{"schema_version":"ocme-site-results-v1","locale":"zh-Hant","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-set"},"kind":"guides","total":3,"pages":1,"page":1,"entries":[{"id":"derivation-mko-set-01","title":{"zh-Hant":"集合：由元素刻畫、外延相等，以及不受限制的建構為何失敗","en":"Sets: characterized by elements, equal by extensionality, and why unrestricted formation fails"},"summary":{"zh-Hant":"在論域 ℤ 中，設 A={x∈ℤ ∣ x²=4}、B={−2,2}、C={2,−2,2}、D={2}。證明 A=B 與 C=B；證明 D≠B，並區分元素 2 與集合 {2}。最後證明：在任何帶有隸屬關係的論域 U 中，都不存在 R∈U 使「y∈R ⇔ y∉y」對所有 y∈U 成立，因此不是每個條件都能形成集合。","en":"In the universe ℤ, let A={x∈ℤ ∣ x²=4}, B={−2,2}, C={2,−2,2} and D={2}. Prove A=B and C=B; prove D≠B and distinguish the element 2 from the set {2}. Finally prove that in any domain U with a membership relation there is no R∈U with 'y∈R ⇔ y∉y' for all y∈U, so not every condition forms a set."},"mko_ids":["mko-set"]},{"id":"derivation-mko-set-membership-01","title":{"zh-Hant":"集合隸屬：x∈A 是命題，並與 ⊆、= 及程式的成員測試區分","en":"Set membership: x∈A is a proposition, kept apart from ⊆, = and a program's membership test"},"summary":{"zh-Hant":"在型別化語境中，設 A={1,2,3}、B={1,2}、𝓕={{1,2},{3}}（元素是整數集合的集合）、E={n∈ℤ ∣ 2∣n}。判定下列命題的真假並給出理由：2∈A；4∈A；B⊆A；A⊆B；B∈𝓕；{1}∈𝓕；1000000∈E；7∈E。並說明 1∈B 與 B∈𝓕 為何不給出 1∈𝓕，以及程式語言的成員測試為何不能取代集合隸屬的語義。","en":"In a typed context let A={1,2,3}, B={1,2}, 𝓕={{1,2},{3}} (a set whose elements are sets of integers) and E={n∈ℤ ∣ 2∣n}. Decide with reasons whether each holds: 2∈A; 4∈A; B⊆A; A⊆B; B∈𝓕; {1}∈𝓕; 1000000∈E; 7∈E. Also explain why 1∈B and B∈𝓕 do not give 1∈𝓕, and why a programming language's membership test cannot replace the semantics of set membership."},"mko_ids":["mko-set-membership"]},{"id":"derivation-mko-set-operations-01","title":{"zh-Hant":"集合運算：在明示全集中計算聯集、交集、差集與補集","en":"Set operations: union, intersection, difference and complement in an explicit universe"},"summary":{"zh-Hant":"全集 U={1,2,3,4,5,6,7,8}，A={2,4,6,8}，B={3,4,5,6}。逐元素算出 A∪B、A∩B、A∖B、B∖A、U∖A、U∖B；說明 A∖B≠B∖A；說明全集換成 U′={1,2,3,4,5,6,7,8,9,10} 時 A 的補集改變；對任意集合證明 U∖(A∪B)=(U∖A)∩(U∖B) 並以本例核對；再以 A∪B=A∪(B∖A) 說明聯集為何有 6 個元素而不是 8 個。","en":"Universe U={1,2,3,4,5,6,7,8}, A={2,4,6,8}, B={3,4,5,6}. Compute A∪B, A∩B, A∖B, B∖A, U∖A and U∖B element by element; show A∖B≠B∖A; show that the complement of A changes when the universe is U′={1,2,3,4,5,6,7,8,9,10}; prove U∖(A∪B)=(U∖A)∩(U∖B) for arbitrary sets and check it on this example; and use A∪B=A∪(B∖A) to explain why the union has 6 elements, not 8."},"mko_ids":["mko-set-operations"]}],"title":"概念推導導讀"}