{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-complex-modulus-conjugate-01",
  "mko_id": "mko-complex-modulus-conjugate",
  "title": {
    "zh-Hant": "複數模與共軛：證明 z·z̄=|z|² 與共軛保持運算",
    "en": "Complex modulus and conjugate: proving z·z̄=|z|² and that conjugation preserves the operations"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-complex-modulus-conjugate-01",
    "source": {
      "mko_id": "mko-complex-modulus-conjugate",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "c4c1f275b4ae1708bd308a1ea5ce57ee90a10921251139a1eb880ffd2101842d",
      "english_sha256": "4d553f442775fa75aa08d2cdf7a109593034fd6b47602575bbec0612a1970cb4"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "mul_assoc",
      "mul_comm",
      "neg_mul",
      "square_nonneg",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "對 z=a+bi、w=c+di（a、b、c、d 為實數）證明：（一）z·z̄=(a²+b²)+0i，即模平方的實數嵌入；（二）共軛保持加法：(z+w)‾=z̄+w̄；（三）共軛保持乘法：(zw)‾=z̄·w̄；（四）由此得 |zw|²=|z|²|w|²。再以 z=3−4i、w=1+2i 精確核對，並區分模平方與模。",
      "en": "For z=a+bi and w=c+di (a, b, c, d real) prove: (1) z·z̄=(a²+b²)+0i, the real embedding of the modulus squared; (2) conjugation preserves addition: (z+w)‾=z̄+w̄; (3) conjugation preserves multiplication: (zw)‾=z̄·w̄; (4) hence |zw|²=|z|²|w|². Then check everything exactly for z=3−4i and w=1+2i, distinguishing the modulus squared from the modulus."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "一般情形：z=a+bi 對應 (a,b)，w=c+di 對應 (c,d)，a、b、c、d 為實數。",
          "en": "General case: z=a+bi corresponds to (a,b) and w=c+di to (c,d), with a, b, c, d real."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "例子：z=3−4i 對應 (3,−4)，w=1+2i 對應 (1,2)。",
          "en": "Example: z=3−4i corresponds to (3,−4) and w=1+2i to (1,2)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "答案以精確整數或有序對表示；非完全平方數的模平方保持原樣，不寫小數近似（MKO 假設 a1）。",
          "en": "Answers are exact integers or ordered pairs; a modulus squared that is not a perfect square is kept as it is, with no decimal approximation (MKO assumption a1)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "mko_modconj_def"
        ],
        "before_tex": "z=a+bi\\leftrightarrow(a,b)",
        "after_tex": "\\overline{z}=a-bi\\leftrightarrow(a,-b),\\ |z|^2=a^2+b^2",
        "relation": "definition",
        "rule": "mko_modconj_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，共軛只把虛部係數變號、實部不變：(a,b) 變成 (a,−b)；模平方是兩個分量的平方和 a²+b²。模平方與模不同：模是它的非負平方根。",
          "en": "By the MKO, conjugation changes the sign of the imaginary coefficient only and leaves the real part alone: (a,b) becomes (a,−b); the modulus squared is the sum of squares of the components, a²+b². The modulus and its square are different notions: the modulus is the nonnegative square root of the modulus squared."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_modconj_def"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出定義，不推出新事實。",
            "en": "This step only states the definitions and derives no new fact."
          }
        },
        "leaf_basis": "mko_modconj_def + def_intro"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "complex_mul_formula"
        ],
        "before_tex": "z\\overline{z}=(a,b)\\cdot(a,-b)",
        "after_tex": "(a^2+b^2,\\,0)",
        "relation": "equivalent",
        "rule": "complex_mul_formula",
        "conditions": [],
        "explanation": {
          "zh-Hant": "以乘法公式（取 c=a、d=−b）計算 z·z̄：實部見 S02.1，虛部係數見 S02.2。",
          "en": "Compute z·z̄ by the multiplication formula with c=a and d=−b: the real part is S02.1 and the imaginary coefficient is S02.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "complex_mul_formula",
            "complex_pair_eq"
          ],
          "explanation": {
            "zh-Hant": "公式是等式，兩個分量分別相等即有序對相等，雙向成立。",
            "en": "The formula is an equation and the pairs are equal because both components are; the step is two-way."
          }
        },
        "leaf_basis": "complex_mul_formula + complex_pair_eq"
      },
      {
        "id": "S02.1",
        "uses": [
          "complex_mul_formula",
          "neg_mul"
        ],
        "before_tex": "\\operatorname{Re}(z\\overline{z})=a\\cdot a-b\\cdot(-b)",
        "after_tex": "a^2+b^2",
        "relation": "equivalent",
        "rule": "neg_mul",
        "conditions": [],
        "explanation": {
          "zh-Hant": "實部 ac−bd 中 c=a、d=−b：a·a−b·(−b)。因 b·(−b)=−b²，減去 −b² 等於加上 b²，得 a²+b²。正號正是來自 i²=−1 與共軛的負號相乘。",
          "en": "In the real part ac−bd put c=a and d=−b: a·a−b·(−b). Since b·(−b)=−b², subtracting −b² adds b², giving a²+b². The plus sign comes from the product of i²=−1 with the conjugate's minus sign."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "neg_mul",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "對所有實數成立的恆等式（N1 以多項式格點核對），雙向成立。",
            "en": "An identity for all reals (checked as a polynomial identity on a grid in N1); two-way."
          }
        },
        "leaf_basis": "neg_mul + add_inverse + power_laws"
      },
      {
        "id": "S02.2",
        "uses": [
          "complex_mul_formula",
          "mul_comm"
        ],
        "before_tex": "\\operatorname{Im}(z\\overline{z})=a\\cdot(-b)+b\\cdot a",
        "after_tex": "0",
        "relation": "equivalent",
        "rule": "mul_comm",
        "conditions": [],
        "explanation": {
          "zh-Hant": "虛部係數 ad+bc 中 d=−b、c=a：a·(−b)+b·a=−ab+ab=0（交換律使 ba=ab）。所以 z·z̄ 的虛部為零，結果是實數。",
          "en": "In the imaginary coefficient ad+bc put d=−b and c=a: a·(−b)+b·a=−ab+ab=0 (commutativity gives ba=ab). So z·z̄ has zero imaginary part and is real."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_comm",
            "neg_mul",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "恆等式（N2），雙向成立。",
            "en": "An identity (N2), two-way."
          }
        },
        "leaf_basis": "mul_comm + neg_mul + add_inverse"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "real_embedding",
          "square_nonneg"
        ],
        "before_tex": "z\\overline{z}=(a^2+b^2,\\,0)",
        "after_tex": "z\\overline{z}=|z|^2\\in\\mathbb{R},\\ |z|^2\\ge 0",
        "relation": "implies",
        "rule": "real_embedding",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(a²+b²,0) 正是實數 a²+b²=|z|² 的嵌入，所以 z·z̄ 等於模平方的實數嵌入，與 MKO 一致。又 a²≥0、b²≥0，所以 |z|²≥0，等號只在 a=b=0 時成立。",
          "en": "(a²+b²,0) is exactly the embedding of the real number a²+b²=|z|², so z·z̄ equals the real embedding of the modulus squared, as the MKO says. Also a²≥0 and b²≥0, so |z|²≥0, with equality only when a=b=0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "real_embedding",
            "square_nonneg"
          ],
          "explanation": {
            "zh-Hant": "非負性用到平方非負；此步不宣稱每個非負實數都是某個整數對的模平方。",
            "en": "Nonnegativity uses that squares are nonnegative; the step does not claim that every nonnegative real is the modulus squared of an integer pair."
          }
        },
        "leaf_basis": "real_embedding + square_nonneg + order_basics"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "complex_addsub",
          "S01"
        ],
        "before_tex": "\\overline{z+w}=\\overline{(a+c)+(b+d)i}",
        "after_tex": "(a+c)-(b+d)i=(a-bi)+(c-di)=\\overline{z}+\\overline{w}",
        "relation": "equivalent",
        "rule": "complex_addsub",
        "conditions": [],
        "explanation": {
          "zh-Hant": "z+w 的虛部係數是 b+d，共軛把它變成 −(b+d)=−b−d；而 z̄+w̄ 逐分量相加的虛部係數也是 −b+(−d)。兩邊實部都是 a+c，所以共軛保持加法。",
          "en": "z+w has imaginary coefficient b+d, which conjugation turns into −(b+d)=−b−d; adding z̄+w̄ componentwise gives imaginary coefficient −b+(−d) as well. Both real parts are a+c, so conjugation preserves addition."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "complex_addsub",
            "complex_pair_eq",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "每個等號都是恆等改寫（−(b+d)=−b−d），兩個分量相等，雙向成立。",
            "en": "Each equals sign is an identity rewrite (−(b+d)=−b−d) and both components agree, so the step is two-way."
          }
        },
        "leaf_basis": "complex_addsub + complex_pair_eq + distributive"
      },
      {
        "id": "S05",
        "uses": [
          "S01",
          "complex_mul_formula"
        ],
        "before_tex": "\\overline{zw}",
        "after_tex": "\\overline{z}\\cdot\\overline{w}",
        "relation": "equivalent",
        "rule": "L_conj_mul",
        "conditions": [],
        "explanation": {
          "zh-Hant": "共軛保持乘法（引理 L_conj_mul）。證明在 S05.1、S05.2：分別算出兩邊的有序對，再逐分量比對。",
          "en": "Conjugation preserves multiplication (Lemma L_conj_mul). The proof is in S05.1 and S05.2: compute the ordered pair of each side and compare components."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_conj_mul"
          ],
          "explanation": {
            "zh-Hant": "引理是等式，兩邊是同一個複數，雙向成立。",
            "en": "The lemma is an equation and both sides are the same complex number, so the step is two-way."
          }
        },
        "leaf_basis": "complex_mul_formula + complex_pair_eq + neg_mul"
      },
      {
        "id": "S05.1",
        "uses": [
          "complex_mul_formula",
          "S01"
        ],
        "before_tex": "\\overline{zw}=\\overline{(ac-bd)+(ad+bc)i}",
        "after_tex": "(ac-bd,\\,-(ad+bc))",
        "relation": "equivalent",
        "rule": "complex_mul_formula",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先依乘法公式寫出 zw=(ac−bd)+(ad+bc)i，再取共軛：實部不變，虛部係數變號。",
          "en": "First write zw=(ac−bd)+(ad+bc)i by the multiplication formula, then conjugate: the real part stays and the imaginary coefficient changes sign."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "complex_mul_formula",
            "mko_modconj_def"
          ],
          "explanation": {
            "zh-Hant": "公式與共軛定義都是等式；再取一次共軛即回到 zw。",
            "en": "The formula and the definition of the conjugate are equations; conjugating again returns zw."
          }
        },
        "leaf_basis": "complex_mul_formula + mko_modconj_def"
      },
      {
        "id": "S05.2",
        "uses": [
          "S05.1",
          "complex_mul_formula",
          "neg_mul"
        ],
        "before_tex": "\\overline{z}\\cdot\\overline{w}=(a,-b)\\cdot(c,-d)",
        "after_tex": "(ac-(-b)(-d),\\,a(-d)+(-b)c)=(ac-bd,\\,-(ad+bc))",
        "relation": "equivalent",
        "rule": "neg_mul",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對 (a,−b) 與 (c,−d) 用乘法公式：實部 ac−(−b)(−d)=ac−bd（兩個負號相乘得正）；虛部係數 a(−d)+(−b)c=−ad−bc=−(ad+bc)。兩個分量都與 S05.1 相同，依有序對相等得 (zw)‾=z̄·w̄。S05.1 與 S05.2 合起來證得引理 L_conj_mul。",
          "en": "Apply the multiplication formula to (a,−b) and (c,−d): real part ac−(−b)(−d)=ac−bd (two minus signs give a plus); imaginary coefficient a(−d)+(−b)c=−ad−bc=−(ad+bc). Both components match S05.1, so by equality of ordered pairs (zw)‾=z̄·w̄. S05.1 and S05.2 together prove Lemma L_conj_mul."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "neg_mul",
            "complex_pair_eq",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "符號規則與分配律都是恆等式（N3、N4 以多項式格點核對），雙向成立。",
            "en": "The sign rules and distributivity are identities (checked on a grid in N3 and N4), so the step is two-way."
          }
        },
        "leaf_basis": "neg_mul + distributive + complex_pair_eq"
      },
      {
        "id": "S06",
        "uses": [
          "S03",
          "S05"
        ],
        "before_tex": "|zw|^2",
        "after_tex": "|z|^2\\,|w|^2",
        "relation": "equivalent",
        "rule": "L_modsq_mul",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由 S03 與 S05 推出模平方可乘（引理 L_modsq_mul），證明在 S06.1、S06.2。N5 另以 (ac−bd)²+(ad+bc)²=(a²+b²)(c²+d²) 的多項式核對作獨立交叉核對。",
          "en": "From S03 and S05, the modulus squared is multiplicative (Lemma L_modsq_mul), proved in S06.1 and S06.2. N5 cross-checks independently with the polynomial identity (ac−bd)²+(ad+bc)²=(a²+b²)(c²+d²)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_modsq_mul"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個實數，雙向成立。",
            "en": "Both sides are the same real number, so the step is two-way."
          }
        },
        "leaf_basis": "real_embedding + mul_assoc + mul_comm + complex_pair_eq"
      },
      {
        "id": "S06.1",
        "uses": [
          "S03",
          "S05",
          "mul_assoc",
          "mul_comm"
        ],
        "before_tex": "(|zw|^2,0)=(zw)\\,\\overline{zw}",
        "after_tex": "(zw)(\\overline{z}\\,\\overline{w})=(z\\overline{z})(w\\overline{w})",
        "relation": "equivalent",
        "rule": "mul_assoc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對複數 zw 套用 S03：(zw)·(zw)‾ 是 |zw|² 的嵌入。依 S05 把 (zw)‾ 換成 z̄·w̄，再用 ℂ 乘法的結合與交換律重排四個因子成 (z z̄)(w w̄)。",
          "en": "Apply S03 to the complex number zw: (zw)·(zw)‾ is the embedding of |zw|². Replace (zw)‾ by z̄·w̄ using S05, then regroup the four factors as (z z̄)(w w̄) by associativity and commutativity of multiplication in ℂ."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S03",
            "S05",
            "mul_assoc",
            "mul_comm"
          ],
          "explanation": {
            "zh-Hant": "等值代換與重排都是等式，反向依序還原。",
            "en": "Replacement of equals and regrouping are equations; undoing them in order restores the original."
          }
        },
        "leaf_basis": "eq_replace + mul_assoc + mul_comm"
      },
      {
        "id": "S06.2",
        "uses": [
          "S06.1",
          "S03",
          "real_embedding",
          "complex_pair_eq"
        ],
        "before_tex": "(z\\overline{z})(w\\overline{w})=(|z|^2,0)(|w|^2,0)",
        "after_tex": "(|z|^2|w|^2,\\,0)\\ \\Rightarrow\\ |zw|^2=|z|^2|w|^2",
        "relation": "implies",
        "rule": "real_embedding",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 S03，z z̄ 與 w w̄ 分別是 |z|² 與 |w|² 的嵌入；兩個嵌入實數相乘得 (|z|²|w|²,0)。比較實部得 |zw|²=|z|²|w|²。S06.1 與 S06.2 合起來證得引理 L_modsq_mul。",
          "en": "By S03, z z̄ and w w̄ are the embeddings of |z|² and |w|²; the product of two embedded reals is (|z|²|w|²,0). Comparing real parts gives |zw|²=|z|²|w|². S06.1 and S06.2 together prove Lemma L_modsq_mul."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "real_embedding",
            "complex_pair_eq"
          ],
          "explanation": {
            "zh-Hant": "比較實部用到有序對相等；此步只取等式的實部，不宣稱其他推論。",
            "en": "Comparing real parts uses equality of ordered pairs; the step takes the real part of the equation and claims nothing else."
          }
        },
        "leaf_basis": "real_embedding + complex_pair_eq"
      },
      {
        "id": "S07",
        "uses": [
          "G2",
          "S01"
        ],
        "before_tex": "z=3-4i\\leftrightarrow(3,-4)",
        "after_tex": "\\overline{z}=3+4i\\leftrightarrow(3,4)",
        "relation": "equivalent",
        "rule": "mko_modconj_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "只把虛部係數 −4 變號成 4，實部 3 不變。寫成 −3−4i 或 −3+4i 都是錯的（前者改了實部、後者等於 −z）。",
          "en": "Only the imaginary coefficient −4 changes sign to 4; the real part 3 is unchanged. Writing −3−4i or −3+4i is wrong (the first changes the real part, the second is −z)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_modconj_def"
          ],
          "explanation": {
            "zh-Hant": "共軛兩次回到原數：(3,4) 的共軛是 (3,−4)=z。",
            "en": "Conjugating twice returns the number: the conjugate of (3,4) is (3,−4)=z."
          }
        },
        "leaf_basis": "mko_modconj_def"
      },
      {
        "id": "S08",
        "uses": [
          "S07",
          "S02",
          "complex_mul_formula"
        ],
        "before_tex": "|z|^2=3^2+(-4)^2,\\ z\\overline{z}=(3,-4)(3,4)",
        "after_tex": "|z|^2=25,\\ z\\overline{z}=(3\\cdot3-(-4)\\cdot4,\\,3\\cdot4+(-4)\\cdot3)=(25,0)",
        "relation": "implies",
        "rule": "complex_mul_formula",
        "conditions": [],
        "explanation": {
          "zh-Hant": "模平方 9+16=25。直接以乘法公式算 z·z̄：實部 9+16=25，虛部係數 12−12=0，得 (25,0)，即 25 的實數嵌入，與 S02、S03 的一般結論一致。",
          "en": "The modulus squared is 9+16=25. Computing z·z̄ directly by the formula: real part 9+16=25, imaginary coefficient 12−12=0, giving (25,0), the real embedding of 25, in line with the general results S02 and S03."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "complex_mul_formula",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "這是一般定理的一個實例（N6 到 N8），用來核對計算，不取代 S02 的證明。",
            "en": "This is one instance of the general theorem (N6 to N8), used to check the computation; it does not replace the proof in S02."
          }
        },
        "leaf_basis": "complex_mul_formula + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S08",
          "modulus_def",
          "sqrt_def"
        ],
        "before_tex": "|z|^2=25",
        "after_tex": "|z|=\\sqrt{25}=5\\neq 25",
        "relation": "implies",
        "rule": "sqrt_def",
        "conditions": [
          "5\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "模是模平方的非負平方根。5≥0 且 5²=25，依平方根的唯一性，|z|=5。模平方 25 與模 5 是不同的數，這正是 MKO 指出的誤解。",
          "en": "The modulus is the nonnegative square root of the modulus squared. Since 5≥0 and 5²=25, uniqueness of the square root gives |z|=5. The modulus squared 25 and the modulus 5 are different numbers, which is the mistake the MKO names."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "modulus_def",
            "sqrt_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "需要 25 是完全平方數，才能寫出精確整數 5；對 w 不成立（S10）。",
            "en": "It needs 25 to be a perfect square for the exact integer 5 to appear; this fails for w (S10)."
          }
        },
        "leaf_basis": "sqrt_def + modulus_def + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "G2",
          "G3",
          "S01"
        ],
        "before_tex": "w=1+2i",
        "after_tex": "\\overline{w}=1-2i,\\ |w|^2=5,\\ |w|=\\sqrt5\\notin\\mathbb{Z}",
        "relation": "implies",
        "rule": "mko_modconj_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "w̄=(1,−2)，|w|²=1+4=5。5 不是完全平方數：非負整數 n≤2 時 n²≤4，n≥3 時 n²≥9，沒有整數平方等於 5。所以 |w|=√5 不是整數；依 G3 保持精確的模平方 5，不寫 2.236 之類的小數。",
          "en": "w̄=(1,−2) and |w|²=1+4=5. 5 is not a perfect square: a nonnegative integer n≤2 has n²≤4 and n≥3 has n²≥9, so no integer squares to 5. Hence |w|=√5 is not an integer; by G3 the exact modulus squared 5 is kept and no decimal such as 2.236 is written."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_modconj_def",
            "order_basics",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "非完全平方的論證只分 n≤2 與 n≥3 兩種情形並涵蓋所有非負整數；本稿不宣稱 √5 的其他性質。",
            "en": "The non-square argument splits into n≤2 and n≥3, which cover all nonnegative integers; nothing further about √5 is claimed."
          }
        },
        "leaf_basis": "mko_modconj_def + order_basics + case_analysis + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "G2",
          "complex_mul_formula"
        ],
        "before_tex": "zw=(3,-4)(1,2)",
        "after_tex": "(3\\cdot1-(-4)\\cdot2,\\,3\\cdot2+(-4)\\cdot1)=(11,2)",
        "relation": "equivalent",
        "rule": "complex_mul_formula",
        "conditions": [],
        "explanation": {
          "zh-Hant": "實部 3+8=11，虛部係數 6−4=2，所以 zw=11+2i。",
          "en": "Real part 3+8=11, imaginary coefficient 6−4=2, so zw=11+2i."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "complex_mul_formula",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "公式的精確求值（N13、N14），兩邊同一個數。",
            "en": "Exact evaluation of the formula (N13, N14); both sides are the same number."
          }
        },
        "leaf_basis": "complex_mul_formula + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S11",
          "S07",
          "S10",
          "S05"
        ],
        "before_tex": "\\overline{zw}\\ \\text{vs}\\ \\overline{z}\\,\\overline{w}=(3,4)(1,-2)",
        "after_tex": "(11,-2)=(3\\cdot1-4\\cdot(-2),\\,3\\cdot(-2)+4\\cdot1)",
        "relation": "implies",
        "rule": "complex_mul_formula",
        "conditions": [],
        "explanation": {
          "zh-Hant": "左邊：(zw)‾=(11,−2)。右邊：z̄·w̄=(3,4)(1,−2)，實部 3+8=11，虛部係數 −6+4=−2。兩邊相同，是 S05 的實例。",
          "en": "Left: (zw)‾=(11,−2). Right: z̄·w̄=(3,4)(1,−2), real part 3+8=11, imaginary coefficient −6+4=−2. The two sides agree, an instance of S05."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "complex_mul_formula",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "實例只核對計算（N15、N16），一般結論來自 S05。",
            "en": "The instance only checks the computation (N15, N16); the general conclusion comes from S05."
          }
        },
        "leaf_basis": "complex_mul_formula + complex_pair_eq + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S11",
          "S08",
          "S10",
          "S06"
        ],
        "before_tex": "|zw|^2=11^2+2^2",
        "after_tex": "125=25\\cdot5=|z|^2|w|^2",
        "relation": "implies",
        "rule": "L_modsq_mul",
        "conditions": [],
        "explanation": {
          "zh-Hant": "121+4=125，而 |z|²|w|²=25·5=125，兩者相同，是 S06 的實例。即使 |w| 不是整數，模平方仍保持精確整數。",
          "en": "121+4=125 and |z|²|w|²=25·5=125 agree, an instance of S06. Even though |w| is not an integer, the modulus squares stay exact integers."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_modsq_mul",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "實例只核對計算（N17）；一般結論來自 S06。",
            "en": "The instance only checks the computation (N17); the general conclusion comes from S06."
          }
        },
        "leaf_basis": "L_modsq_mul + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "G2",
          "S04",
          "S07",
          "S10"
        ],
        "before_tex": "\\overline{z+w}=\\overline{(4,-2)}",
        "after_tex": "(4,2)=(3,4)+(1,-2)=\\overline{z}+\\overline{w}",
        "relation": "implies",
        "rule": "complex_addsub",
        "conditions": [],
        "explanation": {
          "zh-Hant": "z+w=(3+1, −4+2)=(4,−2)，共軛為 (4,2)；z̄+w̄=(3+1, 4−2)=(4,2)。兩邊相同，是 S04 的實例。",
          "en": "z+w=(3+1, −4+2)=(4,−2), whose conjugate is (4,2); z̄+w̄=(3+1, 4−2)=(4,2). The sides agree, an instance of S04."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "complex_addsub",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "實例核對，一般結論來自 S04。",
            "en": "An instance check; the general conclusion comes from S04."
          }
        },
        "leaf_basis": "complex_addsub + numeric_evaluation"
      },
      {
        "id": "S15",
        "uses": [
          "S03",
          "S04",
          "S05",
          "S06"
        ],
        "before_tex": "z=a+bi,\\ w=c+di",
        "after_tex": "z\\overline{z}=(|z|^2,0),\\ \\overline{z+w}=\\overline{z}+\\overline{w},\\ \\overline{zw}=\\overline{z}\\,\\overline{w},\\ |zw|^2=|z|^2|w|^2",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "合併四個一般結論；它們對所有實數分量成立，S07 到 S14 的整數例只是核對。",
          "en": "Combine the four general results; they hold for all real components, and the integer examples of S07 to S14 are only checks."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S03",
            "S04",
            "S05",
            "S06",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：四個一般命題。足夠條件 H：S03、S04、S05、S06 各自已對任意 a、b、c、d 證明；H⇒G 由合取引入。沒有把 G 當作前提，也沒有用例子代替證明。",
            "en": "Goal G: the four general statements. Sufficient condition H: S03, S04, S05 and S06, each proved for arbitrary a, b, c, d; H⇒G by conjunction introduction. G is never used as a premise, and no example stands in for a proof."
          }
        },
        "leaf_basis": "and_intro_elim + forall_intro"
      }
    ],
    "result": {
      "zh-Hant": "對所有實數 a、b、c、d：z·z̄=(a²+b²)+0i 是模平方的實數嵌入且非負；共軛保持加法與乘法；|zw|²=|z|²|w|²。例子 z=3−4i、w=1+2i：z̄=3+4i，|z|²=25，z·z̄=(25,0)，|z|=5（與模平方 25 不同）；w̄=1−2i，|w|²=5（|w|=√5 不是整數，保持精確模平方）；zw=11+2i，(zw)‾=11−2i=z̄·w̄，|zw|²=125=25·5。",
      "en": "For all reals a, b, c, d: z·z̄=(a²+b²)+0i is the real embedding of the modulus squared and is nonnegative; conjugation preserves addition and multiplication; |zw|²=|z|²|w|². For z=3−4i and w=1+2i: z̄=3+4i, |z|²=25, z·z̄=(25,0), |z|=5 (different from the modulus squared 25); w̄=1−2i, |w|²=5 (|w|=√5 is not an integer, so the exact modulus squared is kept); zw=11+2i, (zw)‾=11−2i=z̄·w̄, and |zw|²=125=25·5."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "3−4i 的共軛是 −3−4i",
          "en": "The conjugate of 3−4i is −3−4i"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "共軛只改虛部係數的符號：3+4i。",
          "en": "Conjugation changes only the sign of the imaginary coefficient: 3+4i."
        }
      },
      {
        "claim": {
          "zh-Hant": "|3−4i|=25",
          "en": "|3−4i|=25"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "25 是模平方；模是 √25=5。",
          "en": "25 is the modulus squared; the modulus is √25=5."
        }
      },
      {
        "claim": {
          "zh-Hant": "|1+2i|=2.236 是精確值",
          "en": "|1+2i|=2.236 exactly"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "2.236²=4.999696≠5；精確的模平方是 5，模是 √5。",
          "en": "2.236²=4.999696≠5; the exact modulus squared is 5 and the modulus is √5."
        }
      },
      {
        "claim": {
          "zh-Hant": "(3−4i)(3+4i)=9−16=−7",
          "en": "(3−4i)(3+4i)=9−16=−7"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "i²=−1 使結果為 9+16=25，且 z·z̄ 恆非負。",
          "en": "i²=−1 makes the product 9+16=25, and z·z̄ is always nonnegative."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "mul_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法結合律：(uv)w=u(vw)。",
          "en": "Associativity of multiplication: (uv)w=u(vw)."
        }
      },
      {
        "id": "mul_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。",
          "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)."
        }
      },
      {
        "id": "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": "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": "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": "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": "power_laws",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。",
          "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1."
        }
      },
      {
        "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": "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": "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": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "id": "mko_modconj_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-complex-modulus-conjugate",
          "sha256": "c4c1f275b4ae1708bd308a1ea5ce57ee90a10921251139a1eb880ffd2101842d",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對 z=a+bi，共軛為 a−bi，模平方為 a²+b²，且 z 乘共軛等於模平方的實數嵌入。共軛保持加法與乘法。",
          "en": "For z=a+b*i, the conjugate is a−b*i and modulus squared is a²+b². Multiplying z by its conjugate gives the real embedding of that squared modulus. Conjugation preserves addition and multiplication."
        }
      },
      {
        "id": "complex_mul_formula",
        "kind": "cited",
        "source": {
          "mko_id": "mko-complex-number",
          "sha256": "a0d0b2223ccef6be06ad99f3ba6f05b76e1570d35ae43f7caac2509712c0431a",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "複數乘法由分配律與 i²=−1 導出：(a+bi)(c+di) 的實部為 ac−bd，虛部係數為 ad+bc。",
          "en": "Complex multiplication follows from distributivity and i²=−1: the product (a+bi)(c+di) has real part ac−bd and imaginary coefficient ad+bc."
        }
      },
      {
        "id": "complex_addsub",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逐分量加減：(a+bi)±(c+di)=(a±c)+(b±d)i。",
          "en": "Componentwise addition and subtraction: (a+bi)±(c+di)=(a±c)+(b±d)i."
        }
      },
      {
        "id": "complex_pair_eq",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "對實數 a、b、c、d：a+bi=c+di 當且僅當 a=c 且 b=d。",
          "en": "For reals a, b, c, d: a+bi=c+di if and only if a=c and b=d."
        }
      },
      {
        "id": "real_embedding",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數 r 嵌入為 r+0i，即有序對 (r,0)；兩個嵌入實數的乘積 (r,0)(s,0)=(rs,0)。",
          "en": "A real r is embedded as r+0i, the ordered pair (r,0); the product of two embedded reals is (r,0)(s,0)=(rs,0)."
        }
      },
      {
        "id": "modulus_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "模 |z| 是模平方 a²+b² 的非負平方根：|z|=√(a²+b²)。",
          "en": "The modulus |z| is the nonnegative square root of the modulus squared a²+b²: |z|=√(a²+b²)."
        }
      },
      {
        "id": "L_conj_mul",
        "kind": "lemma",
        "proved_in": [
          "S05.1",
          "S05.2"
        ],
        "statement": {
          "zh-Hant": "對所有複數 z、w：(zw)‾=z̄·w̄。",
          "en": "For all complex numbers z, w: (zw)‾=z̄·w̄."
        }
      },
      {
        "id": "L_modsq_mul",
        "kind": "lemma",
        "proved_in": [
          "S06.1",
          "S06.2"
        ],
        "statement": {
          "zh-Hant": "對所有複數 z、w：|zw|²=|z|²·|w|²。",
          "en": "For all complex numbers z, w: |zw|²=|z|²·|w|²."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02.1",
        "vars": [
          "a",
          "b"
        ],
        "degree": 2,
        "lhs": "a*a-b*(-b)",
        "rhs": "a*a+b*b"
      },
      {
        "id": "N2",
        "step": "S02.2",
        "vars": [
          "a",
          "b"
        ],
        "degree": 1,
        "lhs": "a*(-b)+b*a",
        "rhs": "0"
      },
      {
        "id": "N3",
        "step": "S05.2",
        "vars": [
          "a",
          "b",
          "c",
          "d"
        ],
        "degree": 1,
        "lhs": "a*c-(-b)*(-d)",
        "rhs": "a*c-b*d"
      },
      {
        "id": "N4",
        "step": "S05.2",
        "vars": [
          "a",
          "b",
          "c",
          "d"
        ],
        "degree": 1,
        "lhs": "a*(-d)+(-b)*c",
        "rhs": "-(a*d+b*c)"
      },
      {
        "id": "N5",
        "step": "S06",
        "vars": [
          "a",
          "b",
          "c",
          "d"
        ],
        "degree": 2,
        "lhs": "(a*c-b*d)**2+(a*d+b*c)**2",
        "rhs": "(a*a+b*b)*(c*c+d*d)"
      },
      {
        "id": "N6",
        "step": "S08",
        "lhs": "3**2+(-4)**2",
        "rhs": "25"
      },
      {
        "id": "N7",
        "step": "S08",
        "lhs": "3*3-(-4)*4",
        "rhs": "25"
      },
      {
        "id": "N8",
        "step": "S08",
        "lhs": "3*4+(-4)*3",
        "rhs": "0"
      },
      {
        "id": "N9",
        "step": "S09",
        "lhs": "5**2",
        "rhs": "25"
      },
      {
        "id": "N10",
        "step": "S10",
        "lhs": "1**2+2**2",
        "rhs": "5"
      },
      {
        "id": "N11",
        "step": "S10",
        "lhs": "2**2",
        "relation": "<",
        "rhs": "5"
      },
      {
        "id": "N12",
        "step": "S10",
        "lhs": "5",
        "relation": "<",
        "rhs": "3**2"
      },
      {
        "id": "N13",
        "step": "S11",
        "lhs": "3*1-(-4)*2",
        "rhs": "11"
      },
      {
        "id": "N14",
        "step": "S11",
        "lhs": "3*2+(-4)*1",
        "rhs": "2"
      },
      {
        "id": "N15",
        "step": "S12",
        "lhs": "3*1-4*(-2)",
        "rhs": "11"
      },
      {
        "id": "N16",
        "step": "S12",
        "lhs": "3*(-2)+4*1",
        "rhs": "-2"
      },
      {
        "id": "N17",
        "step": "S13",
        "lhs": "11**2+2**2",
        "rhs": "25*5"
      },
      {
        "id": "N18",
        "lhs": "2.236*2.236",
        "relation": "!=",
        "rhs": "5"
      },
      {
        "id": "N19",
        "lhs": "9-16",
        "relation": "!=",
        "rhs": "25"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "一般證明對實數分量成立；例子用整數有序對（MKO 假設 a1），只算共軛與模平方，不以小數平方根近似冒充精確模。僅在 |z|²=25 為完全平方數時說明 |z|=5。不處理 argument、極式或複可微性（MKO 假設 a2）；不改動 canonical MKO 或英文 edition。",
      "en": "The general proofs hold for real components; the examples use integer ordered pairs (MKO assumption a1) and compute only conjugates and modulus squares, without passing off decimal square-root approximations as exact moduli. Only where |z|²=25 is a perfect square is |z|=5 stated. Arguments, polar form and complex differentiability are not treated (MKO assumption a2); the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "z=a+bi\\leftrightarrow(a,b)",
        "複數與其有序對",
        "a complex number and its ordered pair"
      ],
      [
        "\\overline{z}=a-bi",
        "z 的共軛",
        "the conjugate of z"
      ],
      [
        "|z|^2=a^2+b^2",
        "z 的模平方",
        "the modulus squared of z"
      ],
      [
        "|z|=\\sqrt{a^2+b^2}",
        "z 的模（非負平方根）",
        "the modulus of z (nonnegative square root)"
      ],
      [
        "w=c+di",
        "第二個複數",
        "the second complex number"
      ]
    ],
    "md_error_controls": [
      {
        "title": "共軛只改虛部的符號 / Conjugation changes only the sign of the imaginary part",
        "zh": "MKO 指出的誤解之一是只改實部符號。3−4i 的共軛是 3+4i；寫成 −3−4i 改錯了分量，寫成 −3+4i 則是 −z 而非 z̄。",
        "en": "One mistake the MKO names is changing the sign of the real part. The conjugate of 3−4i is 3+4i; writing −3−4i changes the wrong component, and −3+4i is −z, not z̄."
      },
      {
        "title": "分清模平方與模 / Distinguish the modulus squared from the modulus",
        "zh": "MKO 的另一個誤解是把模平方當成模。|3−4i|²=25 而 |3−4i|=5；|1+2i|²=5 而 |1+2i|=√5 不是整數。寫成 2.236 只是近似：2.236²=4.999696≠5（N18）。本稿依 MKO 保留精確的模平方。",
        "en": "Another mistake identified in the MKO is taking the modulus squared for the modulus. |3−4i|²=25 but |3−4i|=5; |1+2i|²=5 but |1+2i|=√5 is not an integer. Writing 2.236 is only an approximation: 2.236²=4.999696≠5 (N18). Following the MKO, this guide keeps the exact modulus squared."
      },
      {
        "title": "z·z̄ 的實部是 a²+b²，不是 a²−b² / The real part of z·z̄ is a²+b², not a²−b²",
        "zh": "套用平方差 (a+bi)(a−bi)=a²−(bi)² 時，最容易漏掉 (bi)²=b²i²=−b²，於是誤得 a²−b²；對 z=3−4i 會得 9−16=−7。S02.1 顯示正號來自 −b·(−b)=+b²。",
        "en": "Using the difference of squares (a+bi)(a−bi)=a²−(bi)², the easiest slip is forgetting (bi)²=b²i²=−b², which gives the wrong a²−b²; for z=3−4i that is 9−16=−7. S02.1 shows that the plus sign comes from −b·(−b)=+b²."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "ℂ 中乘法的結合與交換律（S06.1 使用）是前提，沒有在本稿由有序對定義推出；本稿不處理複數除法、argument 與極式（MKO 的 formal next）。",
        "en": "Associativity and commutativity of multiplication in ℂ (used in S06.1) are prerequisites and are not derived here from the ordered-pair definition; complex division, arguments and polar form (the MKO's formal next) are not treated."
      },
      {
        "zh-Hant": "N1 到 N5 是多項式恆等式的格點核對，屬交叉核對；一般結論由 S02 到 S06 的代數論證證明。",
        "en": "N1 to N5 are polynomial-identity checks on grids, serving as cross-checks; the general results are proved by the algebra of S02 to S06."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "4d35d95477cfaf8ae77bb1db462786967900da76e6b8489990bb52fed2f61c28",
    "original_source_sha256": "4d35d95477cfaf8ae77bb1db462786967900da76e6b8489990bb52fed2f61c28",
    "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."
  }
}