{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-vector-01",
  "mko_id": "mko-vector",
  "title": {
    "zh-Hant": "固定基底下的平面向量：逐分量運算、相等判定與減法方向",
    "en": "Plane vectors in a fixed basis: componentwise operations, equality and the direction of subtraction"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-vector-01",
    "source": {
      "mko_id": "mko-vector",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "0d8db0f5e7511aaa8401ebe39e808af23991063892ca349e742b0551c91cba3d",
      "english_sha256": "8b23c6e9698bb13ffb12ef863fc1ba12d88c96f880c720e6504681e634fcd99d"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "cartesian_product_def",
      "add_inverse",
      "distributive",
      "neg_mul",
      "numeric_evaluation",
      "neg_vec_def"
    ],
    "goal": {
      "zh-Hant": "在固定標準基底下，設 u=(3,−1)、v=(−2,4)、w=(1,3)、k=−2。逐分量計算 u+v、u−v、v−u、ku、0u、2u+v；求滿足 u+x=v 的向量 x；核對 k(u+v)=ku+kv；並判定 u=w 與 u=(−1,3) 是否成立。",
      "en": "In the fixed standard basis let u=(3,−1), v=(−2,4), w=(1,3) and k=−2. Compute u+v, u−v, v−u, ku, 0u and 2u+v componentwise; find the vector x with u+x=v; check k(u+v)=ku+kv; and decide whether u=w and whether u=(−1,3)."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "基底固定為標準基底；向量是分量對 (z₁,z₂)∈ℝ²，加法、減法與純量乘法逐分量進行，兩向量相等當且僅當兩個對應分量都相等，零向量為 (0,0)。",
          "en": "The basis is the fixed standard basis; a vector is a component pair (z₁,z₂)∈ℝ², addition, subtraction and scalar multiplication act componentwise, two vectors are equal exactly when both corresponding components are equal, and the zero vector is (0,0)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "u=(3,−1)、v=(−2,4)、w=(1,3)、k=−2（整數分量與整數純量，MKO 假設 a1）。",
          "en": "u=(3,−1), v=(−2,4), w=(1,3), k=−2 (integer components and an integer scalar, MKO assumption a1)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "這裡的有序對代表向量分量，不把幾何點、位置向量與任意座標系混為同一對象（MKO 假設 a2）。",
          "en": "Ordered pairs here represent vector components; geometric points, position vectors and arbitrary coordinate systems are not identified as one object (MKO assumption a2)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "G3",
          "mko_vector_ops"
        ],
        "before_tex": "u=(3,-1),\\ v=(-2,4),\\ w=(1,3),\\ k=-2",
        "after_tex": "u+v=(u_1+v_1,u_2+v_2),\\ ku=(ku_1,ku_2),\\ u-v=(u_1-v_1,u_2-v_2)",
        "relation": "definition",
        "rule": "mko_vector_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "基底固定為標準基底，向量就是分量對 (z₁,z₂)∈ℝ²。依 MKO，加法、純量乘法、減法都逐分量進行；兩向量相等當且僅當兩個對應分量都相等；零向量為 (0,0)。這裡的有序對代表向量分量，不當作幾何點。",
          "en": "The basis is fixed as the standard basis, so a vector is a component pair (z₁,z₂)∈ℝ². By the MKO, addition, scalar multiplication and subtraction act componentwise; two vectors are equal exactly when both corresponding components are equal; the zero vector is (0,0). Ordered pairs here are vector components, not geometric points."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "G3",
            "mko_vector_ops"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出定義與資料，不推出新事實；若基底不固定，同一組分量可能代表不同向量。",
            "en": "This step only states the definitions and data and derives no new fact; without a fixed basis, the same components could represent different vectors."
          }
        },
        "leaf_basis": "mko_vector_ops"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "G2"
        ],
        "before_tex": "u+v",
        "after_tex": "(3+(-2),\\,-1+4)=(1,3)",
        "relation": "equivalent",
        "rule": "mko_vector_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐分量相加：第一分量 3+(−2)=1，第二分量 −1+4=3，所以 u+v=(1,3)。不能把分量混在一起，例如只算全部分量的總和 3+(−1)+(−2)+4=4，那不是向量。",
          "en": "Add componentwise: first component 3+(−2)=1, second component −1+4=3, so u+v=(1,3). Components may not be mixed together; for example the total 3+(−1)+(−2)+4=4 of all components is not a vector."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_vector_ops",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個分量對；逐分量重算（N1、N2）。",
            "en": "Both sides are the same component pair; recompute componentwise (N1, N2)."
          }
        },
        "leaf_basis": "mko_vector_ops + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "G2"
        ],
        "before_tex": "u-v",
        "after_tex": "(3-(-2),\\,-1-4)=(5,-5)",
        "relation": "equivalent",
        "rule": "mko_vector_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐分量相減，順序是 u 的分量減 v 的分量：3−(−2)=5，−1−4=−5，所以 u−v=(5,−5)。",
          "en": "Subtract componentwise, with the components of u minus those of v: 3−(−2)=5 and −1−4=−5, so u−v=(5,−5)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_vector_ops",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "逐分量重算（N3、N4）；兩邊是同一個分量對。",
            "en": "Recompute componentwise (N3, N4); both sides are the same component pair."
          }
        },
        "leaf_basis": "mko_vector_ops + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S01",
          "G2"
        ],
        "before_tex": "v-u",
        "after_tex": "(-2-3,\\,4-(-1))=(-5,5)",
        "relation": "equivalent",
        "rule": "mko_vector_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "交換順序：v 的分量減 u 的分量，−2−3=−5，4−(−1)=5，所以 v−u=(−5,5)。",
          "en": "Reverse the order: the components of v minus those of u, −2−3=−5 and 4−(−1)=5, so v−u=(−5,5)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_vector_ops",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "逐分量重算（N5、N6）。",
            "en": "Recompute componentwise (N5, N6)."
          }
        },
        "leaf_basis": "mko_vector_ops + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "S03",
          "S04",
          "L_sub_reverse"
        ],
        "before_tex": "u-v=(5,-5),\\ v-u=(-5,5)",
        "after_tex": "u-v=-(v-u)",
        "relation": "theorem_application",
        "rule": "L_sub_reverse",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_sub_reverse，u−v=−(v−u)。核對本例：−(v−u)=(−1)(−5,5)=(5,−5)，與 S03 的 u−v 相同。交換減法順序得到相反向量，而不是同一個向量。",
          "en": "By Lemma L_sub_reverse, u−v=−(v−u). Check in this example: −(v−u)=(−1)(−5,5)=(5,−5), which equals u−v from S03. Reversing the order of subtraction gives the opposite vector, not the same vector."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_sub_reverse",
            "S03",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "引理在 S05.1、S05.2 對所有向量證明；本例的數字只是核對。",
            "en": "The lemma is proved for all vectors in S05.1 and S05.2; the numbers here are only a check."
          }
        },
        "leaf_basis": "mko_vector_ops + neg_vec_def + distributive + neg_mul"
      },
      {
        "id": "S05.1",
        "uses": [
          "mko_vector_ops",
          "neg_vec_def"
        ],
        "before_tex": "u-v,\\quad -(v-u)",
        "after_tex": "(u_1-v_1,\\,u_2-v_2),\\quad (-(v_1-u_1),\\,-(v_2-u_2))",
        "relation": "equivalent",
        "rule": "mko_vector_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對任意 u、v∈ℝ²，把兩邊逐分量展開：u−v=(u₁−v₁,u₂−v₂)；v−u=(v₁−u₁,v₂−u₂)，再依 neg_vec_def 乘 −1 得 −(v−u)=(−(v₁−u₁),−(v₂−u₂))。",
          "en": "For arbitrary u, v∈ℝ², unfold both sides componentwise: u−v=(u₁−v₁,u₂−v₂); v−u=(v₁−u₁,v₂−u₂), and multiplying by −1 as in neg_vec_def gives −(v−u)=(−(v₁−u₁),−(v₂−u₂))."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_vector_ops",
            "neg_vec_def"
          ],
          "explanation": {
            "zh-Hant": "展開定義是雙向的。",
            "en": "Unfolding definitions is two-way."
          }
        },
        "leaf_basis": "mko_vector_ops + neg_vec_def"
      },
      {
        "id": "S05.2",
        "uses": [
          "S05.1",
          "distributive",
          "neg_mul",
          "add_comm"
        ],
        "before_tex": "-(v_i-u_i)",
        "after_tex": "u_i-v_i\\quad(i=1,2)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對實數 a、b：−(b−a)=(−1)(b+(−a))=(−1)b+(−1)(−a)=−b+a=a+(−b)=a−b（分配律、負號與乘法、加法交換律；N7 做恆等式核對）。取 a=u_i、b=v_i，兩個分量都相等，依向量相等的定義 u−v=−(v−u)。S05.1 與 S05.2 證得引理 L_sub_reverse。",
          "en": "For reals a, b: −(b−a)=(−1)(b+(−a))=(−1)b+(−1)(−a)=−b+a=a+(−b)=a−b (distributivity, signs and multiplication, commutativity of addition; N7 checks the identity). With a=u_i and b=v_i both components agree, so u−v=−(v−u) by the definition of vector equality. S05.1 and S05.2 prove Lemma L_sub_reverse."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "neg_mul",
            "add_comm",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "所用的是對所有實數成立的恆等式，雙向成立；向量相等需要兩個分量都相等，這裡兩者都成立。",
            "en": "These are identities valid for all reals, so they hold both ways; vector equality needs both components to agree, and both do."
          }
        },
        "leaf_basis": "distributive + neg_mul + add_comm + cartesian_product_def"
      },
      {
        "id": "S06",
        "uses": [
          "S03",
          "S04",
          "cartesian_product_def"
        ],
        "before_tex": "u-v=(5,-5),\\ v-u=(-5,5)",
        "after_tex": "u-v\\neq v-u",
        "relation": "implies",
        "rule": "cartesian_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "向量相等需要兩個對應分量都相等。第一分量 5≠−5，已經不相等，所以 u−v≠v−u。減法的方向有意義：u−v 與 v−u 是互為相反的兩個向量，只有在 u=v 時才會相同。",
          "en": "Vector equality needs both corresponding components to agree. The first components are 5≠−5, which already fails, so u−v≠v−u. The direction of subtraction matters: u−v and v−u are opposite vectors, which coincide only when u=v."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "cartesian_product_def",
            "S03",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "一個分量不同就足以推出不等；最後一句的一般說法不在此證明，只是說明，判定只依本例的分量。",
            "en": "One differing component suffices for inequality; the closing general remark is not proved here and serves only as comment, the verdict resting on this example's components."
          }
        },
        "leaf_basis": "cartesian_product_def + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S04",
          "L_solve_vec",
          "G2"
        ],
        "before_tex": "u+x=v",
        "after_tex": "x=v-u=(-5,5)",
        "relation": "equivalent",
        "rule": "L_solve_vec",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_solve_vec，u+x=v 當且僅當 x=v−u；由 S04，x=(−5,5)。代回核對：u+x=(3+(−5),−1+5)=(−2,4)=v。從 u 到 v 的向量是 v−u，不是 u−v。",
          "en": "By Lemma L_solve_vec, u+x=v exactly when x=v−u; by S04, x=(−5,5). Check by substitution: u+x=(3+(−5),−1+5)=(−2,4)=v. The vector taking u to v is v−u, not u−v."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_solve_vec"
          ],
          "explanation": {
            "zh-Hant": "引理兩向在 S07.1、S07.2 證明；代回核對見 N8、N9。",
            "en": "Both directions of the lemma are proved in S07.1 and S07.2; the substitution check is N8 and N9."
          }
        },
        "leaf_basis": "mko_vector_ops + eq_congruence + add_inverse + add_assoc + add_zero"
      },
      {
        "id": "S07.1",
        "uses": [
          "mko_vector_ops",
          "cartesian_product_def",
          "eq_congruence"
        ],
        "before_tex": "u+x=v",
        "after_tex": "x=v-u",
        "relation": "implies",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "⇒：u+x=v 依向量相等拆成 u₁+x₁=v₁ 與 u₂+x₂=v₂。每式兩邊同加 −u_i：x_i=(u_i+x_i)+(−u_i)=v_i−u_i（結合律、交換律、反元素、加法單位元）。兩個分量都等於 v−u 的分量，所以 x=v−u。",
          "en": "⇒: by vector equality, u+x=v splits into u₁+x₁=v₁ and u₂+x₂=v₂. Add −u_i to both sides of each: x_i=(u_i+x_i)+(−u_i)=v_i−u_i (associativity, commutativity, inverses, additive identity). Both components equal those of v−u, so x=v−u."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "cartesian_product_def",
            "eq_congruence",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "這是引理的「⇒」方向，反向另在 S07.2 證明；兩邊同加同一個數總是可行。",
            "en": "This is the '⇒' direction of the lemma, the other being proved in S07.2; adding the same number to both sides is always allowed."
          }
        },
        "leaf_basis": "cartesian_product_def + eq_congruence + add_assoc + add_comm + add_inverse + add_zero"
      },
      {
        "id": "S07.2",
        "uses": [
          "mko_vector_ops",
          "add_inverse"
        ],
        "before_tex": "x=v-u",
        "after_tex": "u+x=v",
        "relation": "implies",
        "rule": "add_inverse",
        "conditions": [],
        "explanation": {
          "zh-Hant": "⇐：若 x=v−u，則 u+x 的第 i 個分量是 u_i+(v_i−u_i)=u_i+(v_i+(−u_i))=v_i+(u_i+(−u_i))=v_i+0=v_i，所以 u+x=v。S07.1 與 S07.2 證得引理 L_solve_vec，同時說明這樣的 x 唯一。",
          "en": "⇐: if x=v−u, the i-th component of u+x is u_i+(v_i−u_i)=u_i+(v_i+(−u_i))=v_i+(u_i+(−u_i))=v_i+0=v_i, so u+x=v. S07.1 and S07.2 prove Lemma L_solve_vec, which also shows such an x is unique."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "add_inverse",
            "add_assoc",
            "add_comm",
            "add_zero"
          ],
          "explanation": {
            "zh-Hant": "只用到實數加法的群性質，對任何 u、v 都成立。",
            "en": "Only the group properties of real addition are used, valid for every u and v."
          }
        },
        "leaf_basis": "add_inverse + add_assoc + add_comm + add_zero + mko_vector_ops"
      },
      {
        "id": "S08",
        "uses": [
          "S01",
          "G2"
        ],
        "before_tex": "ku,\\ k=-2",
        "after_tex": "(-2\\cdot 3,\\,-2\\cdot(-1))=(-6,2)",
        "relation": "equivalent",
        "rule": "mko_vector_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "純量乘法逐分量乘 k：−2·3=−6，−2·(−1)=2，所以 ku=(−6,2)。結果仍是 ℝ² 的向量，不是一個數。負純量給出方向反轉的代數表示：ku=2·(−u)，其中 −u=(−3,1)。",
          "en": "Scalar multiplication multiplies each component by k: −2·3=−6 and −2·(−1)=2, so ku=(−6,2). The result is still a vector in ℝ², not a number. A negative scalar is the algebraic form of reversing direction: ku=2·(−u), where −u=(−3,1)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_vector_ops",
            "numeric_evaluation",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "逐分量重算（N10、N11）；兩邊是同一個向量。",
            "en": "Recompute componentwise (N10, N11); both sides are the same vector."
          }
        },
        "leaf_basis": "mko_vector_ops + numeric_evaluation + neg_mul"
      },
      {
        "id": "S09",
        "uses": [
          "S01",
          "G2",
          "mul_zero"
        ],
        "before_tex": "0u",
        "after_tex": "(0\\cdot 3,\\,0\\cdot(-1))=(0,0)",
        "relation": "equivalent",
        "rule": "mul_zero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "純量 0 乘每個分量都得 0（u·0=0 與乘法交換律），所以 0u=(0,0)，即零向量。MKO 第二段：純量零給出零向量。",
          "en": "The scalar 0 times each component gives 0 (u·0=0 with commutativity), so 0u=(0,0), the zero vector. MKO paragraph two: the zero scalar gives the zero vector."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_zero",
            "mul_comm",
            "mko_vector_ops"
          ],
          "explanation": {
            "zh-Hant": "逐分量相等（N12、N13），兩向成立。",
            "en": "The components agree (N12, N13), so it holds both ways."
          }
        },
        "leaf_basis": "mul_zero + mul_comm + mko_vector_ops"
      },
      {
        "id": "S10",
        "uses": [
          "S01",
          "G2"
        ],
        "before_tex": "2u+v",
        "after_tex": "(6,-2)+(-2,4)=(4,2)",
        "relation": "equivalent",
        "rule": "mko_vector_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先做純量乘法 2u=(6,−2)，再逐分量加 v：6+(−2)=4，−2+4=2，所以 2u+v=(4,2)。要判定 2u+v 是否等於 (4,2)，必須兩個分量都相等，這裡兩者都成立。",
          "en": "First the scalar multiple 2u=(6,−2), then add v componentwise: 6+(−2)=4 and −2+4=2, so 2u+v=(4,2). To decide whether 2u+v equals (4,2), both components must agree, and here both do."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_vector_ops",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "逐分量重算（N14、N15）。",
            "en": "Recompute componentwise (N14, N15)."
          }
        },
        "leaf_basis": "mko_vector_ops + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S02",
          "S08",
          "L_scalar_distrib"
        ],
        "before_tex": "k(u+v),\\ ku+kv",
        "after_tex": "-2(1,3)=(-2,-6)=(-6,2)+(4,-8)",
        "relation": "theorem_application",
        "rule": "L_scalar_distrib",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_scalar_distrib，k(u+v)=ku+kv。核對本例：k(u+v)=−2·(1,3)=(−2,−6)；kv=(−2·(−2),−2·4)=(4,−8)，ku+kv=(−6+4,2+(−8))=(−2,−6)。兩邊逐分量相同（N16、N17）。",
          "en": "By Lemma L_scalar_distrib, k(u+v)=ku+kv. Check in this example: k(u+v)=−2·(1,3)=(−2,−6); kv=(−2·(−2),−2·4)=(4,−8), and ku+kv=(−6+4,2+(−8))=(−2,−6). The two sides agree componentwise (N16, N17)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_scalar_distrib",
            "S02",
            "S08"
          ],
          "explanation": {
            "zh-Hant": "引理在 S11.1 對所有 k、u、v 證明；本例數字只是核對，不取代證明。",
            "en": "The lemma is proved for all k, u, v in S11.1; the numbers here only check it and do not replace the proof."
          }
        },
        "leaf_basis": "distributive + mko_vector_ops + numeric_evaluation"
      },
      {
        "id": "S11.1",
        "uses": [
          "mko_vector_ops",
          "distributive"
        ],
        "before_tex": "k(u+v)",
        "after_tex": "(k(u_1+v_1),k(u_2+v_2))=(ku_1+kv_1,ku_2+kv_2)=ku+kv",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對任意實數 k 與 u、v∈ℝ²：先逐分量展開 u+v，再乘 k 得 (k(u₁+v₁),k(u₂+v₂))；每個分量用實數分配律 k(a+b)=ka+kb，得 (ku₁+kv₁,ku₂+kv₂)，這正是 ku 與 kv 的逐分量和。這證得引理 L_scalar_distrib（N18 做恆等式核對）。",
          "en": "For any real k and u, v∈ℝ²: unfold u+v componentwise and multiply by k to get (k(u₁+v₁),k(u₂+v₂)); real distributivity k(a+b)=ka+kb in each component gives (ku₁+kv₁,ku₂+kv₂), which is exactly the componentwise sum of ku and kv. This proves Lemma L_scalar_distrib (N18 checks the identity)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_vector_ops",
            "distributive",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "每個等號都是恆等式，兩個分量都相等，所以雙向成立。",
            "en": "Each equality is an identity and both components agree, so it holds both ways."
          }
        },
        "leaf_basis": "mko_vector_ops + distributive + cartesian_product_def"
      },
      {
        "id": "S12",
        "uses": [
          "G2",
          "cartesian_product_def"
        ],
        "before_tex": "u=(3,-1),\\ w=(1,3),\\ u_1^2+u_2^2=w_1^2+w_2^2=10",
        "after_tex": "u\\neq w",
        "relation": "implies",
        "rule": "cartesian_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "u 與 w 有相同的平方和：3²+(−1)²=10=1²+3²（N19）。但向量相等只看對應分量：第一分量 3≠1，所以 u≠w。本條目不由分量推論長度；即使某個由分量算出的數量相同，也不能據此判定相等。",
          "en": "u and w have the same sum of squares: 3²+(−1)²=10=1²+3² (N19). But vector equality looks only at corresponding components: the first components are 3≠1, so u≠w. This entry does not infer lengths from components; even when some quantity computed from the components agrees, equality cannot be decided by it."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "cartesian_product_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一個分量不同就足以推出不等（N20）；不宣稱平方和與任何幾何量的關係。",
            "en": "One differing component suffices for inequality (N20); no relation between the sum of squares and any geometric quantity is claimed."
          }
        },
        "leaf_basis": "cartesian_product_def + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "G2",
          "cartesian_product_def"
        ],
        "before_tex": "u=(3,-1),\\ (-1,3)",
        "after_tex": "u\\neq(-1,3)",
        "relation": "implies",
        "rule": "cartesian_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(−1,3) 用了與 u 相同的兩個數，只是交換位置。分量有固定的位置：u 的第一分量 3，(−1,3) 的第一分量 −1，3≠−1，所以兩者不相等。分量和相同也不夠：u 與 (1,1) 的分量和都是 2，但 3≠1。",
          "en": "(−1,3) uses the same two numbers as u in swapped positions. Components have fixed positions: the first component of u is 3 and that of (−1,3) is −1, with 3≠−1, so they are not equal. Equal component sums are not enough either: u and (1,1) both have component sum 2, yet 3≠1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "有序對相等需要逐位置相等；一個位置不同就足以否定。",
            "en": "Ordered-pair equality needs agreement position by position; one differing position suffices to refute it."
          }
        },
        "leaf_basis": "cartesian_product_def + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S02",
          "S03",
          "S04",
          "S06",
          "S07",
          "S08",
          "S09",
          "S10",
          "S11",
          "S12",
          "S13"
        ],
        "before_tex": "\\text{S02 to S13}",
        "after_tex": "u+v=(1,3),\\ u-v=(5,-5),\\ v-u=(-5,5),\\ ku=(-6,2),\\ 0u=(0,0),\\ 2u+v=(4,2),\\ x=(-5,5)",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整：u+v=(1,3)、u−v=(5,−5)、v−u=(−5,5)=−(u−v)、ku=(−6,2)、0u=(0,0)、2u+v=(4,2)；u+x=v 的唯一解是 x=v−u=(−5,5)；k(u+v)=ku+kv=(−2,−6)；u≠w，u≠(−1,3)。",
          "en": "Summary: u+v=(1,3), u−v=(5,−5), v−u=(−5,5)=−(u−v), ku=(−6,2), 0u=(0,0), 2u+v=(4,2); the unique solution of u+x=v is x=v−u=(−5,5); k(u+v)=ku+kv=(−2,−6); u≠w and u≠(−1,3)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S02",
            "S03",
            "S04",
            "S07",
            "S08",
            "S09",
            "S10",
            "S11",
            "S12",
            "S13",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：題目要求的各項計算與判定。足夠條件 H：所列各步的結論；H⇒G 只是合取，H 由 G1、G2 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: the computations and decisions the task asks for. Sufficient condition H: the conclusions of the listed steps; H⇒G is just conjunction, H is proved forwards from G1 and G2, and G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在固定標準基底下：u+v=(1,3)，u−v=(5,−5)，v−u=(−5,5)=−(u−v)，所以 u−v≠v−u；ku=(−6,2)，0u=(0,0)，2u+v=(4,2)；滿足 u+x=v 的唯一向量是 x=v−u=(−5,5)；k(u+v)=ku+kv=(−2,−6)。u≠w（雖然平方和都是 10），u≠(−1,3)（分量交換位置）。",
      "en": "In the fixed standard basis: u+v=(1,3), u−v=(5,−5), v−u=(−5,5)=−(u−v), so u−v≠v−u; ku=(−6,2), 0u=(0,0), 2u+v=(4,2); the unique vector with u+x=v is x=v−u=(−5,5); k(u+v)=ku+kv=(−2,−6). u≠w (although both have sum of squares 10), and u≠(−1,3) (components in swapped positions)."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "u−v=v−u，因為兩者都是 u 與 v 的差",
          "en": "u−v=v−u, since both are the difference of u and v"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "u−v=(5,−5)，v−u=(−5,5)，第一分量 5≠−5。",
          "en": "u−v=(5,−5) and v−u=(−5,5); the first components are 5≠−5."
        }
      },
      {
        "claim": {
          "zh-Hant": "u=w，因為 u₁²+u₂²=w₁²+w₂²=10",
          "en": "u=w, because u₁²+u₂²=w₁²+w₂²=10"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "相等需逐分量相同；第一分量 3≠1。",
          "en": "Equality needs every component to agree; the first components are 3≠1."
        }
      },
      {
        "claim": {
          "zh-Hant": "(3,−1)=(−1,3)，因為兩者用了相同的兩個數",
          "en": "(3,−1)=(−1,3), since both use the same two numbers"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "有序對逐位置比較；3≠−1。",
          "en": "Ordered pairs are compared position by position; 3≠−1."
        }
      },
      {
        "claim": {
          "zh-Hant": "ku 是一個數：(−2)·3+(−2)·(−1)=−4",
          "en": "ku is a number: (−2)·3+(−2)·(−1)=−4"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "ku=(−6,2) 是向量；把分量相加不是純量乘法的一部分。",
          "en": "ku=(−6,2) is a vector; adding its components is not part of scalar multiplication."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "cartesian_product_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "笛卡兒積：A×B 是所有有序對 (a,b)（a∈A、b∈B）的集合；(a,b)=(c,d) ⇔ a=c 且 b=d。",
          "en": "Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d."
        }
      },
      {
        "id": "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": "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": "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_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法交換律：u+v=v+u。",
          "en": "Commutativity of addition: u+v=v+u."
        }
      },
      {
        "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": "add_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。",
          "en": "Associativity of addition: (u+v)+w=u+(v+w)."
        }
      },
      {
        "id": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "id": "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": "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": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "mko_vector_ops",
        "kind": "cited",
        "source": {
          "mko_id": "mko-vector",
          "sha256": "0d8db0f5e7511aaa8401ebe39e808af23991063892ca349e742b0551c91cba3d",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在固定標準基底下，平面向量 u=(u₁,u₂)∈R²；u+v=(u₁+v₁,u₂+v₂)，k u=(k u₁,k u₂)，u−v=(u₁−v₁,u₂−v₂)。向量相等當且僅當兩個對應分量相等；零向量為 (0,0)。",
          "en": "In the fixed standard basis, a plane vector is u=(u₁,u₂)∈R². Define u+v=(u₁+v₁,u₂+v₂), k u=(k u₁,k u₂), and u−v=(u₁−v₁,u₂−v₂). Two vectors are equal exactly when both corresponding components are equal; the zero vector is (0,0)."
        }
      },
      {
        "id": "neg_vec_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "向量的負：−z 表示 (−1)z=(−z₁,−z₂)。",
          "en": "Negative of a vector: −z means (−1)z=(−z₁,−z₂)."
        }
      },
      {
        "id": "L_sub_reverse",
        "kind": "lemma",
        "proved_in": [
          "S05.1",
          "S05.2"
        ],
        "statement": {
          "zh-Hant": "對所有 u、v∈ℝ²：u−v=−(v−u)。",
          "en": "For all u, v∈ℝ²: u−v=−(v−u)."
        }
      },
      {
        "id": "L_solve_vec",
        "kind": "lemma",
        "proved_in": [
          "S07.1",
          "S07.2"
        ],
        "statement": {
          "zh-Hant": "對所有 u、v、x∈ℝ²：u+x=v ⇔ x=v−u。",
          "en": "For all u, v, x∈ℝ²: u+x=v ⇔ x=v−u."
        }
      },
      {
        "id": "L_scalar_distrib",
        "kind": "lemma",
        "proved_in": [
          "S11.1"
        ],
        "statement": {
          "zh-Hant": "對所有實數 k 與 u、v∈ℝ²：k(u+v)=ku+kv。",
          "en": "For every real k and all u, v∈ℝ²: k(u+v)=ku+kv."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "3+(-2)",
        "rhs": "1"
      },
      {
        "id": "N2",
        "step": "S02",
        "lhs": "-1+4",
        "rhs": "3"
      },
      {
        "id": "N3",
        "step": "S03",
        "lhs": "3-(-2)",
        "rhs": "5"
      },
      {
        "id": "N4",
        "step": "S03",
        "lhs": "-1-4",
        "rhs": "-5"
      },
      {
        "id": "N5",
        "step": "S04",
        "lhs": "-2-3",
        "rhs": "-5"
      },
      {
        "id": "N6",
        "step": "S04",
        "lhs": "4-(-1)",
        "rhs": "5"
      },
      {
        "id": "N7",
        "step": "S05.2",
        "vars": [
          "a",
          "b"
        ],
        "degree": 1,
        "lhs": "a-b",
        "rhs": "-(b-a)"
      },
      {
        "id": "N8",
        "step": "S07",
        "lhs": "3+(-5)",
        "rhs": "-2"
      },
      {
        "id": "N9",
        "step": "S07",
        "lhs": "-1+5",
        "rhs": "4"
      },
      {
        "id": "N10",
        "step": "S08",
        "lhs": "-2*3",
        "rhs": "-6"
      },
      {
        "id": "N11",
        "step": "S08",
        "lhs": "-2*(-1)",
        "rhs": "2"
      },
      {
        "id": "N12",
        "step": "S09",
        "lhs": "0*3",
        "rhs": "0"
      },
      {
        "id": "N13",
        "step": "S09",
        "lhs": "0*(-1)",
        "rhs": "0"
      },
      {
        "id": "N14",
        "step": "S10",
        "lhs": "2*3+(-2)",
        "rhs": "4"
      },
      {
        "id": "N15",
        "step": "S10",
        "lhs": "2*(-1)+4",
        "rhs": "2"
      },
      {
        "id": "N16",
        "step": "S11",
        "lhs": "-2*(3+(-2))",
        "rhs": "-2*3+(-2)*(-2)"
      },
      {
        "id": "N17",
        "step": "S11",
        "lhs": "-2*(-1+4)",
        "rhs": "-2*(-1)+(-2)*4"
      },
      {
        "id": "N18",
        "step": "S11.1",
        "vars": [
          "k",
          "a",
          "b"
        ],
        "degree": 1,
        "lhs": "k*(a+b)",
        "rhs": "k*a+k*b"
      },
      {
        "id": "N19",
        "step": "S12",
        "lhs": "3**2+(-1)**2",
        "rhs": "1**2+3**2"
      },
      {
        "id": "N20",
        "step": "S12",
        "lhs": "3",
        "relation": "!=",
        "rhs": "1"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：固定標準基底、實數分量與純量，題目用整數的精確數值。示範逐分量運算與逐分量相等，並證明三個一般小引理（u−v=−(v−u)、u+x=v 的唯一解、純量對向量加法的分配律）。不由分量推論長度、角度、單位向量或基底變換；不處理一般向量空間公理或物理單位；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: a fixed standard basis, real components and scalars, with exact integer values in the task. Shows componentwise operations and componentwise equality, and proves three small general lemmas (u−v=−(v−u), the unique solution of u+x=v, and distributivity of a scalar over vector addition). Does not infer lengths, angles, unit vectors or changes of basis from components; does not treat general vector-space axioms or physical units; does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "u=(u_1,u_2)",
        "以固定標準基底的分量表示的平面向量",
        "a plane vector given by its components in the fixed standard basis"
      ],
      [
        "u+v,\\ u-v",
        "逐分量的和與差",
        "componentwise sum and difference"
      ],
      [
        "ku",
        "純量 k 乘向量 u，結果仍是向量",
        "the scalar k times the vector u, again a vector"
      ],
      [
        "-z",
        "(−1)z=(−z_1,−z_2)",
        "(−1)z=(−z_1,−z_2)"
      ],
      [
        "(0,0)",
        "零向量",
        "the zero vector"
      ]
    ],
    "md_error_controls": [
      {
        "title": "分量有位置，減法有方向 / Components have positions; subtraction has a direction",
        "zh": "MKO 指出的誤解之一是把向量座標當成可交換。(3,−1) 與 (−1,3) 不同（S13）；u−v=(5,−5) 與 v−u=(−5,5) 互為相反向量（S05、S06）。從 u 到 v 要加的是 v−u（S07），寫成 u−v 會得到相反方向。",
        "en": "One misconception named by the MKO is treating vector coordinates as interchangeable. (3,−1) and (−1,3) differ (S13); u−v=(5,−5) and v−u=(−5,5) are opposite vectors (S05, S06). To get from u to v one adds v−u (S07); writing u−v gives the opposite direction."
      },
      {
        "title": "相等要逐分量，不看長度或分量和 / Equality is componentwise, not by length or component sum",
        "zh": "MKO 的第二個誤解是以為向量相等只需長度相同。u 與 w 的平方和都是 10，但 3≠1，所以 u≠w（S12）；u 與 (1,1) 的分量和都是 2，也不相等（S13）。相等的唯一判準是兩個對應分量都相等。",
        "en": "The second misconception identified in the MKO is that equal length is enough to establish vector equality. u and w both have sum of squares 10, but 3≠1, so u≠w (S12); u and (1,1) both have component sum 2 and are not equal either (S13). The only criterion is that both corresponding components agree."
      },
      {
        "title": "純量乘法不是內積 / Scalar multiplication is not an inner product",
        "zh": "MKO 的第三個誤解是把純量乘法當成內積。ku=(−6,2) 是 ℝ² 的向量（S08），不是一個數；把 ku 的分量加起來得到 −4 已經離開本條目的運算。內積是另一種運算，本條目不定義也不使用它。",
        "en": "The third misconception identified in the MKO is mistaking scalar multiplication for an inner product. ku=(−6,2) is a vector in ℝ² (S08), not a number; adding its components to get −4 already leaves the operations of this entry. The inner product is a different operation, which this entry neither defines nor uses."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "N7 與 N18 是各變數次數至多為一的多項式恆等式格點核對，對應 S05.2 與 S11.1 的代數論證；論證本身不依賴它們。",
        "en": "N7 and N18 are grid checks of polynomial identities of degree at most one in each variable matching the algebra in S05.2 and S11.1; the argument does not depend on them."
      },
      {
        "zh-Hant": "本稿不涉及長度、角度、內積或基底變換（MKO 第三段明示另有義務）。",
        "en": "This guide does not involve length, angle, inner product or change of basis (MKO paragraph three says these carry separate obligations)."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "bc749cb55fd9feab11176f4265e52ebf388cff1d90049efe15d0c6b29999d5c1",
    "original_source_sha256": "bc749cb55fd9feab11176f4265e52ebf388cff1d90049efe15d0c6b29999d5c1",
    "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."
  }
}