第 27303 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-anbn-all-splits；資料={"p":3}。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-anbn-all-splits；資料={"p":3}。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-anbn-all-splits]

derived = [[p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope]]

source roles = [3]

derived = [a^3b^3, 6, ALL x>=0,y>=1,x+y<=p]

Pumping ORIGINAL remaining a after i0 split0,1 := sub(3, 1)

Pumping ORIGINAL remaining a after i0 split0,1 = 2

source roles = [3, 0, 1]

derived = [2, 5, [0, 1, 2, a^2b^3, 否], 是]

Pumping ORIGINAL remaining a after i0 split0,2 := sub(3, 2)

Pumping ORIGINAL remaining a after i0 split0,2 = 1

source roles = [3, 0, 2]

derived = [1, 4, [0, 2, 1, a^1b^3, 否], 是]

Pumping ORIGINAL remaining a after i0 split0,3 := sub(3, 3)

Pumping ORIGINAL remaining a after i0 split0,3 = 0

source roles = [3, 0, 3]

derived = [0, 3, [0, 3, 0, a^0b^3, 否], 是]

Pumping ORIGINAL remaining a after i0 split1,1 := sub(3, 1)

Pumping ORIGINAL remaining a after i0 split1,1 = 2

source roles = [3, 1, 1]

derived = [1, 4, [1, 1, 1, a^2b^3, 否], 是]

Pumping ORIGINAL remaining a after i0 split1,2 := sub(3, 2)

Pumping ORIGINAL remaining a after i0 split1,2 = 1

source roles = [3, 1, 2]

derived = [0, 3, [1, 2, 0, a^1b^3, 否], 是]

Pumping ORIGINAL remaining a after i0 split2,1 := sub(3, 1)

Pumping ORIGINAL remaining a after i0 split2,1 = 2

source roles = [3, 2, 1]

derived = [0, 3, [2, 1, 0, a^2b^3, 否], 是]

Pumping ORIGINAL triangular numerator := mul(3, 4)

Pumping ORIGINAL triangular numerator = 12

Pumping ORIGINAL triangular all-splits count := div(12, 2)

Pumping ORIGINAL triangular all-splits count = 6

source roles = [3, [[0, 1, 2, a^2b^3, 否], [0, 2, 1, a^1b^3, 否], [0, 3, 0, a^0b^3, 否], [1, 1, 1, a^2b^3, 否], [1, 2, 0, a^1b^3, 否], [2, 1, 0, a^2b^3, 否]]]

derived = [6, 6, 是]

逐步反向核對

• S017 · 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。
• S016 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL triangular all-splits count，不以小數近似替代。
• S015 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL triangular numerator，不以小數近似替代。
• S014 · 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。
• S013 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split2,1，不以小數近似替代。
• S012 · 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。
• S011 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split1,2，不以小數近似替代。
• S010 · 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。
• S009 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split1,1，不以小數近似替代。
• S008 · 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。
• S007 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split0,3，不以小數近似替代。
• S006 · 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。
• S005 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split0,2，不以小數近似替代。
• 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 — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split0,1，不以小數近似替代。
• 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

(3, a^3b^3, [[0, 1, 2, a^2b^3, 否], [0, 2, 1, a^1b^3, 否], [0, 3, 0, a^0b^3, 否], [1, 1, 1, a^2b^3, 否], [1, 2, 0, a^1b^3, 否], [2, 1, 0, a^2b^3, 否]], 6, 6, 是, every_legal_split_is_defeated_by_i=0)

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。