{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-subset-01",
  "mko_id": "mko-subset",
  "title": {
    "zh-Hant": "子集合：以任意元素證明包含，並以反例否定包含",
    "en": "Subset: proving inclusion from an arbitrary element and refuting it by a counterexample"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-subset-01",
    "source": {
      "mko_id": "mko-subset",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "984a77e67dc9322e36b869843513f1cb76f9527912202782848120c123ce99b3",
      "english_sha256": "af2446cfbbb6eff0aba71545b87591295e79df30d0992a5085cf44be4170d534"
    },
    "mode": "definition_walkthrough",
    "audience_prerequisites": [
      "set_builder",
      "subset_def",
      "int_closure",
      "int_discrete",
      "order_basics"
    ],
    "goal": {
      "zh-Hant": "在論域 ℤ 中，令 A 為 6 的倍數之集、B 為偶數之集、C 為 4 的倍數之集。證明 A⊆B，且這個包含是嚴格的（A≠B）；證明 A⊄C，雖然 A 與 C 有共同元素 12；說明 B⊆B 成立（⊆ 允許相等）；並證明 {6}⊆A ⇔ 6∈A，同時分辨兩個語句的型別角色。",
      "en": "In the universe ℤ, let A be the set of multiples of 6, B the set of even integers and C the set of multiples of 4. Prove A⊆B and that this inclusion is strict (A≠B); prove A⊄C even though A and C share the element 12; note that B⊆B holds (⊆ allows equality); and prove {6}⊆A ⇔ 6∈A while separating the type roles of the two statements."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "論域是整數集合 ℤ；本稿所有集合都是 ℤ 的子集合，因此任兩個都在同一論域中（MKO 假設 a1）。",
          "en": "The universe is the set of integers ℤ; every set in this guide is a subset of ℤ, so any two of them lie in the same universe (MKO assumption a1)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "A={n∈ℤ ∣ 存在 k∈ℤ 使 n=6k}，B={n∈ℤ ∣ 存在 m∈ℤ 使 n=2m}，C={n∈ℤ ∣ 存在 j∈ℤ 使 n=4j}。",
          "en": "A={n∈ℤ ∣ n=6k for some k∈ℤ}, B={n∈ℤ ∣ n=2m for some m∈ℤ}, C={n∈ℤ ∣ n=4j for some j∈ℤ}."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "\\text{multiples of }6,\\ \\text{even integers},\\ \\text{multiples of }4\\ \\text{in }\\mathbb{Z}",
        "after_tex": "A=\\{n\\in\\mathbb{Z}\\mid\\exists k\\in\\mathbb{Z},\\ n=6k\\},\\ B=\\{n\\in\\mathbb{Z}\\mid\\exists m\\in\\mathbb{Z},\\ n=2m\\},\\ C=\\{n\\in\\mathbb{Z}\\mid\\exists j\\in\\mathbb{Z},\\ n=4j\\}",
        "relation": "definition",
        "rule": "def_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "以集合建構式為三個集合取名。三者都由 ℤ 中的元素構成，所以它們在同一論域 ℤ 中，滿足 MKO 假設 a1；之後比較它們的包含關係才有意義。依 set_builder，n∈A 恰好表示 n∈ℤ 且有某個整數 k 使 n=6k；B、C 同理。",
          "en": "Name the three sets with set-builder notation. All three consist of elements of ℤ, so they lie in the same universe ℤ, as MKO assumption a1 requires; only then is comparing them by inclusion meaningful. By set_builder, n∈A means exactly that n∈ℤ and n=6k for some integer k, and likewise for B and C."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "def_intro",
            "set_builder"
          ],
          "explanation": {
            "zh-Hant": "此步只命名，不推出新事實。它核對的是前提：三個集合都從 ℤ 取元素，所以論域相同。",
            "en": "This step only introduces names and derives no new fact. What it checks is the premise: all three sets draw their elements from ℤ, so they share a universe."
          }
        },
        "leaf_basis": "def_intro + set_builder"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "mko_subset_def"
        ],
        "before_tex": "A\\subseteq B",
        "after_tex": "\\forall x\\ (x\\in A\\Rightarrow x\\in B)",
        "relation": "equivalent",
        "rule": "mko_subset_def",
        "conditions": [
          "A,B\\subseteq\\mathbb{Z}"
        ],
        "explanation": {
          "zh-Hant": "依 MKO 的定義展開目標：A⊆B 當且僅當對每個對象 x，x∈A 蘊涵 x∈B。對 ℤ 以外的對象，x∈A 為假，蘊涵自動成立，所以實際上只需對 x∈ℤ 檢查。這一步把一個關於兩個集合的語句，換成一個關於所有元素的全稱語句；要證明它，必須處理任意的 x，而不是挑選出來的幾個元素。",
          "en": "Unfold the goal by the MKO's definition: A⊆B holds exactly when, for every object x, x∈A implies x∈B. For objects outside ℤ, x∈A is false and the implication holds automatically, so in effect only x∈ℤ needs checking. The step turns a statement about two sets into a universal statement about all elements; proving it requires handling an arbitrary x, not a few chosen elements."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_subset_def",
            "subset_def"
          ],
          "explanation": {
            "zh-Hant": "MKO 的定義是「當且僅當」，展開與收回兩個方向都由定義保證。",
            "en": "The MKO's definition is an 'if and only if', so both unfolding and folding back are given by the definition."
          }
        },
        "leaf_basis": "mko_subset_def + subset_def + def_unfold"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "S02"
        ],
        "before_tex": "x\\in A\\ \\ (x\\in\\mathbb{Z}\\ \\text{arbitrary})",
        "after_tex": "\\exists k\\in\\mathbb{Z},\\ x=6k",
        "relation": "equivalent",
        "rule": "set_builder",
        "conditions": [
          "x\\in\\mathbb{Z}"
        ],
        "explanation": {
          "zh-Hant": "取一個任意的 x∈ℤ，並暫時假設 x∈A（這個假設在 S07 釋放）。依 A 的集合建構式，x∈A 等價於「有某個整數 k 使 x=6k」。除了 x∈ℤ 與這個暫時假設之外，對 x 不加任何條件。",
          "en": "Take an arbitrary x∈ℤ and temporarily assume x∈A (this assumption is discharged in S07). By the set-builder definition of A, x∈A is equivalent to 'x=6k for some integer k'. Nothing is assumed about x beyond x∈ℤ and this temporary assumption."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_builder"
          ],
          "explanation": {
            "zh-Hant": "集合建構式的成員條件是雙向的：有這樣的 k 就推出 x∈A，x∈A 也提供這樣的 k。",
            "en": "Membership in a set-builder set is two-way: such a k gives x∈A, and x∈A supplies such a k."
          }
        },
        "leaf_basis": "set_builder + def_unfold"
      },
      {
        "id": "S04",
        "uses": [
          "S03"
        ],
        "before_tex": "\\exists k\\in\\mathbb{Z},\\ x=6k",
        "after_tex": "x=6k_0,\\ k_0\\in\\mathbb{Z}",
        "relation": "implies",
        "rule": "exists_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 ∃消去，給滿足條件的某個整數取新名字 k₀，得 x=6k₀。k₀ 只是那個整數的名字，最後的結論 x∈B 不會提到 k₀，所以這樣做是允許的。",
          "en": "By ∃-elimination, give some integer with this property the fresh name k₀, so x=6k₀. k₀ only names that integer, and the final conclusion x∈B does not mention k₀, so this is allowed."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "exists_elim"
          ],
          "explanation": {
            "zh-Hant": "需要 S03 的存在語句，且 k₀ 是新名字、不出現在結論中。不宣稱 k₀ 是唯一的這種整數（事實上唯一，但本稿不需要）。",
            "en": "It needs the existence statement of S03, and k₀ must be a fresh name that does not appear in the conclusion. It does not claim that k₀ is the only such integer (it is, but this guide does not need that)."
          }
        },
        "leaf_basis": "exists_elim"
      },
      {
        "id": "S05",
        "uses": [
          "S04"
        ],
        "before_tex": "x=6k_0",
        "after_tex": "x=2(3k_0)",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 6 寫成 2·3，再用乘法結合律：6k₀=(2·3)k₀=2(3k₀)。兩邊是同一個整數，所以等式的真假不變。這是本證明唯一的計算，而它對每個 k₀ 都成立（N1 以多項式恆等式核對）。",
          "en": "Write 6 as 2·3 and use associativity of multiplication: 6k₀=(2·3)k₀=2(3k₀). Both sides are the same integer, so the truth of the equation is unchanged. This is the only computation in the proof, and it holds for every k₀ (N1 checks it as a polynomial identity)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_assoc",
            "numeric_evaluation",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "反向：2(3k₀)=(2·3)k₀=6k₀，回到 x=6k₀；以相等的式子代換，雙向成立。",
            "en": "Backward: 2(3k₀)=(2·3)k₀=6k₀, returning to x=6k₀; replacing equals by equals works both ways."
          }
        },
        "leaf_basis": "numeric_evaluation + mul_assoc + eq_replace"
      },
      {
        "id": "S06",
        "uses": [
          "S04",
          "S05",
          "int_closure"
        ],
        "before_tex": "x=2(3k_0),\\ k_0\\in\\mathbb{Z}",
        "after_tex": "x\\in B",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [
          "3k_0\\in\\mathbb{Z}"
        ],
        "explanation": {
          "zh-Hant": "3 與 k₀ 都是整數，依整數的乘法封閉性 3k₀∈ℤ。於是 m=3k₀ 是一個使 x=2m 的整數；依 ∃引入與 B 的集合建構式，x∈B。",
          "en": "Since 3 and k₀ are integers, closure under multiplication gives 3k₀∈ℤ. So m=3k₀ is an integer with x=2m; by ∃-introduction and the set-builder definition of B, x∈B."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "int_closure",
            "exists_intro",
            "set_builder"
          ],
          "explanation": {
            "zh-Hant": "關鍵是見證 m=3k₀ 確實是整數。若見證不是整數（例如 3=2·(3/2)），就不能推出屬於 B。反向 x∈B⇒x∈A 不宣稱：2∈B 但 2∉A（S08）。",
            "en": "The key is that the witness m=3k₀ really is an integer. With a non-integer witness (for example 3=2·(3/2)) membership in B would not follow. The converse x∈B⇒x∈A is not claimed: 2∈B but 2∉A (S08)."
          }
        },
        "leaf_basis": "int_closure + exists_intro + set_builder"
      },
      {
        "id": "S07",
        "uses": [
          "S02",
          "S03",
          "S06"
        ],
        "before_tex": "x\\in A\\ \\text{assumed}\\ \\vdash\\ x\\in B,\\quad x\\in\\mathbb{Z}\\ \\text{arbitrary}",
        "after_tex": "A\\subseteq B",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先依 →引入釋放 S03 的暫時假設，得 x∈A⇒x∈B；此後 x∈A 不再當作已知。S03 到 S06 對 x 只用到 x∈ℤ，沒有挑選特定的 x，所以依 ∀引入得 ∀x∈ℤ (x∈A⇒x∈B)；ℤ 以外的對象如 S02 所述自動成立。再依 S02 收回定義，得 A⊆B。",
          "en": "First discharge the temporary assumption of S03 by →-introduction, giving x∈A⇒x∈B; from here on x∈A is no longer treated as known. S03 to S06 used only x∈ℤ and did not pick a particular x, so ∀-introduction gives ∀x∈ℤ (x∈A⇒x∈B); objects outside ℤ satisfy it automatically, as noted in S02. Folding the definition back by S02 gives A⊆B."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S02",
            "S06",
            "imp_intro",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "目標 G：A⊆B。足夠條件 H：對任意 x∈ℤ，x∈A⇒x∈B；H⇒G 由 S02 的等價。H 由 S03 到 S06 從定義正向證成，x 是任意的，沒有把 G 當作前提。",
            "en": "Goal G: A⊆B. Sufficient condition H: for arbitrary x∈ℤ, x∈A⇒x∈B; H⇒G by the equivalence S02. H is proved forwards from the definitions in S03 to S06 with x arbitrary; G is never used as a premise."
          }
        },
        "leaf_basis": "imp_intro + forall_intro + def_unfold"
      },
      {
        "id": "S08",
        "uses": [
          "S01"
        ],
        "before_tex": "2=2\\cdot 1,\\quad 2\\in A\\iff\\exists k\\in\\mathbb{Z},\\ 6k=2",
        "after_tex": "2\\in B,\\ 2\\notin A",
        "relation": "implies",
        "rule": "L_6k_ne_2",
        "conditions": [],
        "explanation": {
          "zh-Hant": "2=2·1 且 1∈ℤ，所以 2∈B。另一方面，依 A 的定義，2∈A 需要某個整數 k 使 6k=2；引理 L_6k_ne_2（在 S08.1、S08.2 證明）說沒有這樣的整數，依量詞否定，2∉A。",
          "en": "2=2·1 with 1∈ℤ, so 2∈B. On the other hand, by the definition of A, 2∈A would need an integer k with 6k=2; Lemma L_6k_ne_2 (proved in S08.1 and S08.2) says there is no such integer, so by negating the quantifier, 2∉A."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_6k_ne_2",
            "set_builder",
            "exists_intro",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "2∈B 只需一個見證；2∉A 必須排除所有整數 k，這正是引理提供的全稱語句。只試 k=0、1 並不夠，所以引理用分情況涵蓋所有整數。",
            "en": "2∈B needs one witness; 2∉A must rule out every integer k, which is the universal statement the lemma supplies. Trying only k=0 and k=1 would not be enough, which is why the lemma covers all integers by cases."
          }
        },
        "leaf_basis": "set_builder + exists_intro + quantifier_negation + int_discrete + order_basics"
      },
      {
        "id": "S08.1",
        "uses": [
          "int_discrete"
        ],
        "before_tex": "k\\in\\mathbb{Z}",
        "after_tex": "k\\le 0\\ \\lor\\ k\\ge 1",
        "relation": "case_split",
        "rule": "int_discrete",
        "conditions": [
          "k\\in\\mathbb{Z}"
        ],
        "explanation": {
          "zh-Hant": "設 k 是任意整數。依整數的離散性（取 c=0），k≤0 或 k≥1，沒有第三種可能；兩種情況涵蓋所有整數。",
          "en": "Let k be any integer. By discreteness of the integers (with c=0), k≤0 or k≥1, with no third possibility; the two cases cover every integer."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "int_discrete"
          ],
          "explanation": {
            "zh-Hant": "分情況必須窮盡；對實數這兩種情況並不窮盡（例如 k=1/3，而 6·(1/3)=2），所以這裡必須用到 k 是整數。",
            "en": "The cases must be exhaustive; for real numbers they are not (for example k=1/3, and 6·(1/3)=2), so this step must use that k is an integer."
          }
        },
        "leaf_basis": "int_discrete + case_analysis"
      },
      {
        "id": "S08.2",
        "uses": [
          "S08.1",
          "order_basics"
        ],
        "before_tex": "k\\le 0\\ \\lor\\ k\\ge 1",
        "after_tex": "6k\\neq 2",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [
          "6>0"
        ],
        "explanation": {
          "zh-Hant": "情況一：k≤0。兩邊乘以正數 6，得 6k≤0<2，所以 6k≠2。情況二：k≥1。兩邊乘以 6，得 6k≥6>2，所以 6k≠2。兩種情況都得 6k≠2，依分情況論證成立；k 是任意的，S08.1 與 S08.2 合起來證得引理 L_6k_ne_2。",
          "en": "Case 1: k≤0. Multiplying by the positive number 6 gives 6k≤0<2, so 6k≠2. Case 2: k≥1. Multiplying by 6 gives 6k≥6>2, so 6k≠2. Both cases give 6k≠2, so proof by cases applies; since k was arbitrary, S08.1 and S08.2 together prove Lemma L_6k_ne_2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08.1",
            "order_basics",
            "case_analysis",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "乘以 6 保持不等號方向，需要 6>0。結論是單向的：只推出 6k≠2，不宣稱其他。",
            "en": "Multiplying by 6 keeps the direction of the inequality because 6>0. The conclusion is one-way: only 6k≠2 is derived, nothing more."
          }
        },
        "leaf_basis": "order_basics + case_analysis + forall_intro + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S07",
          "S08"
        ],
        "before_tex": "A\\subseteq B,\\ 2\\in B,\\ 2\\notin A",
        "after_tex": "A\\neq B,\\ A\\subsetneq B",
        "relation": "implies",
        "rule": "proper_subset_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依集合相等的判準，A=B 需要兩者有相同元素；2 在 B 中而不在 A 中，所以 A≠B。再加上 S07 的 A⊆B，依嚴格子集合定義得 A⊊B。嚴格性是另外證出的事實，不是 A⊆B 本身的含意。",
          "en": "By the criterion for set equality, A=B would require the same elements; 2 is in B but not in A, so A≠B. Together with A⊆B from S07, the definition of proper subset gives A⊊B. Strictness is an extra fact proved here; it is not part of what A⊆B means."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "S08",
            "set_extensionality",
            "proper_subset_def"
          ],
          "explanation": {
            "zh-Hant": "A⊊B 需要兩件事：A⊆B（S07）與 A≠B（S08 的見證 2）。只有前者不足以得到嚴格包含。",
            "en": "A⊊B needs two facts: A⊆B (S07) and A≠B (the witness 2 from S08). The first alone does not give proper inclusion."
          }
        },
        "leaf_basis": "set_extensionality + proper_subset_def + and_intro_elim"
      },
      {
        "id": "S10",
        "uses": [
          "S01",
          "mko_subset_def"
        ],
        "before_tex": "B\\subseteq B",
        "after_tex": "\\forall x\\ (x\\in B\\Rightarrow x\\in B)",
        "relation": "equivalent",
        "rule": "mko_subset_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把同一定義用在兩個集合都是 B 的情形：B⊆B 等價於「對每個 x，x∈B⇒x∈B」。任何命題都蘊涵自己（假設 x∈B，結論 x∈B 就是這個假設），所以右邊成立，B⊆B 成立。這說明 ⊆ 是非嚴格包含：兩集合相等時包含仍成立，A⊆B 並不要求 A≠B。",
          "en": "Apply the same definition with both sets equal to B: B⊆B is equivalent to 'for every x, x∈B⇒x∈B'. Every statement implies itself (assuming x∈B, the conclusion x∈B is that very assumption), so the right side holds and B⊆B holds. This shows that ⊆ is non-strict: inclusion holds when the two sets are equal, and A⊆B does not require A≠B."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_subset_def",
            "imp_intro"
          ],
          "explanation": {
            "zh-Hant": "與 S02 相同，定義是雙向的；右邊的全稱語句由 →引入直接成立。",
            "en": "As in S02 the definition is two-way; the universal statement on the right holds directly by →-introduction."
          }
        },
        "leaf_basis": "mko_subset_def + imp_intro + forall_intro"
      },
      {
        "id": "S11",
        "uses": [
          "S01",
          "mko_subset_def"
        ],
        "before_tex": "\\neg(A\\subseteq C)",
        "after_tex": "\\exists x\\ (x\\in A\\land x\\notin C)",
        "relation": "equivalent",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "要否定 A⊆C，先依定義寫成 ¬∀x (x∈A⇒x∈C)。依量詞否定得 ∃x ¬(x∈A⇒x∈C)，再依蘊涵的否定得 ∃x (x∈A ∧ x∉C)。所以否定包含只需要一個反例：A 中某個不屬於 C 的元素。",
          "en": "To negate A⊆C, first write it by the definition as ¬∀x (x∈A⇒x∈C). Negating the quantifier gives ∃x ¬(x∈A⇒x∈C), and negating the implication gives ∃x (x∈A ∧ x∉C). So refuting inclusion needs just one counterexample: an element of A that is not in C."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_subset_def",
            "quantifier_negation",
            "neg_imp"
          ],
          "explanation": {
            "zh-Hant": "三個改寫（定義、量詞否定、蘊涵的否定）都是等價，所以整步雙向成立。",
            "en": "The three rewrites (definition, quantifier negation, negated implication) are all equivalences, so the whole step holds both ways."
          }
        },
        "leaf_basis": "mko_subset_def + quantifier_negation + neg_imp"
      },
      {
        "id": "S12",
        "uses": [
          "S01"
        ],
        "before_tex": "6=6\\cdot 1,\\quad 6\\in C\\iff\\exists j\\in\\mathbb{Z},\\ 4j=6",
        "after_tex": "6\\in A,\\ 6\\notin C",
        "relation": "implies",
        "rule": "L_4j_ne_6",
        "conditions": [],
        "explanation": {
          "zh-Hant": "6=6·1 且 1∈ℤ，所以 6∈A。依 C 的定義，6∈C 需要某個整數 j 使 4j=6；引理 L_4j_ne_6（在 S12.1、S12.2 證明）說沒有這樣的整數，所以 6∉C。",
          "en": "6=6·1 with 1∈ℤ, so 6∈A. By the definition of C, 6∈C would need an integer j with 4j=6; Lemma L_4j_ne_6 (proved in S12.1 and S12.2) says there is no such integer, so 6∉C."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_4j_ne_6",
            "set_builder",
            "exists_intro",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "與 S08 相同：屬於只需一個見證，不屬於必須排除所有整數 j。",
            "en": "As in S08: membership needs one witness, while non-membership must rule out every integer j."
          }
        },
        "leaf_basis": "set_builder + exists_intro + quantifier_negation + int_discrete + order_basics"
      },
      {
        "id": "S12.1",
        "uses": [
          "int_discrete"
        ],
        "before_tex": "j\\in\\mathbb{Z}",
        "after_tex": "j\\le 1\\ \\lor\\ j\\ge 2",
        "relation": "case_split",
        "rule": "int_discrete",
        "conditions": [
          "j\\in\\mathbb{Z}"
        ],
        "explanation": {
          "zh-Hant": "設 j 是任意整數。依整數的離散性（取 c=1），j≤1 或 j≥2，兩種情況涵蓋所有整數。",
          "en": "Let j be any integer. By discreteness of the integers (with c=1), j≤1 or j≥2, and the two cases cover every integer."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "int_discrete"
          ],
          "explanation": {
            "zh-Hant": "對實數不窮盡：j=3/2 恰好使 4j=6（N13），所以必須用到 j 是整數。",
            "en": "The cases are not exhaustive for reals: j=3/2 gives exactly 4j=6 (N13), so the step must use that j is an integer."
          }
        },
        "leaf_basis": "int_discrete + case_analysis"
      },
      {
        "id": "S12.2",
        "uses": [
          "S12.1",
          "order_basics"
        ],
        "before_tex": "j\\le 1\\ \\lor\\ j\\ge 2",
        "after_tex": "4j\\neq 6",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [
          "4>0"
        ],
        "explanation": {
          "zh-Hant": "情況一：j≤1，乘以 4>0 得 4j≤4<6。情況二：j≥2，乘以 4 得 4j≥8>6。兩種情況都有 4j≠6；j 是任意的，S12.1 與 S12.2 合起來證得引理 L_4j_ne_6。",
          "en": "Case 1: j≤1; multiplying by 4>0 gives 4j≤4<6. Case 2: j≥2; multiplying by 4 gives 4j≥8>6. Both cases give 4j≠6; since j was arbitrary, S12.1 and S12.2 together prove Lemma L_4j_ne_6."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12.1",
            "order_basics",
            "case_analysis",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要 4>0 才能保持不等號方向；只推出 4j≠6，不宣稱反向。",
            "en": "4>0 is needed to keep the inequality direction; only 4j≠6 is derived, and no converse is claimed."
          }
        },
        "leaf_basis": "order_basics + case_analysis + forall_intro + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S11",
          "S12"
        ],
        "before_tex": "6\\in A\\land 6\\notin C",
        "after_tex": "A\\not\\subseteq C",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "以 x=6 為見證：S12 給出 6∈A 且 6∉C，依 ∃引入得 ∃x (x∈A ∧ x∉C)；依 S11 的等價，A⊄C。一個反例就足以否定包含。",
          "en": "Take x=6 as the witness: S12 gives 6∈A and 6∉C, so ∃-introduction gives ∃x (x∈A ∧ x∉C); by the equivalence S11, A⊄C. One counterexample is enough to refute inclusion."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S11",
            "S12",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "目標 G：A⊄C。足夠條件 H：存在 x 使 x∈A 且 x∉C；H⇒G 由 S11。H 由 S12 的具體見證正向證成，沒有把 G 當作前提。",
            "en": "Goal G: A⊄C. Sufficient condition H: some x has x∈A and x∉C; H⇒G by S11. H is proved forwards by the explicit witness of S12; G is never used as a premise."
          }
        },
        "leaf_basis": "exists_intro + and_intro_elim + logic_chain"
      },
      {
        "id": "S14",
        "uses": [
          "S01",
          "S13"
        ],
        "before_tex": "12=6\\cdot 2,\\ 12=4\\cdot 3",
        "after_tex": "12\\in A\\cap C\\ \\text{and yet}\\ A\\not\\subseteq C",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "12=6·2 且 12=4·3，見證 2、3 都是整數，所以 12∈A 且 12∈C，即 12∈A∩C。但 S13 已證 A⊄C。所以找到共同元素並不能推出包含；包含是對 A 的每一個元素的要求。",
          "en": "12=6·2 and 12=4·3 with integer witnesses 2 and 3, so 12∈A and 12∈C, that is 12∈A∩C. Yet S13 proved A⊄C. So finding a common element does not establish inclusion; inclusion is a requirement on every element of A."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13",
            "set_builder",
            "set_operations_def",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "此步只證明一個共同元素存在，並與 S13 並列；不宣稱共同元素推出任何包含關係。",
            "en": "This step only proves that a common element exists and sets it beside S13; it does not claim that a common element implies any inclusion."
          }
        },
        "leaf_basis": "set_builder + exists_intro + set_operations_def + numeric_evaluation"
      },
      {
        "id": "S15",
        "uses": [
          "S12"
        ],
        "before_tex": "\\{6\\}\\subseteq A",
        "after_tex": "6\\in A",
        "relation": "equivalent",
        "rule": "L_singleton_subset",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_singleton_subset（S=A，a=6），{6}⊆A 與 6∈A 同真同假；S12 已證 6∈A，所以 {6}⊆A 也成立。兩個語句真值相同，卻不是同一個語句：6∈A 比較一個元素與一個集合，{6}⊆A 比較兩個集合。",
          "en": "By Lemma L_singleton_subset (with S=A and a=6), {6}⊆A and 6∈A are true or false together; S12 proved 6∈A, so {6}⊆A holds as well. The two statements have the same truth value but are not the same statement: 6∈A compares an element with a set, while {6}⊆A compares two sets."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_singleton_subset"
          ],
          "explanation": {
            "zh-Hant": "引理的兩個方向分別在 S15.1、S15.2 證明，所以等價雙向成立。",
            "en": "The two directions of the lemma are proved in S15.1 and S15.2, so the equivalence holds both ways."
          }
        },
        "leaf_basis": "singleton_def + subset_def + forall_elim + forall_intro + eq_replace"
      },
      {
        "id": "S15.1",
        "uses": [
          "singleton_def",
          "subset_def"
        ],
        "before_tex": "\\{a\\}\\subseteq S",
        "after_tex": "a\\in S",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "（⇒）依單元素集合定義，a∈{a}，因為 a=a。{a}⊆S 是說 {a} 的每個元素都在 S 中；把這個全稱語句代入 x=a（∀消去），得 a∈S。",
          "en": "(⇒) By the definition of a singleton, a∈{a}, since a=a. {a}⊆S says every element of {a} is in S; instantiating this universal statement at x=a (∀-elimination) gives a∈S."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "singleton_def",
            "subset_def",
            "forall_elim",
            "eq_refl_symm_trans"
          ],
          "explanation": {
            "zh-Hant": "需要 a∈{a}，由等號的自反性 a=a 得到。這一步只證 ⇒ 方向。",
            "en": "It needs a∈{a}, which comes from reflexivity a=a. This step proves only the ⇒ direction."
          }
        },
        "leaf_basis": "singleton_def + eq_refl_symm_trans + subset_def + forall_elim"
      },
      {
        "id": "S15.2",
        "uses": [
          "singleton_def",
          "subset_def"
        ],
        "before_tex": "a\\in S",
        "after_tex": "\\{a\\}\\subseteq S",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "（⇐）設 a∈S。取任意 x∈{a}；依單元素集合定義 x=a，依等值代換 x∈S。x 是任意的，依 ∀引入與子集合定義得 {a}⊆S。S15.1 與 S15.2 合起來證得引理 L_singleton_subset。",
          "en": "(⇐) Suppose a∈S. Take an arbitrary x∈{a}; by the definition of a singleton x=a, and by replacement of equals x∈S. Since x was arbitrary, ∀-introduction and the subset definition give {a}⊆S. S15.1 and S15.2 together prove Lemma L_singleton_subset."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "singleton_def",
            "eq_replace",
            "forall_intro",
            "subset_def"
          ],
          "explanation": {
            "zh-Hant": "這一步只證 ⇐ 方向；與 S15.1 合起來才是等價。除了 x∈{a}，對 x 沒有其他假設。",
            "en": "This step proves only the ⇐ direction; together with S15.1 it gives the equivalence. Nothing is assumed about x except x∈{a}."
          }
        },
        "leaf_basis": "singleton_def + eq_replace + imp_intro + forall_intro + subset_def"
      },
      {
        "id": "S16",
        "uses": [
          "S07",
          "S09",
          "S10",
          "S13",
          "S14",
          "S15"
        ],
        "before_tex": "A\\subseteq B,\\ 2\\in B\\setminus A,\\ 6\\in A\\setminus C,\\ 6\\in A",
        "after_tex": "A\\subsetneq B,\\ A\\not\\subseteq C,\\ B\\subseteq B,\\ \\{6\\}\\subseteq A",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整：A⊊B（S07、S09）；A⊄C，雖然 12∈A∩C（S13、S14）；B⊆B 顯示 ⊆ 允許相等（S10）；{6}⊆A 與 6∈A 同真，但型別角色不同（S15）。",
          "en": "Collect the results: A⊊B (S07, S09); A⊄C even though 12∈A∩C (S13, S14); B⊆B shows that ⊆ allows equality (S10); {6}⊆A and 6∈A are both true but play different type roles (S15)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S07",
            "S09",
            "S10",
            "S13",
            "S14",
            "S15",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G 是題目各結論的合取；每一項都由所列步驟正向證成，此步只做合取，沒有把 G 當作前提。",
            "en": "Goal G is the conjunction of the conclusions of the problem; each conjunct is proved forwards 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": "在論域 ℤ 中，6 的倍數之集 A 是偶數之集 B 的子集合，而且是嚴格子集合（2∈B∖A）；A 不是 4 的倍數之集 C 的子集合（6∈A∖C），雖然 12 同屬兩者；B⊆B 說明 ⊆ 允許相等；{6}⊆A 與 6∈A 等價，但是不同的語句。",
      "en": "In the universe ℤ, the set A of multiples of 6 is a subset, indeed a proper subset, of the set B of even integers (2∈B∖A); A is not a subset of the set C of multiples of 4 (6∈A∖C), even though 12 lies in both; B⊆B shows that ⊆ allows equality; {6}⊆A and 6∈A are equivalent but different statements."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "12∈A 且 12∈C，所以 A⊆C",
          "en": "12∈A and 12∈C, so A⊆C"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "包含要求 A 的每個元素都在 C 中；6∈A 但 6∉C（S12、S13）。",
          "en": "Inclusion requires every element of A to be in C; 6∈A but 6∉C (S12, S13)."
        }
      },
      {
        "claim": {
          "zh-Hant": "6、12、18、24、30 都是偶數，所以 A⊆B 已被證明",
          "en": "6, 12, 18, 24 and 30 are even, so A⊆B has been proved"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "結論雖然為真，但有限個例子不證明全稱語句；證明需要從任意 x∈A 出發（S03 到 S07）。",
          "en": "The conclusion is true, but finitely many examples do not prove a universal statement; the proof must start from an arbitrary x∈A (S03 to S07)."
        }
      },
      {
        "claim": {
          "zh-Hant": "A⊆B 要求 A≠B，所以 B⊆B 不成立",
          "en": "A⊆B requires A≠B, so B⊆B is false"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "⊆ 是非嚴格包含；S10 證明 B⊆B 成立。",
          "en": "⊆ is non-strict; S10 proves B⊆B."
        }
      },
      {
        "claim": {
          "zh-Hant": "6∈B 且 6∈A，所以 B⊆A",
          "en": "6∈B and 6∈A, so B⊆A"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "2∈B 但 2∉A（S08），所以 B⊄A。",
          "en": "2∈B but 2∉A (S08), so B⊄A."
        }
      },
      {
        "claim": {
          "zh-Hant": "6∈A 與 6⊆A 是同一個語句",
          "en": "6∈A and 6⊆A are the same statement"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "6 是元素不是集合，6⊆A 型別不合；與 6∈A 等價的是 {6}⊆A，而且兩者仍是不同語句。",
          "en": "6 is an element, not a set, so 6⊆A is ill-typed; the statement equivalent to 6∈A is {6}⊆A, and even these two are different statements."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "set_builder",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。",
          "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x)."
        }
      },
      {
        "id": "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": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "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": "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": "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": "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": "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": "mul_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法結合律：(uv)w=u(vw)。",
          "en": "Associativity of multiplication: (uv)w=u(vw)."
        }
      },
      {
        "id": "exists_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。",
          "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)."
        }
      },
      {
        "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": "imp_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "→引入（釋放假設）：若在暫時假設 P 的作用域中推出 Q，則得到 P⇒Q；此後 P 不再可當作已知使用。",
          "en": "→-introduction (discharging a hypothesis): if Q is derived inside a scope that temporarily assumes P, then P⇒Q follows, and P may no longer be used as known."
        }
      },
      {
        "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": "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": "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": "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": "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": "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": "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": "eq_refl_symm_trans",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。",
          "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w."
        }
      },
      {
        "id": "mko_subset_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-subset",
          "sha256": "984a77e67dc9322e36b869843513f1cb76f9527912202782848120c123ce99b3",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "給定同一論域中的集合 A、B，A⊆B 當且僅當對每個對象 x，若 x∈A，則 x∈B。",
          "en": "For sets A and B in the same universe, A⊆B if and only if, for every object x, x∈A implies x∈B."
        }
      },
      {
        "id": "int_closure",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數對加法與乘法封閉：若 a、b∈ℤ，則 a+b∈ℤ 且 ab∈ℤ。",
          "en": "The integers are closed under addition and multiplication: if a, b∈ℤ then a+b∈ℤ and ab∈ℤ."
        }
      },
      {
        "id": "int_discrete",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數的離散性：對整數 k 與整數 c，必有 k≤c 或 k≥c+1；c 與 c+1 之間沒有整數。",
          "en": "Discreteness of the integers: for integers k and c, either k≤c or k≥c+1; no integer lies strictly between c and c+1."
        }
      },
      {
        "id": "proper_subset_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "嚴格子集合：A⊊B 當且僅當 A⊆B 且 A≠B。",
          "en": "Proper subset: A⊊B if and only if A⊆B and A≠B."
        }
      },
      {
        "id": "neg_imp",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "蘊涵的否定：¬(P⇒Q) 等價於 P∧¬Q。",
          "en": "Negating an implication: ¬(P⇒Q) is equivalent to P∧¬Q."
        }
      },
      {
        "id": "singleton_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "單元素集合：x∈{a} 當且僅當 x=a。",
          "en": "Singleton: x∈{a} if and only if x=a."
        }
      },
      {
        "id": "L_6k_ne_2",
        "kind": "lemma",
        "proved_in": [
          "S08.1",
          "S08.2"
        ],
        "statement": {
          "zh-Hant": "對每個整數 k，6k≠2。",
          "en": "For every integer k, 6k≠2."
        }
      },
      {
        "id": "L_4j_ne_6",
        "kind": "lemma",
        "proved_in": [
          "S12.1",
          "S12.2"
        ],
        "statement": {
          "zh-Hant": "對每個整數 j，4j≠6。",
          "en": "For every integer j, 4j≠6."
        }
      },
      {
        "id": "L_singleton_subset",
        "kind": "lemma",
        "proved_in": [
          "S15.1",
          "S15.2"
        ],
        "statement": {
          "zh-Hant": "對任何集合 S 與同一論域中的對象 a：{a}⊆S ⇔ a∈S。",
          "en": "For any set S and any object a of the same universe: {a}⊆S ⇔ a∈S."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S05",
        "vars": [
          "k"
        ],
        "degree": 1,
        "lhs": "6*k",
        "rhs": "2*(3*k)"
      },
      {
        "id": "N2",
        "step": "S08",
        "lhs": "2*1",
        "rhs": "2"
      },
      {
        "id": "N3",
        "step": "S08.2",
        "lhs": "6*0",
        "relation": "<",
        "rhs": "2"
      },
      {
        "id": "N4",
        "step": "S08.2",
        "lhs": "6*1",
        "relation": ">",
        "rhs": "2"
      },
      {
        "id": "N5",
        "step": "S12",
        "lhs": "6*1",
        "rhs": "6"
      },
      {
        "id": "N6",
        "step": "S12.2",
        "lhs": "4*1",
        "relation": "<",
        "rhs": "6"
      },
      {
        "id": "N7",
        "step": "S12.2",
        "lhs": "4*2",
        "relation": ">",
        "rhs": "6"
      },
      {
        "id": "N8",
        "step": "S14",
        "lhs": "6*2",
        "rhs": "12"
      },
      {
        "id": "N9",
        "step": "S14",
        "lhs": "4*3",
        "rhs": "12"
      },
      {
        "id": "N10",
        "step": "S08",
        "vars": [
          "k"
        ],
        "range": [
          -20,
          20
        ],
        "lhs": "6*k",
        "relation": "!=",
        "rhs": "2"
      },
      {
        "id": "N11",
        "step": "S08",
        "lhs": "2 % 6",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N12",
        "step": "S12",
        "lhs": "6 % 4",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N13",
        "step": "S12.1",
        "lhs": "4*(3/2)",
        "rhs": "6"
      },
      {
        "id": "N14",
        "step": "S08.1",
        "lhs": "6*(1/3)",
        "rhs": "2"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理同一論域 ℤ 中的集合（MKO 假設 a1），示範子集合定義的展開、證明與否定。整數的乘法封閉性與離散性作為前置知識使用，不從整數公理推出。不綁定 Lean 的 Set.Subset 定義；不改動 canonical MKO 或英文 edition。",
      "en": "Treats only sets in the single universe ℤ (MKO assumption a1) and shows how the subset definition is unfolded, proved and refuted. Closure of the integers under multiplication and their discreteness are used as prerequisites and are not derived from axioms for the integers. It does not bind the Lean definition Set.Subset, and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "\\mathbb{Z}",
        "整數集合，本稿的論域",
        "the integers, the universe here"
      ],
      [
        "A,B,C",
        "6 的倍數、偶數、4 的倍數之集",
        "multiples of 6, even integers, multiples of 4"
      ],
      [
        "A\\subseteq B",
        "A 是 B 的子集合（非嚴格包含）",
        "A is a subset of B (non-strict inclusion)"
      ],
      [
        "A\\subsetneq B",
        "A 是 B 的嚴格子集合",
        "A is a proper subset of B"
      ],
      [
        "x\\in A",
        "元素 x 屬於集合 A",
        "the element x belongs to the set A"
      ],
      [
        "\\{a\\}",
        "只含 a 的單元素集合",
        "the singleton set containing a"
      ]
    ],
    "md_error_controls": [
      {
        "title": "只檢查部分元素不夠 / Checking some elements is not enough",
        "zh": "MKO 指出常見誤解是只檢查部分元素就斷言包含。12 同屬 A 與 C，但 A⊄C（6∈A∖C）。同樣地，驗證 6、12、18、24、30 都是偶數，只說明這五個元素屬於 B，並不證明 A⊆B；本稿的證明從任意的 x∈A 出發（S03 到 S07），用到的只有 x=6k₀ 與恆等式 6k₀=2(3k₀)，所以涵蓋 A 的每一個元素。",
        "en": "The MKO names the mistake of asserting inclusion after checking only some elements. 12 lies in both A and C, yet A⊄C (6∈A∖C). Likewise, verifying that 6, 12, 18, 24, 30 are even shows only that these five elements lie in B and does not prove A⊆B; the proof here starts from an arbitrary x∈A (S03 to S07) and uses only x=6k₀ and the identity 6k₀=2(3k₀), so it covers every element of A."
      },
      {
        "title": "⊆ 是非嚴格包含 / ⊆ is non-strict",
        "zh": "另一個誤解是認為 A⊆B 必須 A≠B。S10 證明 B⊆B 成立，所以 ⊆ 允許兩集合相等。若要說嚴格包含 A⊊B，必須另外證明 A≠B，本稿在 S08、S09 以見證 2∈B∖A 完成；不能從 A⊆B 直接讀出。",
        "en": "Another mistake is to think A⊆B forces A≠B. S10 proves B⊆B, so ⊆ allows the two sets to be equal. To claim proper inclusion A⊊B one must prove A≠B separately, which S08 and S09 do with the witness 2∈B∖A; it cannot be read off from A⊆B."
      },
      {
        "title": "x∈A 與 {x}⊆A 不是同一語句 / x∈A and {x}⊆A are different statements",
        "zh": "MKO 提醒不要把 x∈A 與 {x}⊆A 混為同一語句。S15 證明兩者等價（同真同假），但型別角色不同：6∈A 的左邊是整數，{6}⊆A 的左邊是整數集合。寫成 6⊆A 是型別錯誤（6 不是集合）；在 Lean 的 Set ℤ 中，{6}∈A 也不是合法語句，因為 A 的元素是整數而不是整數集合。",
        "en": "The MKO warns against treating x∈A and {x}⊆A as the same statement. S15 proves they are equivalent (true or false together), but their type roles differ: in 6∈A the left side is an integer, in {6}⊆A it is a set of integers. Writing 6⊆A is a type error (6 is not a set); in Lean's Set ℤ, {6}∈A is not a well-formed statement either, because the elements of A are integers, not sets of integers."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "整數的乘法封閉性與離散性（int_closure、int_discrete）是前置知識，本稿不從整數公理推出。",
        "en": "Closure of the integers under multiplication and their discreteness (int_closure, int_discrete) are prerequisites; this guide does not derive them from axioms for the integers."
      },
      {
        "zh-Hant": "N10 只在 k=−20 到 20 的範圍內核對 6k≠2，是有限檢查，不證明全稱語句；全稱語句由 S08.1、S08.2 證明。N11、N12 以餘數做的交叉核對只對這兩個數有效，論證不依賴它們。",
        "en": "N10 checks 6k≠2 only for k from −20 to 20; it is a finite check and does not prove the universal statement, which is proved in S08.1 and S08.2. N11 and N12 are cross-checks by remainders, valid for those two numbers only; the argument does not depend on them."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "88bf090b03281e737b94f71d45f7b57b3e7962272b65b9b44ff45256ac8be3a0",
    "original_source_sha256": "88bf090b03281e737b94f71d45f7b57b3e7962272b65b9b44ff45256ac8be3a0",
    "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."
  }
}