{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-factorization-01",
  "mko_id": "mko-factorization",
  "title": {
    "zh-Hant": "有理係數多項式因式分解：以係數比對驗證提公因子、二項式乘積與平方差",
    "en": "Polynomial factorization over the rationals: verifying common factors, binomial products and differences of squares by comparing coefficients"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-factorization-01",
    "source": {
      "mko_id": "mko-factorization",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "375bd5e9c171dd33297e9e1954a6c9c2b48d97cebbbc4dd93b44c36108f04015",
      "english_sha256": "a24152abff47aa4af453747775e20a1c3ccccf16231b2c4d63aa1d68f06febf7"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "distributive",
      "add_comm",
      "mul_comm",
      "zero_product",
      "square_nonneg",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在 ℚ[x] 中：(一) 把 p(x)=3x²+3x−18 分解為 3(x+3)(x−2)；(二) 把 q(x)=2x²−50 分解為 2(x−5)(x+5)；每個分解都以展開並比對每個係數驗證。(三) 證明 x²+25≠(x−5)(x+5)，且 x²+25 不能寫成兩個一次因式的乘積。(四) 說明 x³ 與 x 在 x=−1,0,1 取值相同卻不是同一個多項式。",
      "en": "In ℚ[x]: (1) factor p(x)=3x²+3x−18 as 3(x+3)(x−2); (2) factor q(x)=2x²−50 as 2(x−5)(x+5); verify each factorization by expanding and comparing every coefficient. (3) Prove that x²+25≠(x−5)(x+5) and that x²+25 is not a product of two degree-one factors. (4) Explain why x³ and x take equal values at x=−1, 0, 1 and still are different polynomials."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "固定形式變數 x，係數屬於 ℚ，以有限係數資料表示；因子與乘積都去掉尾端零（MKO 假設 a1、a2）。",
          "en": "Fix a formal variable x with coefficients in ℚ, represented by finite coefficient data; trailing zeros are removed from factors and products (MKO assumptions a1, a2)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "p(x)=3x²+3x−18，q(x)=2x²−50，s(x)=x²+25；比較用的單項式 x³ 與 x。",
          "en": "p(x)=3x²+3x−18, q(x)=2x²−50 and s(x)=x²+25; the monomials x³ and x for comparison."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "p=f\\cdot g\\ \\text{(proposed)}",
        "after_tex": "p=f\\cdot g\\iff\\text{every coefficient of }p\\text{ equals that of the expanded }f\\cdot g",
        "relation": "definition",
        "rule": "mko_factorization",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，因式分解候選是一個多項式恆等式：把右邊展開、合併同次項後，每個次方的係數都必須與左邊相同（poly_eq_coeff）。所以驗證一個分解就是做係數比對，而不是代入幾個數值。",
          "en": "By the MKO, a proposed factorization is a polynomial identity: after expanding the right side and collecting like terms, the coefficient of every power must equal that of the left side (poly_eq_coeff). So verifying a factorization means comparing coefficients, not substituting a few numbers."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_factorization",
            "poly_eq_coeff"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出判準；後面每個分解都依這個判準驗證。",
            "en": "This step only states the test; every factorization below is verified by it."
          }
        },
        "leaf_basis": "mko_factorization + poly_eq_coeff"
      },
      {
        "id": "S02",
        "uses": [
          "S01"
        ],
        "before_tex": "(x+a)(x+b)",
        "after_tex": "x^2+(a+b)x+ab",
        "relation": "equivalent",
        "rule": "L_binom",
        "conditions": [
          "a,b\\in\\mathbb{Q}"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_binom，兩個首一一次式的乘積的一次項係數是 a+b、常數項是 ab。這是 MKO 的二項式乘積公式；證明在 S02.1、S02.2。",
          "en": "By Lemma L_binom, the product of two monic degree-one factors has x-coefficient a+b and constant term ab. This is the MKO's binomial-product formula; the proof is S02.1 and S02.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_binom"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個多項式，可雙向代換；引理對所有有理 a、b 成立（N1 以次數界核對）。",
            "en": "Both sides are the same polynomial and may replace each other either way; the lemma holds for all rational a, b (N1 checks it with a degree bound)."
          }
        },
        "leaf_basis": "distributive + mul_comm + like_terms"
      },
      {
        "id": "S02.1",
        "uses": [
          "distributive"
        ],
        "before_tex": "(x+a)(x+b)",
        "after_tex": "x\\cdot x+x\\cdot b+a\\cdot x+a\\cdot b",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先把 (x+a) 分配到 (x+b)：(x+a)(x+b)=x(x+b)+a(x+b)，再各自分配，得四項 x·x、x·b、a·x、a·b。",
          "en": "Distribute (x+a) over (x+b): (x+a)(x+b)=x(x+b)+a(x+b), then distribute each part to get the four terms x·x, x·b, a·x and a·b."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "分配律是恆等式，反向就是提出公因式。",
            "en": "Distributivity is an identity; read backwards it is taking out a common factor."
          }
        },
        "leaf_basis": "distributive"
      },
      {
        "id": "S02.2",
        "uses": [
          "S02.1",
          "like_terms",
          "mul_comm"
        ],
        "before_tex": "x\\cdot x+x\\cdot b+a\\cdot x+a\\cdot b",
        "after_tex": "x^2+(a+b)x+ab",
        "relation": "equivalent",
        "rule": "like_terms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x·x=x²；依乘法交換律 x·b=bx，與 ax 是同次項，合併為 (a+b)x；常數項是 ab。S02.1 與 S02.2 合起來證得引理 L_binom。",
          "en": "x·x=x²; by commutativity x·b=bx, which is a like term with ax, and they combine to (a+b)x; the constant term is ab. S02.1 and S02.2 together prove Lemma L_binom."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "like_terms",
            "mul_comm",
            "add_comm"
          ],
          "explanation": {
            "zh-Hant": "合併同次項與把係數拆開互為反向，都是恆等式。",
            "en": "Combining like terms and splitting the coefficient back are inverse moves, both identities."
          }
        },
        "leaf_basis": "like_terms + mul_comm + add_comm"
      },
      {
        "id": "S03",
        "uses": [
          "G2",
          "distributive"
        ],
        "before_tex": "3x^2+3x-18",
        "after_tex": "3(x^2+x-6)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "三個係數 3、3、−18 都是 3 的倍數，依分配律提出公因子 3：3(x²+x−6)=3x²+3x−18。核對係數：3·1=3、3·1=3、3·(−6)=−18。",
          "en": "The three coefficients 3, 3 and −18 are all multiples of 3, so take out the common factor 3 by distributivity: 3(x²+x−6)=3x²+3x−18. Check the coefficients: 3·1=3, 3·1=3, 3·(−6)=−18."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "poly_eq_coeff",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "展開 3(x²+x−6) 回到原式，三個係數逐一相同（N2、N3），雙向成立。",
            "en": "Expanding 3(x²+x−6) returns the original, with all three coefficients equal (N2, N3), so the step is two-way."
          }
        },
        "leaf_basis": "distributive + poly_eq_coeff + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "S03"
        ],
        "before_tex": "a=3,\\ b=-2",
        "after_tex": "a+b=3+(-2)=1,\\ ab=3\\cdot(-2)=-6",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "要讓 x²+x−6 符合 S02 的形式，需要 a+b=1 且 ab=−6。嘗試 a=3、b=−2：3+(−2)=1、3·(−2)=−6，兩個條件都成立。怎麼找到這組數不重要，重要的是它通過 S05 的展開驗證。",
          "en": "To match x²+x−6 with the form of S02 we need a+b=1 and ab=−6. Try a=3, b=−2: 3+(−2)=1 and 3·(−2)=−6, so both conditions hold. How the pair was found does not matter; what matters is that it passes the expansion check of S05."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "只核對這組候選（N4、N5）；不宣稱這是唯一或由某演算法求得。",
            "en": "Only this candidate pair is checked (N4, N5); it is not claimed to be unique or produced by an algorithm."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "S02",
          "S04"
        ],
        "before_tex": "(x+3)(x-2)",
        "after_tex": "x^2+(3+(-2))x+3\\cdot(-2)=x^2+x-6",
        "relation": "equivalent",
        "rule": "L_binom",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代入引理 L_binom（a=3、b=−2）：(x+3)(x−2)=x²+(3+(−2))x+3·(−2)=x²+x−6。係數 [−6,1,1] 與 x²+x−6 逐一相同。",
          "en": "Apply Lemma L_binom with a=3 and b=−2: (x+3)(x−2)=x²+(3+(−2))x+3·(−2)=x²+x−6. The coefficients [−6,1,1] agree one by one with those of x²+x−6."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_binom",
            "poly_eq_coeff"
          ],
          "explanation": {
            "zh-Hant": "引理是恆等式，代入特定 a、b 後仍雙向成立。",
            "en": "The lemma is an identity and stays two-way after substituting particular a and b."
          }
        },
        "leaf_basis": "L_binom + forall_elim + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S03",
          "S05"
        ],
        "before_tex": "3x^2+3x-18",
        "after_tex": "3(x+3)(x-2)",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 S03 的 3(x²+x−6) 中，把 x²+x−6 換成 S05 的 (x+3)(x−2)，得 p=3(x+3)(x−2)。整體核對：展開 3(x+3)(x−2)=3(x²+x−6)=3x²+3x−18，係數 [−18,3,3] 與 p 相同（N6）。",
          "en": "In 3(x²+x−6) from S03, replace x²+x−6 by (x+3)(x−2) from S05 to get p=3(x+3)(x−2). Overall check: 3(x+3)(x−2)=3(x²+x−6)=3x²+3x−18, with coefficients [−18,3,3] equal to those of p (N6)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_replace",
            "poly_eq_coeff"
          ],
          "explanation": {
            "zh-Hant": "等值代換雙向成立；公因子 3 必須保留，否則首項係數會變成 1。",
            "en": "Replacement of equals works both ways; the factor 3 must be kept, or the leading coefficient would become 1."
          }
        },
        "leaf_basis": "eq_replace + poly_eq_coeff + distributive"
      },
      {
        "id": "S07",
        "uses": [
          "S02"
        ],
        "before_tex": "(x-d)(x+d)",
        "after_tex": "x^2-d^2",
        "relation": "equivalent",
        "rule": "L_diffsq",
        "conditions": [
          "d\\in\\mathbb{Q}"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_diffsq，平方差 x²−d² 可分解為 (x−d)(x+d)。證明在 S07.1，是 L_binom 的特例。",
          "en": "By Lemma L_diffsq, the difference of squares x²−d² factors as (x−d)(x+d). The proof, a special case of L_binom, is S07.1."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_diffsq"
          ],
          "explanation": {
            "zh-Hant": "恆等式雙向成立（N7）。注意它只適用於「差」：x²+d² 不在此列。",
            "en": "The identity is two-way (N7). Note that it applies to a difference only: x²+d² is not covered."
          }
        },
        "leaf_basis": "L_binom + numeric_evaluation"
      },
      {
        "id": "S07.1",
        "uses": [
          "S02"
        ],
        "before_tex": "(x+(-d))(x+d)",
        "after_tex": "x^2+(-d+d)x+(-d)d=x^2-d^2",
        "relation": "equivalent",
        "rule": "L_binom",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 L_binom 中取 a=−d、b=d：一次項係數 −d+d=0，常數項 (−d)·d=−d²，所以 (x−d)(x+d)=x²−d²。這證得引理 L_diffsq。",
          "en": "Take a=−d and b=d in L_binom: the x-coefficient is −d+d=0 and the constant term is (−d)·d=−d², so (x−d)(x+d)=x²−d². This proves Lemma L_diffsq."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_binom",
            "add_inverse",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "代入恆等式並化簡，每步都是等值代換。",
            "en": "Substitution into an identity and simplification; every move replaces equals by equals."
          }
        },
        "leaf_basis": "L_binom + add_inverse + neg_mul"
      },
      {
        "id": "S08",
        "uses": [
          "G2",
          "distributive"
        ],
        "before_tex": "2x^2-50",
        "after_tex": "2(x^2-25)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "係數 2 與 −50 都是 2 的倍數，提出公因子 2：2(x²−25)=2x²−50。",
          "en": "The coefficients 2 and −50 are multiples of 2, so take out the common factor 2: 2(x²−25)=2x²−50."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "展開回到原式：2·1=2、2·(−25)=−50。",
            "en": "Expanding returns the original: 2·1=2 and 2·(−25)=−50."
          }
        },
        "leaf_basis": "distributive + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S07"
        ],
        "before_tex": "x^2-25",
        "after_tex": "(x-5)(x+5)",
        "relation": "equivalent",
        "rule": "L_diffsq",
        "conditions": [
          "25=5^2"
        ],
        "explanation": {
          "zh-Hant": "25=5²，所以 x²−25 是平方差，依 L_diffsq（d=5）得 (x−5)(x+5)。",
          "en": "25=5², so x²−25 is a difference of squares, and L_diffsq with d=5 gives (x−5)(x+5)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_diffsq",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "需要把常數寫成有理數的平方（N8）；恆等式雙向成立。",
            "en": "The constant must be written as the square of a rational (N8); the identity is two-way."
          }
        },
        "leaf_basis": "L_diffsq + forall_elim + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "S08",
          "S09"
        ],
        "before_tex": "2x^2-50",
        "after_tex": "2(x-5)(x+5)",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 S08 中的 x²−25 換成 (x−5)(x+5)，得 q=2(x−5)(x+5)。整體核對：展開得 2(x²−25)=2x²−50，係數 [−50,0,2] 與 q 相同（N9）；一次項係數 0 也必須相符。",
          "en": "Replace x²−25 in S08 by (x−5)(x+5) to get q=2(x−5)(x+5). Overall check: expanding gives 2(x²−25)=2x²−50, with coefficients [−50,0,2] equal to those of q (N9); the x-coefficient 0 must match as well."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_replace",
            "poly_eq_coeff"
          ],
          "explanation": {
            "zh-Hant": "等值代換，雙向成立；三個係數全部比對。",
            "en": "Replacement of equals, two-way; all three coefficients are compared."
          }
        },
        "leaf_basis": "eq_replace + poly_eq_coeff + distributive"
      },
      {
        "id": "S11",
        "uses": [
          "G2",
          "S09",
          "poly_eq_coeff"
        ],
        "before_tex": "x^2+25\\ \\text{vs.}\\ (x-5)(x+5)",
        "after_tex": "(x-5)(x+5)=x^2-25\\neq x^2+25",
        "relation": "implies",
        "rule": "poly_eq_coeff",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由 S09，(x−5)(x+5)=x²−25，常數項是 −25；s=x²+25 的常數項是 25。25≠−25，所以兩者不是同一個多項式。把平方和套用平方差公式是錯的。",
          "en": "By S09, (x−5)(x+5)=x²−25, whose constant coefficient is −25; the constant coefficient of s=x²+25 is 25. Since 25≠−25, they are different polynomials. Applying the difference-of-squares formula to a sum of squares is wrong."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "poly_eq_coeff",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一個係數不同就足以推出不等（N10、N11）；但這只否定這一個候選，是否有其他分解另見 S12。",
            "en": "One differing coefficient suffices for inequality (N10, N11); this rejects only this candidate, and whether any other factorization exists is settled in S12."
          }
        },
        "leaf_basis": "poly_eq_coeff + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "G2",
          "eval_hom",
          "square_nonneg"
        ],
        "before_tex": "x^2+25=(ux+v)(wx+t),\\ u\\neq0,\\ w\\neq0",
        "after_tex": "\\text{impossible in}\\ \\mathbb{Q}[x]",
        "relation": "implies",
        "rule": "L_no_linear_factor",
        "conditions": [
          "u\\neq 0",
          "w\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_no_linear_factor，x²+25 在 ℚ[x] 中不能寫成兩個一次因式的乘積，不只是 (x−5)(x+5) 不行。證明在 S12.1 至 S12.3：若有一次因式，就有有理根，但 c²+25 永遠是正數。",
          "en": "By Lemma L_no_linear_factor, x²+25 cannot be written as a product of two degree-one factors in ℚ[x]; this rules out every such factorization, not just the candidate (x−5)(x+5). The proof is S12.1 to S12.3: a degree-one factor would give a rational root, but c²+25 is always positive."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_no_linear_factor",
            "eval_hom",
            "square_nonneg"
          ],
          "explanation": {
            "zh-Hant": "結論只針對兩個一次因式；把 x²+25 寫成 1·(x²+25) 這類含常數因子的寫法不算新的分解（MKO 假設 a2）。",
            "en": "The conclusion concerns two degree-one factors only; writing x²+25 as 1·(x²+25), with a constant factor, is not a new factorization (MKO assumption a2)."
          }
        },
        "leaf_basis": "L_no_linear_factor + contradiction"
      },
      {
        "id": "S12.1",
        "uses": [
          "eval_hom",
          "mul_inverse_nonzero"
        ],
        "before_tex": "x^2+25=(ux+v)(wx+t),\\ u\\neq 0",
        "after_tex": "c=-\\tfrac{v}{u}\\in\\mathbb{Q},\\quad c^2+25=(uc+v)(wc+t)=0\\cdot(wc+t)=0",
        "relation": "implies",
        "rule": "eval_hom",
        "conditions": [
          "u\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "假設存在這樣的分解。因 u≠0，c=−v/u 是有理數，且 uc+v=−v+v=0。相等的多項式在 c 的值相同，求值又保持乘法（eval_hom），所以 c²+25=(uc+v)(wc+t)=0·(wc+t)=0。",
          "en": "Suppose such a factorization exists. Since u≠0, c=−v/u is rational and uc+v=−v+v=0. Equal polynomials have equal values at c, and evaluation respects products (eval_hom), so c²+25=(uc+v)(wc+t)=0·(wc+t)=0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "eval_hom",
            "mul_inverse_nonzero",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "需要 u≠0 才能取 c=−v/u；這是在反證假設的作用域內推論。",
            "en": "u≠0 is needed to form c=−v/u; this reasoning takes place inside the scope of the assumption to be refuted."
          }
        },
        "leaf_basis": "eval_hom + mul_inverse_nonzero + mul_zero"
      },
      {
        "id": "S12.2",
        "uses": [
          "S12.1",
          "square_nonneg",
          "order_basics"
        ],
        "before_tex": "c\\in\\mathbb{Q}\\subseteq\\mathbb{R}",
        "after_tex": "c^2+25\\geq 25>0",
        "relation": "implies",
        "rule": "square_nonneg",
        "conditions": [],
        "explanation": {
          "zh-Hant": "c 是實數，依 square_nonneg，c²≥0；兩邊加 25 得 c²+25≥25>0，所以 c²+25≠0。",
          "en": "c is a real number, so c²≥0 by square_nonneg; adding 25 to both sides gives c²+25≥25>0, so c²+25≠0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "square_nonneg",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "對每個實數 c 都成立，不是有限點核對；最小值 25 在 c=0 取得（N12）。",
            "en": "It holds for every real c, not just at finitely many points; the minimum 25 is reached at c=0 (N12)."
          }
        },
        "leaf_basis": "square_nonneg + order_basics"
      },
      {
        "id": "S12.3",
        "uses": [
          "S12.1",
          "S12.2"
        ],
        "before_tex": "c^2+25=0\\ \\land\\ c^2+25>0",
        "after_tex": "\\neg\\exists\\,u,v,w,t:\\ x^2+25=(ux+v)(wx+t)",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S12.1 推出 c²+25=0，S12.2 推出 c²+25>0，矛盾。所以假設的分解不存在。S12.1 至 S12.3 合起來證得引理 L_no_linear_factor。同一論證把 c 換成實數也成立，所以 x²+25 連在 ℝ[x] 中也沒有一次因式；可分解性一般依賴係數域，例如 x²−2（見錯誤控制）。",
          "en": "S12.1 gives c²+25=0 and S12.2 gives c²+25>0, a contradiction. So the supposed factorization does not exist. S12.1 to S12.3 together prove Lemma L_no_linear_factor. The same argument works with real c, so x²+25 has no degree-one factor even in ℝ[x]; in general factorability depends on the coefficient field, as with x²−2 (see the error controls)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12.1",
            "S12.2",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "矛盾來自同一個 c 的兩個結論；反證假設在此釋放。",
            "en": "The contradiction comes from two conclusions about the same c; the assumption for contradiction is discharged here."
          }
        },
        "leaf_basis": "contradiction + quantifier_negation"
      },
      {
        "id": "S13",
        "uses": [
          "G2",
          "poly_eq_coeff",
          "eval_hom"
        ],
        "before_tex": "x^3\\ \\text{and}\\ x\\ \\text{at}\\ x=-1,0,1",
        "after_tex": "\\text{equal values, but}\\ x^3\\neq x\\ \\text{in}\\ \\mathbb{Q}[x]",
        "relation": "implies",
        "rule": "poly_eq_coeff",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(−1)³=−1、0³=0、1³=1，所以 x³ 與 x 在這三點取值相同；事實上 x³−x=(x−1)x(x+1) 恰在這三點為零。但 x³ 的 x³ 係數是 1，x 的是 0，所以兩者不是同一個多項式；x=2 時 8≠2。兩者次數至多 3，要用代入值判定恆等至少需要 4 個相異點；三個點的吻合不足以建立恆等式。",
          "en": "(−1)³=−1, 0³=0 and 1³=1, so x³ and x agree at these three points; indeed x³−x=(x−1)x(x+1) vanishes exactly there. But the x³-coefficient is 1 for x³ and 0 for x, so they are different polynomials, and at x=2, 8≠2. Both have degree at most 3, so deciding an identity by values would need at least 4 distinct points; agreement at three points does not establish an identity."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "poly_eq_coeff",
            "eval_hom",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "不等由係數證明，不靠代入；N13 只是三點的有限核對，N14 顯示第四點不同。「次數至多 d 的多項式在 d+1 個點相同即恆等」這個事實本稿沒有證明，只用來說明為何三點不夠。",
            "en": "The inequality is proved by coefficients, not by substitution; N13 is a finite check at three points and N14 shows a fourth point differs. The fact that polynomials of degree at most d agreeing at d+1 points are identical is not proved here and is used only to explain why three points are too few."
          }
        },
        "leaf_basis": "poly_eq_coeff + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S06",
          "S10",
          "S11",
          "S12",
          "S13"
        ],
        "before_tex": "p,\\ q,\\ s,\\ x^3\\ \\text{vs}\\ x",
        "after_tex": "p=3(x+3)(x-2),\\ q=2(x-5)(x+5),\\ s\\neq(x-5)(x+5),\\ s\\ \\text{has no degree-one factor},\\ x^3\\neq x",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整：p=3(x+3)(x−2) 與 q=2(x−5)(x+5) 都通過逐係數比對；x²+25 不等於 (x−5)(x+5)，也沒有兩個一次因式的分解；x³ 與 x 在三點吻合仍是不同的多項式。",
          "en": "Summary: p=3(x+3)(x−2) and q=2(x−5)(x+5) both pass the coefficient-by-coefficient comparison; x²+25 is not (x−5)(x+5) and has no factorization into two degree-one factors; x³ and x agree at three points and are still different polynomials."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S06",
            "S10",
            "S11",
            "S12",
            "S13",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：四項結論。足夠條件 H：S06、S10、S11、S12、S13 各自的結論；H⇒G 只是合取。H 由係數計算與反證正向證成，沒有把 G 當作前提。",
            "en": "Goal G: the four conclusions. Sufficient condition H: the separate conclusions of S06, S10, S11, S12 and S13; H⇒G is just conjunction. H is proved forwards by coefficient computation and contradiction; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在 ℚ[x] 中 3x²+3x−18=3(x+3)(x−2)，2x²−50=2(x−5)(x+5)，兩者都以展開後逐係數比對驗證。x²+25≠(x−5)(x+5)（常數項 25≠−25），而且 x²+25 不是兩個一次因式的乘積（否則 c²+25=0 有有理解）。x³ 與 x 在 −1、0、1 取值相同，但 x³≠x。",
      "en": "In ℚ[x], 3x²+3x−18=3(x+3)(x−2) and 2x²−50=2(x−5)(x+5), each verified by expanding and comparing every coefficient. x²+25≠(x−5)(x+5) (constant terms 25≠−25), and x²+25 is not a product of two degree-one factors (otherwise c²+25=0 would have a rational solution). x³ and x agree at −1, 0 and 1, but x³≠x."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "x²+25=(x−5)(x+5)",
          "en": "x²+25=(x−5)(x+5)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "右邊展開為 x²−25，常數項 −25≠25。",
          "en": "The right side expands to x²−25, and the constant term −25≠25."
        }
      },
      {
        "claim": {
          "zh-Hant": "x³ 與 x 在 x=−1、0、1 取值相同，所以 x³=x 是多項式恆等式",
          "en": "x³ and x agree at x=−1, 0 and 1, so x³=x is a polynomial identity"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x³ 的係數 1≠0；x=2 時 8≠2。",
          "en": "The x³-coefficients are 1≠0; at x=2, 8≠2."
        }
      },
      {
        "claim": {
          "zh-Hant": "3x²+3x−18=(x+3)(x−2)",
          "en": "3x²+3x−18=(x+3)(x−2)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(x+3)(x−2)=x²+x−6，首項係數 1≠3、常數項 −6≠−18；漏掉公因子 3。",
          "en": "(x+3)(x−2)=x²+x−6, with leading coefficient 1≠3 and constant term −6≠−18; the common factor 3 was dropped."
        }
      },
      {
        "claim": {
          "zh-Hant": "x²+x−6=(x−3)(x+2)",
          "en": "x²+x−6=(x−3)(x+2)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(x−3)(x+2)=x²−x−6，一次項係數 −1≠1。",
          "en": "(x−3)(x+2)=x²−x−6, whose x-coefficient −1≠1."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "add_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法交換律：u+v=v+u。",
          "en": "Commutativity of addition: u+v=v+u."
        }
      },
      {
        "id": "mul_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": "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": "square_nonneg",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數平方非負：對所有實數 u，u²≥0，且 u²=0 ⇔ u=0。",
          "en": "Squares are nonnegative: for every real u, u²≥0, and u²=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": "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": "eq_replace",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。",
          "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true."
        }
      },
      {
        "id": "add_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": "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": "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_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": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "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_factorization",
        "kind": "cited",
        "source": {
          "mko_id": "mko-factorization",
          "sha256": "375bd5e9c171dd33297e9e1954a6c9c2b48d97cebbbc4dd93b44c36108f04015",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在 Q[x] 中，p 的因式分解候選 p=f·g 是多項式恆等式：展開並合併同次項後每個係數都相同。本條目示範二項式乘積、提公因子與平方差，不宣稱提供全部多項式的完整分解演算法。",
          "en": "In Q[x], a proposed factorization p=f·g is a polynomial identity: after expansion and collection of like terms, every coefficient agrees. This entry illustrates binomial products, common factors, and differences of squares; it does not provide a complete factorization algorithm for all polynomials."
        }
      },
      {
        "id": "poly_eq_coeff",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "ℚ[x] 中兩個多項式相等，當且僅當對每個 k，x^k 的係數都相等（未列出的係數視為 0）。",
          "en": "Two polynomials in ℚ[x] are equal exactly when, for every k, their coefficients of x^k are equal (coefficients not listed count as 0)."
        }
      },
      {
        "id": "like_terms",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合併同次項：c x^k+d x^k=(c+d)x^k（分配律）；不同次的項不能合併。",
          "en": "Combining like terms: c x^k+d x^k=(c+d)x^k (distributivity); terms of different powers are not combined."
        }
      },
      {
        "id": "eval_hom",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "對 c∈ℚ（或 c∈ℝ），求值 f↦f(c) 滿足 (f+g)(c)=f(c)+g(c) 與 (fg)(c)=f(c)g(c)；相等的多項式在每個 c 的值相同。",
          "en": "For c∈ℚ (or c∈ℝ), evaluation f↦f(c) satisfies (f+g)(c)=f(c)+g(c) and (fg)(c)=f(c)g(c); equal polynomials have equal values at every c."
        }
      },
      {
        "id": "L_binom",
        "kind": "lemma",
        "proved_in": [
          "S02.1",
          "S02.2"
        ],
        "statement": {
          "zh-Hant": "對 a,b∈ℚ，ℚ[x] 中 (x+a)(x+b)=x²+(a+b)x+ab。",
          "en": "For a,b∈ℚ, (x+a)(x+b)=x²+(a+b)x+ab in ℚ[x]."
        }
      },
      {
        "id": "L_diffsq",
        "kind": "lemma",
        "proved_in": [
          "S07.1"
        ],
        "statement": {
          "zh-Hant": "對 d∈ℚ，ℚ[x] 中 (x−d)(x+d)=x²−d²。",
          "en": "For d∈ℚ, (x−d)(x+d)=x²−d² in ℚ[x]."
        }
      },
      {
        "id": "L_no_linear_factor",
        "kind": "lemma",
        "proved_in": [
          "S12.1",
          "S12.2",
          "S12.3"
        ],
        "statement": {
          "zh-Hant": "不存在有理數 u、v、w、t（u≠0、w≠0）使 x²+25=(ux+v)(wx+t)；即 x²+25 在 ℚ[x] 中不是兩個一次因式的乘積。",
          "en": "There are no rationals u, v, w, t with u≠0 and w≠0 such that x²+25=(ux+v)(wx+t); that is, x²+25 is not a product of two degree-one factors in ℚ[x]."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "vars": [
          "x",
          "a",
          "b"
        ],
        "degree": 2,
        "lhs": "(x+a)*(x+b)",
        "rhs": "x**2+(a+b)*x+a*b"
      },
      {
        "id": "N2",
        "step": "S03",
        "lhs": "3*1",
        "rhs": "3"
      },
      {
        "id": "N3",
        "step": "S03",
        "lhs": "3*(-6)",
        "rhs": "-18"
      },
      {
        "id": "N4",
        "step": "S04",
        "lhs": "3+(-2)",
        "rhs": "1"
      },
      {
        "id": "N5",
        "step": "S04",
        "lhs": "3*(-2)",
        "rhs": "-6"
      },
      {
        "id": "N6",
        "step": "S06",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "3*(x+3)*(x-2)",
        "rhs": "3*x**2+3*x-18"
      },
      {
        "id": "N7",
        "step": "S07",
        "vars": [
          "x",
          "d"
        ],
        "degree": 2,
        "lhs": "(x-d)*(x+d)",
        "rhs": "x**2-d**2"
      },
      {
        "id": "N8",
        "step": "S09",
        "lhs": "5**2",
        "rhs": "25"
      },
      {
        "id": "N9",
        "step": "S10",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "2*(x-5)*(x+5)",
        "rhs": "2*x**2-50"
      },
      {
        "id": "N10",
        "step": "S11",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "(x-5)*(x+5)",
        "rhs": "x**2-25"
      },
      {
        "id": "N11",
        "step": "S11",
        "lhs": "25",
        "relation": "!=",
        "rhs": "-25"
      },
      {
        "id": "N12",
        "step": "S12.2",
        "lhs": "0**2+25",
        "rhs": "25"
      },
      {
        "id": "N13",
        "step": "S13",
        "vars": [
          "x"
        ],
        "range": [
          -1,
          1
        ],
        "lhs": "x**3",
        "rhs": "x"
      },
      {
        "id": "N14",
        "step": "S13",
        "lhs": "2**3",
        "relation": "!=",
        "rhs": "2"
      },
      {
        "id": "N15",
        "step": "S13",
        "vars": [
          "x"
        ],
        "degree": 3,
        "lhs": "x**3-x",
        "rhs": "(x-1)*x*(x+1)"
      },
      {
        "id": "N16",
        "lhs": "(0-3)*(0+2)",
        "rhs": "-6"
      },
      {
        "id": "N17",
        "lhs": "(1-3)*(1+2)",
        "relation": "!=",
        "rhs": "1**2+1-6"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內：一個形式變數 x、有理係數、有限係數資料。示範提公因子、二項式乘積與平方差，並以係數比對作為恆等式的判準。不提供一般多項式的完整分解演算法（MKO 明示），a、b 的選取只是嘗試，由隨後的展開驗證。x²−2 在 ℚ 與 ℝ 中可分解性的差別只作說明，不在本稿證明；不改動 canonical MKO 或英文 edition。",
      "en": "Within the MKO's scope: one formal variable x, rational coefficients, finite coefficient data. It illustrates common factors, binomial products and differences of squares, with coefficient comparison as the test for an identity. It gives no complete factorization algorithm for all polynomials (as the MKO states); the choice of a and b is a trial confirmed by the expansion that follows. The difference between factoring x²−2 over ℚ and over ℝ is only remarked on and not proved here; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "\\mathbb{Q}[x]",
        "有理係數的一元多項式全體",
        "all univariate polynomials with rational coefficients"
      ],
      [
        "p=f\\cdot g",
        "因式分解候選：一個多項式恆等式",
        "a proposed factorization: a polynomial identity"
      ],
      [
        "a,\\ b,\\ d",
        "二項式乘積與平方差中的有理常數",
        "rational constants in the binomial product and difference of squares"
      ],
      [
        "p(c)",
        "在 x=c 求值",
        "evaluation at x=c"
      ]
    ],
    "md_error_controls": [
      {
        "title": "平方和不能套平方差 / A sum of squares is not a difference of squares",
        "zh": "MKO 指出的第一個誤解是把平方和套用平方差公式。(x−5)(x+5) 展開是 x²−25，不是 x²+25（S11）；而且 x²+25 根本沒有一次因式（S12），因為 c²+25 對每個實數 c 都是正數。",
        "en": "The first error named in the MKO is to apply the difference-of-squares formula to a sum of squares. (x−5)(x+5) expands to x²−25, not x²+25 (S11); moreover x²+25 has no degree-one factor at all (S12), because c²+25 is positive for every real c."
      },
      {
        "title": "有限個點吻合不是恆等 / Agreement at finitely many points is not an identity",
        "zh": "MKO 指出的第二個誤解是用有限數值吻合冒充沒有次數界證據的恆等式。x³ 與 x 在 −1、0、1 三點相同卻是不同多項式（S13）。本稿的每個分解都用逐係數比對驗證（S06、S10）；格點核對 N1、N6、N7、N9 依次數界設計，只作輔助。",
        "en": "The second error named in the MKO is to pass off finite numerical agreement as an identity without degree-bound evidence. x³ and x agree at −1, 0 and 1 but are different polynomials (S13). Every factorization here is verified by coefficient comparison (S06, S10); the grid checks N1, N6, N7 and N9 are designed with a degree bound and serve only as support."
      },
      {
        "title": "係數域與公因子 / The coefficient field and the common factor",
        "zh": "可分解性取決於係數域（MKO para2）：例如 x²−2 在 ℝ 中是 (x−√2)(x+√2)，在 ℚ 中則沒有一次因式（這依賴 √2 是無理數，本稿未證明）。另外，提出的公因子必須保留：3x²+3x−18 不等於 (x+3)(x−2)，因為後者首項係數是 1。因式分解也不是解方程，用零積性求根時還要指定數域。",
        "en": "Factorability depends on the coefficient field (MKO para2): x²−2 is (x−√2)(x+√2) over ℝ but has no degree-one factor over ℚ (this rests on √2 being irrational, not proved here). Also, a common factor once taken out must be kept: 3x²+3x−18 is not (x+3)(x−2), whose leading coefficient is 1. Factoring is not equation solving either; finding roots with the zero-product property also requires specifying the number system."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "「x²−2 在 ℚ 中沒有一次因式」依賴 √2 的無理性，本稿只作說明，未證明。",
        "en": "The claim that x²−2 has no degree-one factor over ℚ rests on the irrationality of √2; it is only remarked on here, not proved."
      },
      {
        "zh-Hant": "S04 的 a、b 是嘗試得到的候選；本稿不提供一般的因式分解演算法（MKO 明示不宣稱）。",
        "en": "The pair a, b in S04 is a candidate found by trial; this guide gives no general factorization algorithm (the MKO explicitly makes no such claim)."
      },
      {
        "zh-Hant": "N13 只是三個點的有限核對，正是用來展示「有限吻合不足」；N1、N5 至 N10、N15 是依次數界設計的格點核對，作為係數比對的輔助。",
        "en": "N13 is a finite check at three points, used precisely to show that finite agreement is not enough; N1, N5 to N10 and N15 are grid checks designed with a degree bound, supporting the coefficient comparisons."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "96eb7a6a6d15bd4bc31f1e82c3a0f24b1950d37ba0e48a4f3336de4395e33750",
    "original_source_sha256": "96eb7a6a6d15bd4bc31f1e82c3a0f24b1950d37ba0e48a4f3336de4395e33750",
    "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."
  }
}