{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-equivalence-relation-01",
  "mko_id": "mko-equivalence-relation",
  "title": {
    "zh-Hant": "等價關係：在完整有限關係表上逐項核對三公理並輸出等價類",
    "en": "Equivalence relation: checking the three axioms on a complete finite relation and emitting the classes"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-equivalence-relation-01",
    "source": {
      "mko_id": "mko-equivalence-relation",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "ceccd69c4881e4b2eba57eb1a7c6ab6c6616ab09fa42eb8591999522a5948fd8",
      "english_sha256": "f2bdfa53ace4565feb3a764d528c99663030234943ed2139aa6de2f8e878eb57"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "cartesian_product_def",
      "set_extensionality",
      "set_operations_def",
      "relation_def",
      "finite_forall"
    ],
    "goal": {
      "zh-Hant": "在 X={0,1,2,3,4} 上，關係 R 完整列出為 13 個有序對：(0,0)、(1,1)、(2,2)、(3,3)、(4,4)、(0,2)、(2,0)、(1,3)、(3,1)、(1,4)、(4,1)、(3,4)、(4,3)。逐項核對反身、對稱、傳遞，證明 R 是等價關係並輸出等價類；再說明去掉 (3,4)、(4,3) 後的 R′ 雖反身且對稱，卻不是等價關係。",
      "en": "On X={0,1,2,3,4} the relation R is listed completely as 13 ordered pairs: (0,0), (1,1), (2,2), (3,3), (4,4), (0,2), (2,0), (1,3), (3,1), (1,4), (4,1), (3,4), (4,3). Check reflexivity, symmetry and transitivity item by item, prove that R is an equivalence relation and emit its classes; then show that R′, obtained by removing (3,4) and (4,3), is reflexive and symmetric but not an equivalence relation."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "X={0,1,2,3,4}；R={(0,0),(1,1),(2,2),(3,3),(4,4),(0,2),(2,0),(1,3),(3,1),(1,4),(4,1),(3,4),(4,3)}，13 個互異有序對，完整列出。",
          "en": "X={0,1,2,3,4}; R={(0,0),(1,1),(2,2),(3,3),(4,4),(0,2),(2,0),(1,3),(3,1),(1,4),(4,1),(3,4),(4,3)}, 13 distinct ordered pairs, listed completely."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "對照：R′ 是從 R 去掉 (3,4) 與 (4,3) 後的 11 個有序對。",
          "en": "For comparison: R′ consists of the 11 ordered pairs left after removing (3,4) and (4,3) from R."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "relation_def"
        ],
        "before_tex": "R\\ \\text{is an equivalence relation on}\\ X",
        "after_tex": "\\forall x\\,(xRx)\\ \\land\\ \\forall x,y\\,(xRy\\Rightarrow yRx)\\ \\land\\ \\forall x,y,z\\,(xRy\\land yRz\\Rightarrow xRz)",
        "relation": "definition",
        "rule": "mko_equiv_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 展開定義：三項公理反身、對稱、傳遞都必須成立。每一項都是對有限集合的全稱命題，所以可以完整核對；任何一項有一個反例，R 就不是等價關係。",
          "en": "Unfold the definition of the MKO: all three axioms, reflexivity, symmetry and transitivity, must hold. Each is a universal statement over a finite set and can be checked completely; one counterexample to any of them means R is not an equivalence relation."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_equiv_def",
            "relation_def"
          ],
          "explanation": {
            "zh-Hant": "此步只展開定義，列出三項責任，不推出新事實。",
            "en": "This step only unfolds the definition and lists the three obligations; it derives no new fact."
          }
        },
        "leaf_basis": "mko_equiv_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "G1",
          "finite_forall"
        ],
        "before_tex": "\\forall x\\in X:\\ (x,x)\\in R",
        "after_tex": "(0,0),(1,1),(2,2),(3,3),(4,4)\\in R",
        "relation": "equivalent",
        "rule": "finite_forall",
        "conditions": [],
        "explanation": {
          "zh-Hant": "X 有五個元素，反身性等價於五個對角對都在 R 中。逐一查 G1 的列表：五個都在。反身性成立。",
          "en": "X has five elements, so reflexivity is equivalent to all five diagonal pairs lying in R. Look each one up in the list of G1: all five are there. Reflexivity holds."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "finite_forall"
          ],
          "explanation": {
            "zh-Hant": "全稱推出每個實例；五個實例涵蓋 X 全部元素時，反之亦然。",
            "en": "The universal gives each instance; since the five instances cover all of X, the converse holds as well."
          }
        },
        "leaf_basis": "finite_forall + relation_def"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "G1",
          "finite_forall"
        ],
        "before_tex": "\\forall (x,y)\\in R:\\ (y,x)\\in R",
        "after_tex": "(0,2)\\leftrightarrow(2,0),\\ (1,3)\\leftrightarrow(3,1),\\ (1,4)\\leftrightarrow(4,1),\\ (3,4)\\leftrightarrow(4,3);\\ (x,x)\\ \\text{self-reverse}",
        "relation": "equivalent",
        "rule": "finite_forall",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對稱性只需檢查 R 中的每一對（xRy 為假時蘊含自動成立）。13 對中，5 個對角對的反向是自己；其餘 8 對成 4 組互為反向，每組兩對都在列表中。對稱性成立。",
          "en": "Symmetry needs checking only on the pairs of R (when xRy is false the implication holds automatically). Of the 13 pairs, the 5 diagonal pairs are their own reverses; the other 8 form 4 pairs of mutually reversed ordered pairs, with both members of each pair listed. Symmetry holds."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "finite_forall",
            "relation_def"
          ],
          "explanation": {
            "zh-Hant": "13 對全部配對完畢（5+2·4=13），沒有遺漏，所以逐項成立與全稱成立等價。",
            "en": "All 13 pairs are accounted for (5+2·4=13), with none missed, so the itemwise check and the universal statement are equivalent."
          }
        },
        "leaf_basis": "finite_forall + relation_def + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "G1"
        ],
        "before_tex": "X=\\{0,1,2,3,4\\}",
        "after_tex": "B_1=\\{0,2\\},\\ B_2=\\{1,3,4\\},\\ T=(B_1\\times B_1)\\cup(B_2\\times B_2)",
        "relation": "definition",
        "rule": "def_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "為了一次處理全部可串接對，引入兩個候選區塊與區塊關係 T。B_1 與 B_2 不交，聯集是 X。這只是命名，T 是否等於 R 要在 S05 到 S07 證明，不能只看圖猜。",
          "en": "To treat all composable pairs at once, introduce two candidate blocks and the block relation T. B_1 and B_2 are disjoint with union X. This is only naming; whether T equals R must be proved in S05 to S07, not guessed from a picture."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "def_intro"
          ],
          "explanation": {
            "zh-Hant": "定義不推出事實；B_1∩B_2=∅ 可直接看出，之後的引理需要這一點。",
            "en": "A definition derives no fact; B_1∩B_2=∅ is immediate and is needed by the lemma later."
          }
        },
        "leaf_basis": "def_intro + set_operations_def"
      },
      {
        "id": "S05",
        "uses": [
          "S04",
          "cartesian_product_def"
        ],
        "before_tex": "(B_1\\times B_1)\\cup(B_2\\times B_2)",
        "after_tex": "\\{(0,0),(0,2),(2,0),(2,2)\\}\\cup\\{(1,1),(1,3),(1,4),(3,1),(3,3),(3,4),(4,1),(4,3),(4,4)\\}",
        "relation": "equivalent",
        "rule": "cartesian_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依笛卡兒積定義列出：B_1×B_1 有 2·2=4 對，B_2×B_2 有 3·3=9 對。兩個積不交（區塊不交），所以 |T|=4+9=13。",
          "en": "List by the definition of the Cartesian product: B_1×B_1 has 2·2=4 pairs and B_2×B_2 has 3·3=9 pairs. The two products are disjoint (the blocks are disjoint), so |T|=4+9=13."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "cartesian_product_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "列出的集合與定義的集合逐元素相同，雙向成立。",
            "en": "The listed set and the defined set agree element by element, so the step holds both ways."
          }
        },
        "leaf_basis": "cartesian_product_def + finite_counting_basics"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "S05"
        ],
        "before_tex": "(x,y)\\in R",
        "after_tex": "R\\subseteq T",
        "relation": "implies",
        "rule": "subset_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐一檢查 R 的 13 對：五個對角對各在自己的區塊積中；(0,2)、(2,0) 兩端都在 B_1；(1,3)、(3,1)、(1,4)、(4,1)、(3,4)、(4,3) 兩端都在 B_2。每一對都在 T 中。",
          "en": "Check the 13 pairs of R one by one: each diagonal pair lies in its own block product; (0,2) and (2,0) have both ends in B_1; (1,3), (3,1), (1,4), (4,1), (3,4), (4,3) have both ends in B_2. Every pair lies in T."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S05",
            "subset_def"
          ],
          "explanation": {
            "zh-Hant": "只證 R⊆T；反向包含在 S07 另外證明。",
            "en": "Only R⊆T is proved here; the reverse inclusion is proved separately in S07."
          }
        },
        "leaf_basis": "subset_def + finite_forall"
      },
      {
        "id": "S07",
        "uses": [
          "S05",
          "S06",
          "G1"
        ],
        "before_tex": "R\\subseteq T,\\ (x,y)\\in T",
        "after_tex": "T\\subseteq R,\\ \\text{hence}\\ R=T",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S05 列出的 T 的 13 對，每一對都出現在 G1 的 R 列表中，所以 T⊆R。結合 S06 的 R⊆T，依集合外延性 R=T。（另一個核對：兩者都恰有 13 個互異元素。）",
          "en": "Each of the 13 pairs of T listed in S05 appears in the list of R in G1, so T⊆R. With R⊆T from S06, set extensionality gives R=T. (A further check: both have exactly 13 distinct elements.)"
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "兩個包含都由逐元素檢查得到；只有兩個方向都成立才能宣稱相等。",
            "en": "Both inclusions come from element-by-element checks; equality may be claimed only because both directions hold."
          }
        },
        "leaf_basis": "set_extensionality + subset_def + finite_counting_basics"
      },
      {
        "id": "S08",
        "uses": [
          "S04",
          "S07"
        ],
        "before_tex": "R=T=(B_1\\times B_1)\\cup(B_2\\times B_2),\\ B_1\\cap B_2=\\varnothing",
        "after_tex": "xRy\\land yRz\\Rightarrow xRz\\ \\text{for all}\\ x,y,z\\in X",
        "relation": "theorem_application",
        "rule": "L_block_trans",
        "conditions": [
          "B_1\\cap B_2=\\varnothing"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_block_trans（S08.1 到 S08.3 證明），T 是傳遞的；因 R=T，R 也是傳遞的。這一次涵蓋全部 35 個可串接的 (x,y)、(y,z)，不是抽樣。",
          "en": "By Lemma L_block_trans (proved in S08.1 to S08.3), T is transitive; since R=T, R is transitive too. This covers all 35 composable chains (x,y), (y,z) at once, not a sample."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "L_block_trans"
          ],
          "explanation": {
            "zh-Hant": "需要 R=T 完全相等（S07）與區塊不交（S04）；若只有 R⊆T，T 的傳遞性不能轉移到 R。",
            "en": "It needs R=T exactly (S07) and disjoint blocks (S04); with only R⊆T, transitivity of T would not transfer to R."
          }
        },
        "leaf_basis": "L_block_trans + eq_replace"
      },
      {
        "id": "S08.1",
        "uses": [
          "cartesian_product_def",
          "set_operations_def"
        ],
        "before_tex": "(x,y)\\in T,\\ (y,z)\\in T",
        "after_tex": "(x,y)\\in B_i\\times B_i,\\ (y,z)\\in B_j\\times B_j,\\ y\\in B_i\\cap B_j",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取任意 x、y、z 使 (x,y)、(y,z) 都在 T 中。T 是兩個積的聯集，所以有某個 i 使 (x,y)∈B_i×B_i，有某個 j 使 (y,z)∈B_j×B_j。由笛卡兒積定義，y 既在 B_i（第二座標）也在 B_j（第一座標）。",
          "en": "Take any x, y, z with (x,y) and (y,z) in T. T is the union of two products, so (x,y)∈B_i×B_i for some i and (y,z)∈B_j×B_j for some j. By the definition of the Cartesian product, y lies in B_i (as a second coordinate) and in B_j (as a first coordinate)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "set_operations_def",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "只是展開聯集與笛卡兒積的成員資格；此子步不使用 L_block_trans。",
            "en": "It only unfolds membership in a union and a Cartesian product; this sub-step does not use L_block_trans."
          }
        },
        "leaf_basis": "set_operations_def + cartesian_product_def + exists_elim"
      },
      {
        "id": "S08.2",
        "uses": [
          "S08.1",
          "contradiction"
        ],
        "before_tex": "y\\in B_i\\cap B_j,\\ B_1\\cap B_2=\\varnothing",
        "after_tex": "i=j",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [
          "B_1\\cap B_2=\\varnothing"
        ],
        "explanation": {
          "zh-Hant": "若 i≠j，則 {i,j}={1,2}，y∈B_1∩B_2=∅，矛盾。所以 i=j。這裡用到區塊兩兩不交；若區塊相交，這一步會失敗。",
          "en": "If i≠j then {i,j}={1,2} and y∈B_1∩B_2=∅, a contradiction. So i=j. This uses that the blocks are disjoint; with overlapping blocks the step would fail."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08.1",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "反證法的前提是區塊不交；此子步不使用 L_block_trans。",
            "en": "The argument by contradiction rests on disjoint blocks; this sub-step does not use L_block_trans."
          }
        },
        "leaf_basis": "contradiction + set_operations_def"
      },
      {
        "id": "S08.3",
        "uses": [
          "S08.1",
          "S08.2",
          "forall_intro"
        ],
        "before_tex": "x\\in B_i,\\ z\\in B_j,\\ i=j",
        "after_tex": "(x,z)\\in B_i\\times B_i\\subseteq T",
        "relation": "implies",
        "rule": "cartesian_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由 S08.1，x∈B_i、z∈B_j，而 S08.2 給 i=j，所以 x、z 都在 B_i，(x,z)∈B_i×B_i⊆T。x、y、z 是任意取的，依全稱推廣，T 是傳遞的。S08.1 到 S08.3 合起來證得引理 L_block_trans。",
          "en": "By S08.1, x∈B_i and z∈B_j, and S08.2 gives i=j, so x and z both lie in B_i and (x,z)∈B_i×B_i⊆T. Since x, y, z were arbitrary, generalization shows T is transitive. S08.1 to S08.3 together prove Lemma L_block_trans."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08.2",
            "cartesian_product_def",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "證明沒有對 x、y、z 加任何額外假設，所以可以全稱推廣；此子步不使用 L_block_trans。",
            "en": "The proof puts no extra assumption on x, y, z, so generalization applies; this sub-step does not use L_block_trans."
          }
        },
        "leaf_basis": "cartesian_product_def + forall_intro"
      },
      {
        "id": "S09",
        "uses": [
          "S01",
          "S02",
          "S03",
          "S08"
        ],
        "before_tex": "\\text{reflexive, symmetric, transitive}",
        "after_tex": "R\\ \\text{is an equivalence relation on}\\ X",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "反身（S02）、對稱（S03）、傳遞（S08）三項都通過，依 S01 的定義 R 是等價關係。到這裡才允許輸出等價類。",
          "en": "Reflexivity (S02), symmetry (S03) and transitivity (S08) all pass, so by the definition in S01, R is an equivalence relation. Only now may the classes be emitted."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S01",
            "S02",
            "S03",
            "S08"
          ],
          "explanation": {
            "zh-Hant": "目標 G：R 是等價關係。足夠條件 H：三項公理都成立；H⇒G 由定義。H 由 S02、S03、S08 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: R is an equivalence relation. Sufficient condition H: all three axioms hold; H⇒G by the definition. H is proved forwards in S02, S03 and S08; G is never used as a premise."
          }
        },
        "leaf_basis": "mko_equiv_def + and_intro_elim"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "G1",
          "eq_class_def"
        ],
        "before_tex": "[x]=\\{y\\in X: xRy\\}",
        "after_tex": "[0]=[2]=\\{0,2\\},\\ [1]=[3]=[4]=\\{1,3,4\\}",
        "relation": "implies",
        "rule": "eq_class_def",
        "conditions": [
          "R\\ \\text{equivalence relation (S09)}"
        ],
        "explanation": {
          "zh-Hant": "讀出每個 x 對應的行：以 0 開頭的對是 (0,0)、(0,2)，所以 [0]={0,2}；以 2 開頭的是 (2,2)、(2,0)，[2]={0,2}；以 1 開頭的是 (1,1)、(1,3)、(1,4)，[1]={1,3,4}；以 3 開頭的是 (3,3)、(3,1)、(3,4)；以 4 開頭的是 (4,4)、(4,1)、(4,3)，所以 [3]=[4]={1,3,4}。",
          "en": "Read off the row of each x: the pairs starting with 0 are (0,0), (0,2), so [0]={0,2}; with 2, (2,2), (2,0), so [2]={0,2}; with 1, (1,1), (1,3), (1,4), so [1]={1,3,4}; with 3, (3,3), (3,1), (3,4); with 4, (4,4), (4,1), (4,3); so [3]=[4]={1,3,4}."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "eq_class_def"
          ],
          "explanation": {
            "zh-Hant": "輸出等價類的前提是 S09；若三項未全部通過（如 R′），按 MKO 不輸出。",
            "en": "Emitting classes requires S09; if not all three checks passed (as for R′), the MKO says no classes are emitted."
          }
        },
        "leaf_basis": "eq_class_def + set_builder + relation_def"
      },
      {
        "id": "S11",
        "uses": [
          "S10"
        ],
        "before_tex": "\\{0,2\\},\\ \\{1,3,4\\}",
        "after_tex": "\\{0,2\\}\\cap\\{1,3,4\\}=\\varnothing,\\ \\{0,2\\}\\cup\\{1,3,4\\}=X",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個相異的等價類不交，聯集是 X，2+3=5=|X|。它們恰是 S04 的兩個區塊：區塊是由驗證後的關係得到的，不是先由圖猜出的。",
          "en": "The two distinct classes are disjoint with union X, and 2+3=5=|X|. They are exactly the blocks of S04: the blocks come out of the verified relation, not from a guess based on a picture."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "只對這兩個具體集合做交集與聯集的計算；不在此一般地證明等價類必構成分割。",
            "en": "Only the intersection and union of these two concrete sets are computed; the general theorem that classes form a partition is not proved here."
          }
        },
        "leaf_basis": "set_operations_def + finite_counting_basics"
      },
      {
        "id": "S12",
        "uses": [
          "G2",
          "S02",
          "S03"
        ],
        "before_tex": "R'=R\\setminus\\{(3,4),(4,3)\\}",
        "after_tex": "R'\\ \\text{reflexive and symmetric}",
        "relation": "implies",
        "rule": "finite_forall",
        "conditions": [],
        "explanation": {
          "zh-Hant": "R′ 保留全部五個對角對，所以仍反身；去掉的兩對互為反向，其餘 3 組反向配對 (0,2)/(2,0)、(1,3)/(3,1)、(1,4)/(4,1) 都完整保留，所以仍對稱。",
          "en": "R′ keeps all five diagonal pairs, so it is still reflexive; the two removed pairs are mutual reverses, and the remaining 3 pairs of mutually reversed ordered pairs (0,2)/(2,0), (1,3)/(3,1), (1,4)/(4,1) are kept whole, so it is still symmetric."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "finite_forall"
          ],
          "explanation": {
            "zh-Hant": "R′ 的 11 對都已檢查（5+2·3=11）；此步不宣稱 R′ 傳遞。",
            "en": "All 11 pairs of R′ are checked (5+2·3=11); the step does not claim that R′ is transitive."
          }
        },
        "leaf_basis": "finite_forall + relation_def + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "G2",
          "S12"
        ],
        "before_tex": "(3,1)\\in R',\\ (1,4)\\in R'",
        "after_tex": "(3,4)\\notin R'\\Rightarrow R'\\ \\text{not transitive}",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取 x=3、y=1、z=4：3R′1 且 1R′4，但 (3,4) 已被去掉，3R′4 不成立。傳遞性是全稱命題，一個反例就否證它。",
          "en": "Take x=3, y=1, z=4: 3R′1 and 1R′4, but (3,4) was removed, so 3R′4 fails. Transitivity is a universal statement, and one counterexample refutes it."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "需要 R′ 的列表完整，才能確定 (3,4)∉R′。只要一個反例，不需檢查其他串接。",
            "en": "The complete list of R′ is needed to know (3,4)∉R′. One counterexample suffices; no other chains need checking."
          }
        },
        "leaf_basis": "contradiction + quantifier_negation + relation_def"
      },
      {
        "id": "S14",
        "uses": [
          "S01",
          "S12",
          "S13",
          "eq_class_def"
        ],
        "before_tex": "R'\\ \\text{reflexive, symmetric, not transitive}",
        "after_tex": "R'\\ \\text{not an equivalence relation; no classes emitted}",
        "relation": "implies",
        "rule": "mko_equiv_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "缺少傳遞性，依 S01 R′ 不是等價關係，不輸出等價類。若硬算 {y : xR′y}，會得到 {1,3}（x=3）與 {1,4}（x=4），兩者相交於 1 卻不相等，不構成分割。這顯示反身加對稱不足，也顯示看圖分群不可靠。",
          "en": "Without transitivity, R′ is not an equivalence relation by S01, and no classes are emitted. Forcing the computation of {y : xR′y} gives {1,3} for x=3 and {1,4} for x=4, which meet in 1 without being equal, so they do not form a partition. This shows that reflexive plus symmetric is not enough, and that grouping by eye is unreliable."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S01",
            "S13"
          ],
          "explanation": {
            "zh-Hant": "目標 G：R′ 不是等價關係。足夠條件 H：某一項公理失敗；H⇒G 由定義（三項缺一不可）。H 由 S13 正向給出。",
            "en": "Goal G: R′ is not an equivalence relation. Sufficient condition H: some axiom fails; H⇒G by the definition (all three are required). H is supplied forwards by S13."
          }
        },
        "leaf_basis": "mko_equiv_def + de_morgan_logic"
      }
    ],
    "result": {
      "zh-Hant": "R 反身（5 個對角對）、對稱（4 組反向配對）且傳遞（R 等於區塊關係 (B_1×B_1)∪(B_2×B_2)，由引理 L_block_trans 涵蓋全部 35 個可串接對），所以 R 是等價關係，等價類為 {0,2} 與 {1,3,4}。R′ 反身且對稱，但 3R′1、1R′4 而非 3R′4，所以不是等價關係，不輸出等價類。",
      "en": "R is reflexive (5 diagonal pairs), symmetric (4 pairs of mutually reversed ordered pairs) and transitive (R equals the block relation (B_1×B_1)∪(B_2×B_2), and Lemma L_block_trans covers all 35 composable chains), so R is an equivalence relation with classes {0,2} and {1,3,4}. R′ is reflexive and symmetric, but 3R′1 and 1R′4 without 3R′4, so it is not an equivalence relation and no classes are emitted."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "R′ 反身且對稱，所以 R′ 是等價關係",
          "en": "R′ is reflexive and symmetric, so R′ is an equivalence relation"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "3R′1、1R′4，但 (3,4)∉R′，傳遞性失敗。",
          "en": "3R′1 and 1R′4 but (3,4)∉R′, so transitivity fails."
        }
      },
      {
        "claim": {
          "zh-Hant": "對稱且傳遞的關係一定反身",
          "en": "A symmetric and transitive relation is always reflexive"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "在 {0,1,2} 上，S={0,1}×{0,1} 對稱且傳遞（單一區塊），但 (2,2)∉S。",
          "en": "On {0,1,2}, S={0,1}×{0,1} is symmetric and transitive (a single block), but (2,2)∉S."
        }
      },
      {
        "claim": {
          "zh-Hant": "R′ 的串接 (0,2)、(2,0) 推出 (0,0)∈R′，所以 R′ 傳遞",
          "en": "In R′ the chain (0,2), (2,0) gives (0,0)∈R′, so R′ is transitive"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "一組串接成立不代表全部成立；(3,1)、(1,4) 這組失敗。",
          "en": "One chain succeeding does not mean all do; the chain (3,1), (1,4) fails."
        }
      },
      {
        "claim": {
          "zh-Hant": "R′ 的等價類是 {0,2} 與 {1,3,4}",
          "en": "The equivalence classes of R′ are {0,2} and {1,3,4}"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "R′ 不是等價關係，不輸出等價類；硬算得到的 {1,3} 與 {1,4} 也不構成分割。",
          "en": "R′ is not an equivalence relation, so no classes are emitted; the forced sets {1,3} and {1,4} do not form a partition either."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "cartesian_product_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "笛卡兒積：A×B 是所有有序對 (a,b)（a∈A、b∈B）的集合；(a,b)=(c,d) ⇔ a=c 且 b=d。",
          "en": "Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d."
        }
      },
      {
        "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": "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": "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": "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": "def_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。",
          "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable."
        }
      },
      {
        "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": "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": "eq_replace",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。",
          "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true."
        }
      },
      {
        "id": "exists_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃消去：由 ∃x∈D, P(x)，可取一個新名字 a 表示某個滿足 P(a) 的元素；只要最後結論不含 a，就可用 a 推理。",
          "en": "∃-elimination: from ∃x∈D, P(x) we may name some element a with P(a) by a fresh name; reasoning with a is allowed as long as the final conclusion does not mention a."
        }
      },
      {
        "id": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "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": "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": "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": "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": "de_morgan_logic",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "邏輯的德摩根律：¬(P∧Q) ⇔ ¬P∨¬Q；¬(P∨Q) ⇔ ¬P∧¬Q。",
          "en": "De Morgan's laws for logic: ¬(P∧Q) ⇔ ¬P∨¬Q and ¬(P∨Q) ⇔ ¬P∧¬Q."
        }
      },
      {
        "id": "mko_equiv_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-equivalence-relation",
          "sha256": "ceccd69c4881e4b2eba57eb1a7c6ab6c6616ab09fa42eb8591999522a5948fd8",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "關係R在集合X上為等價關係，若每個x都有xRx；xRy推出yRx；xRy且yRz推出xRz。本輪在完整有限載體與完整關係對集合上逐項核對。",
          "en": "A relation R on X is an equivalence relation when every x satisfies xRx, xRy implies yRx, and xRy together with yRz implies xRz. This slice checks each obligation on a complete finite carrier and complete set of relation pairs."
        }
      },
      {
        "id": "relation_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "X 上的關係是子集 R⊆X×X；xRy 表示 (x,y)∈R。完整列出時，未列出的有序對不在 R 中。",
          "en": "A relation on X is a subset R⊆X×X; xRy means (x,y)∈R. When the list is complete, an unlisted ordered pair is not in R."
        }
      },
      {
        "id": "finite_forall",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限全稱：對有限集合 D，「對所有 t∈D，Φ(t)」等價於把 D 的每個元素逐一代入所得命題的合取；列表必須完整。",
          "en": "Finite universal: for a finite set D, 'for all t∈D, Φ(t)' is equivalent to the conjunction of the instances obtained by substituting each element of D; the list must be complete."
        }
      },
      {
        "id": "eq_class_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等價類：R 為等價關係時，[x]={y∈X : xRy}。依 MKO，只有三項公理都通過後才輸出等價類。",
          "en": "Equivalence class: when R is an equivalence relation, [x]={y∈X : xRy}. By the MKO, classes are emitted only after all three axioms pass."
        }
      },
      {
        "id": "L_block_trans",
        "kind": "lemma",
        "proved_in": [
          "S08.1",
          "S08.2",
          "S08.3"
        ],
        "statement": {
          "zh-Hant": "若 B_1、B_2 是 X 的兩兩不交子集，則 T=(B_1×B_1)∪(B_2×B_2) 是傳遞關係：(x,y)∈T 且 (y,z)∈T 推出 (x,z)∈T。",
          "en": "If B_1 and B_2 are pairwise disjoint subsets of X, then T=(B_1×B_1)∪(B_2×B_2) is transitive: (x,y)∈T and (y,z)∈T imply (x,z)∈T."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "5+2*4",
        "rhs": "13"
      },
      {
        "id": "N2",
        "step": "S05",
        "lhs": "2*2+3*3",
        "rhs": "13"
      },
      {
        "id": "N3",
        "step": "S11",
        "lhs": "2+3",
        "rhs": "5"
      },
      {
        "id": "N4",
        "step": "S08",
        "lhs": "2*2**2+3*3**2",
        "rhs": "35"
      },
      {
        "id": "N5",
        "step": "S12",
        "lhs": "13-2",
        "rhs": "5+2*3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只在 MKO 的範圍內：載體標籤 0 到 n−1、關係由互異有序對完整列出（假設 a1）；三項全部通過後才輸出等價類，單一反例足以否證對應公理（假設 a2）。傳遞性以「區塊關係」引理一次涵蓋全部可串接對，而非抽樣。不處理商集與基數比較，不改動 canonical MKO 或英文 edition。",
      "en": "Stays inside the MKO: carrier labelled 0 to n−1 and the relation listed completely as distinct ordered pairs (assumption a1); classes are emitted only after all three checks pass, and one counterexample refutes the corresponding axiom (assumption a2). Transitivity is settled for all composable pairs at once by a block-relation lemma, not by sampling. Quotient sets and cardinality comparison are not treated, and the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "X",
        "載體 {0,1,2,3,4}",
        "carrier {0,1,2,3,4}"
      ],
      [
        "R,\\ R'",
        "完整列出的關係與去掉兩對後的對照關係",
        "the fully listed relation and the comparison relation with two pairs removed"
      ],
      [
        "xRy",
        "(x,y)∈R",
        "(x,y)∈R"
      ],
      [
        "B_1,\\ B_2",
        "區塊 {0,2} 與 {1,3,4}",
        "the blocks {0,2} and {1,3,4}"
      ],
      [
        "T",
        "區塊關係 B_1×B_1 ∪ B_2×B_2",
        "block relation B_1×B_1 ∪ B_2×B_2"
      ],
      [
        "[x]",
        "x 的等價類 {y∈X : xRy}",
        "equivalence class of x, {y∈X : xRy}"
      ]
    ],
    "md_error_controls": [
      {
        "title": "只核反身與對稱不夠 / Reflexive and symmetric are not enough",
        "zh": "MKO 指出的常見誤解是只核反身與對稱就叫等價關係。R′ 兩項都通過（S12），但 3R′1、1R′4 而 (3,4)∉R′（S13）。傳遞性是獨立的第三項責任。",
        "en": "The common mistake named by the MKO is to call a relation an equivalence relation after checking only reflexivity and symmetry. R′ passes both (S12), yet 3R′1 and 1R′4 while (3,4)∉R′ (S13). Transitivity is an independent third obligation."
      },
      {
        "title": "不要看圖分群 / Do not group by looking at the picture",
        "zh": "畫出 R′ 的關係圖，0、2 一群，1、3、4 看起來也連成一群；但 R′ 的「類」{1,3} 與 {1,4} 相交而不相等。等價類只能在三項核對全部通過後由關係本身讀出（S10）。",
        "en": "Draw R′: 0 and 2 form one cluster and 1, 3, 4 also look connected; but the would-be classes {1,3} and {1,4} of R′ overlap without being equal. Classes may be read off the relation itself only after all three checks pass (S10)."
      },
      {
        "title": "傳遞性要涵蓋全部可串接對 / Transitivity must cover every composable chain",
        "zh": "R 的可串接對 (x,y)、(y,z) 共有 2·2²+3·3²=35 組。抽查幾組（例如 (0,2)、(2,0) 推出 (0,0)）不構成證明；本稿用 R=T 與引理 L_block_trans 一次涵蓋全部 35 組。",
        "en": "R has 2·2²+3·3²=35 composable chains (x,y), (y,z). Spot-checking a few (for example (0,2), (2,0) giving (0,0)) is not a proof; this guide uses R=T and Lemma L_block_trans to cover all 35 at once."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿不一般地證明「等價類構成載體的分割」；S11 只對本例的兩個集合核對。",
        "en": "This guide does not prove in general that equivalence classes partition the carrier; S11 checks only the two sets of this example."
      },
      {
        "zh-Hant": "finite_forall、relation_def 與 eq_class_def 是本稿的前提。",
        "en": "finite_forall, relation_def and eq_class_def are premises of this guide."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "90ad3357e3d147bcb375215a2ad86359916fc907de884703d58672c5ebc480e1",
    "original_source_sha256": "90ad3357e3d147bcb375215a2ad86359916fc907de884703d58672c5ebc480e1",
    "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."
  }
}