{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-complex-number-01",
  "mko_id": "mko-complex-number",
  "title": {
    "zh-Hant": "複數：以有序對計算 (2+3i)(4−5i) 並判定相等",
    "en": "Complex numbers: computing (2+3i)(4−5i) as an ordered pair and deciding equality"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-complex-number-01",
    "source": {
      "mko_id": "mko-complex-number",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "a0d0b2223ccef6be06ad99f3ba6f05b76e1570d35ae43f7caac2509712c0431a",
      "english_sha256": "b38702b90e90f16fa680c51c92fc226a64a08dabcfed32e524b7b9465b4e9a75"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "distributive",
      "mul_assoc",
      "mul_comm",
      "add_comm",
      "add_assoc",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "設 z=2+3i、w=4−5i。以有序對（實部, 虛部係數）求 z+w、z−w 與 z·w。乘法先由分配律與 i²=−1 推出一般公式 (a+bi)(c+di)=(ac−bd)+(ad+bc)i，再代入數值，並以直接展開交叉核對；最後判定 z·w 是否等於實數 23。",
      "en": "Let z=2+3i and w=4−5i. Find z+w, z−w and z·w as ordered pairs (real part, imaginary coefficient). For the product, first derive the general formula (a+bi)(c+di)=(ac−bd)+(ad+bc)i from distributivity and i²=−1, then substitute the numbers and cross-check by direct expansion; finally decide whether z·w equals the real number 23."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "z=2+3i，w=4−5i；實部與虛部係數都是整數。",
          "en": "z=2+3i and w=4−5i; the real parts and imaginary coefficients are integers."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "答案以有序對（實部, 虛部係數）表示；分量次序是語義資料，不由圖形位置或未標刻度補值；本稿不做複數除法（MKO 假設 a1、a2）。",
          "en": "Answers are ordered pairs (real part, imaginary coefficient); the order of the components is semantic data and no value is read off a picture or an unmarked scale; complex division is not used (MKO assumptions a1, a2)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "z=2+3i,\\ w=4-5i",
        "after_tex": "z\\leftrightarrow(2,3),\\ w\\leftrightarrow(4,-5)",
        "relation": "definition",
        "rule": "mko_complex_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，複數 a+bi 由實部 a 與虛部係數 b 唯一決定，對應有序對 (a,b)。z 的實部是 2、虛部係數是 3；w=4−5i=4+(−5)i，虛部係數是 −5，負號屬於係數本身，不能丟掉。",
          "en": "By the MKO, a complex number a+bi is determined by its real part a and imaginary coefficient b and corresponds to the ordered pair (a,b). z has real part 2 and imaginary coefficient 3; w=4−5i=4+(−5)i has imaginary coefficient −5: the minus sign belongs to the coefficient and must not be dropped."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_complex_def",
            "complex_pair_eq"
          ],
          "explanation": {
            "zh-Hant": "此步只把給定資料寫成有序對，不推出新事實；對應是一一的（complex_pair_eq），所以之後用有序對計算不會失去資訊。",
            "en": "This step only rewrites the givens as ordered pairs and derives no new fact; the correspondence is one-to-one (complex_pair_eq), so computing with ordered pairs loses no information."
          }
        },
        "leaf_basis": "mko_complex_def + complex_pair_eq"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "complex_addsub"
        ],
        "before_tex": "z+w",
        "after_tex": "(2+4)+(3+(-5))i=6-2i\\leftrightarrow(6,-2)",
        "relation": "equivalent",
        "rule": "complex_addsub",
        "conditions": [],
        "explanation": {
          "zh-Hant": "加法逐分量：實部 2+4=6，虛部係數 3+(−5)=−2，所以 z+w=6−2i，有序對為 (6,−2)。",
          "en": "Addition is componentwise: real part 2+4=6, imaginary coefficient 3+(−5)=−2, so z+w=6−2i, the ordered pair (6,−2)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "complex_addsub",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "反向：(6−2i)−w=(6−4)+(−2−(−5))i=2+3i=z，還原了 z+w=6−2i；兩式表示同一個複數。",
            "en": "Backward: (6−2i)−w=(6−4)+(−2−(−5))i=2+3i=z, which restores z+w=6−2i; both expressions name the same complex number."
          }
        },
        "leaf_basis": "complex_addsub + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "complex_addsub"
        ],
        "before_tex": "z-w",
        "after_tex": "(2-4)+(3-(-5))i=-2+8i\\leftrightarrow(-2,8)",
        "relation": "equivalent",
        "rule": "complex_addsub",
        "conditions": [],
        "explanation": {
          "zh-Hant": "減法也逐分量：實部 2−4=−2，虛部係數 3−(−5)=3+5=8。減去負數等於加上它的相反數，這是最容易弄錯符號的地方。",
          "en": "Subtraction is also componentwise: real part 2−4=−2, imaginary coefficient 3−(−5)=3+5=8. Subtracting a negative number adds its opposite; this is where signs are most easily lost."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "complex_addsub",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "反向：(−2+8i)+w=(−2+4)+(8+(−5))i=2+3i=z，還原了 z−w=−2+8i。",
            "en": "Backward: (−2+8i)+w=(−2+4)+(8+(−5))i=2+3i=z, which restores z−w=−2+8i."
          }
        },
        "leaf_basis": "complex_addsub + add_inverse + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "mko_complex_def",
          "i_squared"
        ],
        "before_tex": "(a+bi)(c+di)",
        "after_tex": "(ac-bd)+(ad+bc)i",
        "relation": "equivalent",
        "rule": "L_complex_mul",
        "conditions": [
          "a,b,c,d\\in\\mathbb{R}"
        ],
        "explanation": {
          "zh-Hant": "MKO 說乘法由分配律與 i²=−1 導出，而不是逐分量相乘。先對任意實數 a、b、c、d 證明一般公式（引理 L_complex_mul），證明在子步 S04.1 到 S04.3；之後的乘積都由這個公式計算。",
          "en": "The MKO says multiplication follows from distributivity and i²=−1; it is not componentwise. First prove the general formula for arbitrary reals a, b, c, d (Lemma L_complex_mul) in sub-steps S04.1 to S04.3; the products below are then computed with it."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_complex_mul"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個複數的兩種寫法；子步中每一步都是等值改寫，所以整條等式雙向成立。",
            "en": "Both sides are two ways of writing the same complex number; every sub-step is a rewriting by equals, so the identity holds in both directions."
          }
        },
        "leaf_basis": "distributive + mul_assoc + mul_comm + i_squared + add_comm"
      },
      {
        "id": "S04.1",
        "uses": [
          "distributive",
          "mul_comm"
        ],
        "before_tex": "(a+bi)(c+di)",
        "after_tex": "ac+a(di)+(bi)c+(bi)(di)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對第二個因子用分配律：(a+bi)(c+di)=(a+bi)c+(a+bi)(di)；再對每一項各用一次分配律（右側分配用交換律轉成左側分配），得到四項 ac、a(di)、(bi)c、(bi)(di)。",
          "en": "Distribute over the second factor: (a+bi)(c+di)=(a+bi)c+(a+bi)(di); then distribute once more in each term (right distribution is turned into left distribution by commutativity), giving the four terms ac, a(di), (bi)c and (bi)(di)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "mul_comm"
          ],
          "explanation": {
            "zh-Hant": "反向是依分配律把四項合併回兩個因子的乘積；分配律是等式，兩個方向都成立。",
            "en": "Backward, the four terms are recombined into the product of the two factors by distributivity; distributivity is an equation, so both directions hold."
          }
        },
        "leaf_basis": "distributive + mul_comm"
      },
      {
        "id": "S04.2",
        "uses": [
          "S04.1",
          "mul_assoc",
          "mul_comm"
        ],
        "before_tex": "ac+a(di)+(bi)c+(bi)(di)",
        "after_tex": "ac+(ad)i+(bc)i+(bd)i^2",
        "relation": "equivalent",
        "rule": "mul_assoc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "用乘法結合律與交換律重排每一項的因子：a(di)=(ad)i，(bi)c=(bc)i，(bi)(di)=(bd)(i·i)=(bd)i²。只改變因子的分組與次序，不改變值。",
          "en": "Regroup and reorder the factors of each term by associativity and commutativity of multiplication: a(di)=(ad)i, (bi)c=(bc)i and (bi)(di)=(bd)(i·i)=(bd)i². Only the grouping and order of the factors change, not the value."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_assoc",
            "mul_comm"
          ],
          "explanation": {
            "zh-Hant": "結合律與交換律是對所有複數成立的恆等式，反向重排即回到原式。",
            "en": "Associativity and commutativity are identities for all complex numbers; reordering backwards returns the original expression."
          }
        },
        "leaf_basis": "mul_assoc + mul_comm"
      },
      {
        "id": "S04.3",
        "uses": [
          "S04.2",
          "i_squared",
          "distributive"
        ],
        "before_tex": "ac+(ad)i+(bc)i+(bd)i^2",
        "after_tex": "(ac-bd)+(ad+bc)i",
        "relation": "equivalent",
        "rule": "i_squared",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代入 i²=−1：(bd)i²=(bd)(−1)=−bd，這一項不含 i，屬於實部。再用加法交換律與結合律把實數項 ac 與 −bd 放在一起，並用分配律把 (ad)i+(bc)i 合併成 (ad+bc)i。S04.1 到 S04.3 合起來證得引理 L_complex_mul；實部中的負號 −bd 正是來自 i²=−1。",
          "en": "Substitute i²=−1: (bd)i²=(bd)(−1)=−bd, a term without i, so it belongs to the real part. Then commutativity and associativity of addition put the real terms ac and −bd together, and distributivity combines (ad)i+(bc)i into (ad+bc)i. S04.1 to S04.3 together prove Lemma L_complex_mul; the minus sign in −bd comes exactly from i²=−1."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "i_squared",
            "distributive",
            "add_comm",
            "add_assoc"
          ],
          "explanation": {
            "zh-Hant": "每一步都把式子換成相等的式子（i² 換成 −1、項的重排與合併），反向依序還原，雙向成立。",
            "en": "Each move replaces an expression by an equal one (i² by −1, reordering and combining terms); undoing them in order restores the original, so the step is two-way."
          }
        },
        "leaf_basis": "i_squared + neg_mul + add_comm + add_assoc + distributive"
      },
      {
        "id": "S05",
        "uses": [
          "S01",
          "S04"
        ],
        "before_tex": "zw=(2+3i)(4+(-5)i)",
        "after_tex": "zw=(2\\cdot 4-3\\cdot(-5))+(2\\cdot(-5)+3\\cdot 4)i",
        "relation": "theorem_application",
        "rule": "L_complex_mul",
        "conditions": [
          "a=2,\\ b=3,\\ c=4,\\ d=-5"
        ],
        "explanation": {
          "zh-Hant": "把 a=2、b=3、c=4、d=−5 代入引理。d 必須取 −5 而不是 5：w 的虛部係數要連同符號一起代入。",
          "en": "Substitute a=2, b=3, c=4, d=−5 into the lemma. d must be −5, not 5: the imaginary coefficient of w is substituted together with its sign."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "L_complex_mul",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "引理對所有實數成立，整數 2、3、4、−5 是實數特例，所以可以代入；此步不宣稱可以由乘積反推出因子。",
            "en": "The lemma holds for all real numbers, including the integers 2, 3, 4, −5, so they may be substituted; the step does not claim that the factors can be recovered from the product."
          }
        },
        "leaf_basis": "L_complex_mul + forall_elim"
      },
      {
        "id": "S06",
        "uses": [
          "S05"
        ],
        "before_tex": "(2\\cdot 4-3\\cdot(-5))+(2\\cdot(-5)+3\\cdot 4)i",
        "after_tex": "23+2i",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "實部：8−(−15)=8+15=23；虛部係數：−10+12=2。所以 zw=23+2i。",
          "en": "Real part: 8−(−15)=8+15=23; imaginary coefficient: −10+12=2. So zw=23+2i."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "整數算式的精確求值，可獨立重算（N5、N6）；兩邊是同一個數，雙向成立。",
            "en": "Exact evaluation of integer expressions, independently recomputable (N5, N6); both sides are the same number, so the step is two-way."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S01",
          "distributive",
          "i_squared"
        ],
        "before_tex": "(2+3i)(4-5i)",
        "after_tex": "8-10i+12i-15i^2=8+2i+15=23+2i",
        "relation": "equivalent",
        "rule": "i_squared",
        "conditions": [],
        "explanation": {
          "zh-Hant": "不用公式，直接展開作交叉核對：分配律給出 2·4+2·(−5i)+3i·4+3i·(−5i)=8−10i+12i−15i²；i²=−1 把 −15i² 變成 +15；合併得 23+2i，與 S06 相同。",
          "en": "Cross-check by direct expansion without the formula: distributivity gives 2·4+2·(−5i)+3i·4+3i·(−5i)=8−10i+12i−15i²; i²=−1 turns −15i² into +15; collecting gives 23+2i, the same as S06."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "i_squared",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "每個等號都是等值改寫；兩條獨立路徑得到同一結果，可抓出抄寫或符號錯誤。",
            "en": "Every equals sign is a rewriting by equals; two independent routes give the same result, which catches copying and sign errors."
          }
        },
        "leaf_basis": "distributive + mul_assoc + mul_comm + i_squared + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S06",
          "complex_pair_eq"
        ],
        "before_tex": "zw=23+2i",
        "after_tex": "zw\\leftrightarrow(23,2)",
        "relation": "equivalent",
        "rule": "complex_pair_eq",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 G2 把答案寫成有序對（實部, 虛部係數）=(23,2)。次序是語義資料：(2,23) 代表 2+23i，是另一個複數。",
          "en": "By G2 the answer is written as the ordered pair (real part, imaginary coefficient)=(23,2). The order is semantic data: (2,23) stands for 2+23i, a different complex number."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "complex_pair_eq",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "a+bi 與 (a,b) 一一對應，所以寫成有序對與寫回 23+2i 互為反向。",
            "en": "a+bi and (a,b) correspond one-to-one, so writing the pair and writing back 23+2i are inverse moves."
          }
        },
        "leaf_basis": "complex_pair_eq"
      },
      {
        "id": "S09",
        "uses": [
          "S08",
          "complex_pair_eq"
        ],
        "before_tex": "(23,2)\\ \\text{vs}\\ (2,23)",
        "after_tex": "(23,2)\\neq(2,23)",
        "relation": "implies",
        "rule": "complex_pair_eq",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個有序對相等必須實部相等且虛部係數相等；23≠2，所以 (23,2)≠(2,23)。交換實部與虛部會得到另一個複數，這正是 MKO 指出的常見誤解。",
          "en": "Two ordered pairs are equal only if the real parts agree and the imaginary coefficients agree; 23≠2, so (23,2)≠(2,23). Swapping the real and imaginary parts gives a different complex number, which is the common mistake the MKO names."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "complex_pair_eq",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一個分量不同就足以推出不等；反向（不等推出某個指定分量不同）並未宣稱。",
            "en": "One differing component suffices for inequality; the converse (inequality forces a particular component to differ) is not claimed."
          }
        },
        "leaf_basis": "complex_pair_eq + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "real_embedding"
        ],
        "before_tex": "23\\in\\mathbb{R}",
        "after_tex": "23=23+0i\\leftrightarrow(23,0)",
        "relation": "definition",
        "rule": "real_embedding",
        "conditions": [],
        "explanation": {
          "zh-Hant": "要判定 zw 是否等於實數 23，先把 23 嵌入複數：23=23+0i，有序對為 (23,0)。比較必須在同一個集合 ℂ 中進行。",
          "en": "To decide whether zw equals the real number 23, first embed 23 in the complex numbers: 23=23+0i, the ordered pair (23,0). The comparison must take place in one set, ℂ."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "real_embedding"
          ],
          "explanation": {
            "zh-Hant": "此步只是嵌入的寫法，不推出新事實。",
            "en": "This step only writes out the embedding and derives no new fact."
          }
        },
        "leaf_basis": "real_embedding"
      },
      {
        "id": "S11",
        "uses": [
          "S08",
          "S10",
          "complex_pair_eq"
        ],
        "before_tex": "zw=23",
        "after_tex": "23=23\\ \\land\\ 2=0",
        "relation": "equivalent",
        "rule": "complex_pair_eq",
        "conditions": [],
        "explanation": {
          "zh-Hant": "zw=23 等價於 (23,2)=(23,0)，再等價於兩個分量分別相等：23=23 且 2=0。",
          "en": "zw=23 is equivalent to (23,2)=(23,0), which in turn is equivalent to both components agreeing: 23=23 and 2=0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "complex_pair_eq"
          ],
          "explanation": {
            "zh-Hant": "有序對相等的條件本身是雙向的，所以這一步兩個方向都成立。",
            "en": "The equality condition for ordered pairs is itself two-way, so the step holds in both directions."
          }
        },
        "leaf_basis": "complex_pair_eq + and_intro_elim"
      },
      {
        "id": "S12",
        "uses": [
          "S11"
        ],
        "before_tex": "23=23\\ \\land\\ 2=0",
        "after_tex": "zw\\neq 23",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "2=0 為假，所以合取為假；依 S11 的等價，zw=23 不成立。這正是 MKO 第三段：虛部係數不為零的複數不能等於任何實數，不論實部是否相同。",
          "en": "2=0 is false, so the conjunction is false, and by the equivalence S11, zw=23 fails. This is the MKO's third paragraph: a complex number with nonzero imaginary coefficient equals no real number, whatever its real part."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "contradiction",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "只用到一個分量 2≠0；實部相同（23=23）並不足以推出相等。",
            "en": "Only one component, 2≠0, is used; equal real parts (23=23) are not enough for equality."
          }
        },
        "leaf_basis": "contradiction + and_intro_elim + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S04",
          "G1"
        ],
        "before_tex": "wz=(4+(-5)i)(2+3i)",
        "after_tex": "wz=(4\\cdot 2-(-5)\\cdot 3)+(4\\cdot 3+(-5)\\cdot 2)i=23+2i",
        "relation": "theorem_application",
        "rule": "L_complex_mul",
        "conditions": [
          "a=4,\\ b=-5,\\ c=2,\\ d=3"
        ],
        "explanation": {
          "zh-Hant": "反過來算 wz 作另一個核對：引理給出實部 8+15=23、虛部係數 12−10=2，與 zw 相同，和 ℂ 乘法的交換律一致。",
          "en": "Compute wz in the other order as a further check: the lemma gives real part 8+15=23 and imaginary coefficient 12−10=2, the same as zw, consistent with commutativity of multiplication in ℂ."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_complex_mul",
            "mul_comm",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "這只是一組數的一致性核對，不用來證明交換律；交換律在本稿是前提。",
            "en": "This is a consistency check for one pair of numbers and does not prove commutativity, which is a prerequisite of this guide."
          }
        },
        "leaf_basis": "L_complex_mul + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S02",
          "S03",
          "S08",
          "S12"
        ],
        "before_tex": "z=2+3i,\\ w=4-5i",
        "after_tex": "z+w\\leftrightarrow(6,-2),\\ z-w\\leftrightarrow(-2,8),\\ zw\\leftrightarrow(23,2),\\ zw\\neq 23",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "合併四個結論：和 (6,−2)、差 (−2,8)、積 (23,2)，而且積不等於實數 23。",
          "en": "Combine the four conclusions: sum (6,−2), difference (−2,8), product (23,2), and the product is not the real number 23."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S02",
            "S03",
            "S08",
            "S12",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：三個運算結果與相等判定。足夠條件 H：S02、S03、S08、S12 四個已證結論；H⇒G 由合取引入。H 由給定資料正向推出，沒有把 G 當作前提。",
            "en": "Goal G: the three results and the equality decision. Sufficient condition H: the four proved conclusions S02, S03, S08 and S12; H⇒G by conjunction introduction. H is derived forwards from the givens; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "z+w=6−2i，即 (6,−2)；z−w=−2+8i，即 (−2,8)；z·w=23+2i，即 (23,2)，由一般公式 (a+bi)(c+di)=(ac−bd)+(ad+bc)i 與直接展開兩條路徑得到相同結果。因虛部係數 2≠0，z·w 不等於實數 23。",
      "en": "z+w=6−2i, that is (6,−2); z−w=−2+8i, that is (−2,8); z·w=23+2i, that is (23,2), obtained both from the general formula (a+bi)(c+di)=(ac−bd)+(ad+bc)i and by direct expansion. Since the imaginary coefficient 2≠0, z·w is not the real number 23."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "(2+3i)(4−5i)=8−15i，因為乘法逐分量進行",
          "en": "(2+3i)(4−5i)=8−15i, because multiplication is componentwise"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "乘法由分配律與 i²=−1 導出，正確結果是 23+2i。",
          "en": "Multiplication follows from distributivity and i²=−1; the correct product is 23+2i."
        }
      },
      {
        "claim": {
          "zh-Hant": "(2+3i)(4−5i)=−7+2i",
          "en": "(2+3i)(4−5i)=−7+2i"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "這是把 i² 當成 +1 的結果；−15i²=+15，實部是 8+15=23。",
          "en": "This takes i² as +1; in fact −15i²=+15 and the real part is 8+15=23."
        }
      },
      {
        "claim": {
          "zh-Hant": "zw 的有序對是 (2,23)",
          "en": "The ordered pair of zw is (2,23)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "次序是（實部, 虛部係數），zw=23+2i 對應 (23,2)，而 (2,23)≠(23,2)。",
          "en": "The order is (real part, imaginary coefficient); zw=23+2i corresponds to (23,2), and (2,23)≠(23,2)."
        }
      },
      {
        "claim": {
          "zh-Hant": "23+2i 的實部是 23，所以它等於實數 23",
          "en": "23+2i has real part 23, so it equals the real number 23"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "23 嵌入為 (23,0)，相等還需要 2=0，不成立。",
          "en": "23 embeds as (23,0), and equality would also need 2=0, which fails."
        }
      }
    ],
    "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": "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": "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": "add_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。",
          "en": "Associativity of addition: (u+v)+w=u+(v+w)."
        }
      },
      {
        "id": "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": "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": "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": "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": "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": "mko_complex_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-complex-number",
          "sha256": "a0d0b2223ccef6be06ad99f3ba6f05b76e1570d35ae43f7caac2509712c0431a",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "複數 z=a+bi 由實部 a 與虛部係數 b 唯一決定，可視為複平面有序對 (a,b)。加減逐分量進行，乘法由分配律與 i²=−1 導出。",
          "en": "A complex number z=a+b*i is uniquely determined by its real part a and imaginary coefficient b and can be viewed as the ordered pair (a,b) in the complex plane. Addition and subtraction are componentwise; multiplication follows from distributivity and i²=−1."
        }
      },
      {
        "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；因此 a+bi 與有序對 (a,b) 一一對應。",
          "en": "For reals a, b, c, d: a+bi=c+di if and only if a=c and b=d; hence a+bi corresponds one-to-one to the ordered pair (a,b)."
        }
      },
      {
        "id": "complex_addsub",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逐分量加減：(a+bi)+(c+di)=(a+c)+(b+d)i，(a+bi)−(c+di)=(a−c)+(b−d)i。",
          "en": "Componentwise addition and subtraction: (a+bi)+(c+di)=(a+c)+(b+d)i and (a+bi)−(c+di)=(a−c)+(b−d)i."
        }
      },
      {
        "id": "i_squared",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "虛數單位滿足 i²=i·i=−1。",
          "en": "The imaginary unit satisfies i²=i·i=−1."
        }
      },
      {
        "id": "real_embedding",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數 r 嵌入為複數 r+0i，對應有序對 (r,0)。",
          "en": "A real number r is embedded as the complex number r+0i, the ordered pair (r,0)."
        }
      },
      {
        "id": "L_complex_mul",
        "kind": "lemma",
        "proved_in": [
          "S04.1",
          "S04.2",
          "S04.3"
        ],
        "statement": {
          "zh-Hant": "對所有實數 a、b、c、d：(a+bi)(c+di)=(ac−bd)+(ad+bc)i；以有序對表示即 (a,b)·(c,d)=(ac−bd, ad+bc)。",
          "en": "For all reals a, b, c, d: (a+bi)(c+di)=(ac−bd)+(ad+bc)i; in ordered pairs, (a,b)·(c,d)=(ac−bd, ad+bc)."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "2+4",
        "rhs": "6"
      },
      {
        "id": "N2",
        "step": "S02",
        "lhs": "3+(-5)",
        "rhs": "-2"
      },
      {
        "id": "N3",
        "step": "S03",
        "lhs": "2-4",
        "rhs": "-2"
      },
      {
        "id": "N4",
        "step": "S03",
        "lhs": "3-(-5)",
        "rhs": "8"
      },
      {
        "id": "N5",
        "step": "S06",
        "lhs": "2*4-3*(-5)",
        "rhs": "23"
      },
      {
        "id": "N6",
        "step": "S06",
        "lhs": "2*(-5)+3*4",
        "rhs": "2"
      },
      {
        "id": "N7",
        "step": "S07",
        "lhs": "8-15*(-1)",
        "rhs": "23"
      },
      {
        "id": "N8",
        "step": "S07",
        "lhs": "-10+12",
        "rhs": "2"
      },
      {
        "id": "N9",
        "step": "S13",
        "lhs": "4*2-(-5)*3",
        "rhs": "23"
      },
      {
        "id": "N10",
        "step": "S13",
        "lhs": "4*3+(-5)*2",
        "rhs": "2"
      },
      {
        "id": "N11",
        "step": "S12",
        "lhs": "2",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N12",
        "lhs": "8-15",
        "relation": "!=",
        "rhs": "23"
      },
      {
        "id": "N13",
        "step": "S04",
        "lhs": "0*0-1*1",
        "rhs": "-1"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 範圍內工作：整數實部與虛部係數（視為實數特例），加、減、乘三種運算，答案以有序對表示；不做複數除法、共軛、模或極式。ℂ 中加法與乘法的交換、結合與分配律作為前提使用，不在本稿由有序對構造推出。不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: integer real parts and imaginary coefficients (regarded as real numbers), the three operations addition, subtraction and multiplication, and answers as ordered pairs; no complex division, conjugate, modulus or polar form. The commutative, associative and distributive laws of addition and multiplication in ℂ are used as prerequisites and are not derived here from the ordered-pair construction. The canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "z=a+bi",
        "實部 a、虛部係數 b 的複數",
        "the complex number with real part a and imaginary coefficient b"
      ],
      [
        "(a,b)",
        "z 對應的有序對，次序為（實部, 虛部係數）",
        "the ordered pair of z, in the order (real part, imaginary coefficient)"
      ],
      [
        "i",
        "虛數單位，滿足 i^2=-1",
        "the imaginary unit, with i^2=-1"
      ],
      [
        "z,\\ w",
        "本稿的兩個給定複數",
        "the two given complex numbers"
      ]
    ],
    "md_error_controls": [
      {
        "title": "忘記 i²=−1 / Forgetting i²=−1",
        "zh": "MKO 指出的誤解之一是乘法時忘記 i²=−1。若把 −15i² 當成 −15（等於把 i² 當成 1），會得到 (8−15)+2i=−7+2i；若乾脆丟掉 i² 項，會得到 8+2i。正確做法是 −15i²=−15·(−1)=+15，實部為 23。S04.3 顯示一般公式中的 −bd 正是這個負號。",
        "en": "One mistake the MKO names is forgetting i²=−1 in multiplication. Treating −15i² as −15 (that is, taking i² as 1) gives (8−15)+2i=−7+2i; dropping the i² term gives 8+2i. Correctly, −15i²=−15·(−1)=+15 and the real part is 23. S04.3 shows that the −bd in the general formula is exactly this sign."
      },
      {
        "title": "交換實部與虛部 / Swapping the real and imaginary parts",
        "zh": "MKO 的另一個誤解是交換實部與虛部。答案 (23,2) 表示 23+2i；寫成 (2,23) 就成了 2+23i，是另一個數（S09）。有序對的次序是語義資料，不能依圖上位置猜。",
        "en": "Another mistake identified in the MKO is swapping the real and imaginary parts. The answer (23,2) means 23+2i; writing (2,23) gives 2+23i, a different number (S09). The order of the pair is semantic data and is not guessed from positions in a picture."
      },
      {
        "title": "乘法不是逐分量 / Multiplication is not componentwise",
        "zh": "加減是逐分量的，但乘法不是：(2,3)·(4,−5) 不等於 (2·4, 3·(−5))=(8,−15)。正確公式的實部 ac−bd 混合了兩個實部與兩個虛部係數，虛部係數 ad+bc 是交叉項之和。",
        "en": "Addition and subtraction are componentwise, but multiplication is not: (2,3)·(4,−5) is not (2·4, 3·(−5))=(8,−15). In the correct formula the real part ac−bd mixes both real parts and both imaginary coefficients, and the imaginary coefficient ad+bc is the sum of the cross terms."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "ℂ 中加法與乘法的交換、結合、分配律是本稿的前提，沒有由有序對的定義推出；本稿也不處理複數除法、共軛、模與極式（MKO 的 formal next）。",
        "en": "The commutative, associative and distributive laws in ℂ are prerequisites of this guide and are not derived from the ordered-pair definition; complex division, conjugates, moduli and polar form (the MKO's formal next) are not treated."
      },
      {
        "zh-Hant": "N13 用引理重算 i·i=−1 只是一致性核對；引理本身用到了 i²=−1，所以它不是 i²=−1 的證明。",
        "en": "N13, recomputing i·i=−1 with the lemma, is only a consistency check; the lemma itself uses i²=−1, so it is not a proof of i²=−1."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "311fc9976ea86edccc67e94b6680145f12b64abe38568601a6bbabf8fd324b42",
    "original_source_sha256": "311fc9976ea86edccc67e94b6680145f12b64abe38568601a6bbabf8fd324b42",
    "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."
  }
}