{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-ratio-proportion-01",
  "mko_id": "mko-ratio-proportion",
  "title": {
    "zh-Hant": "比與比例：解 3:4=x:10、代回核對與檢查正比例資料",
    "en": "Ratio and proportion: solving 3:4=x:10, checking back, and testing data for direct proportion"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-ratio-proportion-01",
    "source": {
      "mko_id": "mko-ratio-proportion",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "b0200b9dd1e1ceb4c2b34f37f0279a1abaf31c3021957de6cd9ea72919da47ec",
      "english_sha256": "f41a9985bd51a22f5108587c130e1ecaa9d44b5c929b8653cb23ff7ac3021771"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "mul_inverse_nonzero",
      "zero_product",
      "frac_rules",
      "eq_congruence",
      "mul_assoc",
      "mul_comm",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在有理數中：(1) 證明對 b≠0、d≠0，比例 a:b=x:d 恰有一個有理解 x=ad/b；(2) 用它解 3:4=x:10 並代回核對；(3) 說明 4:3=x:10 的解不同，比的順序有意義；(4) 判定資料 (2,5)、(4,10)、(6,15)、(8,21) 是否符合同一個正比例 y=kx。",
      "en": "Over the rationals: (1) prove that for b≠0 and d≠0 the proportion a:b=x:d has exactly one rational solution, x=ad/b; (2) use it to solve 3:4=x:10 and check back; (3) show that 4:3=x:10 has a different solution, so the order of a ratio matters; (4) decide whether the data (2,5), (4,10), (6,15), (8,21) fit a single direct proportion y=kx."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "a、b、c、d、x 是有理數，且 b≠0、d≠0（MKO 假設 a1）。",
          "en": "a, b, c, d and x are rational numbers with b≠0 and d≠0 (MKO assumption a1)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "數值比 a:b 的值是 a/b；比例 a:b=c:d 表示 a/b=c/d。",
          "en": "The value of the ratio a:b is a/b; the proportion a:b=c:d means a/b=c/d."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "資料表 (xᵢ,yᵢ)：(2,5)、(4,10)、(6,15)、(8,21)。「資料符合正比例 y=kx」的意思是：存在同一個有理數 k，使每一筆都有 yᵢ=k·xᵢ。",
          "en": "The data (xᵢ,yᵢ): (2,5), (4,10), (6,15), (8,21). 'The data fit a direct proportion y=kx' means that one rational number k satisfies yᵢ=k·xᵢ for every data point."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_ratio_def"
        ],
        "before_tex": "a:b=x:d",
        "after_tex": "\\frac{a}{b}=\\frac{x}{d}\\quad (b\\neq 0,\\ d\\neq 0)",
        "relation": "definition",
        "rule": "mko_ratio_def",
        "conditions": [
          "b\\neq 0",
          "d\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依 MKO，比 a:b 的值是 a/b，所以比例 a:b=x:d 就是有理數等式 a/b=x/d。兩個後項 b、d 必須非零，兩個商才有定義。",
          "en": "By the MKO the value of a:b is a/b, so the proportion a:b=x:d is the equation of rationals a/b=x/d. Both consequents b and d must be nonzero for the quotients to be defined."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_ratio_def"
          ],
          "explanation": {
            "zh-Hant": "此步只展開定義；條件 b≠0、d≠0 之後每一步都要用到。",
            "en": "This step only unfolds the definition; the conditions b≠0 and d≠0 are needed in every later step."
          }
        },
        "leaf_basis": "mko_ratio_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "L_cross"
        ],
        "before_tex": "\\frac{a}{b}=\\frac{x}{d}",
        "after_tex": "ad=bx",
        "relation": "equivalent",
        "rule": "L_cross",
        "conditions": [
          "b\\neq 0",
          "d\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_cross（c=x），比例等價於交叉乘積相等 ad=bx。這是 MKO 中「等價於 ad=bc」的內容，引理在 S02.1、S02.2 證明。",
          "en": "By Lemma L_cross with c=x, the proportion is equivalent to equal cross products, ad=bx. This is the clause 'equivalent to ad=bc' of the MKO; the lemma is proved in S02.1 and S02.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_cross"
          ],
          "explanation": {
            "zh-Hant": "引理兩個方向都有證明，且都需要 b≠0、d≠0。",
            "en": "Both directions of the lemma are proved, and both need b≠0 and d≠0."
          }
        },
        "leaf_basis": "L_cross"
      },
      {
        "id": "S02.1",
        "uses": [
          "eq_congruence",
          "mul_inverse_nonzero"
        ],
        "before_tex": "\\frac{a}{b}=\\frac{c}{d}",
        "after_tex": "ad=bc",
        "relation": "implies",
        "rule": "eq_congruence",
        "conditions": [
          "b\\neq 0",
          "d\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "兩邊同乘 bd：左邊 a·(1/b)·b·d=a·1·d=ad，右邊 c·(1/d)·b·d=b·c·(d·(1/d))=bc（結合律、交換律、b·(1/b)=1、d·(1/d)=1）。",
          "en": "Multiply both sides by bd: the left side is a·(1/b)·b·d=a·1·d=ad and the right side is c·(1/d)·b·d=b·c·(d·(1/d))=bc (associativity, commutativity, b·(1/b)=1 and d·(1/d)=1)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "eq_congruence",
            "mul_inverse_nonzero",
            "mul_assoc",
            "mul_comm"
          ],
          "explanation": {
            "zh-Hant": "化簡需要 1/b、1/d 存在。這是 ⇒ 方向。",
            "en": "The simplification needs 1/b and 1/d to exist. This is the ⇒ direction."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero + mul_assoc + mul_comm + mul_one"
      },
      {
        "id": "S02.2",
        "uses": [
          "zero_product",
          "eq_congruence",
          "frac_rules"
        ],
        "before_tex": "ad=bc",
        "after_tex": "\\frac{a}{b}=\\frac{c}{d}",
        "relation": "implies",
        "rule": "eq_congruence",
        "conditions": [
          "bd\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "因 b≠0、d≠0，依零乘積性質 bd≠0。兩邊同乘 1/(bd)：左邊 (ad)/(bd)=a/b（消去非零的 d），右邊 (bc)/(bd)=c/d（消去非零的 b）。S02.1 與 S02.2 合起來證得引理 L_cross。",
          "en": "Since b≠0 and d≠0, the zero-product property gives bd≠0. Multiply both sides by 1/(bd): the left side is (ad)/(bd)=a/b (cancelling the nonzero d) and the right side is (bc)/(bd)=c/d (cancelling the nonzero b). S02.1 and S02.2 together prove Lemma L_cross."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "zero_product",
            "frac_rules",
            "eq_congruence"
          ],
          "explanation": {
            "zh-Hant": "這是 ⇐ 方向，需要 bd≠0；若容許 b=0 或 d=0，ad=bc 推不回任何比例（見錯誤控制一）。",
            "en": "This is the ⇐ direction and needs bd≠0; if b=0 or d=0 were allowed, ad=bc would not lead back to any proportion (see error control 1)."
          }
        },
        "leaf_basis": "eq_congruence + zero_product + frac_rules"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "L_solve"
        ],
        "before_tex": "ad=bx",
        "after_tex": "x=\\frac{ad}{b}",
        "relation": "equivalent",
        "rule": "L_solve",
        "conditions": [
          "b\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依等號對稱，ad=bx 即 bx=ad。依引理 L_solve（u=ad），因 b≠0，bx=ad ⇔ x=ad/b。",
          "en": "By symmetry of equality, ad=bx is bx=ad. By Lemma L_solve with u=ad, since b≠0, bx=ad ⇔ x=ad/b."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_solve",
            "eq_refl_symm_trans"
          ],
          "explanation": {
            "zh-Hant": "引理兩向都在 S03.1 證明；需要 b≠0。",
            "en": "Both directions of the lemma are proved in S03.1; b≠0 is needed."
          }
        },
        "leaf_basis": "L_solve + eq_refl_symm_trans"
      },
      {
        "id": "S03.1",
        "uses": [
          "mul_inverse_nonzero",
          "eq_congruence"
        ],
        "before_tex": "bx=u",
        "after_tex": "x=\\frac{u}{b}",
        "relation": "equivalent",
        "rule": "mul_inverse_nonzero",
        "conditions": [
          "b\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "⇒：兩邊同乘 1/b，(1/b)(bx)=((1/b)b)x=1·x=x，右邊 u·(1/b)=u/b。⇐：兩邊同乘 b，b·(u/b)=u·((1/b)b)=u。b=0 時 1/b 不存在，兩個方向都不能做。此子步證得引理 L_solve。",
          "en": "⇒: multiply both sides by 1/b: (1/b)(bx)=((1/b)b)x=1·x=x, and the right side becomes u·(1/b)=u/b. ⇐: multiply both sides by b: b·(u/b)=u·((1/b)b)=u. If b=0 there is no 1/b and neither direction can be taken. This sub-step proves Lemma L_solve."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_inverse_nonzero",
            "eq_congruence",
            "mul_assoc"
          ],
          "explanation": {
            "zh-Hant": "乘 1/b 與乘 b 互為反運算，前提 b≠0。",
            "en": "Multiplying by 1/b and by b undo each other, given b≠0."
          }
        },
        "leaf_basis": "mul_inverse_nonzero + eq_congruence + mul_assoc + mul_comm + mul_one"
      },
      {
        "id": "S04",
        "uses": [
          "S01",
          "S02",
          "S03"
        ],
        "before_tex": "a:b=x:d",
        "after_tex": "\\{x\\in\\mathbb{Q}\\mid a:b=x:d\\}=\\left\\{\\frac{ad}{b}\\right\\}",
        "relation": "implies",
        "rule": "logic_chain",
        "conditions": [
          "b\\neq 0",
          "d\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "S01 到 S03 是一串等價：a:b=x:d ⇔ x=ad/b。所以解集恰為 {ad/b}。存在：ad/b 是有理數（有理數對乘法封閉，且 b≠0 時可除）。唯一：任何解都等於 ad/b。這就是 MKO 第三段所說的唯一有理解。",
          "en": "S01 to S03 form a chain of equivalences: a:b=x:d ⇔ x=ad/b. So the solution set is exactly {ad/b}. Existence: ad/b is rational (rationals are closed under multiplication, and division by b≠0 is allowed). Uniqueness: every solution equals ad/b. This is the unique rational solution described in the third paragraph of the MKO."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S03",
            "logic_chain",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "因為每一步都是等價，所以同時得到存在與唯一；若其中有單向步驟，就只能得到其中一半。",
            "en": "Because every link is an equivalence, existence and uniqueness both follow; a one-way link would give only half of this."
          }
        },
        "leaf_basis": "logic_chain + set_extensionality + iff_intro"
      },
      {
        "id": "S05",
        "uses": [
          "S04"
        ],
        "before_tex": "3:4=x:10",
        "after_tex": "x=\\frac{3\\cdot 10}{4}=\\frac{30}{4}=\\frac{15}{2}",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [
          "4\\neq 0",
          "10\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "代入 a=3、b=4、d=10；條件 4≠0、10≠0 成立。依 S04，唯一解是 x=3·10/4=30/4；約去公因數 2 得 15/2。",
          "en": "Substitute a=3, b=4, d=10; the conditions 4≠0 and 10≠0 hold. By S04 the unique solution is x=3·10/4=30/4, and cancelling the common factor 2 gives 15/2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "forall_elim",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "套用 S04 前先確認兩個後項非零；30/4=15/2 是約分（N1）。",
            "en": "Before applying S04 both consequents are checked to be nonzero; 30/4=15/2 is a reduction (N1)."
          }
        },
        "leaf_basis": "forall_elim + frac_rules + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "L_cross"
        ],
        "before_tex": "x=\\frac{15}{2}",
        "after_tex": "\\frac{15/2}{10}=\\frac{15}{20}=\\frac{3}{4},\\quad 3\\cdot 10=30=4\\cdot\\frac{15}{2}",
        "relation": "implies",
        "rule": "L_cross",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代回核對：(15/2)/10=15/20=3/4，與 3:4 的值相同；交叉乘積 3·10=30、4·(15/2)=30 也相等。代回只確認 15/2 是解；「只有這個解」來自 S04 的等價鏈。",
          "en": "Check back: (15/2)/10=15/20=3/4, the same as the value of 3:4; the cross products 3·10=30 and 4·(15/2)=30 also agree. Substitution only confirms that 15/2 is a solution; that it is the only one comes from the chain of equivalences in S04."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "L_cross",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "這是獨立的數值核對（N2、N3），不取代 S04 的唯一性論證。",
            "en": "This is an independent numerical check (N2, N3); it does not replace the uniqueness argument of S04."
          }
        },
        "leaf_basis": "L_cross + frac_rules + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S04",
          "S05",
          "L_cross"
        ],
        "before_tex": "4:3=x:10",
        "after_tex": "x=\\frac{40}{3}\\neq\\frac{15}{2}",
        "relation": "implies",
        "rule": "L_cross",
        "conditions": [
          "3\\neq 0",
          "10\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "把比的前後項對調：4:3 的後項是 3≠0，所以它是合法的比，依 S04 解得 x=4·10/3=40/3。40/3 與 15/2 的交叉乘積 40·2=80、3·15=45 不相等，所以兩個解不同。又 4/3 與 3/4 的交叉乘積 4·4=16≠9=3·3，所以 4:3≠3:4。比的順序是資料的一部分。",
          "en": "Swap the terms of the ratio: the consequent of 4:3 is 3≠0, so it is a valid ratio, and S04 gives x=4·10/3=40/3. The cross products of 40/3 and 15/2 are 40·2=80 and 3·15=45, which differ, so the two solutions differ. Also 4/3 and 3/4 have cross products 4·4=16≠9=3·3, so 4:3≠3:4. The order of a ratio is part of its data."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "L_cross",
            "contrapositive"
          ],
          "explanation": {
            "zh-Hant": "反向比 b:a 需要原前項 a≠0；這裡 a=3≠0。不等的結論用 L_cross 的逆否：交叉乘積不等推出值不等。",
            "en": "The reversed ratio b:a needs the original antecedent a≠0; here a=3≠0. The inequality uses L_cross in contrapositive form: unequal cross products give unequal values."
          }
        },
        "leaf_basis": "L_cross + contrapositive + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "G3",
          "L_solve"
        ],
        "before_tex": "y_i=k\\,x_i\\quad (x_i\\neq 0)",
        "after_tex": "k=\\frac{y_i}{x_i}",
        "relation": "equivalent",
        "rule": "L_solve",
        "conditions": [
          "x_i\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "四筆資料的 xᵢ 是 2、4、6、8，都非零。依 L_solve（b=xᵢ、u=yᵢ），yᵢ=k·xᵢ ⇔ k=yᵢ/xᵢ。所以資料符合正比例，當且僅當四個比值 yᵢ/xᵢ 是同一個數。",
          "en": "The xᵢ of the four data points are 2, 4, 6 and 8, all nonzero. By L_solve with b=xᵢ and u=yᵢ, yᵢ=k·xᵢ ⇔ k=yᵢ/xᵢ. So the data fit a direct proportion exactly when the four ratios yᵢ/xᵢ are one and the same number."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_solve",
            "G3"
          ],
          "explanation": {
            "zh-Hant": "每一筆都是等價變形；需要 xᵢ≠0。",
            "en": "Each data point is transformed by an equivalence; xᵢ≠0 is needed."
          }
        },
        "leaf_basis": "L_solve + mul_comm"
      },
      {
        "id": "S09",
        "uses": [
          "S08",
          "L_cross"
        ],
        "before_tex": "\\frac{5}{2},\\ \\frac{10}{4},\\ \\frac{15}{6}",
        "after_tex": "\\frac{5}{2}=\\frac{10}{4}=\\frac{15}{6}",
        "relation": "implies",
        "rule": "L_cross",
        "conditions": [],
        "explanation": {
          "zh-Hant": "前三筆的比值：10/4 與 5/2 的交叉乘積 10·2=20=4·5；15/6 與 5/2 的交叉乘積 15·2=30=6·5。依 L_cross，三個比值都等於 5/2，所以 k=5/2 適合前三筆。",
          "en": "Ratios of the first three points: 10/4 and 5/2 have cross products 10·2=20=4·5, and 15/6 and 5/2 have 15·2=30=6·5. By L_cross all three ratios equal 5/2, so k=5/2 fits the first three points."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "L_cross",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "逐點比較同一個 k；這只涉及前三筆，不推論表外或第四筆。",
            "en": "Each point is compared with the same k; this concerns only the first three points and says nothing about the fourth or about points outside the table."
          }
        },
        "leaf_basis": "L_cross + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "S08",
          "S09",
          "L_cross"
        ],
        "before_tex": "\\frac{21}{8}\\ \\text{vs}\\ \\frac{5}{2}",
        "after_tex": "21\\cdot 2=42\\neq 40=8\\cdot 5,\\ \\text{so no single}\\ k",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第四筆的比值 21/8 與 5/2 的交叉乘積 42≠40，所以 21/8≠5/2。若有同一個 k 適合全部四筆，依 S08，k=5/2（第一筆）且 k=21/8（第四筆），推出 5/2=21/8，矛盾。所以整張表不符合任何正比例 y=kx，雖然前三筆符合。",
          "en": "The fourth ratio 21/8 and 5/2 have cross products 42≠40, so 21/8≠5/2. If one k fitted all four points, S08 would give k=5/2 (first point) and k=21/8 (fourth point), hence 5/2=21/8, a contradiction. So the table fits no direct proportion y=kx, even though the first three points do."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "S09",
            "contradiction",
            "L_cross"
          ],
          "explanation": {
            "zh-Hant": "矛盾來自「同一個 k」必須逐點成立；也可直接看 (5/2)·8=20≠21（N10）。",
            "en": "The contradiction comes from requiring one k at every point; equivalently (5/2)·8=20≠21 (N10)."
          }
        },
        "leaf_basis": "contradiction + L_cross + eq_refl_symm_trans + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S01",
          "mko_ratio_def"
        ],
        "before_tex": "0:0=x:5",
        "after_tex": "\\text{not a proportion, although}\\ 0\\cdot 5=0\\cdot x\\ \\text{for every}\\ x",
        "relation": "implies",
        "rule": "mko_ratio_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "非實例：0:0 的後項是 0，依 MKO 它不是數值比，0/0 沒有定義，所以 0:0=x:5 不是比例。若硬算交叉乘積，0·5=0·x 對每個 x 都成立，似乎「每個 x 都是解」；這只說明 L_cross 的前提 b≠0 不能省略，交叉乘積相等不能讓零後項變得合法。",
          "en": "Non-example: 0:0 has consequent 0, so by the MKO it is not a numerical ratio and 0/0 is undefined; 0:0=x:5 is not a proportion. Forcing the cross products gives 0·5=0·x for every x, as if every x were a solution; this only shows that the hypothesis b≠0 of L_cross cannot be dropped, and equal cross products cannot make a zero consequent legal."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_ratio_def",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "結論只是「這不是比例」；不對 x 宣稱任何解集。",
            "en": "The conclusion is only that this is not a proportion; no solution set for x is claimed."
          }
        },
        "leaf_basis": "mko_ratio_def + mul_inverse_nonzero + mul_zero"
      },
      {
        "id": "S12",
        "uses": [
          "S04",
          "S05",
          "S07",
          "S10"
        ],
        "before_tex": "S04,\\ S05,\\ S07,\\ S10",
        "after_tex": "x=\\frac{ad}{b}\\ \\text{unique};\\ 3:4=x:10\\Leftrightarrow x=\\frac{15}{2};\\ 4:3=x:10\\Leftrightarrow x=\\frac{40}{3};\\ \\text{table not proportional}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "四個目標分別由 S04（唯一解公式）、S05 與 S06（3:4=x:10 的解與核對）、S07（順序）與 S10（資料判定）完成。",
          "en": "The four goals are completed by S04 (the unique-solution formula), S05 and S06 (solving 3:4=x:10 and checking), S07 (order) and S10 (the data test)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S04",
            "S05",
            "S07",
            "S10",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：(1) 到 (4)。足夠條件 H：S04、S05、S07、S10 各自成立；H⇒G 由合取引入。H 由定義、引理與精確計算正向證成，沒有把 G 當作前提。",
            "en": "Goal G: (1) to (4). Sufficient condition H: S04, S05, S07 and S10 each hold; H⇒G by conjunction introduction. H is proved forwards from the definition, the lemmas and exact computation; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "對 b≠0、d≠0，a:b=x:d 恰有一解 x=ad/b。3:4=x:10 的解是 15/2（代回 3·10=30=4·15/2）；4:3=x:10 的解是 40/3，比的順序有意義。資料前三筆符合 y=(5/2)x，但第四筆 21/8≠5/2，所以整張表不是正比例。0:0 不是比，交叉乘積不能讓它合法。",
      "en": "For b≠0 and d≠0, a:b=x:d has exactly one solution, x=ad/b. The solution of 3:4=x:10 is 15/2 (check: 3·10=30=4·15/2); that of 4:3=x:10 is 40/3, so the order of a ratio matters. The first three data points fit y=(5/2)x, but the fourth gives 21/8≠5/2, so the table is not directly proportional. 0:0 is not a ratio, and cross products cannot make it legal."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "因為 0·5=0·x，所以 0:0=x:5 對每個 x 成立",
          "en": "Since 0·5=0·x, the proportion 0:0=x:5 holds for every x"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "0:0 的後項為 0，不是比；L_cross 需要後項非零。",
          "en": "0:0 has consequent 0 and is not a ratio; L_cross needs nonzero consequents."
        }
      },
      {
        "claim": {
          "zh-Hant": "3:4=4:3",
          "en": "3:4=4:3"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "交叉乘積 3·3=9≠16=4·4。",
          "en": "Cross products 3·3=9≠16=4·4."
        }
      },
      {
        "claim": {
          "zh-Hant": "前三筆資料都滿足 y=(5/2)x，所以整張表是正比例",
          "en": "The first three data points satisfy y=(5/2)x, so the whole table is directly proportional"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "第四筆 (5/2)·8=20≠21。",
          "en": "At the fourth point, (5/2)·8=20≠21."
        }
      },
      {
        "claim": {
          "zh-Hant": "邊長 1、2 的 60° 平行四邊形與 2×4 長方形對應邊成比例，所以兩者相似",
          "en": "A parallelogram with sides 1 and 2 and a 60° angle has sides proportional to a 2×4 rectangle, so the two are similar"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "角不相等（60° 與 90°）；數值比例不推出幾何相似。",
          "en": "The angles differ (60° and 90°); a numerical proportion does not give geometric similarity."
        }
      }
    ],
    "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": "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": "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": "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_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": "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": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "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": "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": "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": "forall_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。",
          "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds."
        }
      },
      {
        "id": "contrapositive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逆否命題：P⇒Q 與 ¬Q⇒¬P 等價。逆命題 Q⇒P 一般不與 P⇒Q 等價。",
          "en": "Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q."
        }
      },
      {
        "id": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "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_ratio_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-ratio-proportion",
          "sha256": "b0200b9dd1e1ceb4c2b34f37f0279a1abaf31c3021957de6cd9ea72919da47ec",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對有理數 a、b、c、d 且 b≠0、d≠0，數值比 a:b 的值為 a/b；比例 a:b=c:d 表示 a/b=c/d，等價於 ad=bc。",
          "en": "For rational numbers a, b, c, and d with b≠0 and d≠0, the value of the numerical ratio a:b is a/b. The proportion a:b=c:d means a/b=c/d and is equivalent to ad=bc."
        }
      },
      {
        "id": "L_cross",
        "kind": "lemma",
        "proved_in": [
          "S02.1",
          "S02.2"
        ],
        "statement": {
          "zh-Hant": "對有理數 a、b、c、d，b≠0、d≠0：a/b=c/d ⇔ ad=bc。",
          "en": "For rationals a, b, c, d with b≠0 and d≠0: a/b=c/d ⇔ ad=bc."
        }
      },
      {
        "id": "L_solve",
        "kind": "lemma",
        "proved_in": [
          "S03.1"
        ],
        "statement": {
          "zh-Hant": "對有理數 u 與 b≠0：bx=u ⇔ x=u/b。",
          "en": "For a rational u and b≠0: bx=u ⇔ x=u/b."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S05",
        "lhs": "3*10/4",
        "rhs": "15/2"
      },
      {
        "id": "N2",
        "step": "S06",
        "lhs": "(15/2)/10",
        "rhs": "3/4"
      },
      {
        "id": "N3",
        "step": "S06",
        "lhs": "4*(15/2)",
        "rhs": "3*10"
      },
      {
        "id": "N4",
        "step": "S07",
        "lhs": "4*10/3",
        "rhs": "40/3"
      },
      {
        "id": "N5",
        "step": "S07",
        "lhs": "40*2",
        "relation": "!=",
        "rhs": "3*15"
      },
      {
        "id": "N6",
        "step": "S07",
        "lhs": "4*4",
        "relation": "!=",
        "rhs": "3*3"
      },
      {
        "id": "N7",
        "step": "S09",
        "lhs": "10*2",
        "rhs": "4*5"
      },
      {
        "id": "N8",
        "step": "S09",
        "lhs": "15*2",
        "rhs": "6*5"
      },
      {
        "id": "N9",
        "step": "S10",
        "lhs": "21*2",
        "relation": "!=",
        "rhs": "8*5"
      },
      {
        "id": "N10",
        "step": "S10",
        "lhs": "(5/2)*8",
        "relation": "!=",
        "rhs": "21"
      },
      {
        "id": "N11",
        "step": "S11",
        "vars": [
          "x"
        ],
        "degree": 1,
        "lhs": "0*5",
        "rhs": "0*x"
      },
      {
        "id": "N12",
        "lhs": "1*4",
        "rhs": "2*2"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內：有理數的數值比，後項 b、d 非零（假設 a1）。資料是不帶單位的純數；若描述量，須先指定單位與可比較性（假設 a2）。本稿不由數值比例推出幾何相似或因果關係；不改動 canonical MKO 或英文 edition。",
      "en": "Inside the MKO's scope: numerical ratios over the rationals with nonzero consequents b and d (assumption a1). The data are pure numbers without units; ratios of quantities would first need their units and comparability stated (assumption a2). The guide draws no geometric similarity or causation from numerical proportion, and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "a:b",
        "數值比，值為 a/b（b≠0）",
        "a numerical ratio, with value a/b (b≠0)"
      ],
      [
        "a:b=c:d",
        "比例，意即 a/b=c/d",
        "a proportion, meaning a/b=c/d"
      ],
      [
        "x",
        "比例中的未知項",
        "the unknown term of a proportion"
      ],
      [
        "k",
        "正比例 y=kx 的比例常數",
        "the constant of a direct proportion y=kx"
      ],
      [
        "(x_i,y_i)",
        "第 i 筆資料",
        "the i-th data point"
      ]
    ],
    "md_error_controls": [
      {
        "title": "交叉乘積不能讓零後項合法 / Cross products cannot legalize a zero consequent",
        "zh": "MKO 提醒：交叉乘積相等不能讓零分母的表示變得合法。0:0=x:5 的「交叉乘積」0·5=0·x 對每個 x 都成立，3:0=6:0 的 3·0=0·6 也成立，但 0:0、3:0、6:0 都不是比。L_cross 的兩個方向（S02.1、S02.2）都用到 b≠0、d≠0；先確認後項非零，再比較交叉乘積。",
        "en": "The MKO warns that equal cross products cannot make a representation with a zero denominator legal. For 0:0=x:5 the 'cross products' 0·5=0·x agree for every x, and for 3:0=6:0, 3·0=0·6 holds too, but 0:0, 3:0 and 6:0 are not ratios. Both directions of L_cross (S02.1, S02.2) use b≠0 and d≠0: first check that the consequents are nonzero, then compare cross products."
      },
      {
        "title": "相同比例不自動建立幾何相似 / Equal ratios do not by themselves give geometric similarity",
        "zh": "MKO 指出相同比例也不自動建立幾何相似。例如邊長 1、2、夾角 60° 的平行四邊形，與邊長 2、4 的長方形，對應邊成比例（1:2=2:4），但一個有 60° 角、一個全是 90° 角，兩者不相似。數值比例只是關於數的陳述；相似或因果需要另外的條件與證明（假設 a2）。",
        "en": "The MKO also notes that equal ratios do not automatically establish geometric similarity. For example, a parallelogram with sides 1 and 2 and an angle of 60°, and a rectangle with sides 2 and 4, have proportional corresponding sides (1:2=2:4), but one has a 60° angle and the other only 90° angles, so they are not similar. A numerical proportion is a statement about numbers; similarity or causation needs further conditions and proof (assumption a2)."
      },
      {
        "title": "順序有意義，有限資料要逐點檢查 / Order matters, and finite data must be checked point by point",
        "zh": "3:4 與 4:3 不同（交叉乘積 9≠16），把比例寫反會得到 40/3 而不是 15/2；反向比還需要原前項非零。判定正比例時，要對每一筆資料檢查同一個 k：前三筆都給 k=5/2，但第四筆 (8,21) 不符合，所以不能從一點或幾點推論全部輸入（MKO 第三段）。",
        "en": "3:4 and 4:3 differ (cross products 9≠16), and writing the proportion backwards gives 40/3 instead of 15/2; a reversed ratio also needs the original antecedent to be nonzero. When testing for direct proportion, the same k must be checked at every data point: the first three points all give k=5/2, but the fourth, (8,21), does not, so one or a few points cannot settle all inputs (third paragraph of the MKO)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "錯誤控制二的平行四邊形例子用到「相似圖形的對應角相等」這一幾何事實，本稿未證明它，只用來說明數值比例不足以推出相似。",
        "en": "The parallelogram example in error control 2 uses the geometric fact that similar figures have equal corresponding angles; this guide does not prove it and uses it only to show that numerical proportion is not enough for similarity."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "6da57afccee31bb4320b5e7da91728489dd9ba8a361fc46d45ebaea5514eb65e",
    "original_source_sha256": "6da57afccee31bb4320b5e7da91728489dd9ba8a361fc46d45ebaea5514eb65e",
    "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."
  }
}