{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-set-01",
  "mko_id": "mko-set",
  "title": {
    "zh-Hant": "集合：由元素刻畫、外延相等，以及不受限制的建構為何失敗",
    "en": "Sets: characterized by elements, equal by extensionality, and why unrestricted formation fails"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-set-01",
    "source": {
      "mko_id": "mko-set",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "4f736510823c29158d78264f57fa9f96c5c2107ddc99999622656fdacc7a5728",
      "english_sha256": "f1024a1b7bcb63048a9ba5c6ca166a8236e5d37f63c4c64d3985f769c7ae742c"
    },
    "mode": "definition_walkthrough",
    "audience_prerequisites": [
      "set_extensionality",
      "set_builder",
      "subset_def",
      "zero_product",
      "distributive",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在論域 ℤ 中，設 A={x∈ℤ ∣ x²=4}、B={−2,2}、C={2,−2,2}、D={2}。證明 A=B 與 C=B；證明 D≠B，並區分元素 2 與集合 {2}。最後證明：在任何帶有隸屬關係的論域 U 中，都不存在 R∈U 使「y∈R ⇔ y∉y」對所有 y∈U 成立，因此不是每個條件都能形成集合。",
      "en": "In the universe ℤ, let A={x∈ℤ ∣ x²=4}, B={−2,2}, C={2,−2,2} and D={2}. Prove A=B and C=B; prove D≠B and distinguish the element 2 from the set {2}. Finally prove that in any domain U with a membership relation there is no R∈U with 'y∈R ⇔ y∉y' for all y∈U, so not every condition forms a set."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "論域與型別：前半在 ℤ 中工作，所討論的集合都是整數的集合；x∈S 的左邊是整數，右邊是整數的集合。列舉記號：x∈{a,b,c} ⇔ x=a ∨ x=b ∨ x=c，兩個或一個元素時同理。",
          "en": "Universe and types: the first part works in ℤ, and the sets discussed are sets of integers; in x∈S the left side is an integer and the right side a set of integers. List notation: x∈{a,b,c} ⇔ x=a ∨ x=b ∨ x=c, and likewise with two elements or one."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "A={x∈ℤ ∣ x²=4}，B={−2,2}，C={2,−2,2}，D={2}。",
          "en": "A={x∈ℤ ∣ x²=4}, B={−2,2}, C={2,−2,2} and D={2}."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "羅素論證的設定：U 是任意一個對象的論域，帶有二元關係 ∈（對 y,z∈U，y∈z 是命題）；不假設任何集合論公理；採古典邏輯。",
          "en": "Setting for Russell's argument: U is any domain of objects with a binary relation ∈ (for y,z∈U, y∈z is a proposition); no axioms of set theory are assumed; classical logic is used."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "\\text{a set}\\ S\\ \\text{in the universe}\\ \\mathbb{Z}",
        "after_tex": "A=B\\iff\\forall x\\in\\mathbb{Z}\\,(x\\in A\\iff x\\in B)",
        "relation": "definition",
        "rule": "mko_set",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，在指定論域中集合由「哪些對象是它的元素」刻畫，元素完全相同的集合相等。論域是 ℤ（G1），所以要證兩個集合相等，就要對任意整數 x 證明 x∈A ⇔ x∈B；要證不相等，只需找到一個整數屬於其中一個而不屬於另一個。",
          "en": "By the MKO, within a specified universe a set is characterized by which objects are its elements, and sets with exactly the same elements are equal. The universe is ℤ (G1), so proving two sets equal means proving x∈A ⇔ x∈B for an arbitrary integer x, and proving them unequal means finding one integer in one set but not the other."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_set",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出判準，不推出新事實。",
            "en": "This step only states the criterion and derives no new fact."
          }
        },
        "leaf_basis": "mko_set + set_extensionality"
      },
      {
        "id": "S02",
        "uses": [
          "G2"
        ],
        "before_tex": "x\\in A",
        "after_tex": "x^2=4",
        "relation": "equivalent",
        "rule": "set_builder",
        "conditions": [
          "x\\in\\mathbb{Z}"
        ],
        "explanation": {
          "zh-Hant": "依集合建構式，x∈A ⇔ x∈ℤ 且 x²=4。在論域 ℤ 中 x∈ℤ 已成立，所以 x∈A ⇔ x²=4。A 是在既有集合 ℤ 中以條件分出元素，這是受限的建構方式。",
          "en": "By set-builder notation, x∈A ⇔ x∈ℤ and x²=4. In the universe ℤ, x∈ℤ already holds, so x∈A ⇔ x²=4. A is carved out of the existing set ℤ by a condition, which is a restricted form of set formation."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_builder",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "建構式的定義本身是雙向的；拿掉已成立的合取項 x∈ℤ 也是雙向的。",
            "en": "The set-builder definition is itself two-way, and dropping the conjunct x∈ℤ, which already holds, is two-way as well."
          }
        },
        "leaf_basis": "set_builder + and_intro_elim"
      },
      {
        "id": "S03",
        "uses": [
          "S02"
        ],
        "before_tex": "x^2=4",
        "after_tex": "x^2-4=0",
        "relation": "equivalent",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊同加 −4；反向是兩邊同加 4。",
          "en": "Add −4 to both sides; backwards, add 4 to both sides."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "加 −4 與加 4 互為反運算，在 ℤ 中沒有附加條件。",
            "en": "Adding −4 and adding 4 undo each other, with no side condition in ℤ."
          }
        },
        "leaf_basis": "eq_congruence + add_inverse + add_assoc + add_zero"
      },
      {
        "id": "S04",
        "uses": [
          "S03"
        ],
        "before_tex": "x^2-4=0",
        "after_tex": "(x-2)(x+2)=0",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依分配律 (x−2)(x+2)=x²+2x−2x−4=x²−4，對每個整數成立；等值代換不改變等式的真假。",
          "en": "By distributivity (x−2)(x+2)=x²+2x−2x−4=x²−4 for every integer; replacing equals does not change whether the equation holds."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "恆等式的代換雙向成立；N1 是獨立的多項式核對。",
            "en": "Replacement by an identity is two-way; N1 is an independent polynomial check."
          }
        },
        "leaf_basis": "distributive + neg_mul + add_inverse + eq_replace"
      },
      {
        "id": "S05",
        "uses": [
          "S04"
        ],
        "before_tex": "(x-2)(x+2)=0",
        "after_tex": "x-2=0\\ \\lor\\ x+2=0",
        "relation": "equivalent",
        "rule": "zero_product",
        "conditions": [],
        "explanation": {
          "zh-Hant": "ℤ 是整域，依零乘積性質：兩個整數的積為 0，當且僅當至少一個因子為 0。",
          "en": "ℤ is an integral domain, so by the zero-product property a product of two integers is 0 exactly when at least one factor is 0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "zero_product"
          ],
          "explanation": {
            "zh-Hant": "兩個方向都包含在 zero_product 中。",
            "en": "Both directions are contained in zero_product."
          }
        },
        "leaf_basis": "zero_product"
      },
      {
        "id": "S06",
        "uses": [
          "S05"
        ],
        "before_tex": "x-2=0\\ \\lor\\ x+2=0",
        "after_tex": "x=2\\ \\lor\\ x=-2",
        "relation": "equivalent",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第一支兩邊同加 2，第二支兩邊同加 −2；兩支各自是雙向等價，所以析取前後同真同假。",
          "en": "Add 2 to both sides of the first disjunct and −2 to both sides of the second; each is a two-way equivalence, so the disjunction before and after is true or false together."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "add_inverse",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "代回：2−2=0、(−2)+2=0。",
            "en": "Substituting back: 2−2=0 and (−2)+2=0."
          }
        },
        "leaf_basis": "eq_congruence + add_inverse + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "G1",
          "G2"
        ],
        "before_tex": "x=2\\ \\lor\\ x=-2",
        "after_tex": "x\\in B",
        "relation": "equivalent",
        "rule": "list_set",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依析取交換律改寫為 x=−2 ∨ x=2；依列舉記號（G1）這正是 x∈B。",
          "en": "By commutativity of disjunction rewrite as x=−2 ∨ x=2; by list notation (G1) this is exactly x∈B."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "or_laws",
            "list_set"
          ],
          "explanation": {
            "zh-Hant": "交換律與列舉記號的定義都是雙向的。",
            "en": "Commutativity and the definition of list notation are both two-way."
          }
        },
        "leaf_basis": "or_laws + list_set"
      },
      {
        "id": "S08",
        "uses": [
          "S02",
          "S03",
          "S04",
          "S05",
          "S06",
          "S07"
        ],
        "before_tex": "\\text{for arbitrary}\\ x\\in\\mathbb{Z}:\\ x\\in A\\iff x\\in B",
        "after_tex": "A=B",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S02 到 S07 對任意整數 x 成立，依推論的傳遞串起來得 x∈A ⇔ x∈B；依全稱推廣與集合相等的判準，A=B。作為對照的非元素：3²=9≠4，所以 3∉A；又 3≠−2、3≠2，所以 3∉B，兩邊一致。",
          "en": "S02 to S07 hold for an arbitrary integer x; chaining them gives x∈A ⇔ x∈B, and by generalization and the criterion for set equality, A=B. As a contrasting non-element: 3²=9≠4, so 3∉A; also 3≠−2 and 3≠2, so 3∉B, and the two sides agree."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S07",
            "logic_chain",
            "forall_intro",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "需要每個整數都比較過：x 是任取的，沒有額外假設。x=3 只是例子，不是證明的一部分。",
            "en": "Every integer must be covered: x was arbitrary, with no extra assumption. x=3 is only an example and not part of the proof."
          }
        },
        "leaf_basis": "logic_chain + forall_intro + set_extensionality"
      },
      {
        "id": "S09",
        "uses": [
          "L_C",
          "G2"
        ],
        "before_tex": "\\forall x\\in\\mathbb{Z}:\\ x\\in C\\iff x\\in B",
        "after_tex": "C=B",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_C（在 S09.1、S09.2 證明），每個整數 x 都有 x∈C ⇔ x∈B；依集合相等的判準，C=B。C 的寫法有三個符號、順序也不同，但集合只看元素，順序與重複都不產生新集合。",
          "en": "By Lemma L_C (proved in S09.1 and S09.2), every integer x has x∈C ⇔ x∈B; by the criterion for set equality, C=B. C is written with three symbols in a different order, but a set depends only on its elements, so order and repetition do not create a new set."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_C",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "需要對每個 x 的等價；引理對任意 x 證成。",
            "en": "The equivalence is needed for every x; the lemma is proved for an arbitrary x."
          }
        },
        "leaf_basis": "set_extensionality + list_set + or_laws"
      },
      {
        "id": "S09.1",
        "uses": [
          "G2",
          "list_set"
        ],
        "before_tex": "x\\in C",
        "after_tex": "x=2\\lor x=-2\\lor x=2",
        "relation": "equivalent",
        "rule": "list_set",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依列舉記號展開 C={2,−2,2}：x∈C ⇔ x=2 ∨ x=−2 ∨ x=2。",
          "en": "Unfold C={2,−2,2} by list notation: x∈C ⇔ x=2 ∨ x=−2 ∨ x=2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "list_set"
          ],
          "explanation": {
            "zh-Hant": "列舉記號的定義是雙向的。",
            "en": "The definition of list notation is two-way."
          }
        },
        "leaf_basis": "list_set"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "or_laws"
        ],
        "before_tex": "x=2\\lor x=-2\\lor x=2",
        "after_tex": "x=-2\\lor x=2,\\ \\text{i.e.}\\ x\\in B",
        "relation": "equivalent",
        "rule": "or_laws",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依析取的交換與結合改寫為 (x=2∨x=2)∨x=−2；依冪等律 x=2∨x=2 ⇔ x=2，得 x=2∨x=−2；再交換得 x=−2∨x=2，依列舉記號即 x∈B。S09.1 與 S09.2 合起來證得引理 L_C。",
          "en": "By commutativity and associativity of disjunction rewrite as (x=2∨x=2)∨x=−2; by idempotence x=2∨x=2 ⇔ x=2, giving x=2∨x=−2; commuting gives x=−2∨x=2, which by list notation is x∈B. S09.1 and S09.2 together prove Lemma L_C."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "or_laws",
            "list_set"
          ],
          "explanation": {
            "zh-Hant": "交換、結合、冪等都是雙向等價。",
            "en": "Commutativity, associativity and idempotence are all two-way equivalences."
          }
        },
        "leaf_basis": "or_laws + list_set"
      },
      {
        "id": "S10",
        "uses": [
          "G1",
          "G2",
          "list_set"
        ],
        "before_tex": "-2\\in B,\\ -2\\notin D",
        "after_tex": "D\\neq B",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依列舉記號，−2∈B（第一支 −2=−2 成立）；而 −2∈D ⇔ −2=2，但 −2≠2，所以 −2∉D。若 D=B，兩者元素相同，就會有 −2∈D，矛盾。所以 D={2}≠B。",
          "en": "By list notation −2∈B (the first disjunct −2=−2 holds), while −2∈D ⇔ −2=2, and −2≠2, so −2∉D. If D=B the two would have the same elements and −2∈D would follow, a contradiction. So D={2}≠B."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "list_set",
            "set_extensionality",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "一個屬於一邊而不屬於另一邊的元素就足以否定相等；不宣稱兩者沒有共同元素（2 屬於兩者）。",
            "en": "One element in one set but not the other is enough to refute equality; it is not claimed that they share no element (2 is in both)."
          }
        },
        "leaf_basis": "contradiction + set_extensionality + list_set + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "G1",
          "G2",
          "list_set",
          "typed_membership"
        ],
        "before_tex": "2\\ \\text{versus}\\ \\{2\\}",
        "after_tex": "\\{2\\}\\subseteq B\\iff 2\\in B;\\quad \\{2\\}\\in B\\ \\text{is not a true statement}",
        "relation": "implies",
        "rule": "subset_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "2 是整數，{2}=D 是整數的集合，兩者不是同一個對象。它們的正確關係是：依子集定義，{2}⊆B ⇔ 對每個 y，y∈{2} ⇒ y∈B；因 y∈{2} ⇔ y=2，這等價於 2∈B。2∈B 成立，所以 {2}⊆B。另一方面，依 typed_membership，B 的元素是整數，{2} 不是整數，所以「{2}∈B」在本語境中不是為真的命題（在型別化的讀法中它甚至不是良構的）。混淆 2 與 {2}，就會把 ⊆ 與 ∈ 混為一談。",
          "en": "2 is an integer and {2}=D is a set of integers; they are not the same object. Their correct relation: by the definition of subset, {2}⊆B ⇔ for every y, y∈{2} ⇒ y∈B; since y∈{2} ⇔ y=2, this is equivalent to 2∈B. As 2∈B holds, {2}⊆B. On the other hand, by typed_membership the elements of B are integers and {2} is not an integer, so '{2}∈B' is not a true statement in this context (in the typed reading it is not even well formed). Confusing 2 with {2} leads to confusing ⊆ with ∈."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "subset_def",
            "list_set",
            "typed_membership"
          ],
          "explanation": {
            "zh-Hant": "{2}⊆B ⇔ 2∈B 由子集定義與列舉記號雙向成立；「{2}∈B 不為真」只依賴型別區分，不需要其他公理。",
            "en": "{2}⊆B ⇔ 2∈B holds both ways by the definition of subset and list notation; '{2}∈B is not true' depends only on the type distinction and needs no further axiom."
          }
        },
        "leaf_basis": "subset_def + list_set + typed_membership"
      },
      {
        "id": "S12",
        "uses": [
          "G3"
        ],
        "before_tex": "\\exists R\\in U\\ \\forall y\\in U:\\ (y\\in R\\iff y\\notin y)",
        "after_tex": "\\text{fix such an}\\ R\\in U",
        "relation": "implies",
        "rule": "exists_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "轉到 MKO 的第二個誤解：「任意條件都能不受限制地定義集合」。若條件「y∉y」能在論域 U 中定義一個集合，就會有某個 R∈U，其元素恰好是不屬於自身的對象。為了反證，暫時假設這樣的 R 存在，並給它一個新名字 R。",
          "en": "Turn to the second mistake identified in the MKO, that any condition defines a set without restriction. If the condition 'y∉y' defined a set in the domain U, there would be some R∈U whose elements are exactly the objects that are not elements of themselves. For a proof by contradiction, temporarily assume such an R exists and give it the fresh name R."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "exists_elim"
          ],
          "explanation": {
            "zh-Hant": "這是暫時假設，只在 S12 到 S14 的反證範圍內使用；最後的結論 S15 不含見證 R 的自由出現。",
            "en": "This is a temporary assumption, used only inside the proof by contradiction S12 to S14; the witness R has no free occurrence in the final conclusion S15."
          }
        },
        "leaf_basis": "exists_elim"
      },
      {
        "id": "S13",
        "uses": [
          "S12"
        ],
        "before_tex": "\\forall y\\in U:\\ (y\\in R\\iff y\\notin y)",
        "after_tex": "R\\in R\\iff R\\notin R",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [
          "R\\in U"
        ],
        "explanation": {
          "zh-Hant": "S12 的性質對每個 y∈U 成立，而 R 本身就在 U 中，所以可以取 y=R，得 R∈R ⇔ R∉R。",
          "en": "The property in S12 holds for every y∈U, and R itself is in U, so we may take y=R and get R∈R ⇔ R∉R."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "代入需要 R∈U，這正是「R 是論域中的一個集合」的意思。",
            "en": "The instantiation needs R∈U, which is exactly what 'R is a set of the domain' means."
          }
        },
        "leaf_basis": "forall_elim"
      },
      {
        "id": "S14",
        "uses": [
          "S13",
          "excluded_middle"
        ],
        "before_tex": "R\\in R\\iff R\\notin R",
        "after_tex": "R\\in R\\ \\lor\\ R\\notin R",
        "relation": "case_split",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依排中律，R∈R 或 R∉R。兩種情形在 S14.1、S14.2 中都由 S13 導出矛盾，兩支涵蓋所有可能。",
          "en": "By excluded middle, R∈R or R∉R. In S14.1 and S14.2 each case yields a contradiction from S13, and the two branches cover every possibility."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13",
            "excluded_middle",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "兩個情形涵蓋全部可能，依據是排中律。",
            "en": "The two cases cover every possibility, by excluded middle."
          }
        },
        "leaf_basis": "case_analysis + excluded_middle"
      },
      {
        "id": "S14.1",
        "uses": [
          "S13",
          "S14"
        ],
        "before_tex": "R\\in R",
        "after_tex": "R\\notin R,\\ \\text{contradiction}",
        "relation": "implies",
        "rule": "modus_ponens",
        "conditions": [],
        "explanation": {
          "zh-Hant": "情形一：R∈R。由 S13 的「⇒」方向得 R∉R，與 R∈R 同時成立，矛盾。",
          "en": "Case one: R∈R. The '⇒' direction of S13 gives R∉R, which together with R∈R is a contradiction."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13",
            "modus_ponens",
            "iff_intro"
          ],
          "explanation": {
            "zh-Hant": "只用到雙條件的一個方向。",
            "en": "Only one direction of the biconditional is used."
          }
        },
        "leaf_basis": "modus_ponens + iff_intro"
      },
      {
        "id": "S14.2",
        "uses": [
          "S13",
          "S14"
        ],
        "before_tex": "R\\notin R",
        "after_tex": "R\\in R,\\ \\text{contradiction}",
        "relation": "implies",
        "rule": "modus_ponens",
        "conditions": [],
        "explanation": {
          "zh-Hant": "情形二：R∉R。由 S13 的「⇐」方向得 R∈R，與 R∉R 同時成立，矛盾。",
          "en": "Case two: R∉R. The '⇐' direction of S13 gives R∈R, which together with R∉R is a contradiction."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13",
            "modus_ponens",
            "iff_intro"
          ],
          "explanation": {
            "zh-Hant": "只用到雙條件的另一個方向。",
            "en": "Only the other direction of the biconditional is used."
          }
        },
        "leaf_basis": "modus_ponens + iff_intro"
      },
      {
        "id": "S15",
        "uses": [
          "S12",
          "S14",
          "S14.1",
          "S14.2"
        ],
        "before_tex": "\\text{the assumption of S12 leads to a contradiction}",
        "after_tex": "\\neg\\exists R\\in U\\ \\forall y\\in U:\\ (y\\in R\\iff y\\notin y)",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S12 的暫時假設在兩個情形下都導出矛盾，依反證法，這樣的 R 不存在。這個結論對任何論域 U 與任何隸屬關係都成立，沒有用到集合論公理。所以不存在 U 內的對象 R∈U，其對 U 中各 y 的隸屬關係恰好由 y∉y 刻畫：不能要求每個條件都由 U 內的一個集合表示。這不排除在通常集合論中以受限分離形成 R={y∈U ∣ y∉y}⊆U；此時 R 不屬於 U。相對地，S02 用的是在既有集合 ℤ 中分出元素的受限形式。",
          "en": "The temporary assumption of S12 leads to a contradiction in both cases, so by proof by contradiction no such R exists. This holds for every domain U and every membership relation and uses no axiom of set theory. Thus no object R∈U represents the condition y∉y for all y∈U: one cannot require every condition to be represented by a set within U. This does not rule out forming the relative subset R={y∈U | y∉y}⊆U by separation in ordinary set theory; that R is not an element of U. By contrast, S02 used the restricted form that carves elements out of the existing set ℤ."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12",
            "S14",
            "contradiction",
            "exists_elim"
          ],
          "explanation": {
            "zh-Hant": "反證的結論不含見證名字 R 的自由出現，符合 exists_elim 的限制；不宣稱任何特定公理系統中哪些條件可以形成集合（MKO 不選定 ZF 或 ZFC）。",
            "en": "The witness name R has no free occurrence in the conclusion, as exists_elim requires; nothing is claimed about which conditions form sets in any particular axiom system (the MKO chooses neither ZF nor ZFC)."
          }
        },
        "leaf_basis": "contradiction + exists_elim"
      },
      {
        "id": "S16",
        "uses": [
          "S08",
          "S09",
          "S10",
          "S11",
          "S15"
        ],
        "before_tex": "\\text{conclusions of S08, S09, S10, S11, S15}",
        "after_tex": "A=B=C,\\ D\\neq B,\\ \\{2\\}\\subseteq B,\\ \\neg\\exists R\\in U\\ \\forall y\\in U:\\ (y\\in R\\iff y\\notin y)",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "合起來：以條件寫成的 A、以列舉寫成的 B、帶重複的 C 是同一個集合；D={2} 不等於 B；元素 2 與集合 {2} 以 ∈ 與 ⊆ 分別連到 B；不存在 R∈U 對所有 y∈U 滿足 y∈R ⇔ y∉y。",
          "en": "Together: A written by a condition, B written as a list and C written with a repetition are one and the same set; D={2} differs from B; the element 2 and the set {2} relate to B by ∈ and by ⊆ respectively; no R∈U satisfies y∈R iff y∉y for every y∈U."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S08",
            "S09",
            "S10",
            "S11",
            "S15",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：題目要求的四個結論。足夠條件 H：S08、S09、S10、S11、S15；H⇒G 只是並列。H 由 G1 到 G3 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: the four conclusions the problem asks for. Sufficient condition H: S08, S09, S10, S11 and S15; H⇒G just lists them. H is proved forwards from G1 to G3; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在論域 ℤ 中，A={x∈ℤ ∣ x²=4}、B={−2,2}、C={2,−2,2} 有相同元素，所以 A=B=C；D={2}≠B，因為 −2∈B 而 −2∉D。2 是整數，{2} 是集合：2∈B 與 {2}⊆B 成立，而「{2}∈B」不為真。在任何帶有隸屬關係的論域 U 中，都不存在 R∈U 對所有 y∈U 滿足 y∈R ⇔ y∉y；這禁止不受限制地要求每個條件都由 U 內的集合表示，並不禁止在通常集合論中以受限分離形成 R={y∈U ∣ y∉y}⊆U（此時 R∉U）。",
      "en": "In the universe ℤ, A={x∈ℤ ∣ x²=4}, B={−2,2} and C={2,−2,2} have the same elements, so A=B=C; D={2}≠B because −2∈B and −2∉D. 2 is an integer and {2} is a set: 2∈B and {2}⊆B hold, while '{2}∈B' is not true. For any domain U with a membership relation, no R∈U satisfies y∈R iff y∉y for every y∈U. This rules out requiring every condition to be represented without restriction by a set within U; it does not prohibit the relative subset R={y∈U | y∉y}⊆U formed by separation in ordinary set theory (which has R∉U)."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "{2,−2,2} 有三個元素，所以與 {−2,2} 不同",
          "en": "{2,−2,2} has three elements, so it differs from {−2,2}"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x∈{2,−2,2} ⇔ x=−2∨x=2，與 B 的元素相同，依外延性相等（S09）。",
          "en": "x∈{2,−2,2} ⇔ x=−2∨x=2, the same elements as B, so they are equal by extensionality (S09)."
        }
      },
      {
        "claim": {
          "zh-Hant": "2 與 {2} 是同一個東西",
          "en": "2 and {2} are the same thing"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "2 是整數，{2} 是整數的集合；2∈B 成立而「{2}∈B」不為真（S11）。",
          "en": "2 is an integer and {2} is a set of integers; 2∈B holds while '{2}∈B' is not true (S11)."
        }
      },
      {
        "claim": {
          "zh-Hant": "對任何條件 P(y)，{y ∣ P(y)} 都是集合",
          "en": "For every condition P(y), {y ∣ P(y)} is a set"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "取 P(y) 為 y∉y，S12 到 S15 導出矛盾。",
          "en": "Take P(y) to be y∉y; S12 to S15 derive a contradiction."
        }
      },
      {
        "claim": {
          "zh-Hant": "A 用條件寫、B 用列舉寫，所以 A≠B",
          "en": "A is written with a condition and B as a list, so A≠B"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "集合由元素決定，與寫法無關；S08 證明 A=B。",
          "en": "A set is determined by its elements, not by how it is written; S08 proves A=B."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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_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": "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": "zero_product",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "零乘積性質：在 ℤ、ℚ、ℝ、ℂ（或任何整域）中，uv=0 ⇔ u=0 或 v=0。",
          "en": "Zero-product property: in ℤ, ℚ, ℝ, ℂ (or any integral domain), uv=0 ⇔ u=0 or v=0."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "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": "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": "eq_congruence",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。",
          "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule."
        }
      },
      {
        "id": "add_inverse",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。",
          "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)."
        }
      },
      {
        "id": "add_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。",
          "en": "Associativity of addition: (u+v)+w=u+(v+w)."
        }
      },
      {
        "id": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "id": "neg_mul",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。",
          "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv."
        }
      },
      {
        "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": "logic_chain",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "推論的傳遞：由 P⇒Q 與 Q⇒R 得 P⇒R；⇔ 同理。",
          "en": "Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔."
        }
      },
      {
        "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": "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": "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": "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": "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": "modus_ponens",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "肯定前件：由 P 與 P⇒Q 得 Q。",
          "en": "Modus ponens: from P and P⇒Q infer Q."
        }
      },
      {
        "id": "iff_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "雙條件：P⇒Q 與 Q⇒P 都成立時，P⇔Q 成立；反之亦然。",
          "en": "Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold."
        }
      },
      {
        "id": "mko_set",
        "kind": "cited",
        "source": {
          "mko_id": "mko-set",
          "sha256": "4f736510823c29158d78264f57fa9f96c5c2107ddc99999622656fdacc7a5728",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在指定的論域或型別中，集合 A 由哪些對象是其元素來刻畫；具有完全相同元素的集合視為相同。",
          "en": "Within a specified universe or type, a set A is characterized by which objects are its elements; sets with exactly the same elements are regarded as equal."
        }
      },
      {
        "id": "list_set",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "列舉式集合：x∈{a,b,c} ⇔ x=a ∨ x=b ∨ x=c；兩個或一個元素時同理。",
          "en": "List notation: x∈{a,b,c} ⇔ x=a ∨ x=b ∨ x=c; likewise with two elements or one."
        }
      },
      {
        "id": "or_laws",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "析取的交換、結合與冪等：P∨Q ⇔ Q∨P；(P∨Q)∨R ⇔ P∨(Q∨R)；P∨P ⇔ P。",
          "en": "Commutativity, associativity and idempotence of disjunction: P∨Q ⇔ Q∨P; (P∨Q)∨R ⇔ P∨(Q∨R); P∨P ⇔ P."
        }
      },
      {
        "id": "excluded_middle",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "排中律（古典邏輯）：對任何命題 P，P∨¬P。",
          "en": "Excluded middle (classical logic): for every proposition P, P∨¬P."
        }
      },
      {
        "id": "typed_membership",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "型別化的隸屬：在 G1 的語境中，x∈S 只在 x 是整數、S 是整數的集合時才是良構的命題；整數的集合本身不是整數。",
          "en": "Typed membership: in the context of G1, x∈S is a well-formed proposition only when x is an integer and S is a set of integers; a set of integers is not itself an integer."
        }
      },
      {
        "id": "L_C",
        "kind": "lemma",
        "proved_in": [
          "S09.1",
          "S09.2"
        ],
        "statement": {
          "zh-Hant": "對每個 x∈ℤ：x∈C ⇔ x∈B。",
          "en": "For every x∈ℤ: x∈C ⇔ x∈B."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S04",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "x*x-4",
        "rhs": "(x-2)*(x+2)"
      },
      {
        "id": "N2",
        "step": "S06",
        "lhs": "2-2",
        "rhs": "0"
      },
      {
        "id": "N3",
        "step": "S06",
        "lhs": "-2+2",
        "rhs": "0"
      },
      {
        "id": "N4",
        "step": "S08",
        "lhs": "(-2)**2",
        "rhs": "4"
      },
      {
        "id": "N5",
        "step": "S08",
        "lhs": "2**2",
        "rhs": "4"
      },
      {
        "id": "N6",
        "step": "S08",
        "lhs": "3**2",
        "relation": "!=",
        "rhs": "4"
      },
      {
        "id": "N7",
        "step": "S08",
        "lhs": "3",
        "relation": "!=",
        "rhs": "-2"
      },
      {
        "id": "N8",
        "step": "S08",
        "lhs": "3",
        "relation": "!=",
        "rhs": "2"
      },
      {
        "id": "N9",
        "step": "S10",
        "lhs": "-2",
        "relation": "!=",
        "rhs": "2"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "依 MKO：在指定論域中，集合由其元素刻畫，具有相同元素的集合相等。前半在論域 ℤ 中工作，集合都是整數的集合；後半的羅素論證只用純邏輯，對任何帶有隸屬關係的論域都成立，不選定 ZF、ZFC 或其他公理系統。不處理集合運算的一般理論；不改動 canonical MKO 或英文 edition。",
      "en": "Follows the MKO: within a specified universe a set is characterized by its elements, and sets with the same elements are equal. The first part works in the universe ℤ, where all sets are sets of integers; the second part, Russell's argument, uses pure logic and holds for every domain with a membership relation, without choosing ZF, ZFC or any other axiom system. It does not treat the general theory of set operations and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "x\\in A",
        "x 是 A 的元素",
        "x is an element of A"
      ],
      [
        "\\{a,b,c\\}",
        "列舉式集合：x∈{a,b,c} ⇔ x=a∨x=b∨x=c",
        "list notation: x∈{a,b,c} ⇔ x=a∨x=b∨x=c"
      ],
      [
        "\\{x\\in\\mathbb{Z}\\mid P(x)\\}",
        "在 ℤ 中以條件 P 分出的集合",
        "the set carved out of ℤ by the condition P"
      ],
      [
        "A=B",
        "集合相等：元素完全相同",
        "set equality: exactly the same elements"
      ],
      [
        "U",
        "任意帶有隸屬關係的論域（羅素論證用）",
        "any domain with a membership relation (for Russell's argument)"
      ]
    ],
    "md_error_controls": [
      {
        "title": "元素與只含它的集合 / An element and the set containing only it",
        "zh": "MKO 指出的常見誤解是把元素 x 與只含 x 的集合混為一談。2 是整數，{2} 是整數的集合：2∈B 與 {2}⊆B 都成立而且互相等價（S11），但「{2}∈B」不為真，{2} 也不等於 B（S10）。讀到 2 或 {2} 等記號時，先辨認它在本語境中表示整數元素，還是整數的集合，再決定用 ∈ 還是 ⊆。",
        "en": "The MKO names the common mistake of confusing an element x with the set containing only x. 2 is an integer and {2} is a set of integers: 2∈B and {2}⊆B both hold and are equivalent (S11), but '{2}∈B' is not true and {2} is not equal to B (S10). On meeting notation such as 2 or {2}, first identify whether it denotes an integer element or a set of integers in this context, then decide between ∈ and ⊆."
      },
      {
        "title": "不是每個條件都能形成集合 / Not every condition forms a set",
        "zh": "MKO 的另一個誤解是認為任意條件都能不受限制地定義集合。S12 到 S15 證明：對任何論域 U，不存在 R∈U 對所有 y∈U 滿足 y∈R ⇔ y∉y。這不排除在通常集合論中以受限分離形成 R={y∈U ∣ y∉y}⊆U；此時 R∉U。所以集合的形成需要原則，例如只在已知集合中分出元素（S02 的 {x∈ℤ ∣ x²=4}）；本稿不選定具體公理系統。",
        "en": "Another mistake identified in the MKO is believing that any condition defines a set without restriction. S12 to S15 prove that for any domain U, no R∈U satisfies y∈R iff y∉y for every y∈U. This does not exclude forming the relative subset R={y∈U | y∉y}⊆U by separation in ordinary set theory; that R has R∉U. So set formation needs principles, for example carving elements out of a known set (as with {x∈ℤ ∣ x²=4} in S02); this guide does not choose a particular axiom system."
      },
      {
        "title": "順序與重複不產生新集合 / Order and repetition do not create new sets",
        "zh": "C={2,−2,2} 寫了三個符號，但依外延性 C=B={−2,2}（S09）。若要區分順序或重數，需要另一種對象，例如有序三元組 (2,−2,2) 或多重集合；它們不是本 MKO 所說的集合。",
        "en": "C={2,−2,2} is written with three symbols, but by extensionality C=B={−2,2} (S09). Distinguishing order or multiplicity needs a different kind of object, such as the ordered triple (2,−2,2) or a multiset; those are not sets in the sense of this MKO."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "前半的型別化語境（typed_membership）是本稿的前提，對應 MKO 的「指定論域或型別」；在不分型別的集合論中，2 與 {2} 的關係取決於所選公理與數的編碼，本稿不處理。",
        "en": "The typed context of the first part (typed_membership) is a premise of this guide, matching the MKO's 'specified universe or type'; in untyped set theory the relation between 2 and {2} depends on the chosen axioms and the coding of numbers, which this guide does not treat."
      },
      {
        "zh-Hant": "羅素論證只證明不存在 U 內的 R∈U 對 U 中所有 y 表示條件 y∉y，並未否定受限分離形成的子集 R⊆U；哪些條件可以形成集合由具體公理系統決定（MKO 第三段），本稿不選定。",
        "en": "Russell's argument shows only that no internal R∈U represents y∉y for every y∈U, and does not rule out a relative subset R⊆U formed by separation; which conditions do form sets is fixed by a concrete axiom system (the MKO's third paragraph), which this guide does not choose."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "81bb053444cb9fae0c721ca4c93b2655272c5d7fc28922dbed489183ee255de8",
    "original_source_sha256": "81bb053444cb9fae0c721ca4c93b2655272c5d7fc28922dbed489183ee255de8",
    "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."
  }
}