第 27346 題：正規語言Pumping引理

原題與已知條件

exists p forall w exists split forall i量詞與split constraints責任。操作=pump-boundary；資料={"mode":"single_split","model":{"p":21,"word":["a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","b"],"x_length":20,"y_length":1}}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — exists p forall w exists split forall i量詞與split constraints責任。操作=pump-boundary；資料={"mode":"single_split","model":{"p":21,"word":["a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","b"],"x_length":20,"y_length":1}}。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-pumping-lemma-regular-languages

逐步推導

source roles = [pump-boundary]

derived = [[mode, p, word, x_length, y_length, z_length, split_legal, pumping_quantifiers_resolved_claim, boundary]]

source roles = [21, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 20, 1]

derived = [1, 是, 否]

逐步反向核對

• S002 · premise_and_result_audit — 回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。
• S001 · premise_and_result_audit — 回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。

結論

(single_split, 21, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 20, 1, 1, 是, 否, one_decomposition_does_not_resolve_exists_split_forall_i)

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。