{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-set-operations-01",
  "mko_id": "mko-set-operations",
  "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"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-set-operations-01",
    "source": {
      "mko_id": "mko-set-operations",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "e857b27da045df1a33d58d7cd56b319d6c8ec23162a8cfc3c15f7c70d30f317b",
      "english_sha256": "3cad86e57536464f92b934935e0803aa9aee386b3c235758ae9da575c56f3ac3"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "set_operations_def",
      "set_extensionality",
      "subset_def",
      "finite_set_listing",
      "finite_counting_basics",
      "de_morgan_logic"
    ],
    "goal": {
      "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."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "全集 U={1,2,3,4,5,6,7,8}（MKO 假設 a2：使用補集前已明示全集）。",
          "en": "Universe U={1,2,3,4,5,6,7,8} (MKO assumption a2: the universe is stated before any complement is used)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "A={2,4,6,8}，B={3,4,5,6}。",
          "en": "A={2,4,6,8} and B={3,4,5,6}."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "比較用的另一個全集 U′={1,2,3,4,5,6,7,8,9,10}。",
          "en": "A second universe for comparison: U′={1,2,3,4,5,6,7,8,9,10}."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "\\text{a universe and two sets}",
        "after_tex": "U=\\{1,2,3,4,5,6,7,8\\},\\ A=\\{2,4,6,8\\},\\ B=\\{3,4,5,6\\}",
        "relation": "definition",
        "rule": "def_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先寫出全集 U，再寫出 A、B。依 MKO，補集必須相對於明示的全集，所以全集在任何補集計算之前就固定。集合以列舉給出，依 finite_set_listing，x∈A 恰好表示 x 是 2、4、6、8 之一。",
          "en": "State the universe U first, then A and B. By the MKO, a complement must be relative to an explicit universe, so the universe is fixed before any complement is computed. The sets are given by lists, so by finite_set_listing, x∈A means exactly that x is one of 2, 4, 6, 8."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "def_intro",
            "finite_set_listing"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出資料，不推出新事實；全集已明示，符合假設 a2。",
            "en": "This step only records the data and derives no new fact; the universe is explicit, as assumption a2 requires."
          }
        },
        "leaf_basis": "def_intro + finite_set_listing"
      },
      {
        "id": "S02",
        "uses": [
          "S01"
        ],
        "before_tex": "A=\\{2,4,6,8\\},\\ B=\\{3,4,5,6\\},\\ U=\\{1,2,3,4,5,6,7,8\\}",
        "after_tex": "A\\subseteq U,\\ B\\subseteq U",
        "relation": "implies",
        "rule": "subset_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "A 的元素恰好是 2、4、6、8 四個，每一個都在 U 的列舉中；B 的元素 3、4、5、6 也都在 U 中。A、B 是有限列舉，逐一檢查全部元素就是對「每個元素」的完整檢查，所以依子集合定義 A⊆U、B⊆U，符合假設 a1：A、B 在同一論域中。",
          "en": "The elements of A are exactly 2, 4, 6, 8, and each appears in the list of U; the elements 3, 4, 5, 6 of B are also in U. Since A and B are finite lists, checking all their listed elements is a complete check of 'every element', so by the subset definition A⊆U and B⊆U, as assumption a1 requires: A and B lie in one universe."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "subset_def",
            "finite_set_listing"
          ],
          "explanation": {
            "zh-Hant": "這是完整窮舉，不是抽樣：A、B 的每個元素都被檢查。反向 U⊆A 不宣稱（1∈U 但 1∉A）。",
            "en": "This is a complete enumeration, not a sample: every element of A and of B is checked. The converse U⊆A is not claimed (1∈U but 1∉A)."
          }
        },
        "leaf_basis": "subset_def + finite_set_listing + forall_intro"
      },
      {
        "id": "S03",
        "uses": [
          "S02"
        ],
        "before_tex": "A\\subseteq U,\\ B\\subseteq U",
        "after_tex": "A\\cup B,\\ A\\cap B,\\ A\\setminus B,\\ B\\setminus A\\subseteq U",
        "relation": "implies",
        "rule": "L_ops_in_U",
        "conditions": [
          "A\\subseteq U",
          "B\\subseteq U"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_ops_in_U（在 S03.1、S03.2 證明），四個運算結果都是 U 的子集合。因此 U 以外的對象都不屬於它們；要列出它們，只需逐一檢查 U 的八個元素。",
          "en": "By Lemma L_ops_in_U (proved in S03.1 and S03.2), all four resulting sets are subsets of U. So no object outside U belongs to them, and to list them it suffices to test the eight elements of U one by one."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "L_ops_in_U"
          ],
          "explanation": {
            "zh-Hant": "需要 A⊆U 與 B⊆U（S02）；若 A 有 U 以外的元素，只檢查 U 會漏掉聯集的元素。",
            "en": "It needs A⊆U and B⊆U (S02); if A had an element outside U, testing only U would miss elements of the union."
          }
        },
        "leaf_basis": "set_operations_def + subset_def + case_analysis"
      },
      {
        "id": "S03.1",
        "uses": [
          "set_operations_def",
          "subset_def"
        ],
        "before_tex": "x\\in A\\cup B",
        "after_tex": "x\\in U",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [
          "A\\subseteq U",
          "B\\subseteq U"
        ],
        "explanation": {
          "zh-Hant": "設 x∈A∪B；依聯集定義，x∈A 或 x∈B。情況一 x∈A：由 A⊆U 得 x∈U。情況二 x∈B：由 B⊆U 得 x∈U。兩種情況都得 x∈U，所以 A∪B⊆U。",
          "en": "Let x∈A∪B; by the definition of union, x∈A or x∈B. Case 1, x∈A: A⊆U gives x∈U. Case 2, x∈B: B⊆U gives x∈U. Both cases give x∈U, so A∪B⊆U."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "set_operations_def",
            "subset_def",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "兩種情況涵蓋「或」的全部可能；每種情況各用一個包含假設。",
            "en": "The two cases cover every possibility of the 'or'; each case uses one of the inclusion hypotheses."
          }
        },
        "leaf_basis": "set_operations_def + case_analysis + subset_def + forall_elim"
      },
      {
        "id": "S03.2",
        "uses": [
          "set_operations_def",
          "subset_def"
        ],
        "before_tex": "x\\in A\\cap B\\ \\lor\\ x\\in A\\setminus B\\ \\lor\\ x\\in B\\setminus A",
        "after_tex": "x\\in U",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [
          "A\\subseteq U",
          "B\\subseteq U"
        ],
        "explanation": {
          "zh-Hant": "x∈A∩B 與 x∈A∖B 都含有合取項 x∈A；取出它再用 A⊆U，得 x∈U。x∈B∖A 含有 x∈B，用 B⊆U 得 x∈U。S03.1 與 S03.2 合起來證得引理 L_ops_in_U。",
          "en": "Both x∈A∩B and x∈A∖B contain the conjunct x∈A; extracting it and using A⊆U gives x∈U. x∈B∖A contains x∈B, and B⊆U gives x∈U. S03.1 and S03.2 together prove Lemma L_ops_in_U."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "set_operations_def",
            "and_intro_elim",
            "subset_def"
          ],
          "explanation": {
            "zh-Hant": "只用到從合取中取出一項；不宣稱反向（U 的元素不必在這些集合中）。",
            "en": "It only extracts one conjunct; no converse is claimed (elements of U need not lie in these sets)."
          }
        },
        "leaf_basis": "set_operations_def + and_intro_elim + subset_def + forall_elim"
      },
      {
        "id": "S04",
        "uses": [
          "S01",
          "S03"
        ],
        "before_tex": "x\\in U",
        "after_tex": "1{:}(F,F),\\ 2{:}(T,F),\\ 3{:}(F,T),\\ 4{:}(T,T),\\ 5{:}(F,T),\\ 6{:}(T,T),\\ 7{:}(F,F),\\ 8{:}(T,F)",
        "relation": "implies",
        "rule": "finite_set_listing",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對 U 的每個元素記錄一對真值（x∈A，x∈B）：例如 2 在 A 的列舉中、不在 B 的列舉中，記為 (T,F)；4 同在兩者，記為 (T,T)；1 都不在，記為 (F,F)。八列涵蓋 U 的全部元素；依 S03，之後每個運算都只需讀這張表。",
          "en": "For each element of U record a pair of truth values (x∈A, x∈B): for example 2 is in the list of A and not in that of B, recorded as (T,F); 4 is in both, (T,T); 1 is in neither, (F,F). The eight rows cover all of U, and by S03 every operation below only needs this table."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "finite_set_listing",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "每一列都可以對照列舉重查；表格完整涵蓋 U，所以沒有遺漏的元素。",
            "en": "Each row can be rechecked against the lists; the table covers all of U, so no element is missed."
          }
        },
        "leaf_basis": "finite_set_listing + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "S03",
          "S04"
        ],
        "before_tex": "A\\cup B",
        "after_tex": "\\{2,3,4,5,6,8\\}",
        "relation": "equivalent",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x∈A∪B ⇔ x∈A 或 x∈B。表中至少有一個 T 的列是 2、3、4、5、6、8，所以 A∪B={2,3,4,5,6,8}。4 與 6 同時屬於兩集，但在聯集中只算一個元素。",
          "en": "x∈A∪B ⇔ x∈A or x∈B. The rows with at least one T are 2, 3, 4, 5, 6, 8, so A∪B={2,3,4,5,6,8}. 4 and 6 belong to both sets but each counts as one element of the union."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_operations_def",
            "set_extensionality",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "兩邊對每個對象的成員判定相同：在 U 中逐列對照，在 U 外兩邊都不含它（S03）；依集合相等判準，雙向成立。",
            "en": "Both sides decide membership the same way for every object: row by row in U, and outside U neither contains it (S03); by set equality the step holds both ways."
          }
        },
        "leaf_basis": "set_operations_def + set_extensionality + finite_set_listing"
      },
      {
        "id": "S06",
        "uses": [
          "S03",
          "S04"
        ],
        "before_tex": "A\\cap B",
        "after_tex": "\\{4,6\\}",
        "relation": "equivalent",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x∈A∩B ⇔ x∈A 且 x∈B。表中兩個都是 T 的列只有 4、6，所以 A∩B={4,6}。對照 S05：2 在 A∪B 中卻不在 A∩B 中，「或」與「且」給出不同的集合。",
          "en": "x∈A∩B ⇔ x∈A and x∈B. The only rows with T in both places are 4 and 6, so A∩B={4,6}. Compare with S05: 2 is in A∪B but not in A∩B, so 'or' and 'and' give different sets."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_operations_def",
            "set_extensionality",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "同 S05，逐元素雙向對照。",
            "en": "As in S05, membership is compared element by element in both directions."
          }
        },
        "leaf_basis": "set_operations_def + set_extensionality + finite_set_listing"
      },
      {
        "id": "S07",
        "uses": [
          "S03",
          "S04"
        ],
        "before_tex": "A\\setminus B",
        "after_tex": "\\{2,8\\}",
        "relation": "equivalent",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x∈A∖B ⇔ x∈A 且 x∉B，即表中 (T,F) 的列：2、8。所以 A∖B={2,8}。",
          "en": "x∈A∖B ⇔ x∈A and x∉B, that is the rows (T,F): 2 and 8. So A∖B={2,8}."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_operations_def",
            "set_extensionality",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "同 S05，逐元素雙向對照。",
            "en": "As in S05, membership is compared element by element in both directions."
          }
        },
        "leaf_basis": "set_operations_def + set_extensionality + finite_set_listing"
      },
      {
        "id": "S08",
        "uses": [
          "S03",
          "S04"
        ],
        "before_tex": "B\\setminus A",
        "after_tex": "\\{3,5\\}",
        "relation": "equivalent",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x∈B∖A ⇔ x∈B 且 x∉A，即表中 (F,T) 的列：3、5。所以 B∖A={3,5}。這裡讀的是第一個真值為 F、第二個為 T 的列，方向與 S07 相反。",
          "en": "x∈B∖A ⇔ x∈B and x∉A, that is the rows (F,T): 3 and 5. So B∖A={3,5}. The rows read here have F first and T second, the opposite direction to S07."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_operations_def",
            "set_extensionality",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "同 S05，逐元素雙向對照。",
            "en": "As in S05, membership is compared element by element in both directions."
          }
        },
        "leaf_basis": "set_operations_def + set_extensionality + finite_set_listing"
      },
      {
        "id": "S09",
        "uses": [
          "S07",
          "S08"
        ],
        "before_tex": "A\\setminus B=\\{2,8\\},\\ B\\setminus A=\\{3,5\\}",
        "after_tex": "A\\setminus B\\neq B\\setminus A",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "2∈A∖B 但 2∉B∖A，兩集元素不同，依集合相等判準 A∖B≠B∖A。本例中兩者甚至沒有共同元素。差集有方向性。",
          "en": "2∈A∖B but 2∉B∖A, so the sets have different elements and by the criterion for equality A∖B≠B∖A. In this example they have no common element at all. Set difference has a direction."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "S08",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "一個只屬於其中一邊的元素就足以否定相等。不宣稱對所有集合 A∖B 與 B∖A 都不等：A=B 時兩者都是 ∅。",
            "en": "One element in only one of the sets is enough to refute equality. It is not claimed that A∖B and B∖A differ for all sets: when A=B both are ∅."
          }
        },
        "leaf_basis": "set_extensionality + finite_set_listing"
      },
      {
        "id": "S10",
        "uses": [
          "S01",
          "S04"
        ],
        "before_tex": "U\\setminus A,\\ U\\setminus B",
        "after_tex": "U\\setminus A=\\{1,3,5,7\\},\\ U\\setminus B=\\{1,2,7,8\\}",
        "relation": "equivalent",
        "rule": "mko_set_ops",
        "conditions": [
          "U\\ \\text{explicit}"
        ],
        "explanation": {
          "zh-Hant": "依 MKO，A 的補集是相對於明示全集 U 的 U∖A。x∈U∖A ⇔ x∈U 且 x∉A：表中第一個真值為 F 的列是 1、3、5、7。同理，U∖B 是第二個真值為 F 的列 1、2、7、8。個數核對：U 是 A 與 U∖A 的不相交聯集，所以 |U∖A|=8−4=4。",
          "en": "By the MKO, the complement of A is U∖A, relative to the explicit universe U. x∈U∖A ⇔ x∈U and x∉A: the rows with F in the first place are 1, 3, 5, 7. Likewise U∖B consists of the rows with F in the second place, 1, 2, 7, 8. Count check: U is the disjoint union of A and U∖A, so |U∖A|=8−4=4."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_set_ops",
            "set_operations_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "U∖A 由成員條件逐元素決定，雙向對照；U 以外的對象不屬於 U∖A。",
            "en": "U∖A is decided element by element by its membership condition, compared both ways; no object outside U belongs to U∖A."
          }
        },
        "leaf_basis": "set_operations_def + set_extensionality + finite_set_listing + finite_counting_basics"
      },
      {
        "id": "S11",
        "uses": [
          "G3",
          "S01",
          "S10"
        ],
        "before_tex": "U'\\setminus A",
        "after_tex": "U'\\setminus A=\\{1,3,5,7,9,10\\}\\neq U\\setminus A",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [
          "A\\subseteq U'"
        ],
        "explanation": {
          "zh-Hant": "換成全集 U′：A⊆U′ 仍成立，x∈U′∖A ⇔ x∈U′ 且 x∉A，得 U′∖A={1,3,5,7,9,10}，共 10−4=6 個元素。9 屬於 U′∖A 卻不屬於 U∖A（9∉U），所以兩個「補集」不同。A 的列舉沒變，變的只是全集；不指定全集就寫「A 的補集」，答案不確定。",
          "en": "Switch to the universe U′: A⊆U′ still holds, and x∈U′∖A ⇔ x∈U′ and x∉A, giving U′∖A={1,3,5,7,9,10}, with 10−4=6 elements. 9 belongs to U′∖A but not to U∖A (9∉U), so the two 'complements' differ. The list of A did not change, only the universe did; writing 'the complement of A' without a universe has no determinate answer."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "set_operations_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "需要 U′ 明示；單一見證 9 足以否定兩個補集相等。",
            "en": "U′ must be explicit; the single witness 9 is enough to refute equality of the two complements."
          }
        },
        "leaf_basis": "set_operations_def + set_extensionality + finite_set_listing"
      },
      {
        "id": "S12",
        "uses": [
          "set_operations_def"
        ],
        "before_tex": "U\\setminus(A\\cup B)",
        "after_tex": "(U\\setminus A)\\cap(U\\setminus B)",
        "relation": "equivalent",
        "rule": "L_de_morgan_set",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_de_morgan_set（在 S12.1 到 S12.3 對任意集合證明），兩者是同一集合。這是一般論證，不依賴本例的數字。",
          "en": "By Lemma L_de_morgan_set (proved for arbitrary sets in S12.1 to S12.3), the two are the same set. This is a general argument and does not depend on the numbers of this example."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_de_morgan_set"
          ],
          "explanation": {
            "zh-Hant": "引理證明的是集合相等，由逐元素的雙向等價得到。",
            "en": "The lemma proves an equality of sets, obtained from an element-wise two-way equivalence."
          }
        },
        "leaf_basis": "set_operations_def + de_morgan_logic + and_intro_elim + set_extensionality"
      },
      {
        "id": "S12.1",
        "uses": [
          "set_operations_def"
        ],
        "before_tex": "x\\in U\\setminus(A\\cup B)",
        "after_tex": "x\\in U\\land\\neg(x\\in A\\lor x\\in B)",
        "relation": "equivalent",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依差集定義，x∈U∖(A∪B) ⇔ x∈U 且 x∉A∪B；再依聯集定義，x∉A∪B 就是 ¬(x∈A 或 x∈B)。",
          "en": "By the definition of difference, x∈U∖(A∪B) ⇔ x∈U and x∉A∪B; by the definition of union, x∉A∪B is ¬(x∈A or x∈B)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "兩個定義展開都是雙向的。",
            "en": "Both unfoldings of definitions are two-way."
          }
        },
        "leaf_basis": "set_operations_def + def_unfold"
      },
      {
        "id": "S12.2",
        "uses": [
          "S12.1",
          "de_morgan_logic"
        ],
        "before_tex": "x\\in U\\land\\neg(x\\in A\\lor x\\in B)",
        "after_tex": "x\\in U\\land x\\notin A\\land x\\notin B",
        "relation": "equivalent",
        "rule": "de_morgan_logic",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依邏輯的德摩根律 ¬(P∨Q) ⇔ ¬P∧¬Q，取 P 為 x∈A、Q 為 x∈B。否定把「或」變成「且」。",
          "en": "By De Morgan's law for logic, ¬(P∨Q) ⇔ ¬P∧¬Q, with P the statement x∈A and Q the statement x∈B. Negation turns 'or' into 'and'."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "de_morgan_logic"
          ],
          "explanation": {
            "zh-Hant": "德摩根律本身是等價，雙向成立。",
            "en": "De Morgan's law is itself an equivalence, so the step holds both ways."
          }
        },
        "leaf_basis": "de_morgan_logic"
      },
      {
        "id": "S12.3",
        "uses": [
          "S12.2",
          "set_operations_def"
        ],
        "before_tex": "x\\in U\\land x\\notin A\\land x\\notin B",
        "after_tex": "x\\in(U\\setminus A)\\cap(U\\setminus B)",
        "relation": "equivalent",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 x∈U 重複一次（P ⇔ P∧P）並重組合取：(x∈U 且 x∉A) 且 (x∈U 且 x∉B)，即 x∈U∖A 且 x∈U∖B，也就是 x∈(U∖A)∩(U∖B)。x 是任意的，依集合相等判準，S12.1 到 S12.3 合起來證得引理 L_de_morgan_set。",
          "en": "Duplicate x∈U (P ⇔ P∧P) and regroup the conjunction: (x∈U and x∉A) and (x∈U and x∉B), that is x∈U∖A and x∈U∖B, which is x∈(U∖A)∩(U∖B). Since x was arbitrary, set equality applies, and S12.1 to S12.3 together prove Lemma L_de_morgan_set."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "and_intro_elim",
            "set_operations_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "合取的重組與重複都是雙向的；每個 x 兩邊同真同假，所以兩集合相等。",
            "en": "Regrouping and duplicating conjuncts are two-way; for each x both sides are true or false together, so the sets are equal."
          }
        },
        "leaf_basis": "and_intro_elim + set_operations_def + forall_intro + set_extensionality"
      },
      {
        "id": "S13",
        "uses": [
          "S05",
          "S10",
          "S12"
        ],
        "before_tex": "U\\setminus\\{2,3,4,5,6,8\\},\\quad \\{1,3,5,7\\}\\cap\\{1,2,7,8\\}",
        "after_tex": "\\{1,7\\}=\\{1,7\\}",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "以本例核對引理：由 S05，U∖(A∪B) 是 U 中不在 {2,3,4,5,6,8} 的元素，即 {1,7}；由 S10，(U∖A)∩(U∖B) 是 {1,3,5,7} 與 {1,2,7,8} 的共同元素，也是 {1,7}。兩邊相同，與 S12 一致。",
          "en": "Check the lemma on this example: by S05, U∖(A∪B) consists of the elements of U not in {2,3,4,5,6,8}, namely {1,7}; by S10, (U∖A)∩(U∖B) consists of the common elements of {1,3,5,7} and {1,2,7,8}, also {1,7}. The two agree, consistent with S12."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S10",
            "S12",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "這是一個例子的核對，可以獨立重算；它不取代 S12 的一般證明，一個例子也不能證明德摩根律。",
            "en": "This is a check on one example and can be recomputed independently; it does not replace the general proof in S12, and one example could not prove De Morgan's law."
          }
        },
        "leaf_basis": "set_operations_def + finite_set_listing + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S05",
          "S08"
        ],
        "before_tex": "A\\cup B",
        "after_tex": "A\\cup(B\\setminus A)\\ \\text{with}\\ A\\cap(B\\setminus A)=\\varnothing",
        "relation": "equivalent",
        "rule": "L_union_split",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_union_split（在 S14.1、S14.2 證明），A∪B 可以拆成 A 與 B∖A 兩個不相交的部分。本例：{2,4,6,8}∪{3,5}={2,3,4,5,6,8}，與 S05 一致。",
          "en": "By Lemma L_union_split (proved in S14.1 and S14.2), A∪B splits into the two disjoint parts A and B∖A. In this example {2,4,6,8}∪{3,5}={2,3,4,5,6,8}, in agreement with S05."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_union_split"
          ],
          "explanation": {
            "zh-Hant": "引理給出集合相等，雙向成立；不相交另由 S14.2 給出。",
            "en": "The lemma gives an equality of sets, so the step holds both ways; disjointness is given separately by S14.2."
          }
        },
        "leaf_basis": "set_operations_def + case_analysis + set_extensionality + contradiction"
      },
      {
        "id": "S14.1",
        "uses": [
          "set_operations_def"
        ],
        "before_tex": "x\\in A\\lor(x\\in B\\land x\\notin A)",
        "after_tex": "x\\in A\\lor x\\in B",
        "relation": "equivalent",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "左邊是 x∈A∪(B∖A) 的定義展開。對 x∈A 分情況：若 x∈A，兩邊都成立；若 x∉A，左邊化為 x∈B∧x∉A，即 x∈B，右邊也化為 x∈B。兩種情況下兩邊同真同假，所以 A∪(B∖A)=A∪B。",
          "en": "The left side unfolds x∈A∪(B∖A). Split on x∈A: if x∈A, both sides hold; if x∉A, the left side reduces to x∈B∧x∉A, that is x∈B, and the right side also reduces to x∈B. In both cases the two sides agree, so A∪(B∖A)=A∪B."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "case_analysis",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "每種情況下兩邊都等價，且兩種情況窮盡，所以整體等價雙向成立。",
            "en": "In each case the two sides are equivalent, and the cases are exhaustive, so the equivalence holds both ways."
          }
        },
        "leaf_basis": "case_analysis + set_operations_def + set_extensionality"
      },
      {
        "id": "S14.2",
        "uses": [
          "set_operations_def"
        ],
        "before_tex": "x\\in A\\cap(B\\setminus A)",
        "after_tex": "A\\cap(B\\setminus A)=\\varnothing",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若 x∈A∩(B∖A)，則 x∈A 且 x∈B 且 x∉A，前後矛盾。所以沒有這樣的 x，A∩(B∖A)=∅。S14.1 與 S14.2 合起來證得引理 L_union_split。",
          "en": "If x∈A∩(B∖A), then x∈A and x∈B and x∉A, a contradiction. So there is no such x, and A∩(B∖A)=∅. S14.1 and S14.2 together prove Lemma L_union_split."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "set_operations_def",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "只需兩個定義展開與矛盾；不宣稱 A∩B=∅（本例 A∩B={4,6}）。",
            "en": "Only the two unfoldings and the contradiction are needed; it is not claimed that A∩B=∅ (here A∩B={4,6})."
          }
        },
        "leaf_basis": "set_operations_def + contradiction + quantifier_negation"
      },
      {
        "id": "S15",
        "uses": [
          "S08",
          "S14"
        ],
        "before_tex": "|A\\cup B|=|A\\cup(B\\setminus A)|",
        "after_tex": "|A\\cup B|=|A|+|B\\setminus A|=4+2=6",
        "relation": "implies",
        "rule": "finite_counting_basics",
        "conditions": [
          "A\\cap(B\\setminus A)=\\varnothing"
        ],
        "explanation": {
          "zh-Hant": "A 與 B∖A 不相交（S14），依有限計數，|A∪(B∖A)|=|A|+|B∖A|=4+2=6。若直接把 |A|+|B|=8 當作聯集的大小，就把 4、6 各算了兩次；差 8−6=2 正是 |A∩B|。",
          "en": "A and B∖A are disjoint (S14), so finite counting gives |A∪(B∖A)|=|A|+|B∖A|=4+2=6. Taking |A|+|B|=8 as the size of the union would count 4 and 6 twice; the difference 8−6=2 is exactly |A∩B|."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S14",
            "finite_counting_basics"
          ],
          "explanation": {
            "zh-Hant": "加法公式只適用於不相交的集合，所以必須先有 S14.2 的 A∩(B∖A)=∅；直接把 A、B 的大小相加不合法，因為 A∩B≠∅。",
            "en": "The addition rule applies only to disjoint sets, so S14.2's A∩(B∖A)=∅ is needed first; adding the sizes of A and B directly is not allowed, because A∩B≠∅."
          }
        },
        "leaf_basis": "finite_counting_basics + numeric_evaluation"
      },
      {
        "id": "S16",
        "uses": [
          "S05",
          "S06",
          "S07",
          "S08",
          "S09",
          "S10",
          "S11",
          "S13",
          "S15"
        ],
        "before_tex": "\\text{membership table of }U",
        "after_tex": "A\\cup B=\\{2,3,4,5,6,8\\},\\ A\\cap B=\\{4,6\\},\\ A\\setminus B=\\{2,8\\},\\ B\\setminus A=\\{3,5\\},\\ U\\setminus A=\\{1,3,5,7\\},\\ U\\setminus B=\\{1,2,7,8\\}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整 S05 到 S15 的結果：六個運算結果如上；A∖B≠B∖A；U′∖A≠U∖A；德摩根律在本例成立；|A∪B|=6。",
          "en": "Collect the results of S05 to S15: the six results are as shown; A∖B≠B∖A; U′∖A≠U∖A; De Morgan's law holds in this example; |A∪B|=6."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S05",
            "S06",
            "S07",
            "S08",
            "S09",
            "S10",
            "S11",
            "S13",
            "S15",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G 是題目各項結論的合取；每一項都由所列步驟從已知正向得到，此步只做合取，沒有把 G 當作前提。",
            "en": "Goal G is the conjunction of the problem's conclusions; each conjunct is obtained forwards from the givens by the listed steps, and this step only forms the conjunction; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在全集 U={1,2,3,4,5,6,7,8} 中：A∪B={2,3,4,5,6,8}，A∩B={4,6}，A∖B={2,8}，B∖A={3,5}，U∖A={1,3,5,7}，U∖B={1,2,7,8}。A∖B≠B∖A；全集改為 U′ 時 A 的補集變成 {1,3,5,7,9,10}；U∖(A∪B)=(U∖A)∩(U∖B) 對任意集合成立，本例兩邊都是 {1,7}；|A∪B|=|A|+|B∖A|=6。",
      "en": "In the universe U={1,2,3,4,5,6,7,8}: A∪B={2,3,4,5,6,8}, A∩B={4,6}, A∖B={2,8}, B∖A={3,5}, U∖A={1,3,5,7}, U∖B={1,2,7,8}. A∖B≠B∖A; with the universe U′ the complement of A becomes {1,3,5,7,9,10}; U∖(A∪B)=(U∖A)∩(U∖B) holds for all sets, and both sides are {1,7} here; |A∪B|=|A|+|B∖A|=6."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "A∖B=B∖A",
          "en": "A∖B=B∖A"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "2∈A∖B 但 2∉B∖A（S09）。",
          "en": "2∈A∖B but 2∉B∖A (S09)."
        }
      },
      {
        "claim": {
          "zh-Hant": "A∩B={2,3,4,5,6,8}",
          "en": "A∩B={2,3,4,5,6,8}"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "那是聯集；2∉B，所以 2∉A∩B，正確答案是 {4,6}。",
          "en": "That is the union; 2∉B, so 2∉A∩B, and the correct answer is {4,6}."
        }
      },
      {
        "claim": {
          "zh-Hant": "A∪B 有 |A|+|B|=8 個元素",
          "en": "A∪B has |A|+|B|=8 elements"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "4 與 6 屬於兩集但只算一次；|A∪B|=6（S15）。",
          "en": "4 and 6 belong to both sets but count once; |A∪B|=6 (S15)."
        }
      },
      {
        "claim": {
          "zh-Hant": "不必說明全集，A={2,4,6,8} 的補集就是 {1,3,5,7}",
          "en": "Without naming a universe, the complement of A={2,4,6,8} is {1,3,5,7}"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "在 U′ 中補集是 {1,3,5,7,9,10}（S11）；補集依賴全集。",
          "en": "In U′ the complement is {1,3,5,7,9,10} (S11); the complement depends on the universe."
        }
      },
      {
        "claim": {
          "zh-Hant": "U∖(A∪B)=(U∖A)∪(U∖B)",
          "en": "U∖(A∪B)=(U∖A)∪(U∖B)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "右邊是 {1,2,3,5,7,8}，左邊是 {1,7}；正確的德摩根律右邊是交集（S12）。",
          "en": "The right side is {1,2,3,5,7,8} and the left side is {1,7}; the correct De Morgan law has an intersection on the right (S12)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "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": "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": "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": "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": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "forall_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。",
          "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds."
        }
      },
      {
        "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": "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_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": "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": "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_set_ops",
        "kind": "cited",
        "source": {
          "mko_id": "mko-set-operations",
          "sha256": "e857b27da045df1a33d58d7cd56b319d6c8ec23162a8cfc3c15f7c70d30f317b",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對同一論域中的集合 A、B，聯集收納至少屬於其中一集的元素，交集收納同時屬於兩集的元素，差集 A∖B 收納屬於 A 但不屬於 B 的元素；補集必須相對明示全集 U，以 U∖A 表示。",
          "en": "For sets A and B in one domain, the union contains elements belonging to at least one set, the intersection contains elements belonging to both, and A∖B contains elements belonging to A but not B. A complement must be relative to an explicit universe U and is written U∖A."
        }
      },
      {
        "id": "finite_set_listing",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "列舉集合：x 屬於以列舉方式給出的有限集合，當且僅當 x 等於所列對象之一；列舉的次序與重複不改變集合，例如 {4,4,6}={4,6}。",
          "en": "Listed sets: x belongs to a finite set given by a list exactly when x equals one of the listed objects; the order and repetitions of the list do not change the set, for example {4,4,6}={4,6}."
        }
      },
      {
        "id": "L_ops_in_U",
        "kind": "lemma",
        "proved_in": [
          "S03.1",
          "S03.2"
        ],
        "statement": {
          "zh-Hant": "若 A⊆U 且 B⊆U，則 A∪B、A∩B、A∖B、B∖A 都是 U 的子集合。",
          "en": "If A⊆U and B⊆U, then A∪B, A∩B, A∖B and B∖A are subsets of U."
        }
      },
      {
        "id": "L_de_morgan_set",
        "kind": "lemma",
        "proved_in": [
          "S12.1",
          "S12.2",
          "S12.3"
        ],
        "statement": {
          "zh-Hant": "對任意集合 U、A、B：U∖(A∪B)=(U∖A)∩(U∖B)。",
          "en": "For any sets U, A, B: U∖(A∪B)=(U∖A)∩(U∖B)."
        }
      },
      {
        "id": "L_union_split",
        "kind": "lemma",
        "proved_in": [
          "S14.1",
          "S14.2"
        ],
        "statement": {
          "zh-Hant": "對任意集合 A、B：A∪B=A∪(B∖A)，且 A∩(B∖A)=∅。",
          "en": "For any sets A, B: A∪B=A∪(B∖A), and A∩(B∖A)=∅."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S15",
        "lhs": "4+2",
        "rhs": "6"
      },
      {
        "id": "N2",
        "step": "S15",
        "lhs": "4+4",
        "relation": "!=",
        "rhs": "6"
      },
      {
        "id": "N3",
        "step": "S15",
        "lhs": "4+4-6",
        "rhs": "2"
      },
      {
        "id": "N4",
        "step": "S10",
        "lhs": "8-4",
        "rhs": "4"
      },
      {
        "id": "N5",
        "step": "S11",
        "lhs": "10-4",
        "rhs": "6"
      },
      {
        "id": "N6",
        "step": "S13",
        "lhs": "8-6",
        "rhs": "2"
      },
      {
        "id": "N7",
        "step": "S16",
        "lhs": "2+2+2",
        "rhs": "6"
      },
      {
        "id": "N8",
        "step": "S10",
        "lhs": "8-4",
        "rhs": "4"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理一個明示全集中的有限集合（MKO 假設 a1、a2）。運算由成員條件定義並逐元素計算；德摩根律以一般論證證明，具體例子只作核對。不綁定 Lean 的 Set.mem_union 等聲明；不改動 canonical MKO 或英文 edition。",
      "en": "Treats finite sets in one explicit universe (MKO assumptions a1 and a2). The operations are defined by membership conditions and computed element by element; De Morgan's law is proved by a general argument, and the example only checks it. It does not bind Lean declarations such as Set.mem_union, and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "U,\\ U'",
        "兩個明示全集",
        "two explicit universes"
      ],
      [
        "A\\cup B",
        "聯集：至少屬於其中一集",
        "union: in at least one of the sets"
      ],
      [
        "A\\cap B",
        "交集：同時屬於兩集",
        "intersection: in both sets"
      ],
      [
        "A\\setminus B",
        "差集：屬於 A 但不屬於 B",
        "difference: in A but not in B"
      ],
      [
        "U\\setminus A",
        "A 相對於全集 U 的補集",
        "the complement of A relative to U"
      ],
      [
        "|X|",
        "有限集合 X 的元素個數",
        "the number of elements of a finite set X"
      ],
      [
        "(T,F)",
        "成員表的一列：x∈A 為真、x∈B 為假",
        "a row of the membership table: x∈A true, x∈B false"
      ]
    ],
    "md_error_controls": [
      {
        "title": "聯集與交集互換 / Swapping union and intersection",
        "zh": "MKO 指出常見誤解是把聯集和交集互換。聯集對應「或」（至少屬於其一），交集對應「且」（同時屬於兩者）。本例 2∈A∪B 但 2∉A∩B，聯集有 6 個元素、交集只有 2 個；算錯時，可回到成員表檢查讀的是「至少一個 T」還是「兩個都是 T」。",
        "en": "The MKO names the mistake of swapping union and intersection. Union corresponds to 'or' (in at least one set), intersection to 'and' (in both). Here 2∈A∪B but 2∉A∩B; the union has 6 elements and the intersection only 2. To catch the error, go back to the membership table and check whether the rule reads 'at least one T' or 'both T'."
      },
      {
        "title": "差集有方向 / Difference has a direction",
        "zh": "MKO 提醒 A∖B 與 B∖A 通常不同。本例 A∖B={2,8} 而 B∖A={3,5}（S07 到 S09），兩者甚至不相交。只有在特殊情況（例如 A=B，兩者都是 ∅）才相等。",
        "en": "The MKO warns that A∖B and B∖A usually differ. Here A∖B={2,8} while B∖A={3,5} (S07 to S09), and they are even disjoint. They are equal only in special cases (for example A=B, when both are ∅)."
      },
      {
        "title": "補集要先指定全集 / A complement needs a universe first",
        "zh": "MKO 提醒不要在未指定全集時直接列出補集。同一個 A={2,4,6,8}，在 U 中的補集是 {1,3,5,7}，在 U′ 中是 {1,3,5,7,9,10}（S10、S11）。所以必須寫成 U∖A 並先列出 U。",
        "en": "The MKO warns against listing a complement without a specified universe. The same A={2,4,6,8} has complement {1,3,5,7} in U but {1,3,5,7,9,10} in U′ (S10, S11). So one must write U∖A and list U first."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "S10、S11 的個數核對用到「A⊆U 時 U 是 A 與 U∖A 的不相交聯集」，這可由 S14 同樣的論證得到，本稿未另立引理。",
        "en": "The count checks in S10 and S11 use that U is the disjoint union of A and U∖A when A⊆U; this follows by the same argument as S14, and no separate lemma is stated here."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "c42753a14041612b07055d8ace7c5c9df4c243cd4ab72e79efd18921fc2f0b46",
    "original_source_sha256": "c42753a14041612b07055d8ace7c5c9df4c243cd4ab72e79efd18921fc2f0b46",
    "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."
  }
}