1bc15b14ad4233a3dd207513d33ddbf056621b8b44de5","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166009,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[1,0,1,0,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[1, 0, 1, 0, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([1, 0, 1, 0, 0], [[0, 2], [1, 3], [1, 4], [3, 4]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 1, 0, 0], [[0, 2], [1, 3], [1, 4], [3, 4]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0010","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"128080b1918c2ce2f07b9b335edae45f04875736a37b35d3eabf21cd4fb3b7d9","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166010,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[0,1,1,0,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[0, 1, 1, 0, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([0, 1, 1, 0, 0], [[0, 3], [0, 4], [1, 2], [3, 4]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. 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], [0, 4], [1, 2], [3, 4]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0011","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7d7a692bb43839472ceefd349bc8c7bf156556b0a28c3e23a94e7f7764e01396","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166011,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[1,1,1,0,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[1, 1, 1, 0, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([1, 1, 1, 0, 0], [[0, 1], [0, 2], [1, 2], [3, 4]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 1, 0, 0], [[0, 1], [0, 2], [1, 2], [3, 4]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0012","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f044e49f7e32aff8d53b377db0b87a63799347e1b8aee38ff32143a32cda962","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166012,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[0,0,0,1,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[0, 0, 0, 1, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([0, 0, 0, 1, 0], [[0, 1], [0, 2], [0, 4], [1, 2], [1, 4], [2, 4]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 0, 1, 0], [[0, 1], [0, 2], [0, 4], [1, 2], [1, 4], [2, 4]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0013","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"89502ec7f6ea2d28766d76d9694788ec4d16dfdc6f2ae67ecd863e7ce8ca6f48","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166013,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[1,0,0,1,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[1, 0, 0, 1, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([1, 0, 0, 1, 0], [[0, 3], [1, 2], [1, 4], [2, 4]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 0, 1, 0], [[0, 3], [1, 2], [1, 4], [2, 4]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0014","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"547254cb52da68f3a0dcab9d6fc00a30ef6c1d35de2fc0ce17ac6d174b7cf708","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166014,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[0,1,0,1,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[0, 1, 0, 1, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([0, 1, 0, 1, 0], [[0, 2], [0, 4], [1, 3], [2, 4]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 0, 1, 0], [[0, 2], [0, 4], [1, 3], [2, 4]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0015","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7a6d670853964978387f6f00720eaaa7218d855029b0c8ee4be52953e80ff17","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166015,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[1,1,0,1,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[1, 1, 0, 1, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([1, 1, 0, 1, 0], [[0, 1], [0, 3], [1, 3], [2, 4]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 0, 1, 0], [[0, 1], [0, 3], [1, 3], [2, 4]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0016","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4b6c737c65f660f6cf7f821bc693907836eace1589e9be5946711cf297f9b447","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166016,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[0,0,1,1,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[0, 0, 1, 1, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([0, 0, 1, 1, 0], [[0, 1], [0, 4], [1, 4], [2, 3]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1, 1, 0], [[0, 1], [0, 4], [1, 4], [2, 3]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0017","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"241e2a5a9eb590d5e6a44f7addf60d6b75f82da6cea7a4d070fee076ee82d56c","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166017,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[1,0,1,1,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[1, 0, 1, 1, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([1, 0, 1, 1, 0], [[0, 2], [0, 3], [1, 4], [2, 3]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 1, 1, 0], [[0, 2], [0, 3], [1, 4], [2, 3]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0018","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7d1179832e6bd918e02bb934cee221e9d1fb7ebac1288c30a28f12429bc019bd","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166018,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[0,1,1,1,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[0, 1, 1, 1, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([0, 1, 1, 1, 0], [[0, 4], [1, 2], [1, 3], [2, 3]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 1, 1, 0], [[0, 4], [1, 2], [1, 3], [2, 3]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0019","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d2a74a33486b44f4688c67f3d1535884bd0f0b7279003a3d3da9865d92d15bb5","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166019,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[1,1,1,1,0],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[1, 1, 1, 1, 0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([1, 1, 1, 1, 0], [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 1, 1, 0], [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0020","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"652a98b83e2f564b5d6db54b48478805d3ade8158e38e2aef2b01a8c15bba84b","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166020,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[0,0,0,0,1],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[0, 0, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([0, 0, 0, 0, 1], [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 0, 0, 1], [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0021","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"695310a47e54bffeadc95ab080d402a8bd202f1efb3d2b8cd85d831ccaa2b420","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166021,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[1,0,0,0,1],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[1, 0, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([1, 0, 0, 0, 1], [[0, 4], [1, 2], [1, 3], [2, 3]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 0, 0, 1], [[0, 4], [1, 2], [1, 3], [2, 3]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0022","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"58e00eb09378b144e975acc9069cafea17ac79fdbade87b3214296571af8b3fc","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166022,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[0,1,0,0,1],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[0, 1, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([0, 1, 0, 0, 1], [[0, 2], [0, 3], [1, 4], [2, 3]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 0, 0, 1], [[0, 2], [0, 3], [1, 4], [2, 3]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0023","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"46e45ed737c0b66179ab5844b6f0ae6ff0047fefda6d94490b95deb786e20640","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166023,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[1,1,0,0,1],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[1, 1, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([1, 1, 0, 0, 1], [[0, 1], [0, 4], [1, 4], [2, 3]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 0, 0, 1], [[0, 1], [0, 4], [1, 4], [2, 3]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0024","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"46bca451c4ad73bcb2f6571be4634a54777af0ae67c51cbbf7918f18bbd0a0d9","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166024,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[0,0,1,0,1],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[0, 0, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([0, 0, 1, 0, 1], [[0, 1], [0, 3], [1, 3], [2, 4]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1, 0, 1], [[0, 1], [0, 3], [1, 3], [2, 4]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0025","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"472b02779ecade545b2108f27fac5f3358920ade90ed34b5aa2d7419d00fe42f","family_id":"qf-density16-inj-all-collision-pairs","seed":202610166025,"semantic_template":"qf-density16-inj-all-collision-pairs/{\"images\":[1,0,1,0,1],\"m\":2,\"n\":5}"},"fields":{"title":"Every Distinct-Input Collision and Injectivity","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(2), complete images=[1, 0, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Give (original images, all pairs x<x2 with equal outputs, injectivity eligibility), with collision pairs in original-input lexicographic order.","answer":{"kind":"expression","value":"([1, 0, 1, 0, 1], [[0, 2], [0, 4], [1, 3], [2, 4]], No)","explanation":"Validate the complete table first, then enumerate distinct-input pairs; no collisions means injective, including the empty source. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 1, 0, 1], [[0, 2], [0, 4], [1, 3], [2, 4]], No)."},"skills":["Every Distinct-Input Collision and Injectivity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0026","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bbe1c30bf8d6b4f76ef5e7e4afdc565d68e963b83db01a193be6359b4cd1c310","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166026,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":0,\"images\":[0],\"m\":1,\"n\":1}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(1)={0,…,0}→Fin(1), complete images=[0]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=0, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([0], [], [], [], Yes)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0], [], [], [], Yes)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0027","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"84a0191ab2a230b31743b2ff6df415aee28cf83e6056f9957aacb66cbce00362","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166027,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":0,\"images\":[0,0,1],\"m\":2,\"n\":3}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(3)={0,…,2}→Fin(2), complete images=[0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=0, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([0, 0, 1], [[0, 1]], [1, 2], [], Yes)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1], [[0, 1]], [1, 2], [], Yes)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0028","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f981fc6e87016b996ade270220dc449377c3d460ee1ada691f2b470a2928d7c","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166028,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":0,\"images\":[0,0,1,1],\"m\":2,\"n\":4}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(4)={0,…,3}→Fin(2), complete images=[0, 0, 1, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=0, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([0, 0, 1, 1], [[0, 1], [2, 3]], [1, 2, 3], [[2, 3]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1, 1], [[0, 1], [2, 3]], [1, 2, 3], [[2, 3]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0029","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5dff2c823c461f61026530ee615b63f6b39b924f5231d7debbb29d3355b94365","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166029,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":0,\"images\":[0,0,0,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 0, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=0, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([0, 0, 0, 0, 1], [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]], [1, 2, 3, 4], [[1, 2], [1, 3], [2, 3]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 0, 0, 1], [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]], [1, 2, 3, 4], [[1, 2], [1, 3], [2, 3]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0030","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"92fc7a7c0b0c940632cd3a3497647cb35a75f4d2f42c8adb3065ad14df462985","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166030,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":1,\"images\":[1,0,0,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 0, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=1, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([1, 0, 0, 0, 1], [[0, 4], [1, 2], [1, 3], [2, 3]], [0, 2, 3, 4], [[0, 4], [2, 3]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 0, 0, 1], [[0, 4], [1, 2], [1, 3], [2, 3]], [0, 2, 3, 4], [[0, 4], [2, 3]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0031","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3fbf0340586830fa481c5b9cb8c162647e9d38b69fa4aaf0f30801583535d9db","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166031,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":2,\"images\":[2,0,0,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 0, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=2, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([2, 0, 0, 0, 1], [[1, 2], [1, 3], [2, 3]], [0, 1, 3, 4], [[1, 3]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 0, 0, 1], [[1, 2], [1, 3], [2, 3]], [0, 1, 3, 4], [[1, 3]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0032","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f5ddd667df1d7865c3c8002c82a26a602bb8e23d8e9311b0d404ab3e9da746c","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166032,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":3,\"images\":[0,1,0,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 1, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=3, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([0, 1, 0, 0, 1], [[0, 2], [0, 3], [1, 4], [2, 3]], [0, 1, 2, 4], [[0, 2], [1, 4]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 0, 0, 1], [[0, 2], [0, 3], [1, 4], [2, 3]], [0, 1, 2, 4], [[0, 2], [1, 4]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0033","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3df22d8df106a7568335b320b23040109aa49862f29310691f8278d7b07fb732","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166033,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":4,\"images\":[1,1,0,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 1, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=4, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([1, 1, 0, 0, 1], [[0, 1], [0, 4], [1, 4], [2, 3]], [0, 1, 2, 3], [[0, 1], [2, 3]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 0, 0, 1], [[0, 1], [0, 4], [1, 4], [2, 3]], [0, 1, 2, 3], [[0, 1], [2, 3]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0034","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9cd600a6a44a8b0c85144fc2995651761063d8d292e703a49f38f76d15fc9e6c","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166034,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":0,\"images\":[2,1,0,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 1, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=0, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([2, 1, 0, 0, 1], [[1, 4], [2, 3]], [1, 2, 3, 4], [[1, 4], [2, 3]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 0, 0, 1], [[1, 4], [2, 3]], [1, 2, 3, 4], [[1, 4], [2, 3]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0035","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e765770b9fff36e1e2fdfcfa1619075b21249890bf8da26e3055585beda7bee7","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166035,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":1,\"images\":[0,2,0,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 2, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=1, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([0, 2, 0, 0, 1], [[0, 2], [0, 3], [2, 3]], [0, 2, 3, 4], [[0, 2], [0, 3], [2, 3]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 0, 0, 1], [[0, 2], [0, 3], [2, 3]], [0, 2, 3, 4], [[0, 2], [0, 3], [2, 3]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0036","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9cd47df98ca4b88b4b2f3db6744dd394e1a541e1b0f75d3d57a47e0d7a7ec76a","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166036,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":2,\"images\":[1,2,0,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 2, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=2, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([1, 2, 0, 0, 1], [[0, 4], [2, 3]], [0, 1, 3, 4], [[0, 4]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 0, 0, 1], [[0, 4], [2, 3]], [0, 1, 3, 4], [[0, 4]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0037","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f6f7aa866fb7ae2a5c24533b3660c49b225dfa167aeb1010c1475b4827458a84","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166037,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":3,\"images\":[2,2,0,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 2, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=3, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([2, 2, 0, 0, 1], [[0, 1], [2, 3]], [0, 1, 2, 4], [[0, 1]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 0, 0, 1], [[0, 1], [2, 3]], [0, 1, 2, 4], [[0, 1]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0038","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"86f2ca91716470d222c9e37fc18c994dc3e115721be077dd008d2501c7d430cc","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166038,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":4,\"images\":[0,0,1,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 0, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=4, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([0, 0, 1, 0, 1], [[0, 1], [0, 3], [1, 3], [2, 4]], [0, 1, 2, 3], [[0, 1], [0, 3], [1, 3]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1, 0, 1], [[0, 1], [0, 3], [1, 3], [2, 4]], [0, 1, 2, 3], [[0, 1], [0, 3], [1, 3]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0039","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"27f9cf0a32652e27e1b94e79aa85c348ce56165bc9bd6850ef4d45adb992e177","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166039,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":0,\"images\":[1,0,1,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 0, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=0, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([1, 0, 1, 0, 1], [[0, 2], [0, 4], [1, 3], [2, 4]], [1, 2, 3, 4], [[1, 3], [2, 4]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 1, 0, 1], [[0, 2], [0, 4], [1, 3], [2, 4]], [1, 2, 3, 4], [[1, 3], [2, 4]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0040","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"53fc790798eaeb031800596006a0c3bbabbdd9437b139459d7751e71e4b721c1","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166040,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":1,\"images\":[2,0,1,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 0, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=1, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([2, 0, 1, 0, 1], [[1, 3], [2, 4]], [0, 2, 3, 4], [[2, 4]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 1, 0, 1], [[1, 3], [2, 4]], [0, 2, 3, 4], [[2, 4]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0041","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bd5c1b4a7a88a0ad26cd59d5626cd3e10d250ed5cf271fca77251ad0160a7545","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166041,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":2,\"images\":[0,1,1,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 1, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=2, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([0, 1, 1, 0, 1], [[0, 3], [1, 2], [1, 4], [2, 4]], [0, 1, 3, 4], [[0, 3], [1, 4]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 1, 0, 1], [[0, 3], [1, 2], [1, 4], [2, 4]], [0, 1, 3, 4], [[0, 3], [1, 4]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0042","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a2fd47509b4e7639baaeea4ff9c976835dcddb7d333a2c7a4eb39912ee368e4d","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166042,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":3,\"images\":[1,1,1,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 1, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=3, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([1, 1, 1, 0, 1], [[0, 1], [0, 2], [0, 4], [1, 2], [1, 4], [2, 4]], [0, 1, 2, 4], [[0, 1], [0, 2], [0, 4], [1, 2], [1, 4], [2, 4]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 1, 0, 1], [[0, 1], [0, 2], [0, 4], [1, 2], [1, 4], [2, 4]], [0, 1, 2, 4], [[0, 1], [0, 2], [0, 4], [1, 2], [1, 4], [2, 4]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0043","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"887b5cac7d226ff24db0053bdc28f9715108b3955030e1ae93e0be5b7b243e92","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166043,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":4,\"images\":[2,1,1,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 1, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=4, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([2, 1, 1, 0, 1], [[1, 2], [1, 4], [2, 4]], [0, 1, 2, 3], [[1, 2]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 1, 0, 1], [[1, 2], [1, 4], [2, 4]], [0, 1, 2, 3], [[1, 2]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0044","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7019cbfb05338bf3c5d164cdab1305151103fbdeccd5d2812b74b9066a469701","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166044,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":0,\"images\":[0,2,1,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 2, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=0, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([0, 2, 1, 0, 1], [[0, 3], [2, 4]], [1, 2, 3, 4], [[2, 4]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 1, 0, 1], [[0, 3], [2, 4]], [1, 2, 3, 4], [[2, 4]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0045","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5be24ceb2b935f16027c9db0783e30a70e5b0f521a391f21d877a84a1365df63","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166045,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":1,\"images\":[1,2,1,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 2, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=1, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([1, 2, 1, 0, 1], [[0, 2], [0, 4], [2, 4]], [0, 2, 3, 4], [[0, 2], [0, 4], [2, 4]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 1, 0, 1], [[0, 2], [0, 4], [2, 4]], [0, 2, 3, 4], [[0, 2], [0, 4], [2, 4]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0046","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"abeaf26aab862e76fc6e23f42abc3d3130c0cbef0387b5d847b7d133692a3654","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166046,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":2,\"images\":[2,2,1,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 2, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=2, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([2, 2, 1, 0, 1], [[0, 1], [2, 4]], [0, 1, 3, 4], [[0, 1]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 1, 0, 1], [[0, 1], [2, 4]], [0, 1, 3, 4], [[0, 1]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0047","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2cf37083d32675d6c27642a0b79687138902c2fcdd9612b98a52e4530ec155cf","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166047,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":3,\"images\":[0,0,2,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 0, 2, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=3, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([0, 0, 2, 0, 1], [[0, 1], [0, 3], [1, 3]], [0, 1, 2, 4], [[0, 1]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 2, 0, 1], [[0, 1], [0, 3], [1, 3]], [0, 1, 2, 4], [[0, 1]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0048","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b757eb23f6924a0a066a91ee022809f92114b4cddb8cd32eaf3beb7d53c7c140","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166048,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":4,\"images\":[1,0,2,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 0, 2, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=4, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([1, 0, 2, 0, 1], [[0, 4], [1, 3]], [0, 1, 2, 3], [[1, 3]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 2, 0, 1], [[0, 4], [1, 3]], [0, 1, 2, 3], [[1, 3]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0049","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a86bbea1ddec6acd8c0dad8b492ae37c2e152bc87981830ea17b83ba7970bd0c","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166049,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":0,\"images\":[2,0,2,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 0, 2, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=0, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([2, 0, 2, 0, 1], [[0, 2], [1, 3]], [1, 2, 3, 4], [[1, 3]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 2, 0, 1], [[0, 2], [1, 3]], [1, 2, 3, 4], [[1, 3]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0050","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"75d3d900f602e7cdafd3ee50303725277cfbef00604267332d02395175550536","family_id":"qf-density16-inj-remove-input-collision-change","seed":202610166050,"semantic_template":"qf-density16-inj-remove-input-collision-change/{\"at\":1,\"images\":[0,1,2,0,1],\"m\":3,\"n\":5}"},"fields":{"title":"Input Deletion with Original Collision Labels Retained","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 1, 2, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Delete original input at=1, keeping the codomain. Give (original images, original collisions, retained original input labels, later collisions, later injectivity), without renumbering inputs.","answer":{"kind":"expression","value":"([0, 1, 2, 0, 1], [[0, 3], [1, 4]], [0, 2, 3, 4], [[0, 3]], No)","explanation":"Remove only the specified original label and compare equal outputs using all retained original labels. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 0, 1], [[0, 3], [1, 4]], [0, 2, 3, 4], [[0, 3]], No)."},"skills":["Input Deletion with Original Collision Labels Retained"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0051","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eef6035383219922869b61388233c052086eb08eec8b0564f2acb344ab9381c3","family_id":"qf-density16-inj-append-output-qualification","seed":202610166051,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[],\"m\":0,\"n\":0,\"value\":0}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(0)=∅→Fin(0), complete images=[]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=0. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([], No, outside-codomain, not-evaluated, not-evaluated)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([], No, outside-codomain, not-evaluated, not-evaluated)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0052","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f580beafa4a27ec3d94310215b559d4167a6bd06a0bffb1e010be0ab25afb410","family_id":"qf-density16-inj-append-output-qualification","seed":202610166052,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[],\"m\":1,\"n\":0,\"value\":0}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(0)=∅→Fin(1), complete images=[]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=0. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([], Yes, [0], [], Yes)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([], Yes, [0], [], Yes)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0053","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf550fa535cba4b63b6d904b3feab920c033515c9a404435eb9f356748b8695c","family_id":"qf-density16-inj-append-output-qualification","seed":202610166053,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[0,1],\"m\":3,\"n\":2,\"value\":2}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(2)={0,…,1}→Fin(3), complete images=[0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=2. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([0, 1], Yes, [0, 1, 2], [], Yes)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1], Yes, [0, 1, 2], [], Yes)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0054","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7e7a31a53353bb90ce096dc6e699cd01981f97d3e82cbdc971321965178d46b2","family_id":"qf-density16-inj-append-output-qualification","seed":202610166054,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[0,1],\"m\":2,\"n\":2,\"value\":0}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(2)={0,…,1}→Fin(2), complete images=[0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=0. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([0, 1], Yes, [0, 1, 0], [[0, 2]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1], Yes, [0, 1, 0], [[0, 2]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0055","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7da63d2647474027d36ba496e2b4e01675c7a356e5bb107b55e66d31b52c6757","family_id":"qf-density16-inj-append-output-qualification","seed":202610166055,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[0,1],\"m\":2,\"n\":2,\"value\":-1}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(2)={0,…,1}→Fin(2), complete images=[0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=-1. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([0, 1], No, outside-codomain, not-evaluated, not-evaluated)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1], No, outside-codomain, not-evaluated, not-evaluated)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0056","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"56613c764efc3d619cd7b70d2c907e2ea0eb22b7b37ebe3cdc56cf3007c3591a","family_id":"qf-density16-inj-append-output-qualification","seed":202610166056,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[0,1,2,3,4,5,6,7],\"m\":8,\"n\":8,\"value\":7}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(8)={0,…,7}→Fin(8), complete images=[0, 1, 2, 3, 4, 5, 6, 7]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=7. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5, 6, 7], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 7], [[7, 8]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], Yes, [0, 1, 2, 3, 4, 5, 6, 7, 7], [[7, 8]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0057","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"93b1180471cf8bf09733c96eb803c055de41d3cf0f0240db2df5b54607588565","family_id":"qf-density16-inj-append-output-qualification","seed":202610166057,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[0,0,0,0,1],\"m\":3,\"n\":5,\"value\":1}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 0, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=1. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([0, 0, 0, 0, 1], Yes, [0, 0, 0, 0, 1, 1], [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3], [4, 5]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 0, 0, 1], Yes, [0, 0, 0, 0, 1, 1], [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3], [4, 5]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0058","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b716b5225000788d1ceef091f1ae9ca1dc8d7501f1a08604a19457fb9664574","family_id":"qf-density16-inj-append-output-qualification","seed":202610166058,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[1,0,0,0,1],\"m\":3,\"n\":5,\"value\":2}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 0, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=2. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([1, 0, 0, 0, 1], Yes, [1, 0, 0, 0, 1, 2], [[0, 4], [1, 2], [1, 3], [2, 3]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 0, 0, 1], Yes, [1, 0, 0, 0, 1, 2], [[0, 4], [1, 2], [1, 3], [2, 3]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0059","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"290581d5df58f1f203bcdec61789ddd8eee265d94fa3fea9138bcc1e6c30f163","family_id":"qf-density16-inj-append-output-qualification","seed":202610166059,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[2,0,0,0,1],\"m\":3,\"n\":5,\"value\":-1}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 0, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=-1. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([2, 0, 0, 0, 1], No, outside-codomain, not-evaluated, not-evaluated)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 0, 0, 1], No, outside-codomain, not-evaluated, not-evaluated)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0060","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f0f3d75140786576952fc7ed44911a9d32e2711d7778e8937287b2278fb91abd","family_id":"qf-density16-inj-append-output-qualification","seed":202610166060,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[0,1,0,0,1],\"m\":3,\"n\":5,\"value\":0}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 1, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=0. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([0, 1, 0, 0, 1], Yes, [0, 1, 0, 0, 1, 0], [[0, 2], [0, 3], [0, 5], [1, 4], [2, 3], [2, 5], [3, 5]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 0, 0, 1], Yes, [0, 1, 0, 0, 1, 0], [[0, 2], [0, 3], [0, 5], [1, 4], [2, 3], [2, 5], [3, 5]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0061","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a935dd52ebf3a4c386a3430d2d810644f22f5669d8a92fd9ba1b35dc4366bb39","family_id":"qf-density16-inj-append-output-qualification","seed":202610166061,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[1,1,0,0,1],\"m\":3,\"n\":5,\"value\":1}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 1, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=1. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([1, 1, 0, 0, 1], Yes, [1, 1, 0, 0, 1, 1], [[0, 1], [0, 4], [0, 5], [1, 4], [1, 5], [2, 3], [4, 5]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 0, 0, 1], Yes, [1, 1, 0, 0, 1, 1], [[0, 1], [0, 4], [0, 5], [1, 4], [1, 5], [2, 3], [4, 5]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0062","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"799d51e15b2e5435667f8f33c0b41fa463f320022d6568fb9415da8b64442851","family_id":"qf-density16-inj-append-output-qualification","seed":202610166062,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[2,1,0,0,1],\"m\":3,\"n\":5,\"value\":2}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 1, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=2. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([2, 1, 0, 0, 1], Yes, [2, 1, 0, 0, 1, 2], [[0, 5], [1, 4], [2, 3]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 0, 0, 1], Yes, [2, 1, 0, 0, 1, 2], [[0, 5], [1, 4], [2, 3]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0063","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7ff7c8be82234bb83ab33056f7c5a051fdd8918b9a1eb4811f44aa554efd24ae","family_id":"qf-density16-inj-append-output-qualification","seed":202610166063,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[0,2,0,0,1],\"m\":3,\"n\":5,\"value\":-1}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 2, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=-1. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([0, 2, 0, 0, 1], No, outside-codomain, not-evaluated, not-evaluated)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 0, 0, 1], No, outside-codomain, not-evaluated, not-evaluated)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0064","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7a031b51ff2809010276c8f47a266f5fba3814c41b8b3c862dda27af4fbad3c2","family_id":"qf-density16-inj-append-output-qualification","seed":202610166064,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[1,2,0,0,1],\"m\":3,\"n\":5,\"value\":0}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 2, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=0. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([1, 2, 0, 0, 1], Yes, [1, 2, 0, 0, 1, 0], [[0, 4], [2, 3], [2, 5], [3, 5]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 0, 0, 1], Yes, [1, 2, 0, 0, 1, 0], [[0, 4], [2, 3], [2, 5], [3, 5]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0065","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"55728573046adfc69ea1ee4b77ffed2076463f98dbbec0014aa709559bfeec63","family_id":"qf-density16-inj-append-output-qualification","seed":202610166065,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[2,2,0,0,1],\"m\":3,\"n\":5,\"value\":1}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 2, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=1. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([2, 2, 0, 0, 1], Yes, [2, 2, 0, 0, 1, 1], [[0, 1], [2, 3], [4, 5]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 0, 0, 1], Yes, [2, 2, 0, 0, 1, 1], [[0, 1], [2, 3], [4, 5]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0066","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5c48a1271e285387ad4fcfbf44a71c74b1db6c29d8baf8bb5ba139af5322c85","family_id":"qf-density16-inj-append-output-qualification","seed":202610166066,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[0,0,1,0,1],\"m\":3,\"n\":5,\"value\":2}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 0, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=2. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([0, 0, 1, 0, 1], Yes, [0, 0, 1, 0, 1, 2], [[0, 1], [0, 3], [1, 3], [2, 4]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1, 0, 1], Yes, [0, 0, 1, 0, 1, 2], [[0, 1], [0, 3], [1, 3], [2, 4]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0067","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"99d77c01442543e83ee6d760d803b7d5bc50921ff55094cd54d961457a0388b2","family_id":"qf-density16-inj-append-output-qualification","seed":202610166067,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[1,0,1,0,1],\"m\":3,\"n\":5,\"value\":-1}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 0, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=-1. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([1, 0, 1, 0, 1], No, outside-codomain, not-evaluated, not-evaluated)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 1, 0, 1], No, outside-codomain, not-evaluated, not-evaluated)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0068","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a50e07f9f604f1f146aa265229e2d6959d9fded51fc6762ce6b6da39f9bb4acb","family_id":"qf-density16-inj-append-output-qualification","seed":202610166068,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[2,0,1,0,1],\"m\":3,\"n\":5,\"value\":0}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 0, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=0. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([2, 0, 1, 0, 1], Yes, [2, 0, 1, 0, 1, 0], [[1, 3], [1, 5], [2, 4], [3, 5]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 1, 0, 1], Yes, [2, 0, 1, 0, 1, 0], [[1, 3], [1, 5], [2, 4], [3, 5]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0069","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"21fde17edad93f0c4b26b6e580da0bc0acc629c0cd394ffcd9ac0f847d0f29dd","family_id":"qf-density16-inj-append-output-qualification","seed":202610166069,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[0,1,1,0,1],\"m\":3,\"n\":5,\"value\":1}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 1, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=1. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([0, 1, 1, 0, 1], Yes, [0, 1, 1, 0, 1, 1], [[0, 3], [1, 2], [1, 4], [1, 5], [2, 4], [2, 5], [4, 5]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 1, 0, 1], Yes, [0, 1, 1, 0, 1, 1], [[0, 3], [1, 2], [1, 4], [1, 5], [2, 4], [2, 5], [4, 5]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0070","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d7abc33099ae5602d92894275b4ce82bfa25ceb9707adb4e10ae1e73b6f1b616","family_id":"qf-density16-inj-append-output-qualification","seed":202610166070,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[1,1,1,0,1],\"m\":3,\"n\":5,\"value\":2}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 1, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=2. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([1, 1, 1, 0, 1], Yes, [1, 1, 1, 0, 1, 2], [[0, 1], [0, 2], [0, 4], [1, 2], [1, 4], [2, 4]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 1, 0, 1], Yes, [1, 1, 1, 0, 1, 2], [[0, 1], [0, 2], [0, 4], [1, 2], [1, 4], [2, 4]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0071","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"55edb0093f1d35d729e6bb02768315bffb2a6f73b1d68b4297e5c6f61036be2e","family_id":"qf-density16-inj-append-output-qualification","seed":202610166071,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[2,1,1,0,1],\"m\":3,\"n\":5,\"value\":-1}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 1, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=-1. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([2, 1, 1, 0, 1], No, outside-codomain, not-evaluated, not-evaluated)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 1, 0, 1], No, outside-codomain, not-evaluated, not-evaluated)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0072","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b79e01d293f95bd63097972a1767a488c2c45d13faac0d2c0dfbc96b57f9b03","family_id":"qf-density16-inj-append-output-qualification","seed":202610166072,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[0,2,1,0,1],\"m\":3,\"n\":5,\"value\":0}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 2, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=0. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([0, 2, 1, 0, 1], Yes, [0, 2, 1, 0, 1, 0], [[0, 3], [0, 5], [2, 4], [3, 5]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 1, 0, 1], Yes, [0, 2, 1, 0, 1, 0], [[0, 3], [0, 5], [2, 4], [3, 5]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0073","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b019e319aa749c5362c776697774c9b1b1a3f1417e1049dbc88807f32da33945","family_id":"qf-density16-inj-append-output-qualification","seed":202610166073,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[1,2,1,0,1],\"m\":3,\"n\":5,\"value\":1}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 2, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=1. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([1, 2, 1, 0, 1], Yes, [1, 2, 1, 0, 1, 1], [[0, 2], [0, 4], [0, 5], [2, 4], [2, 5], [4, 5]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 1, 0, 1], Yes, [1, 2, 1, 0, 1, 1], [[0, 2], [0, 4], [0, 5], [2, 4], [2, 5], [4, 5]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0074","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fdaab13fbac07f3dc0c9dfab626eeb6aeb60f56f6abd3136058c00cbddaf88ad","family_id":"qf-density16-inj-append-output-qualification","seed":202610166074,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[2,2,1,0,1],\"m\":3,\"n\":5,\"value\":2}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 2, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=2. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([2, 2, 1, 0, 1], Yes, [2, 2, 1, 0, 1, 2], [[0, 1], [0, 5], [1, 5], [2, 4]], No)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 1, 0, 1], Yes, [2, 2, 1, 0, 1, 2], [[0, 1], [0, 5], [1, 5], [2, 4]], No)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0075","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"56c7c2afffb0b994fc3c1e4f30a6ef80d227464981d518b8a99796360a0052c0","family_id":"qf-density16-inj-append-output-qualification","seed":202610166075,"semantic_template":"qf-density16-inj-append-output-qualification/{\"images\":[0,0,2,0,1],\"m\":3,\"n\":5,\"value\":-1}"},"fields":{"title":"Append Candidate Guard before New Table and Collisions","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 0, 2, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. At the new input label n, append candidate value=-1. Give (original images, new-value codomain eligibility, new images or outside-codomain, new collisions or not-evaluated, new injectivity or not-evaluated). A valid append can create nine input labels.","answer":{"kind":"expression","value":"([0, 0, 2, 0, 1], No, outside-codomain, not-evaluated, not-evaluated)","explanation":"Require the candidate to lie in the original codomain before constructing the extended table or testing its properties; an outside value supplies no new map. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 2, 0, 1], No, outside-codomain, not-evaluated, not-evaluated)."},"skills":["Append Candidate Guard before New Table and Collisions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0076","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dbce92416d1fbce9fba334a7dc733c79935c66ff047e84e603be9225a6117ff9","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166076,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[],\"m\":0,\"n\":0,\"subset\":[]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(0)=∅→Fin(0), complete images=[]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([], [], [], Yes)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([], [], [], Yes)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0077","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"34ce937f769d514498dd56764452f4f06cbca557b9c61788a6136fc3165f1783","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166077,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[0,0,1],\"m\":2,\"n\":3,\"subset\":[0,2]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(3)={0,…,2}→Fin(2), complete images=[0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[2, 0] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([0, 0, 1], [0, 2], [], Yes)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1], [0, 2], [], Yes)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0078","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b505ccacb183816be392e74ccbcbc95aa239022fc4d7910539858b537f720829","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166078,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[0,0,1],\"m\":2,\"n\":3,\"subset\":[0,1]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(3)={0,…,2}→Fin(2), complete images=[0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[1, 0] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([0, 0, 1], [0, 1], [[0, 1]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1], [0, 1], [[0, 1]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0079","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e7f18464d73785647714bc5838f31985c36fd71577474f605dc991a6fa5acaa","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166079,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[0,0,0,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 0, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([0, 0, 0, 0, 1], [0, 1, 2, 3, 4], [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 0, 0, 1], [0, 1, 2, 3, 4], [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0080","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0501cf05620a9c3c571b2e2f70196d784506ca98d862cc3678d2e3489233e8f1","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166080,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[1,0,0,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 0, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([1, 0, 0, 0, 1], [0, 1, 2, 3, 4], [[0, 4], [1, 2], [1, 3], [2, 3]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 0, 0, 1], [0, 1, 2, 3, 4], [[0, 4], [1, 2], [1, 3], [2, 3]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0081","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"93d2acaa3220425cdc8d83e207ec6df56715f2ac30e0c9b9667916463f3ef237","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166081,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[2,0,0,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 0, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([2, 0, 0, 0, 1], [0, 1, 2, 3, 4], [[1, 2], [1, 3], [2, 3]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 0, 0, 1], [0, 1, 2, 3, 4], [[1, 2], [1, 3], [2, 3]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0082","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"465150009c54550e7721c1f3a2cb86a2e8be39d5b53202e91b3e7b9ab76da75f","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166082,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[0,1,0,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 1, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([0, 1, 0, 0, 1], [0, 1, 2, 3, 4], [[0, 2], [0, 3], [1, 4], [2, 3]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 0, 0, 1], [0, 1, 2, 3, 4], [[0, 2], [0, 3], [1, 4], [2, 3]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0083","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"09331c181b097f42c1a16c3954ec40b205786113e76d8cfba99654ddfafb3777","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166083,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[1,1,0,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 1, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([1, 1, 0, 0, 1], [0, 1, 2, 3, 4], [[0, 1], [0, 4], [1, 4], [2, 3]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 0, 0, 1], [0, 1, 2, 3, 4], [[0, 1], [0, 4], [1, 4], [2, 3]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0084","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c0033908bf563b5a1e3ebd8e15200d14fb4b169875cd33fe17688bd1453c492","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166084,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[2,1,0,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 1, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([2, 1, 0, 0, 1], [0, 1, 2, 3, 4], [[1, 4], [2, 3]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 0, 0, 1], [0, 1, 2, 3, 4], [[1, 4], [2, 3]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0085","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"94c3f51a65d7da2601973396f93d93cf551e12c9c03793131536898ded242c52","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166085,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[0,2,0,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 2, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([0, 2, 0, 0, 1], [0, 1, 2, 3, 4], [[0, 2], [0, 3], [2, 3]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 0, 0, 1], [0, 1, 2, 3, 4], [[0, 2], [0, 3], [2, 3]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0086","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a7c0151c904299ad54fd8c80941717869aabb581be5dea741c8daf8ccabbc589","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166086,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[1,2,0,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 2, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([1, 2, 0, 0, 1], [0, 1, 2, 3, 4], [[0, 4], [2, 3]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 0, 0, 1], [0, 1, 2, 3, 4], [[0, 4], [2, 3]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0087","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dd14b401b3607326eafeb0db4c95f60379da0f4c7d854a4810bb21ea76beefac","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166087,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[2,2,0,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 2, 0, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([2, 2, 0, 0, 1], [0, 1, 2, 3, 4], [[0, 1], [2, 3]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 0, 0, 1], [0, 1, 2, 3, 4], [[0, 1], [2, 3]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0088","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"05abf03f86caf9bb233b751607d9476ad355778271406fcd0d848f62403f9a07","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166088,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[0,0,1,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 0, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([0, 0, 1, 0, 1], [0, 1, 2, 3, 4], [[0, 1], [0, 3], [1, 3], [2, 4]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 1, 0, 1], [0, 1, 2, 3, 4], [[0, 1], [0, 3], [1, 3], [2, 4]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0089","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5ca45fe7b3fde4a2b839122ddf9f863c9471c612fb5b3cee7afb50dcb14c8ffc","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166089,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[1,0,1,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 0, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([1, 0, 1, 0, 1], [0, 1, 2, 3, 4], [[0, 2], [0, 4], [1, 3], [2, 4]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 1, 0, 1], [0, 1, 2, 3, 4], [[0, 2], [0, 4], [1, 3], [2, 4]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0090","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5fb7a65d9b0e77bc2c2cdd55addbbd9fd9bc9c8831a1933c4e247da24c17bfef","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166090,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[2,0,1,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 0, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([2, 0, 1, 0, 1], [0, 1, 2, 3, 4], [[1, 3], [2, 4]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 1, 0, 1], [0, 1, 2, 3, 4], [[1, 3], [2, 4]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0091","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c21a5a82f55557b182899631975a1338df37b9fc15e22de022c380ddd1eaa01","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166091,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[0,1,1,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 1, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([0, 1, 1, 0, 1], [0, 1, 2, 3, 4], [[0, 3], [1, 2], [1, 4], [2, 4]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 1, 0, 1], [0, 1, 2, 3, 4], [[0, 3], [1, 2], [1, 4], [2, 4]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0092","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"78754440be70bd50e4200ff8c50783d525153363ca7c0f2b54ea5bc05b1df2ed","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166092,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[1,1,1,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 1, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([1, 1, 1, 0, 1], [0, 1, 2, 3, 4], [[0, 1], [0, 2], [0, 4], [1, 2], [1, 4], [2, 4]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 1, 1, 0, 1], [0, 1, 2, 3, 4], [[0, 1], [0, 2], [0, 4], [1, 2], [1, 4], [2, 4]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0093","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"369c38c6db3f8c2096cb24c8d0f7487151e8db9ebf027238281b0f6f54e5c7f4","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166093,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[2,1,1,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 1, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([2, 1, 1, 0, 1], [0, 1, 2, 3, 4], [[1, 2], [1, 4], [2, 4]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 1, 1, 0, 1], [0, 1, 2, 3, 4], [[1, 2], [1, 4], [2, 4]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0094","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7522cd14eca368dcb5f93e900b4bf610a805e5741b05f020e020481397d8cd3","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166094,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[0,2,1,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 2, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([0, 2, 1, 0, 1], [0, 1, 2, 3, 4], [[0, 3], [2, 4]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 2, 1, 0, 1], [0, 1, 2, 3, 4], [[0, 3], [2, 4]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0095","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f9b33383774a0d96de9f272de25f550d924f74fa36ec858c91edc8ff1dc52e4","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166095,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[1,2,1,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 2, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([1, 2, 1, 0, 1], [0, 1, 2, 3, 4], [[0, 2], [0, 4], [2, 4]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 2, 1, 0, 1], [0, 1, 2, 3, 4], [[0, 2], [0, 4], [2, 4]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0096","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b77dbed516d3e89da3fd78e66e85d8fa4a02529956d99c106bf0d9f9daacef5","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166096,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[2,2,1,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 2, 1, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([2, 2, 1, 0, 1], [0, 1, 2, 3, 4], [[0, 1], [2, 4]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 2, 1, 0, 1], [0, 1, 2, 3, 4], [[0, 1], [2, 4]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0097","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"82107a7fcbe4b1bc9739d11ba6a6e5e9e7319b20b203b7ae4668844883b5d8c0","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166097,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[0,0,2,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 0, 2, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([0, 0, 2, 0, 1], [0, 1, 2, 3, 4], [[0, 1], [0, 3], [1, 3]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 0, 2, 0, 1], [0, 1, 2, 3, 4], [[0, 1], [0, 3], [1, 3]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0098","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b61d4aedefa11dde2695603cf8d1962d6f92d85083e315ad6b4dca31e296dc9e","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166098,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[1,0,2,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[1, 0, 2, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([1, 0, 2, 0, 1], [0, 1, 2, 3, 4], [[0, 4], [1, 3]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([1, 0, 2, 0, 1], [0, 1, 2, 3, 4], [[0, 4], [1, 3]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0099","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"29d329f14b38d7880cb15f394f2974fc9c7c7028ab2b839d40c7be3e946279bd","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166099,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[2,0,2,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[2, 0, 2, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([2, 0, 2, 0, 1], [0, 1, 2, 3, 4], [[0, 2], [1, 3]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([2, 0, 2, 0, 1], [0, 1, 2, 3, 4], [[0, 2], [1, 3]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261016-r06-injective-0100","batch_id":"batch-20261012-advance-20261016-r06-injective-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f123d2c08e20b14b5326d9ea591fd07eb1038357b68767359e07cad8ea2f75d5","family_id":"qf-density16-inj-restriction-subset-witness","seed":202610166100,"semantic_template":"qf-density16-inj-restriction-subset-witness/{\"images\":[0,1,2,0,1],\"m\":3,\"n\":5,\"subset\":[0,1,2,3,4]}"},"fields":{"title":"Typed Source Restriction and Complete Collision Witnesses","stem":"Finite types start at 0: Fin(5)={0,…,4}→Fin(3), complete images=[0, 1, 2, 0, 1]. Check the whole original total table before any candidate evaluation; empty-function properties depend on the declared codomain. Original subset=[4, 2, 0, 1, 3] is declared a distinct set of legal source labels. Give (original images, sorted subset, every within-subset collision, restricted injectivity).","answer":{"kind":"expression","value":"([0, 1, 2, 0, 1], [0, 1, 2, 3, 4], [[0, 3], [1, 4]], No)","explanation":"Check the entire original table before comparing the restricted original labels; sorting does not merge distinct inputs or invent renumbered ones. These exact finite checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 0, 1], [0, 1, 2, 3, 4], [[0, 3], [1, 4]], No)."},"skills":["Typed Source Restriction and Complete Collision Witnesses"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}