{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-real-vector-space-01",
  "mko_id": "mko-real-vector-space",
  "title": {
    "zh-Hant": "實向量空間：在 ℝ³ 核對公理實例並證明一條分配律",
    "en": "Real vector space: auditing axiom instances in ℝ³ and proving one distributive law"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-real-vector-space-01",
    "source": {
      "mko_id": "mko-real-vector-space",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "d57c98eb1c006f761ecaaf2870e2a09b30e5a29627351005b7e3bb7b30b33356",
      "english_sha256": "b41b524022de7b68103686353c9d8b1f730143e3cb9c5de0be7d936a81343cd8"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "distributive",
      "mul_assoc",
      "mul_one",
      "add_inverse",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在 ℝ³ 中取 u=(1,−2,3)、v=(4,0,−1)、a=2、b=−3。（一）確認維度相同；（二）精確核對 a(u+v)=au+av、(a+b)u=au+bu、(ab)u=a(bu)、1u=u、u+(−u)=0 五個公理實例；（三）對所有實數 a 與所有 u、v∈ℝ³ 證明 a(u+v)=au+av；（四）以 ℝ² 上改造的純量乘法說明實例通過不等於公理成立。",
      "en": "In ℝ³ take u=(1,−2,3), v=(4,0,−1), a=2, b=−3. (1) Confirm the dimensions agree; (2) check exactly the five axiom instances a(u+v)=au+av, (a+b)u=au+bu, (ab)u=a(bu), 1u=u and u+(−u)=0; (3) prove a(u+v)=au+av for every real a and all u, v∈ℝ³; (4) use a modified scalar multiplication on ℝ² to show that successful instances do not establish an axiom."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "ℝ³ 是有序實數三元組的集合；加法逐分量：(x₁,x₂,x₃)+(y₁,y₂,y₃)=(x₁+y₁,x₂+y₂,x₃+y₃)；純量乘法逐分量：a(x₁,x₂,x₃)=(ax₁,ax₂,ax₃)；零向量為 (0,0,0)；−(x₁,x₂,x₃)=(−x₁,−x₂,−x₃)。加法只對相同維度的向量定義。",
          "en": "ℝ³ is the set of ordered triples of reals; addition is componentwise: (x₁,x₂,x₃)+(y₁,y₂,y₃)=(x₁+y₁,x₂+y₂,x₃+y₃); scalar multiplication is componentwise: a(x₁,x₂,x₃)=(ax₁,ax₂,ax₃); the zero vector is (0,0,0); −(x₁,x₂,x₃)=(−x₁,−x₂,−x₃). Addition is defined only between vectors of the same dimension."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "u=(1,−2,3)，v=(4,0,−1)，a=2，b=−3。",
          "en": "u=(1,−2,3), v=(4,0,−1), a=2, b=−3."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "對照結構：ℝ² 配上通常的逐分量加法，但純量乘法改為 a⊙(x,y)=(ax,y)，即只縮放第一個分量。",
          "en": "Comparison structure: ℝ² with the usual componentwise addition, but with scalar multiplication changed to a⊙(x,y)=(ax,y), which scales only the first component."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "mko_rvs_def"
        ],
        "before_tex": "(\\mathbb{R}^3,+,\\cdot)",
        "after_tex": "\\text{audit: } a(u+v)=au+av,\\ (a+b)u=au+bu,\\ (ab)u=a(bu),\\ 1u=u,\\ u+(-u)=0",
        "relation": "definition",
        "rule": "vector_space_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，實向量空間要有向量加法、零向量、加法逆元與實數純量乘法，並對所有向量與所有純量滿足結合、交換、分配、單位純量與純量乘法相容等公理。本稿挑出其中五條，先在給定資料上核實例。",
          "en": "By the MKO, a real vector space needs vector addition, a zero vector, additive inverses and real scalar multiplication satisfying, for all vectors and all scalars, the associative, commutative, distributive, unit-scalar and compatibility axioms. This guide picks five of them and first checks instances on the given data."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_rvs_def",
            "vector_space_axioms"
          ],
          "explanation": {
            "zh-Hant": "此步只列出要核的公理，不推出新事實。每條公理都是對所有元素量化的命題，這一點在 S08 與 S15 會用到。",
            "en": "This step only lists the axioms to audit and derives no new fact. Each axiom is a statement quantified over all elements, which S08 and S15 will use."
          }
        },
        "leaf_basis": "vector_space_axioms + mko_rvs_def"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "u=(1,-2,3),\\ v=(4,0,-1)",
        "after_tex": "u,v\\in\\mathbb{R}^3,\\ u+v\\ \\text{defined}",
        "relation": "implies",
        "rule": "coord_ops",
        "conditions": [
          "\\dim u=\\dim v=3"
        ],
        "explanation": {
          "zh-Hant": "u 與 v 都有三個分量，屬於同一個 ℝ³，所以 u+v 有定義。若一個向量在 ℝ² 而另一個在 ℝ³，例如 (1,2)+(1,2,3)，模型中沒有這個運算。",
          "en": "u and v both have three components and lie in the same ℝ³, so u+v is defined. If one vector were in ℝ² and the other in ℝ³, as in (1,2)+(1,2,3), the model has no such operation."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "coord_ops"
          ],
          "explanation": {
            "zh-Hant": "需要的是兩向量維度相同；此步不宣稱任意兩個有序組都可相加。",
            "en": "What is needed is equal dimension; the step does not claim that any two tuples can be added."
          }
        },
        "leaf_basis": "coord_ops"
      },
      {
        "id": "S03",
        "uses": [
          "S02"
        ],
        "before_tex": "u+v",
        "after_tex": "(1+4,\\,-2+0,\\,3+(-1))=(5,-2,2)",
        "relation": "equivalent",
        "rule": "coord_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐分量相加：1+4=5，−2+0=−2，3+(−1)=2。",
          "en": "Add componentwise: 1+4=5, −2+0=−2, 3+(−1)=2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "coord_ops",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "反向：(5,−2,2)+(−v)=(5−4, −2−0, 2+1)=(1,−2,3)=u，還原 u+v=(5,−2,2)。",
            "en": "Backward: (5,−2,2)+(−v)=(5−4, −2−0, 2+1)=(1,−2,3)=u, which restores u+v=(5,−2,2)."
          }
        },
        "leaf_basis": "coord_ops + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "G2"
        ],
        "before_tex": "a(u+v)=2(5,-2,2)",
        "after_tex": "(10,-4,4)",
        "relation": "equivalent",
        "rule": "coord_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "純量 2 乘到每一個分量：2·5=10，2·(−2)=−4，2·2=4。只乘第一個分量是常見錯誤。",
          "en": "The scalar 2 multiplies every component: 2·5=10, 2·(−2)=−4, 2·2=4. Scaling only the first component is a common mistake."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "coord_ops",
            "mul_inverse_nonzero",
            "two_nonzero"
          ],
          "explanation": {
            "zh-Hant": "反向：因 2≠0，乘以 1/2 回到 (5,−2,2)；兩式表示同一向量。",
            "en": "Backward: since 2≠0, multiplying by 1/2 returns (5,−2,2); both sides name the same vector."
          }
        },
        "leaf_basis": "coord_ops + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "G2"
        ],
        "before_tex": "au,\\ av",
        "after_tex": "au=(2,-4,6),\\ av=(8,0,-2)",
        "relation": "equivalent",
        "rule": "coord_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "分別縮放：2·(1,−2,3)=(2,−4,6)，2·(4,0,−1)=(8,0,−2)。",
          "en": "Scale each vector: 2·(1,−2,3)=(2,−4,6) and 2·(4,0,−1)=(8,0,−2)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "coord_ops",
            "mul_inverse_nonzero",
            "two_nonzero"
          ],
          "explanation": {
            "zh-Hant": "同 S04：乘 1/2 分別還原 u 與 v，雙向成立。",
            "en": "As in S04, multiplying by 1/2 restores u and v, so the step is two-way."
          }
        },
        "leaf_basis": "coord_ops + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S05"
        ],
        "before_tex": "au+av",
        "after_tex": "(2+8,\\,-4+0,\\,6+(-2))=(10,-4,4)",
        "relation": "equivalent",
        "rule": "coord_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐分量相加：2+8=10，−4+0=−4，6+(−2)=4。",
          "en": "Add componentwise: 2+8=10, −4+0=−4, 6+(−2)=4."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "coord_ops",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "反向：(10,−4,4)−av=(10−8, −4−0, 4+2)=(2,−4,6)=au。",
            "en": "Backward: (10,−4,4)−av=(10−8, −4−0, 4+2)=(2,−4,6)=au."
          }
        },
        "leaf_basis": "coord_ops + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S04",
          "S06"
        ],
        "before_tex": "a(u+v)=(10,-4,4),\\ au+av=(10,-4,4)",
        "after_tex": "a(u+v)=au+av\\ \\text{for these } a,u,v",
        "relation": "implies",
        "rule": "eq_refl_symm_trans",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊都等於 (10,−4,4)，依等號的對稱與傳遞，這組資料上的實例成立。這只是一個實例：MKO 假設 a2 提醒，實例通過不等於公理已證成。",
          "en": "Both sides equal (10,−4,4), so by symmetry and transitivity of equality the instance holds for these data. It is only an instance: MKO assumption a2 warns that a passing instance does not establish the axiom."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S06",
            "eq_refl_symm_trans",
            "tuple_eq"
          ],
          "explanation": {
            "zh-Hant": "需要三個分量逐一相等；此步只宣稱這一組 a、u、v，不宣稱對所有向量成立。",
            "en": "All three components must agree; the step claims the equation only for this a, u, v, not for all vectors."
          }
        },
        "leaf_basis": "eq_refl_symm_trans + tuple_eq"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "distributive"
        ],
        "before_tex": "a\\in\\mathbb{R},\\ u,v\\in\\mathbb{R}^3\\ \\text{arbitrary}",
        "after_tex": "\\forall a\\in\\mathbb{R}\\ \\forall u,v\\in\\mathbb{R}^3:\\ a(u+v)=au+av",
        "relation": "implies",
        "rule": "L_coord_distrib",
        "conditions": [],
        "explanation": {
          "zh-Hant": "要把 S07 的實例提升為公理，必須對任意實數 a 與任意 u=(u₁,u₂,u₃)、v=(v₁,v₂,v₃) 證明。證明在子步 S08.1 到 S08.3（引理 L_coord_distrib）：向量等式化為三個實數分配律。",
          "en": "To raise the instance of S07 to the axiom, the law must be proved for an arbitrary real a and arbitrary u=(u₁,u₂,u₃), v=(v₁,v₂,v₃). The proof is in sub-steps S08.1 to S08.3 (Lemma L_coord_distrib): the vector equation reduces to three real distributive laws."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_coord_distrib"
          ],
          "explanation": {
            "zh-Hant": "引理在子步中對任意元素證明，沒有使用 S07 的具體數值；S07 只是這個一般結論的一個特例，反過來不能由 S07 推出一般結論。",
            "en": "The lemma is proved in the sub-steps for arbitrary elements without using the numbers of S07; S07 is one special case of the general conclusion, and the general conclusion cannot be inferred from S07."
          }
        },
        "leaf_basis": "coord_ops + distributive + tuple_eq + forall_intro"
      },
      {
        "id": "S08.1",
        "uses": [
          "G1",
          "coord_ops"
        ],
        "before_tex": "a(u+v)",
        "after_tex": "(a(u_1+v_1),\\,a(u_2+v_2),\\,a(u_3+v_3))",
        "relation": "equivalent",
        "rule": "coord_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取任意實數 a 與任意 u、v∈ℝ³，對它們不加任何額外假設。先展開加法 u+v=(u₁+v₁, u₂+v₂, u₃+v₃)，再展開純量乘法，每個分量乘 a。",
          "en": "Take an arbitrary real a and arbitrary u, v∈ℝ³, with no extra assumption on them. Unfold the addition u+v=(u₁+v₁, u₂+v₂, u₃+v₃), then the scalar multiplication, multiplying each component by a."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "coord_ops",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "只是依定義展開；收回定義即回到 a(u+v)，雙向成立。",
            "en": "This only unfolds definitions; folding them back returns a(u+v), so the step is two-way."
          }
        },
        "leaf_basis": "coord_ops + def_unfold"
      },
      {
        "id": "S08.2",
        "uses": [
          "S08.1",
          "distributive"
        ],
        "before_tex": "(a(u_1+v_1),\\,a(u_2+v_2),\\,a(u_3+v_3))",
        "after_tex": "(au_1+av_1,\\,au_2+av_2,\\,au_3+av_3)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "每個分量都是實數，對第 k 個分量（k=1,2,3）使用實數分配律 a(u_k+v_k)=au_k+av_k。這是本證明唯一的實質步驟：向量的分配律來自實數的分配律。",
          "en": "Each component is a real number; apply the real distributive law a(u_k+v_k)=au_k+av_k to the k-th component (k=1,2,3). This is the only substantive move of the proof: the vector distributive law comes from the real one."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "tuple_eq"
          ],
          "explanation": {
            "zh-Hant": "每個分量都是等式改寫，有序組逐分量相等，雙向成立。",
            "en": "Each component is rewritten by an equation and tuples are equal componentwise, so the step is two-way."
          }
        },
        "leaf_basis": "distributive + tuple_eq"
      },
      {
        "id": "S08.3",
        "uses": [
          "S08.2",
          "coord_ops",
          "forall_intro"
        ],
        "before_tex": "(au_1+av_1,\\,au_2+av_2,\\,au_3+av_3)",
        "after_tex": "au+av",
        "relation": "equivalent",
        "rule": "coord_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "收回定義：au=(au₁,au₂,au₃)，av=(av₁,av₂,av₃)，兩者逐分量相加正是左式。因 a、u、v 任意，依 ∀ 引入，a(u+v)=au+av 對所有實數 a 與所有 u,v∈ℝ³ 成立。S08.1 到 S08.3 合起來證得引理 L_coord_distrib。",
          "en": "Fold the definitions back: au=(au₁,au₂,au₃) and av=(av₁,av₂,av₃), and their componentwise sum is exactly the left-hand tuple. Since a, u, v were arbitrary, ∀-introduction gives a(u+v)=au+av for every real a and all u, v∈ℝ³. S08.1 to S08.3 together prove Lemma L_coord_distrib."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "coord_ops",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "收回與展開定義互為反向；∀ 引入只要求 a、u、v 沒有被特別取值，本證明確實沒有。",
            "en": "Folding and unfolding definitions undo each other; ∀-introduction only requires that a, u, v were not given special values, and the proof gave them none."
          }
        },
        "leaf_basis": "coord_ops + def_unfold + forall_intro"
      },
      {
        "id": "S09",
        "uses": [
          "G2"
        ],
        "before_tex": "(a+b)u",
        "after_tex": "(-1)(1,-2,3)=(-1,2,-3)",
        "relation": "equivalent",
        "rule": "coord_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先在實數中算 a+b=2+(−3)=−1（這裡的 + 是實數加法），再做純量乘法：(−1)·(1,−2,3)=(−1,2,−3)。",
          "en": "First compute a+b=2+(−3)=−1 in the reals (this + is real addition), then scale: (−1)·(1,−2,3)=(−1,2,−3)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "coord_ops",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "反向：再乘 −1 得回 u，因為 (−1)(−1)=1。",
            "en": "Backward: multiplying by −1 again returns u, since (−1)(−1)=1."
          }
        },
        "leaf_basis": "coord_ops + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "S05",
          "S09"
        ],
        "before_tex": "au+bu",
        "after_tex": "(2,-4,6)+(-3,6,-9)=(-1,2,-3)=(a+b)u",
        "relation": "implies",
        "rule": "coord_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "bu=(−3)·(1,−2,3)=(−3,6,−9)。這裡的 + 是向量加法，逐分量：(2−3, −4+6, 6−9)=(−1,2,−3)，與 S09 相同，所以 (a+b)u=au+bu 這個實例成立。",
          "en": "bu=(−3)·(1,−2,3)=(−3,6,−9). Here + is vector addition, componentwise: (2−3, −4+6, 6−9)=(−1,2,−3), the same as S09, so the instance (a+b)u=au+bu holds."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S09",
            "tuple_eq",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "只宣稱這組 a、b、u。左式的 + 是實數加法、右式的 + 是向量加法，兩者不同，這條公理正是在連接它們。",
            "en": "Only this a, b, u is claimed. The + on the left is real addition and the + on the right is vector addition; they differ, and this axiom is exactly what links them."
          }
        },
        "leaf_basis": "coord_ops + tuple_eq + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "G2"
        ],
        "before_tex": "(ab)u\\ \\text{and}\\ a(bu)",
        "after_tex": "(-6)(1,-2,3)=(-6,12,-18)=2(-3,6,-9)",
        "relation": "implies",
        "rule": "coord_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "ab=2·(−3)=−6，所以 (ab)u=(−6,12,−18)；bu=(−3,6,−9)，a(bu)=2·(−3,6,−9)=(−6,12,−18)。兩者相同，純量乘法相容律的實例成立。",
          "en": "ab=2·(−3)=−6, so (ab)u=(−6,12,−18); bu=(−3,6,−9) and a(bu)=2·(−3,6,−9)=(−6,12,−18). They agree, so the instance of the compatibility law holds."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "coord_ops",
            "mul_assoc",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "每個分量是 (ab)u_k=a(bu_k)，即實數乘法結合律的實例；此步只核這組資料。",
            "en": "Each component is (ab)u_k=a(bu_k), an instance of associativity of real multiplication; the step checks these data only."
          }
        },
        "leaf_basis": "coord_ops + mul_assoc + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "1u,\\ u+(-u)",
        "after_tex": "1u=(1,-2,3)=u,\\ u+(-u)=(0,0,0)",
        "relation": "implies",
        "rule": "coord_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "1·(1,−2,3)=(1,−2,3)=u；−u=(−1,2,−3)，u+(−u)=(1−1, −2+2, 3−3)=(0,0,0)，即零向量。",
          "en": "1·(1,−2,3)=(1,−2,3)=u; −u=(−1,2,−3) and u+(−u)=(1−1, −2+2, 3−3)=(0,0,0), the zero vector."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "coord_ops",
            "mul_one",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "兩個實例分別由實數的 1·x=x 與 x+(−x)=0 逐分量得到；此步只核 u 這一個向量。",
            "en": "The two instances come componentwise from the real facts 1·x=x and x+(−x)=0; the step checks the single vector u only."
          }
        },
        "leaf_basis": "coord_ops + mul_one + add_inverse"
      },
      {
        "id": "S13",
        "uses": [
          "G3"
        ],
        "before_tex": "(1+1)\\odot(1,0)\\ \\text{vs}\\ 1\\odot(1,0)+1\\odot(1,0)",
        "after_tex": "(2,0)=(1,0)+(1,0)",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 G3 的對照結構中核 (a+b)⊙u=a⊙u+b⊙u 的一個實例：a=b=1，u=(1,0)。左邊 2⊙(1,0)=(2·1, 0)=(2,0)；右邊 1⊙(1,0)+1⊙(1,0)=(1,0)+(1,0)=(2,0)。實例通過。",
          "en": "In the comparison structure of G3, check one instance of (a+b)⊙u=a⊙u+b⊙u: a=b=1, u=(1,0). Left: 2⊙(1,0)=(2·1, 0)=(2,0); right: 1⊙(1,0)+1⊙(1,0)=(1,0)+(1,0)=(2,0). The instance passes."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "def_unfold",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "只核一個向量 u，而它的第二分量是 0，⊙ 對第二分量的異常在這裡看不出來。",
            "en": "Only one vector u is checked, and its second component is 0, so the anomaly of ⊙ in the second component is invisible here."
          }
        },
        "leaf_basis": "def_unfold + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "G3"
        ],
        "before_tex": "(1+1)\\odot(0,1)\\ \\text{vs}\\ 1\\odot(0,1)+1\\odot(0,1)",
        "after_tex": "(0,1)\\neq(0,2)",
        "relation": "implies",
        "rule": "tuple_eq",
        "conditions": [],
        "explanation": {
          "zh-Hant": "換成 u=(0,1)：左邊 2⊙(0,1)=(2·0, 1)=(0,1)；右邊 (0,1)+(0,1)=(0,2)。第二分量 1≠2，所以兩向量不等，這條公理在此失敗。",
          "en": "Take u=(0,1): left 2⊙(0,1)=(2·0, 1)=(0,1); right (0,1)+(0,1)=(0,2). The second components differ, 1≠2, so the vectors differ and the axiom fails here."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "tuple_eq",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一個分量不同就足以推出兩向量不等；反向並未宣稱。",
            "en": "One differing component suffices for the vectors to differ; no converse is claimed."
          }
        },
        "leaf_basis": "def_unfold + tuple_eq + numeric_evaluation"
      },
      {
        "id": "S15",
        "uses": [
          "S08",
          "S13",
          "S14"
        ],
        "before_tex": "\\exists a,b,u:\\ (a+b)\\odot u\\neq a\\odot u+b\\odot u",
        "after_tex": "(\\mathbb{R}^2,+,\\odot)\\ \\text{is not a real vector space}",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "公理 (a+b)u=au+bu 是對所有純量與所有向量的全稱命題。S14 給出一個反例，依量詞否定，這個全稱命題為假，所以對照結構不是實向量空間，儘管 S13 的實例通過。對比：ℝ³ 的 a(u+v)=au+av 在 S08 已對所有元素證明；S07、S10 到 S12 的其他實例只是實例。",
          "en": "The axiom (a+b)u=au+bu is a universal statement over all scalars and all vectors. S14 gives a counterexample, so by quantifier negation the universal statement is false and the comparison structure is not a real vector space, even though the instance in S13 passed. By contrast, a(u+v)=au+av in ℝ³ was proved for all elements in S08; the other checks in S07 and S10 to S12 are instances only."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S08",
            "S14",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "目標 G：對照結構不是向量空間。足夠條件 H：某條公理有反例（H⇒G 由定義與量詞否定）。H 由 S14 直接計算得到，沒有把 G 當作前提。",
            "en": "Goal G: the comparison structure is not a vector space. Sufficient condition H: some axiom has a counterexample (H⇒G by the definition and quantifier negation). H is obtained by direct computation in S14; G is never used as a premise."
          }
        },
        "leaf_basis": "quantifier_negation + mko_rvs_def"
      }
    ],
    "result": {
      "zh-Hant": "在 ℝ³ 中，u 與 v 同維度可相加；a(u+v)=au+av=(10,−4,4)，(a+b)u=au+bu=(−1,2,−3)，(ab)u=a(bu)=(−6,12,−18)，1u=u，u+(−u)=(0,0,0)，五個實例全部精確成立。其中 a(u+v)=au+av 已對所有實數 a 與所有 u,v∈ℝ³ 證明。對照結構 (ℝ², +, ⊙) 在 u=(1,0) 通過 (a+b)⊙u=a⊙u+b⊙u，卻在 u=(0,1) 失敗，所以不是實向量空間：實例通過不等於公理成立。",
      "en": "In ℝ³, u and v have the same dimension and can be added; a(u+v)=au+av=(10,−4,4), (a+b)u=au+bu=(−1,2,−3), (ab)u=a(bu)=(−6,12,−18), 1u=u and u+(−u)=(0,0,0): all five instances hold exactly. Of these, a(u+v)=au+av is proved for every real a and all u, v∈ℝ³. The comparison structure (ℝ², +, ⊙) passes (a+b)⊙u=a⊙u+b⊙u at u=(1,0) but fails at u=(0,1), so it is not a real vector space: successful instances do not establish an axiom."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "(1,2)+(1,2,3)=(2,4,3)",
          "en": "(1,2)+(1,2,3)=(2,4,3)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "兩向量維度不同（2 與 3），向量加法沒有定義；補零是另加的約定。",
          "en": "The vectors have different dimensions (2 and 3), so vector addition is undefined; padding with zeros is an extra convention."
        }
      },
      {
        "claim": {
          "zh-Hant": "因為 a(u+v)=au+av 在 a=2、u=(1,−2,3)、v=(4,0,−1) 成立，所以分配律對 ℝ³ 已證",
          "en": "Since a(u+v)=au+av holds for a=2, u=(1,−2,3), v=(4,0,−1), the distributive law is proved for ℝ³"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "一個實例只支持那組資料；一般證明在 S08 另外對任意 a、u、v 完成。",
          "en": "One instance supports only those data; the general proof is done separately in S08 for arbitrary a, u, v."
        }
      },
      {
        "claim": {
          "zh-Hant": "(ℝ², +, ⊙) 以 a⊙(x,y)=(ax,y) 是向量空間，因為 1⊙u=u 與 (ab)⊙u=a⊙(b⊙u) 對所有 u 成立",
          "en": "(ℝ², +, ⊙) with a⊙(x,y)=(ax,y) is a vector space, because 1⊙u=u and (ab)⊙u=a⊙(b⊙u) hold for all u"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(a+b)⊙u=a⊙u+b⊙u 在 a=b=1、u=(0,1) 失敗：(0,1)≠(0,2)。",
          "en": "(a+b)⊙u=a⊙u+b⊙u fails at a=b=1, u=(0,1): (0,1)≠(0,2)."
        }
      },
      {
        "claim": {
          "zh-Hant": "2(5,−2,2)=(10,−2,2)",
          "en": "2(5,−2,2)=(10,−2,2)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "純量乘法縮放每一個分量，正確結果是 (10,−4,4)。",
          "en": "Scalar multiplication scales every component; the correct result is (10,−4,4)."
        }
      }
    ],
    "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_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "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": "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": "vector_space_axioms",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "向量空間公理（實數或其他域 F 上）：加法成交換群；純量乘法滿足 a(u+v)=au+av、(a+b)u=au+bu、(ab)u=a(bu)、1u=u。",
          "en": "Vector space axioms (over ℝ or another field F): vectors form a commutative group under addition; scalar multiplication satisfies a(u+v)=au+av, (a+b)u=au+bu, (ab)u=a(bu) and 1u=u."
        }
      },
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "two_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數中 0<1，故 0<1+1=2，所以 2≠0。",
          "en": "In the reals 0<1, hence 0<1+1=2, so 2≠0."
        }
      },
      {
        "id": "eq_refl_symm_trans",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。",
          "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w."
        }
      },
      {
        "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": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "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": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "mko_rvs_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-real-vector-space",
          "sha256": "d57c98eb1c006f761ecaaf2870e2a09b30e5a29627351005b7e3bb7b30b33356",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "實向量空間 V 具有向量加法、零向量、加法逆元與實數純量乘法，並滿足結合、交換、分配、單位純量及純量乘法相容等公理。本輪題庫只核固定維度實座標空間的精確公理實例。",
          "en": "A real vector space V has vector addition, a zero vector, additive inverses and real scalar multiplication, satisfying associativity, commutativity, distributivity, the unit-scalar law and scalar-multiplication compatibility. Exercises here check only exact axiom instances in fixed-dimensional real coordinate spaces."
        }
      },
      {
        "id": "coord_ops",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "固定維度 n 的實座標空間 ℝⁿ 中，向量加法與純量乘法逐分量進行；不同維度的向量相加沒有定義。",
          "en": "In the real coordinate space ℝⁿ of fixed dimension n, vector addition and scalar multiplication act componentwise; the sum of vectors of different dimensions is undefined."
        }
      },
      {
        "id": "tuple_eq",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "兩個同維度的有序組相等，當且僅當每個對應分量相等。",
          "en": "Two ordered tuples of the same length are equal if and only if all corresponding components are equal."
        }
      },
      {
        "id": "L_coord_distrib",
        "kind": "lemma",
        "proved_in": [
          "S08.1",
          "S08.2",
          "S08.3"
        ],
        "statement": {
          "zh-Hant": "對所有實數 a 與所有 u,v∈ℝ³：a(u+v)=au+av。",
          "en": "For every real a and all u, v∈ℝ³: a(u+v)=au+av."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "1+4",
        "rhs": "5"
      },
      {
        "id": "N2",
        "step": "S03",
        "lhs": "3+(-1)",
        "rhs": "2"
      },
      {
        "id": "N3",
        "step": "S04",
        "lhs": "2*(-2)",
        "rhs": "-4"
      },
      {
        "id": "N4",
        "step": "S06",
        "lhs": "2+8",
        "rhs": "10"
      },
      {
        "id": "N5",
        "step": "S06",
        "lhs": "6+(-2)",
        "rhs": "4"
      },
      {
        "id": "N6",
        "step": "S08.2",
        "vars": [
          "a",
          "x",
          "y"
        ],
        "degree": 2,
        "lhs": "a*(x+y)",
        "rhs": "a*x+a*y"
      },
      {
        "id": "N7",
        "step": "S09",
        "lhs": "2+(-3)",
        "rhs": "-1"
      },
      {
        "id": "N8",
        "step": "S10",
        "lhs": "-4+(-3)*(-2)",
        "rhs": "2"
      },
      {
        "id": "N9",
        "step": "S10",
        "lhs": "6+(-3)*3",
        "rhs": "-3"
      },
      {
        "id": "N10",
        "step": "S11",
        "lhs": "(2*(-3))*3",
        "rhs": "2*((-3)*3)"
      },
      {
        "id": "N11",
        "step": "S11",
        "lhs": "(2*(-3))*(-2)",
        "rhs": "12"
      },
      {
        "id": "N12",
        "step": "S12",
        "lhs": "3+(-3)",
        "rhs": "0"
      },
      {
        "id": "N13",
        "step": "S14",
        "lhs": "1+1",
        "relation": "!=",
        "rhs": "1"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理固定維度的實座標空間 ℝ³（以及作為反例的 ℝ² 結構），整數向量與整數純量作為實數特例（MKO 假設 a1）。一般證明只做分配律 a(u+v)=au+av 一條；其餘公理在本稿只核實例，不宣稱已對 ℝ³ 完整證明（MKO 假設 a2）。不改動 canonical MKO 或英文 edition。",
      "en": "Treats only the fixed-dimensional real coordinate space ℝ³ (and an ℝ² structure as a counterexample), with integer-coordinate vectors and integer scalars regarded as real vectors and real scalars, respectively (MKO assumption a1). The general proof covers only the distributive law a(u+v)=au+av; the other axioms are checked here as instances only, and no complete proof for ℝ³ is claimed (MKO assumption a2). The canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "\\mathbb{R}^3",
        "有序實數三元組的集合",
        "the set of ordered triples of reals"
      ],
      [
        "u,\\ v",
        "給定向量",
        "the given vectors"
      ],
      [
        "a,\\ b",
        "給定實數純量",
        "the given real scalars"
      ],
      [
        "0=(0,0,0)",
        "ℝ³ 的零向量",
        "the zero vector of ℝ³"
      ],
      [
        "u_k",
        "向量 u 的第 k 個分量，k=1,2,3",
        "the k-th component of u, k=1,2,3"
      ],
      [
        "a\\odot(x,y)",
        "對照結構中改造的純量乘法 (ax,y)",
        "the modified scalar multiplication (ax,y) of the comparison structure"
      ]
    ],
    "md_error_controls": [
      {
        "title": "不同維度不能相加 / Vectors of different dimensions cannot be added",
        "zh": "MKO 指出的誤解之一是把不同維度的向量直接相加。(1,2)+(1,2,3) 在模型中沒有定義；把 (1,2) 補成 (1,2,0) 再相加是另加的約定，不是 ℝⁿ 的向量加法。S02 必須先確認維度相同。",
        "en": "One mistake the MKO names is adding vectors of different dimensions directly. (1,2)+(1,2,3) is undefined in the model; padding (1,2) to (1,2,0) first is an extra convention, not vector addition in ℝⁿ. S02 must confirm equal dimensions first."
      },
      {
        "title": "少數例子不是公理證明 / A few examples do not prove an axiom",
        "zh": "MKO 的另一個誤解是用少數例子冒充一般公理證明。S13 的實例通過，但 S14 在另一個向量上失敗；所以 S07、S10 到 S12 的實例也不能代替證明。本稿唯一對全部元素證明的是 S08，它把向量等式化為對任意分量成立的實數分配律。",
        "en": "Another mistake identified in the MKO is passing off a few examples as a proof of a general axiom. The instance in S13 passes, but S14 fails on another vector; so the instances in S07 and S10 to S12 cannot replace a proof either. The only statement proved here for all elements is S08, which reduces the vector equation to the real distributive law for arbitrary components."
      },
      {
        "title": "兩種加法要分清 / Keep the two additions apart",
        "zh": "(a+b)u 中的 + 是實數加法，au+bu 中的 + 是向量加法。S09 先在實數中算 a+b=−1 再縮放，S10 則逐分量相加向量；混用會把兩邊算成同一件事而失去核對意義。",
        "en": "In (a+b)u the + is real addition; in au+bu it is vector addition. S09 computes a+b=−1 in the reals before scaling, while S10 adds vectors componentwise; confusing them removes the distinction the check is meant to test."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿只對 ℝ³ 一般證明 a(u+v)=au+av。具體核對的五個實例是 a(u+v)=au+av、(a+b)u=au+bu、(ab)u=a(bu)、1u=u 與 u+(−u)=0。加法結合律、交換律與加法零向量律沒有在此另作公理實例核對；其餘一般公理證明也未在此寫出，雖可逐分量化為實數性質。",
        "en": "Only a(u+v)=au+av is proved generally for ℝ³ here. The five concrete instances checked are a(u+v)=au+av, (a+b)u=au+bu, (ab)u=a(bu), 1u=u and u+(−u)=0. Associativity, commutativity and the additive zero-vector law are not separately checked as axiom instances here. The other general axiom proofs are also unwritten, although they can be reduced componentwise to properties of the reals."
      },
      {
        "zh-Hant": "N6 是多項式恆等式在格點上的核對，對應 S08.2 的實數分配律；它是交叉核對，S08 的證明不依賴它。",
        "en": "N6 is a polynomial-identity check on a grid matching the real distributive law of S08.2; it is a cross-check, and the proof in S08 does not depend on it."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "b6ac042765621bf0059bdacec69bbd5ab8eea063304339c44a944552c9a7b347",
    "original_source_sha256": "04c133c4b49ec86f0f5dcac73a472cdc65fcde21915a34e8bd8506fbe9184960",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M052"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}