{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-equivalence-class-01",
  "mko_id": "mko-equivalence-class",
  "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"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-equivalence-class-01",
    "source": {
      "mko_id": "mko-equivalence-class",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "6ead5a95cd13a163d636e78db505bb8f7aad6744f9e25afee8148c838a3a44d9",
      "english_sha256": "6d92bfefd809a7377beec0c3584a0746a97568a9481e11f85ed0bd5ad27ee177"
    },
    "mode": "definition_walkthrough",
    "audience_prerequisites": [
      "set_builder",
      "set_extensionality",
      "subset_def",
      "division_algorithm",
      "eq_refl_symm_trans",
      "finite_counting_basics"
    ],
    "goal": {
      "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."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "載體 X={0,1,2,3,4,5}（MKO 假設 a1）。",
          "en": "The carrier X={0,1,2,3,4,5} (MKO assumption a1)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "R 恰為 12 個有序對：(0,0)、(0,3)、(3,0)、(3,3)、(1,1)、(1,4)、(4,1)、(4,4)、(2,2)、(2,5)、(5,2)、(5,5)；這是完整的關係對集合。",
          "en": "R consists of exactly 12 ordered pairs: (0,0), (0,3), (3,0), (3,3), (1,1), (1,4), (4,1), (4,4), (2,2), (2,5), (5,2), (5,5); this is the complete set of relation pairs."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "定義：R 通過反身、對稱、傳遞的完整核對後，[a]={x∈X ∣ aRx}，K(R) 是所有 [a] 所成的集合（MKO 敘述）。",
          "en": "Definition: after R passes complete checks of reflexivity, symmetry and transitivity, [a]={x∈X ∣ aRx}, and K(R) is the set of all [a] (MKO statement)."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "對照關係：S=R∪{(0,1),(1,0)}，共 14 個有序對。",
          "en": "Contrast relation: S=R∪{(0,1),(1,0)}, 14 ordered pairs in all."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "R=\\{(0,0),(0,3),(3,0),(3,3),(1,1),(1,4),(4,1),(4,4),(2,2),(2,5),(5,2),(5,5)\\}",
        "after_tex": "R=\\{(a,b)\\in X\\times X\\mid r(a)=r(b)\\}",
        "relation": "equivalent",
        "rule": "L_rel_desc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_rel_desc（在 S01.1、S01.2 證明），明列的 12 個對恰好是「除以 3 餘數相同」的全部對。這個描述讓之後的三個公理能對全部對一次核完，而不是抽樣。",
          "en": "By Lemma L_rel_desc (proved in S01.1 and S01.2), the 12 listed pairs are exactly all pairs with the same remainder on division by 3. This description lets the three axioms be checked for all pairs at once, rather than by sampling."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_rel_desc",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "兩個包含都在子步證明，所以集合相等雙向成立。",
            "en": "Both inclusions are proved in the sub-steps, so the set equality holds both ways."
          }
        },
        "leaf_basis": "set_extensionality + division_algorithm + finite_counting_basics"
      },
      {
        "id": "S01.1",
        "uses": [
          "G2",
          "division_algorithm"
        ],
        "before_tex": "(a,b)\\in R",
        "after_tex": "r(a)=r(b)",
        "relation": "implies",
        "rule": "division_algorithm",
        "conditions": [],
        "explanation": {
          "zh-Hant": "（⊆）餘數：r(0)=r(3)=0、r(1)=r(4)=1、r(2)=r(5)=2（3=1·3+0、4=1·3+1、5=1·3+2，N4 到 N6）。明列的對都在 {0,3}、{1,4}、{2,5} 之一的內部，所以每對的兩個元素餘數相同。",
          "en": "(⊆) Remainders: r(0)=r(3)=0, r(1)=r(4)=1, r(2)=r(5)=2 (3=1·3+0, 4=1·3+1, 5=1·3+2; N4 to N6). Every listed pair lies inside one of {0,3}, {1,4}, {2,5}, so the two entries of each pair have the same remainder."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "division_algorithm"
          ],
          "explanation": {
            "zh-Hant": "12 個對逐一可核；只證一個包含。",
            "en": "All 12 pairs can be checked one by one; only one inclusion is proved here."
          }
        },
        "leaf_basis": "division_algorithm + numeric_evaluation"
      },
      {
        "id": "S01.2",
        "uses": [
          "S01.1",
          "G2",
          "finite_counting_basics"
        ],
        "before_tex": "(a,b)\\in X\\times X,\\ r(a)=r(b)",
        "after_tex": "(a,b)\\in R",
        "relation": "implies",
        "rule": "finite_counting_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "（⊇）每個餘數 0、1、2 恰有 2 個元素，所以餘數相同的有序對共 3·2²=12 個（N1），即 {0,3}²、{1,4}²、{2,5}² 中的對。G2 列出 12 個互異的對，由 S01.1 它們都在這 12 個之中；一個 12 元集合的 12 元子集就是它本身，所以每個餘數相同的對都在 R 中。S01.1 與 S01.2 合起來證得引理 L_rel_desc。",
          "en": "(⊇) Each remainder 0, 1, 2 has exactly 2 elements, so there are 3·2²=12 ordered pairs with equal remainders (N1), namely the pairs in {0,3}², {1,4}² and {2,5}². G2 lists 12 distinct pairs, all among these 12 by S01.1; a 12-element subset of a 12-element set is the whole set, so every pair with equal remainders is in R. S01.1 and S01.2 together prove Lemma L_rel_desc."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01.1",
            "finite_counting_basics"
          ],
          "explanation": {
            "zh-Hant": "需要 G2 的 12 個對互異（逐一可見）與計數；若漏列一對，計數就會揭露。",
            "en": "It needs the 12 pairs of G2 to be distinct (visible one by one) and the count; a missing pair would show up in the count."
          }
        },
        "leaf_basis": "finite_counting_basics + subset_def"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "eq_refl_symm_trans"
        ],
        "before_tex": "a\\in X",
        "after_tex": "aRa",
        "relation": "implies",
        "rule": "eq_refl_symm_trans",
        "conditions": [],
        "explanation": {
          "zh-Hant": "反身：對每個 a，r(a)=r(a)，由 S01 得 aRa。對應 G2 中的 6 個對角對 (0,0)、(1,1)、(2,2)、(3,3)、(4,4)、(5,5)。",
          "en": "Reflexivity: for every a, r(a)=r(a), so S01 gives aRa. These are the 6 diagonal pairs (0,0), (1,1), (2,2), (3,3), (4,4), (5,5) of G2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "eq_refl_symm_trans"
          ],
          "explanation": {
            "zh-Hant": "對全部 6 個元素成立，不是抽樣。",
            "en": "It holds for all 6 elements, not for a sample."
          }
        },
        "leaf_basis": "eq_refl_symm_trans + forall_intro"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "eq_refl_symm_trans"
        ],
        "before_tex": "aRb",
        "after_tex": "bRa",
        "relation": "implies",
        "rule": "eq_refl_symm_trans",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對稱：aRb 表示 r(a)=r(b)，由等號對稱 r(b)=r(a)，所以 bRa。",
          "en": "Symmetry: aRb means r(a)=r(b); symmetry of equality gives r(b)=r(a), so bRa."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "eq_refl_symm_trans"
          ],
          "explanation": {
            "zh-Hant": "對全部 12 個對一次成立。",
            "en": "It holds for all 12 pairs at once."
          }
        },
        "leaf_basis": "eq_refl_symm_trans"
      },
      {
        "id": "S04",
        "uses": [
          "S01",
          "eq_refl_symm_trans"
        ],
        "before_tex": "aRb,\\ bRc",
        "after_tex": "aRc",
        "relation": "implies",
        "rule": "eq_refl_symm_trans",
        "conditions": [],
        "explanation": {
          "zh-Hant": "傳遞：r(a)=r(b) 且 r(b)=r(c)，由等號傳遞 r(a)=r(c)，所以 aRc。這一次處理了全部 6·2·2=24 個可接續的對（每個 b 有 2 個進入對與 2 個離開對，N8）。由 S02 到 S04，R 是已認證的等價關係（MKO 假設 a1）。",
          "en": "Transitivity: r(a)=r(b) and r(b)=r(c) give r(a)=r(c) by transitivity of equality, so aRc. This handles all 6·2·2=24 composable pairs at once (each b has 2 incoming and 2 outgoing pairs; N8). By S02 to S04, R is a certified equivalence relation (MKO assumption a1)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "eq_refl_symm_trans",
            "equiv_rel_def"
          ],
          "explanation": {
            "zh-Hant": "認證依賴 S01 的完整描述；若只核幾個三元組，不能排除 S 那種失敗（S09）。",
            "en": "The certification rests on the complete description of S01; checking only a few triples could not rule out a failure like that of S (S09)."
          }
        },
        "leaf_basis": "eq_refl_symm_trans + equiv_rel_def"
      },
      {
        "id": "S05",
        "uses": [
          "G2",
          "G3",
          "S04"
        ],
        "before_tex": "[a]=\\{x\\in X\\mid aRx\\}",
        "after_tex": "[0]=[3]=\\{0,3\\},\\ [1]=[4]=\\{1,4\\},\\ [2]=[5]=\\{2,5\\}",
        "relation": "equivalent",
        "rule": "set_builder",
        "conditions": [],
        "explanation": {
          "zh-Hant": "R 已認證，才依定義求類：[a] 收集所有以 a 為第一座標的對的第二座標。以 0 開頭的對是 (0,0)、(0,3)，所以 [0]={0,3}；以 3 開頭的是 (3,0)、(3,3)，[3]={0,3}；同理 [1]=[4]={1,4}，[2]=[5]={2,5}。",
          "en": "Only now that R is certified are the classes formed from the definition: [a] collects the second entries of all pairs whose first entry is a. The pairs starting with 0 are (0,0), (0,3), so [0]={0,3}; those starting with 3 are (3,0), (3,3), so [3]={0,3}; likewise [1]=[4]={1,4} and [2]=[5]={2,5}."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_builder",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "集合建構式雙向：x∈[a] ⇔ (a,x)∈R，可逐一對照 G2 的列表。",
            "en": "Set-builder membership is two-way: x∈[a] ⇔ (a,x)∈R, checkable against the list of G2 one pair at a time."
          }
        },
        "leaf_basis": "set_builder + def_unfold"
      },
      {
        "id": "S06",
        "uses": [
          "S03",
          "S04"
        ],
        "before_tex": "aRb",
        "after_tex": "[a]=[b]",
        "relation": "implies",
        "rule": "L_same_class",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_same_class（在 S06.1、S06.2 證明），相關的代表元產生同一類（MKO 段落 1）。實例：0R3，所以 [0]=[3]，與 S05 一致。",
          "en": "By Lemma L_same_class (proved in S06.1 and S06.2), related representatives give the same class (MKO paragraph 1). Instance: 0R3, so [0]=[3], in agreement with S05."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S04",
            "L_same_class"
          ],
          "explanation": {
            "zh-Hant": "引理需要對稱與傳遞；不宣稱反向（反向 [a]=[b] ⇒ aRb 也成立，但本步不需要）。",
            "en": "The lemma needs symmetry and transitivity; the converse ([a]=[b] ⇒ aRb, also true) is not needed or claimed here."
          }
        },
        "leaf_basis": "set_extensionality + subset_def + equiv_rel_def"
      },
      {
        "id": "S06.1",
        "uses": [
          "S04",
          "G3"
        ],
        "before_tex": "aRb,\\ x\\in[b]",
        "after_tex": "x\\in[a]",
        "relation": "implies",
        "rule": "subset_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x∈[b] 表示 bRx。由 aRb、bRx 與傳遞性得 aRx，所以 x∈[a]。故 [b]⊆[a]。",
          "en": "x∈[b] means bRx. From aRb, bRx and transitivity we get aRx, so x∈[a]. Hence [b]⊆[a]."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "subset_def"
          ],
          "explanation": {
            "zh-Hant": "只用傳遞性；另一個包含在 S06.2。",
            "en": "Only transitivity is used; the other inclusion is S06.2."
          }
        },
        "leaf_basis": "subset_def + set_builder"
      },
      {
        "id": "S06.2",
        "uses": [
          "S03",
          "S04",
          "S06.1",
          "G3"
        ],
        "before_tex": "aRb,\\ x\\in[a]",
        "after_tex": "x\\in[b],\\ \\text{so}\\ [a]=[b]",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由對稱 bRa；x∈[a] 表示 aRx；傳遞得 bRx，x∈[b]，故 [a]⊆[b]。結合 S06.1 的 [b]⊆[a]，依集合相等 [a]=[b]。S06.1 與 S06.2 合起來證得引理 L_same_class。",
          "en": "Symmetry gives bRa; x∈[a] means aRx; transitivity gives bRx, so x∈[b] and [a]⊆[b]. With [b]⊆[a] from S06.1, set equality gives [a]=[b]. S06.1 and S06.2 together prove Lemma L_same_class."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S04",
            "S06.1",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "這裡需要對稱；兩個包含都需要傳遞。",
            "en": "Symmetry is needed here, and both inclusions need transitivity."
          }
        },
        "leaf_basis": "set_extensionality + subset_def + set_builder"
      },
      {
        "id": "S07",
        "uses": [
          "S03",
          "S04",
          "S06"
        ],
        "before_tex": "x\\in[a]\\cap[b]",
        "after_tex": "[a]=[b];\\ \\text{hence}\\ [a]=[b]\\ \\lor\\ [a]\\cap[b]=\\varnothing",
        "relation": "implies",
        "rule": "L_same_class",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若 x∈[a]∩[b]，則 aRx、bRx；對稱得 xRb，傳遞得 aRb，依 S06，[a]=[b]。逆否：[a]≠[b] ⇒ [a]∩[b]=∅。所以任兩類相等或不交；例如 [0]∩[1]=∅。",
          "en": "If x∈[a]∩[b], then aRx and bRx; symmetry gives xRb and transitivity gives aRb, so [a]=[b] by S06. Contrapositive: [a]≠[b] ⇒ [a]∩[b]=∅. So any two classes are equal or disjoint; for example [0]∩[1]=∅."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S04",
            "S06",
            "contrapositive"
          ],
          "explanation": {
            "zh-Hant": "論證在「傳遞得 aRb」這一步用到傳遞性；S10 正顯示沒有傳遞性時這一步失敗。",
            "en": "The argument uses transitivity at 'transitivity gives aRb'; S10 shows exactly this step failing without transitivity."
          }
        },
        "leaf_basis": "contrapositive + equiv_rel_def + set_operations_def"
      },
      {
        "id": "S08",
        "uses": [
          "S02",
          "S05",
          "S07"
        ],
        "before_tex": "a\\in[a]\\ \\text{for every}\\ a",
        "after_tex": "K(R)=\\{\\{0,3\\},\\{1,4\\},\\{2,5\\}\\},\\ 2+2+2=6=|X|",
        "relation": "implies",
        "rule": "mko_eqclass_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由反身性 aRa，a∈[a]，所以各類覆蓋 X。由 S05，六個 [a] 只有三個相異集合，兩兩不交（S07），大小 2+2+2=6=|X|（N2）。所以 K(R)={{0,3},{1,4},{2,5}} 是 X 的分割。商集作為獨立節點另立（MKO 段落 2）。",
          "en": "By reflexivity aRa, so a∈[a] and the classes cover X. By S05 the six sets [a] are only three distinct sets, pairwise disjoint (S07), of sizes 2+2+2=6=|X| (N2). So K(R)={{0,3},{1,4},{2,5}} is a partition of X. The quotient set as a node of its own is separate (MKO paragraph 2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S05",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "覆蓋需要反身性，不交需要 S07；計數只是一致性核對。",
            "en": "Covering needs reflexivity and disjointness needs S07; the count is a consistency check."
          }
        },
        "leaf_basis": "set_operations_def + finite_counting_basics + and_intro_elim"
      },
      {
        "id": "S09",
        "uses": [
          "G4",
          "S01"
        ],
        "before_tex": "S=R\\cup\\{(0,1),(1,0)\\}",
        "after_tex": "(3,0),(0,1)\\in S,\\ (3,1)\\notin S:\\ S\\ \\text{not transitive}",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S 含全部 6 個對角對，所以反身；新加的兩對互為反向，所以仍對稱。但 (3,0)∈S、(0,1)∈S，而 (3,1) 不在 R 中（r(3)=0≠1=r(1)，N3），也不是新加的對，所以 (3,1)∉S。傳遞性失敗，S 不是等價關係，不能用來形成 canonical classes。",
          "en": "S contains all 6 diagonal pairs, so it is reflexive; the two added pairs are reverses of each other, so it is still symmetric. But (3,0)∈S and (0,1)∈S, while (3,1) is not in R (r(3)=0≠1=r(1), N3) and is not an added pair, so (3,1)∉S. Transitivity fails, S is not an equivalence relation, and it must not be used to form canonical classes."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G4",
            "S01",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "一個失敗的三元組 (3,0,1) 就足以否定傳遞性。",
            "en": "One failing triple (3,0,1) suffices to refute transitivity."
          }
        },
        "leaf_basis": "quantifier_negation + set_operations_def"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "G4"
        ],
        "before_tex": "[a]_S=\\{x\\mid aSx\\}",
        "after_tex": "[0]_S=\\{0,1,3\\},\\ [3]_S=\\{0,3\\},\\ [1]_S=\\{0,1,4\\}",
        "relation": "implies",
        "rule": "set_builder",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若硬用 S 的對求「類」：以 0 開頭的對是 (0,0)、(0,1)、(0,3)，所以 [0]_S={0,1,3}；[3]_S={0,3}；[1]_S={0,1,4}。[0]_S∩[3]_S={0,3}≠∅ 但 [0]_S≠[3]_S；[0]_S∩[1]_S={0,1} 但兩者不等。這些集合不構成分割，S07 的論證正是在需要傳遞性處斷掉。",
          "en": "If 'classes' are forced from the pairs of S: the pairs starting with 0 are (0,0), (0,1), (0,3), so [0]_S={0,1,3}; [3]_S={0,3}; [1]_S={0,1,4}. [0]_S∩[3]_S={0,3}≠∅ yet [0]_S≠[3]_S, and [0]_S∩[1]_S={0,1} yet they differ. These sets do not form a partition, and the argument of S07 breaks exactly where it needs transitivity."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "set_builder"
          ],
          "explanation": {
            "zh-Hant": "集合可由 S 的列表逐一重算；相交而不相等的兩個集合直接違反「相等或不交」。",
            "en": "The sets can be recomputed from the list of S; two sets that meet without being equal directly violate 'equal or disjoint'."
          }
        },
        "leaf_basis": "set_builder + set_operations_def"
      },
      {
        "id": "S11",
        "uses": [
          "S01",
          "S05"
        ],
        "before_tex": "\\text{visual grouping}\\ \\{0,1,2\\},\\ \\{3,4,5\\}",
        "after_tex": "\\{0,1,2\\}\\notin K(R)\\ \\text{since}\\ (0,1)\\notin R",
        "relation": "implies",
        "rule": "mko_eqclass_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依數字相鄰的外觀分群會得到 {0,1,2}、{3,4,5}。但 (0,1)∉R，所以 1∉[0]={0,3}，{0,1,2} 不是類。類只能由同一關係的對求出（MKO 假設 a2）。",
          "en": "Grouping by how close the numbers look gives {0,1,2} and {3,4,5}. But (0,1)∉R, so 1∉[0]={0,3}, and {0,1,2} is not a class. Classes come only from the pairs of the same relation (MKO assumption a2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S05"
          ],
          "explanation": {
            "zh-Hant": "一個不在 R 中的對就足以否定這個猜測的分群。",
            "en": "One pair not in R suffices to refute the guessed grouping."
          }
        },
        "leaf_basis": "set_builder + quantifier_negation"
      },
      {
        "id": "S12",
        "uses": [
          "S04",
          "S05",
          "S07",
          "S08",
          "S10"
        ],
        "before_tex": "\\text{S04, S05, S07, S08, S10}",
        "after_tex": "K(R)=\\{\\{0,3\\},\\{1,4\\},\\{2,5\\}\\};\\ S\\ \\text{yields no classes}",
        "relation": "implies",
        "rule": "mko_eqclass_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "R 經完整認證（S04），其類為 {0,3}、{1,4}、{2,5}（S05），相等或不交並覆蓋 X（S07、S08）。S 未通過傳遞性，所得集合相交而不相等（S10），不能作為等價類。",
          "en": "R is fully certified (S04), its classes are {0,3}, {1,4}, {2,5} (S05), and they are equal or disjoint and cover X (S07, S08). S fails transitivity, and the sets it yields meet without being equal (S10), so they cannot serve as equivalence classes."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S04",
            "S05",
            "S07",
            "S08",
            "S10"
          ],
          "explanation": {
            "zh-Hant": "目標 G：本稿目標的認證、類、性質與對照。足夠條件 H：S04、S05、S07、S08、S10；H⇒G 由合取。H 由 G1 到 G4 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: the certification, classes, properties and contrast stated in the goal. Sufficient condition H: S04, S05, S07, S08 and S10; H⇒G by conjunction. H is proved forwards from G1 to G4; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "明列的 12 個對恰為「除以 3 餘數相同」的全部對，因此 R 完整通過反身、對稱、傳遞。其類為 [0]=[3]={0,3}、[1]=[4]={1,4}、[2]=[5]={2,5}；aRb ⇒ [a]=[b]，任兩類相等或不交，且覆蓋 X，K(R)={{0,3},{1,4},{2,5}}。S=R∪{(0,1),(1,0)} 反身且對稱但不傳遞（(3,0)、(0,1)∈S，(3,1)∉S），其「類」{0,1,3} 與 {0,3} 相交而不相等，不能作為等價類。",
      "en": "The 12 listed pairs are exactly all pairs with equal remainders on division by 3, so R fully passes reflexivity, symmetry and transitivity. Its classes are [0]=[3]={0,3}, [1]=[4]={1,4}, [2]=[5]={2,5}; aRb ⇒ [a]=[b], any two classes are equal or disjoint, and they cover X, so K(R)={{0,3},{1,4},{2,5}}. S=R∪{(0,1),(1,0)} is reflexive and symmetric but not transitive ((3,0), (0,1)∈S, (3,1)∉S), and its 'classes' {0,1,3} and {0,3} meet without being equal, so they cannot be equivalence classes."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "S 反身且對稱，所以它的「類」分割 X",
          "en": "S is reflexive and symmetric, so its 'classes' partition X"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "S 不傳遞；[0]_S={0,1,3} 與 [3]_S={0,3} 相交而不相等（S10）。",
          "en": "S is not transitive; [0]_S={0,1,3} and [3]_S={0,3} meet without being equal (S10)."
        }
      },
      {
        "claim": {
          "zh-Hant": "{0,1,3} 與 {0,3} 可以同時保留為等價類",
          "en": "{0,1,3} and {0,3} can both be kept as equivalence classes"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "等價類相等或不交；兩者相交於 {0,3} 卻不相等。",
          "en": "Equivalence classes are equal or disjoint; these meet in {0,3} but differ."
        }
      },
      {
        "claim": {
          "zh-Hant": "{0,1,2} 是一個類，因為它們是連續的數",
          "en": "{0,1,2} is a class because they are consecutive numbers"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(0,1)∉R，所以 1∉[0]（S11）。",
          "en": "(0,1)∉R, so 1∉[0] (S11)."
        }
      },
      {
        "claim": {
          "zh-Hant": "抽查三元組 (0,3,0) 與 (1,4,1) 都符合傳遞，所以任何這樣的關係都已認證",
          "en": "The sampled triples (0,3,0) and (1,4,1) satisfy transitivity, so any such relation is certified"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "S 也通過這兩個三元組，但 (3,0,1) 失敗；認證要對全部可接續的對（S04 以完整描述一次處理 24 個）。",
          "en": "S also passes these two triples, yet (3,0,1) fails; certification must cover all composable pairs (S04 handles all 24 at once through the complete description)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "set_builder",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。",
          "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x)."
        }
      },
      {
        "id": "set_extensionality",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合相等：A=B 當且僅當 A⊆B 且 B⊆A，即兩者有相同元素。",
          "en": "Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements."
        }
      },
      {
        "id": "subset_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "子集：A⊆B 表示 A 的每個元素都是 B 的元素，即 ∀x (x∈A⇒x∈B)。",
          "en": "Subset: A⊆B means every element of A is an element of B, i.e. ∀x (x∈A⇒x∈B)."
        }
      },
      {
        "id": "division_algorithm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "帶餘除法：對整數 a 與正整數 b，存在唯一的整數 q、r 使 a=bq+r 且 0≤r<b。",
          "en": "Division algorithm: for an integer a and a positive integer b there are unique integers q, r with a=bq+r and 0≤r<b."
        }
      },
      {
        "id": "eq_refl_symm_trans",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。",
          "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w."
        }
      },
      {
        "id": "finite_counting_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限計數：互斥有限集合 A、B 有 |A∪B|=|A|+|B|；|A×B|=|A|·|B|；有限集之間存在雙射 ⇔ 元素個數相同。",
          "en": "Finite counting: for disjoint finite sets A, B, |A∪B|=|A|+|B|; |A×B|=|A|·|B|; there is a bijection between finite sets ⇔ they have the same number of elements."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "contrapositive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逆否命題：P⇒Q 與 ¬Q⇒¬P 等價。逆命題 Q⇒P 一般不與 P⇒Q 等價。",
          "en": "Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q."
        }
      },
      {
        "id": "set_operations_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合運算：x∈A∪B ⇔ x∈A 或 x∈B；x∈A∩B ⇔ x∈A 且 x∈B；x∈A∖B ⇔ x∈A 且 x∉B；在全集 U 中 Aᶜ=U∖A。",
          "en": "Set operations: x∈A∪B ⇔ x∈A or x∈B; x∈A∩B ⇔ x∈A and x∈B; x∈A∖B ⇔ x∈A and x∉B; in a universe U, Aᶜ=U∖A."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "mko_eqclass_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-equivalence-class",
          "sha256": "6ead5a95cd13a163d636e78db505bb8f7aad6744f9e25afee8148c838a3a44d9",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "給完整有限載體 X 及已認證等價關係 R，[a]={x 屬 X:aRx}。若 aRb 則 [a]=[b]；任兩等價類相等或不交。不得從抽樣關係或未通過等價公理的 relation 形成 canonical classes。",
          "en": "Given a complete finite carrier X and a certified equivalence relation R, define [a]={x in X:aRx}. If aRb then [a]=[b]; any two equivalence classes are equal or disjoint. Sampled relations or relations that fail the equivalence axioms must not be used to form canonical classes."
        }
      },
      {
        "id": "equiv_rel_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等價關係：反身（對所有 a，aRa）、對稱（aRb⇒bRa）、傳遞（aRb 且 bRc ⇒ aRc）。",
          "en": "Equivalence relation: reflexive (aRa for all a), symmetric (aRb⇒bRa) and transitive (aRb and bRc ⇒ aRc)."
        }
      },
      {
        "id": "L_rel_desc",
        "kind": "lemma",
        "proved_in": [
          "S01.1",
          "S01.2"
        ],
        "statement": {
          "zh-Hant": "R={(a,b)∈X×X ∣ r(a)=r(b)}，其中 r(a) 是 a 除以 3 的餘數。",
          "en": "R={(a,b)∈X×X ∣ r(a)=r(b)}, where r(a) is the remainder of a on division by 3."
        }
      },
      {
        "id": "L_same_class",
        "kind": "lemma",
        "proved_in": [
          "S06.1",
          "S06.2"
        ],
        "statement": {
          "zh-Hant": "對等價關係 R：aRb ⇒ [a]=[b]。",
          "en": "For an equivalence relation R: aRb ⇒ [a]=[b]."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S01.2",
        "lhs": "3*2**2",
        "rhs": "12"
      },
      {
        "id": "N2",
        "step": "S08",
        "lhs": "2+2+2",
        "rhs": "6"
      },
      {
        "id": "N3",
        "step": "S09",
        "lhs": "3%3",
        "relation": "!=",
        "rhs": "1%3"
      },
      {
        "id": "N4",
        "step": "S01.1",
        "lhs": "3%3",
        "rhs": "0%3"
      },
      {
        "id": "N5",
        "step": "S01.1",
        "lhs": "4%3",
        "rhs": "1%3"
      },
      {
        "id": "N6",
        "step": "S01.1",
        "lhs": "5%3",
        "rhs": "2%3"
      },
      {
        "id": "N7",
        "lhs": "12+2",
        "rhs": "14"
      },
      {
        "id": "N8",
        "step": "S04",
        "lhs": "6*2*2",
        "rhs": "24"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只在 MKO 範圍內工作：載體 0 到 5 與完整關係對集合明示，且先完整通過反身、對稱、傳遞（假設 a1）；類使用同一關係與載體，不以圖像分群猜測（假設 a2）。不建立商集節點或商環；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: the carrier 0 to 5 and the complete set of relation pairs are explicit and first pass full reflexivity, symmetry and transitivity checks (assumption a1); the classes use that same relation and carrier, not a visual guess at clustering (assumption a2). It does not set up the quotient-set node or quotient rings, and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "X",
        "載體 {0,1,2,3,4,5}",
        "the carrier {0,1,2,3,4,5}"
      ],
      [
        "aRb",
        "(a,b)∈R",
        "(a,b)∈R"
      ],
      [
        "r(a)",
        "a 除以 3 的餘數",
        "the remainder of a on division by 3"
      ],
      [
        "[a]",
        "a 的等價類 {x∈X ∣ aRx}",
        "the class {x∈X ∣ aRx} of a"
      ],
      [
        "K(R)",
        "全部等價類所成的集合",
        "the set of all equivalence classes"
      ]
    ],
    "md_error_controls": [
      {
        "title": "先認證等價關係，再形成等價類 / Certify equivalence before forming equivalence classes",
        "zh": "MKO 指出的誤解：從未認證的 relation 直接分群。S 看起來與 R 只差兩個對，反身與對稱都成立，但傳遞性失敗（S09），所得集合不構成分割（S10）。必須先完整通過三個公理（S02 到 S04），才求類。",
        "en": "The MKO's misconception: grouping directly from an uncertified relation. S differs from R by only two pairs and is reflexive and symmetric, yet transitivity fails (S09) and the sets it yields do not form a partition (S10). All three axioms must be passed completely (S02 to S04) before forming classes."
      },
      {
        "title": "不保留相交而不相等的類 / Do not keep classes that meet without being equal",
        "zh": "MKO 的第二個誤解。對等價關係，S07 證明相交的兩類必相等；若手上有 {0,1,3} 與 {0,3} 這樣相交卻不相等的集合，表示關係不是等價關係或計算有誤，應回頭檢查，而不是兩者都保留。",
        "en": "The MKO's second misconception. For an equivalence relation, S07 proves that two classes that meet are equal; holding sets like {0,1,3} and {0,3}, which meet but differ, signals that the relation is not an equivalence or that a computation is wrong, and calls for a recheck rather than keeping both."
      },
      {
        "title": "不以外觀分群 / Do not group by appearance",
        "zh": "MKO 假設 a2：類使用同一關係，不以圖像分群猜測。數字相鄰的 {0,1,2} 不是類，因為 (0,1)∉R（S11）。",
        "en": "MKO assumption a2: classes use the same relation, not a visual guess at clustering. The consecutive numbers {0,1,2} do not form a class, because (0,1)∉R (S11)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "商集 X/R 作為獨立節點及有限環的商環（MKO 的下一步）不在本稿範圍；本稿只求出 K(R) 並證明其分割性質。",
        "en": "The quotient set X/R as a node of its own and quotient rings of finite rings (the MKO's next step) are outside this guide; it only computes K(R) and proves its partition properties."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "a49fd6fdffc0b18e8964d08495131e69cf39153f1e976af4811cc47509502b53",
    "original_source_sha256": "a49fd6fdffc0b18e8964d08495131e69cf39153f1e976af4811cc47509502b53",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}