{"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-equivalence-class"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-equivalence-class-01","title":{"zh-Hant":"等價類：在完整有限關係上先認證、再求類，並以非等價關係作對照","en":"Equivalence class: certify a complete finite relation first, then form classes, with a non-equivalence as contrast"},"summary":{"zh-Hant":"在 X={0,1,2,3,4,5} 上，R 由 12 個明列的有序對給出。先完整認證 R 是等價關係，再依定義 [a]={x∈X ∣ aRx} 求出全部類；證明 aRb ⇒ [a]=[b]、任兩類相等或不交、各類覆蓋 X，得 K(R)={{0,3},{1,4},{2,5}}。最後以 S=R∪{(0,1),(1,0)} 說明未通過傳遞性的關係不能形成等價類。","en":"On X={0,1,2,3,4,5}, R is given by 12 explicitly listed ordered pairs. First certify completely that R is an equivalence relation, then compute every class from the definition [a]={x∈X ∣ aRx}; prove that aRb ⇒ [a]=[b], that any two classes are equal or disjoint and that the classes cover X, giving K(R)={{0,3},{1,4},{2,5}}. Finally use S=R∪{(0,1),(1,0)} to show that a relation failing transitivity cannot form equivalence classes."},"mko_ids":["mko-equivalence-class"]}],"title":"概念推導導讀"}