R0041-21340 · q-20261014-advance-20261108-r06-unique-prime-factorization-0015
完整bounded integer factorization與exact primality。操作=upf-factorization-audit；資料={"n":282,"factors":[2,3,47]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。
G001 — 完整bounded integer factorization與exact primality。操作=upf-factorization-audit；資料={"n":282,"factors":[2,3,47]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。
操作前
source operation=upf-factorization-audit;original snapshot and mathematics retained
操作後
explicit ordered source output fields=[n, factors, all_factors_prime, factor_product, valid_factorization, certificate]
條件：無
原題歷史介面標記保留；此處直接列有序輸出欄位，不依賴JavaScript。欄位或tag不是額外數學證據，固定No保留原聲稱範圍。
規則：source_ordered_interface_explicit_without_javascript · 依賴：G001
操作前
upf candidate factor index0:ORIGINAL candidate factor=2;ALL possible trial divisors2..floor(sqrt(v))=[]
操作後
ALL [divisor,remainder] rows=[];prime=是
條件：無
合數若大於1必有不超過平方根的因數；完整區間無整除才證prime，2的區間為空仍是prime。乘積相等不能替代此責任。
規則：upf_every_trial_divisor_primality_not_product_only · 依賴：G001, S001
操作前
upf candidate complete product at original factor0 := mul(1, 2)
操作後
upf candidate complete product at original factor0 = 2
條件：無
依原題角色與已核中間量精確代值，保留原順序、正負號及量詞條件。
規則：exact_rational_arithmetic · 依賴：G001, S002
操作前
upf candidate factor index1:ORIGINAL candidate factor=3;ALL possible trial divisors2..floor(sqrt(v))=[]
操作後
ALL [divisor,remainder] rows=[];prime=是
條件：無
合數若大於1必有不超過平方根的因數；完整區間無整除才證prime，2的區間為空仍是prime。乘積相等不能替代此責任。
規則：upf_every_trial_divisor_primality_not_product_only · 依賴：G001, S003
操作前
upf candidate complete product at original factor1 := mul(2, 3)
操作後
upf candidate complete product at original factor1 = 6
條件：無
依原題角色與已核中間量精確代值，保留原順序、正負號及量詞條件。
規則：exact_rational_arithmetic · 依賴：G001, S004
操作前
upf candidate factor index2:ORIGINAL candidate factor=47;ALL possible trial divisors2..floor(sqrt(v))=[2, 3, 4, 5, 6]
操作後
ALL [divisor,remainder] rows=[[2, 1], [3, 2], [4, 3], [5, 2], [6, 5]];prime=是
條件：無
合數若大於1必有不超過平方根的因數；完整區間無整除才證prime，2的區間為空仍是prime。乘積相等不能替代此責任。
規則：upf_every_trial_divisor_primality_not_product_only · 依賴：G001, S005
操作前
upf candidate complete product at original factor2 := mul(6, 47)
操作後
upf candidate complete product at original factor2 = 282
條件：無
依原題角色與已核中間量精確代值，保留原順序、正負號及量詞條件。
規則：exact_rational_arithmetic · 依賴：G001, S006
操作前
upf candidate:ORIGINAL factors=[2, 3, 47];ALL primality flags=[是, 是, 是];original n=282
操作後
empty product convention1;actual complete product=282;all_prime=是;valid_prime_product=是
條件：無
保留原factor順序。空列表每項為prime是vacuous True但乘積1不等n>1；合數乘積即使吻合亦無prime factorization證書。SAFE是來源投影限制，不是定理真假判斷。
規則：upf_candidate_prime_and_complete_product_separate · 依賴：G001, S007
S008 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。S007 · premise_and_result_audit — 以精確有理數重算 upf candidate complete product at original factor2，不以小數近似替代。S006 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。S005 · premise_and_result_audit — 以精確有理數重算 upf candidate complete product at original factor1，不以小數近似替代。S004 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。S003 · premise_and_result_audit — 以精確有理數重算 upf candidate complete product at original factor0，不以小數近似替代。S002 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。S001 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。(282, [2, 3, 47], 是, 282, 是, existence_requires_complete_prime_product)
依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。
覆核日期：2026-10-26
原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。
input commit: 0f7be61e2c22317a4075561dc822c87d881a50dboriginal question SHA-256: ceb97098c8d3e561e8d3a1caa6d99fd89d20603359ac3523469ffa51ff5e19c8candidate record SHA-256: ccf0aa5c6bbf332a93a4106f825fe78018443670ff6a41e2487b5e46618e09afsemantic review receipt SHA-256: 8f134403487694af4c70d2e00ec84bab542043ef4b3c54517c04f3c409c0262aquestion-first witness SHA-256: ad772845088ed8e99f87f90d4264b05ea97be35775305a02011af5ee35f0f4ceLean: NOT_RUN · human review: NOT_PERFORMED