第 27371 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-dfa-loop-witness；資料={"modulus":21,"word_length":21}。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-dfa-loop-witness；資料={"modulus":21,"word_length":21}。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-dfa-loop-witness]

derived = [[modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified]]

Pumping ORIGINAL unary z length := sub(21, 21)

Pumping ORIGINAL unary z length = 0

source roles = [21, 21]

derived = [[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 17], [17, 18], [18, 19], [19, 20], [20, 0]], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 0], [0, 21, 0], 0, 21, 0]

source roles = [21, 21, 0]

derived = [EVERY integer i>=0: pumped length=i*modulus+z, z is divisible by modulus, state remains0 after each loop;accepted, 是]

逐步反向核對

• S004 · 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。
• S003 · 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。
• S002 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL unary z length，不以小數近似替代。
• 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

(21, 21, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 0], [0, 21, 0], 0, 21, 0, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, 是)

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。