{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-system-linear-equations-01",
  "mko_id": "mko-system-linear-equations",
  "title": {
    "zh-Hant": "二元聯立一次方程組：唯一解公式、兩列核對與 D=0 的一致性判定",
    "en": "Linear systems in two variables: the unique-solution formula, checking both rows, and consistency when D=0"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-system-linear-equations-01",
    "source": {
      "mko_id": "mko-system-linear-equations",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "62dc0d59fbb66d23a8ef47c7b2725a56713d925c7f2c8ef0a380c46552972403",
      "english_sha256": "7fc6d87498d9c2a7943b1715c26c7d3e2baf79d0f54d4beb9019b628da0010c4"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "eq_congruence",
      "distributive",
      "mul_inverse_nonzero",
      "mul_zero",
      "numeric_evaluation",
      "solution_def"
    ],
    "goal": {
      "zh-Hant": "(I) 解 2x+3y=7、x−y=1：證明 D=ae−bd≠0 時的唯一解公式，求解並代入兩列核對；(II) 判定 x+2y=3、2x+4y=7；(III) 判定 x+2y=3、2x+4y=6；並以 (IV) 0x+0y=1、0x+0y=2 說明 D 與兩個 Cramer 分子都為零時不能斷定有無限多解。",
      "en": "(I) Solve 2x+3y=7, x−y=1: prove the unique-solution formula when D=ae−bd≠0, solve, and check by substituting into both rows; (II) decide x+2y=3, 2x+4y=7; (III) decide x+2y=3, 2x+4y=6; and use (IV) 0x+0y=1, 0x+0y=2 to show that D and both Cramer numerators being zero does not imply infinitely many solutions."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "未知數 x、y 為實數；一般系統為 ax+by=c、dx+ey=f，係數與右側常數為實數；L(x,y)=(ax+by−c, dx+ey−f)，z=(0,0)，D=ae−bd。",
          "en": "The unknowns x, y are real; the general system is ax+by=c, dx+ey=f with real coefficients and right sides; L(x,y)=(ax+by−c, dx+ey−f), z=(0,0) and D=ae−bd."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "系統 I：2x+3y=7、x−y=1，即 a=2、b=3、c=7、d=1、e=−1、f=1。",
          "en": "System I: 2x+3y=7, x−y=1, that is a=2, b=3, c=7, d=1, e=−1, f=1."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "系統 II：x+2y=3、2x+4y=7；系統 III：x+2y=3、2x+4y=6；系統 IV：0x+0y=1、0x+0y=2（零係數列容許作為限制，MKO 假設 a2）。",
          "en": "System II: x+2y=3, 2x+4y=7; System III: x+2y=3, 2x+4y=6; System IV: 0x+0y=1, 0x+0y=2 (zero-coefficient rows are allowed as constraints, MKO assumption a2)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "solution_def",
          "mko_linear_system"
        ],
        "before_tex": "L(x,y)=z",
        "after_tex": "ax+by=c\\ \\land\\ dx+ey=f",
        "relation": "equivalent",
        "rule": "solution_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，解集由 L(x,y)=z 刻畫。L(x,y)=(ax+by−c, dx+ey−f) 與 z=(0,0) 相等，當且僅當兩個分量都是 0，即 ax+by−c=0 且 dx+ey−f=0，兩邊同加常數就是兩個方程同時成立。所以「解」永遠指同時滿足兩列的 (x,y)。",
          "en": "By the MKO, the solution set is described by L(x,y)=z. L(x,y)=(ax+by−c, dx+ey−f) equals z=(0,0) exactly when both components are 0, i.e. ax+by−c=0 and dx+ey−f=0, which after adding the constants to both sides says both equations hold at once. So 'solution' always means an (x,y) satisfying both rows."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "cartesian_product_def",
            "eq_congruence"
          ],
          "explanation": {
            "zh-Hant": "有序對相等是雙向的；兩邊同加或同減常數也是雙向的。",
            "en": "Equality of ordered pairs is two-way, and so is adding or subtracting a constant on both sides."
          }
        },
        "leaf_basis": "solution_def + cartesian_product_def + eq_congruence"
      },
      {
        "id": "S02",
        "uses": [
          "G2"
        ],
        "before_tex": "D=ae-bd",
        "after_tex": "D=2\\cdot(-1)-3\\cdot 1=-5\\neq 0",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代入 a=2、b=3、d=1、e=−1：D=2·(−1)−3·1=−2−3=−5，不為零。",
          "en": "Substitute a=2, b=3, d=1, e=−1: D=2·(−1)−3·1=−2−3=−5, which is nonzero."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確整數計算，可重算（N1）。",
            "en": "An exact integer computation, recomputable (N1)."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "G2",
          "L_cramer"
        ],
        "before_tex": "D=-5\\neq 0",
        "after_tex": "x=\\frac{7\\cdot(-1)-3\\cdot 1}{-5}=2,\\quad y=\\frac{2\\cdot 1-7\\cdot 1}{-5}=1",
        "relation": "theorem_application",
        "rule": "L_cramer",
        "conditions": [
          "D\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "因 D≠0，引理 L_cramer 適用：系統 I 恰有一解，x=(ce−bf)/D=(−7−3)/(−5)=2，y=(af−cd)/D=(2−7)/(−5)=1。所以唯一解是 (2,1)。引理在子步 S03.1 到 S03.4 證明。",
          "en": "Since D≠0, Lemma L_cramer applies: System I has exactly one solution, x=(ce−bf)/D=(−7−3)/(−5)=2 and y=(af−cd)/D=(2−7)/(−5)=1. So the unique solution is (2,1). The lemma is proved in sub-steps S03.1 to S03.4."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "L_cramer"
          ],
          "explanation": {
            "zh-Hant": "唯一的前提是 D≠0（S02）；數值見 N2、N3。D=0 時這一步不可用。",
            "en": "The only hypothesis is D≠0 (S02); the values are N2 and N3. The step is unavailable when D=0."
          }
        },
        "leaf_basis": "eq_congruence + distributive + mul_inverse_nonzero + frac_rules"
      },
      {
        "id": "S03.1",
        "uses": [
          "eq_congruence",
          "distributive"
        ],
        "before_tex": "ax+by=c\\ \\land\\ dx+ey=f",
        "after_tex": "(ae-bd)x=ce-bf",
        "relation": "implies",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "設 (x,y) 是解。第一列兩邊乘 e、第二列兩邊乘 b，再相減：e(ax+by)−b(dx+ey)=ec−bf。左邊展開為 aex+bey−bdx−bey=(ae−bd)x（N4 核對這個恆等式）。所以任何解都滿足 (ae−bd)x=ce−bf。",
          "en": "Let (x,y) be a solution. Multiply row one by e and row two by b, then subtract: e(ax+by)−b(dx+ey)=ec−bf. The left side expands to aex+bey−bdx−bey=(ae−bd)x (N4 checks this identity). So every solution satisfies (ae−bd)x=ce−bf."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "eq_congruence",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "這是必要條件：只推出 x 必須滿足的方程，反向不宣稱（D=0 時這一行可能是 0=0，什麼也不限制）。",
            "en": "This is a necessary condition: it only derives an equation x must satisfy, and no converse is claimed (when D=0 this line may read 0=0 and restrict nothing)."
          }
        },
        "leaf_basis": "eq_congruence + distributive + mul_comm + add_inverse"
      },
      {
        "id": "S03.2",
        "uses": [
          "eq_congruence",
          "distributive"
        ],
        "before_tex": "ax+by=c\\ \\land\\ dx+ey=f",
        "after_tex": "(ae-bd)y=af-cd",
        "relation": "implies",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "同理消去 x：第二列乘 a、第一列乘 d，再相減：a(dx+ey)−d(ax+by)=af−dc。左邊展開為 adx+aey−adx−bdy=(ae−bd)y（N5）。所以任何解都滿足 (ae−bd)y=af−cd。",
          "en": "Likewise eliminate x: multiply row two by a and row one by d, then subtract: a(dx+ey)−d(ax+by)=af−dc. The left side expands to adx+aey−adx−bdy=(ae−bd)y (N5). So every solution satisfies (ae−bd)y=af−cd."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "eq_congruence",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "同 S03.1，只是必要條件，不宣稱反向。",
            "en": "As in S03.1, this is only a necessary condition; no converse is claimed."
          }
        },
        "leaf_basis": "eq_congruence + distributive + mul_comm + add_inverse"
      },
      {
        "id": "S03.3",
        "uses": [
          "S03.1",
          "S03.2",
          "mul_inverse_nonzero"
        ],
        "before_tex": "Dx=ce-bf,\\quad Dy=af-cd,\\quad D\\neq 0",
        "after_tex": "x=\\frac{ce-bf}{D},\\quad y=\\frac{af-cd}{D}",
        "relation": "implies",
        "rule": "mul_inverse_nonzero",
        "conditions": [
          "D\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "因 D≠0，1/D 存在，兩式同乘 1/D 得 x 與 y 的值。所以任何解都等於這一對數：解至多一個（唯一性）。若 D=0，1/D 不存在，這一步不能做，這正是 MKO 說「不得在 D=0 時套唯一解除法」的原因。",
          "en": "Since D≠0, 1/D exists, and multiplying both equations by 1/D gives the values of x and y. So every solution equals this pair: there is at most one solution (uniqueness). If D=0 there is no 1/D and the step cannot be taken, which is exactly why the MKO forbids the unique-solution division when D=0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03.1",
            "S03.2",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "此步只給唯一性；這一對數確實是解（存在性）另在 S03.4 證明。",
            "en": "This step gives uniqueness only; that the pair really is a solution (existence) is proved separately in S03.4."
          }
        },
        "leaf_basis": "mul_inverse_nonzero + eq_congruence + mul_assoc + mul_one"
      },
      {
        "id": "S03.4",
        "uses": [
          "frac_rules",
          "distributive"
        ],
        "before_tex": "x_0=\\frac{ce-bf}{D},\\ y_0=\\frac{af-cd}{D}",
        "after_tex": "ax_0+by_0=c\\ \\land\\ dx_0+ey_0=f",
        "relation": "implies",
        "rule": "frac_rules",
        "conditions": [
          "D\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "存在性：代回第一列，ax₀+by₀=(a(ce−bf)+b(af−cd))/D=(ace−abf+abf−bcd)/D=c(ae−bd)/D=c；代回第二列，dx₀+ey₀=(d(ce−bf)+e(af−cd))/D=(cde−bdf+aef−cde)/D=f(ae−bd)/D=f（N6、N7 核對兩個恆等式）。S03.1 到 S03.4 合起來證得引理 L_cramer：D≠0 時恰有一解。",
          "en": "Existence: substituting into row one, ax₀+by₀=(a(ce−bf)+b(af−cd))/D=(ace−abf+abf−bcd)/D=c(ae−bd)/D=c; into row two, dx₀+ey₀=(d(ce−bf)+e(af−cd))/D=(cde−bdf+aef−cde)/D=f(ae−bd)/D=f (N6 and N7 check the two identities). S03.1 to S03.4 together prove Lemma L_cramer: when D≠0 there is exactly one solution."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "frac_rules",
            "distributive",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "同分母相加與約去 D 都需要 D≠0。唯一性（S03.3）加存在性（此步）才是「恰有一解」。",
            "en": "Adding over the common denominator and cancelling D both need D≠0. Uniqueness (S03.3) together with existence (this step) gives 'exactly one solution'."
          }
        },
        "leaf_basis": "frac_rules + distributive + mul_comm + add_inverse + mul_inverse_nonzero"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "G2",
          "solution_def"
        ],
        "before_tex": "(x,y)=(2,1)",
        "after_tex": "2\\cdot 2+3\\cdot 1=7\\ \\land\\ 2-1=1",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "獨立核對：把 (2,1) 代入兩列，第一列 4+3=7，第二列 2−1=1，兩列都成立。這一步不依賴公式，可以抓出代入公式時的計算錯誤。",
          "en": "Independent check: substitute (2,1) into both rows; row one gives 4+3=7 and row two gives 2−1=1, so both rows hold. This step does not depend on the formula and can catch arithmetic slips made in applying it."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "solution_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "必須代入兩列（N8、N9）；唯一性另由 S03.3 保證，不由這次代入推出。",
            "en": "Both rows must be checked (N8, N9); uniqueness comes from S03.3, not from this substitution."
          }
        },
        "leaf_basis": "numeric_evaluation + solution_def"
      },
      {
        "id": "S05",
        "uses": [
          "G2",
          "solution_def"
        ],
        "before_tex": "(x,y)=(5,-1)",
        "after_tex": "2\\cdot 5+3\\cdot(-1)=7,\\ \\ 5-(-1)=6\\neq 1",
        "relation": "implies",
        "rule": "solution_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "只代入一列的陷阱：(5,−1) 使第一列成立（10−3=7），但第二列給 5−(−1)=6≠1。依解的定義必須兩列同時成立，所以 (5,−1) 不是系統 I 的解。這也與 S03 的唯一解 (2,1) 一致。",
          "en": "The one-row trap: (5,−1) satisfies row one (10−3=7), but row two gives 5−(−1)=6≠1. By the definition of a solution both rows must hold at once, so (5,−1) is not a solution of System I. This agrees with the unique solution (2,1) of S03."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "solution_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一列失敗就足以排除（N10、N11）。",
            "en": "Failure of one row is enough to exclude it (N10, N11)."
          }
        },
        "leaf_basis": "solution_def + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "G3",
          "mko_linear_system"
        ],
        "before_tex": "x+2y=3,\\ 2x+4y=7",
        "after_tex": "D=1\\cdot 4-2\\cdot 2=0",
        "relation": "implies",
        "rule": "mko_linear_system",
        "conditions": [],
        "explanation": {
          "zh-Hant": "系統 II 的 D=1·4−2·2=0，所以 L_cramer 不適用，公式中的除法沒有定義。依 MKO，D=0 時必須另外檢查一致性。（它的 Cramer 分子 ce−bf=3·4−2·7=−2≠0，但本稿不以分子作判準，而用 S07 的列運算直接判定。）",
          "en": "System II has D=1·4−2·2=0, so L_cramer does not apply and the division in the formula is undefined. By the MKO, consistency must be checked separately when D=0. (Its Cramer numerator ce−bf=3·4−2·7=−2≠0, but this guide does not use numerators as the criterion; S07 decides directly by a row operation.)"
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_linear_system",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "D 的計算可重算（N12、N22）；此步只排除公式，不判定解集。",
            "en": "D can be recomputed (N12, N22); this step only rules out the formula and does not yet decide the solution set."
          }
        },
        "leaf_basis": "mko_linear_system + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "L_row_replace"
        ],
        "before_tex": "x+2y=3\\ \\land\\ 2x+4y=7",
        "after_tex": "x+2y=3\\ \\land\\ 0x+0y=1",
        "relation": "equivalent",
        "rule": "L_row_replace",
        "conditions": [
          "\\lambda=2"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_row_replace，取 λ=2，把第二列換成第二列減第一列的 2 倍：(2−2·1)x+(4−2·2)y=7−2·3，即 0x+0y=1。新系統與原系統解集相同。",
          "en": "By Lemma L_row_replace with λ=2, replace row two by row two minus 2 times row one: (2−2·1)x+(4−2·2)y=7−2·3, that is 0x+0y=1. The new system has the same solution set as the original."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_row_replace"
          ],
          "explanation": {
            "zh-Hant": "引理兩向在 S07.1 證明：減去 2 倍第一列，與加回 2 倍第一列，互為反運算。",
            "en": "Both directions of the lemma are proved in S07.1: subtracting 2 times row one and adding it back undo each other."
          }
        },
        "leaf_basis": "eq_congruence + distributive + numeric_evaluation"
      },
      {
        "id": "S07.1",
        "uses": [
          "eq_congruence",
          "distributive"
        ],
        "before_tex": "ax+by=c\\ \\land\\ dx+ey=f",
        "after_tex": "ax+by=c\\ \\land\\ (d-\\lambda a)x+(e-\\lambda b)y=f-\\lambda c",
        "relation": "equivalent",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "⇒：設兩列成立。第一列兩邊乘 λ 得 λ(ax+by)=λc；從第二列兩邊減去它：(dx+ey)−λ(ax+by)=f−λc，依分配律整理為 (d−λa)x+(e−λb)y=f−λc；第一列保留不變。⇐：設新系統成立，把第一列的 λ 倍加回新第二列，得 dx+ey=f。兩向都只用「對等式兩邊做同一運算」。這證得引理 L_row_replace。",
          "en": "⇒: assume both rows. Multiplying row one by λ gives λ(ax+by)=λc; subtracting this from both sides of row two gives (dx+ey)−λ(ax+by)=f−λc, which distributivity rearranges to (d−λa)x+(e−λb)y=f−λc; row one is kept. ⇐: assume the new system and add λ times row one back to the new row two, giving dx+ey=f. Both directions only apply the same operation to both sides of an equation. This proves Lemma L_row_replace."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "兩向都有理由；關鍵是第一列被保留，所以可以把它的 λ 倍加回去。",
            "en": "Both directions have reasons; the key is that row one is kept, so its λ multiple can be added back."
          }
        },
        "leaf_basis": "eq_congruence + distributive + add_inverse + add_assoc"
      },
      {
        "id": "S08",
        "uses": [
          "S07",
          "mul_zero",
          "add_zero"
        ],
        "before_tex": "0x+0y=1",
        "after_tex": "0=1\\ \\text{for every}\\ (x,y),\\ \\text{solution set of II}=\\varnothing",
        "relation": "implies",
        "rule": "mul_zero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對每個 (x,y)，0x+0y=0+0=0（乘以零、加法單位元），所以 0x+0y=1 等於要求 0=1，沒有任何 (x,y) 滿足。依 S07 的等價，系統 II 無解。MKO 假設 a2：0=c≠0 的零列不一致。",
          "en": "For every (x,y), 0x+0y=0+0=0 (multiplying by zero, additive identity), so 0x+0y=1 demands 0=1, which no (x,y) satisfies. By the equivalence in S07, System II has no solution. MKO assumption a2: a zero row 0=c with c≠0 is inconsistent."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mul_zero",
            "add_zero",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "需要 S07 的解集等價；只因 D=0 不能推出無解，必須看到不一致的零列。",
            "en": "It needs the equivalence of solution sets from S07; D=0 alone does not imply no solution, an inconsistent zero row must actually appear."
          }
        },
        "leaf_basis": "mul_zero + add_zero + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "G3",
          "L_row_replace",
          "mul_zero",
          "add_zero"
        ],
        "before_tex": "x+2y=3\\ \\land\\ 2x+4y=6",
        "after_tex": "x+2y=3\\ \\land\\ 0x+0y=0\\ \\Leftrightarrow\\ x+2y=3",
        "relation": "equivalent",
        "rule": "L_row_replace",
        "conditions": [
          "\\lambda=2"
        ],
        "explanation": {
          "zh-Hant": "系統 III 的 D 也是 1·4−2·2=0。取 λ=2 做同樣的列運算：第二列變成 0x+0y=6−2·3=0。對每個 (x,y)，0x+0y=0 恆成立，不加任何限制（MKO 假設 a2），所以系統 III 等價於單一方程 x+2y=3。",
          "en": "System III also has D=1·4−2·2=0. The same row operation with λ=2 turns row two into 0x+0y=6−2·3=0. For every (x,y), 0x+0y=0 always holds and imposes no restriction (MKO assumption a2), so System III is equivalent to the single equation x+2y=3."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_row_replace",
            "mul_zero",
            "add_zero",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "列運算雙向保持解集；與恆真條件做合取不改變真假，雙向成立。",
            "en": "The row operation preserves the solution set both ways, and conjoining an always-true condition does not change truth, so it holds both ways."
          }
        },
        "leaf_basis": "eq_congruence + distributive + mul_zero + add_zero + and_intro_elim"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "set_extensionality"
        ],
        "before_tex": "\\{(x,y)\\in\\mathbb{R}^2\\mid x+2y=3\\}",
        "after_tex": "\\{(3-2t,\\,t)\\mid t\\in\\mathbb{R}\\}",
        "relation": "equivalent",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "⊆：若 x+2y=3，取 t=y，則 x=3−2t（兩邊同減 2y），所以 (x,y)=(3−2t,t)。⊇：對任意實數 t，(3−2t)+2t=3（N17），所以 (3−2t,t) 是解。兩個方向都成立，依集合相等兩集合相同。",
          "en": "⊆: if x+2y=3, take t=y; then x=3−2t (subtract 2y from both sides), so (x,y)=(3−2t,t). ⊇: for every real t, (3−2t)+2t=3 (N17), so (3−2t,t) is a solution. Both directions hold, so the two sets are equal."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_extensionality",
            "eq_congruence"
          ],
          "explanation": {
            "zh-Hant": "兩個包含都已證明，所以是集合相等。",
            "en": "Both inclusions are proved, so the sets are equal."
          }
        },
        "leaf_basis": "set_extensionality + eq_congruence + add_inverse + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S10",
          "cartesian_product_def"
        ],
        "before_tex": "\\{(3-2t,\\,t)\\mid t\\in\\mathbb{R}\\}",
        "after_tex": "\\text{infinitely many solutions, e.g.}\\ (3,0),\\ (1,1)",
        "relation": "implies",
        "rule": "cartesian_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "不同的 t 給出不同的解，因為第二分量就是 t（有序對相等需要第二分量相等）。ℝ 有無限多個元素，所以系統 III 有無限多解。例：t=0 得 (3,0)，t=1 得 (1,1)；兩者都滿足原系統的兩列：3+0=3、6+0=6；1+2=3、2+4=6（N18 對一般 t 核對第二列）。",
          "en": "Different values of t give different solutions, because the second component is t itself (equal ordered pairs need equal second components). ℝ has infinitely many elements, so System III has infinitely many solutions. For example t=0 gives (3,0) and t=1 gives (1,1); both satisfy both original rows: 3+0=3, 6+0=6; 1+2=3, 2+4=6 (N18 checks row two for general t)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "需要參數化是單射（不同 t 不同解）；例子只是示範，結論來自參數化。",
            "en": "It needs the parametrization to be injective (distinct t, distinct solutions); the examples only illustrate, and the conclusion comes from the parametrization."
          }
        },
        "leaf_basis": "cartesian_product_def + injective_surjective_def + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "G3",
          "mul_zero",
          "add_zero"
        ],
        "before_tex": "0x+0y=1,\\ 0x+0y=2",
        "after_tex": "D=0,\\ ce-bf=0,\\ af-cd=0,\\ \\text{yet no solution}",
        "relation": "implies",
        "rule": "mul_zero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "系統 IV：a=b=d=e=0、c=1、f=2。D=0·0−0·0=0，ce−bf=1·0−0·2=0，af−cd=0·2−1·0=0，三者都是零。但第一列對每個 (x,y) 都讀作 0=1，為假，所以系統無解。可見「D 與兩個 Cramer 分子都為零」不能推出無限多解，必須先檢查不一致的零列（MKO 第三段）。",
          "en": "System IV: a=b=d=e=0, c=1, f=2. D=0·0−0·0=0, ce−bf=1·0−0·2=0, af−cd=0·2−1·0=0, all three zero. Yet row one reads 0=1 for every (x,y), which is false, so the system has no solution. Hence 'D and both Cramer numerators are zero' does not imply infinitely many solutions; inconsistent zero rows must be checked first (MKO paragraph three)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mul_zero",
            "add_zero",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一個具體反例就足以否定這個一般說法（N19 到 N21）；此步不給出 D=0 時的一般判準。",
            "en": "One concrete counterexample suffices to refute the general claim (N19 to N21); this step gives no general criterion for D=0."
          }
        },
        "leaf_basis": "mul_zero + add_zero + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S03",
          "S04",
          "S08",
          "S10",
          "S11",
          "S12"
        ],
        "before_tex": "\\text{S03, S04, S08, S10, S11, S12}",
        "after_tex": "\\text{I}:\\ \\{(2,1)\\};\\ \\text{II}:\\ \\varnothing;\\ \\text{III}:\\ \\{(3-2t,t)\\mid t\\in\\mathbb{R}\\};\\ \\text{IV}:\\ \\varnothing",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整：系統 I 的 D=−5≠0，唯一解 (2,1)，已代入兩列核對；系統 II 的 D=0，列運算出現 0=1，無解；系統 III 的 D=0，列運算得 0=0，解集為 {(3−2t,t) ∣ t∈ℝ}，無限多解；系統 IV 三個行列式值都是零，但無解。",
          "en": "Summary: System I has D=−5≠0 and the unique solution (2,1), checked in both rows; System II has D=0 and a row operation produces 0=1, so no solution; System III has D=0 and the row operation gives 0=0, with solution set {(3−2t,t) ∣ t∈ℝ}, infinitely many; System IV has all three determinant values zero yet no solution."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S03",
            "S04",
            "S08",
            "S10",
            "S11",
            "S12",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：四個系統的判定。足夠條件 H：所列各步的結論；H⇒G 只是合取，H 由 G1 到 G3 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: the verdicts on the four systems. Sufficient condition H: the conclusions of the listed steps; H⇒G is just conjunction, H is proved forwards from G1 to G3, and G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "D≠0 時系統恰有一解 x=(ce−bf)/D、y=(af−cd)/D（已完整證明）。系統 I（D=−5）的唯一解是 (2,1)，兩列都已核對；(5,−1) 只滿足一列，不是解。系統 II（D=0）化為 0=1，無解；系統 III（D=0）化為 x+2y=3，解集 {(3−2t,t) ∣ t∈ℝ}，無限多解；系統 IV 的 D 與兩個 Cramer 分子都為零，卻無解。",
      "en": "When D≠0 the system has exactly one solution x=(ce−bf)/D, y=(af−cd)/D (fully proved). System I (D=−5) has the unique solution (2,1), checked in both rows; (5,−1) satisfies only one row and is not a solution. System II (D=0) reduces to 0=1 and has no solution; System III (D=0) reduces to x+2y=3, with solution set {(3−2t,t) ∣ t∈ℝ}, infinitely many; System IV has D and both Cramer numerators zero, yet no solution."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "(5,−1) 滿足 2x+3y=7，所以是系統 I 的解",
          "en": "(5,−1) satisfies 2x+3y=7, so it is a solution of System I"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "第二列 5−(−1)=6≠1；解必須同時滿足兩列。",
          "en": "Row two gives 5−(−1)=6≠1; a solution must satisfy both rows."
        }
      },
      {
        "claim": {
          "zh-Hant": "系統 II 的解是 x=(3·4−2·7)/0",
          "en": "The solution of System II is x=(3·4−2·7)/0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "除以 0 沒有定義；系統 II 經列運算出現 0=1，無解。",
          "en": "Division by 0 is undefined; a row operation turns System II into 0=1, so it has no solution."
        }
      },
      {
        "claim": {
          "zh-Hant": "系統 IV 的 D 與兩個 Cramer 分子都為零，所以有無限多解",
          "en": "System IV has D and both Cramer numerators zero, so it has infinitely many solutions"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "第一列對每個 (x,y) 都是 0=1，為假，所以無解。",
          "en": "Row one reads 0=1 for every (x,y), which is false, so there is no solution."
        }
      },
      {
        "claim": {
          "zh-Hant": "系統 III 的解只有 (3,0)",
          "en": "The only solution of System III is (3,0)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(1,1) 也滿足兩列；解集是 {(3−2t,t) ∣ t∈ℝ}，有無限多個。",
          "en": "(1,1) also satisfies both rows; the solution set is {(3−2t,t) ∣ t∈ℝ}, which is infinite."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "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": "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": "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": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "mul_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": "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": "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": "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_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "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": "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": "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": "mko_linear_system",
        "kind": "cited",
        "source": {
          "mko_id": "mko-system-linear-equations",
          "sha256": "62dc0d59fbb66d23a8ef47c7b2725a56713d925c7f2c8ef0a380c46552972403",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "令 x、y 為實數未知數，要求 a x+b y=c 與 d x+e y=f 同時成立。令 L(x,y)=(a x+b y−c,d x+e y−f)，z=(0,0)，則解集滿足 L(x,y)=z。D=ae−bd≠0 時唯一解為 x=(ce−bf)/D、y=(af−cd)/D；D=0 時須檢查一致性，可能無解或無限多解。",
          "en": "For real unknowns x and y, require a x+b y=c and d x+e y=f simultaneously. Define L(x,y)=(a x+b y−c,d x+e y−f) and z=(0,0), so the solution set satisfies L(x,y)=z. If D=ae−bd≠0, the unique solution is x=(ce−bf)/D and y=(af−cd)/D. If D=0, check consistency: there may be no solution or infinitely many solutions."
        }
      },
      {
        "id": "solution_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "(x,y)∈ℝ² 是系統的解，當且僅當 L(x,y)=z，即兩個方程同時成立（有序對相等）。",
          "en": "(x,y)∈ℝ² is a solution of the system exactly when L(x,y)=z, that is, when both equations hold at once (equality of ordered pairs)."
        }
      },
      {
        "id": "L_cramer",
        "kind": "lemma",
        "proved_in": [
          "S03.1",
          "S03.2",
          "S03.3",
          "S03.4"
        ],
        "statement": {
          "zh-Hant": "若 D=ae−bd≠0，則系統 ax+by=c、dx+ey=f 恰有一解，即 x=(ce−bf)/D、y=(af−cd)/D。",
          "en": "If D=ae−bd≠0, the system ax+by=c, dx+ey=f has exactly one solution, namely x=(ce−bf)/D, y=(af−cd)/D."
        }
      },
      {
        "id": "L_row_replace",
        "kind": "lemma",
        "proved_in": [
          "S07.1"
        ],
        "statement": {
          "zh-Hant": "對任意實數 λ，系統 {ax+by=c, dx+ey=f} 與 {ax+by=c, (d−λa)x+(e−λb)y=f−λc} 有相同的解集。",
          "en": "For every real λ, the systems {ax+by=c, dx+ey=f} and {ax+by=c, (d−λa)x+(e−λb)y=f−λc} have the same solution set."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "2*(-1)-3*1",
        "rhs": "-5"
      },
      {
        "id": "N2",
        "step": "S03",
        "lhs": "(7*(-1)-3*1)/(-5)",
        "rhs": "2"
      },
      {
        "id": "N3",
        "step": "S03",
        "lhs": "(2*1-7*1)/(-5)",
        "rhs": "1"
      },
      {
        "id": "N4",
        "step": "S03.1",
        "vars": [
          "a",
          "b",
          "d",
          "e",
          "x",
          "y"
        ],
        "degree": 2,
        "lhs": "e*(a*x+b*y)-b*(d*x+e*y)",
        "rhs": "(a*e-b*d)*x"
      },
      {
        "id": "N5",
        "step": "S03.2",
        "vars": [
          "a",
          "b",
          "d",
          "e",
          "x",
          "y"
        ],
        "degree": 2,
        "lhs": "a*(d*x+e*y)-d*(a*x+b*y)",
        "rhs": "(a*e-b*d)*y"
      },
      {
        "id": "N6",
        "step": "S03.4",
        "vars": [
          "a",
          "b",
          "c",
          "d",
          "e",
          "f"
        ],
        "degree": 2,
        "lhs": "a*(c*e-b*f)+b*(a*f-c*d)",
        "rhs": "c*(a*e-b*d)"
      },
      {
        "id": "N7",
        "step": "S03.4",
        "vars": [
          "a",
          "b",
          "c",
          "d",
          "e",
          "f"
        ],
        "degree": 2,
        "lhs": "d*(c*e-b*f)+e*(a*f-c*d)",
        "rhs": "f*(a*e-b*d)"
      },
      {
        "id": "N8",
        "step": "S04",
        "lhs": "2*2+3*1",
        "rhs": "7"
      },
      {
        "id": "N9",
        "step": "S04",
        "lhs": "2-1",
        "rhs": "1"
      },
      {
        "id": "N10",
        "step": "S05",
        "lhs": "2*5+3*(-1)",
        "rhs": "7"
      },
      {
        "id": "N11",
        "step": "S05",
        "lhs": "5-(-1)",
        "relation": "!=",
        "rhs": "1"
      },
      {
        "id": "N12",
        "step": "S06",
        "lhs": "1*4-2*2",
        "rhs": "0"
      },
      {
        "id": "N13",
        "step": "S07",
        "lhs": "7-2*3",
        "rhs": "1"
      },
      {
        "id": "N14",
        "step": "S07",
        "lhs": "2-2*1",
        "rhs": "0"
      },
      {
        "id": "N15",
        "step": "S07",
        "lhs": "4-2*2",
        "rhs": "0"
      },
      {
        "id": "N16",
        "step": "S09",
        "lhs": "6-2*3",
        "rhs": "0"
      },
      {
        "id": "N17",
        "step": "S10",
        "vars": [
          "t"
        ],
        "degree": 1,
        "lhs": "(3-2*t)+2*t",
        "rhs": "3"
      },
      {
        "id": "N18",
        "step": "S11",
        "vars": [
          "t"
        ],
        "degree": 1,
        "lhs": "2*(3-2*t)+4*t",
        "rhs": "6"
      },
      {
        "id": "N19",
        "step": "S12",
        "lhs": "0*0-0*0",
        "rhs": "0"
      },
      {
        "id": "N20",
        "step": "S12",
        "lhs": "1*0-0*2",
        "rhs": "0"
      },
      {
        "id": "N21",
        "step": "S12",
        "lhs": "0*2-1*0",
        "rhs": "0"
      },
      {
        "id": "N22",
        "step": "S06",
        "lhs": "3*4-2*7",
        "rhs": "-2"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：實數未知數 x、y 的兩個方程；四組具體例子的係數為整數，一般唯一解公式則對 G1 的實數係數證明。完整證明 D≠0 時恰有一解及其公式；D=0 時以列運算（保持解集的等價變形）逐例判定無解或無限多解，並檢查零列。不給出 D=0 時一般的秩判準證明；不從示意直線交點猜解；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: two equations with real unknowns x, y; the four worked examples have integer coefficients, while the general unique-solution formula is proved for the real coefficients of G1. Proves completely that D≠0 gives exactly one solution with the stated formula; for D=0 decides each example by row operations (equivalences that preserve the solution set) as having no solution or infinitely many, checking zero rows. Does not prove a general rank criterion for D=0; does not guess solutions from sketched line intersections; does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "ax+by=c,\\ dx+ey=f",
        "兩列方程（係數 a 到 f 為實數）",
        "the two rows (real coefficients a to f)"
      ],
      [
        "L(x,y)",
        "(ax+by−c, dx+ey−f)",
        "(ax+by−c, dx+ey−f)"
      ],
      [
        "z",
        "零向量 (0,0)",
        "the zero vector (0,0)"
      ],
      [
        "D",
        "ae−bd，二階行列式",
        "ae−bd, the 2×2 determinant"
      ],
      [
        "r_1,\\ r_2",
        "第一列、第二列方程",
        "the first and second row equations"
      ],
      [
        "r_2-\\lambda r_1",
        "第二列減去第一列的 λ 倍",
        "row two minus λ times row one"
      ]
    ],
    "md_error_controls": [
      {
        "title": "只代入一列不夠 / Checking one row is not enough",
        "zh": "MKO 指出的誤解之一是只代入一列。(5,−1) 滿足 2x+3y=7，卻使 x−y=6≠1（S05）。驗證候選解必須代入兩列，且若候選值含分母，要先確認分母非零。",
        "en": "One misconception named by the MKO is substituting into one row only. (5,−1) satisfies 2x+3y=7 yet gives x−y=6≠1 (S05). A candidate must be checked in both rows, and any denominators in it must first be confirmed nonzero."
      },
      {
        "title": "D=0 時不能除 / No division when D=0",
        "zh": "MKO 的第二個誤解是在 D=0 時套用除法。系統 II 與 III 的 D 都是 0，結果卻一個無解、一個有無限多解（S08、S11）；公式除以 0 沒有定義，也分辨不出這兩種情況。必須用保持解集的列運算或一致性檢查來判定。",
        "en": "The second misconception identified in the MKO is dividing when D=0. Systems II and III both have D=0, yet one has no solution and the other infinitely many (S08, S11); dividing by 0 is undefined and could not tell these cases apart. They must be decided by solution-preserving row operations or a consistency check."
      },
      {
        "title": "不要忽略 0=非零 的零列 / Do not ignore a zero row 0=nonzero",
        "zh": "MKO 的第三個誤解是忽略不一致零列。0x+0y=c 在 c≠0 時對任何 (x,y) 都不成立，整個系統因此無解（S08、S12）。系統 IV 的 D 與兩個 Cramer 分子都為零，若只看這三個數會誤判為無限多解。",
        "en": "The third misconception identified in the MKO is ignoring an inconsistent zero row. 0x+0y=c with c≠0 fails for every (x,y), so the whole system has no solution (S08, S12). System IV has D and both Cramer numerators zero; looking only at these three numbers would wrongly suggest infinitely many solutions."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "D=0 時一般的秩與增廣矩陣一致性判準未在本稿證明；系統 II、III、IV 是逐例以列運算或零列判定。",
        "en": "The general rank and augmented-matrix consistency criterion for D=0 is not proved here; Systems II, III and IV are each decided by row operations or zero rows."
      },
      {
        "zh-Hant": "N4 到 N7 是多項式恆等式的格點核對（每個變數次數至多 2），對應 S03.1 到 S03.4 的代數展開；論證不依賴它們。",
        "en": "N4 to N7 are grid checks of polynomial identities (degree at most 2 in each variable) matching the expansions in S03.1 to S03.4; the argument does not depend on them."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "0f098bbbe0d8574f864b3569777f182c32f28dd1c5a192d61ca7d28c7748ec19",
    "original_source_sha256": "2f1efd35aa77f21aa1e0568cb94972c9eea89944a758fa7c0c901c5d58cc423a",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M011"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}