{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-inverse-function-01",
  "mko_id": "mko-inverse-function",
  "title": {
    "zh-Hant": "反函數：交換有限雙射表並求非零斜率仿射函數的反函數",
    "en": "Inverse functions: exchanging a finite bijection table and inverting an affine map with nonzero slope"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-inverse-function-01",
    "source": {
      "mko_id": "mko-inverse-function",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "bad163b9a218429518937fb4e86abe0e8707ebda0fc06c7e27f625184bc05c4f",
      "english_sha256": "c8aec9f36446191c0529002aef5c9e758f57dd5f6a6f73d01d62310e907d6bea"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "function_def",
      "injective_surjective_def",
      "composition_def",
      "frac_rules",
      "distributive",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "（一）A={1,2,3}、B={4,5,6}，p(1)=5、p(2)=6、p(3)=4。先確認 p 是雙射，再交換有序對得到候選 q，驗證 q 是總函數 B→A，並逐側檢查 q(p(x))=x 與 p(q(y))=y。（二）f:ℝ→ℝ，f(x)=−4x+6（m=−4≠0）：求反函數 g(y)=uy+v，u、v 以最簡、正分母的有理數表示，並對所有實數驗證兩側恆等式。（三）說明 h(1)=4、h(2)=4、h(3)=6 的交換表不是 B→A 的函數，以及 f^{-1} 不是 1/f。",
      "en": "(1) A={1,2,3}, B={4,5,6}, p(1)=5, p(2)=6, p(3)=4. First confirm that p is a bijection, then exchange its ordered pairs to get a candidate q, validate q as a total function B→A, and check q(p(x))=x and p(q(y))=y separately. (2) f:ℝ→ℝ, f(x)=−4x+6 (m=−4≠0): find the inverse g(y)=uy+v with u, v reduced rationals with positive denominators, and verify both identities for all reals. (3) Show that the exchanged table of h(1)=4, h(2)=4, h(3)=6 is not a function B→A, and that f^{-1} is not 1/f."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "A={1,2,3}、B={4,5,6}；p:A→B，p(1)=5、p(2)=6、p(3)=4。",
          "en": "A={1,2,3}, B={4,5,6}; p:A→B with p(1)=5, p(2)=6, p(3)=4."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "f:ℝ→ℝ，f(x)=−4x+6，即 m=−4、b=6。",
          "en": "f:ℝ→ℝ, f(x)=−4x+6, that is m=−4 and b=6."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "h:A→B，h(1)=4、h(2)=4、h(3)=6。",
          "en": "h:A→B with h(1)=4, h(2)=4, h(3)=6."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "驗收標準（MKO a2）：候選 g 是雙射 f:A→B 的反函數，當且僅當 g 是總函數 B→A，且對每個 x∈A 有 g(f(x))=x、對每個 y∈B 有 f(g(y))=y。",
          "en": "Acceptance test (MKO a2): a candidate g is the inverse of a bijection f:A→B exactly when g is a total function B→A, g(f(x))=x for every x∈A, and f(g(y))=y for every y∈B."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "mko_bijective_fiber"
        ],
        "before_tex": "p(1)=5,\\ p(2)=6,\\ p(3)=4",
        "after_tex": "p^{-1}(\\{4\\})=\\{3\\},\\ p^{-1}(\\{5\\})=\\{1\\},\\ p^{-1}(\\{6\\})=\\{2\\}\\ \\Rightarrow\\ p\\ \\text{bijective}",
        "relation": "implies",
        "rule": "mko_bijective_fiber",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 假設 a1 要求先確認 p 是宣告 A→B 的雙射。掃描 A 的三個元素：目標 4 只由 3 達到、5 只由 1 達到、6 只由 2 達到。B 的每個目標的纖維都恰有一個元素，所以 p 是雙射。",
          "en": "MKO assumption a1 requires first establishing that p is a bijection on the declared A→B. Scanning the three elements of A: target 4 is reached only from 3, 5 only from 1, and 6 only from 2. Every target of B has a fiber with exactly one element, so p is bijective."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_bijective_fiber",
            "set_builder"
          ],
          "explanation": {
            "zh-Hant": "需要掃描全部三個輸入並檢查宣告陪域的全部三個目標；只用纖維刻畫的 ⇐ 方向。",
            "en": "It needs all three inputs scanned and all three targets of the declared codomain checked; only the ⇐ direction of the fiber characterization is used."
          }
        },
        "leaf_basis": "mko_bijective_fiber + set_builder + numeric_evaluation"
      },
      {
        "id": "S02",
        "uses": [
          "G1"
        ],
        "before_tex": "p=\\{(1,5),(2,6),(3,4)\\}",
        "after_tex": "q=p^{\\mathsf T}=\\{(5,1),(6,2),(4,3)\\}",
        "relation": "definition",
        "rule": "swap_table",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 p 寫成有序對的集合，交換每一對的兩個坐標，得到候選 q。此時 q 只是 B×A 的一個子集合（關係）；它是否為函數 B→A 必須另外檢查（a2）。",
          "en": "Write p as a set of ordered pairs and exchange the two coordinates of each pair to get the candidate q. At this point q is only a subset (a relation) of B×A; whether it is a function B→A has to be checked separately (a2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "swap_table",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "只是構造一個集合並命名，不宣稱 q 是函數。",
            "en": "This only constructs and names a set; it does not claim q is a function."
          }
        },
        "leaf_basis": "swap_table + cartesian_product_def"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "S02"
        ],
        "before_tex": "q=\\{(4,3),(5,1),(6,2)\\}",
        "after_tex": "q:B\\to A\\ \\text{total};\\ q(4)=3,\\ q(5)=1,\\ q(6)=2",
        "relation": "implies",
        "rule": "function_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 swap_table，b 在 q 中作為第一坐標出現的次數等於 b 在 p 中作為值出現的次數，也就是纖維 p^{-1}({b}) 的大小，由 S01 都是 1。所以 4、5、6 各恰好出現一次：每個 b 有值（總）且只有一個值（單值），值 3、1、2 都在 A 中。q 是總函數 B→A。",
          "en": "By swap_table, the number of times b occurs as a first coordinate in q equals the number of times b occurs as a value of p, which is the size of the fiber p^{-1}({b}), equal to 1 by S01. So 4, 5, 6 each occur exactly once: every b has a value (total) and only one (single-valued), and the values 3, 1, 2 lie in A. q is a total function B→A."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "swap_table",
            "function_def"
          ],
          "explanation": {
            "zh-Hant": "總性需要每個纖維至少一個元素（p 滿射），單值性需要至多一個元素（p 單射），兩者都由 S01 提供。若 p 不是雙射，這一步就會失敗（見 S15）。",
            "en": "Totality needs every fiber to have at least one element (p surjective) and single-valuedness needs at most one (p injective); both come from S01. If p were not bijective this step would fail (see S15)."
          }
        },
        "leaf_basis": "function_def + swap_table + set_builder"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S03"
        ],
        "before_tex": "x\\in A=\\{1,2,3\\}",
        "after_tex": "q(p(x))=x\\ \\ \\forall x\\in A",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "A 恰有三個元素，逐一檢查就是窮盡所有情形的證明，不是抽樣：q(p(1))=q(5)=1，q(p(2))=q(6)=2，q(p(3))=q(4)=3。所以 q∘p=id_A。",
          "en": "A has exactly three elements, so checking each one is a proof by exhaustive cases, not a sample: q(p(1))=q(5)=1, q(p(2))=q(6)=2, q(p(3))=q(4)=3. So q∘p=id_A."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "case_analysis",
            "composition_def",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "合成有定義，因為 p 的值都在 B，也就是 q 的定義域；三個情形窮盡 A。這只是一側的恆等式，另一側在 S05。",
            "en": "The composition is defined because the values of p lie in B, the domain of q; the three cases exhaust A. This is one identity only; the other side is S05."
          }
        },
        "leaf_basis": "case_analysis + composition_def + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S03"
        ],
        "before_tex": "y\\in B=\\{4,5,6\\}",
        "after_tex": "p(q(y))=y\\ \\ \\forall y\\in B",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第二個恆等式必須另外檢查，因為代表公式 g(f(x))=x 只呈現其中一側：p(q(4))=p(3)=4，p(q(5))=p(1)=5，p(q(6))=p(2)=6。B 的三個元素全部檢查過，所以 p∘q=id_B。",
          "en": "The second identity has to be checked separately, since the representative formula g(f(x))=x shows only one side: p(q(4))=p(3)=4, p(q(5))=p(1)=5, p(q(6))=p(2)=6. All three elements of B were checked, so p∘q=id_B."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "case_analysis",
            "composition_def",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "q 的值都在 A，p∘q 有定義；三個情形窮盡 B。",
            "en": "The values of q lie in A, so p∘q is defined; the three cases exhaust B."
          }
        },
        "leaf_basis": "case_analysis + composition_def + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S03",
          "S04",
          "S05",
          "G4"
        ],
        "before_tex": "q:B\\to A\\ \\text{total},\\ q\\circ p=\\mathrm{id}_A,\\ p\\circ q=\\mathrm{id}_B",
        "after_tex": "q=p^{-1}",
        "relation": "implies",
        "rule": "mko_inverse_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "G4 的三個要求都已成立：q 是總函數（S03）、q∘p=id_A（S04）、p∘q=id_B（S05）。而且 q(b) 正是 S01 中 b 的唯一原像，與 MKO 的定義一致。所以 q 是 p 的反函數。",
          "en": "All three requirements of G4 hold: q is a total function (S03), q∘p=id_A (S04) and p∘q=id_B (S05). Moreover q(b) is exactly the unique preimage of b found in S01, matching the MKO's definition. So q is the inverse of p."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S04",
            "S05",
            "mko_inverse_def"
          ],
          "explanation": {
            "zh-Hant": "需要三個條件同時成立；只有其中一側恆等式不夠（見錯誤控制）。",
            "en": "All three conditions are needed together; one identity alone is not enough (see the error controls)."
          }
        },
        "leaf_basis": "mko_inverse_def + and_intro_elim"
      },
      {
        "id": "S07",
        "uses": [
          "G2"
        ],
        "before_tex": "y=-4x+6",
        "after_tex": "x=\\frac{y-6}{-4}",
        "relation": "equivalent",
        "rule": "L_solve_affine",
        "conditions": [
          "-4\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "轉到仿射例子。對任意實數 y，依引理 L_solve_affine（m=−4、b=6）：y=−4x+6 ⇔ x=(y−6)/(−4)。引理在 S07.1、S07.2 證明，需要 m=−4≠0。",
          "en": "Turn to the affine example. For any real y, by Lemma L_solve_affine (m=−4, b=6): y=−4x+6 ⇔ x=(y−6)/(−4). The lemma is proved in S07.1 and S07.2 and needs m=−4≠0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_solve_affine"
          ],
          "explanation": {
            "zh-Hant": "引理兩個方向都在子步中證明，條件 −4≠0 成立。",
            "en": "Both directions of the lemma are proved in the sub-steps, and the condition −4≠0 holds."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero"
      },
      {
        "id": "S07.1",
        "uses": [
          "eq_congruence"
        ],
        "before_tex": "y=mx+b",
        "after_tex": "y-b=mx",
        "relation": "equivalent",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊同減 b：(mx+b)−b=mx（加法結合律與反元素）。反向是兩邊同加 b。",
          "en": "Subtract b from both sides: (mx+b)−b=mx (associativity and inverses for addition). The backward move is to add b to both sides."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "減 b 與加 b 互為反運算，對任何實數 b 都可做。",
            "en": "Subtracting b and adding b undo each other and are possible for every real b."
          }
        },
        "leaf_basis": "eq_congruence + add_assoc + add_comm + add_inverse + add_zero"
      },
      {
        "id": "S07.2",
        "uses": [
          "S07.1",
          "mul_inverse_nonzero"
        ],
        "before_tex": "y-b=mx",
        "after_tex": "x=\\frac{y-b}{m}",
        "relation": "equivalent",
        "rule": "mul_inverse_nonzero",
        "conditions": [
          "m\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "因 m≠0，1/m 存在；兩邊同乘 1/m 得 (y−b)/m=x。反向是兩邊同乘 m。若 m=0，此步不能做，而且 f(x)=b 是常數函數，根本不是雙射。S07.1 與 S07.2 合起來證得引理 L_solve_affine。",
          "en": "Since m≠0, 1/m exists; multiplying both sides by 1/m gives (y−b)/m=x. The backward move is to multiply both sides by m. If m=0 the step cannot be taken, and then f(x)=b is constant and not a bijection at all. S07.1 and S07.2 together prove Lemma L_solve_affine."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_inverse_nonzero",
            "eq_congruence"
          ],
          "explanation": {
            "zh-Hant": "乘 1/m 與乘 m 互為反運算，前提 m≠0，正是 MKO a3 的條件。",
            "en": "Multiplying by 1/m and by m undo each other, given m≠0, which is exactly the condition of MKO a3."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero + mul_assoc + mul_comm + mul_one"
      },
      {
        "id": "S08",
        "uses": [
          "S07",
          "mko_bijective_fiber"
        ],
        "before_tex": "\\forall y\\in\\mathbb{R}:\\ \\bigl(f(x)=y\\iff x=\\tfrac{y-6}{-4}\\bigr)",
        "after_tex": "f^{-1}(\\{y\\})=\\{\\tfrac{y-6}{-4}\\}\\ \\Rightarrow\\ f\\ \\text{bijective}",
        "relation": "implies",
        "rule": "mko_bijective_fiber",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對每個實數 y：S07 的 ⇐ 方向說明實數 (y−6)/(−4) 在 y 的纖維中（存在），⇒ 方向說明纖維中任何元素都等於它（唯一）。所以每個纖維恰有一個元素，f 是雙射，滿足 a1 的前提。",
          "en": "For every real y: the ⇐ direction of S07 shows that the real number (y−6)/(−4) lies in the fiber of y (existence), and the ⇒ direction shows that every element of the fiber equals it (uniqueness). So every fiber has exactly one element and f is bijective, meeting the precondition of a1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "mko_bijective_fiber",
            "iff_intro"
          ],
          "explanation": {
            "zh-Hant": "兩個方向都要用：只有 ⇐ 只給滿射，只有 ⇒ 只給單射。",
            "en": "Both directions are used: ⇐ alone gives only surjectivity and ⇒ alone gives only injectivity."
          }
        },
        "leaf_basis": "mko_bijective_fiber + set_builder + iff_intro"
      },
      {
        "id": "S09",
        "uses": [
          "S08"
        ],
        "before_tex": "y\\in\\mathbb{R}",
        "after_tex": "g(y)=\\frac{y-6}{-4}",
        "relation": "definition",
        "rule": "mko_inverse_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，反函數把每個 y 送到它的唯一原像；由 S08，這個原像是 (y−6)/(−4)。分母 −4≠0，對每個實數 y 都有定義，所以 g 是總函數 ℝ→ℝ。",
          "en": "By the MKO, the inverse sends each y to its unique preimage; by S08 that preimage is (y−6)/(−4). The denominator −4≠0, so it is defined for every real y and g is a total function ℝ→ℝ."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "mko_inverse_def",
            "function_def"
          ],
          "explanation": {
            "zh-Hant": "只是依唯一原像命名函數；兩側恆等式仍在 S11、S12 另行驗證。",
            "en": "This only names the function by the unique preimage; both identities are still verified separately in S11 and S12."
          }
        },
        "leaf_basis": "mko_inverse_def + function_def"
      },
      {
        "id": "S10",
        "uses": [
          "S09"
        ],
        "before_tex": "\\frac{y-6}{-4}",
        "after_tex": "-\\frac{1}{4}y+\\frac{3}{2}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "-4\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "改寫成 g(y)=uy+v 的形式，得 u=−1/4、v=3/2，細節在 S10.1、S10.2。這與 MKO 的公式一致：u=1/m=1/(−4)=−1/4，v=−b/m=−6/(−4)=3/2。兩個係數都已約到最簡，分母 4 與 2 為正。",
          "en": "Rewrite in the form g(y)=uy+v to get u=−1/4 and v=3/2; details are in S10.1 and S10.2. This agrees with the MKO formula: u=1/m=1/(−4)=−1/4 and v=−b/m=−6/(−4)=3/2. Both coefficients are reduced, with positive denominators 4 and 2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "S10.1、S10.2 兩個子步都是對所有 y 成立的恆等改寫，所以兩個式子定義同一個函數。",
            "en": "Both sub-steps S10.1 and S10.2 are identities valid for all y, so the two expressions define the same function."
          }
        },
        "leaf_basis": "frac_rules + neg_mul"
      },
      {
        "id": "S10.1",
        "uses": [
          "S09",
          "frac_rules"
        ],
        "before_tex": "\\frac{y-6}{-4}",
        "after_tex": "\\frac{y}{-4}-\\frac{6}{-4}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "-4\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "同分母分式的加減：(a−c)/d=a/d−c/d（d=−4≠0）。反向是把兩個同分母分式合併。",
          "en": "Fractions with a common denominator: (a−c)/d=a/d−c/d with d=−4≠0. The backward move combines the two fractions again."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "分式加法規則雙向成立，前提只有分母不為零。",
            "en": "The fraction addition rule holds both ways; its only condition is a nonzero denominator."
          }
        },
        "leaf_basis": "frac_rules"
      },
      {
        "id": "S10.2",
        "uses": [
          "S10.1",
          "neg_mul"
        ],
        "before_tex": "\\frac{y}{-4}-\\frac{6}{-4}",
        "after_tex": "-\\frac{1}{4}y+\\frac{3}{2}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把負分母移到分子：y/(−4)=−y/4=(−1/4)y；6/(−4)=−6/4=−3/2（以 gcd(6,4)=2 約分），減去 −3/2 即加 3/2。所以 u=−1/4、v=3/2，gcd(1,4)=1、gcd(3,2)=1，分母為正。",
          "en": "Move the negative denominator to the numerator: y/(−4)=−y/4=(−1/4)y; 6/(−4)=−6/4=−3/2 (reduced by gcd(6,4)=2), and subtracting −3/2 means adding 3/2. So u=−1/4 and v=3/2, with gcd(1,4)=1, gcd(3,2)=1 and positive denominators."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "分子分母同乘 −1 與約分都是等值改寫，可逆。",
            "en": "Multiplying numerator and denominator by −1 and reducing are rewritings of equal values and can be undone."
          }
        },
        "leaf_basis": "frac_rules + neg_mul + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "G2",
          "S10"
        ],
        "before_tex": "x\\in\\mathbb{R}\\ \\text{arbitrary}",
        "after_tex": "g(f(x))=-\\tfrac{1}{4}(-4x+6)+\\tfrac{3}{2}=x",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "任取實數 x：g(f(x))=−(1/4)(−4x+6)+3/2=x−3/2+3/2=x（分配律與符號規則）。x 沒有任何附加條件，所以對所有實數成立，這是代數恆等式而不是有限核對。",
          "en": "Take any real x: g(f(x))=−(1/4)(−4x+6)+3/2=x−3/2+3/2=x (distributivity and sign rules). No extra condition was placed on x, so this holds for all reals; it is an algebraic identity, not a finite check."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "composition_def",
            "distributive",
            "neg_mul",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要 f(x)∈ℝ 位於 g 的定義域，合成才有定義；推廣到所有 x 只因 x 是任意取的。",
            "en": "It needs f(x)∈ℝ to lie in the domain of g so the composition is defined; the generalization to all x relies only on x being arbitrary."
          }
        },
        "leaf_basis": "composition_def + distributive + neg_mul + add_inverse + forall_intro"
      },
      {
        "id": "S12",
        "uses": [
          "G2",
          "S10"
        ],
        "before_tex": "y\\in\\mathbb{R}\\ \\text{arbitrary}",
        "after_tex": "f(g(y))=-4\\bigl(-\\tfrac{1}{4}y+\\tfrac{3}{2}\\bigr)+6=y",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "另一側：任取實數 y，f(g(y))=−4(−(1/4)y+3/2)+6=y−6+6=y。兩側恆等式都對所有實數成立。",
          "en": "The other side: for any real y, f(g(y))=−4(−(1/4)y+3/2)+6=y−6+6=y. Both identities hold for all reals."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "composition_def",
            "distributive",
            "neg_mul",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "與 S11 對稱；此側不能由 S11 推出，所以單獨證明。",
            "en": "Symmetric to S11; this side does not follow from S11 and is proved on its own."
          }
        },
        "leaf_basis": "composition_def + distributive + neg_mul + add_inverse + forall_intro"
      },
      {
        "id": "S13",
        "uses": [
          "S09",
          "S11",
          "S12",
          "G4"
        ],
        "before_tex": "g\\ \\text{total},\\ g\\circ f=\\mathrm{id}_{\\mathbb{R}},\\ f\\circ g=\\mathrm{id}_{\\mathbb{R}}",
        "after_tex": "f^{-1}(y)=-\\tfrac{1}{4}y+\\tfrac{3}{2}",
        "relation": "implies",
        "rule": "mko_inverse_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "g 是總函數（S09），兩側恆等式成立（S11、S12），依 G4 g 是 f 的反函數。具體例：f(2)=−2，而 g(−2)=1/2+3/2=2，回到 2。",
          "en": "g is a total function (S09) and both identities hold (S11, S12), so by G4 g is the inverse of f. Concrete instance: f(2)=−2, and g(−2)=1/2+3/2=2, returning to 2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "S11",
            "S12",
            "mko_inverse_def"
          ],
          "explanation": {
            "zh-Hant": "三個驗收條件都已逐一成立；具體例只是示意，不是證明的一部分。",
            "en": "All three acceptance conditions hold one by one; the concrete instance is an illustration, not part of the proof."
          }
        },
        "leaf_basis": "mko_inverse_def + and_intro_elim + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S13",
          "G2"
        ],
        "before_tex": "f^{-1}(2)=1,\\ \\frac{1}{f(2)}=-\\frac{1}{2}",
        "after_tex": "f^{-1}\\neq\\frac{1}{f}",
        "relation": "implies",
        "rule": "function_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "f^{-1}(2)=−(1/4)·2+3/2=1，但 f(2)=−2，所以 1/f(2)=−1/2。兩個函數在同一輸入 2 的值不同，依函數相等的定義，f^{-1}≠1/f。此外 f(3/2)=0，1/f 在 x=3/2 根本沒有定義，而 f^{-1} 對每個實數都有定義。上標 −1 表示合成意義下的反，不是倒數。",
          "en": "f^{-1}(2)=−(1/4)·2+3/2=1, but f(2)=−2, so 1/f(2)=−1/2. The two functions take different values at the same input 2, so by the definition of equality of functions f^{-1}≠1/f. Moreover f(3/2)=0, so 1/f is not even defined at x=3/2, while f^{-1} is defined at every real. The superscript −1 means inverse under composition, not reciprocal."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "function_def",
            "numeric_evaluation",
            "S13"
          ],
          "explanation": {
            "zh-Hant": "否定兩函數相等只需一個取值不同的輸入；本步不宣稱它們在每一點都不同。",
            "en": "Refuting equality of two functions needs only one input where they differ; the step does not claim they differ at every point."
          }
        },
        "leaf_basis": "function_def + numeric_evaluation + frac_rules"
      },
      {
        "id": "S15",
        "uses": [
          "G3"
        ],
        "before_tex": "h^{\\mathsf T}=\\{(4,1),(4,2),(6,3)\\}",
        "after_tex": "h^{\\mathsf T}\\ \\text{not a function}\\ B\\to A",
        "relation": "implies",
        "rule": "function_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "交換 h 的表：4 同時對應 1 與 2（不是單值），5 沒有對應（不是總的），函數的兩個要求都不滿足。這對應 h 的纖維概況 (2,0,1)：h 不是雙射，依 a1 不能用交換表當作反函數。事實上沒有任何 r:B→A 能滿足 h(r(y))=y 對所有 y∈B，因為取 y=5 需要 h(r(5))=5，而 5 不是 h 的值。",
          "en": "Exchange the table of h: 4 is paired with both 1 and 2 (not single-valued) and 5 with nothing (not total), so both requirements of a function fail. This matches the fiber profile (2,0,1) of h: h is not bijective, and by a1 its exchanged table cannot serve as an inverse. In fact no r:B→A can satisfy h(r(y))=y for every y∈B, since y=5 would need h(r(5))=5, and 5 is not a value of h."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "function_def",
            "swap_table",
            "G3"
          ],
          "explanation": {
            "zh-Hant": "任一個要求失敗就足以否定「是函數」；這裡兩個都失敗。結論只關於 h，不影響 p、f。",
            "en": "Failure of either requirement suffices to refute 'is a function'; here both fail. The conclusion concerns h only and does not affect p or f."
          }
        },
        "leaf_basis": "function_def + swap_table + set_builder"
      },
      {
        "id": "S16",
        "uses": [
          "S06",
          "S13",
          "S14",
          "S15"
        ],
        "before_tex": "\\text{S06, S13, S14, S15}",
        "after_tex": "q=p^{-1};\\ f^{-1}(y)=-\\tfrac{1}{4}y+\\tfrac{3}{2};\\ h^{\\mathsf T}\\ \\text{rejected};\\ f^{-1}\\neq 1/f",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "整理三部分：有限雙射表交換後得到 p 的反函數 q；仿射函數的反函數係數 u=−1/4、v=3/2；非雙射的交換表被拒絕，反函數也與倒數區分開。",
          "en": "Collect the three parts: exchanging the finite bijection table gives the inverse q of p; the affine inverse has coefficients u=−1/4 and v=3/2; the exchanged table of the non-bijective h is rejected, and the inverse is distinguished from the reciprocal."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S06",
            "S13",
            "S14",
            "S15",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G 是 goal 的三部分。足夠條件 H 為 S06、S13、S14、S15 的合取，各自已從已知正向證成；H⇒G 只用合取引入。",
            "en": "Goal G is the three parts of the goal. The sufficient condition H is the conjunction of S06, S13, S14 and S15, each proved forwards from the givens; H⇒G uses only conjunction introduction."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "p 的纖維概況為 (1,1,1)，是雙射；交換表 q 是總函數 B→A，q(4)=3、q(5)=1、q(6)=2，且 q∘p=id_A、p∘q=id_B，所以 q=p^{-1}。f(x)=−4x+6 是雙射，反函數為 f^{-1}(y)=(y−6)/(−4)=−(1/4)y+3/2，即 u=−1/4、v=3/2，兩側恆等式對所有實數成立。h 的交換表不是 B→A 的函數（4 有兩個值、5 沒有值）；f^{-1}(2)=1 而 1/f(2)=−1/2，所以 f^{-1}≠1/f。",
      "en": "p has fiber profile (1,1,1) and is bijective; the exchanged table q is a total function B→A with q(4)=3, q(5)=1, q(6)=2, and q∘p=id_A, p∘q=id_B, so q=p^{-1}. f(x)=−4x+6 is bijective with inverse f^{-1}(y)=(y−6)/(−4)=−(1/4)y+3/2, i.e. u=−1/4 and v=3/2, and both identities hold for all reals. The exchanged table of h is not a function B→A (4 has two values, 5 has none); f^{-1}(2)=1 while 1/f(2)=−1/2, so f^{-1}≠1/f."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "g(y)=1/(−4y+6) 是 f 的反函數",
          "en": "g(y)=1/(−4y+6) is the inverse of f"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "這是 1/f，不是反函數：f(0)=6，而 1/(−4·6+6)=−1/18≠0，所以 g(f(0))≠0。",
          "en": "This is 1/f, not the inverse: f(0)=6, and 1/(−4·6+6)=−1/18≠0, so g(f(0))≠0."
        }
      },
      {
        "claim": {
          "zh-Hant": "h 的交換表 {(4,1),(4,2),(6,3)} 是 h 的反函數",
          "en": "The exchanged table {(4,1),(4,2),(6,3)} of h is the inverse of h"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "它不是 B→A 的函數：4 有兩個值，5 沒有值；h 不是雙射。",
          "en": "It is not a function B→A: 4 has two values and 5 has none; h is not bijective."
        }
      },
      {
        "claim": {
          "zh-Hant": "r(4)=1、r(5)=2、r(6)=1 滿足 r(p′(x))=x，所以 r 是 p′ 的反函數",
          "en": "r(4)=1, r(5)=2, r(6)=1 satisfies r(p′(x))=x, so r is the inverse of p′"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "另一側不成立：p′(r(6))=p′(1)=4≠6。",
          "en": "The other side fails: p′(r(6))=p′(1)=4≠6."
        }
      },
      {
        "claim": {
          "zh-Hant": "f(x)=−4x+6 的反函數是 g(y)=(y+6)/(−4)",
          "en": "The inverse of f(x)=−4x+6 is g(y)=(y+6)/(−4)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "符號錯：f(0)=6，而 (6+6)/(−4)=−3≠0；正確的是 (y−6)/(−4)。",
          "en": "Sign error: f(0)=6, and (6+6)/(−4)=−3≠0; the correct inverse is (y−6)/(−4)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "function_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "函數：f:A→B 對每個 a∈A 指定恰好一個 f(a)∈B。兩函數相等 ⇔ 定義域、對應域相同且對每個 a 都有相同值。",
          "en": "Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a."
        }
      },
      {
        "id": "injective_surjective_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "單射：f(a)=f(a′) ⇒ a=a′。滿射：每個 b∈B 都有 a∈A 使 f(a)=b。雙射：既單又滿，此時有唯一反函數 f⁻¹ 使 f⁻¹∘f=id_A、f∘f⁻¹=id_B。",
          "en": "Injective: f(a)=f(a′) ⇒ a=a′. Surjective: every b∈B has some a∈A with f(a)=b. Bijective: both; then there is a unique inverse f⁻¹ with f⁻¹∘f=id_A and f∘f⁻¹=id_B."
        }
      },
      {
        "id": "composition_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合成：對 f:A→B 與 g:B→C，g∘f:A→C 定義為 (g∘f)(a)=g(f(a))。",
          "en": "Composition: for f:A→B and g:B→C, g∘f:A→C is defined by (g∘f)(a)=g(f(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": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "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": "cartesian_product_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "笛卡兒積：A×B 是所有有序對 (a,b)（a∈A、b∈B）的集合；(a,b)=(c,d) ⇔ a=c 且 b=d。",
          "en": "Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d."
        }
      },
      {
        "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": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "eq_congruence",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。",
          "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule."
        }
      },
      {
        "id": "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": "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_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法交換律：u+v=v+u。",
          "en": "Commutativity of addition: u+v=v+u."
        }
      },
      {
        "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": "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_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": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "id": "iff_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "雙條件：P⇒Q 與 Q⇒P 都成立時，P⇔Q 成立；反之亦然。",
          "en": "Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold."
        }
      },
      {
        "id": "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": "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": "mko_inverse_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-inverse-function",
          "sha256": "bad163b9a218429518937fb4e86abe0e8707ebda0fc06c7e27f625184bc05c4f",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "若 f:A→B 是雙射，反函數 g:B→A 由每個 b 的唯一原像定義，滿足 g(f(x))=x（x∈A）及 f(g(y))=y（y∈B）。有限表可以交換有序對後建立 g；若 f 非雙射，交換表不能自動成為這個 B→A 的反函數。",
          "en": "If f:A→B is bijective, its inverse g:B→A assigns each b its unique preimage and satisfies g(f(x))=x for x∈A and f(g(y))=y for y∈B. A finite bijection table can be inverted by exchanging ordered pairs. For nonbijective f, the exchanged table is not automatically an inverse on all of B."
        }
      },
      {
        "id": "mko_bijective_fiber",
        "kind": "cited",
        "source": {
          "mko_id": "mko-bijective",
          "sha256": "3f8aacffa7cd2d06121165a4baccd488e1af92185dc745b6ee2cd5733c59c73c",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "總函數 f:A→B 為雙射，當且僅當它同時單射與滿射；等價地，每個 b∈B 的纖維恰有一個元素。",
          "en": "A total function f:A→B is bijective exactly when it is both injective and surjective, or equivalently when each fiber of b∈B has exactly one element."
        }
      },
      {
        "id": "swap_table",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "交換表：關係 R⊆A×B 的交換為 R^T={(b,a) ∣ (a,b)∈R}⊆B×A。b 在 R^T 中作為第一坐標出現的次數，等於 b 在 R 中作為第二坐標出現的次數。",
          "en": "Exchanged table: the exchange of a relation R⊆A×B is R^T={(b,a) ∣ (a,b)∈R}⊆B×A. The number of times b occurs as a first coordinate in R^T equals the number of times it occurs as a second coordinate in R."
        }
      },
      {
        "id": "L_solve_affine",
        "kind": "lemma",
        "proved_in": [
          "S07.1",
          "S07.2"
        ],
        "statement": {
          "zh-Hant": "對實數 m≠0、b、x、y：y=mx+b ⇔ x=(y−b)/m。",
          "en": "For reals m≠0, b, x, y: y=mx+b ⇔ x=(y−b)/m."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S10",
        "lhs": "1/(-4)",
        "rhs": "-1/4"
      },
      {
        "id": "N2",
        "step": "S10",
        "lhs": "-6/(-4)",
        "rhs": "3/2"
      },
      {
        "id": "N3",
        "step": "S10",
        "vars": [
          "y"
        ],
        "degree": 1,
        "lhs": "(y-6)/(-4)",
        "rhs": "-1/4*y+3/2"
      },
      {
        "id": "N4",
        "step": "S11",
        "vars": [
          "x"
        ],
        "degree": 1,
        "lhs": "-1/4*(-4*x+6)+3/2",
        "rhs": "x"
      },
      {
        "id": "N5",
        "step": "S12",
        "vars": [
          "y"
        ],
        "degree": 1,
        "lhs": "-4*(-1/4*y+3/2)+6",
        "rhs": "y"
      },
      {
        "id": "N6",
        "step": "S13",
        "lhs": "-4*2+6",
        "rhs": "-2"
      },
      {
        "id": "N7",
        "step": "S13",
        "lhs": "-1/4*(-2)+3/2",
        "rhs": "2"
      },
      {
        "id": "N8",
        "step": "S14",
        "lhs": "-1/4*2+3/2",
        "rhs": "1"
      },
      {
        "id": "N9",
        "step": "S14",
        "lhs": "1/(-4*2+6)",
        "rhs": "-1/2"
      },
      {
        "id": "N10",
        "step": "S14",
        "lhs": "1",
        "relation": "!=",
        "rhs": "-1/2"
      },
      {
        "id": "N11",
        "step": "S14",
        "lhs": "-4*(3/2)+6",
        "rhs": "0"
      },
      {
        "id": "N12",
        "step": "S10.2",
        "lhs": "gcd(3,2)",
        "rhs": "1"
      },
      {
        "id": "N13",
        "step": "S14",
        "lhs": "1/(-4*6+6)",
        "rhs": "-1/18"
      },
      {
        "id": "N14",
        "step": "S10",
        "lhs": "(6+6)/(-4)",
        "rhs": "-3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "依 MKO 假設：有限表先確認是宣告 A→B 的雙射（a1）；候選先驗為總函數再查兩側恆等式（a2）；仿射例子為 f:ℝ→ℝ、m≠0，係數以最簡正分母有理數表示（a3）。仿射部分以代數恆等式對所有實數證明，不是抽樣。不處理一般單側反函數理論、反函數唯一性或機器證明；不改動 canonical MKO 或英文 edition。",
      "en": "Follows the MKO assumptions: a finite table is first shown to be a bijection on the declared A→B (a1); a candidate is first validated as a total function and then both identities are checked (a2); the affine example is f:ℝ→ℝ with m≠0, coefficients as reduced rationals with positive denominators (a3). The affine part is proved for all reals by algebraic identities, not by sampling. General one-sided inverse theory, uniqueness of inverses and machine proofs are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "p,q",
        "有限雙射表與交換後的表",
        "the finite bijection table and the exchanged table"
      ],
      [
        "f,g",
        "仿射函數 f(x)=−4x+6 與其反函數",
        "the affine map f(x)=−4x+6 and its inverse"
      ],
      [
        "h",
        "非雙射的有限表",
        "a non-bijective finite table"
      ],
      [
        "R^{\\mathsf T}",
        "交換關係 {(b,a) ∣ (a,b)∈R}",
        "the exchanged relation {(b,a) ∣ (a,b)∈R}"
      ],
      [
        "\\mathrm{id}_A",
        "A 上的恆等函數",
        "the identity function on A"
      ]
    ],
    "md_error_controls": [
      {
        "title": "反函數不是倒數 / The inverse is not the reciprocal",
        "zh": "MKO 的誤解之一是把 f^{-1} 當成 1/f。上標 −1 表示對合成的反：f^{-1}(2)=1，但 1/f(2)=−1/2（S14）。而且 1/f 在 x=3/2 沒有定義（f(3/2)=0），f^{-1} 卻對每個實數都有定義。",
        "en": "One misconception identified in the MKO is reading f^{-1} as 1/f. The superscript −1 means inverse under composition: f^{-1}(2)=1 but 1/f(2)=−1/2 (S14). Also 1/f is undefined at x=3/2 (f(3/2)=0), while f^{-1} is defined at every real."
      },
      {
        "title": "非雙射的交換表不是反函數 / Exchanging a non-bijective table does not give an inverse",
        "zh": "另一個誤解是不管是否雙射都交換表。h 的纖維概況是 (2,0,1)，交換後 4 對應兩個值、5 沒有值，不是 B→A 的函數（S15）。MKO 要求先確認雙射（a1），再交換。",
        "en": "The other misconception is to exchange a table whether or not it is bijective. h has fiber profile (2,0,1); after exchanging, 4 has two values and 5 has none, so it is not a function B→A (S15). The MKO requires establishing bijectivity first (a1) and only then exchanging."
      },
      {
        "title": "兩側恆等式都要查 / Both identities must be checked",
        "zh": "只有一側成立時不是反函數。例：p′:{1,2}→{4,5,6}，p′(1)=4、p′(2)=5；候選 r:{4,5,6}→{1,2}，r(4)=1、r(5)=2、r(6)=1 是總函數，且對每個 x∈{1,2} 都有 r(p′(x))=x，但 p′(r(6))=p′(1)=4≠6。r 只是左反函數。本稿對 q 與 g 都分別驗證兩側（S04、S05、S11、S12）。",
        "en": "If only one side holds, the candidate is not an inverse. Example: p′:{1,2}→{4,5,6} with p′(1)=4, p′(2)=5; the candidate r:{4,5,6}→{1,2} with r(4)=1, r(5)=2, r(6)=1 is a total function and r(p′(x))=x for every x∈{1,2}, but p′(r(6))=p′(1)=4≠6. r is only a left inverse. This guide checks both sides for q and for g (S04, S05, S11, S12)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿不另證反函數的唯一性；只驗證構造出的 q 與 g 符合 MKO 的兩側定義。",
        "en": "This guide does not separately prove that inverses are unique; it verifies only that the constructed q and g satisfy the MKO's two-sided definition."
      },
      {
        "zh-Hant": "N3 至 N5 是一次多項式在格點上的恆等式核對，與 S10 至 S12 的代數推導互相獨立。",
        "en": "N3 to N5 are identity checks of degree-one polynomials on a grid, independent of the algebra in S10 to S12."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "a9ff58a9eb164d3ba3885c3f347bc130c608c3b605ef14409ff4d974ddfcb2e6",
    "original_source_sha256": "a9ff58a9eb164d3ba3885c3f347bc130c608c3b605ef14409ff4d974ddfcb2e6",
    "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."
  }
}