{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0001","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"39531d223baea0ee7d0f58a6b44621dec83153c1e5ffefa0bc2bbc0b2a5063aa","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092001,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[1,-2,[3,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[1, -2, 3/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([1, -2, 3/2], 2, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([1, -2, 3/2], 2, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0002","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a045a285ca1acc4649e86cec1f9663e023c9fc6cca5e1541a588539b175a914f","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092002,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[2,-3,[4,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[2, -3, 4/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([2, -3, 2], 3, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([2, -3, 2], 3, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0003","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"03b29646dd7298a47a47c949ce741efa2de3f9da6d1f27d0889cddeceefce502","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092003,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[3,-4,[5,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[3, -4, 5/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([3, -4, 5/2], 4, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([3, -4, 5/2], 4, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0004","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8895ec9d62e4df11dfe126b188d397b6b814b0ec24f44f813d71383e6b724c77","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092004,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[4,-5,[6,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[4, -5, 6/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([4, -5, 3], 5, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([4, -5, 3], 5, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0005","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"51ec9aae26db6c6cad036e60f93bc8a71b1bc61a1fed2e08244a3eb65a50a5bc","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092005,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[5,-6,[7,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[5, -6, 7/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([5, -6, 7/2], 6, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([5, -6, 7/2], 6, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0006","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"39307029cc54e1fd8ac4a2de75097dd2d23bdc5e485e6230152acbf37320c048","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092006,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[6,-7,[8,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[6, -7, 8/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([6, -7, 4], 7, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([6, -7, 4], 7, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0007","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a8cfe6ff7512f9f256270ff910476a6f53c6b09597599918abb0565f68c4b591","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092007,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[7,-8,[9,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[7, -8, 9/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([7, -8, 9/2], 8, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([7, -8, 9/2], 8, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0008","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b2123965c27060e93cc5f8e8b1210eff0341ecf21b6ba2095ff97a2f902fc206","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092008,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[8,-9,[10,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[8, -9, 10/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([8, -9, 5], 9, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([8, -9, 5], 9, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0009","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4898e62fa0c8096b5ffce3efaa3440f5cf91436a31a7dce76df42112a600d7cc","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092009,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[9,-10,[11,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[9, -10, 11/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([9, -10, 11/2], 10, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([9, -10, 11/2], 10, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0010","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8cf02925fc45864168a82626cbe467199516e965cd17d73999c14020a0eec270","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092010,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[10,-11,[12,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[10, -11, 12/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([10, -11, 6], 11, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([10, -11, 6], 11, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0011","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7ea0bee74107c484412dd12b3169c7639bbadf4a287738928c49e1e2cc4d11df","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092011,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[11,-12,[13,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[11, -12, 13/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([11, -12, 13/2], 12, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([11, -12, 13/2], 12, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0012","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4de944a22cc4224c20d96eceffac9e02b41d87d13859598f3d3bb41a97160dfe","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092012,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[12,-13,[14,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[12, -13, 14/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([12, -13, 7], 13, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([12, -13, 7], 13, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0013","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68efd38f4015ce72d077328b3f39b0726dc23eefa027b0f3868a3f9015facd9e","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092013,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[13,-14,[15,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[13, -14, 15/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([13, -14, 15/2], 14, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([13, -14, 15/2], 14, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0014","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3fce0b01b245ca1bedcd1a00a3c0c9dedfb49fa2ff6e70423958608c8db017e7","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092014,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[14,-15,[16,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[14, -15, 16/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([14, -15, 8], 15, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([14, -15, 8], 15, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0015","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"96ff823839311022aca738e2de73673e6a0bfb23b781a3af58ffbc79bbb4fb75","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092015,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[15,-16,[17,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[15, -16, 17/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([15, -16, 17/2], 16, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([15, -16, 17/2], 16, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0016","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe2b5519e96a725caa2eb221e18a949fb614b3e2d1641bd9f8d5f0fcdff327de","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092016,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[16,-17,[18,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[16, -17, 18/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([16, -17, 9], 17, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([16, -17, 9], 17, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0017","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ac452383b262a8506a7dfb17107dd52fa3694aba4dabedfe17f1caef235757f","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092017,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[17,-18,[19,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[17, -18, 19/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([17, -18, 19/2], 18, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([17, -18, 19/2], 18, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0018","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"67b19856ebf8f79a8c19b7c9fa02b05fd07bcb08138901c918f23fb7bb443a45","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092018,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[18,-19,[20,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[18, -19, 20/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([18, -19, 10], 19, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([18, -19, 10], 19, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0019","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4fc4320c070acaf7513e1f3ad76bedd1a5367f159f4ed15d7af24e56323f6658","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092019,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[19,-20,[21,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[19, -20, 21/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([19, -20, 21/2], 20, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([19, -20, 21/2], 20, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0020","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"19af2a60a659124ab6069aa0efac3555b3d22a9e29f8fcd3c28c51b7371f8ee7","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092020,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[20,-21,[22,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[20, -21, 22/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([20, -21, 11], 21, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([20, -21, 11], 21, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0021","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c9f77f19e0286cf7192d76ce0f8163e1bf9fcb2c4f1627491ad17a550463d393","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092021,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[21,-22,[23,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[21, -22, 23/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([21, -22, 23/2], 22, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([21, -22, 23/2], 22, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0022","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e29220fac11893e54d8bdfe5b7796a5a34e089196a49a98e17bfb5d7c7ebbb82","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092022,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[22,-23,[24,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[22, -23, 24/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([22, -23, 12], 23, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([22, -23, 12], 23, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0023","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"925f8e04425fa270a8a46a80b3a2c708b1f5a8721bbbaaff157701e7d3654de0","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092023,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[23,-24,[25,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[23, -24, 25/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([23, -24, 25/2], 24, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([23, -24, 25/2], 24, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0024","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b20e17758bac1252047eab1a1386a81c4ca20c9b1ce1649163bc9e665776d3a3","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092024,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[24,-25,[26,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[24, -25, 26/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([24, -25, 13], 25, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([24, -25, 13], 25, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0025","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ea1868e91187f4ed71ea33803610509cb458b1811644228e019ec5717c263630","family_id":"qf-mod40-bw-periodic-boundedness","seed":202611092025,"semantic_template":"qf-mod40-bw-periodic-boundedness/{\"period_values\":[25,-26,[27,2]]}"},"fields":{"title":"A Whole Declared Periodic Sequence Has an Exact Absolute Bound","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[25, -26, 27/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Give (period_values, absolute_bound, whole_sequence_bounded, certificate).","answer":{"kind":"expression","value":"([25, -26, 27/2], 26, Yes, whole_infinite_periodic_sequence_bounded)","explanation":"Take the largest absolute value in the complete period as a bound for every term of the infinite sequence. This finite periodic certificate does not prove the theorem for arbitrary sequences. The exact result, retaining the requested order, is ([25, -26, 27/2], 26, Yes, whole_infinite_periodic_sequence_bounded)."},"skills":["A Whole Declared Periodic Sequence Has an Exact Absolute Bound"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0026","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c22c23ef7eb251e4c775c5ed532f8e8493434dab99096d2089f8090a443df3d1","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092026,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":1,\"period_values\":[1,-2,[3,2]],\"residue\":0}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[1, -2, 3/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=0 and subsequence parameter j=1≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(0, 1, 3, 1, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (0, 1, 3, 1, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0027","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7fcc61cba44052a5fc52674b9bedb71b3bdd7aba0eef940c00438ddb26be262c","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092027,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":2,\"period_values\":[2,-3,[4,2]],\"residue\":1}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[2, -3, 4/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=1 and subsequence parameter j=2≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(1, 2, 7, -3, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (1, 2, 7, -3, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0028","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ec17f5d5d57ad4fc21fb3e24cd1a375a2f16c6f08b837de7e0715a891558ea49","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092028,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":3,\"period_values\":[3,-4,[5,2]],\"residue\":2}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[3, -4, 5/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=2 and subsequence parameter j=3≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(2, 3, 11, 5/2, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (2, 3, 11, 5/2, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0029","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cfae45c993e9826442ce243358222f6deead4f859f709dd5d1ba2ab0eab9939f","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092029,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":4,\"period_values\":[4,-5,[6,2]],\"residue\":0}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[4, -5, 6/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=0 and subsequence parameter j=4≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(0, 4, 12, 4, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (0, 4, 12, 4, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0030","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d70796082c001fd1f687a5c86a96122ccab9d36e945c8b91b9cdf15c0c8e59f3","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092030,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":5,\"period_values\":[5,-6,[7,2]],\"residue\":1}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[5, -6, 7/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=1 and subsequence parameter j=5≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(1, 5, 16, -6, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (1, 5, 16, -6, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0031","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6917ad7fc7b2f21685e9c6c49237cae83149c83d074f443b0be5d3fc7f8aa47a","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092031,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":6,\"period_values\":[6,-7,[8,2]],\"residue\":2}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[6, -7, 8/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=2 and subsequence parameter j=6≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(2, 6, 20, 4, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (2, 6, 20, 4, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0032","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"98dba41b0b978f539dee2b01df247a733d340c95cee5e278d3a7b9223f2580d0","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092032,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":7,\"period_values\":[7,-8,[9,2]],\"residue\":0}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[7, -8, 9/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=0 and subsequence parameter j=7≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(0, 7, 21, 7, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (0, 7, 21, 7, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0033","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"13b39ddca4650d65542d6992ce63db943dd23ce87dd26a49bebb2b969778b54d","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092033,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":8,\"period_values\":[8,-9,[10,2]],\"residue\":1}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[8, -9, 10/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=1 and subsequence parameter j=8≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(1, 8, 25, -9, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (1, 8, 25, -9, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0034","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"522887b448224950dae9ab4847bee52a201643cc3a6a1f18c19e7abd219a2e3e","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092034,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":9,\"period_values\":[9,-10,[11,2]],\"residue\":2}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[9, -10, 11/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=2 and subsequence parameter j=9≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(2, 9, 29, 11/2, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (2, 9, 29, 11/2, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0035","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e5d25d2debff8a972dd7fd06449bba809c11583e6af177ee881ecd4ccebbeb9a","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092035,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":10,\"period_values\":[10,-11,[12,2]],\"residue\":0}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[10, -11, 12/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=0 and subsequence parameter j=10≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(0, 10, 30, 10, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (0, 10, 30, 10, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0036","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"703e742d4ddcc858baa88ffa564ea851143a421aa4d9adcd6b68272fe60c3a7e","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092036,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":11,\"period_values\":[11,-12,[13,2]],\"residue\":1}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[11, -12, 13/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=1 and subsequence parameter j=11≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(1, 11, 34, -12, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (1, 11, 34, -12, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0037","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"93bfc9679532a8ad39e9a0fec33d86f5b0bcc50bf7e05b498748d745bf02461f","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092037,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":12,\"period_values\":[12,-13,[14,2]],\"residue\":2}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[12, -13, 14/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=2 and subsequence parameter j=12≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(2, 12, 38, 7, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (2, 12, 38, 7, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0038","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1870168da1b1b4cbf5b139c2bd114e4917bfd3b7a9a8c7f503379716a4aef528","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092038,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":13,\"period_values\":[13,-14,[15,2]],\"residue\":0}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[13, -14, 15/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=0 and subsequence parameter j=13≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(0, 13, 39, 13, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (0, 13, 39, 13, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0039","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d310a223c80dda881f547c7a725a6428497775684be02be7459650342d0983f2","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092039,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":14,\"period_values\":[14,-15,[16,2]],\"residue\":1}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[14, -15, 16/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=1 and subsequence parameter j=14≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(1, 14, 43, -15, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (1, 14, 43, -15, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0040","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a3ede3783d12dd3ed3d8d97f4c156d80a1c0c2d6bd7339e9ab0a0207d314d33","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092040,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":15,\"period_values\":[15,-16,[17,2]],\"residue\":2}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[15, -16, 17/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=2 and subsequence parameter j=15≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(2, 15, 47, 17/2, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (2, 15, 47, 17/2, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0041","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c8bf81c81879e25e3c4405b899042ced49d34fc7285c76c1f0739ad81e103d90","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092041,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":16,\"period_values\":[16,-17,[18,2]],\"residue\":0}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[16, -17, 18/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=0 and subsequence parameter j=16≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(0, 16, 48, 16, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (0, 16, 48, 16, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0042","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e0fe9cdcb2b82632b70664b66d6a34c638e983026fab18516f7e9c26a1415e9","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092042,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":17,\"period_values\":[17,-18,[19,2]],\"residue\":1}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[17, -18, 19/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=1 and subsequence parameter j=17≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(1, 17, 52, -18, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (1, 17, 52, -18, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0043","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f041287b2da4d586e39024a3b8f37a7da67de23d1aa01aa7c4c9a3fd53dd1a1b","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092043,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":18,\"period_values\":[18,-19,[20,2]],\"residue\":2}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[18, -19, 20/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=2 and subsequence parameter j=18≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(2, 18, 56, 10, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (2, 18, 56, 10, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0044","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"09785d393c292ca69075ccf47dc6e9c5b36332f709691b8aa8931ee5270758f1","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092044,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":19,\"period_values\":[19,-20,[21,2]],\"residue\":0}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[19, -20, 21/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=0 and subsequence parameter j=19≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(0, 19, 57, 19, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (0, 19, 57, 19, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0045","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d8d1f31ee621f615feb008af48db44043e016fea5ba5c07e6db8451f30853cbd","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092045,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":20,\"period_values\":[20,-21,[22,2]],\"residue\":1}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[20, -21, 22/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=1 and subsequence parameter j=20≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(1, 20, 61, -21, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (1, 20, 61, -21, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0046","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"38d2f303f34c7133cfcdcf143fd67c0159d84ddf81ba338e9cf8aafbc2437fed","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092046,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":21,\"period_values\":[21,-22,[23,2]],\"residue\":2}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[21, -22, 23/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=2 and subsequence parameter j=21≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(2, 21, 65, 23/2, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (2, 21, 65, 23/2, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0047","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"496e6be29e0db638ccdceb0abf63e8db448ad2db8ed22db80d74a41035384f84","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092047,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":22,\"period_values\":[22,-23,[24,2]],\"residue\":0}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[22, -23, 24/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=0 and subsequence parameter j=22≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(0, 22, 66, 22, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (0, 22, 66, 22, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0048","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"555adfc8d11b384ee3b09fa8cab5538de57ae0a65816202d2b399395966842d2","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092048,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":23,\"period_values\":[23,-24,[25,2]],\"residue\":1}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[23, -24, 25/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=1 and subsequence parameter j=23≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(1, 23, 70, -24, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (1, 23, 70, -24, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0049","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bbe824edf48e4b6ffe84fc31c2b702e220489788a45071c8dedc29a33faecbde","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092049,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":24,\"period_values\":[24,-25,[26,2]],\"residue\":2}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[24, -25, 26/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=2 and subsequence parameter j=24≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(2, 24, 74, 13, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (2, 24, 74, 13, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0050","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"565b4e2671422a1a0b8070de9a99fc9fb7872d41715af9e84150fecc9b2203e0","family_id":"qf-mod40-bw-periodic-subsequence","seed":202611092050,"semantic_template":"qf-mod40-bw-periodic-subsequence/{\"j\":25,\"period_values\":[25,-26,[27,2]],\"residue\":0}"},"fields":{"title":"A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence","stem":"Use the whole infinite real periodic sequence with nonempty period_values=[25, -26, 27/2], starting at index0 and repeating that full block forever. This is a declared infinite law, not an observed finite prefix. Use residue=0 and subsequence parameter j=25≥0. Give (residue, j, k_j, value, constant_subsequence_certificate, certificate).","answer":{"kind":"expression","value":"(0, 25, 75, 25, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)","explanation":"With period length t, k(j)=residue+t*j is strictly increasing on all nonnegative j, and every selected value is the same period entry. The displayed j only evaluates this infinite index law. The exact result, retaining the requested order, is (0, 25, 75, 25, Yes, k_j_residue_plus_period_times_j_strictly_increasing_constant_subsequence)."},"skills":["A Fixed Residue Gives an Infinite Strictly Increasing Constant Subsequence"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0051","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"579fb4627072f8db137763df4517e0313f7c3884443ca8879209249aa242906f","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092051,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":1,\"epsilon\":[1,10],\"limit\":[1,2],\"parity\":0,\"shift\":2}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=1/2, c=1>0, shift=2≥1, parity=0 and epsilon=1/10>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(0, k(j)=2j+0, 1/2, 1/10, 5, 7/12, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (0, k(j)=2j+0, 1/2, 1/10, 5, 7/12, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0052","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de1f130871404832454e8492e73723200f7b053567454bcad741aa380cc965d0","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092052,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":2,\"epsilon\":[1,11],\"limit\":[2,3],\"parity\":1,\"shift\":3}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=2/3, c=2>0, shift=3≥1, parity=1 and epsilon=1/11>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(1, k(j)=2j+1, 2/3, 1/11, 10, 7/12, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (1, k(j)=2j+1, 2/3, 1/11, 10, 7/12, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0053","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"252086248f046988758fffbbf9eb76e927fb27127b8df9ba68eb342bc16ebd98","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092053,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":3,\"epsilon\":[1,12],\"limit\":[3,4],\"parity\":0,\"shift\":4}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=3/4, c=3>0, shift=4≥1, parity=0 and epsilon=1/12>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(0, k(j)=2j+0, 3/4, 1/12, 17, 63/76, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (0, k(j)=2j+0, 3/4, 1/12, 17, 63/76, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0054","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65344dec300991fb38bf859e74abe51fdc76f2a87bd554100adc78f5b08a4abd","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092054,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":4,\"epsilon\":[1,13],\"limit\":[4,5],\"parity\":1,\"shift\":5}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=4/5, c=4>0, shift=5≥1, parity=1 and epsilon=1/13>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(1, k(j)=2j+1, 4/5, 1/13, 24, 98/135, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (1, k(j)=2j+1, 4/5, 1/13, 24, 98/135, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0055","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dd8daa607a61511ab6cc3fd4fa83678aa34c96d9a4ebbd7238cc04bf7f6bf343","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092055,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":5,\"epsilon\":[1,14],\"limit\":[5,6],\"parity\":0,\"shift\":6}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=5/6, c=5>0, shift=6≥1, parity=0 and epsilon=1/14>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(0, k(j)=2j+0, 5/6, 1/14, 33, 65/72, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (0, k(j)=2j+0, 5/6, 1/14, 33, 65/72, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0056","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa60b3d5691674b7257488e2a9ff15a6e51832edbf91f1df7e5a3506b454ff0a","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092056,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":6,\"epsilon\":[1,15],\"limit\":[6,7],\"parity\":1,\"shift\":7}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=6/7, c=6>0, shift=7≥1, parity=1 and epsilon=1/15>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(1, k(j)=2j+1, 6/7, 1/15, 42, 255/322, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (1, k(j)=2j+1, 6/7, 1/15, 42, 255/322, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0057","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5e342e71921bb27af607c5be52c89ce497f92513d832f9ed77859c7835c2f9a","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092057,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":7,\"epsilon\":[1,16],\"limit\":[7,8],\"parity\":0,\"shift\":8}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=7/8, c=7>0, shift=8≥1, parity=0 and epsilon=1/16>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(0, k(j)=2j+0, 7/8, 1/16, 53, 427/456, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (0, k(j)=2j+0, 7/8, 1/16, 53, 427/456, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0058","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d173d22f7f612eefc5160d00748e1636b18322a76da3caa2b7d23432a6bbbc02","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092058,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":8,\"epsilon\":[1,17],\"limit\":[8,9],\"parity\":1,\"shift\":9}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=8/9, c=8>0, shift=9≥1, parity=1 and epsilon=1/17>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(1, k(j)=2j+1, 8/9, 1/17, 64, 172/207, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (1, k(j)=2j+1, 8/9, 1/17, 64, 172/207, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0059","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"35d97923a16983d593b51b2559fde742a2139dffb853b59ef4826981fad8f451","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092059,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":9,\"epsilon\":[1,18],\"limit\":[9,10],\"parity\":0,\"shift\":10}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=9/10, c=9>0, shift=10≥1, parity=0 and epsilon=1/18>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(0, k(j)=2j+0, 9/10, 1/18, 77, 783/820, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (0, k(j)=2j+0, 9/10, 1/18, 77, 783/820, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0060","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a7d459d4e1ffe0784fe78a611a3889fcfc38f68867b60a8ced5dcdd7459c793","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092060,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":10,\"epsilon\":[1,19],\"limit\":[10,11],\"parity\":1,\"shift\":11}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=10/11, c=10>0, shift=11≥1, parity=1 and epsilon=1/19>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(1, k(j)=2j+1, 10/11, 1/19, 90, 905/1056, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (1, k(j)=2j+1, 10/11, 1/19, 90, 905/1056, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0061","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b2aa74b1877b1b4127637d58cb5023378985cc021d828c0a26f58d7f080499e","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092061,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":11,\"epsilon\":[1,20],\"limit\":[11,12],\"parity\":0,\"shift\":12}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=11/12, c=11>0, shift=12≥1, parity=0 and epsilon=1/20>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(0, k(j)=2j+0, 11/12, 1/20, 105, 143/148, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (0, k(j)=2j+0, 11/12, 1/20, 105, 143/148, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0062","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e0a3297160b37a130d6127db031e126b851112370747e81d6d47351ce7a4f652","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092062,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":12,\"epsilon\":[1,21],\"limit\":[12,13],\"parity\":1,\"shift\":13}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=12/13, c=12>0, shift=13≥1, parity=1 and epsilon=1/21>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(1, k(j)=2j+1, 12/13, 1/21, 120, 1446/1651, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (1, k(j)=2j+1, 12/13, 1/21, 120, 1446/1651, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0063","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5aa7c9e1ead6ab4bea4156e7fac1fe0312ea2df56068f22f6572a8a6ce8572fc","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092063,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":13,\"epsilon\":[1,22],\"limit\":[13,14],\"parity\":0,\"shift\":14}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=13/14, c=13>0, shift=14≥1, parity=0 and epsilon=1/22>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(0, k(j)=2j+0, 13/14, 1/22, 137, 1963/2016, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (0, k(j)=2j+0, 13/14, 1/22, 137, 1963/2016, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0064","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"088338f9fccf8b52bc8e0c06bc372a2cbe8ceddaea5af560afcd26c4b2bd8dd2","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092064,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":14,\"epsilon\":[1,23],\"limit\":[14,15],\"parity\":1,\"shift\":15}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=14/15, c=14>0, shift=15≥1, parity=1 and epsilon=1/23>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(1, k(j)=2j+1, 14/15, 1/23, 154, 721/810, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (1, k(j)=2j+1, 14/15, 1/23, 154, 721/810, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0065","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f9441d5dc84682a7ddd05d996b83870131d23f2239b66aec72d131c77f76dc5e","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092065,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":15,\"epsilon\":[1,24],\"limit\":[15,16],\"parity\":0,\"shift\":16}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=15/16, c=15>0, shift=16≥1, parity=0 and epsilon=1/24>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(0, k(j)=2j+0, 15/16, 1/24, 173, 2835/2896, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (0, k(j)=2j+0, 15/16, 1/24, 173, 2835/2896, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0066","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"92dd4c8ab1e5d0cfd645ceb5afa9d2a575471c4cd505f0e164535976816282c0","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092066,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":16,\"epsilon\":[1,25],\"limit\":[16,17],\"parity\":1,\"shift\":17}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=16/17, c=16>0, shift=17≥1, parity=1 and epsilon=1/25>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(1, k(j)=2j+1, 16/17, 1/25, 192, 3080/3417, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (1, k(j)=2j+1, 16/17, 1/25, 192, 3080/3417, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0067","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"09a71787a6352b9f820b656da4eb233c01492b511d546565d2595a6b2c014f05","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092067,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":17,\"epsilon\":[1,26],\"limit\":[17,18],\"parity\":0,\"shift\":18}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=17/18, c=17>0, shift=18≥1, parity=0 and epsilon=1/26>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(0, k(j)=2j+0, 17/18, 1/26, 213, 1309/1332, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (0, k(j)=2j+0, 17/18, 1/26, 213, 1309/1332, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0068","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7f3e523e795d8e27f51683a02db74637cb348fdc3e60c6ab0cc2953ef2eca660","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092068,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":18,\"epsilon\":[1,27],\"limit\":[18,19],\"parity\":1,\"shift\":19}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=18/19, c=18>0, shift=19≥1, parity=1 and epsilon=1/27>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(1, k(j)=2j+1, 18/19, 1/27, 234, 4221/4636, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (1, k(j)=2j+1, 18/19, 1/27, 234, 4221/4636, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0069","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d2834896ae65e0b31b3384d35723df4070166c5a1b469754ae51d1ee74a38bd","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092069,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":19,\"epsilon\":[1,28],\"limit\":[19,20],\"parity\":0,\"shift\":20}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=19/20, c=19>0, shift=20≥1, parity=0 and epsilon=1/28>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(0, k(j)=2j+0, 19/20, 1/28, 257, 5263/5340, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (0, k(j)=2j+0, 19/20, 1/28, 257, 5263/5340, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0070","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f701c2ad55ce6136b01eebc3976f06c2782b493457b1b92521479ae2dc22748d","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092070,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":20,\"epsilon\":[1,29],\"limit\":[20,21],\"parity\":1,\"shift\":21}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=20/21, c=20>0, shift=21≥1, parity=1 and epsilon=1/29>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(1, k(j)=2j+1, 20/21, 1/29, 280, 1870/2037, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (1, k(j)=2j+1, 20/21, 1/29, 280, 1870/2037, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0071","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d023744bb96b6e38b7f434c4b10686cc8d8f47fd44ee158cc472e1e3fa1cd71a","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092071,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":21,\"epsilon\":[1,30],\"limit\":[21,22],\"parity\":0,\"shift\":22}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=21/22, c=21>0, shift=22≥1, parity=0 and epsilon=1/30>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(0, k(j)=2j+0, 21/22, 1/30, 305, 6867/6952, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (0, k(j)=2j+0, 21/22, 1/30, 305, 6867/6952, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0072","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1431378588183160441ec488cd9ed8592638b0b888c4f055c4c9818292ec33ba","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092072,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":22,\"epsilon\":[1,31],\"limit\":[22,23],\"parity\":1,\"shift\":23}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=22/23, c=22>0, shift=23≥1, parity=1 and epsilon=1/31>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(1, k(j)=2j+1, 22/23, 1/31, 330, 7271/7866, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (1, k(j)=2j+1, 22/23, 1/31, 330, 7271/7866, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0073","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"690a995171bd1988b69a88f1c4f5b033098ddad1934c2caa47e3752c7ea52fb2","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092073,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":23,\"epsilon\":[1,32],\"limit\":[23,24],\"parity\":0,\"shift\":24}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=23/24, c=23>0, shift=24≥1, parity=0 and epsilon=1/32>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(0, k(j)=2j+0, 23/24, 1/32, 357, 2921/2952, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (0, k(j)=2j+0, 23/24, 1/32, 357, 2921/2952, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0074","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"169e2d25f8bc8e03788b4b7566d0329b694d14ab9df1c4ac9e290b7c3ff5a36f","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092074,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":24,\"epsilon\":[1,33],\"limit\":[24,25],\"parity\":1,\"shift\":25}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=24/25, c=24>0, shift=25≥1, parity=1 and epsilon=1/33>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(1, k(j)=2j+1, 24/25, 1/33, 384, 9228/9925, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (1, k(j)=2j+1, 24/25, 1/33, 384, 9228/9925, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0075","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b9f6238908c2f19a6b1b7339b722824d2185be5eec93d7c988284174b602d454","family_id":"qf-mod40-bw-oscillatory-tail","seed":202611092075,"semantic_template":"qf-mod40-bw-oscillatory-tail/{\"coefficient\":25,\"epsilon\":[1,34],\"limit\":[25,26],\"parity\":0,\"shift\":26}"},"fields":{"title":"Parity Subsequences Need a Strict All-Later Epsilon Witness","stem":"Use a_n=L+(−1)^n*c/(n+shift) for every n≥0, L=25/26, c=25>0, shift=26≥1, parity=0 and epsilon=1/34>0. The infinite index rule is k(j)=2j+parity for every j≥0. Give (parity, index_rule, limit, epsilon, J, value_at_k_J, strict_tail_certificate, certificate).","answer":{"kind":"expression","value":"(0, k(j)=2j+0, 25/26, 1/34, 413, 10975/11076, Yes, forall_j_ge_J_symbolic_subsequence_tail)","explanation":"Find the least nonnegative J such that c/(2J+parity+shift)<epsilon. Equality still fails the strict inequality. Report a at k(J), not a_J; increasing denominators establish the same bound for every j≥J. Both parity subsequences converge to L. The exact result, retaining the requested order, is (0, k(j)=2j+0, 25/26, 1/34, 413, 10975/11076, Yes, forall_j_ge_J_symbolic_subsequence_tail)."},"skills":["Parity Subsequences Need a Strict All-Later Epsilon Witness"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0076","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d8e77b40e3af0b1af090b5ce2681b6da402ac6b5a76f6e136dad414717c3456a","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092076,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[0,1,0,2]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[0, 1, 0, 2] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([0, 1, 0, 2], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([0, 1, 0, 2], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0077","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5fdefd4e6947194398c54c630d389cdc0c100a622827c927fa9e91d65a5a314","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092077,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[1,2,1,3]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[1, 2, 1, 3] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([1, 2, 1, 3], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([1, 2, 1, 3], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0078","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b1cac6f35c3212410f9211024889df53a8ad472dd3f4e487f561dbb4f260cb2b","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092078,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[2,3,2,4]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[2, 3, 2, 4] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([2, 3, 2, 4], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([2, 3, 2, 4], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0079","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"541256cb8ee00a6c3adefe05f77a23da9200422374253332aed51a7c2b91a314","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092079,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[3,4,3,5]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[3, 4, 3, 5] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([3, 4, 3, 5], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([3, 4, 3, 5], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0080","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0dabdc017f71cfd6b32b490797781a7d6192d5d2102d22b39a21279e9ce7fea4","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092080,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[4,5,4,6]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[4, 5, 4, 6] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([4, 5, 4, 6], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([4, 5, 4, 6], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0081","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f361574ddea4085b42222e78fa267183b1d95f381bdaa214b2661803741b940","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092081,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[5,6,5,7]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[5, 6, 5, 7] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([5, 6, 5, 7], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([5, 6, 5, 7], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0082","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"33bcc6f0a88390cdc6356e181ca1f1c178031bbc4eb3b8c73345ac68f1f9c5b0","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092082,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[6,7,6,8]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[6, 7, 6, 8] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([6, 7, 6, 8], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([6, 7, 6, 8], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0083","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"856c5c3f15a8db67e81a4d159313b32e3a833189fa61e1d0dfd319e90e4c54d8","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092083,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[7,8,7,9]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[7, 8, 7, 9] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([7, 8, 7, 9], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([7, 8, 7, 9], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0084","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7d7d63335db55abb8c87acf720e1a6c8bc091158e5be4a5a36fdffae3f635d59","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092084,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[8,9,8,10]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[8, 9, 8, 10] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([8, 9, 8, 10], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([8, 9, 8, 10], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0085","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3dd79cc12c284750919bd8110c24f69b4a2f1fc97cdaef2ea77d823c91c73944","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092085,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[9,10,9,11]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[9, 10, 9, 11] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([9, 10, 9, 11], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([9, 10, 9, 11], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0086","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3791c3818e27b2f2dda9c15db4fe152a8f4b91b82665e62295eeb04ac55ebefd","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092086,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[10,11,10,12]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[10, 11, 10, 12] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([10, 11, 10, 12], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([10, 11, 10, 12], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0087","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4564c9fb2c76e407f291ca102593cd3f6aa40d8b42876c837cf7b02583b338f2","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092087,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[11,12,11,13]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[11, 12, 11, 13] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([11, 12, 11, 13], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([11, 12, 11, 13], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0088","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"99eb61f661d575622aa9294c827e24d0a2131ec85275c60a61f26bb0f91f71ac","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092088,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[12,13,12,14]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[12, 13, 12, 14] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([12, 13, 12, 14], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([12, 13, 12, 14], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0089","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ec96af4a17f221dbafcfece4cfb58a6b84b96f75281127207ca52e792a156da","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092089,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[13,14,13,15]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[13, 14, 13, 15] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([13, 14, 13, 15], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([13, 14, 13, 15], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0090","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b068b563c8dd899abec8f4d54e36f1601a0db8e9e07a1fcaaab8ee57e97837aa","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092090,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[14,15,14,16]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[14, 15, 14, 16] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([14, 15, 14, 16], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([14, 15, 14, 16], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0091","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b4eca453628b433e1675b417b3e218cce0a0eb0dfe1fc0de3d75c410a1daaae","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092091,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[15,16,15,17]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[15, 16, 15, 17] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([15, 16, 15, 17], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([15, 16, 15, 17], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0092","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"48dafaf5c92a57161caf2c13d79d7fad17dfa28c2f06856b1e79a03e54165c98","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092092,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[16,17,16,18]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[16, 17, 16, 18] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([16, 17, 16, 18], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([16, 17, 16, 18], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0093","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b4269fc40b3e5fdc237db87aa3a1f83238fa288f7397af86999127a92207c3f","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092093,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[17,18,17,19]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[17, 18, 17, 19] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([17, 18, 17, 19], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([17, 18, 17, 19], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0094","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1f5fb06bd3ea6a0a2d4dd0f5b8ef6e26960c2d475349f0d5fbd0023bc17cc9c7","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092094,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[18,19,18,20]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[18, 19, 18, 20] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([18, 19, 18, 20], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([18, 19, 18, 20], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0095","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b0da5fb03b4f6c110f76b243ef067d6acb36ab84182d1dfc96c619b5350397af","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092095,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[19,20,19,21]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[19, 20, 19, 21] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([19, 20, 19, 21], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([19, 20, 19, 21], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0096","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef7bc76cb9dfacb544493f7bdac884586753ca4e55aa12fcf6d195d5f87ab9eb","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092096,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[20,21,20,22]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[20, 21, 20, 22] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([20, 21, 20, 22], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([20, 21, 20, 22], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0097","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3411d052bae973dde63ce9b67fd362e5e80f44e57f253b58bca0f57a13d5885","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092097,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[21,22,21,23]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[21, 22, 21, 23] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([21, 22, 21, 23], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([21, 22, 21, 23], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0098","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2228ec92cf4e300c6cff67b4d5f4009a9256d758386269b40fb6214de64b263d","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092098,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[22,23,22,24]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[22, 23, 22, 24] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([22, 23, 22, 24], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([22, 23, 22, 24], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0099","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e5f0b7d28f363f188c778f4fc03fd2f69c4343334d27b2f3ed6e3562f7b73c39","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092099,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[23,24,23,25]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[23, 24, 23, 25] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([23, 24, 23, 25], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([23, 24, 23, 25], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-0100","batch_id":"batch-20261014-advance-20261109-r02-bolzano-weierstrass-theorem-real-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d161d605cb9b69bd7bfca72884601a7a39db1776aa55054b8523bcb0af911ecd","family_id":"qf-mod40-bw-finite-prefix-boundary","seed":202611092100,"semantic_template":"qf-mod40-bw-finite-prefix-boundary/{\"values\":[24,25,24,26]}"},"fields":{"title":"A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate","stem":"Only the finite prefix values=[24, 25, 24, 26] is given; no infinite continuation is specified. Give (prefix, whole_sequence_boundedness_certified, infinite_subsequence_certified, convergence_certified, boundary).","answer":{"kind":"expression","value":"([24, 25, 24, 26], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)","explanation":"The three No flags withhold whole-sequence boundedness, an infinite subsequence and convergence certificates. They do not say that every continuation is unbounded or divergent. Bolzano-Weierstrass requires a whole bounded real sequence, an infinite strictly increasing index map and a convergent subsequence; it does not require the original sequence to converge or its subsequential limit to be unique. The exact result, retaining the requested order, is ([24, 25, 24, 26], No, No, No, finite_prefix_cannot_certify_whole_sequence_boundedness_or_infinite_convergent_subsequence)."},"skills":["A Finite Prefix Supplies No Whole-Sequence or Infinite-Subsequence Certificate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}