{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-power-root-01",
  "mko_id": "mko-power-root",
  "title": {
    "zh-Hant": "自然數冪與主平方根：由遞迴算冪，並證明 √(a²)=|a|",
    "en": "Natural-number powers and principal square roots: computing powers by recursion and proving √(a²)=|a|"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-power-root-01",
    "source": {
      "mko_id": "mko-power-root",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "29380d066f7612e50db0cbd16c05855d0464ed8f50858e7f348e62e899557c19",
      "english_sha256": "52d750ba77161d6a82d15e4c2d59f1983c52a8affd327d39f3f1b6c3756c664f"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "distributive",
      "neg_mul",
      "zero_product",
      "order_basics",
      "square_nonneg",
      "abs_rules",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "只用遞迴定義 a^0=1、a^(n+1)=a^n·a 算出 (−2)^n（n=0,1,2,3,4）與 0^0、0^3；證明非負平方根的唯一性，並證明對每個實數 a，√(a²)=|a|；再求 √49、√((−5)²)、方程 y²=9 的實數解集，並說明 √(−4) 在實數中沒有定義。",
      "en": "Using only the recursive definition a^0=1, a^(n+1)=a^n·a, compute (−2)^n for n=0,1,2,3,4 and 0^0, 0^3; prove that a nonnegative square root is unique and that √(a²)=|a| for every real a; then find √49, √((−5)²) and the real solution set of y²=9, and explain why √(−4) is not defined in the reals."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "a∈ℝ，n∈ℕ（ℕ 含 0）；冪由 a^0=1（本代數語境也包括 0^0=1）與 a^(n+1)=a^n·a 遞迴定義（MKO 陳述與假設 a1）。",
          "en": "a∈ℝ and n∈ℕ (ℕ contains 0); powers are defined recursively by a^0=1 (including 0^0=1 in this algebraic setting) and a^(n+1)=a^n·a (MKO statement and assumption a1)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "對 r≥0，主平方根 √r 是滿足 y≥0 且 y²=r 的實數 y；MKO 斷言這樣的 y 唯一。r<0 時本語境不定義 √r（MKO 假設 a2）。",
          "en": "For r≥0, the principal square root √r is the real y with y≥0 and y²=r; the MKO asserts that such a y is unique. For r<0, √r is not defined in this setting (MKO assumption a2)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "題目資料：底數 −2 與 0；根號輸入 49、(−5)²、9、−4。",
          "en": "Data: the bases −2 and 0; the root inputs 49, (−5)², 9 and −4."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "a^n,\\quad a\\in\\mathbb{R},\\ n\\in\\mathbb{N}",
        "after_tex": "a^0=1,\\qquad a^{n+1}=a^n\\cdot a",
        "relation": "definition",
        "rule": "mko_power_root",
        "conditions": [
          "n\\in\\mathbb{N}"
        ],
        "explanation": {
          "zh-Hant": "依 MKO，冪以遞迴定義：指數為 0 時是空乘積 1（包括 0^0=1，這是本代數語境的慣例），指數加一時再乘一次底數。所以 n>0 時 a^n 是 n 個 a 的乘積，而不是 a·n。指數限於自然數；負指數、非整數指數與複數分支不在範圍內。",
          "en": "By the MKO, powers are defined recursively: exponent 0 gives the empty product 1 (including 0^0=1, a convention of this algebraic setting), and raising the exponent by one multiplies once more by the base. So for n>0, a^n is a product of n copies of a, not a·n. Exponents are natural numbers only; negative and noninteger exponents and complex branches are out of scope."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_power_root"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出定義；後面每個冪的值都只由這兩條遞迴算出，不另外假設指數律。",
            "en": "This step only states the definition; every power below is computed from these two clauses alone, with no exponent law assumed."
          }
        },
        "leaf_basis": "mko_power_root + def_intro"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "G3"
        ],
        "before_tex": "(-2)^1",
        "after_tex": "(-2)^0\\cdot(-2)=1\\cdot(-2)=-2",
        "relation": "equivalent",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "用遞迴的第二條（n=0）：(−2)^1=(−2)^0·(−2)；再用第一條 (−2)^0=1，以及 1·u=u，得 −2。",
          "en": "Use the second clause with n=0: (−2)^1=(−2)^0·(−2); then the first clause (−2)^0=1 and 1·u=u give −2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "def_unfold",
            "mul_one"
          ],
          "explanation": {
            "zh-Hant": "每個等號都是把式子換成相等的式子，倒著讀就回到 (−2)^1；兩式表示同一個數。",
            "en": "Each equality replaces an expression by an equal one, and reading backwards returns to (−2)^1; both sides name the same number."
          }
        },
        "leaf_basis": "def_unfold + mko_power_root + mul_one"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "S02"
        ],
        "before_tex": "(-2)^2",
        "after_tex": "(-2)^1\\cdot(-2)=(-2)(-2)=4",
        "relation": "equivalent",
        "rule": "neg_mul",
        "conditions": [],
        "explanation": {
          "zh-Hant": "遞迴（n=1）：(−2)^2=(−2)^1·(−2)=(−2)(−2)。依 (−u)(−v)=uv，(−2)(−2)=2·2=4。",
          "en": "Recursion with n=1: (−2)^2=(−2)^1·(−2)=(−2)(−2). By (−u)(−v)=uv, (−2)(−2)=2·2=4."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "def_unfold",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "等值代換，雙向成立；N2 可重算。",
            "en": "A replacement of equals, valid both ways; N2 recomputes it."
          }
        },
        "leaf_basis": "def_unfold + neg_mul + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S01",
          "S03"
        ],
        "before_tex": "(-2)^3",
        "after_tex": "(-2)^2\\cdot(-2)=4\\cdot(-2)=-8",
        "relation": "equivalent",
        "rule": "neg_mul",
        "conditions": [],
        "explanation": {
          "zh-Hant": "遞迴（n=2）：(−2)^3=(−2)^2·(−2)=4·(−2)。依 u(−v)=−(uv)，4·(−2)=−8。注意這不是 (−2)·3=−6：冪是重複乘底數，不是乘以指數。",
          "en": "Recursion with n=2: (−2)^3=(−2)^2·(−2)=4·(−2). By u(−v)=−(uv), 4·(−2)=−8. This is not (−2)·3=−6: a power multiplies the base repeatedly, it does not multiply by the exponent."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "def_unfold",
            "neg_mul",
            "mul_comm"
          ],
          "explanation": {
            "zh-Hant": "等值代換，雙向成立；N3、N4 重算，N5 確認 −8≠−6。",
            "en": "A replacement of equals, valid both ways; N3 and N4 recompute it, and N5 confirms −8≠−6."
          }
        },
        "leaf_basis": "def_unfold + neg_mul + mul_comm + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "S01",
          "S04"
        ],
        "before_tex": "(-2)^4",
        "after_tex": "(-2)^3\\cdot(-2)=(-8)(-2)=16",
        "relation": "equivalent",
        "rule": "neg_mul",
        "conditions": [],
        "explanation": {
          "zh-Hant": "遞迴（n=3）：(−2)^4=(−8)(−2)=16。於是 (−2)^n 在 n=0,1,2,3,4 依次為 1、−2、4、−8、16：這幾個奇次冪為負、偶次冪為正，與 MKO para0 的描述一致。這只是五個指數的計算，不當作對所有 n 的證明。",
          "en": "Recursion with n=3: (−2)^4=(−8)(−2)=16. So (−2)^n for n=0,1,2,3,4 is 1, −2, 4, −8, 16: these odd powers are negative and these even powers positive, as MKO para0 describes. This is a computation for five exponents and is not offered as a proof for every n."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "def_unfold",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "等值代換，雙向成立；正負交替的一般命題本稿不宣稱已證。",
            "en": "A replacement of equals, valid both ways; the general sign-alternation statement is not claimed as proved here."
          }
        },
        "leaf_basis": "def_unfold + neg_mul + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S01",
          "G3"
        ],
        "before_tex": "0^0,\\ 0^1,\\ 0^2,\\ 0^3",
        "after_tex": "0^0=1,\\ 0^1=1\\cdot 0=0,\\ 0^2=0\\cdot 0=0,\\ 0^3=0\\cdot 0=0",
        "relation": "equivalent",
        "rule": "mul_zero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "0^0=1 直接來自定義的第一條（MKO 明示的代數慣例）。之後 0^1=0^0·0=1·0=0，0^2=0^1·0=0，0^3=0^2·0=0（u·0=0）。所以零底數在正冪下仍為零，而 n=0 的值 1 是慣例給的，不是由任何極限決定。",
          "en": "0^0=1 comes straight from the first clause of the definition (the algebraic convention stated in the MKO). Then 0^1=0^0·0=1·0=0, 0^2=0^1·0=0 and 0^3=0^2·0=0 (u·0=0). So a zero base gives zero for positive exponents, and the value 1 at n=0 is fixed by convention, not by any limit."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "def_unfold",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "每個等號都是定義或 u·0=0 的代換，雙向成立；0^0=1 不能用來推出極限結論。",
            "en": "Each equality is the definition or the substitution u·0=0, valid both ways; 0^0=1 may not be used to derive any statement about limits."
          }
        },
        "leaf_basis": "def_unfold + mko_power_root + mul_zero + mul_one"
      },
      {
        "id": "S07",
        "uses": [
          "G2"
        ],
        "before_tex": "y_1\\geq 0,\\ y_2\\geq 0,\\ y_1^2=y_2^2",
        "after_tex": "y_1=y_2",
        "relation": "implies",
        "rule": "L_root_unique",
        "conditions": [
          "y_1\\geq 0",
          "y_2\\geq 0"
        ],
        "explanation": {
          "zh-Hant": "MKO 說主平方根是「唯一」滿足 y≥0、y²=r 的實數；這個唯一性可以證明。依引理 L_root_unique，兩個非負數平方相等時本身相等。證明在 S07.1 至 S07.3。",
          "en": "The MKO says the principal square root is the unique real with y≥0 and y²=r; this uniqueness can be proved. By Lemma L_root_unique, two nonnegative numbers with equal squares are equal. The proof is S07.1 to S07.3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_root_unique",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "唯一性需要非負條件；沒有它，3 與 −3 平方相等卻不相等。反向（相等推出平方相等）顯然但未使用。",
            "en": "Uniqueness needs the nonnegativity condition; without it, 3 and −3 have equal squares but differ. The converse (equal numbers have equal squares) is obvious and not used."
          }
        },
        "leaf_basis": "L_root_unique + order_basics"
      },
      {
        "id": "S07.1",
        "uses": [
          "distributive"
        ],
        "before_tex": "y_1^2=y_2^2",
        "after_tex": "(y_1-y_2)(y_1+y_2)=0",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "展開 (y_1−y_2)(y_1+y_2)=y_1²+y_1y_2−y_2y_1−y_2²=y_1²−y_2²（分配律、乘法交換律、加法反元素）。所以 y_1²=y_2² 等價於 y_1²−y_2²=0，即 (y_1−y_2)(y_1+y_2)=0。",
          "en": "Expand (y_1−y_2)(y_1+y_2)=y_1²+y_1y_2−y_2y_1−y_2²=y_1²−y_2² (distributivity, commutativity of multiplication, additive inverses). So y_1²=y_2² is equivalent to y_1²−y_2²=0, that is (y_1−y_2)(y_1+y_2)=0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "mul_comm",
            "add_inverse",
            "eq_congruence"
          ],
          "explanation": {
            "zh-Hant": "兩邊同減 y_2² 與同加 y_2² 互為反運算，展開式是恆等式（N10），所以雙向成立。",
            "en": "Subtracting y_2² and adding y_2² undo each other, and the expansion is an identity (N10), so the step is two-way."
          }
        },
        "leaf_basis": "distributive + mul_comm + add_inverse + eq_congruence"
      },
      {
        "id": "S07.2",
        "uses": [
          "S07.1",
          "zero_product"
        ],
        "before_tex": "(y_1-y_2)(y_1+y_2)=0",
        "after_tex": "y_1-y_2=0\\ \\lor\\ y_1+y_2=0",
        "relation": "equivalent",
        "rule": "zero_product",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依零乘積性質，實數乘積為 0 當且僅當至少一個因子為 0。",
          "en": "By the zero-product property, a product of reals is 0 exactly when at least one factor is 0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "zero_product"
          ],
          "explanation": {
            "zh-Hant": "零乘積性質本身是雙向的（ℝ 是整域）。",
            "en": "The zero-product property is itself two-way (ℝ is an integral domain)."
          }
        },
        "leaf_basis": "zero_product"
      },
      {
        "id": "S07.3",
        "uses": [
          "S07.2",
          "order_basics"
        ],
        "before_tex": "y_1-y_2=0\\ \\lor\\ y_1+y_2=0",
        "after_tex": "y_1=y_2",
        "relation": "case_split",
        "rule": "case_analysis",
        "conditions": [
          "y_1\\geq 0",
          "y_2\\geq 0"
        ],
        "explanation": {
          "zh-Hant": "情形一：y_1−y_2=0，兩邊加 y_2 得 y_1=y_2。情形二：y_1+y_2=0，則 y_1=−y_2；因 y_2≥0，−y_2≤0，所以 y_1≤0，又 y_1≥0，故 y_1=0，進而 y_2=−y_1=0，也有 y_1=y_2。兩種情形都得 y_1=y_2。S07.1 至 S07.3 合起來證得引理 L_root_unique。",
          "en": "Case one: y_1−y_2=0; adding y_2 gives y_1=y_2. Case two: y_1+y_2=0, so y_1=−y_2; since y_2≥0, −y_2≤0, so y_1≤0, and with y_1≥0 this gives y_1=0, hence y_2=−y_1=0 and again y_1=y_2. Both cases give y_1=y_2. S07.1 to S07.3 together prove Lemma L_root_unique."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07.2",
            "case_analysis",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "情形二正是非負條件發揮作用之處；若允許 y_2<0，例如 y_1=3、y_2=−3，情形二成立但 y_1≠y_2。",
            "en": "Case two is exactly where nonnegativity is used; if y_2<0 were allowed, for example y_1=3 and y_2=−3, case two would hold with y_1≠y_2."
          }
        },
        "leaf_basis": "case_analysis + order_basics + add_inverse"
      },
      {
        "id": "S08",
        "uses": [
          "abs_rules"
        ],
        "before_tex": "|a|\\quad(a\\in\\mathbb{R})",
        "after_tex": "|a|\\geq 0\\ \\land\\ |a|^2=a^2",
        "relation": "implies",
        "rule": "L_abs_sq",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_abs_sq，|a| 是非負數，且平方等於 a²。依三一律 a≥0 或 a<0 兩種情形涵蓋全部實數，證明在 S08.1、S08.2。",
          "en": "By Lemma L_abs_sq, |a| is nonnegative and its square is a². By trichotomy the two cases a≥0 and a<0 cover every real; the proof is S08.1 and S08.2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_abs_sq",
            "abs_rules",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "兩個情形必須涵蓋全部實數；三一律保證沒有遺漏。",
            "en": "The two cases must cover every real; trichotomy ensures none is missed."
          }
        },
        "leaf_basis": "L_abs_sq + abs_rules + case_analysis"
      },
      {
        "id": "S08.1",
        "uses": [
          "abs_rules"
        ],
        "before_tex": "a\\geq 0",
        "after_tex": "|a|=a\\geq 0,\\quad |a|^2=a^2",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [
          "a\\geq 0"
        ],
        "explanation": {
          "zh-Hant": "a≥0 時 |a|=a，所以 |a|≥0，且 |a|²=a²（等值代換）。",
          "en": "If a≥0 then |a|=a, so |a|≥0 and |a|²=a² (replacement of equals)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "只在 a≥0 的情形使用 |a|=a；另一情形見 S08.2。",
            "en": "|a|=a is used only in the case a≥0; the other case is S08.2."
          }
        },
        "leaf_basis": "abs_rules + eq_replace"
      },
      {
        "id": "S08.2",
        "uses": [
          "abs_rules",
          "neg_mul",
          "order_basics"
        ],
        "before_tex": "a<0",
        "after_tex": "|a|=-a>0,\\quad |a|^2=(-a)(-a)=a^2",
        "relation": "implies",
        "rule": "neg_mul",
        "conditions": [
          "a<0"
        ],
        "explanation": {
          "zh-Hant": "a<0 時 |a|=−a。由 a<0 兩邊加 −a 得 0<−a，所以 |a|>0。又 (−a)(−a)=a·a=a²（neg_mul）。S08.1 與 S08.2 合起來證得引理 L_abs_sq。",
          "en": "If a<0 then |a|=−a. Adding −a to both sides of a<0 gives 0<−a, so |a|>0. Also (−a)(−a)=a·a=a² (neg_mul). S08.1 and S08.2 together prove Lemma L_abs_sq."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "neg_mul",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要 a<0 才能用 |a|=−a；恆等式 (−a)(−a)=a² 對所有 a 成立（N11）。",
            "en": "|a|=−a needs a<0; the identity (−a)(−a)=a² holds for every a (N11)."
          }
        },
        "leaf_basis": "abs_rules + neg_mul + order_basics"
      },
      {
        "id": "S09",
        "uses": [
          "G2",
          "S07",
          "S08",
          "square_nonneg"
        ],
        "before_tex": "\\sqrt{a^2}\\quad(a\\in\\mathbb{R})",
        "after_tex": "\\sqrt{a^2}=|a|",
        "relation": "implies",
        "rule": "L_root_unique",
        "conditions": [
          "a^2\\geq 0"
        ],
        "explanation": {
          "zh-Hant": "先確認 √(a²) 有意義：依 square_nonneg，a²≥0，所以 G2 的定義適用於 r=a²。y=|a| 滿足 y≥0 與 y²=a²（S08），所以至少有一個這樣的 y；依 S07，這樣的非負 y 至多一個。因此 √(a²) 存在且等於 |a|。注意結論是 |a| 而不是 a。",
          "en": "First check that √(a²) makes sense: by square_nonneg, a²≥0, so the definition in G2 applies with r=a². The number y=|a| satisfies y≥0 and y²=a² (S08), so at least one such y exists; by S07 there is at most one such nonnegative y. Hence √(a²) exists and equals |a|. Note that the result is |a|, not a."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S07",
            "S08",
            "square_nonneg",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "目標 G：√(a²)=|a|。足夠條件 H：a²≥0、|a| 是一個非負且平方為 a² 的數、而且這樣的數唯一；H⇒G 由主平方根的定義。H 由 square_nonneg、S08 與 S07 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: √(a²)=|a|. Sufficient condition H: a²≥0, |a| is a nonnegative number whose square is a², and such a number is unique; H⇒G by the definition of the principal root. H is proved forwards by square_nonneg, S08 and S07; G is never used as a premise."
          }
        },
        "leaf_basis": "L_root_unique + L_abs_sq + square_nonneg + sqrt_def"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "G3"
        ],
        "before_tex": "\\sqrt{49}",
        "after_tex": "\\sqrt{7^2}=|7|=7",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [
          "7\\geq 0"
        ],
        "explanation": {
          "zh-Hant": "49=7²，代入 S09（a=7）得 √49=|7|=7。−7 的平方也是 49，但 −7<0，不符合 y≥0，所以不是主平方根。",
          "en": "49=7², so S09 with a=7 gives √49=|7|=7. The square of −7 is also 49, but −7<0 fails y≥0, so it is not the principal root."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "forall_elim",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "需要 49 寫成某數的平方（N13）；−7 被非負條件排除（N14 顯示它的平方也是 49）。",
            "en": "It needs 49 written as a square (N13); −7 is excluded by the nonnegativity condition (N14 shows its square is also 49)."
          }
        },
        "leaf_basis": "forall_elim + abs_rules + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S09",
          "G3"
        ],
        "before_tex": "\\sqrt{(-5)^2}",
        "after_tex": "|-5|=5",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [
          "-5<0"
        ],
        "explanation": {
          "zh-Hant": "代入 S09（a=−5）：√((−5)²)=|−5|=5。直接核對：(−5)²=25，5≥0 且 5²=25。若寫成 √((−5)²)=−5，右邊是負數，違反主平方根的 y≥0。",
          "en": "Substitute a=−5 in S09: √((−5)²)=|−5|=5. Direct check: (−5)²=25, 5≥0 and 5²=25. Writing √((−5)²)=−5 would put a negative number on the right, violating y≥0 for the principal root."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "forall_elim",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "這是 MKO 指出的誤解所在：√(a²) 不能無條件寫成 a；a<0 時兩者不同（N15 至 N17）。",
            "en": "This illustrates the misconception identified in the MKO: √(a²) may not be written as a without conditions; for a<0 they differ (N15 to N17)."
          }
        },
        "leaf_basis": "forall_elim + abs_rules + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "G3",
          "distributive"
        ],
        "before_tex": "y^2=9",
        "after_tex": "(y-3)(y+3)=0",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊同減 9 得 y²−9=0；展開 (y−3)(y+3)=y²+3y−3y−9=y²−9，所以 y²=9 等價於 (y−3)(y+3)=0。",
          "en": "Subtract 9 from both sides to get y²−9=0; expanding (y−3)(y+3)=y²+3y−3y−9=y²−9 shows that y²=9 is equivalent to (y−3)(y+3)=0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "eq_congruence",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "同減 9 與同加 9 互為反運算，展開式是恆等式（N18），雙向成立。",
            "en": "Subtracting 9 and adding 9 undo each other, and the expansion is an identity (N18), so the step is two-way."
          }
        },
        "leaf_basis": "distributive + eq_congruence + add_inverse"
      },
      {
        "id": "S13",
        "uses": [
          "S12",
          "zero_product"
        ],
        "before_tex": "(y-3)(y+3)=0",
        "after_tex": "y=3\\ \\lor\\ y=-3,\\quad\\text{solution set}\\ \\{-3,3\\}",
        "relation": "equivalent",
        "rule": "zero_product",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依零乘積性質，y−3=0 或 y+3=0，即 y=3 或 y=−3。所以方程 y²=9 的實數解集是 {−3,3}，有兩個元素。",
          "en": "By the zero-product property, y−3=0 or y+3=0, that is y=3 or y=−3. So the real solution set of y²=9 is {−3,3}, with two elements."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "zero_product",
            "eq_congruence"
          ],
          "explanation": {
            "zh-Hant": "零乘積性質雙向成立；代回 3²=9、(−3)²=9（N19、N20）。",
            "en": "The zero-product property is two-way; substituting back gives 3²=9 and (−3)²=9 (N19, N20)."
          }
        },
        "leaf_basis": "zero_product + eq_congruence"
      },
      {
        "id": "S14",
        "uses": [
          "S09",
          "S13"
        ],
        "before_tex": "\\{-3,3\\}",
        "after_tex": "\\sqrt{9}=3,\\quad \\{-3,3\\}=\\{-\\sqrt{9},\\sqrt{9}\\}",
        "relation": "implies",
        "rule": "L_root_unique",
        "conditions": [],
        "explanation": {
          "zh-Hant": "解集中只有 3 滿足 y≥0，依 S09（a=3），√9=|3|=3。所以方程有兩個解 ±√9，但符號 √9 只表示一個數 3。「方程的解」與「主平方根」是兩件事。",
          "en": "Only 3 in the solution set satisfies y≥0, and by S09 with a=3, √9=|3|=3. So the equation has two solutions ±√9, but the symbol √9 denotes one number, 3. The solutions of an equation and the principal root are different things."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "S13",
            "L_root_unique"
          ],
          "explanation": {
            "zh-Hant": "需要非負條件從兩個解中選出一個；把 √9 寫成 ±3 違反唯一性。",
            "en": "The nonnegativity condition is what selects one of the two solutions; writing √9 as ±3 contradicts uniqueness."
          }
        },
        "leaf_basis": "L_root_unique + forall_elim + order_basics"
      },
      {
        "id": "S15",
        "uses": [
          "G2",
          "G3",
          "square_nonneg"
        ],
        "before_tex": "y^2=-4,\\ y\\in\\mathbb{R}",
        "after_tex": "\\text{no real solution};\\ \\sqrt{-4}\\ \\text{undefined here}",
        "relation": "implies",
        "rule": "square_nonneg",
        "conditions": [
          "-4<0"
        ],
        "explanation": {
          "zh-Hant": "對每個實數 y，y²≥0>−4，所以 y²≠−4，方程 y²=−4 沒有實數解。G2 也只對 r≥0 定義 √r，所以 √(−4) 在本語境沒有定義。寫成 −2 是錯的：(−2)²=4≠−4，而且 −2<0。",
          "en": "For every real y, y²≥0>−4, so y²≠−4 and the equation y²=−4 has no real solution. G2 also defines √r only for r≥0, so √(−4) is undefined in this setting. Writing −2 is wrong: (−2)²=4≠−4, and −2<0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "square_nonneg",
            "G2",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "結論是「在實數中無定義」；本稿不討論複數分支，也不採用把負輸入補成 0 的程式語義（MKO para2）。",
            "en": "The conclusion is 'undefined in the reals'; complex branches are not discussed, and the software convention of returning 0 for negative input (MKO para2) is not adopted."
          }
        },
        "leaf_basis": "square_nonneg + order_basics + sqrt_def"
      }
    ],
    "result": {
      "zh-Hant": "由遞迴：(−2)^0、(−2)^1、(−2)^2、(−2)^3、(−2)^4 依次為 1、−2、4、−8、16；0^0=1（慣例）、0^3=0。非負平方根唯一；對每個實數 a，√(a²)=|a|。因此 √49=7，√((−5)²)=5（不是 −5），y²=9 的解集是 {−3,3} 而 √9=3，√(−4) 在實數中沒有定義。",
      "en": "By the recursion, (−2)^0, (−2)^1, (−2)^2, (−2)^3, (−2)^4 are 1, −2, 4, −8, 16; 0^0=1 (convention) and 0^3=0. A nonnegative square root is unique, and √(a²)=|a| for every real a. Hence √49=7, √((−5)²)=5 (not −5), the solution set of y²=9 is {−3,3} while √9=3, and √(−4) is undefined in the reals."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "√9=±3",
          "en": "√9=±3"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "主平方根要求 y≥0 且唯一，√9=3；±3 是方程 y²=9 的兩個解。",
          "en": "The principal root requires y≥0 and is unique, so √9=3; ±3 are the two solutions of y²=9."
        }
      },
      {
        "claim": {
          "zh-Hant": "√((−5)²)=−5",
          "en": "√((−5)²)=−5"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "√((−5)²)=√25=5=|−5|；−5<0 不符合主平方根的條件。",
          "en": "√((−5)²)=√25=5=|−5|; −5<0 fails the condition for the principal root."
        }
      },
      {
        "claim": {
          "zh-Hant": "(−2)³=(−2)·3=−6",
          "en": "(−2)³=(−2)·3=−6"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "冪是重複乘底數：(−2)³=4·(−2)=−8≠−6。",
          "en": "A power multiplies the base repeatedly: (−2)³=4·(−2)=−8≠−6."
        }
      },
      {
        "claim": {
          "zh-Hant": "因為 0^0=1，數列 0^n 的極限是 1",
          "en": "Since 0^0=1, the sequence 0^n has limit 1"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "n≥1 時 0^n=0，數列極限是 0；0^0=1 是代數慣例，不決定極限。",
          "en": "For n≥1, 0^n=0, so the limit is 0; 0^0=1 is an algebraic convention and does not decide limits."
        }
      },
      {
        "claim": {
          "zh-Hant": "√(−4)=−2",
          "en": "√(−4)=−2"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(−2)²=4≠−4，且 −2<0；沒有實數平方等於 −4，√(−4) 在實數中無定義。",
          "en": "(−2)²=4≠−4 and −2<0; no real square equals −4, so √(−4) is undefined in the reals."
        }
      }
    ],
    "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": "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": "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": "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": "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": "abs_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "絕對值（實數）：|u|=u（u≥0）、|u|=−u（u<0）；|u|≥0；|uv|=|u||v|；三角不等式 |u+v|≤|u|+|v|；|u|<r ⇔ −r<u<r。",
          "en": "Absolute value (reals): |u|=u if u≥0 and |u|=−u if u<0; |u|≥0; |uv|=|u||v|; triangle inequality |u+v|≤|u|+|v|; |u|<r ⇔ −r<u<r."
        }
      },
      {
        "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_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。",
          "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable."
        }
      },
      {
        "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": "mul_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。",
          "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)."
        }
      },
      {
        "id": "mul_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": "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": "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": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "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": "sqrt_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "主平方根：對 a≥0，√a 是唯一滿足 b≥0 且 b²=a 的實數 b。負數在實數中沒有平方根。",
          "en": "Principal square root: for a≥0, √a is the unique real b with b≥0 and b²=a. Negative numbers have no real square root."
        }
      },
      {
        "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": "mko_power_root",
        "kind": "cited",
        "source": {
          "mko_id": "mko-power-root",
          "sha256": "29380d066f7612e50db0cbd16c05855d0464ed8f50858e7f348e62e899557c19",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對實數 a 與自然數 n，採空乘積慣例 a^0=1（本代數語境也包括 0^0=1），並由 a^(n+1)=a^n·a 定義冪。對 r≥0，主平方根 √r 是唯一滿足 y≥0 且 y²=r 的實數 y。",
          "en": "For a real number a and a natural number n, use the empty-product convention a^0=1 (including 0^0=1 in this algebraic setting) and define powers recursively by a^(n+1)=a^n·a. For r≥0, the principal square root √r is the unique real number y satisfying y≥0 and y²=r."
        }
      },
      {
        "id": "L_root_unique",
        "kind": "lemma",
        "proved_in": [
          "S07.1",
          "S07.2",
          "S07.3"
        ],
        "statement": {
          "zh-Hant": "若實數 y_1、y_2 都 ≥0 且 y_1²=y_2²，則 y_1=y_2。",
          "en": "If reals y_1, y_2 are both ≥0 and y_1²=y_2², then y_1=y_2."
        }
      },
      {
        "id": "L_abs_sq",
        "kind": "lemma",
        "proved_in": [
          "S08.1",
          "S08.2"
        ],
        "statement": {
          "zh-Hant": "對每個實數 a，|a|≥0 且 |a|²=a²。",
          "en": "For every real a, |a|≥0 and |a|²=a²."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "(-2)**1",
        "rhs": "-2"
      },
      {
        "id": "N2",
        "step": "S03",
        "lhs": "(-2)*(-2)",
        "rhs": "4"
      },
      {
        "id": "N3",
        "step": "S04",
        "lhs": "4*(-2)",
        "rhs": "-8"
      },
      {
        "id": "N4",
        "step": "S04",
        "lhs": "(-2)**3",
        "rhs": "-8"
      },
      {
        "id": "N5",
        "step": "S04",
        "lhs": "(-2)*3",
        "relation": "!=",
        "rhs": "(-2)**3"
      },
      {
        "id": "N6",
        "step": "S05",
        "lhs": "(-8)*(-2)",
        "rhs": "16"
      },
      {
        "id": "N7",
        "step": "S05",
        "lhs": "(-2)**4",
        "rhs": "16"
      },
      {
        "id": "N8",
        "step": "S06",
        "lhs": "0**0",
        "rhs": "1"
      },
      {
        "id": "N9",
        "step": "S06",
        "lhs": "0**3",
        "rhs": "0"
      },
      {
        "id": "N10",
        "step": "S07.1",
        "vars": [
          "y1",
          "y2"
        ],
        "degree": 2,
        "lhs": "(y1-y2)*(y1+y2)",
        "rhs": "y1**2-y2**2"
      },
      {
        "id": "N11",
        "step": "S08.2",
        "vars": [
          "a"
        ],
        "degree": 2,
        "lhs": "(-a)*(-a)",
        "rhs": "a**2"
      },
      {
        "id": "N12",
        "step": "S08",
        "vars": [
          "a"
        ],
        "range": [
          -20,
          20
        ],
        "lhs": "abs(a)**2",
        "rhs": "a**2"
      },
      {
        "id": "N13",
        "step": "S10",
        "lhs": "7**2",
        "rhs": "49"
      },
      {
        "id": "N14",
        "step": "S10",
        "lhs": "(-7)**2",
        "rhs": "49"
      },
      {
        "id": "N15",
        "step": "S11",
        "lhs": "(-5)**2",
        "rhs": "25"
      },
      {
        "id": "N16",
        "step": "S11",
        "lhs": "abs(-5)",
        "rhs": "5"
      },
      {
        "id": "N17",
        "step": "S11",
        "lhs": "5**2",
        "rhs": "25"
      },
      {
        "id": "N18",
        "step": "S12",
        "vars": [
          "y"
        ],
        "degree": 2,
        "lhs": "(y-3)*(y+3)",
        "rhs": "y**2-9"
      },
      {
        "id": "N19",
        "step": "S13",
        "lhs": "3**2",
        "rhs": "9"
      },
      {
        "id": "N20",
        "step": "S13",
        "lhs": "(-3)**2",
        "rhs": "9"
      },
      {
        "id": "N21",
        "step": "S15",
        "lhs": "-4",
        "relation": "<",
        "rhs": "0"
      },
      {
        "id": "N22",
        "step": "S15",
        "lhs": "(-2)**2",
        "relation": "!=",
        "rhs": "-4"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：實數底數、自然數指數（含 0），主平方根只對 r≥0 的實數定義。不處理負指數、非整數指數或複數分支；0^0=1 只作為本代數語境的慣例，不推出任何極限規則。對一般 r≥0 的平方根存在性（例如 √2）不在本稿證明，本稿只需 r=a² 的情形；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: real bases, natural-number exponents (including 0), and the principal square root only for real r≥0. Negative and noninteger exponents and complex branches are not treated; 0^0=1 is used only as a convention of this algebraic setting and yields no rule for limits. Existence of a square root for a general r≥0 (for example √2) is not proved here, since the guide only needs the case r=a²; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "a^n",
        "實數 a 的 n 次冪（n∈ℕ）",
        "the n-th power of the real a (n∈ℕ)"
      ],
      [
        "\\sqrt{r}",
        "r≥0 的主平方根：唯一滿足 y≥0、y²=r 的實數",
        "the principal square root of r≥0: the unique real y with y≥0 and y²=r"
      ],
      [
        "|a|",
        "絕對值：a≥0 時為 a，a<0 時為 −a",
        "absolute value: a if a≥0, −a if a<0"
      ],
      [
        "y_1,\\ y_2",
        "唯一性證明中的兩個非負候選根",
        "two nonnegative candidate roots in the uniqueness proof"
      ]
    ],
    "md_error_controls": [
      {
        "title": "主平方根不是 ± 值 / The principal square root is not a ± value",
        "zh": "MKO 指出的第一個誤解是把主平方根寫成 ± 值。方程 y²=9 有兩個解 ±3（S13），但 √9 由 y≥0 與唯一性（S07）確定為 3 一個數（S14）。需要兩個解時要寫 ±√9，而不是說 √9 等於 ±3。",
        "en": "The first error named in the MKO is to write the principal root as a ± value. The equation y²=9 has two solutions ±3 (S13), but √9 is fixed as the single number 3 by y≥0 and uniqueness (S07, S14). When both solutions are meant, write ±√9; do not say √9 equals ±3."
      },
      {
        "title": "√(a²)=|a|，不是 a / √(a²)=|a|, not a",
        "zh": "MKO 指出的第二個誤解是把 √(a²) 當成 a。a=−5 時 √((−5)²)=√25=5=|−5|，而 −5 不滿足 y≥0（S11）。只有 a≥0 時才可寫 √(a²)=a。",
        "en": "The second error named in the MKO is to take √(a²) to be a. For a=−5, √((−5)²)=√25=5=|−5|, while −5 fails y≥0 (S11). Only for a≥0 may one write √(a²)=a."
      },
      {
        "title": "0^0=1 是代數慣例；a^n 不是 a·n / 0^0=1 is an algebraic convention; a^n is not a·n",
        "zh": "MKO 指出的第三個誤解是把 0^0=1 套到極限。數列 0^n 的值依次為 1、0、0、0，以此類推，n≥1 時都是 0，所以極限是 0，不是 1；n=0 的值只是慣例（S06）。另外 (−2)^3=−8，不是 (−2)·3=−6（S04）。",
        "en": "The third error named in the MKO is to carry 0^0=1 over to limits. The sequence 0^n takes the values 1, 0, 0, 0 and so on, equal to 0 for every n≥1, so its limit is 0, not 1; the value at n=0 is only a convention (S06). Also, (−2)^3=−8, not (−2)·3=−6 (S04)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿證明唯一性，並對 r=a² 給出存在性（取 y=|a|）。對一般 r≥0（例如 r=2）的存在性需要實數完備性，本稿沒有證明，也沒有用到。",
        "en": "This guide proves uniqueness and gives existence for r=a² (take y=|a|). Existence for a general r≥0 (for example r=2) needs the completeness of the reals; it is neither proved nor used here."
      },
      {
        "zh-Hant": "N12 只是 a 從 −20 到 20 的有限點核對；|a|²=a² 的一般證明是 S08。",
        "en": "N12 is a finite check for a from −20 to 20 only; the general proof of |a|²=a² is S08."
      },
      {
        "zh-Hant": "Mathlib Real.sqrt 對負輸入回傳 0 的語義與本稿「無定義」不同，橋接需要 r≥0 的前提（MKO para2），不在本稿範圍。",
        "en": "Mathlib's Real.sqrt returns 0 on negative input, unlike 'undefined' here; bridging them needs the premise r≥0 (MKO para2) and is outside this guide."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "500af794d5ac8f50bed01f893eca7a76224302cea15cd5ef78efee08f8a81236",
    "original_source_sha256": "500af794d5ac8f50bed01f893eca7a76224302cea15cd5ef78efee08f8a81236",
    "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."
  }
}