第 27308 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":8}。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":8}。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 = [8]

derived = [a^8b^8, 16, ALL x>=0,y>=1,x+y<=p]

Pumping ORIGINAL remaining a after i0 split0,1 := sub(8, 1)

Pumping ORIGINAL remaining a after i0 split0,1 = 7

source roles = [8, 0, 1]

derived = [7, 15, [0, 1, 7, a^7b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split0,2 := sub(8, 2)

Pumping ORIGINAL remaining a after i0 split0,2 = 6

source roles = [8, 0, 2]

derived = [6, 14, [0, 2, 6, a^6b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split0,3 := sub(8, 3)

Pumping ORIGINAL remaining a after i0 split0,3 = 5

source roles = [8, 0, 3]

derived = [5, 13, [0, 3, 5, a^5b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split0,4 := sub(8, 4)

Pumping ORIGINAL remaining a after i0 split0,4 = 4

source roles = [8, 0, 4]

derived = [4, 12, [0, 4, 4, a^4b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split0,5 := sub(8, 5)

Pumping ORIGINAL remaining a after i0 split0,5 = 3

source roles = [8, 0, 5]

derived = [3, 11, [0, 5, 3, a^3b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split0,6 := sub(8, 6)

Pumping ORIGINAL remaining a after i0 split0,6 = 2

source roles = [8, 0, 6]

derived = [2, 10, [0, 6, 2, a^2b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split0,7 := sub(8, 7)

Pumping ORIGINAL remaining a after i0 split0,7 = 1

source roles = [8, 0, 7]

derived = [1, 9, [0, 7, 1, a^1b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split0,8 := sub(8, 8)

Pumping ORIGINAL remaining a after i0 split0,8 = 0

source roles = [8, 0, 8]

derived = [0, 8, [0, 8, 0, a^0b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split1,1 := sub(8, 1)

Pumping ORIGINAL remaining a after i0 split1,1 = 7

source roles = [8, 1, 1]

derived = [6, 14, [1, 1, 6, a^7b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split1,2 := sub(8, 2)

Pumping ORIGINAL remaining a after i0 split1,2 = 6

source roles = [8, 1, 2]

derived = [5, 13, [1, 2, 5, a^6b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split1,3 := sub(8, 3)

Pumping ORIGINAL remaining a after i0 split1,3 = 5

source roles = [8, 1, 3]

derived = [4, 12, [1, 3, 4, a^5b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split1,4 := sub(8, 4)

Pumping ORIGINAL remaining a after i0 split1,4 = 4

source roles = [8, 1, 4]

derived = [3, 11, [1, 4, 3, a^4b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split1,5 := sub(8, 5)

Pumping ORIGINAL remaining a after i0 split1,5 = 3

source roles = [8, 1, 5]

derived = [2, 10, [1, 5, 2, a^3b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split1,6 := sub(8, 6)

Pumping ORIGINAL remaining a after i0 split1,6 = 2

source roles = [8, 1, 6]

derived = [1, 9, [1, 6, 1, a^2b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split1,7 := sub(8, 7)

Pumping ORIGINAL remaining a after i0 split1,7 = 1

source roles = [8, 1, 7]

derived = [0, 8, [1, 7, 0, a^1b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split2,1 := sub(8, 1)

Pumping ORIGINAL remaining a after i0 split2,1 = 7

source roles = [8, 2, 1]

derived = [5, 13, [2, 1, 5, a^7b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split2,2 := sub(8, 2)

Pumping ORIGINAL remaining a after i0 split2,2 = 6

source roles = [8, 2, 2]

derived = [4, 12, [2, 2, 4, a^6b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split2,3 := sub(8, 3)

Pumping ORIGINAL remaining a after i0 split2,3 = 5

source roles = [8, 2, 3]

derived = [3, 11, [2, 3, 3, a^5b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split2,4 := sub(8, 4)

Pumping ORIGINAL remaining a after i0 split2,4 = 4

source roles = [8, 2, 4]

derived = [2, 10, [2, 4, 2, a^4b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split2,5 := sub(8, 5)

Pumping ORIGINAL remaining a after i0 split2,5 = 3

source roles = [8, 2, 5]

derived = [1, 9, [2, 5, 1, a^3b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split2,6 := sub(8, 6)

Pumping ORIGINAL remaining a after i0 split2,6 = 2

source roles = [8, 2, 6]

derived = [0, 8, [2, 6, 0, a^2b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split3,1 := sub(8, 1)

Pumping ORIGINAL remaining a after i0 split3,1 = 7

source roles = [8, 3, 1]

derived = [4, 12, [3, 1, 4, a^7b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split3,2 := sub(8, 2)

Pumping ORIGINAL remaining a after i0 split3,2 = 6

source roles = [8, 3, 2]

derived = [3, 11, [3, 2, 3, a^6b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split3,3 := sub(8, 3)

Pumping ORIGINAL remaining a after i0 split3,3 = 5

source roles = [8, 3, 3]

derived = [2, 10, [3, 3, 2, a^5b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split3,4 := sub(8, 4)

Pumping ORIGINAL remaining a after i0 split3,4 = 4

source roles = [8, 3, 4]

derived = [1, 9, [3, 4, 1, a^4b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split3,5 := sub(8, 5)

Pumping ORIGINAL remaining a after i0 split3,5 = 3

source roles = [8, 3, 5]

derived = [0, 8, [3, 5, 0, a^3b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split4,1 := sub(8, 1)

Pumping ORIGINAL remaining a after i0 split4,1 = 7

source roles = [8, 4, 1]

derived = [3, 11, [4, 1, 3, a^7b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split4,2 := sub(8, 2)

Pumping ORIGINAL remaining a after i0 split4,2 = 6

source roles = [8, 4, 2]

derived = [2, 10, [4, 2, 2, a^6b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split4,3 := sub(8, 3)

Pumping ORIGINAL remaining a after i0 split4,3 = 5

source roles = [8, 4, 3]

derived = [1, 9, [4, 3, 1, a^5b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split4,4 := sub(8, 4)

Pumping ORIGINAL remaining a after i0 split4,4 = 4

source roles = [8, 4, 4]

derived = [0, 8, [4, 4, 0, a^4b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split5,1 := sub(8, 1)

Pumping ORIGINAL remaining a after i0 split5,1 = 7

source roles = [8, 5, 1]

derived = [2, 10, [5, 1, 2, a^7b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split5,2 := sub(8, 2)

Pumping ORIGINAL remaining a after i0 split5,2 = 6

source roles = [8, 5, 2]

derived = [1, 9, [5, 2, 1, a^6b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split5,3 := sub(8, 3)

Pumping ORIGINAL remaining a after i0 split5,3 = 5

source roles = [8, 5, 3]

derived = [0, 8, [5, 3, 0, a^5b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split6,1 := sub(8, 1)

Pumping ORIGINAL remaining a after i0 split6,1 = 7

source roles = [8, 6, 1]

derived = [1, 9, [6, 1, 1, a^7b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split6,2 := sub(8, 2)

Pumping ORIGINAL remaining a after i0 split6,2 = 6

source roles = [8, 6, 2]

derived = [0, 8, [6, 2, 0, a^6b^8, 否], 是]

Pumping ORIGINAL remaining a after i0 split7,1 := sub(8, 1)

Pumping ORIGINAL remaining a after i0 split7,1 = 7

source roles = [8, 7, 1]

derived = [0, 8, [7, 1, 0, a^7b^8, 否], 是]

Pumping ORIGINAL triangular numerator := mul(8, 9)

Pumping ORIGINAL triangular numerator = 72

Pumping ORIGINAL triangular all-splits count := div(72, 2)

Pumping ORIGINAL triangular all-splits count = 36

source roles = [8, [[0, 1, 7, a^7b^8, 否], [0, 2, 6, a^6b^8, 否], [0, 3, 5, a^5b^8, 否], [0, 4, 4, a^4b^8, 否], [0, 5, 3, a^3b^8, 否], [0, 6, 2, a^2b^8, 否], [0, 7, 1, a^1b^8, 否], [0, 8, 0, a^0b^8, 否], [1, 1, 6, a^7b^8, 否], [1, 2, 5, a^6b^8, 否], [1, 3, 4, a^5b^8, 否], [1, 4, 3, a^4b^8, 否], [1, 5, 2, a^3b^8, 否], [1, 6, 1, a^2b^8, 否], [1, 7, 0, a^1b^8, 否], [2, 1, 5, a^7b^8, 否], [2, 2, 4, a^6b^8, 否], [2, 3, 3, a^5b^8, 否], [2, 4, 2, a^4b^8, 否], [2, 5, 1, a^3b^8, 否], [2, 6, 0, a^2b^8, 否], [3, 1, 4, a^7b^8, 否], [3, 2, 3, a^6b^8, 否], [3, 3, 2, a^5b^8, 否], [3, 4, 1, a^4b^8, 否], [3, 5, 0, a^3b^8, 否], [4, 1, 3, a^7b^8, 否], [4, 2, 2, a^6b^8, 否], [4, 3, 1, a^5b^8, 否], [4, 4, 0, a^4b^8, 否], [5, 1, 2, a^7b^8, 否], [5, 2, 1, a^6b^8, 否], [5, 3, 0, a^5b^8, 否], [6, 1, 1, a^7b^8, 否], [6, 2, 0, a^6b^8, 否], [7, 1, 0, a^7b^8, 否]]]

derived = [36, 36, 是]

逐步反向核對

• S077 · 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。
• S076 · 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。
• S075 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL triangular numerator，不以小數近似替代。
• S074 · 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。
• S073 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split7,1，不以小數近似替代。
• S072 · 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。
• S071 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split6,2，不以小數近似替代。
• S070 · 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。
• S069 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split6,1，不以小數近似替代。
• S068 · 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。
• S067 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split5,3，不以小數近似替代。
• S066 · 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。
• S065 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split5,2，不以小數近似替代。
• S064 · 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。
• S063 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split5,1，不以小數近似替代。
• S062 · 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。
• S061 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split4,4，不以小數近似替代。
• S060 · 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。
• S059 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split4,3，不以小數近似替代。
• S058 · 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。
• S057 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split4,2，不以小數近似替代。
• S056 · 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。
• S055 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split4,1，不以小數近似替代。
• S054 · 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。
• S053 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split3,5，不以小數近似替代。
• S052 · 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。
• S051 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split3,4，不以小數近似替代。
• S050 · 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。
• S049 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split3,3，不以小數近似替代。
• S048 · 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。
• S047 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split3,2，不以小數近似替代。
• S046 · 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。
• S045 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split3,1，不以小數近似替代。
• S044 · 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。
• S043 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split2,6，不以小數近似替代。
• S042 · 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。
• S041 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split2,5，不以小數近似替代。
• S040 · 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。
• S039 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split2,4，不以小數近似替代。
• S038 · 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。
• S037 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split2,3，不以小數近似替代。
• S036 · 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。
• S035 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split2,2，不以小數近似替代。
• S034 · 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。
• S033 · 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。
• S032 · 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。
• S031 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split1,7，不以小數近似替代。
• S030 · 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。
• S029 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split1,6，不以小數近似替代。
• S028 · 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。
• S027 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split1,5，不以小數近似替代。
• S026 · 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。
• S025 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split1,4，不以小數近似替代。
• S024 · 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。
• S023 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split1,3，不以小數近似替代。
• S022 · 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。
• S021 · 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。
• S020 · 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。
• S019 · 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。
• S018 · 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。
• S017 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split0,8，不以小數近似替代。
• S016 · 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。
• S015 · premise_and_result_audit — 以精確有理數重算 Pumping ORIGINAL remaining a after i0 split0,7，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 split0,6，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 split0,5，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 split0,4，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

(8, a^8b^8, [[0, 1, 7, a^7b^8, 否], [0, 2, 6, a^6b^8, 否], [0, 3, 5, a^5b^8, 否], [0, 4, 4, a^4b^8, 否], [0, 5, 3, a^3b^8, 否], [0, 6, 2, a^2b^8, 否], [0, 7, 1, a^1b^8, 否], [0, 8, 0, a^0b^8, 否], [1, 1, 6, a^7b^8, 否], [1, 2, 5, a^6b^8, 否], [1, 3, 4, a^5b^8, 否], [1, 4, 3, a^4b^8, 否], [1, 5, 2, a^3b^8, 否], [1, 6, 1, a^2b^8, 否], [1, 7, 0, a^1b^8, 否], [2, 1, 5, a^7b^8, 否], [2, 2, 4, a^6b^8, 否], [2, 3, 3, a^5b^8, 否], [2, 4, 2, a^4b^8, 否], [2, 5, 1, a^3b^8, 否], [2, 6, 0, a^2b^8, 否], [3, 1, 4, a^7b^8, 否], [3, 2, 3, a^6b^8, 否], [3, 3, 2, a^5b^8, 否], [3, 4, 1, a^4b^8, 否], [3, 5, 0, a^3b^8, 否], [4, 1, 3, a^7b^8, 否], [4, 2, 2, a^6b^8, 否], [4, 3, 1, a^5b^8, 否], [4, 4, 0, a^4b^8, 否], [5, 1, 2, a^7b^8, 否], [5, 2, 1, a^6b^8, 否], [5, 3, 0, a^5b^8, 否], [6, 1, 1, a^7b^8, 否], [6, 2, 0, a^6b^8, 否], [7, 1, 0, a^7b^8, 否]], 36, 36, 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。