{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0001","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"473da8b25bdbaef681d3ed6a42b00b89e1e422f42aa988ac5af3ebb3f657f905","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195001,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,3,4,5],\"n\":6,\"steps\":[[0,1],[1,2],[2,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0002","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f92474abaf017a9593918d2f27b7dcd4c3d786db189ae0b3b94c35484f16821","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195002,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[0,3],[1,2],[2,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0003","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6c070d3d0c013d18f239850cc7cc51a4f610f329d9cda55e39587ba2c811700c","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195003,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0004","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"23ff8e27a79186611061bfe99f7e61f2c3d4049afae9c45c3e1a7e26edb2b8d6","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195004,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,3,4,5],\"n\":6,\"steps\":[[0,1],[0,3],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0005","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cbae0c190d7fa0e99724e2cad760975f2e5eb97b21ed402d26b1aacaa296a50b","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195005,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0006","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b154608bf753b86bff20f1e7958aacaa8c7384cd1a6ee2d5dcf6ceddabd2143b","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195006,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[0,3],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0007","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"41a76f6256c65aeb0425fa4eae296805293bb0ed562eec63cea966f956e18d95","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195007,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,3,4,5],\"n\":6,\"steps\":[[0,1],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0008","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"46dad1a21e366b97a0fd5c02e3d5730b90fb6995471af81d4bfd0eb6f165e27b","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195008,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[0,3],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0009","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"16a6a3c64b54378817393edefa577ff14401cfa0e9f1943c0634492c4988534e","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195009,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[1,2],[2,1],[2,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0010","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"02be6ebbf8733353d88a08d1ef20e0a26d15127a54096b2ea0ca1d131e47a266","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195010,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,3,4,5],\"n\":6,\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0011","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5f67f49f360367ab13a5b03bd6232c96f04e4723621108246d0c9969b81f3830","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195011,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[1,2],[2,1],[2,5],[3,4]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0012","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b8b53e2847620ac3b19e9061e9a6329b60c37cb66c78a8faf83dc5a5b951bbe9","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195012,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5],[3,4]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0013","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9bd66d831cfbd8491ec1a837cb66d04ed11de9e2dda7b51807190f1a802f16bf","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195013,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,3,4,5],\"n\":6,\"steps\":[[0,1],[1,2],[2,1],[2,5],[4,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0014","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ee695727d8364b7befac4ebf1e79e2e29aadbcd149af40ba023439577a187711","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195014,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5],[4,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0015","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e55fb7b18655684d97f0e013e3c30e337a19b42c8fd6362e21b31f01850134d3","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195015,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[1,2],[2,1],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0016","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"41d47a69462ccf4244833b6d9f0c7f402d4e0142e5889b773b8cef6e449dd8f3","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195016,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,3,4,5],\"n\":6,\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0017","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2caa40869b09796fc0cae89a67ccf004464d30c183d976ec68ef353848576e7e","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195017,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[1,1],[1,2],[2,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0018","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca70eefa74405259c944b442c06b9068aef0de3ae0f2d70dc7c80a21d3ce1b35","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195018,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0019","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd48166223d220567673af90885ee0982c7c4e40377facb8d90dfbf67aba0e3e","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195019,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,3,4,5],\"n\":6,\"steps\":[[0,1],[1,1],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0020","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b71ab08e00ce9ba8326806638e2fe2051d0e1d655afd692c3a9b7632fdaaa07a","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195020,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0021","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4b36e105dd3ce7fe0d922ae9090f6f18c22910d13733fb69bd6674b2edadb7ad","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195021,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[1,1],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0022","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9f55dd9f5c50ea47cf28e1b3409b49ae14a56b34eb5e47e04daef1900e8591b9","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195022,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,3,4,5],\"n\":6,\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0023","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a8b485dc47df73d278ade8711f703c2d4da55baaf7a0af43b7fcb800c6f38ba0","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195023,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[1,1],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0024","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b2fcd0df6b0af84acd0dff4e1f70bf47d86a7cdc54429384cc427f3d9b27f746","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195024,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,5],\"n\":6,\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 5], Yes, [[0, 3]], No, No, No)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0025","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2eccf9f7a7fb6dac534893447750f2398fb53708bc0db1a85d4e972a7c4d0c07","family_id":"qf-ext19-correctness-invariant-audit","seed":202610195025,"semantic_template":"qf-ext19-correctness-invariant-audit/{\"halt\":[5],\"init\":[0],\"invariant\":[0,1,2,3,4,5],\"n\":6,\"steps\":[[0,1],[1,1],[1,2],[2,1],[2,5]]}"},"fields":{"title":"Inductiveness versus Merely Covering Reachable States","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Proposed invariant=[0, 1, 2, 3, 4, 5]. Give (invariant, contains Init?, every Step edge from an invariant nonhalt to outside the invariant, step-preserved?, inductive?, covers all reachable states?). Inspect all invariant states for preservation, including unreachable ones.","answer":{"kind":"expression","value":"([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)","explanation":"Initial validity and one-step preservation establish inductiveness; reachable-state coverage alone is a different, weaker observation. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], Yes, [], Yes, Yes, Yes)."},"skills":["Inductiveness versus Merely Covering Reachable States"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0026","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"906e92eaeaa09edcb5e96ea250c7dbd9fe0966981d647fc77d336428c4428801","family_id":"qf-ext19-correctness-partial-audit","seed":202610195026,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [], [5], No, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0027","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ccb343e22a1eb54671a1a83758b8ded28aef1e73e2fb149bac39ecbab657c21","family_id":"qf-ext19-correctness-partial-audit","seed":202610195027,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0028","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"78e42ca3c05e380b199a2ed82a10eeb2dc6a4b7458c093a5eda69c0528e06528","family_id":"qf-ext19-correctness-partial-audit","seed":202610195028,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [], [5], No, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0029","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8cab64fb366a04dfea1d3ec86d926436fa381a3a3d9e800db1ec40615707e892","family_id":"qf-ext19-correctness-partial-audit","seed":202610195029,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0030","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7df538f44410cb36df17103669eb4c2eaa15d8538583a6d582ec4793f9c600ab","family_id":"qf-ext19-correctness-partial-audit","seed":202610195030,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [], [5], No, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0031","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ea9e3a65647b978901f2bb3aed8c0ea7ced72fd69933207014d4d1ff278cf3c","family_id":"qf-ext19-correctness-partial-audit","seed":202610195031,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0032","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2567becbeba804c676c008e1c5de31b519532cde8a6e6a3b90132762f7cf6842","family_id":"qf-ext19-correctness-partial-audit","seed":202610195032,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [], [5], No, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0033","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"25e537687616ae28439748d5000badf20b2e2997e0fde151f66bd39f140e2f3b","family_id":"qf-ext19-correctness-partial-audit","seed":202610195033,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0034","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"44f9cc648d46f21f9843841585f8f03f4248d0b0145917d8893d325312c9568a","family_id":"qf-ext19-correctness-partial-audit","seed":202610195034,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,1],[2,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [], [5], No, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0035","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"17ff6e09eb3aa1e40e442bd570742b8e4992d36bde72698009158ac666a98149","family_id":"qf-ext19-correctness-partial-audit","seed":202610195035,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0036","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"df9a18c23ef6b11aca4238e9c90a6222480e08bba1d2b32ea78ca7ad536a775f","family_id":"qf-ext19-correctness-partial-audit","seed":202610195036,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,1],[2,5],[3,4]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [], [5], No, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0037","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a7b775f1fbb57ea25191fd7f2b2bc209abe0460ba92fc3e6625da25ba3cc4127","family_id":"qf-ext19-correctness-partial-audit","seed":202610195037,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5],[3,4]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0038","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"25c99fb5eb17fb413754c2ac9233cdf566ce68b79c374bb726f92b8ea9c6a60f","family_id":"qf-ext19-correctness-partial-audit","seed":202610195038,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,1],[2,5],[4,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [], [5], No, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0039","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"77d3301c1d1a226a38c0ef5a46628623f11adb6326ef52c8c1d9a78539233956","family_id":"qf-ext19-correctness-partial-audit","seed":202610195039,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5],[4,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0040","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3f9127b9a3b6c0d953483e1af34d687310cef5ebc20eaca5f6bbe1a350e4484","family_id":"qf-ext19-correctness-partial-audit","seed":202610195040,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,1],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [], [5], No, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0041","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"95d212bd4934a3c74c3dd0112c2c0c08ec51b494a21b2c4d149081d535f32982","family_id":"qf-ext19-correctness-partial-audit","seed":202610195041,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0042","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2ede11dea513049fadc28cbc1cd655ba390b64d89c1243d653b826a55b3cf67c","family_id":"qf-ext19-correctness-partial-audit","seed":202610195042,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,1],[1,2],[2,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [], [5], No, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0043","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de607e65d048405ca0a1f9a92b53d4ffabdf61c767f56fd7b1acf90474f480e1","family_id":"qf-ext19-correctness-partial-audit","seed":202610195043,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0044","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b269a6b16a4befa2864549e247e391a779eb3531b8786aad2394d2e307bfda02","family_id":"qf-ext19-correctness-partial-audit","seed":202610195044,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,1],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [], [5], No, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0045","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"00354e92b87d9f4d857786721181d185cbbfda2b09f90e92f1942ac0a4dc44bc","family_id":"qf-ext19-correctness-partial-audit","seed":202610195045,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0046","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"53211f0a26eb169138e308f6321ea3157999f7968a6f84320e3f2b85b74f6466","family_id":"qf-ext19-correctness-partial-audit","seed":202610195046,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,1],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [], [5], No, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0047","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a3dedd0cb2ea587c95b205ac3bd9ed31dac0064bf344fef21f6e2e9f64f0c43c","family_id":"qf-ext19-correctness-partial-audit","seed":202610195047,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0048","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"caf2c202f02e6db730b697606f84f92d0c795bc5608da026a572c0bf3aebb133","family_id":"qf-ext19-correctness-partial-audit","seed":202610195048,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,1],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [], [5], No, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0049","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b824c382c89ee112437bc013b68abff7240a58804b1658fda050218ff14d93aa","family_id":"qf-ext19-correctness-partial-audit","seed":202610195049,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0050","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dcb75391663af4e8e2f1bb472cdc7cf31c05d4918d1acf0ed1c84f1c2c0d83b2","family_id":"qf-ext19-correctness-partial-audit","seed":202610195050,"semantic_template":"qf-ext19-correctness-partial-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,1],[1,2],[2,1],[2,5]]}"},"fields":{"title":"Partial Correctness and Its Vacuous Case","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad reachable halts, partial correctness?, no reachable halt?).","answer":{"kind":"expression","value":"([5], [], [5], No, No)","explanation":"Partial correctness checks only reached halts; it may hold vacuously while no run terminates successfully. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No)."},"skills":["Partial Correctness and Its Vacuous Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0051","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f0dab7538804cebcfea1998bec4cc9a9591feaa4865e5a4183f4818248e0c64c","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195051,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[6,5,4,3,2,0],\"steps\":[[0,1],[1,2],[2,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[6, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([6, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([6, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0052","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"032124fc5b0e304ee46727642b8cb59b8a7c40c09a13a8b5eac09357ffaa5dd1","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195052,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[7,5,4,3,2,0],\"steps\":[[0,1],[0,3],[1,2],[2,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[7, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([7, 5, 4, 3, 2, 0], [], Yes, [3], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([7, 5, 4, 3, 2, 0], [], Yes, [3], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0053","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3aab141775d5cf84dc1fa8afaafaa407a6e878c866e5bef9f99b363674279b9a","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195053,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[8,5,4,3,2,0],\"steps\":[[0,1],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[8, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([8, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([8, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0054","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ba3e4941dab5ba30eb717cd938649e95d7a0e56046e64e9fbf6d5e21be1ecc56","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195054,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[9,5,4,3,2,0],\"steps\":[[0,1],[0,3],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[9, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([9, 5, 4, 3, 2, 0], [], Yes, [4], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([9, 5, 4, 3, 2, 0], [], Yes, [4], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0055","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a83a28927a042e2d06350517075836e35d1362d2bb041159ff03809cf5156dee","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195055,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[10,5,4,3,2,0],\"steps\":[[0,1],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[10, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([10, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([10, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0056","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fdb2aca05b31b1d04402baf3bea3c572dc36fc85ef38bebe942846631e2160f0","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195056,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[11,5,4,3,2,0],\"steps\":[[0,1],[0,3],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[11, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([11, 5, 4, 3, 2, 0], [], Yes, [3], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([11, 5, 4, 3, 2, 0], [], Yes, [3], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0057","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"764649f8ca2b7daeb0738e06ce551a25e65b1079ca922714280a28468b131f9d","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195057,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[12,5,4,3,2,0],\"steps\":[[0,1],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[12, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([12, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([12, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0058","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"16940c0c0eabbd17275a964c7c4d21891a2f3a7800c4f48967d2a8b82378bd96","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195058,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[13,5,4,3,2,0],\"steps\":[[0,1],[0,3],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[13, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([13, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([13, 5, 4, 3, 2, 0], [], Yes, [], Yes, Yes)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0059","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e28a8579bd51de3401191a3688ffd953e12f4aba0afc05f23285320d0d4820b8","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195059,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[14,5,4,3,2,0],\"steps\":[[0,1],[1,2],[2,1],[2,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[14, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([14, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([14, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0060","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"98b3b7bbd7d00dfee593efc8166125001280e4b21d94c713fd9f37fbd6980bfc","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195060,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[15,5,4,3,2,0],\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[15, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([15, 5, 4, 3, 2, 0], [[2, 1]], No, [3], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([15, 5, 4, 3, 2, 0], [[2, 1]], No, [3], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0061","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c104ce688864f89f669a673a94a9fee9cbb49c1de79a901ca54c864d7ff2847a","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195061,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[16,5,4,3,2,0],\"steps\":[[0,1],[1,2],[2,1],[2,5],[3,4]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[16, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([16, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([16, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0062","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"98f5cb531b915d80a8b60c3c3cc0165b2c6efcefa209cccc53effce9a3e8fd67","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195062,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[17,5,4,3,2,0],\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5],[3,4]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[17, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([17, 5, 4, 3, 2, 0], [[2, 1]], No, [4], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([17, 5, 4, 3, 2, 0], [[2, 1]], No, [4], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0063","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"63828f94e1c857f1a82b957d880349cb5708cf2424493486fef9de74c3c08a10","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195063,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[18,5,4,3,2,0],\"steps\":[[0,1],[1,2],[2,1],[2,5],[4,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[18, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([18, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([18, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0064","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a91dcd1f492b80521a18dfb7067da9105dd4a96c5e6b624f5261522005caa48","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195064,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[19,5,4,3,2,0],\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5],[4,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[19, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([19, 5, 4, 3, 2, 0], [[2, 1]], No, [3], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([19, 5, 4, 3, 2, 0], [[2, 1]], No, [3], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0065","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1fff0e18dc22dff94e2cadf9e82daa544f2da618246812f0134b7c6eb2d3ee81","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195065,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[20,5,4,3,2,0],\"steps\":[[0,1],[1,2],[2,1],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[20, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([20, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([20, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0066","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9da362901e88db1edf43f3d8df4599772842a40602db59a7bd3d2efe9e24898b","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195066,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[21,5,4,3,2,0],\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[21, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([21, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([21, 5, 4, 3, 2, 0], [[2, 1]], No, [], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0067","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"85e94622800edc210dda40cc4c2de69a6347c46d344e1e3fe2dd4a45d7787dcb","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195067,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[22,5,4,3,2,0],\"steps\":[[0,1],[1,1],[1,2],[2,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[22, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([22, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([22, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0068","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ddbd759889e0c3e3df2dfe051e5489cfa7a74536b69f363b9469622a87eefc25","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195068,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[23,5,4,3,2,0],\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[23, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([23, 5, 4, 3, 2, 0], [[1, 1]], No, [3], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([23, 5, 4, 3, 2, 0], [[1, 1]], No, [3], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0069","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0499b17205a5bfa70db28aee981a18cd0a2bba0e8ab374c5430fb1e711a47fe1","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195069,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[24,5,4,3,2,0],\"steps\":[[0,1],[1,1],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[24, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([24, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([24, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0070","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aeeebbe760fecd8f3a03c5b8feaf73f19f714908b756a2a36b236eda0d63bdb9","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195070,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[25,5,4,3,2,0],\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[25, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([25, 5, 4, 3, 2, 0], [[1, 1]], No, [4], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([25, 5, 4, 3, 2, 0], [[1, 1]], No, [4], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0071","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"02f7194530afb2e91d53db523e1ea1f3aa183c874dbdc80220be01fab4b1b441","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195071,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[26,5,4,3,2,0],\"steps\":[[0,1],[1,1],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[26, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([26, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([26, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0072","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"19bda48b0f2f1a3c1e95bc5df41640565e196d1615c4d1f5987b1011429e65bd","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195072,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[27,5,4,3,2,0],\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[27, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([27, 5, 4, 3, 2, 0], [[1, 1]], No, [3], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([27, 5, 4, 3, 2, 0], [[1, 1]], No, [3], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0073","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1c63228df8860d2dd9e1996d8da85112f7b1110d5f84269ccbf4a8c96e06df1a","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195073,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[28,5,4,3,2,0],\"steps\":[[0,1],[1,1],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[28, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([28, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([28, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0074","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b18be02175cebe93c378b856d9de7d8294e806a398edeabbe19986015f6f805d","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195074,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[29,5,4,3,2,0],\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[29, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([29, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([29, 5, 4, 3, 2, 0], [[1, 1]], No, [], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0075","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4432c0fcd96f7ccb7e52c75c0ea8d3a888d59817bd23282e18f067cdbf48d6d1","family_id":"qf-ext19-correctness-ranking-audit","seed":202610195075,"semantic_template":"qf-ext19-correctness-ranking-audit/{\"halt\":[5],\"init\":[0],\"n\":6,\"rank\":[30,5,4,3,2,0],\"steps\":[[0,1],[1,1],[1,2],[2,1],[2,5]]}"},"fields":{"title":"Ranking Decrease, Stuck States, and Actual Termination","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Natural ranking vector=[30, 5, 4, 3, 2, 0]. Give (rank vector, reachable nonhalt Step edges failing strict decrease, ranking valid?, reachable stuck states, ranking valid and no stuck states?, actual all-path termination?).","answer":{"kind":"expression","value":"([30, 5, 4, 3, 2, 0], [[1, 1], [2, 1]], No, [], No, No)","explanation":"Check ranking decrease on every reachable nonhalt transition; vacuous edge decrease alone does not eliminate a stuck nonhalt. A failed ranking need not mean the model fails to terminate. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([30, 5, 4, 3, 2, 0], [[1, 1], [2, 1]], No, [], No, No)."},"skills":["Ranking Decrease, Stuck States, and Actual Termination"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0076","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aefc1e87f22be026d7576f445e10b868db916ea9460e45d72ec83370374961c2","family_id":"qf-ext19-correctness-total-separation","seed":202610195076,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [], [5], No, No, [], No, Yes, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], No, Yes, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0077","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ee89f1dd854d4228787130a109cd8b4f17a8a60af94bc0a8d09b2f0c2823d9b","family_id":"qf-ext19-correctness-total-separation","seed":202610195077,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No, [3], No, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [3], No, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0078","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f8f7075c76dcdded81924a67b6ca55082a26f78992d3c1fef99f42334d316008","family_id":"qf-ext19-correctness-total-separation","seed":202610195078,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [], [5], No, No, [], No, Yes, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], No, Yes, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0079","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b10f271c23a20e39e44d0134de6c52b6342e537ef0c54bae3e07fa4bdfa4e6ae","family_id":"qf-ext19-correctness-total-separation","seed":202610195079,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No, [4], No, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [4], No, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0080","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5da771c6acdcfd50f5f97cabc96d8a848a2e1bbb6af26b53066954c3c6cabf4","family_id":"qf-ext19-correctness-total-separation","seed":202610195080,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [], [5], No, No, [], No, Yes, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], No, Yes, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0081","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cdc515a88ad6b277b4041667f5933deb8d1b0e1b705dbdff95d21960e8aaf7f8","family_id":"qf-ext19-correctness-total-separation","seed":202610195081,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No, [3], No, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [3], No, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0082","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4bff773696e624058a09cd1f8119a91e263fca15bd8ca822df0558b45f7eb982","family_id":"qf-ext19-correctness-total-separation","seed":202610195082,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [], [5], No, No, [], No, Yes, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], No, Yes, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0083","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ad0de822f633b051777b3d3bfb708b3b2ae65d17a7a7b7322576b48a4a765af9","family_id":"qf-ext19-correctness-total-separation","seed":202610195083,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No, [], No, Yes, Yes)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [], No, Yes, Yes)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0084","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5772475f9476de1c8b923fe40998485070cdcfccd88d359979c2afd565ed094e","family_id":"qf-ext19-correctness-total-separation","seed":202610195084,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,1],[2,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [], [5], No, No, [], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0085","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ba1cd252af6027a66b50a3e6853b001197f5b1b5b479e052ad774e314710c53","family_id":"qf-ext19-correctness-total-separation","seed":202610195085,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No, [3], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [3], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0086","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8c6be668c6059eba6be3e51de6bf1617e09ffdb140af24f35f0e3929a65b733d","family_id":"qf-ext19-correctness-total-separation","seed":202610195086,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,1],[2,5],[3,4]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [], [5], No, No, [], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0087","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"179fc1818ea204c1366719b1c48d8b25224d82d8ffd68a76aeaad2918e22b976","family_id":"qf-ext19-correctness-total-separation","seed":202610195087,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5],[3,4]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No, [4], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [4], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0088","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d19481cc2dec51096c273aa01a151234514f878eced62e0bdadce3b917124db","family_id":"qf-ext19-correctness-total-separation","seed":202610195088,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,1],[2,5],[4,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [], [5], No, No, [], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0089","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e9d425350b441074489db8103d82a62083c3b41fbfcd31a41deeed788e15f7a","family_id":"qf-ext19-correctness-total-separation","seed":202610195089,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5],[4,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No, [3], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [3], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0090","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a7b75a9e86c7127b55fe86b879ba81a8cdd7053d02f83782525ca022a5749cd3","family_id":"qf-ext19-correctness-total-separation","seed":202610195090,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,2],[2,1],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [], [5], No, No, [], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0091","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"69a1d90f167cdba3ceeb60f24e36a4c242fd0dd061d49ea7b9353e35ec49fa2d","family_id":"qf-ext19-correctness-total-separation","seed":202610195091,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,2],[2,1],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [2, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No, [], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0092","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ddaf7123bd91202943fddd10b5644317e2dbc7e50f69edf8fc3eed9f67ea15e7","family_id":"qf-ext19-correctness-total-separation","seed":202610195092,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,1],[1,2],[2,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [], [5], No, No, [], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0093","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff4ef04328d0c4dc4f18bc02ce8ffe4c8204daf7147b4aa54a281ac60872b59d","family_id":"qf-ext19-correctness-total-separation","seed":202610195093,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No, [3], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [3], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0094","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ed9f984a0fcb2395ec903d4a656b731bb347be9b4a506a6d166decf155866ff5","family_id":"qf-ext19-correctness-total-separation","seed":202610195094,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,1],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [], [5], No, No, [], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0095","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d82aa5ec09bdb72f49b4722d1550428f9e7efe65a2ac4358066ae7a258d62942","family_id":"qf-ext19-correctness-total-separation","seed":202610195095,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5],[3,4]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No, [4], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [4], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0096","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c06f24f96b33bc40d19a7cd91f6c0d108075081e1ccb2fed15745205840b9aa","family_id":"qf-ext19-correctness-total-separation","seed":202610195096,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,1],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [], [5], No, No, [], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0097","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8aa856b23a53042e9618bafb21c288761d5307be5e224f12367416dc8e177b66","family_id":"qf-ext19-correctness-total-separation","seed":202610195097,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5],[4,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No, [3], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [3], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0098","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"37c81b4a22108df20b677ca9c68df1b7777a3c7872c140b2d4656c7b1502208a","family_id":"qf-ext19-correctness-total-separation","seed":202610195098,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,1],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [], [5], No, No, [], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0099","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b186cb462849b8e546300ad203789d2b1f08e7262a6e491ac5be05ded3093141","family_id":"qf-ext19-correctness-total-separation","seed":202610195099,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[5],\"steps\":[[0,1],[0,3],[1,1],[1,2],[2,5],[3,4],[4,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [0, 3], [3, 4], [4, 5], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[5]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [5], [], Yes, No, [], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [5], [], Yes, No, [], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261019-r05-algorithm-correctness-0100","batch_id":"batch-20261013-advance-20261019-r05-algorithm-correctness-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"17445e65ceb109c27c079cca0dd4fbd0f6530bd218af5377803c9918b2bf5711","family_id":"qf-ext19-correctness-total-separation","seed":202610195100,"semantic_template":"qf-ext19-correctness-total-separation/{\"halt\":[5],\"init\":[0],\"n\":6,\"post\":[],\"steps\":[[0,1],[1,1],[1,2],[2,1],[2,5]]}"},"fields":{"title":"Partial Correctness, Termination, and Total Correctness","stem":"Use the complete finite, possibly nondeterministic transition model S=Fin(6)={0,…,5}, Init=[0], Halt=[5], Step=[[0, 1], [1, 2], [2, 5], [2, 1], [1, 1]]. Normalize sets and ordered transition pairs. Execution stops at a declared Halt state, ignoring all of its outgoing edges. A stuck nonhalting state is not a successful halt; finite prefixes prove no general termination. Postcondition set=[]. Give (reachable halts, postcondition, bad halts, partial correctness?, vacuous partial case?, stuck states, nonhalting cycle?, termination?, total correctness?).","answer":{"kind":"expression","value":"([5], [], [5], No, No, [], Yes, No, No)","explanation":"Total correctness requires both partial correctness and all-path termination; neither one successful trace nor vacuous partial correctness is enough. Finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is ([5], [], [5], No, No, [], Yes, No, No)."},"skills":["Partial Correctness, Termination, and Total Correctness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}