第 20308 題：鴿籠原理

原題與已知條件

完整有限集合大小與total map。操作=pigeon-collision-witness；資料={"mapping":[2,4,1,3,0,2,4],"codomain_count":5}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 完整有限集合大小與total map。操作=pigeon-collision-witness；資料={"mapping":[2,4,1,3,0,2,4],"codomain_count":5}。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-pigeonhole-principle

逐步推導

source operation=pigeon-collision-witness;original snapshot and mathematics retained

explicit ordered source output fields=[domain_count, codomain_count, collision, has_collision, certificate]

ORIGINAL complete source-indexed total map=[2, 4, 1, 3, 0, 2, 4];source Fin(7);ACTUAL target Fin(5)

every source has ONE in-range image;all target labels0..4 retained even if unused;empty source allowed

next LEX ordered original source pair=[0, 1] with left<right;original images=[2, 4]

same target image=否;every earlier LEFT-then-RIGHT pair already checked unequal

next LEX ordered original source pair=[0, 2] with left<right;original images=[2, 1]

same target image=否;every earlier LEFT-then-RIGHT pair already checked unequal

next LEX ordered original source pair=[0, 3] with left<right;original images=[2, 3]

same target image=否;every earlier LEFT-then-RIGHT pair already checked unequal

next LEX ordered original source pair=[0, 4] with left<right;original images=[2, 0]

same target image=否;every earlier LEFT-then-RIGHT pair already checked unequal

next LEX ordered original source pair=[0, 5] with left<right;original images=[2, 2]

same target image=是;every earlier LEFT-then-RIGHT pair already checked unequal

ORIGINAL full domain size=7>target size=5;actual first collision=[0, 5, 2]

two DISTINCT original sources have the SAME original image;exact instance witness and finite size theorem agree

逐步反向核對

• S008 · 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。
• 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 — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• 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 — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• 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 — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• 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, 5, [0, 5, 2], 是, complete_total_map_collision)

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。