第 20458 題：子數列

原題與已知條件

自然數索引與symbolic嚴格遞增證書。操作=subseq-finite-prefix-boundary；資料={"indices":[7,10,13,16,19]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 自然數索引與symbolic嚴格遞增證書。操作=subseq-finite-prefix-boundary；資料={"indices":[7,10,13,16,19]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

定義與適用範圍

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。

• mko-subsequence

逐步推導

source operation=subseq-finite-prefix-boundary;original snapshot and mathematics retained

explicit ordered source output fields=[indices, finite_prefix_increasing, infinite_certificate_supplied, boundary]

subsequence finite prefix adjacent index1 difference := sub(10, 7)

subsequence finite prefix adjacent index1 difference = 3

ORIGINAL finite adjacent positions=[0, 1];actual indices=[7, 10]

exact forward difference=3;strictly increasing here=是

subsequence finite prefix adjacent index2 difference := sub(13, 10)

subsequence finite prefix adjacent index2 difference = 3

ORIGINAL finite adjacent positions=[1, 2];actual indices=[10, 13]

exact forward difference=3;strictly increasing here=是

subsequence finite prefix adjacent index3 difference := sub(16, 13)

subsequence finite prefix adjacent index3 difference = 3

ORIGINAL finite adjacent positions=[2, 3];actual indices=[13, 16]

exact forward difference=3;strictly increasing here=是

subsequence finite prefix adjacent index4 difference := sub(19, 16)

subsequence finite prefix adjacent index4 difference = 3

ORIGINAL finite adjacent positions=[3, 4];actual indices=[16, 19]

exact forward difference=3;strictly increasing here=是

ORIGINAL finite index list=[7, 10, 13, 16, 19];ALL adjacent strict flags=[是, 是, 是, 是]

this finite prefix increasing=是;INFINITE certificate supplied by original data=否;finite_prefix_not_infinite_subsequence_certificate

逐步反向核對

• S010 · 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。
• S009 · 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。
• S008 · premise_and_result_audit — 以精確有理數重算 subsequence finite prefix adjacent index4 difference，不以小數近似替代。
• S007 · 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。
• S006 · premise_and_result_audit — 以精確有理數重算 subsequence finite prefix adjacent index3 difference，不以小數近似替代。
• S005 · 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。
• S004 · premise_and_result_audit — 以精確有理數重算 subsequence finite prefix adjacent index2 difference，不以小數近似替代。
• S003 · 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。
• S002 · premise_and_result_audit — 以精確有理數重算 subsequence finite prefix adjacent index1 difference，不以小數近似替代。
• 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。

結論

([7, 10, 13, 16, 19], 是, 否, finite_increasing_prefix_not_infinite_subsequence_certificate)

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。

本題覆核與來源綁定

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。