{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-equation-01",
  "mko_id": "mko-equation",
  "title": {
    "zh-Hant": "方程式：在定義域內解 x²/(x−1)=1/(x−1)",
    "en": "Equation: solving x²/(x−1)=1/(x−1) inside its domain"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-equation-01",
    "source": {
      "mko_id": "mko-equation",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "6f8d6ca379da35727c14f320d99d0bc8c53c47ad478d0f462d30a4127aea0a78",
      "english_sha256": "de50265f38fde7fafcc45aa22db603721664273dbafb01a18c77a7d51b393bdd"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "mul_inverse_nonzero",
      "eq_congruence",
      "distributive",
      "zero_product",
      "set_builder",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在允許範圍 ℝ 中，求方程 x²/(x−1)=1/(x−1) 的解集；並說明為什麼兩邊乘以 x−1 後得到的方程 x²=1 有兩個解，而原方程只有一個。",
      "en": "With allowed domain ℝ, find the solution set of the equation x²/(x−1)=1/(x−1), and explain why the equation x²=1 obtained by multiplying both sides by x−1 has two solutions while the original has only one."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "允許範圍與運算：變數 x 取值於 ℝ；使用實數的加、減、乘、除，v/u 只在 u≠0 時有定義（mul_inverse_nonzero）。",
          "en": "Allowed domain and operations: the variable x ranges over ℝ; real addition, subtraction, multiplication and division are used, and v/u is defined only when u≠0 (mul_inverse_nonzero)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "方程：L(x)=R(x)，其中 L(x)=x²/(x−1)，R(x)=1/(x−1)。",
          "en": "The equation: L(x)=R(x), where L(x)=x²/(x−1) and R(x)=1/(x−1)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "解集的定義（MKO）：𝒮={x∈ℝ ∣ L(x) 與 R(x) 都有定義且 L(x)=R(x)}。",
          "en": "Definition of the solution set (MKO): 𝒮={x∈ℝ ∣ L(x) and R(x) are both defined and L(x)=R(x)}."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "G3"
        ],
        "before_tex": "L(x)=R(x)\\ \\text{over}\\ \\mathbb{R}",
        "after_tex": "\\mathcal{S}=\\{x\\in\\mathbb{R}\\mid L(x),R(x)\\ \\text{defined}\\ \\land\\ L(x)=R(x)\\}",
        "relation": "definition",
        "rule": "mko_equation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，方程的解集不是「讓式子看起來相等的數」，而是同時滿足兩個條件的允許 x：兩側都有定義，而且值相等。這裡允許範圍是 ℝ，運算是實數運算（G1）。之後每一步都要同時追蹤「有定義」與「相等」這兩個條件。",
          "en": "By the MKO, the solution set is not 'the numbers that make the formulas look equal' but the allowed x satisfying two conditions at once: both sides are defined, and their values are equal. Here the allowed domain is ℝ with real operations (G1). Every later step must keep track of both conditions, 'defined' and 'equal'."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "G3",
            "mko_equation"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出解集的定義，不推出新事實。",
            "en": "This step only writes out the definition of the solution set and derives no new fact."
          }
        },
        "leaf_basis": "mko_equation + set_builder"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "L(x)\\ \\text{and}\\ R(x)\\ \\text{defined}",
        "after_tex": "x-1\\neq 0",
        "relation": "equivalent",
        "rule": "mul_inverse_nonzero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "L(x)=x²/(x−1) 與 R(x)=1/(x−1) 各只含一個除法，除數都是 x−1；分子 x² 與 1 對每個實數都有定義。依 mul_inverse_nonzero，v/u 有定義當且僅當 u≠0，所以兩側都有定義 ⇔ x−1≠0。",
          "en": "L(x)=x²/(x−1) and R(x)=1/(x−1) each contain one division, both with divisor x−1; the numerators x² and 1 are defined for every real. By mul_inverse_nonzero, v/u is defined exactly when u≠0, so both sides are defined ⇔ x−1≠0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "若 x−1≠0，1/(x−1) 存在，兩側都有定義；若 x−1=0，兩側都含除以 0，都沒有定義。兩個方向都有理由。",
            "en": "If x−1≠0 then 1/(x−1) exists and both sides are defined; if x−1=0 both sides contain a division by 0 and neither is defined. Both directions have reasons."
          }
        },
        "leaf_basis": "mul_inverse_nonzero + def_unfold"
      },
      {
        "id": "S03",
        "uses": [
          "S02"
        ],
        "before_tex": "x-1\\neq 0",
        "after_tex": "x\\neq 1;\\quad D=\\{x\\in\\mathbb{R}\\mid x\\neq 1\\}",
        "relation": "equivalent",
        "rule": "L_zero_diff",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_zero_diff（u=x、c=1），x−1=0 ⇔ x=1；兩邊同取否定得 x−1≠0 ⇔ x≠1。所以兩側都有定義的 x 恰是 D={x∈ℝ ∣ x≠1}，而且 𝒮⊆D。",
          "en": "By Lemma L_zero_diff (u=x, c=1), x−1=0 ⇔ x=1; negating both sides gives x−1≠0 ⇔ x≠1. So the x at which both sides are defined form exactly D={x∈ℝ ∣ x≠1}, and 𝒮⊆D."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_zero_diff"
          ],
          "explanation": {
            "zh-Hant": "等價的兩邊同取否定仍是等價；引理的兩個方向都在子步證明。",
            "en": "Negating both sides of an equivalence gives an equivalence; both directions of the lemma are proved in the sub-steps."
          }
        },
        "leaf_basis": "eq_congruence + add_assoc + add_inverse + add_zero"
      },
      {
        "id": "S03.1",
        "uses": [
          "eq_congruence"
        ],
        "before_tex": "u-c=0",
        "after_tex": "(u-c)+c=0+c",
        "relation": "equivalent",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊同加 c。反向是兩邊同加 −c，再依 S03.2 的同類化簡。",
          "en": "Add c to both sides. The backward move is to add −c to both sides and simplify in the same way as S03.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "加 c 與加 −c 互為反運算，對任何實數都可做，沒有附加條件。",
            "en": "Adding c and adding −c undo each other and can be done for every real, with no side condition."
          }
        },
        "leaf_basis": "eq_congruence + add_inverse"
      },
      {
        "id": "S03.2",
        "uses": [
          "S03.1"
        ],
        "before_tex": "(u-c)+c=0+c",
        "after_tex": "u=c",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "左邊 (u+(−c))+c=u+((−c)+c)=u+0=u（u−c 的意義、結合律、交換律、反元素、單位元）；右邊 0+c=c。S03.1 與 S03.2 合起來證得引理 L_zero_diff。",
          "en": "The left side is (u+(−c))+c=u+((−c)+c)=u+0=u (meaning of u−c, associativity, commutativity, inverses, identity); the right side is 0+c=c. S03.1 and S03.2 together prove Lemma L_zero_diff."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "add_assoc",
            "add_comm",
            "add_inverse",
            "add_zero"
          ],
          "explanation": {
            "zh-Hant": "只把式子換成相等的式子，等式真假不變，雙向成立。",
            "en": "Expressions are only replaced by equal ones, so whether the equation holds is unchanged and the step is two-way."
          }
        },
        "leaf_basis": "add_assoc + add_comm + add_inverse + add_zero + eq_replace"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "G2"
        ],
        "before_tex": "\\frac{x^2}{x-1}=\\frac{1}{x-1},\\ x\\in D",
        "after_tex": "x^2=1,\\ x\\in D",
        "relation": "equivalent",
        "rule": "L_den",
        "conditions": [
          "x-1\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "對 x∈D，x−1≠0。依引理 L_den（a=x²、b=1、u=x−1）：x²/(x−1)=1/(x−1) ⇔ x²=1。這個等價只在 D 中成立：它的條件 u≠0 正是 x∈D。",
          "en": "For x∈D, x−1≠0. By Lemma L_den (a=x², b=1, u=x−1): x²/(x−1)=1/(x−1) ⇔ x²=1. This equivalence holds only in D: its hypothesis u≠0 is exactly x∈D."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_den",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "反向是兩邊同乘 1/(x−1)，需要 x−1≠0；在 D 之外（x=1）反向不能做，所以這裡一直帶著「x∈D」。",
            "en": "The backward move multiplies both sides by 1/(x−1) and needs x−1≠0; outside D (at x=1) it cannot be made, which is why 'x∈D' is carried along here."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero + mul_assoc + mul_one"
      },
      {
        "id": "S04.1",
        "uses": [
          "eq_congruence",
          "mul_inverse_nonzero"
        ],
        "before_tex": "\\frac{a}{u}=\\frac{b}{u}",
        "after_tex": "\\frac{a}{u}\\cdot u=\\frac{b}{u}\\cdot u",
        "relation": "equivalent",
        "rule": "eq_congruence",
        "conditions": [
          "u\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "向前：兩邊同乘 u。向後：兩邊同乘 1/u，需要 u≠0，因為 ((a/u)u)(1/u)=(a/u)(u·(1/u))=a/u。所以在 u≠0 時兩式等價。",
          "en": "Forwards: multiply both sides by u. Backwards: multiply both sides by 1/u, which needs u≠0, because ((a/u)u)(1/u)=(a/u)(u·(1/u))=a/u. So the two equations are equivalent when u≠0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "mul_inverse_nonzero",
            "mul_assoc"
          ],
          "explanation": {
            "zh-Hant": "向後乘以 1/u 需要 u≠0。若 u=0，a/u 與 b/u 原本就未定義，不能先乘 0 再把它們算成 0。另對已定義的實數式，乘以 0 雖會把兩邊變成 0，卻會失去原式資訊；這是另一個警告，不是本步在 u=0 的求值。",
            "en": "Backward multiplication by 1/u requires u≠0. If u=0, a/u and b/u are already undefined; they cannot be multiplied by zero and then evaluated as zero. Separately, multiplying two defined real expressions by zero loses their original information. That warning is not an evaluation of this step at u=0."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero + mul_assoc"
      },
      {
        "id": "S04.2",
        "uses": [
          "S04.1"
        ],
        "before_tex": "\\frac{a}{u}\\cdot u=\\frac{b}{u}\\cdot u",
        "after_tex": "a=b",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [
          "u\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依 a/u=a·(1/u)：(a·(1/u))u=a((1/u)u)=a·1=a；右邊同理得 b。S04.1 與 S04.2 合起來證得引理 L_den。",
          "en": "Using a/u=a·(1/u): (a·(1/u))u=a((1/u)u)=a·1=a; the right side gives b in the same way. S04.1 and S04.2 together prove Lemma L_den."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_assoc",
            "mul_comm",
            "mul_inverse_nonzero",
            "mul_one"
          ],
          "explanation": {
            "zh-Hant": "等值代換雙向成立；(1/u)u=1 需要 u≠0。",
            "en": "The replacement of equals holds both ways; (1/u)u=1 needs u≠0."
          }
        },
        "leaf_basis": "mul_assoc + mul_comm + mul_inverse_nonzero + mul_one + eq_replace"
      },
      {
        "id": "S05",
        "uses": [
          "S04"
        ],
        "before_tex": "x^2=1",
        "after_tex": "x^2-1=0",
        "relation": "equivalent",
        "rule": "L_zero_diff",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_zero_diff（u=x²、c=1）：x²−1=0 ⇔ x²=1。把方程整理成「某式=0」，才能使用零乘積性質。",
          "en": "By Lemma L_zero_diff (u=x², c=1): x²−1=0 ⇔ x²=1. Bringing the equation to the form 'expression = 0' is what allows the zero-product property."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_zero_diff"
          ],
          "explanation": {
            "zh-Hant": "引理兩向都已證；這一步對每個實數成立，不依賴 x∈D。",
            "en": "Both directions of the lemma are proved; this step holds for every real and does not depend on x∈D."
          }
        },
        "leaf_basis": "eq_congruence + add_inverse + add_assoc + add_zero"
      },
      {
        "id": "S06",
        "uses": [
          "S05"
        ],
        "before_tex": "x^2-1=0",
        "after_tex": "(x-1)(x+1)=0",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依分配律：(x−1)(x+1)=x·x+x·1−1·x−1·1=x²+x−x−1=x²−1。兩式對每個實數相等，代換後等式真假不變。",
          "en": "By distributivity: (x−1)(x+1)=x·x+x·1−1·x−1·1=x²+x−x−1=x²−1. The two expressions are equal for every real, so replacing one by the other does not change whether the equation holds."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "這是對所有實數成立的恆等式（N4 是獨立的多項式核對），等值代換雙向成立。",
            "en": "This is an identity for all reals (N4 is an independent polynomial check), and the replacement of equals is two-way."
          }
        },
        "leaf_basis": "distributive + neg_mul + add_inverse + eq_replace"
      },
      {
        "id": "S07",
        "uses": [
          "S06"
        ],
        "before_tex": "(x-1)(x+1)=0",
        "after_tex": "x-1=0\\ \\lor\\ x+1=0",
        "relation": "equivalent",
        "rule": "zero_product",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依零乘積性質（ℝ 沒有零因子）：兩個實數的積為 0，當且僅當至少一個因子為 0。",
          "en": "By the zero-product property (ℝ has no zero divisors): a product of two reals is 0 exactly when at least one factor is 0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "zero_product"
          ],
          "explanation": {
            "zh-Hant": "向前需要 ℝ 沒有零因子；向後用 u·0=0。兩個方向都包含在 zero_product 中。",
            "en": "Forwards needs that ℝ has no zero divisors; backwards uses u·0=0. Both directions are contained in zero_product."
          }
        },
        "leaf_basis": "zero_product"
      },
      {
        "id": "S08",
        "uses": [
          "S07"
        ],
        "before_tex": "x-1=0\\ \\lor\\ x+1=0",
        "after_tex": "x=1\\ \\lor\\ x=-1",
        "relation": "equivalent",
        "rule": "L_zero_diff",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_zero_diff：x−1=0 ⇔ x=1；又 x+1=x−(−1)，所以 x+1=0 ⇔ x=−1。兩個析取支各自等價變形，析取前後同真同假。",
          "en": "By Lemma L_zero_diff: x−1=0 ⇔ x=1; also x+1=x−(−1), so x+1=0 ⇔ x=−1. Each disjunct is transformed by an equivalence, so the disjunction before and after is true or false together."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_zero_diff",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "代回：1−1=0、(−1)+1=0；兩支都是雙向等價。",
            "en": "Substituting back: 1−1=0 and (−1)+1=0; both branches are two-way equivalences."
          }
        },
        "leaf_basis": "eq_congruence + add_inverse + add_zero + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S01",
          "S03",
          "S04",
          "S05",
          "S06",
          "S07",
          "S08"
        ],
        "before_tex": "x\\in D:\\ L(x)=R(x)",
        "after_tex": "\\mathcal{S}=\\{x\\in\\mathbb{R}\\mid x\\in D\\ \\land\\ (x=1\\lor x=-1)\\}",
        "relation": "equivalent",
        "rule": "logic_chain",
        "conditions": [
          "x\\in D"
        ],
        "explanation": {
          "zh-Hant": "S04 到 S08 每一步都是 x∈D 條件下的等價，依推論的傳遞串起來：對 x∈D，L(x)=R(x) ⇔ x=1 ∨ x=−1。再配合 S01 與 S03（有定義 ⇔ x∈D），得 𝒮={x∈ℝ ∣ x∈D 且 (x=1 或 x=−1)}。",
          "en": "Each of S04 to S08 is an equivalence under the condition x∈D; chaining them gives, for x∈D, L(x)=R(x) ⇔ x=1 ∨ x=−1. Together with S01 and S03 ('defined' ⇔ x∈D), 𝒮={x∈ℝ ∣ x∈D and (x=1 or x=−1)}."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "logic_chain",
            "S04",
            "S08"
          ],
          "explanation": {
            "zh-Hant": "每一環都是雙向的，而且 S04 只在 x∈D 中使用；所以串起來的等價也只在 D 中成立，並把 x∈D 保留在集合條件裡。",
            "en": "Every link is two-way, and S04 is used only for x∈D; so the chained equivalence holds only in D, and x∈D is kept as part of the set condition."
          }
        },
        "leaf_basis": "logic_chain + set_builder"
      },
      {
        "id": "S10",
        "uses": [
          "S03",
          "S09"
        ],
        "before_tex": "x=1",
        "after_tex": "1\\notin D,\\ \\text{so}\\ 1\\notin\\mathcal{S}",
        "relation": "implies",
        "rule": "set_builder",
        "conditions": [],
        "explanation": {
          "zh-Hant": "候選值 x=1：1−1=0，所以 1∉D，L(1) 與 R(1) 都含除以 0，都沒有定義。依 S09 的集合條件，1∉𝒮。這個候選值滿足 x²=1，但不屬於原方程的解集。",
          "en": "Candidate x=1: 1−1=0, so 1∉D, and L(1) and R(1) both involve division by 0 and are undefined. By the set condition of S09, 1∉𝒮. This candidate satisfies x²=1 but is not in the solution set of the original equation."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S09",
            "set_builder"
          ],
          "explanation": {
            "zh-Hant": "只需 1 不滿足合取項 x∈D；不必（也無法）計算兩側的值。",
            "en": "It is enough that 1 fails the conjunct x∈D; the two sides need not (and cannot) be computed."
          }
        },
        "leaf_basis": "set_builder + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S03",
          "S09"
        ],
        "before_tex": "x=-1",
        "after_tex": "-1\\in D,\\ \\text{so}\\ -1\\in\\mathcal{S}",
        "relation": "implies",
        "rule": "set_builder",
        "conditions": [
          "-1-1\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "候選值 x=−1：−1−1=−2≠0，所以 −1∈D；而且 x=−1 滿足析取的第二支。依 S09 的集合條件，兩個合取項都成立，所以 −1∈𝒮。",
          "en": "Candidate x=−1: −1−1=−2≠0, so −1∈D; and x=−1 satisfies the second disjunct. By the set condition of S09 both conjuncts hold, so −1∈𝒮."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S09",
            "set_builder"
          ],
          "explanation": {
            "zh-Hant": "兩個合取項都檢查了：x∈D 與 x=1∨x=−1。",
            "en": "Both conjuncts were checked: x∈D and x=1∨x=−1."
          }
        },
        "leaf_basis": "set_builder + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S09",
          "S10",
          "S11"
        ],
        "before_tex": "x\\in\\mathcal{S}\\iff x\\in D\\land(x=1\\lor x=-1)",
        "after_tex": "\\mathcal{S}=\\{-1\\}",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對任意 x：若 x∈𝒮，則 x∈D 且 x=1 或 x=−1；x=1 與 x∈D 不相容（S10），所以 x=−1。反之 x=−1 時 x∈𝒮（S11）。兩個包含合起來，依集合相等，𝒮={−1}。",
          "en": "For any x: if x∈𝒮 then x∈D and x=1 or x=−1; x=1 is incompatible with x∈D (S10), so x=−1. Conversely x=−1 gives x∈𝒮 (S11). The two inclusions together give 𝒮={−1} by set equality."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "S11",
            "set_extensionality",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "需要兩個包含：𝒮⊆{−1} 用 S10 排除 x=1，{−1}⊆𝒮 用 S11。",
            "en": "Both inclusions are needed: 𝒮⊆{−1} uses S10 to exclude x=1, and {−1}⊆𝒮 uses S11."
          }
        },
        "leaf_basis": "set_extensionality + case_analysis"
      },
      {
        "id": "S13",
        "uses": [
          "S12",
          "G2"
        ],
        "before_tex": "x=-1\\ \\text{in}\\ L,\\ R",
        "after_tex": "L(-1)=\\frac{1}{-2}=-\\tfrac{1}{2}=R(-1)",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，驗證候選解要把值代入原方程，兩側分別計算：L(−1)=(−1)²/(−1−1)=1/(−2)=−1/2；R(−1)=1/(−1−1)=−1/2。兩側都有定義且相等，與 S12 一致。",
          "en": "By the MKO, a candidate is checked by substituting it into the original equation and computing each side: L(−1)=(−1)²/(−1−1)=1/(−2)=−1/2 and R(−1)=1/(−1−1)=−1/2. Both sides are defined and equal, in agreement with S12."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12",
            "G2",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "這是對 x=−1 的獨立核對；它確認 −1 是解，但不能單獨證明沒有其他解，那由 S04 到 S12 的等價鏈負責。",
            "en": "This is an independent check of x=−1; it confirms that −1 is a solution but cannot by itself show there are no others, which is the job of the chain of equivalences S04 to S12."
          }
        },
        "leaf_basis": "numeric_evaluation + frac_rules"
      },
      {
        "id": "S14",
        "uses": [
          "S05",
          "S06",
          "S07",
          "S08",
          "S12"
        ],
        "before_tex": "\\{x\\in\\mathbb{R}\\mid x^2=1\\}",
        "after_tex": "\\{-1,1\\}\\supsetneq\\mathcal{S}=\\{-1\\}",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若忽略定義域，把兩邊乘以 x−1 後在整個 ℝ 上解 x²=1，S05 到 S08 的等價（它們不依賴 x∈D）給出 {−1,1}。多出來的 1 正是乘數 x−1 為 0 的地方：乘以 x−1 能把原方程的解帶到新方程，但在 x−1=0 處不能乘 1/(x−1) 回去，所以新方程多了一個不屬於原方程的解。",
          "en": "If the domain is ignored and x²=1 is solved on all of ℝ after multiplying by x−1, the equivalences S05 to S08 (which do not depend on x∈D) give {−1,1}. The extra 1 is exactly where the multiplier x−1 is 0: multiplying by x−1 carries solutions of the original equation to the new one, but where x−1=0 one cannot multiply by 1/(x−1) to come back, so the new equation gains a solution that the original does not have."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S08",
            "S12",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "兩集合只差 1，而 1∉D；這解釋了差異，但不宣稱乘以任何式子都只會多出乘數的零點以外沒有別的影響。",
            "en": "The two sets differ only by 1, and 1∉D; this explains the difference here, without claiming anything general about every possible multiplier."
          }
        },
        "leaf_basis": "set_extensionality + logic_chain"
      },
      {
        "id": "S15",
        "uses": [
          "S12",
          "S13",
          "S14"
        ],
        "before_tex": "\\mathcal{S}=\\{-1\\},\\ \\{x\\mid x^2=1\\}=\\{-1,1\\}",
        "after_tex": "\\text{answer: }\\mathcal{S}=\\{-1\\};\\ 1\\ \\text{excluded since}\\ 1\\notin D",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "原方程 x²/(x−1)=1/(x−1) 在 ℝ 中的解集是 {−1}；x²=1 的另一個解 1 因兩側在 x=1 沒有定義而被排除。",
          "en": "The solution set of x²/(x−1)=1/(x−1) in ℝ is {−1}; the other solution 1 of x²=1 is excluded because both sides are undefined at x=1."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S12",
            "S13",
            "S14",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：求 𝒮 並說明 x²=1 多出的解。足夠條件 H：S12（𝒮={−1}）與 S14（多出的解來自 x−1=0）；H⇒G 直接成立。H 由 G1 到 G3 正向證成，沒有把 G 當作前提；S13 是獨立核對。",
            "en": "Goal G: find 𝒮 and explain the extra solution of x²=1. Sufficient condition H: S12 (𝒮={−1}) and S14 (the extra solution comes from x−1=0); H⇒G is immediate. H is proved forwards from G1 to G3 without using G as a premise; S13 is an independent check."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在 ℝ 中，x²/(x−1)=1/(x−1) 的解集是 {−1}。兩側只在 x≠1 時有定義；在這個範圍內方程等價於 x²=1，即 x=1 或 x=−1，而 x=1 不在定義域內。兩邊乘以 x−1 後的方程 x²=1 有解 {−1,1}，多出的 1 正是乘數 x−1 為零之處。",
      "en": "In ℝ, the solution set of x²/(x−1)=1/(x−1) is {−1}. Both sides are defined only for x≠1; there the equation is equivalent to x²=1, that is x=1 or x=−1, and x=1 is outside the domain. The equation x²=1 obtained by multiplying by x−1 has solutions {−1,1}; the extra 1 is exactly where the multiplier x−1 vanishes."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "原方程的解集是 {−1,1}",
          "en": "The solution set of the original equation is {−1,1}"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x=1 時 1−1=0，兩側都沒有定義，1∉𝒮（S10）。",
          "en": "At x=1, 1−1=0 and neither side is defined, so 1∉𝒮 (S10)."
        }
      },
      {
        "claim": {
          "zh-Hant": "x=1 是解，因為 1²=1",
          "en": "x=1 is a solution because 1²=1"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "那是在檢查 x²=1，不是原方程；原方程在 x=1 兩側都沒有定義。",
          "en": "That checks x²=1, not the original equation; the original has both sides undefined at x=1."
        }
      },
      {
        "claim": {
          "zh-Hant": "兩邊同乘一個式子永遠不改變解集",
          "en": "Multiplying both sides by the same expression never changes the solution set"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x=2 兩邊乘 0 得 0=0，解集從 {2} 變成 ℝ；本題乘以 x−1 也多出解 1（S14）。",
          "en": "Multiplying x=2 by 0 gives 0=0, and the solution set grows from {2} to ℝ; here multiplying by x−1 also adds the solution 1 (S14)."
        }
      },
      {
        "claim": {
          "zh-Hant": "x²/(x−1) 本身就是一個方程",
          "en": "x²/(x−1) is itself an equation"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "它沒有等號，只是一個表達式（方程的一側）；沒有另一側就沒有解集。",
          "en": "It has no equality sign; it is an expression (one side of the equation), and without a second side there is no solution set."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "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": "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": "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": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "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_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_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "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": "add_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法交換律：u+v=v+u。",
          "en": "Commutativity of addition: u+v=v+u."
        }
      },
      {
        "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": "mul_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。",
          "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)."
        }
      },
      {
        "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": "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_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": "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": "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": "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_equation",
        "kind": "cited",
        "source": {
          "mko_id": "mko-equation",
          "sha256": "6f8d6ca379da35727c14f320d99d0bc8c53c47ad478d0f462d30a4127aea0a78",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "給定兩個具有共同語境的表達式 L(x)、R(x)，方程式 L(x)=R(x) 的解集由所有使兩側皆有定義且值相等的允許 x 組成。",
          "en": "Given two expressions L(x) and R(x) in a common context, the solution set of L(x)=R(x) consists of all allowed x for which both sides are defined and equal."
        }
      },
      {
        "id": "L_zero_diff",
        "kind": "lemma",
        "proved_in": [
          "S03.1",
          "S03.2"
        ],
        "statement": {
          "zh-Hant": "對實數 u、c：u−c=0 ⇔ u=c。",
          "en": "For reals u and c: u−c=0 ⇔ u=c."
        }
      },
      {
        "id": "L_den",
        "kind": "lemma",
        "proved_in": [
          "S04.1",
          "S04.2"
        ],
        "statement": {
          "zh-Hant": "對實數 a、b 與 u≠0：a/u=b/u ⇔ a=b。",
          "en": "For reals a, b and u≠0: a/u=b/u ⇔ a=b."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "1-1",
        "rhs": "0"
      },
      {
        "id": "N2",
        "step": "S11",
        "lhs": "-1-1",
        "rhs": "-2"
      },
      {
        "id": "N3",
        "step": "S11",
        "lhs": "-2",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N4",
        "step": "S06",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "x*x-1",
        "rhs": "(x-1)*(x+1)"
      },
      {
        "id": "N5",
        "step": "S08",
        "lhs": "-1+1",
        "rhs": "0"
      },
      {
        "id": "N6",
        "step": "S13",
        "lhs": "(-1)**2/(-1-1)",
        "rhs": "-1/2"
      },
      {
        "id": "N7",
        "step": "S13",
        "lhs": "1/(-1-1)",
        "rhs": "-1/2"
      },
      {
        "id": "N8",
        "step": "S14",
        "lhs": "1**2",
        "rhs": "1"
      },
      {
        "id": "N9",
        "step": "S14",
        "lhs": "(-1)**2",
        "rhs": "1"
      },
      {
        "id": "N10",
        "lhs": "1+1",
        "rhs": "2"
      },
      {
        "id": "N11",
        "lhs": "2*0",
        "rhs": "0"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "依 MKO 的定義：解集由使兩側都有定義且相等的允許 x 組成。只處理一個含分式的實數方程，示範定義域、乘以非零式的等價變形、零乘積、排除候選值與代回核對。不處理一般有理方程的理論或複數解；不改動 canonical MKO 或英文 edition。",
      "en": "Follows the MKO's definition: the solution set consists of the allowed x at which both sides are defined and equal. Treats one real equation with fractions, illustrating the domain, equivalence under multiplication by a nonzero expression, the zero-product property, rejecting a candidate and checking by substitution. It does not treat the general theory of rational equations or complex solutions, and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "L(x)=\\frac{x^2}{x-1}",
        "左側表達式",
        "the left-hand expression"
      ],
      [
        "R(x)=\\frac{1}{x-1}",
        "右側表達式",
        "the right-hand expression"
      ],
      [
        "D",
        "兩側都有定義的 x 所成的集合",
        "the set of x at which both sides are defined"
      ],
      [
        "\\mathcal{S}",
        "原方程的解集",
        "the solution set of the original equation"
      ],
      [
        "u\\neq 0",
        "非零條件；除以 u 或使用 1/u 都需要它",
        "the nonzero condition needed to divide by u or to use 1/u"
      ]
    ],
    "md_error_controls": [
      {
        "title": "乘以可能為零的式子 / Multiplying by an expression that may be zero",
        "zh": "MKO 列出的誤解包括「乘零後仍宣稱解集不變」與「除以可能為零的式子」。兩邊乘以 x−1 只在 x−1≠0 時是等價變形（S04）；在 x=1 處乘數為 0，變形不可逆，x²=1 因而多出解 1（S14）。正確做法是先寫出定義域 D={x ∣ x≠1}，並在整條等價鏈中保留 x∈D。",
        "en": "The MKO lists the mistakes of claiming the solution set is unchanged after multiplying by zero and of dividing by an expression that may be zero. Multiplying both sides by x−1 is an equivalence only where x−1≠0 (S04); at x=1 the multiplier is 0, the step cannot be reversed, and x²=1 gains the solution 1 (S14). The correct procedure is to write the domain D={x ∣ x≠1} first and keep x∈D throughout the chain of equivalences."
      },
      {
        "title": "求得候選值後要代回原式 / Substitute candidates into the original equation",
        "zh": "MKO 提醒求得候選值後要代回原式檢查。候選值 1 代入原式時兩側都是除以 0，沒有定義，所以不是解（S10）；候選值 −1 代入後兩側都是 −1/2（S13）。只把候選值代入 x²=1 檢查會錯把 1 當成解，因為 x²=1 已經不是原方程。",
        "en": "The MKO warns that candidates must be substituted back into the original equation. Substituting 1 makes both sides a division by 0, undefined, so 1 is not a solution (S10); substituting −1 makes both sides −1/2 (S13). Checking only in x²=1 would wrongly accept 1, because x²=1 is no longer the original equation."
      },
      {
        "title": "約分也不能丟掉定義域 / Cancelling does not remove the domain",
        "zh": "類似的方程 (x²−1)/(x−1)=2，若先約分寫成 x+1=2，會得 x=1；但 x=1 時左側除以 0，沒有定義。約分 (x²−1)/(x−1)=x+1 只在 x≠1 時成立，所以這個方程的解集是空集。只化簡一側時，必須連同它成立的條件一起帶著。",
        "en": "For the similar equation (x²−1)/(x−1)=2, cancelling first to x+1=2 gives x=1; but at x=1 the left side divides by 0 and is undefined. The cancellation (x²−1)/(x−1)=x+1 holds only for x≠1, so this equation's solution set is empty. Simplifying one side must carry along the condition under which the simplification holds."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿只處理一個具體方程；一般有理方程（多個分母、高次分子）需要同樣的定義域追蹤，但不在此證明。",
        "en": "This guide treats one concrete equation; general rational equations (several denominators, higher-degree numerators) need the same tracking of the domain, which is not proved here."
      },
      {
        "zh-Hant": "零乘積性質與分式規則以規則表中的前提使用，不在此由域公理推出。",
        "en": "The zero-product property and the fraction rules are used as premises from the glossary and are not derived here from the field axioms."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "70e4d5766cd19e87cc7a8b9f5a847402b534f1a14a56165b9364b027e1664496",
    "original_source_sha256": "105cc4d24c32a42632c2fa35bdefc5573f9ebff81adc1112f41ccac5a8a97832",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M006"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}