{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261029-r01-finite-integral-domain-0001","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a224560193cff5e69da0056e5b08dd94192d1a284521c48fbd02f8a5d665fab","family_id":"qf-mod29-integraldomain-qualification","seed":202610291001,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"multiplication\":[[0,0,0,0],[0,1,2,3],[0,2,0,2],[0,3,2,1]],\"n\":4}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(4)={0,…,3}, with addition=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]] and multiplication=[[0, 0, 0, 0], [0, 1, 2, 3], [0, 2, 0, 2], [0, 3, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(4, 0, 1, [[2, 2]], No, commutative_ring_with_zero_divisors)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (4, 0, 1, [[2, 2]], No, commutative_ring_with_zero_divisors)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0002","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d048ad9e32c6e798ed831aba0503b087cf12e07f385762f32ad920c6d13fded7","family_id":"qf-mod29-integraldomain-qualification","seed":202610291002,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 1, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 1, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0003","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dc773369996bb2d50aa96534c122fe1f1a140b9729d22fa85ca819f75d9d809b","family_id":"qf-mod29-integraldomain-qualification","seed":202610291003,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,4,1,3],[0,3,1,4,2],[0,4,3,2,1]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 1, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 1, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0004","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e00288db0756216c07423a4e4ea0cbbf56541be2bf262a7314c9993a5958780","family_id":"qf-mod29-integraldomain-qualification","seed":202610291004,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,1,4,3],[0,3,4,2,1],[0,4,3,1,2]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 1, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 1, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0005","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"22357aec774a524dc84f0ab7b1589d393b4afd6256a75ffa555fc9155f667fb8","family_id":"qf-mod29-integraldomain-qualification","seed":202610291005,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 1, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 1, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0006","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"04a0c0eb2d94d874613fe42d4db8f4ca79de31f26be98b50688585cd1b20f6f4","family_id":"qf-mod29-integraldomain-qualification","seed":202610291006,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3],[1,2,3,0],[2,3,0,1],[3,0,1,2]],\"multiplication\":[[0,0,0,0],[0,3,2,1],[0,2,0,2],[0,1,2,3]],\"n\":4}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(4)={0,…,3}, with addition=[[0, 1, 2, 3], [1, 2, 3, 0], [2, 3, 0, 1], [3, 0, 1, 2]] and multiplication=[[0, 0, 0, 0], [0, 3, 2, 1], [0, 2, 0, 2], [0, 1, 2, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(4, 0, 3, [[2, 2]], No, commutative_ring_with_zero_divisors)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (4, 0, 3, [[2, 2]], No, commutative_ring_with_zero_divisors)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0007","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c855f5e3881248840fd8a7089433ed3312711b1d66dddd7793f097166cd569f","family_id":"qf-mod29-integraldomain-qualification","seed":202610291007,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 2, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 2, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0008","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ee8a00d3002bdb8383e6ab7abf8cd12a275fa3830040634e13cca9faa5d2cbaa","family_id":"qf-mod29-integraldomain-qualification","seed":202610291008,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,3,1,4,2],[0,1,2,3,4],[0,4,3,2,1],[0,2,4,1,3]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 2, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 2, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0009","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b91088107a320f2f0ce50963509a0707bda6054a5593d25c772f12bc6f133075","family_id":"qf-mod29-integraldomain-qualification","seed":202610291009,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 2, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 2, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0010","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b9fd136394d41444b8019a257149b6fa98bac0f0c9250faa615380dd6f8b3729","family_id":"qf-mod29-integraldomain-qualification","seed":202610291010,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 2, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 2, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0011","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b0a01f0396a227f3f23b3151dbf7f8cc3d3597f33202bccb47570983ce8b8033","family_id":"qf-mod29-integraldomain-qualification","seed":202610291011,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[1,0,3,2],[0,1,2,3],[3,2,0,1],[2,3,1,0]],\"multiplication\":[[1,1,0,0],[1,1,1,1],[0,1,3,2],[0,1,2,3]],\"n\":4}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(4)={0,…,3}, with addition=[[1, 0, 3, 2], [0, 1, 2, 3], [3, 2, 0, 1], [2, 3, 1, 0]] and multiplication=[[1, 1, 0, 0], [1, 1, 1, 1], [0, 1, 3, 2], [0, 1, 2, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(4, 1, 3, [[0, 0]], No, commutative_ring_with_zero_divisors)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (4, 1, 3, [[0, 0]], No, commutative_ring_with_zero_divisors)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0012","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"66bb77da9f1300994a7c212a0a1a5cac6b7457489da8f6c05d5380a55f80a6bf","family_id":"qf-mod29-integraldomain-qualification","seed":202610291012,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 2, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 2, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0013","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"29c0acb9c60a1832b05bb6f7379349f7a1ba5894395c88826b39394ab0b6d871","family_id":"qf-mod29-integraldomain-qualification","seed":202610291013,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 3, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 3, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0014","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f3ec44e7f230c944be277a6a7d256b6bfb2af8566b32a9910da9aba99c048d89","family_id":"qf-mod29-integraldomain-qualification","seed":202610291014,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 3, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 3, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0015","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a9eaf049521dc604db00221524c36e262366e07df6a9efac09061a1ea1566d75","family_id":"qf-mod29-integraldomain-qualification","seed":202610291015,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 3, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 3, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0016","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"688d9b9f2ae566812134e18e30e6516771c44179122b1f2c8303c04dd03559f8","family_id":"qf-mod29-integraldomain-qualification","seed":202610291016,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[3,2,0,1],[2,3,1,0],[0,1,2,3],[1,0,3,2]],\"multiplication\":[[1,0,2,3],[0,1,2,3],[2,2,2,2],[3,3,2,2]],\"n\":4}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(4)={0,…,3}, with addition=[[3, 2, 0, 1], [2, 3, 1, 0], [0, 1, 2, 3], [1, 0, 3, 2]] and multiplication=[[1, 0, 2, 3], [0, 1, 2, 3], [2, 2, 2, 2], [3, 3, 2, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(4, 2, 1, [[3, 3]], No, commutative_ring_with_zero_divisors)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (4, 2, 1, [[3, 3]], No, commutative_ring_with_zero_divisors)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0017","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3b9912072b5a8b461916e828ffc1d0df34f38b970f8a613a28b23cb59ad6ff58","family_id":"qf-mod29-integraldomain-qualification","seed":202610291017,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 3, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 3, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0018","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3ac0583ba9ca88eef028da130a40bc1069df69ca7512bccc4e3e005c276c8db3","family_id":"qf-mod29-integraldomain-qualification","seed":202610291018,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 3, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 3, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0019","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91a874892f6ff749726d317f4f105f1e3445896af97481af6a0d89764d049d20","family_id":"qf-mod29-integraldomain-qualification","seed":202610291019,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,3,4,2,1],[0,4,3,1,2],[0,2,1,4,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 4, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 4, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0020","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"86dcd64af41f8f7b13952a0130d928e61e3d1e62b46f870db6d250f3c3f01706","family_id":"qf-mod29-integraldomain-qualification","seed":202610291020,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 4, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 4, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0021","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3cbc33c9f8de7626a13e0e389743100571b54b42e14e9ed57e646b372aeced93","family_id":"qf-mod29-integraldomain-qualification","seed":202610291021,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[3,2,1,0],[2,0,3,1],[1,3,0,2],[0,1,2,3]],\"multiplication\":[[3,0,0,3],[0,1,2,3],[0,2,1,3],[3,3,3,3]],\"n\":4}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(4)={0,…,3}, with addition=[[3, 2, 1, 0], [2, 0, 3, 1], [1, 3, 0, 2], [0, 1, 2, 3]] and multiplication=[[3, 0, 0, 3], [0, 1, 2, 3], [0, 2, 1, 3], [3, 3, 3, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(4, 3, 1, [[0, 0]], No, commutative_ring_with_zero_divisors)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (4, 3, 1, [[0, 0]], No, commutative_ring_with_zero_divisors)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0022","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a27656daf0ce8aba929bfbfdd707155812e77e4cde49c92b8d95a9fde747e127","family_id":"qf-mod29-integraldomain-qualification","seed":202610291022,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 4, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 4, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0023","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"932bc66376534272f59d0444cabd798dcbed093996b4e44ccc989a2b089d6dc7","family_id":"qf-mod29-integraldomain-qualification","seed":202610291023,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 4, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 4, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0024","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36f8f0ff2d708c8c27861f2e86d94a29c738f14aac72b89ed4b95088b1e31be3","family_id":"qf-mod29-integraldomain-qualification","seed":202610291024,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 0, 4, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 0, 4, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0025","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"27a7f28441bb039010b5d66ad28ecd763d361afb9d86c07853028b601c914a93","family_id":"qf-mod29-integraldomain-qualification","seed":202610291025,"semantic_template":"qf-mod29-integraldomain-qualification/{\"addition\":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]],\"multiplication\":[[0,1,2,3,4],[1,1,1,1,1],[2,1,4,0,3],[3,1,0,4,2],[4,1,3,2,0]],\"n\":5}"},"fields":{"title":"An Integral Domain Requires Every Nonzero Pair to Avoid Zero","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] and multiplication=[[0, 1, 2, 3, 4], [1, 1, 1, 1, 1], [2, 1, 4, 0, 3], [3, 1, 0, 4, 2], [4, 1, 3, 2, 0]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (n,zero,one,zero_divisor_pairs,domain?,classification). List all ordered nonzero [a,b] with a*b=zero in increasing a then b order. Classification is finite_integral_domain or commutative_ring_with_zero_divisors.","answer":{"kind":"expression","value":"(5, 1, 0, [], Yes, finite_integral_domain)","explanation":"Use the actual additive zero, not its presumed label. A complete absence of such pairs qualifies this finite integral domain. Invertibility is a subsequent theorem using finiteness, not an extra definition clause. The exact result, retaining the requested order, is (5, 1, 0, [], Yes, finite_integral_domain)."},"skills":["An Integral Domain Requires Every Nonzero Pair to Avoid Zero"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0026","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ec270db0baca20e58c716db27c3527a716910fb3a2cebdf9749bcd42022013fb","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291026,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,4,1,3],[0,3,1,4,2],[0,4,3,2,1]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0027","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"df973bc8d8af67405159f5617900d5b5ed6ff18d3a34eb227a7027eb6e450be6","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291027,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0028","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f25df5c49f8775ecf3daeba8df9f440e53f554e82af6cd7c5c08b11484158bf5","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291028,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,4,1,3],[0,3,1,4,2],[0,4,3,2,1]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0029","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a9c3d371c3f67a35df2b91b81911f60788310be64200ee937b975dbd6331583e","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291029,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,1,4,3],[0,3,4,2,1],[0,4,3,1,2]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0030","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ad3b7810c541128deba05845f958dc624dbdcc99dde8fdc90ed937cb7cfdcd4a","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291030,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0031","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"64473522a4a7d95a71f8d621ad83375c71678e12ac88c71678702f3b8f774f87","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291031,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,1,4,3],[0,3,4,2,1],[0,4,3,1,2]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0032","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d97b0021b3d03e88b9e6f4db390e7e9e2962592b1220eb6f487c7fef982afd9","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291032,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0033","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d1096e3dbb22ecceae21da3604fb0b2bc687b926f23736217ceeb8db80e340a1","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291033,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,3,1,4,2],[0,1,2,3,4],[0,4,3,2,1],[0,2,4,1,3]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0034","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf090ddb857e19812e18514fa08f0c54191149cc7fe75a07439a1b95ecc52a33","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291034,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0035","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91f114e72f6fde7cbcd30c579751eca05fc0893d518aa9d491e48dc172c335ac","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291035,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0036","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f38179fe9bf9c3982606f70cf8cc31a9853132ec2413585034d4c2befbb5ef84","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291036,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"multiplication\":[[0,0,0,0,0],[0,3,1,4,2],[0,1,2,3,4],[0,4,3,2,1],[0,2,4,1,3]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0037","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"09184c645ca654f3d1aabec2dac99e63e9d5aa62f097df4c3b9442a0758351d7","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291037,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0038","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c2475eb6b9a58784fb09d3ec3065db679980365ecc1e7cacc33a570a50acfdbb","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291038,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0039","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d7ab2f399ed43d8ab42cd5b153f5e18f7a126b1cbc6e8bfbfdff3a1ed0ff00f9","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291039,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0040","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"98b673cb45b9bf34afdb6a56f2206a89d780fb9c65d3f760e04cb5bcd9331314","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291040,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0041","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2360e8617a2f222456c22c4fcafe9d8bf88fcbe58d9d38c480a122cd44ea12f3","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291041,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"multiplication\":[[0,0,0,0,0],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0042","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b7998b794b265d9d92c1b7023a800438e6ff38f84b98847d028d5b419f69dc64","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291042,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0043","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4a0fc6031be955259315b9f7dd72ac610ad0f97e122379c30981d3f12316f0ca","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291043,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0044","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"130713d232c019ccb9766b9fb0a99506d51a16433b270fe1addd60c10a5d7f0e","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291044,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,3,4,2,1],[0,4,3,1,2],[0,2,1,4,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0045","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d02738b48700e411adc9f405c440a926d2cc4da724dfe3426c374e316479299e","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291045,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0046","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b5e5aa28c5862ab2ab1e821d98ec0384f3e64eff1b5ced18184e86a6788626b","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291046,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"multiplication\":[[0,0,0,0,0],[0,3,4,2,1],[0,4,3,1,2],[0,2,1,4,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0047","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b692e5f8e0c65ea75a08093b81b0d3da3139ae36249f3e2b7670467f285e4458","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291047,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0048","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e7cdadb512521fe52376aad7abf95a54c6ff9fc81fb21470d90ab8f8df0cd811","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291048,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0049","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"27dc96701b5754baafa7efec74a78a6f66f0da0d7814a4178b26ce5e74e468ad","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291049,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0050","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"35104e40445d33ff6dc3d17eca2f2734dfcaf08cbae9df11a9357bb3e1624038","family_id":"qf-mod29-integraldomain-zero-divisor-audit","seed":202610291050,"semantic_template":"qf-mod29-integraldomain-zero-divisor-audit/{\"addition\":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]],\"multiplication\":[[0,1,2,3,4],[1,1,1,1,1],[2,1,4,0,3],[3,1,0,4,2],[4,1,3,2,0]],\"n\":5}"},"fields":{"title":"Audit All Nonzero Zero-Divisor Pairs","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] and multiplication=[[0, 1, 2, 3, 4], [1, 1, 1, 1, 1], [2, 1, 4, 0, 3], [3, 1, 0, 4, 2], [4, 1, 3, 2, 0]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Give (zero_divisor_pairs,no_nonzero_zero_divisors?), with pairs in increasing first- then second-label order.","answer":{"kind":"expression","value":"([], Yes)","explanation":"Search every ordered pair of nonzero elements in the complete multiplication table. A single pair yielding zero refutes the no-zero-divisor obligation; sampled successes do not discharge it. The exact result, retaining the requested order, is ([], Yes)."},"skills":["Audit All Nonzero Zero-Divisor Pairs"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0051","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1d655efe41c87da3a86d1ec32e0d7a3d231b095eeafaa2cd69e2e6fe84a6233d","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291051,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":1,\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"b\":1,\"c\":2,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,4,1,3],[0,3,1,4,2],[0,4,3,2,1]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=1, with b=1, c=2. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(1, 1, 2, 1, 2, No, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (1, 1, 2, 1, 2, No, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0052","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bcf929928b182bd0375a308a47bd33eb88c0d7c98b9fb2585b9e97316e82e2b9","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291052,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":2,\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"b\":2,\"c\":2,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=2, with b=2, c=2. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(2, 2, 2, 3, 3, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (2, 2, 2, 3, 3, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0053","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e8fa24e7333e842484175c7cb1e56a778cdeb287f7a8c98c037289f747b42ac","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291053,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":2,\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"b\":2,\"c\":2,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,4,1,3],[0,3,1,4,2],[0,4,3,2,1]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=2, with b=2, c=2. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(2, 2, 2, 4, 4, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (2, 2, 2, 4, 4, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0054","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"84cd1cf84a5d317167369d398c354752840d9ad91685f1c9ca262a99e7a7b9e3","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291054,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":2,\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"b\":2,\"c\":2,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,1,4,3],[0,3,4,2,1],[0,4,3,1,2]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=2, with b=2, c=2. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(2, 2, 2, 1, 1, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (2, 2, 2, 1, 1, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0055","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"011431967e7684e72aed7417119a563868ec5007356a4bebec14d980121fe8ed","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291055,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":1,\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"b\":0,\"c\":1,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=1, with b=0, c=1. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(1, 0, 1, 0, 1, No, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (1, 0, 1, 0, 1, No, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0056","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c920a4e112b2c09717b9bf52bfea0e14586aae3d61db8eeff9cab201d224c6a1","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291056,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":4,\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"b\":1,\"c\":1,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,1,4,3],[0,3,4,2,1],[0,4,3,1,2]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=4, with b=1, c=1. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(4, 1, 1, 4, 4, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (4, 1, 1, 4, 4, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0057","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cc43adf7de2a516c32b1733b97ec82851518f54740e6a5914c2e8473d996c75f","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291057,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":3,\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"b\":1,\"c\":1,\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=3, with b=1, c=1. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(3, 1, 1, 2, 2, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (3, 1, 1, 2, 2, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0058","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"156dea458277231d4ffc53be6e387cec65d865b21bc5738ae2bf8e64babcbe55","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291058,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":3,\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"b\":4,\"c\":4,\"multiplication\":[[0,0,0,0,0],[0,3,1,4,2],[0,1,2,3,4],[0,4,3,2,1],[0,2,4,1,3]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=3, with b=4, c=4. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(3, 4, 4, 1, 1, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (3, 4, 4, 1, 1, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0059","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7764cbfaabb7bfff9a81904a52dab898d1140afd63d03b9d46d45d09387dec34","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291059,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":2,\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"b\":4,\"c\":0,\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=2, with b=4, c=0. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(2, 4, 0, 4, 0, No, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (2, 4, 0, 4, 0, No, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0060","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"26052543549f68163c63f08cda985e5a111ebdbfa751a4199e8b6f54ce96c510","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291060,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":3,\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"b\":0,\"c\":0,\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=3, with b=0, c=0. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(3, 0, 0, 0, 0, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (3, 0, 0, 0, 0, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0061","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"66718976b83c1f5475134b5933ecc1d4575badae77ec74dbe9cb5ba8ef9c1649","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291061,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":1,\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"b\":2,\"c\":2,\"multiplication\":[[0,0,0,0,0],[0,3,1,4,2],[0,1,2,3,4],[0,4,3,2,1],[0,2,4,1,3]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=1, with b=2, c=2. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(1, 2, 2, 1, 1, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (1, 2, 2, 1, 1, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0062","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b08f198898443e156593bd660076ccbca5b8435a48236973ceff3d3f9351845","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291062,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":1,\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"b\":4,\"c\":4,\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=1, with b=4, c=4. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(1, 4, 4, 3, 3, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (1, 4, 4, 3, 3, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0063","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b04a32afc1e2cb23af76229ef4a2b405207031e201112df487fd9cd5395d0f19","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291063,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":3,\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"b\":2,\"c\":4,\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=3, with b=2, c=4. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(3, 2, 4, 2, 4, No, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (3, 2, 4, 2, 4, No, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0064","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2384c47e0fe2ce8a53092ea3063a646cf1f9c622a7d1efd6f03c8e325fbf1555","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291064,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":1,\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"b\":2,\"c\":2,\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=1, with b=2, c=2. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(1, 2, 2, 4, 4, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (1, 2, 2, 4, 4, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0065","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"edb7d446c4d0ea581f8eb61fbae7554b1fa1b448b72cc46717f225c1aeb80fbc","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291065,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":1,\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"b\":0,\"c\":0,\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=1, with b=0, c=0. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(1, 0, 0, 0, 0, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (1, 0, 0, 0, 0, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0066","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4dbb52ffbd2c187991aeef46c19da1cfa59180757a69a46fe3f79b14e153cf39","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291066,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":1,\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"b\":3,\"c\":3,\"multiplication\":[[0,0,0,0,0],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=1, with b=3, c=3. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(1, 3, 3, 1, 1, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (1, 3, 3, 1, 1, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0067","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"98d268f13365c1dcb91c6bf45b561816fea5861da3198a30ee71a4855d0d583d","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291067,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":3,\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"b\":4,\"c\":1,\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=3, with b=4, c=1. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(3, 4, 1, 4, 1, No, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (3, 4, 1, 4, 1, No, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0068","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f5203cce823a238fd0c7dc020d2be7036a0b8f01ff54a33bc45b91d3d3d3d3ad","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291068,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":4,\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"b\":2,\"c\":2,\"multiplication\":[[0,0,0,0,0],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=4, with b=2, c=2. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(4, 2, 2, 3, 3, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (4, 2, 2, 3, 3, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0069","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"493f3d7e6e3d90f69530e4882ae61cdc434ab6918086c8a863bd1ae755e381da","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291069,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":2,\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"b\":3,\"c\":3,\"multiplication\":[[0,0,0,0,0],[0,3,4,2,1],[0,4,3,1,2],[0,2,1,4,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=2, with b=3, c=3. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(2, 3, 3, 1, 1, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (2, 3, 3, 1, 1, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0070","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ca72b088d6d78d4c1957e443abf3d11e39b539af33c4de27d9ec184c6b30f5f","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291070,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":2,\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"b\":0,\"c\":0,\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=2, with b=0, c=0. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(2, 0, 0, 0, 0, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (2, 0, 0, 0, 0, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0071","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"523544727a064eaa183d6d180e6b7134be7ca9812330f14168bcad64c3238c8c","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291071,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":4,\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"b\":4,\"c\":2,\"multiplication\":[[0,0,0,0,0],[0,3,4,2,1],[0,4,3,1,2],[0,2,1,4,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=4, with b=4, c=2. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(4, 4, 2, 4, 2, No, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (4, 4, 2, 4, 2, No, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0072","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ebf11444ba6cdc3b6df813935cca49f67af1173ca78ab5920820b215fb1e865","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291072,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":2,\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"b\":2,\"c\":2,\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=2, with b=2, c=2. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(2, 2, 2, 1, 1, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (2, 2, 2, 1, 1, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0073","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ebbde64e7187d3da52d76244e078069eb7758b573db4ce345af17ab2c40daf4","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291073,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":1,\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"b\":1,\"c\":1,\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=1, with b=1, c=1. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(1, 1, 1, 2, 2, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (1, 1, 1, 2, 2, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0074","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bcdec6d80f374cdd5ec6d15971d248fc03d1e62604b4e4a39ecf8edb2992a813","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291074,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":1,\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"b\":1,\"c\":1,\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=1, with b=1, c=1. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(1, 1, 1, 4, 4, Yes, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (1, 1, 1, 4, 4, Yes, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0075","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f8c263f27e9e973d33041e4b17f0c4f421c26357408d7af942835ac073459471","family_id":"qf-mod29-integraldomain-cancellation","seed":202610291075,"semantic_template":"qf-mod29-integraldomain-cancellation/{\"a\":0,\"addition\":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]],\"b\":1,\"c\":0,\"multiplication\":[[0,1,2,3,4],[1,1,1,1,1],[2,1,4,0,3],[3,1,0,4,2],[4,1,3,2,0]],\"n\":5}"},"fields":{"title":"Nonzero Cancellation Has an Implication, Not an Equality Premise Flag","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] and multiplication=[[0, 1, 2, 3, 4], [1, 1, 1, 1, 1], [2, 1, 4, 0, 3], [3, 1, 0, 4, 2], [4, 1, 3, 2, 0]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require a qualified integral domain and nonzero multiplier a=0, with b=1, c=0. Give (a,b,c,ab,ac,premise_equal?,cancellation_valid?,nonzero_cancellation_in_domain).","answer":{"kind":"expression","value":"(0, 1, 0, 1, 0, No, Yes, nonzero_cancellation_in_domain)","explanation":"Compute ab and ac in the same table. cancellation_valid is the implication ab=ac implies b=c; it can be Yes even when the separately reported premise_equal is No. Zero multipliers and rings with nonzero zero divisors are rejected, not repaired. The exact result, retaining the requested order, is (0, 1, 0, 1, 0, No, Yes, nonzero_cancellation_in_domain)."},"skills":["Nonzero Cancellation Has an Implication, Not an Equality Premise Flag"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0076","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e02c5cf777643a89a256c1c943380914f33740bc7cdcad7fcb59b9195ec573c5","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291076,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,4,1,3],[0,3,1,4,2],[0,4,3,2,1]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=1. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(1, 1, 1, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (1, 1, 1, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0077","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6f0ad7eb542ca574aed07560b7689f774f8a83b67d4e10ea5b65dbddde790107","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291077,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"element\":2,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=2. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(2, 1, 4, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (2, 1, 4, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0078","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ea4eed5c10297b27ef651106a8eec2a155f10a2ad859dfb7f3c2cfa47d5d4d6e","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291078,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"element\":2,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,4,1,3],[0,3,1,4,2],[0,4,3,2,1]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=2. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(2, 1, 3, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (2, 1, 3, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0079","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"41cab6137b1ea796588324e88ad37d7ca5de0981c38e4db6b22ffaa6b638ee6f","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291079,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"element\":2,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,1,4,3],[0,3,4,2,1],[0,4,3,1,2]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=2. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(2, 1, 2, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (2, 1, 2, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0080","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"09f51e7f26da3bb020bfd2fa62595ef83599a276708ed095274fb4a14d1a0762","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291080,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=1. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(1, 1, 1, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (1, 1, 1, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0081","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1182df78f47ba4e7a4e53dd1855342c07127d9a0fd3627e579af867cdb5de126","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291081,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"element\":4,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,1,4,3],[0,3,4,2,1],[0,4,3,1,2]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=4. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(4, 1, 3, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (4, 1, 3, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0082","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bfafaee1474ec47bc078a2b032c1af8ec247fd99cf9ea66e79cd8ac6330ec6ce","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291082,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"element\":3,\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=3. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(3, 2, 1, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (3, 2, 1, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0083","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d7d98b7d0f36e0e6b2fe0797ddfa7229131b42604907f772ae8d91b3bfec11f6","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291083,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"element\":3,\"multiplication\":[[0,0,0,0,0],[0,3,1,4,2],[0,1,2,3,4],[0,4,3,2,1],[0,2,4,1,3]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=3. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(3, 2, 3, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (3, 2, 3, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0084","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cc51f62a9edae9dfa05de11f13a6c9b015491d770ff04ca786089051d62d958b","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291084,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"element\":2,\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=2. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(2, 2, 2, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (2, 2, 2, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0085","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a3a86361c627dd44e4719b53a331e5abd76f062adaaf9f5bc6449823043f9778","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291085,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"element\":3,\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=3. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(3, 2, 4, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (3, 2, 4, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0086","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e36619d28ee9e0f7e5f3a86aaedff42d309545382a38948f8e1ae89624c0e171","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291086,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,3,1,4,2],[0,1,2,3,4],[0,4,3,2,1],[0,2,4,1,3]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=1. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(1, 2, 4, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (1, 2, 4, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0087","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"422192583b4aa5d957338412aaa8af419fe9eb3681cb78a5ecf75d517071ddc8","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291087,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=1. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(1, 2, 1, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (1, 2, 1, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0088","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d68b11c33ce04a3c2c451e65220abdb6077e6e7c3f83fd170e3e2e7a80ccfd6f","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291088,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"element\":3,\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=3. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(3, 3, 3, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (3, 3, 3, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0089","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7958eadc9ea8afd9831954dada3268139c6809c60fc6019655b795c13e7b1d4a","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291089,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=1. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(1, 3, 4, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (1, 3, 4, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0090","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e22701d6c048690c637df892d60ce3360d4983f2aa2b5e71408bd1e178ac63b","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291090,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=1. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(1, 3, 2, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (1, 3, 2, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0091","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2548c00095fb78c34a4d7c4443da7153f379c7cdd652064c1cb171ff20776d9b","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291091,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=1. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(1, 3, 1, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (1, 3, 1, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0092","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f7b9fbab53a2f1164438681c33bdfad5b3083b9a5f9a762e06c89986af1de0e","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291092,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"element\":3,\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=3. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(3, 3, 3, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (3, 3, 3, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0093","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e4690215e1e4d77af68098c1c15f609248c74f62de250f327c65071a0af62e2d","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291093,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"element\":4,\"multiplication\":[[0,0,0,0,0],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=4. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(4, 3, 2, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (4, 3, 2, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0094","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff2af089f73d61d59541d9d01cd2b7b60153a3a7bd30ce4fda9226799b31b632","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291094,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"element\":2,\"multiplication\":[[0,0,0,0,0],[0,3,4,2,1],[0,4,3,1,2],[0,2,1,4,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=2. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(2, 4, 1, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (2, 4, 1, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0095","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"21c230f271671272fafc10fcfb12719a07c6bcbfe52438ad4acd6c6b384e9fe9","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291095,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"element\":2,\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=2. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(2, 4, 2, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (2, 4, 2, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0096","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"97326d94ce4fe620113a0611380e4730f830b3ca3a4758e486f834111ebba09c","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291096,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"element\":4,\"multiplication\":[[0,0,0,0,0],[0,3,4,2,1],[0,4,3,1,2],[0,2,1,4,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=4. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(4, 4, 4, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (4, 4, 4, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0097","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"507702aa0cd284c06b48843845fa1dca3f8b5fa6325cd14f25248398f6aa8176","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291097,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"element\":2,\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=2. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(2, 4, 3, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (2, 4, 3, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0098","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bd93d706188cf08e5d5b70c8e923367045a055b7819eede6e8dcc5b420f9f85b","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291098,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=1. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(1, 4, 3, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (1, 4, 3, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0099","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a4acb6baae18d9cf694a5789e962dceee0b041b561423a7f317f5bd26ff195e2","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291099,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=1. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(1, 4, 1, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (1, 4, 1, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261029-r01-finite-integral-domain-0100","batch_id":"batch-20261014-advance-20261029-r01-finite-integral-domain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91fa07581557986256a756618b32fe88084b3eda9658d030c147f89a70e528eb","family_id":"qf-mod29-integraldomain-finite-inverse","seed":202610291100,"semantic_template":"qf-mod29-integraldomain-finite-inverse/{\"addition\":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]],\"element\":0,\"multiplication\":[[0,1,2,3,4],[1,1,1,1,1],[2,1,4,0,3],[3,1,0,4,2],[4,1,3,2,0]],\"n\":5}"},"fields":{"title":"Keep the Finite Step from Integral Domain to Invertibility","stem":"Use the same qualified complete finite commutative unital ring on Fin(5)={0,…,4}, with addition=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] and multiplication=[[0, 1, 2, 3, 4], [1, 1, 1, 1, 1], [2, 1, 4, 0, 3], [3, 1, 0, 4, 2], [4, 1, 3, 2, 0]] in carrier-label order. All ring axioms and distinct two-sided identities already hold; the zero and one labels need not be 0 and 1. No nonzero-zero-divisor or field premise is silently added. Additionally require an integral domain and nonzero element=0. Give (element,one,inverse,found?,finite_injective_multiplication_implies_surjective).","answer":{"kind":"expression","value":"(0, 0, 0, Yes, finite_injective_multiplication_implies_surjective)","explanation":"Multiplication by a nonzero element is injective by cancellation. Because this carrier is finite it is also surjective, so the actual identity has a preimage. Return the first inverse in increasing carrier-label order; do not extend this finite step to arbitrary infinite domains. The exact result, retaining the requested order, is (0, 0, 0, Yes, finite_injective_multiplication_implies_surjective)."},"skills":["Keep the Finite Step from Integral Domain to Invertibility"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}