{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-event-independence-01",
  "mko_id": "mko-event-independence",
  "title": {
    "zh-Hant": "事件獨立：以交集乘積判定，並核對條件式、補集與零機率事件",
    "en": "Event independence: deciding by the product rule, with the conditional form, complements and a zero-probability event"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-event-independence-01",
    "source": {
      "mko_id": "mko-event-independence",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "9b6f08c178cbc612effac540bd933bfde0d1b9ebc7708cc7ab928ac4505ad563",
      "english_sha256": "aec2eb901fdb80f97ffdcf96a423b5c1c1bd366fc0f88344974d9133c21ff8e6"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "prob_axioms",
      "set_operations_def",
      "frac_rules",
      "mul_inverse_nonzero",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "公正骰子模型：Ω={1,2,3,4,5,6}，每個結果的質量為 1/6。設 A={2,4,6}、B={1,2,3,4}、D={1,3,5}、Z=∅。（一）以乘積定義判定 A 與 B 是否獨立，並在 P(B)>0 下核對 P(A|B)=P(A)。（二）證明 A 與 B 獨立推出 A 與 Bᶜ 獨立，並以數字核對。（三）判定互斥事件 A、D 是否獨立。（四）判定 A 與零機率事件 Z 是否獨立，並說明條件式為何不能使用。",
      "en": "Fair-die model: Ω={1,2,3,4,5,6}, each outcome with mass 1/6. Let A={2,4,6}, B={1,2,3,4}, D={1,3,5} and Z=∅. (1) Decide by the product definition whether A and B are independent, and with P(B)>0 check P(A|B)=P(A). (2) Prove that independence of A and B implies independence of A and Bᶜ, and check it numerically. (3) Decide whether the mutually exclusive events A and D are independent. (4) Decide whether A and the zero-probability event Z are independent, and explain why the conditional form cannot be used."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "Ω={1,2,3,4,5,6}，事件族為 Ω 的全部子集，每個結果的質量 m(ω)=1/6，P(E)=Σ_{ω∈E} m(ω)。",
          "en": "Ω={1,2,3,4,5,6}, the events are all subsets of Ω, each outcome has mass m(ω)=1/6, and P(E)=Σ_{ω∈E} m(ω)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "A={2,4,6}、B={1,2,3,4}、D={1,3,5}、Z=∅，四者都是同一個 Ω 的子集。",
          "en": "A={2,4,6}, B={1,2,3,4}, D={1,3,5} and Z=∅, all subsets of the same Ω."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "\\Omega=\\{1,2,3,4,5,6\\},\\ m(\\omega)=\\tfrac16",
        "after_tex": "P(E)=\\sum_{\\omega\\in E}m(\\omega)=\\tfrac{|E|}{6},\\ \\ P(\\Omega)=6\\cdot\\tfrac16=1",
        "relation": "definition",
        "rule": "point_mass_sum",
        "conditions": [],
        "explanation": {
          "zh-Hant": "所有事件都在同一個模型中計算機率：P(E) 是 E 中質量之和；因每個質量都是 1/6，P(E)=|E|/6。總質量 6·(1/6)=1，所以這是歸一化的機率模型。MKO 要求兩個事件屬於同一機率空間，這一步把它固定下來。",
          "en": "Every event is measured in this one model: P(E) is the sum of the masses in E, and since each mass is 1/6, P(E)=|E|/6. The total mass is 6·(1/6)=1, so the model is normalized. The MKO requires both events to live in one probability space; this step fixes it."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "point_mass_sum"
          ],
          "explanation": {
            "zh-Hant": "只寫出模型並核對總和為 1（N1）；之後所有機率都由這個 P 計算，不能混用另一個模型的數字。",
            "en": "It only states the model and checks that the total is 1 (N1); every later probability is computed with this P, and numbers from another model may not be mixed in."
          }
        },
        "leaf_basis": "point_mass_sum + prob_axioms + numeric_evaluation"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "G2"
        ],
        "before_tex": "P(A),\\ A=\\{2,4,6\\}",
        "after_tex": "P(A)=\\tfrac36=\\tfrac12",
        "relation": "equivalent",
        "rule": "point_mass_sum",
        "conditions": [],
        "explanation": {
          "zh-Hant": "A 有 3 個結果，P(A)=3·(1/6)=3/6=1/2。",
          "en": "A has 3 outcomes, so P(A)=3·(1/6)=3/6=1/2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "point_mass_sum",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "反向：1/2=3/6 是 A 的三個質量之和，兩式表示同一個數（N2）。",
            "en": "Backward: 1/2=3/6 is the sum of the three masses of A, so both expressions name the same number (N2)."
          }
        },
        "leaf_basis": "point_mass_sum + frac_rules + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "G2"
        ],
        "before_tex": "P(B),\\ B=\\{1,2,3,4\\}",
        "after_tex": "P(B)=\\tfrac46=\\tfrac23",
        "relation": "equivalent",
        "rule": "point_mass_sum",
        "conditions": [],
        "explanation": {
          "zh-Hant": "B 有 4 個結果，P(B)=4/6=2/3（分子分母同除 2）。",
          "en": "B has 4 outcomes, so P(B)=4/6=2/3 (divide numerator and denominator by 2)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "point_mass_sum",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "約分是等值改寫，2/3=4/6，雙向成立（N3）。",
            "en": "Reducing a fraction is an equal rewriting, 2/3=4/6, valid both ways (N3)."
          }
        },
        "leaf_basis": "point_mass_sum + frac_rules + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "G2"
        ],
        "before_tex": "A\\cap B",
        "after_tex": "\\{2,4\\}",
        "relation": "equivalent",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "交集由同時屬於 A 與 B 的結果組成：2、4 屬於兩者；6 不在 B；1、3 不在 A；5 兩者都不在。",
          "en": "The intersection consists of outcomes in both A and B: 2 and 4 are in both; 6 is not in B; 1 and 3 are not in A; 5 is in neither."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_operations_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "逐一核對 Ω 的六個結果，兩邊含有相同元素，依集合外延性相等。",
            "en": "Checking each of the six outcomes of Ω, both sides contain the same elements, so they are equal by extensionality."
          }
        },
        "leaf_basis": "set_operations_def + set_extensionality"
      },
      {
        "id": "S05",
        "uses": [
          "S01",
          "S04"
        ],
        "before_tex": "P(A\\cap B)=P(\\{2,4\\})",
        "after_tex": "P(A\\cap B)=\\tfrac26=\\tfrac13",
        "relation": "equivalent",
        "rule": "point_mass_sum",
        "conditions": [],
        "explanation": {
          "zh-Hant": "A∩B 有 2 個結果，機率 2/6=1/3。",
          "en": "A∩B has 2 outcomes, so its probability is 2/6=1/3."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "point_mass_sum",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "同 S03，約分雙向成立（N4）。",
            "en": "As in S03, the reduction is valid both ways (N4)."
          }
        },
        "leaf_basis": "point_mass_sum + frac_rules + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S02",
          "S03"
        ],
        "before_tex": "P(A)P(B)",
        "after_tex": "\\tfrac12\\cdot\\tfrac23=\\tfrac13",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "分數相乘：(1/2)(2/3)=2/6=1/3。",
          "en": "Multiply fractions: (1/2)(2/3)=2/6=1/3."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確有理數運算，兩邊同值（N5），雙向成立。",
            "en": "Exact rational arithmetic; both sides have the same value (N5), so it holds both ways."
          }
        },
        "leaf_basis": "frac_rules + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S05",
          "S06",
          "mko_event_indep_def"
        ],
        "before_tex": "P(A\\cap B)=\\tfrac13=P(A)P(B)",
        "after_tex": "A,\\ B\\ \\text{independent}",
        "relation": "equivalent",
        "rule": "mko_event_indep_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 的主資格，獨立就是交集機率等於兩個機率的乘積。S05 與 S06 給出兩邊都是 1/3，所以 A 與 B 獨立。判定用的是精確有理數相等，不是近似。",
          "en": "By the MKO's main criterion, independence means the probability of the intersection equals the product of the two probabilities. S05 and S06 give 1/3 on both sides, so A and B are independent. The decision uses exact rational equality, not an approximation."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_event_indep_def",
            "S05",
            "S06"
          ],
          "explanation": {
            "zh-Hant": "MKO 的定義是「當且僅當」，所以這一步雙向成立：獨立也推回乘積等式。",
            "en": "The MKO's definition is an 'iff', so the step holds both ways: independence also gives back the product equation."
          }
        },
        "leaf_basis": "mko_event_indep_def + def_unfold + eq_refl_symm_trans"
      },
      {
        "id": "S08",
        "uses": [
          "S03",
          "S05",
          "cond_prob_def"
        ],
        "before_tex": "P(B)=\\tfrac23>0",
        "after_tex": "P(A\\mid B)=\\frac{P(A\\cap B)}{P(B)}=\\frac{1/3}{2/3}=\\tfrac12",
        "relation": "implies",
        "rule": "cond_prob_def",
        "conditions": [
          "P(B)>0"
        ],
        "explanation": {
          "zh-Hant": "因 P(B)=2/3>0，條件機率有定義：P(A|B)=(1/3)/(2/3)=(1/3)(3/2)=1/2。直觀上，已知結果在 {1,2,3,4} 中時，偶數 2、4 佔一半。",
          "en": "Since P(B)=2/3>0 the conditional probability is defined: P(A|B)=(1/3)/(2/3)=(1/3)(3/2)=1/2. Intuitively, given that the outcome is in {1,2,3,4}, the even outcomes 2 and 4 make up half."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S05",
            "cond_prob_def",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "前提是 P(B)>0；沒有它，這個除法不被允許（見 S14）。數值由 N6 核對。",
            "en": "The premise is P(B)>0; without it the division is not allowed (see S14). The value is checked by N6."
          }
        },
        "leaf_basis": "cond_prob_def + frac_rules + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S02",
          "S07",
          "S08"
        ],
        "before_tex": "P(A\\mid B)=\\tfrac12,\\ P(A)=\\tfrac12,\\ P(B)>0",
        "after_tex": "P(A\\mid B)=P(A),\\ \\text{consistent with}\\ P(A\\cap B)=P(A)P(B)",
        "relation": "implies",
        "rule": "L_cond_equiv",
        "conditions": [
          "P(B)>0"
        ],
        "explanation": {
          "zh-Hant": "S08 給 P(A|B)=1/2=P(A)。依引理 L_cond_equiv（在 P(B)>0 下乘積式與條件式等價），這與 S07 的乘積判定一致：知道 B 發生不改變 A 的機率。引理在子步 S09.1、S09.2 證明。",
          "en": "S08 gives P(A|B)=1/2=P(A). By Lemma L_cond_equiv (with P(B)>0 the product form and the conditional form are equivalent), this agrees with the product decision of S07: learning that B occurred does not change the probability of A. The lemma is proved in sub-steps S09.1 and S09.2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "L_cond_equiv"
          ],
          "explanation": {
            "zh-Hant": "引理需要 P(B)>0，S03 已確認。主資格仍是 S07 的乘積式；條件式只是在正分母時的等價核對。",
            "en": "The lemma needs P(B)>0, established in S03. The main criterion remains the product form of S07; the conditional form is only an equivalent check when the denominator is positive."
          }
        },
        "leaf_basis": "L_cond_equiv + eq_refl_symm_trans"
      },
      {
        "id": "S09.1",
        "uses": [
          "cond_prob_def"
        ],
        "before_tex": "P(A\\mid B)=P(A)",
        "after_tex": "\\frac{P(A\\cap B)}{P(B)}=P(A)",
        "relation": "equivalent",
        "rule": "cond_prob_def",
        "conditions": [
          "P(B)>0"
        ],
        "explanation": {
          "zh-Hant": "在 P(B)>0 下，依條件機率定義把 P(A|B) 換成 P(A∩B)/P(B)。若 P(B)=0，左式根本沒有定義，這一步不能做。",
          "en": "With P(B)>0, replace P(A|B) by P(A∩B)/P(B) according to the definition. If P(B)=0 the left side is not defined at all and the step cannot be taken."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "cond_prob_def",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "展開與折回定義互為反操作，前提 P(B)>0。",
            "en": "Unfolding and folding the definition undo each other, given P(B)>0."
          }
        },
        "leaf_basis": "cond_prob_def + def_unfold"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "mul_inverse_nonzero"
        ],
        "before_tex": "\\frac{P(A\\cap B)}{P(B)}=P(A)",
        "after_tex": "P(A\\cap B)=P(A)P(B)",
        "relation": "equivalent",
        "rule": "mul_inverse_nonzero",
        "conditions": [
          "P(B)\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "兩邊同乘 P(B)；反向是兩邊同乘 1/P(B)，需要 P(B)≠0，由 P(B)>0 保證。S09.1 與 S09.2 合起來證得引理 L_cond_equiv。",
          "en": "Multiply both sides by P(B); the backward move multiplies by 1/P(B), which needs P(B)≠0, guaranteed by P(B)>0. S09.1 and S09.2 together prove Lemma L_cond_equiv."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_inverse_nonzero",
            "eq_congruence"
          ],
          "explanation": {
            "zh-Hant": "乘 P(B) 與乘 1/P(B) 互為反運算，條件 P(B)≠0。",
            "en": "Multiplying by P(B) and by 1/P(B) undo each other, under the condition P(B)≠0."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero + mul_assoc + mul_one"
      },
      {
        "id": "S10",
        "uses": [
          "S07"
        ],
        "before_tex": "P(A\\cap B)=P(A)P(B)",
        "after_tex": "P(A\\cap B^c)=P(A)P(B^c)",
        "relation": "implies",
        "rule": "L_complement",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 指出事件與補集的獨立可由同一等式推導。依引理 L_complement，由 S07 得 A 與 Bᶜ 也獨立。引理在子步 S10.1 到 S10.3 證明，只用可加性與分配律，不需要 P(B)>0。",
          "en": "The MKO notes that independence of an event and a complement follows from the same equation. By Lemma L_complement, S07 gives that A and Bᶜ are independent too. The lemma is proved in sub-steps S10.1 to S10.3 using only additivity and distributivity, without needing P(B)>0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "L_complement"
          ],
          "explanation": {
            "zh-Hant": "需要 S07 的乘積等式；反向（由 A⊥Bᶜ 推回 A⊥B）也成立，因為 (Bᶜ)ᶜ=B，但本步不使用。",
            "en": "It needs the product equation of S07; the converse (from A⊥Bᶜ back to A⊥B) is also true since (Bᶜ)ᶜ=B, but this step does not use it."
          }
        },
        "leaf_basis": "prob_axioms + distributive"
      },
      {
        "id": "S10.1",
        "uses": [
          "set_operations_def"
        ],
        "before_tex": "A",
        "after_tex": "A=(A\\cap B)\\cup(A\\cap B^c),\\ \\ (A\\cap B)\\cap(A\\cap B^c)=\\varnothing",
        "relation": "equivalent",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "A 的每個結果要嘛在 B 中、要嘛在 Bᶜ 中，恰好一種，所以 A 是 A∩B 與 A∩Bᶜ 的不交聯集。",
          "en": "Each outcome of A lies either in B or in Bᶜ, exactly one of the two, so A is the disjoint union of A∩B and A∩Bᶜ."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_operations_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "x∈A ⇔ x∈A∩B 或 x∈A∩Bᶜ，兩邊元素相同；不交性來自 B∩Bᶜ=∅。",
            "en": "x∈A ⇔ x∈A∩B or x∈A∩Bᶜ, so both sides have the same elements; disjointness comes from B∩Bᶜ=∅."
          }
        },
        "leaf_basis": "set_operations_def + set_extensionality"
      },
      {
        "id": "S10.2",
        "uses": [
          "S10.1",
          "prob_axioms"
        ],
        "before_tex": "A=(A\\cap B)\\sqcup(A\\cap B^c)",
        "after_tex": "P(A)=P(A\\cap B)+P(A\\cap B^c)",
        "relation": "implies",
        "rule": "prob_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "不交事件的機率相加（機率公理），所以 P(A) 等於兩部分之和；移項得 P(A∩Bᶜ)=P(A)−P(A∩B)。",
          "en": "Probabilities of disjoint events add (probability axioms), so P(A) is the sum of the two parts; rearranging gives P(A∩Bᶜ)=P(A)−P(A∩B)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10.1",
            "prob_axioms"
          ],
          "explanation": {
            "zh-Hant": "可加性需要兩部分不交，S10.1 已確認；若兩部分重疊，和會重複計算。",
            "en": "Additivity needs the two parts to be disjoint, established in S10.1; if they overlapped the sum would double count."
          }
        },
        "leaf_basis": "prob_axioms + eq_congruence"
      },
      {
        "id": "S10.3",
        "uses": [
          "S10.2",
          "prob_axioms",
          "distributive"
        ],
        "before_tex": "P(A\\cap B^c)=P(A)-P(A\\cap B)",
        "after_tex": "P(A\\cap B^c)=P(A)-P(A)P(B)=P(A)(1-P(B))=P(A)P(B^c)",
        "relation": "implies",
        "rule": "distributive",
        "conditions": [
          "P(A\\cap B)=P(A)P(B)"
        ],
        "explanation": {
          "zh-Hant": "代入假設 P(A∩B)=P(A)P(B)，提出公因子 P(A)，再用 P(Bᶜ)=1−P(B)（機率公理的推論）。S10.1 到 S10.3 合起來證得引理 L_complement。",
          "en": "Substitute the hypothesis P(A∩B)=P(A)P(B), factor out P(A), and use P(Bᶜ)=1−P(B) (a consequence of the probability axioms). S10.1 to S10.3 together prove Lemma L_complement."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10.2",
            "distributive",
            "prob_axioms"
          ],
          "explanation": {
            "zh-Hant": "用到的假設只有乘積等式本身；推導中沒有除法，所以不需要 P(B)>0 或 P(A)>0。",
            "en": "The only hypothesis used is the product equation itself; there is no division, so neither P(B)>0 nor P(A)>0 is needed."
          }
        },
        "leaf_basis": "distributive + prob_axioms + eq_replace"
      },
      {
        "id": "S11",
        "uses": [
          "S01",
          "S02",
          "S03",
          "S10"
        ],
        "before_tex": "B^c=\\{5,6\\},\\ A\\cap B^c=\\{6\\}",
        "after_tex": "P(A\\cap B^c)=\\tfrac16=\\tfrac12\\cdot\\tfrac13=P(A)P(B^c)",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "直接核對：Bᶜ={5,6}，P(Bᶜ)=1/3；A∩Bᶜ={6}，機率 1/6；乘積 (1/2)(1/3)=1/6。與 S10 的一般推論一致，也與 S10.2 的 P(A)−P(A∩B)=1/2−1/3=1/6 一致。",
          "en": "Direct check: Bᶜ={5,6}, P(Bᶜ)=1/3; A∩Bᶜ={6} has probability 1/6; the product is (1/2)(1/3)=1/6. This agrees with the general result of S10 and with S10.2: P(A)−P(A∩B)=1/2−1/3=1/6."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S10",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "這是對這組數字的獨立重算（N7、N8），不是 L_complement 的證明。",
            "en": "This is an independent recomputation for these numbers (N7, N8), not a proof of L_complement."
          }
        },
        "leaf_basis": "numeric_evaluation + set_operations_def + frac_rules"
      },
      {
        "id": "S12",
        "uses": [
          "S01",
          "S02",
          "G2",
          "mko_event_indep_def"
        ],
        "before_tex": "A\\cap D=\\{2,4,6\\}\\cap\\{1,3,5\\}=\\varnothing",
        "after_tex": "P(A\\cap D)=0\\neq\\tfrac14=P(A)P(D)\\ \\Rightarrow\\ A,D\\ \\text{not independent}",
        "relation": "implies",
        "rule": "mko_event_indep_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "A 與 D 互斥：沒有結果同時是偶數與奇數，P(A∩D)=P(∅)=0。但 P(D)=3/6=1/2，P(A)P(D)=1/4≠0。乘積等式不成立，所以 A 與 D 不獨立。互斥反而表示強烈相依：知道 D 發生就知道 A 不發生。",
          "en": "A and D are mutually exclusive: no outcome is both even and odd, so P(A∩D)=P(∅)=0. But P(D)=3/6=1/2 and P(A)P(D)=1/4≠0. The product equation fails, so A and D are not independent. Mutual exclusion is in fact strong dependence: knowing D occurred tells us A did not."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_event_indep_def",
            "S02",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "否定獨立只需乘積等式在這一對事件不成立（N9）。",
            "en": "Denying independence needs only that the product equation fails for this pair (N9)."
          }
        },
        "leaf_basis": "mko_event_indep_def + point_mass_sum + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S12",
          "mko_event_indep_def"
        ],
        "before_tex": "A\\cap D=\\varnothing,\\ P(A)>0,\\ P(D)>0",
        "after_tex": "P(A\\cap D)=0<P(A)P(D)\\ \\Rightarrow\\ A,D\\ \\text{not independent}",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "P(A)>0",
          "P(D)>0"
        ],
        "explanation": {
          "zh-Hant": "一般化：任何兩個互斥且機率皆為正的事件都不獨立，因為兩個正數的乘積為正，而交集機率為 0。條件「機率皆為正」不可省略：若其中一個機率為 0，見 S14。",
          "en": "Generalization: any two mutually exclusive events with positive probabilities are not independent, because the product of two positive numbers is positive while the intersection has probability 0. The condition of positive probabilities cannot be dropped: if one of them is 0, see S14."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "prob_axioms"
          ],
          "explanation": {
            "zh-Hant": "需要 u>0、w>0 ⇒ uw>0（順序公理）與 P(∅)=0；不宣稱反向，即不宣稱不獨立的事件必互斥。",
            "en": "It needs u>0, w>0 ⇒ uw>0 (order axioms) and P(∅)=0; no converse is claimed, that is, dependent events need not be mutually exclusive."
          }
        },
        "leaf_basis": "order_basics + prob_axioms + mko_event_indep_def"
      },
      {
        "id": "S14",
        "uses": [
          "S01",
          "S02",
          "G2",
          "mko_event_indep_def",
          "cond_prob_def"
        ],
        "before_tex": "Z=\\varnothing,\\ P(Z)=0",
        "after_tex": "P(A\\cap Z)=0=P(A)\\cdot 0=P(A)P(Z)\\ \\Rightarrow\\ A,Z\\ \\text{independent};\\ P(A\\mid Z)\\ \\text{undefined}",
        "relation": "implies",
        "rule": "mko_event_indep_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在本模型中，每個非空事件至少有質量 1/6，所以唯一的零機率事件是 Z=∅。A∩Z=∅，P(A∩Z)=0，而 P(A)P(Z)=(1/2)·0=0，乘積等式成立，所以依定義 A 與 Z 獨立。另一方面 P(Z)=0，P(A|Z) 依條件機率定義不存在；MKO 規定此時不得以未定義的條件機率代替乘積定義。",
          "en": "In this model every nonempty event has mass at least 1/6, so the only zero-probability event is Z=∅. A∩Z=∅ gives P(A∩Z)=0, and P(A)P(Z)=(1/2)·0=0, so the product equation holds and by the definition A and Z are independent. On the other hand P(Z)=0, so P(A|Z) does not exist by the definition of conditional probability; the MKO forbids replacing the product definition by an undefined conditional probability in this case."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_event_indep_def",
            "cond_prob_def",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "判定只用乘積定義與 u·0=0（N10）；沒有使用 P(A|Z)，也不宣稱它等於任何數。",
            "en": "The decision uses only the product definition and u·0=0 (N10); it does not use P(A|Z) and does not claim it equals any number."
          }
        },
        "leaf_basis": "mko_event_indep_def + mul_zero + prob_axioms"
      },
      {
        "id": "S15",
        "uses": [
          "S07",
          "S09",
          "S10",
          "S11",
          "S12",
          "S14"
        ],
        "before_tex": "\\text{S07},\\ \\text{S09},\\ \\text{S10},\\ \\text{S12},\\ \\text{S14}",
        "after_tex": "A\\perp B;\\ A\\perp B^c;\\ A\\not\\perp D;\\ A\\perp Z",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "合併四個任務：A 與 B 獨立（乘積式，並以條件式核對）；A 與 Bᶜ 獨立（一般引理加數字核對）；互斥的 A、D 不獨立；A 與零機率事件 Z 依乘積定義獨立，而 P(A|Z) 無定義。",
          "en": "Combine the four tasks: A and B are independent (product form, checked by the conditional form); A and Bᶜ are independent (general lemma plus a numeric check); the mutually exclusive A and D are not independent; A and the zero-probability event Z are independent by the product definition, while P(A|Z) is undefined."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S07",
            "S10",
            "S12",
            "S14",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G 是四個判定的合取；足夠條件 H 是 S07、S10、S12、S14，H⇒G 由合取引入，H 由已知正向證得，沒有把 G 當作前提。",
            "en": "Goal G is the conjunction of the four decisions; the sufficient condition H is S07, S10, S12 and S14, with H⇒G by conjunction introduction; H is proved forwards from the givens and G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在公正骰子模型中，P(A∩B)=1/3=P(A)P(B)，所以 A={2,4,6} 與 B={1,2,3,4} 獨立，且因 P(B)>0，P(A|B)=1/2=P(A)。由同一等式推出 A 與 Bᶜ 獨立，數字核對 1/6=(1/2)(1/3)。互斥的 A 與 D={1,3,5} 不獨立（0≠1/4）。A 與零機率事件 ∅ 依乘積定義獨立，而條件機率 P(A|∅) 無定義、不得使用。",
      "en": "In the fair-die model P(A∩B)=1/3=P(A)P(B), so A={2,4,6} and B={1,2,3,4} are independent, and since P(B)>0, P(A|B)=1/2=P(A). The same equation implies that A and Bᶜ are independent, checked as 1/6=(1/2)(1/3). The mutually exclusive A and D={1,3,5} are not independent (0≠1/4). A and the zero-probability event ∅ are independent by the product definition, while the conditional probability P(A|∅) is undefined and may not be used."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "A={2,4,6} 與 D={1,3,5} 沒有共同結果，所以兩者獨立",
          "en": "A={2,4,6} and D={1,3,5} share no outcome, so they are independent"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "P(A∩D)=0，但 P(A)P(D)=1/4；乘積等式不成立。",
          "en": "P(A∩D)=0 but P(A)P(D)=1/4; the product equation fails."
        }
      },
      {
        "claim": {
          "zh-Hant": "P(A|∅) 無定義，所以無法判定 A 與 ∅ 是否獨立",
          "en": "P(A|∅) is undefined, so independence of A and ∅ cannot be decided"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "主資格是乘積定義：P(A∩∅)=0=P(A)·0，所以 A 與 ∅ 獨立。",
          "en": "The main criterion is the product definition: P(A∩∅)=0=P(A)·0, so A and ∅ are independent."
        }
      },
      {
        "claim": {
          "zh-Hant": "P(B)=0 時可以約定 P(A|B)=P(A)，再用條件式判定獨立",
          "en": "When P(B)=0 we may agree that P(A|B)=P(A) and then decide independence by the conditional form"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "MKO 明文禁止以未定義的條件機率取代乘積定義；P(B)=0 時只能用 P(A∩B)=P(A)P(B)。",
          "en": "The MKO explicitly forbids replacing the product definition by an undefined conditional probability; when P(B)=0 only P(A∩B)=P(A)P(B) may be used."
        }
      },
      {
        "claim": {
          "zh-Hant": "兩枚硬幣例子中 E₁、E₂、E₃ 兩兩獨立，所以三者相互獨立",
          "en": "In the two-coin example E₁, E₂, E₃ are pairwise independent, so the three are mutually independent"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "P(E₁∩E₂∩E₃)=1/4，而 P(E₁)P(E₂)P(E₃)=1/8。",
          "en": "P(E₁∩E₂∩E₃)=1/4 while P(E₁)P(E₂)P(E₃)=1/8."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "prob_axioms",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "機率公理：P(A)≥0；P(Ω)=1；兩兩互斥事件的（可數）聯集機率等於各機率之和。由此 P(∅)=0、P(Aᶜ)=1−P(A)。",
          "en": "Probability axioms: P(A)≥0; P(Ω)=1; the probability of a (countable) union of pairwise disjoint events is the sum of their probabilities. Hence P(∅)=0 and P(Aᶜ)=1−P(A)."
        }
      },
      {
        "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": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "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": "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": "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": "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": "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": "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": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "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": "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": "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": "mul_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "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": "mko_event_indep_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-event-independence",
          "sha256": "9b6f08c178cbc612effac540bd933bfde0d1b9ebc7708cc7ab928ac4505ad563",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在同一機率空間中，A 與 B 獨立當且僅當 P(A∩B)=P(A)P(B)。若 P(B)>0，這等價於 P(A|B)=P(A)；P(B)=0 時不得以未定義的條件機率取代乘積定義。",
          "en": "In one probability space, A and B are independent iff P(A∩B)=P(A)P(B). If P(B)>0, this is equivalent to P(A|B)=P(A). For P(B)=0, an undefined conditional probability must not replace the product definition."
        }
      },
      {
        "id": "cond_prob_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-conditional-probability",
          "sha256": "7b0f4824aa1f3e4c64ea306c5bed04843566b8e4bbdf96d94dbc1367dbb68419",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在有限樣本空間中，對 P(B)>0 定義 P(A|B)=P(A∩B)/P(B)；若 P(B)=0，這個條件機率不定義，即使 B 非空。",
          "en": "In a finite sample space, for P(B)>0 define P(A|B)=P(A∩B)/P(B); if P(B)=0 this conditional probability is left undefined, even when B is nonempty."
        }
      },
      {
        "id": "point_mass_sum",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限模型中事件的機率是其中各結果質量之和：P(E)=Σ_{ω∈E} m(ω)；空和為 0。",
          "en": "In a finite model the probability of an event is the sum of the masses of its outcomes: P(E)=Σ_{ω∈E} m(ω); the empty sum is 0."
        }
      },
      {
        "id": "L_cond_equiv",
        "kind": "lemma",
        "proved_in": [
          "S09.1",
          "S09.2"
        ],
        "statement": {
          "zh-Hant": "若 P(B)>0，則 P(A∩B)=P(A)P(B) ⇔ P(A|B)=P(A)。",
          "en": "If P(B)>0, then P(A∩B)=P(A)P(B) ⇔ P(A|B)=P(A)."
        }
      },
      {
        "id": "L_complement",
        "kind": "lemma",
        "proved_in": [
          "S10.1",
          "S10.2",
          "S10.3"
        ],
        "statement": {
          "zh-Hant": "在同一機率空間中，若 P(A∩B)=P(A)P(B)，則 P(A∩Bᶜ)=P(A)P(Bᶜ)。",
          "en": "In one probability space, if P(A∩B)=P(A)P(B), then P(A∩Bᶜ)=P(A)P(Bᶜ)."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S01",
        "lhs": "6*(1/6)",
        "rhs": "1"
      },
      {
        "id": "N2",
        "step": "S02",
        "lhs": "3/6",
        "rhs": "1/2"
      },
      {
        "id": "N3",
        "step": "S03",
        "lhs": "4/6",
        "rhs": "2/3"
      },
      {
        "id": "N4",
        "step": "S05",
        "lhs": "2/6",
        "rhs": "1/3"
      },
      {
        "id": "N5",
        "step": "S06",
        "lhs": "(1/2)*(2/3)",
        "rhs": "1/3"
      },
      {
        "id": "N6",
        "step": "S08",
        "lhs": "(1/3)/(2/3)",
        "rhs": "1/2"
      },
      {
        "id": "N7",
        "step": "S11",
        "lhs": "(1/2)*(1-2/3)",
        "rhs": "1/6"
      },
      {
        "id": "N8",
        "step": "S11",
        "lhs": "1/2-1/3",
        "rhs": "1/6"
      },
      {
        "id": "N9",
        "step": "S12",
        "lhs": "(1/2)*(3/6)",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N10",
        "step": "S14",
        "lhs": "(1/2)*0",
        "rhs": "0"
      },
      {
        "id": "N11",
        "lhs": "(1/2)*(1/2)",
        "rhs": "1/4"
      },
      {
        "id": "N12",
        "lhs": "1/4",
        "relation": "!=",
        "rhs": "(1/2)**3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：樣本空間、精確有理質量與事件子集完整明示（假設 a1）；只處理兩個事件的獨立（假設 a2），三事件的例子只出現在錯誤控制與負向控制中，用來說明兩兩獨立不能冒充相互獨立。不處理無限樣本空間或隨機變數的獨立（後者屬下一節點），不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: the sample space, exact rational masses and event subsets are fully explicit (assumption a1); only independence of two events is treated (assumption a2), and a three-event example appears only in the error and negative controls, to show that pairwise independence does not stand in for mutual independence. Infinite sample spaces and independence of random variables (the next node) are not treated, and the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "\\Omega=\\{1,2,3,4,5,6\\}",
        "樣本空間",
        "the sample space"
      ],
      [
        "m(\\omega)=\\tfrac16",
        "每個結果的質量",
        "the mass of each outcome"
      ],
      [
        "P(E)",
        "事件 E 的機率，E 中質量之和",
        "the probability of E, the sum of masses in E"
      ],
      [
        "A,B,D,Z",
        "四個給定事件；Z=∅",
        "the four given events; Z=∅"
      ],
      [
        "B^c",
        "B 在 Ω 中的補集",
        "the complement of B in Ω"
      ],
      [
        "P(A\\mid B)",
        "條件機率，只在 P(B)>0 時有定義",
        "conditional probability, defined only when P(B)>0"
      ]
    ],
    "md_error_controls": [
      {
        "title": "互斥不代表獨立 / Mutual exclusion does not imply independence",
        "zh": "MKO 指出的第一個誤解是把互斥當獨立。互斥是集合關係 A∩D=∅；獨立是機率關係 P(A∩D)=P(A)P(D)。S12、S13 顯示：兩個機率為正的互斥事件必不獨立，因為 0<P(A)P(D)。",
        "en": "The first misconception identified in the MKO is to treat mutually exclusive events as independent. Mutual exclusion is the set relation A∩D=∅; independence is the probability relation P(A∩D)=P(A)P(D). S12 and S13 show that two mutually exclusive events of positive probability are never independent, because 0<P(A)P(D)."
      },
      {
        "title": "P(B)=0 時用乘積定義 / Use the product definition when P(B)=0",
        "zh": "條件式 P(A|B)=P(A) 只在 P(B)>0 時與獨立等價（S09）。P(B)=0 時 P(A|B) 無定義，不能「算出來」再比較，也不能自行定義為 P(A)。乘積定義仍然可用，並給出：零機率事件與每個事件都獨立（S14）。",
        "en": "The conditional form P(A|B)=P(A) is equivalent to independence only when P(B)>0 (S09). When P(B)=0, P(A|B) is undefined: it cannot be 'computed' and compared, nor set to P(A) by fiat. The product definition still applies and shows that a zero-probability event is independent of every event (S14)."
      },
      {
        "title": "兩兩獨立不保證相互獨立 / Pairwise independence does not ensure mutual independence",
        "zh": "本稿只處理兩個事件。對三個事件，兩兩獨立不能推出相互獨立：擲兩枚公正硬幣，Ω′={HH,HT,TH,TT}，E₁=第一枚正面、E₂=第二枚正面、E₃=兩枚相同。每兩個的交集都是 {HH}，機率 1/4=(1/2)(1/2)；但 P(E₁∩E₂∩E₃)=1/4≠1/8=(1/2)³。",
        "en": "This guide handles only two events. For three events, pairwise independence does not imply mutual independence: toss two fair coins, Ω′={HH,HT,TH,TT}, E₁=first is heads, E₂=second is heads, E₃=both agree. Every pairwise intersection is {HH} with probability 1/4=(1/2)(1/2), but P(E₁∩E₂∩E₃)=1/4≠1/8=(1/2)³."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "三事件的相互獨立只在錯誤控制中以反例說明，沒有完整的定義與定理；這屬於 MKO 範圍之外。",
        "en": "Mutual independence of three events appears only as a counterexample in the error controls, without a full definition or theorem; it lies outside the MKO's scope."
      },
      {
        "zh-Hant": "N11、N12 屬於兩枚硬幣模型 Ω′，與骰子模型 Ω 分開；兩者的數字不可混用。",
        "en": "N11 and N12 belong to the two-coin model Ω′, separate from the die model Ω; their numbers must not be mixed."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "95a388554abac792bff1076b0a008b4da36c317d8313513927b13a553a4460d7",
    "original_source_sha256": "95a388554abac792bff1076b0a008b4da36c317d8313513927b13a553a4460d7",
    "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."
  }
}