{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-norm-induced-metric-01",
  "mko_id": "mko-norm-induced-metric",
  "title": {
    "zh-Hant": "範數誘導度量：由範數公理證明 d(x,y)=‖x−y‖ 是度量，並在 ℝ³ 上以 l1 與 l∞ 精確計算",
    "en": "Norm-induced metric: proving from the norm axioms that d(x,y)=‖x−y‖ is a metric, with exact l1 and l∞ computations in ℝ³"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-norm-induced-metric-01",
    "source": {
      "mko_id": "mko-norm-induced-metric",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "579d0833bcf98eaea0df0559e17c92ecaaa60c6760d51154ff23dcc92d83da4f",
      "english_sha256": "b7e34d77c8ec2850d00096ccb3e031e9c6b20a860974880e45e7fe95d8bbb5bb"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "vector_space_axioms",
      "abs_rules",
      "order_basics",
      "metric_def",
      "finite_sum_rules"
    ],
    "goal": {
      "zh-Hant": "設 (V,‖·‖) 是實賦範向量空間。證明 d(x,y)=‖x−y‖ 滿足非負、分離、對稱與三角不等式；證明 ℝⁿ 上的 l1 與 l∞ 公式都是範數，所以都誘導度量；再在 ℝ³ 上對 x=(1,−2,3)、y=(4,0,−1)、z=(2,5,1) 精確計算 d1 與 d∞。",
      "en": "Let (V,‖·‖) be a real normed vector space. Prove that d(x,y)=‖x−y‖ is nonnegative, separates points, is symmetric and satisfies the triangle inequality; prove that the l1 and l∞ formulas on ℝⁿ are norms, so both induce metrics; then compute d1 and d∞ exactly in ℝ³ for x=(1,−2,3), y=(4,0,−1), z=(2,5,1)."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "(V,‖·‖) 是實賦範向量空間：(N1) ‖v‖≥0 且 ‖v‖=0 ⇔ v=0；(N2) ‖tv‖=|t|‖v‖；(N3) ‖u+v‖≤‖u‖+‖v‖，對所有 u、v∈V、t∈ℝ。",
          "en": "(V,‖·‖) is a real normed vector space: (N1) ‖v‖≥0 and ‖v‖=0 ⇔ v=0; (N2) ‖tv‖=|t|‖v‖; (N3) ‖u+v‖≤‖u‖+‖v‖, for all u, v∈V and t∈ℝ."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "對 x、y∈V 定義 d(x,y)=‖x−y‖。",
          "en": "For x, y∈V define d(x,y)=‖x−y‖."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "對正整數 n，在 ℝⁿ 上：‖v‖₁=Σ_{i=1}^{n}|v_i|，‖v‖∞=max_{1≤i≤n}|v_i|。例子在 ℝ³：x=(1,−2,3)、y=(4,0,−1)、z=(2,5,1)。",
          "en": "For a positive integer n, on ℝⁿ: ‖v‖₁=Σ_{i=1}^{n}|v_i| and ‖v‖∞=max_{1≤i≤n}|v_i|. Example in ℝ³: x=(1,−2,3), y=(4,0,−1), z=(2,5,1)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "metric_def"
        ],
        "before_tex": "d(x,y)=\\|x-y\\|",
        "after_tex": "d\\ge0,\\ d(x,y)=0\\Leftrightarrow x=y,\\ d(x,y)=d(y,x),\\ d(x,z)\\le d(x,y)+d(y,z)",
        "relation": "definition",
        "rule": "mko_nim_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 與度量的定義，要證的四項是：非負、分離、對稱、三角不等式，每一項都對所有 x、y、z∈V 成立。證明只用 G1 的三條範數公理，不用任何具體點。",
          "en": "By the MKO and the definition of a metric, four properties must be proved, each for all x, y, z∈V: nonnegativity, separation, symmetry and the triangle inequality. The proof uses only the three norm axioms of G1, no particular points."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_nim_def",
            "metric_def"
          ],
          "explanation": {
            "zh-Hant": "此步只列出要證的責任，不推出新事實。",
            "en": "This step only lists the obligations and derives no new fact."
          }
        },
        "leaf_basis": "mko_nim_def + metric_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "G2",
          "vector_space_axioms"
        ],
        "before_tex": "x,y\\in V",
        "after_tex": "x-y=x+(-1)y\\in V,\\ d:V\\times V\\to\\mathbb{R}",
        "relation": "implies",
        "rule": "vector_space_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "向量空間對加法與純量乘法封閉，所以 x−y=x+(−1)y 仍在 V 中，‖x−y‖ 是實數。d 是 V×V 到 ℝ 的函數。這需要 x、y 在同一個空間（假設 a1）。",
          "en": "A vector space is closed under addition and scalar multiplication, so x−y=x+(−1)y lies in V and ‖x−y‖ is a real number. d is a function V×V→ℝ. This needs x and y in the same space (assumption a1)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "vector_space_axioms",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "只確認 d 有定義；若兩點不在同一空間，x−y 沒有意義。",
            "en": "It only confirms that d is defined; if the two points were in different spaces, x−y would be meaningless."
          }
        },
        "leaf_basis": "vector_space_axioms + function_def"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "G1"
        ],
        "before_tex": "d(x,y)=\\|x-y\\|",
        "after_tex": "d(x,y)\\ge 0",
        "relation": "implies",
        "rule": "mko_norm_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "N1 對向量 v=x−y 給出 ‖x−y‖≥0，所以 d(x,y)≥0。",
          "en": "N1 applied to the vector v=x−y gives ‖x−y‖≥0, so d(x,y)≥0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_norm_def"
          ],
          "explanation": {
            "zh-Hant": "只用 N1 的第一半；對任意 x、y 成立。",
            "en": "It uses only the first half of N1 and holds for every x, y."
          }
        },
        "leaf_basis": "mko_norm_def"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "G1"
        ],
        "before_tex": "d(x,y)=0",
        "after_tex": "x=y",
        "relation": "equivalent",
        "rule": "mko_norm_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "d(x,y)=0 ⇔ ‖x−y‖=0 ⇔ x−y=0（N1 的第二半）⇔ x=y（S04.1）。這是分離性：距離為零恰好在兩點相同時。",
          "en": "d(x,y)=0 ⇔ ‖x−y‖=0 ⇔ x−y=0 (second half of N1) ⇔ x=y (S04.1). This is separation: the distance is zero exactly when the points coincide."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_norm_def",
            "vector_space_axioms"
          ],
          "explanation": {
            "zh-Hant": "N1 的第二半本身是當且僅當，子步 S04.1 證明的 x−y=0⇔x=y 也是，所以整條鏈雙向成立。",
            "en": "The second half of N1 is itself an iff, and so is x−y=0⇔x=y, proved in sub-step S04.1, so the whole chain holds both ways."
          }
        },
        "leaf_basis": "mko_norm_def + vector_space_axioms"
      },
      {
        "id": "S04.1",
        "uses": [
          "vector_space_axioms"
        ],
        "before_tex": "x-y=0",
        "after_tex": "x=y",
        "relation": "equivalent",
        "rule": "vector_space_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊加上 y：(x−y)+y=x+((−y)+y)=x+0=x，右邊 0+y=y，所以 x=y。反向：x=y 時 x−y=y−y=0。",
          "en": "Add y to both sides: (x−y)+y=x+((−y)+y)=x+0=x, and on the right 0+y=y, so x=y. Backward: if x=y then x−y=y−y=0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "vector_space_axioms"
          ],
          "explanation": {
            "zh-Hant": "加 y 與減 y 互為反運算，兩個方向都由加法群公理給出。",
            "en": "Adding y and subtracting y undo each other; both directions come from the additive group axioms."
          }
        },
        "leaf_basis": "vector_space_axioms + eq_congruence"
      },
      {
        "id": "S05",
        "uses": [
          "S02",
          "G1"
        ],
        "before_tex": "d(x,y)=\\|x-y\\|",
        "after_tex": "d(x,y)=\\|y-x\\|=d(y,x)",
        "relation": "equivalent",
        "rule": "mko_norm_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對稱性由兩個子步得到：S05.1 把 x−y 寫成 (−1)(y−x)，S05.2 用 N2（t=−1）得到 ‖(−1)(y−x)‖=‖y−x‖。",
          "en": "Symmetry comes from two sub-steps: S05.1 writes x−y as (−1)(y−x), and S05.2 uses N2 with t=−1 to get ‖(−1)(y−x)‖=‖y−x‖."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_norm_def",
            "vector_space_axioms",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個實數（子步 S05.1、S05.2 用向量空間公理與 N2 給出）；交換 x、y 再做一次就倒回來。",
            "en": "Both sides are the same real number (sub-steps S05.1 and S05.2 give this from the vector space axioms and N2); swapping x and y and repeating the argument goes back."
          }
        },
        "leaf_basis": "mko_norm_def + vector_space_axioms + abs_rules"
      },
      {
        "id": "S05.1",
        "uses": [
          "vector_space_axioms"
        ],
        "before_tex": "x-y",
        "after_tex": "(-1)(y-x)",
        "relation": "equivalent",
        "rule": "vector_space_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(−1)(y−x)=(−1)y+(−1)(−1)x=−y+x=x−y，用到純量對向量加法的分配、(ab)u=a(bu) 與加法交換。",
          "en": "(−1)(y−x)=(−1)y+(−1)(−1)x=−y+x=x−y, using distributivity of a scalar over vector addition, (ab)u=a(bu) and commutativity of addition."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "vector_space_axioms"
          ],
          "explanation": {
            "zh-Hant": "同一向量的兩種寫法，雙向成立。",
            "en": "Two expressions for the same vector, so the step is two-way."
          }
        },
        "leaf_basis": "vector_space_axioms + neg_mul"
      },
      {
        "id": "S05.2",
        "uses": [
          "S05.1",
          "G1"
        ],
        "before_tex": "\\|(-1)(y-x)\\|",
        "after_tex": "|-1|\\,\\|y-x\\|=\\|y-x\\|",
        "relation": "equivalent",
        "rule": "mko_norm_def",
        "conditions": [
          "t=-1"
        ],
        "explanation": {
          "zh-Hant": "N2 取 t=−1：‖(−1)(y−x)‖=|−1|·‖y−x‖=‖y−x‖。這一步真的需要 N2；只有 N1 與 N3 的函數不一定對稱（見錯誤控制）。",
          "en": "N2 with t=−1: ‖(−1)(y−x)‖=|−1|·‖y−x‖=‖y−x‖. This step really needs N2; a function satisfying only N1 and N3 need not give a symmetric distance (see the error controls)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_norm_def",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "N2 是等式，|−1|=1，兩邊是同一個數。",
            "en": "N2 is an equation and |−1|=1, so both sides are the same number."
          }
        },
        "leaf_basis": "mko_norm_def + abs_rules + mul_one"
      },
      {
        "id": "S06",
        "uses": [
          "S02",
          "G1"
        ],
        "before_tex": "d(x,z)=\\|x-z\\|",
        "after_tex": "d(x,z)\\le d(x,y)+d(y,z)",
        "relation": "implies",
        "rule": "mko_norm_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "三角不等式由兩個子步得到：S06.1 把 x−z 拆成 (x−y)+(y−z)，S06.2 對這個和用 N3。",
          "en": "The triangle inequality comes from two sub-steps: S06.1 splits x−z as (x−y)+(y−z), and S06.2 applies N3 to this sum."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_norm_def",
            "vector_space_axioms"
          ],
          "explanation": {
            "zh-Hant": "需要向量恆等式（子步 S06.1）與 N3（子步 S06.2）；這是不等式，不宣稱反向。",
            "en": "It needs the vector identity (sub-step S06.1) and N3 (sub-step S06.2); this is an inequality, and no converse is claimed."
          }
        },
        "leaf_basis": "mko_norm_def + vector_space_axioms"
      },
      {
        "id": "S06.1",
        "uses": [
          "vector_space_axioms"
        ],
        "before_tex": "x-z",
        "after_tex": "(x-y)+(y-z)",
        "relation": "equivalent",
        "rule": "vector_space_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(x−y)+(y−z)=x+((−y)+y)+(−z)=x+0−z=x−z，用到加法結合、逆元與零元。",
          "en": "(x−y)+(y−z)=x+((−y)+y)+(−z)=x+0−z=x−z, using associativity, inverses and the zero vector."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "vector_space_axioms"
          ],
          "explanation": {
            "zh-Hant": "同一向量的兩種寫法，雙向成立。",
            "en": "Two expressions for the same vector, so the step is two-way."
          }
        },
        "leaf_basis": "vector_space_axioms"
      },
      {
        "id": "S06.2",
        "uses": [
          "S06.1",
          "G1"
        ],
        "before_tex": "\\|(x-y)+(y-z)\\|",
        "after_tex": "\\le\\|x-y\\|+\\|y-z\\|",
        "relation": "implies",
        "rule": "mko_norm_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "N3 取 u=x−y、v=y−z：‖u+v‖≤‖u‖+‖v‖。結合 S06.1，d(x,z)≤d(x,y)+d(y,z)。",
          "en": "N3 with u=x−y and v=y−z: ‖u+v‖≤‖u‖+‖v‖. Combined with S06.1, d(x,z)≤d(x,y)+d(y,z)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_norm_def",
            "S06.1"
          ],
          "explanation": {
            "zh-Hant": "N3 是單向不等式；等號一般不成立，不宣稱。",
            "en": "N3 is a one-way inequality; equality does not hold in general and is not claimed."
          }
        },
        "leaf_basis": "mko_norm_def"
      },
      {
        "id": "S07",
        "uses": [
          "S03",
          "S04",
          "S05",
          "S06",
          "metric_def"
        ],
        "before_tex": "\\text{nonnegativity, separation, symmetry, triangle}",
        "after_tex": "d(x,y)=\\|x-y\\|\\ \\text{is a metric on}\\ V",
        "relation": "implies",
        "rule": "metric_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "四項都對任意 x、y、z 證成，沒有對它們加任何額外假設，依全稱推廣與度量定義，d 是 V 上的度量。這是全稱證明，沒有使用任何有限樣本。",
          "en": "All four properties are proved for arbitrary x, y, z with no extra assumption on them, so by generalization and the definition of a metric, d is a metric on V. This is a universal proof using no finite sample."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S03",
            "S04",
            "S05",
            "S06",
            "metric_def",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "目標 G：d 是度量。足夠條件 H：四項性質對所有點成立；H⇒G 由度量定義。H 由 S03 到 S06 正向證成。",
            "en": "Goal G: d is a metric. Sufficient condition H: the four properties hold for all points; H⇒G by the definition of a metric. H is proved forwards in S03 to S06."
          }
        },
        "leaf_basis": "metric_def + forall_intro + and_intro_elim"
      },
      {
        "id": "S08",
        "uses": [
          "G3"
        ],
        "before_tex": "\\|v\\|_1=\\sum_{i=1}^{n}|v_i|",
        "after_tex": "\\|\\cdot\\|_1\\ \\text{satisfies N1, N2, N3}",
        "relation": "theorem_application",
        "rule": "L_l1_norm",
        "conditions": [],
        "explanation": {
          "zh-Hant": "引理 L_l1_norm（S08.1 到 S08.3 證明）：l1 公式對每個座標用絕對值的性質，再對有限和相加，得到三條範數公理。這是對所有向量的代數證書（MKO 的命名範數資格）。",
          "en": "Lemma L_l1_norm (proved in S08.1 to S08.3): the l1 formula applies the properties of absolute value to each coordinate and adds over the finite sum, giving the three norm axioms. This is an algebraic certificate for all vectors (the qualification of the named norm in the MKO)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "L_l1_norm"
          ],
          "explanation": {
            "zh-Hant": "只需 l1 的公式與引理；對所有 n 與所有向量成立。",
            "en": "It needs only the l1 formula and the lemma; it holds for every n and every vector."
          }
        },
        "leaf_basis": "L_l1_norm"
      },
      {
        "id": "S08.1",
        "uses": [
          "G3",
          "abs_rules",
          "order_basics"
        ],
        "before_tex": "\\sum_{i}|v_i|",
        "after_tex": "\\sum_i|v_i|\\ge0,\\ \\sum_i|v_i|=0\\Leftrightarrow v=0",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "每個 |v_i|≥0，非負數之和非負。若和為 0，每一項都介於 0 與整個和之間，所以每個 |v_i|=0，即每個 v_i=0，v=0；反之 v=0 時和為 0。這是 N1。",
          "en": "Each |v_i|≥0, and a sum of nonnegative numbers is nonnegative. If the sum is 0, each term lies between 0 and the whole sum, so every |v_i|=0, that is every v_i=0 and v=0; conversely v=0 gives the sum 0. This is N1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "關鍵是非負項的和為零迫使每項為零；此子步不使用 L_l1_norm。",
            "en": "The key is that a zero sum of nonnegative terms forces every term to be zero; this sub-step does not use L_l1_norm."
          }
        },
        "leaf_basis": "abs_rules + order_basics + finite_sum_rules"
      },
      {
        "id": "S08.2",
        "uses": [
          "G3",
          "abs_rules",
          "finite_sum_rules"
        ],
        "before_tex": "\\sum_i|t\\,v_i|",
        "after_tex": "|t|\\sum_i|v_i|",
        "relation": "equivalent",
        "rule": "abs_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "|t v_i|=|t||v_i|，再把常數 |t| 提出有限和。這是 N2。",
          "en": "|t v_i|=|t||v_i|, then the constant |t| comes out of the finite sum. This is N2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "abs_rules",
            "finite_sum_rules"
          ],
          "explanation": {
            "zh-Hant": "兩個等式都可倒推；此子步不使用 L_l1_norm。",
            "en": "Both equalities can be reversed; this sub-step does not use L_l1_norm."
          }
        },
        "leaf_basis": "abs_rules + finite_sum_rules"
      },
      {
        "id": "S08.3",
        "uses": [
          "G3",
          "abs_rules",
          "order_basics"
        ],
        "before_tex": "\\sum_i|u_i+v_i|",
        "after_tex": "\\le\\sum_i(|u_i|+|v_i|)=\\sum_i|u_i|+\\sum_i|v_i|",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "每個座標有 |u_i+v_i|≤|u_i|+|v_i|（實數三角不等式）。把 n 個不等式相加（u<v 推出 u+w<v+w，反覆使用並允許等號）得到和的不等式，最後拆開和。這是 N3。S08.1 到 S08.3 合起來證得引理 L_l1_norm。",
          "en": "Each coordinate satisfies |u_i+v_i|≤|u_i|+|v_i| (the real triangle inequality). Adding the n inequalities (u<v gives u+w<v+w, used repeatedly, allowing equality) gives the inequality of sums, and the sum then splits. This is N3. S08.1 to S08.3 together prove Lemma L_l1_norm."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "不等式逐項相加只保持方向，不宣稱反向；此子步不使用 L_l1_norm。",
            "en": "Adding inequalities term by term preserves their direction, and no converse is claimed; this sub-step does not use L_l1_norm."
          }
        },
        "leaf_basis": "abs_rules + order_basics + finite_sum_rules"
      },
      {
        "id": "S09",
        "uses": [
          "G3"
        ],
        "before_tex": "\\|v\\|_\\infty=\\max_i|v_i|",
        "after_tex": "\\|\\cdot\\|_\\infty\\ \\text{satisfies N1, N2, N3}",
        "relation": "theorem_application",
        "rule": "L_linf_norm",
        "conditions": [],
        "explanation": {
          "zh-Hant": "引理 L_linf_norm（S09.1 到 S09.3 證明）：l∞ 公式用有限最大值的性質逐條驗證範數公理。",
          "en": "Lemma L_linf_norm (proved in S09.1 to S09.3): the l∞ formula satisfies the norm axioms by the properties of a finite maximum."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "L_linf_norm"
          ],
          "explanation": {
            "zh-Hant": "只需 l∞ 的公式與引理；對所有向量成立。",
            "en": "It needs only the l∞ formula and the lemma; it holds for every vector."
          }
        },
        "leaf_basis": "L_linf_norm"
      },
      {
        "id": "S09.1",
        "uses": [
          "G3",
          "abs_rules",
          "max_rules"
        ],
        "before_tex": "\\max_i|v_i|",
        "after_tex": "\\max_i|v_i|\\ge0,\\ \\max_i|v_i|=0\\Leftrightarrow v=0",
        "relation": "implies",
        "rule": "max_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "最大值是某個 |v_k|≥0。若最大值為 0，則每個 |v_i|≤0，又 |v_i|≥0，所以每個 v_i=0；反之 v=0 時最大值為 0。這是 N1。",
          "en": "The maximum equals some |v_k|≥0. If the maximum is 0, every |v_i|≤0, and also |v_i|≥0, so every v_i=0; conversely v=0 gives maximum 0. This is N1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "max_rules",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "用到最大值是上界且被取到；此子步不使用 L_linf_norm。",
            "en": "It uses that the maximum is an upper bound and is attained; this sub-step does not use L_linf_norm."
          }
        },
        "leaf_basis": "max_rules + abs_rules + order_basics"
      },
      {
        "id": "S09.2",
        "uses": [
          "G3",
          "abs_rules",
          "max_rules"
        ],
        "before_tex": "\\max_i|t\\,v_i|",
        "after_tex": "|t|\\max_i|v_i|",
        "relation": "equivalent",
        "rule": "max_rules",
        "conditions": [
          "|t|\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "|t v_i|=|t||v_i|。設最大值在 k 取到：對每個 i，|v_i|≤|v_k|，乘以非負數 |t| 保持不等號，所以 |t||v_i|≤|t||v_k|，最大值是 |t||v_k|=|t|max|v_i|。這是 N2。",
          "en": "|t v_i|=|t||v_i|. Let the maximum be attained at k: for every i, |v_i|≤|v_k|, and multiplying by the nonnegative |t| keeps the inequality, so |t||v_i|≤|t||v_k|; the maximum is |t||v_k|=|t|max|v_i|. This is N2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "max_rules",
            "abs_rules",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個數；條件是 |t|≥0（乘以負數會反轉不等號）。此子步不使用 L_linf_norm。",
            "en": "Both sides are the same number; the condition is |t|≥0 (multiplying by a negative number would reverse inequalities). This sub-step does not use L_linf_norm."
          }
        },
        "leaf_basis": "max_rules + abs_rules + order_basics"
      },
      {
        "id": "S09.3",
        "uses": [
          "G3",
          "abs_rules",
          "max_rules"
        ],
        "before_tex": "|u_i+v_i|\\le|u_i|+|v_i|\\le\\|u\\|_\\infty+\\|v\\|_\\infty",
        "after_tex": "\\|u+v\\|_\\infty\\le\\|u\\|_\\infty+\\|v\\|_\\infty",
        "relation": "implies",
        "rule": "max_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對每個 i：|u_i+v_i|≤|u_i|+|v_i|（實數三角不等式），而 |u_i|≤‖u‖∞、|v_i|≤‖v‖∞。所以 ‖u‖∞+‖v‖∞ 是所有 |u_i+v_i| 的上界，最大值也不超過它。這是 N3。S09.1 到 S09.3 合起來證得引理 L_linf_norm。",
          "en": "For every i: |u_i+v_i|≤|u_i|+|v_i| (the real triangle inequality), with |u_i|≤‖u‖∞ and |v_i|≤‖v‖∞. So ‖u‖∞+‖v‖∞ is an upper bound of every |u_i+v_i|, and the maximum does not exceed it. This is N3. S09.1 to S09.3 together prove Lemma L_linf_norm."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "max_rules",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "上界論證只給單向不等式；此子步不使用 L_linf_norm。",
            "en": "The upper-bound argument gives a one-way inequality; this sub-step does not use L_linf_norm."
          }
        },
        "leaf_basis": "abs_rules + max_rules + order_basics"
      },
      {
        "id": "S10",
        "uses": [
          "S07",
          "S08",
          "S09"
        ],
        "before_tex": "\\|\\cdot\\|_1,\\ \\|\\cdot\\|_\\infty\\ \\text{norms on}\\ \\mathbb{R}^n",
        "after_tex": "d_1(x,y)=\\|x-y\\|_1,\\ d_\\infty(x,y)=\\|x-y\\|_\\infty\\ \\text{are metrics}",
        "relation": "theorem_application",
        "rule": "mko_nim_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S08、S09 確認兩者都是 ℝⁿ 上的範數，於是 (ℝⁿ,‖·‖₁) 與 (ℝⁿ,‖·‖∞) 都是賦範空間，S07 的一般定理分別給出兩個度量 d1、d∞。",
          "en": "S08 and S09 confirm that both are norms on ℝⁿ, so (ℝⁿ,‖·‖₁) and (ℝⁿ,‖·‖∞) are normed spaces, and the general theorem S07 gives the two metrics d1 and d∞."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "S08",
            "S09"
          ],
          "explanation": {
            "zh-Hant": "S07 的前提是範數公理，S08、S09 正好提供；每個度量只與誘導它的那個範數配對。",
            "en": "S07 needs the norm axioms, which S08 and S09 supply; each metric is paired only with the norm that induces it."
          }
        },
        "leaf_basis": "mko_nim_def + L_l1_norm + L_linf_norm"
      },
      {
        "id": "S11",
        "uses": [
          "S10",
          "G3"
        ],
        "before_tex": "x-y=(1-4,\\,-2-0,\\,3-(-1))=(-3,-2,4)",
        "after_tex": "d_1(x,y)=3+2+4=9,\\ d_\\infty(x,y)=\\max(3,2,4)=4",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐座標相減得 x−y=(−3,−2,4)，絕對值 3、2、4。l1 取和得 9，l∞ 取最大值得 4。",
          "en": "Subtract coordinatewise: x−y=(−3,−2,4), with absolute values 3, 2, 4. l1 takes the sum, 9, and l∞ the maximum, 4."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確整數計算，可重算（N1、N2）。",
            "en": "Exact integer computation, recomputable (N1, N2)."
          }
        },
        "leaf_basis": "numeric_evaluation + abs_rules"
      },
      {
        "id": "S12",
        "uses": [
          "S05",
          "S11",
          "G3"
        ],
        "before_tex": "y-x=(3,2,-4)",
        "after_tex": "d_1(y,x)=9=d_1(x,y)",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "y−x=(3,2,−4) 的絕對值也是 3、2、4，所以 d1(y,x)=9，與 S05 的一般對稱性一致。這只是一個例子，不是對稱性的證明。",
          "en": "y−x=(3,2,−4) also has absolute values 3, 2, 4, so d1(y,x)=9, consistent with the general symmetry of S05. This is one example, not a proof of symmetry."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一般結論來自 S05；此步只是例子。",
            "en": "The general conclusion comes from S05; this step is only an example."
          }
        },
        "leaf_basis": "numeric_evaluation + abs_rules"
      },
      {
        "id": "S13",
        "uses": [
          "S10",
          "G3"
        ],
        "before_tex": "x-z=(-1,-7,2),\\ y-z=(2,-5,-2)",
        "after_tex": "d_1(x,z)=10,\\ d_\\infty(x,z)=7,\\ d_1(y,z)=9,\\ d_\\infty(y,z)=5",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x−z=(1−2,−2−5,3−1)=(−1,−7,2)：l1 為 1+7+2=10，l∞ 為 7。y−z=(4−2,0−5,−1−1)=(2,−5,−2)：l1 為 2+5+2=9，l∞ 為 5。",
          "en": "x−z=(1−2,−2−5,3−1)=(−1,−7,2): l1 gives 1+7+2=10 and l∞ gives 7. y−z=(4−2,0−5,−1−1)=(2,−5,−2): l1 gives 2+5+2=9 and l∞ gives 5."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確整數計算，可重算（N4 到 N7）。",
            "en": "Exact integer computation, recomputable (N4 to N7)."
          }
        },
        "leaf_basis": "numeric_evaluation + abs_rules"
      },
      {
        "id": "S14",
        "uses": [
          "S06",
          "S11",
          "S13"
        ],
        "before_tex": "d_1(x,z)=10,\\ d_\\infty(x,z)=7",
        "after_tex": "10\\le 9+9,\\ 7\\le 4+5",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在這三點上，d1 的三角不等式為 10≤18，d∞ 為 7≤9，都與 S06 一致。這是例子的核對，不能取代 S06 的全稱證明。",
          "en": "At these three points the triangle inequality reads 10≤18 for d1 and 7≤9 for d∞, both consistent with S06. This checks an example and cannot replace the universal proof of S06."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "有限點的核對只支持這些點；一般結論來自 S06。",
            "en": "A check at finitely many points supports only those points; the general conclusion comes from S06."
          }
        },
        "leaf_basis": "order_basics + numeric_evaluation"
      },
      {
        "id": "S15",
        "uses": [
          "S11"
        ],
        "before_tex": "d_1(x,y)=9,\\ d_\\infty(x,y)=4",
        "after_tex": "d_1(x,y)\\neq d_\\infty(x,y)",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "同一對點在兩個命名範數下距離不同（9≠4），所以每次計算須明示所用範數，且兩點須在同一維度（假設 a1）。同一度量的三角不等式不能不加說明地混用範數；若另有範數比較，則可導出混合界，例如 ‖v‖∞≤‖v‖₁ 給出 d∞(x,z)≤d1(x,y)+d∞(y,z)。",
          "en": "The same pair of points has different distances under the two named norms (9≠4), so each computation must identify its norm and keep both points in the same dimension (assumption a1). A mixed-norm bound needs an additional comparison estimate, not just the triangle inequality for one metric. Here ‖v‖∞≤‖v‖₁ gives d∞(x,z)≤d1(x,y)+d∞(y,z)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "只比較兩個精確數，說明度量依賴所選範數。",
            "en": "It only compares two exact numbers, showing that the metric depends on the chosen norm."
          }
        },
        "leaf_basis": "numeric_evaluation"
      }
    ],
    "result": {
      "zh-Hant": "對任何實賦範向量空間，d(x,y)=‖x−y‖ 非負（N1）、分離（N1 與 x−y=0⇔x=y）、對稱（N2 取 t=−1）、滿足三角不等式（x−z=(x−y)+(y−z) 與 N3），所以是度量。ℝⁿ 上的 l1 與 l∞ 都是範數，因此 d1、d∞ 都是度量。例子：d1(x,y)=9、d∞(x,y)=4、d1(x,z)=10、d∞(x,z)=7、d1(y,z)=9、d∞(y,z)=5。",
      "en": "For every real normed vector space, d(x,y)=‖x−y‖ is nonnegative (N1), separates points (N1 with x−y=0⇔x=y), is symmetric (N2 with t=−1) and satisfies the triangle inequality (x−z=(x−y)+(y−z) and N3), so it is a metric. The l1 and l∞ formulas on ℝⁿ are norms, so d1 and d∞ are metrics. Example: d1(x,y)=9, d∞(x,y)=4, d1(x,z)=10, d∞(x,z)=7, d1(y,z)=9, d∞(y,z)=5."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "d(x,y)=‖x‖₁−‖y‖₁ 是 ℝ³ 上的度量",
          "en": "d(x,y)=‖x‖₁−‖y‖₁ is a metric on ℝ³"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "取 x=0、y=(1,0,0) 得 0−1=−1<0，違反非負。",
          "en": "x=0 and y=(1,0,0) give 0−1=−1<0, violating nonnegativity."
        }
      },
      {
        "claim": {
          "zh-Hant": "ℝ³ 上的離散度量由某個範數誘導",
          "en": "The discrete metric on ℝ³ is induced by some norm"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "誘導度量滿足 d(2v,0)=2d(v,0)，離散度量對 v≠0 給出 1≠2。",
          "en": "An induced metric satisfies d(2v,0)=2d(v,0), but the discrete metric gives 1≠2 for v≠0."
        }
      },
      {
        "claim": {
          "zh-Hant": "已在 x、y、z 三點核對三角不等式（10≤18、7≤9），所以 d1、d∞ 是度量",
          "en": "The triangle inequality was checked at x, y, z (10≤18, 7≤9), so d1 and d∞ are metrics"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "有限樣本只支持這些點；度量性質來自 S06 至 S10 的全稱證明。",
          "en": "A finite sample supports only those points; the metric property comes from the universal proof in S06 to S10."
        }
      },
      {
        "claim": {
          "zh-Hant": "(1,−2,3)∈ℝ³ 與 (1,2)∈ℝ² 的 l1 距離是 4",
          "en": "The l1 distance between (1,−2,3)∈ℝ³ and (1,2)∈ℝ² is 4"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "兩點不在同一空間，x−y 沒有定義（假設 a1）。",
          "en": "The points are not in the same space, so x−y is undefined (assumption a1)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "abs_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "絕對值（實數）：|u|=u（u≥0）、|u|=−u（u<0）；|u|≥0；|uv|=|u||v|；三角不等式 |u+v|≤|u|+|v|；|u|<r ⇔ −r<u<r。",
          "en": "Absolute value (reals): |u|=u if u≥0 and |u|=−u if u<0; |u|≥0; |uv|=|u||v|; triangle inequality |u+v|≤|u|+|v|; |u|<r ⇔ −r<u<r."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "metric_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "度量：d:X×X→ℝ 滿足 d(x,y)≥0 且 d(x,y)=0 ⇔ x=y；d(x,y)=d(y,x)；三角不等式 d(x,z)≤d(x,y)+d(y,z)。",
          "en": "Metric: d:X×X→ℝ with d(x,y)≥0 and d(x,y)=0 ⇔ x=y; d(x,y)=d(y,x); triangle inequality d(x,z)≤d(x,y)+d(y,z)."
        }
      },
      {
        "id": "finite_sum_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限和：Σ_{i=1}^{0} a_i=0、Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}；Σ(a_i+b_i)=Σa_i+Σb_i；Σ c·a_i=c·Σa_i。",
          "en": "Finite sums: Σ_{i=1}^{0} a_i=0 and Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}; Σ(a_i+b_i)=Σa_i+Σb_i; Σ c·a_i=c·Σa_i."
        }
      },
      {
        "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": "function_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "函數：f:A→B 對每個 a∈A 指定恰好一個 f(a)∈B。兩函數相等 ⇔ 定義域、對應域相同且對每個 a 都有相同值。",
          "en": "Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a."
        }
      },
      {
        "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": "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": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "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": "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": "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": "mko_nim_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-norm-induced-metric",
          "sha256": "579d0833bcf98eaea0df0559e17c92ecaaa60c6760d51154ff23dcc92d83da4f",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "若(V,norm)為已認證的實賦範向量空間，則d(x,y)=norm(x-y)滿足非負、分離、對稱與三角不等式。本輪只在R^n的命名l1與linfinity範數作精確座標切片。",
          "en": "If (V,norm) is a certified real normed vector space, d(x,y)=norm(x−y) is nonnegative, separates points, is symmetric and satisfies the triangle inequality. This slice uses exact coordinate calculations only for the named l1 and linfinity norms on R^n."
        }
      },
      {
        "id": "mko_norm_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-real-normed-vector-space",
          "sha256": "8f2547c8a3312254b34134e96a3083f313e6e7334ab0a87a1ba73cc147931964",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "範數把向量送到非負實數，且norm(v)=0當且僅當v=0、norm(a v)=|a| norm(v)、norm(u+v)不超過norm(u)+norm(v)。本輪只使用R^n上的l1與linfinity命名公式。",
          "en": "A norm sends vectors to nonnegative real numbers, with norm(v)=0 iff v=0, norm(a v)=|a| norm(v), and norm(u+v)≤norm(u)+norm(v). This slice uses only the named l1 and linfinity formulas on R^n."
        }
      },
      {
        "id": "max_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限個實數的最大值：max_i w_i 是其中某一個 w_k，且對每個 i 有 w_i≤max_i w_i；若所有 w_i≤M 則 max_i w_i≤M。",
          "en": "Maximum of finitely many reals: max_i w_i equals some w_k, and w_i≤max_i w_i for every i; if every w_i≤M then max_i w_i≤M."
        }
      },
      {
        "id": "L_l1_norm",
        "kind": "lemma",
        "proved_in": [
          "S08.1",
          "S08.2",
          "S08.3"
        ],
        "statement": {
          "zh-Hant": "ℝⁿ 上的 ‖v‖₁=Σ|v_i| 滿足 N1、N2、N3。",
          "en": "On ℝⁿ, ‖v‖₁=Σ|v_i| satisfies N1, N2 and N3."
        }
      },
      {
        "id": "L_linf_norm",
        "kind": "lemma",
        "proved_in": [
          "S09.1",
          "S09.2",
          "S09.3"
        ],
        "statement": {
          "zh-Hant": "ℝⁿ 上的 ‖v‖∞=max|v_i| 滿足 N1、N2、N3。",
          "en": "On ℝⁿ, ‖v‖∞=max|v_i| satisfies N1, N2 and N3."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S11",
        "lhs": "abs(1-4)+abs(-2-0)+abs(3-(-1))",
        "rhs": "9"
      },
      {
        "id": "N2",
        "step": "S11",
        "lhs": "max(abs(1-4),abs(-2-0),abs(3-(-1)))",
        "rhs": "4"
      },
      {
        "id": "N3",
        "step": "S12",
        "lhs": "abs(4-1)+abs(0-(-2))+abs(-1-3)",
        "rhs": "9"
      },
      {
        "id": "N4",
        "step": "S13",
        "lhs": "abs(1-2)+abs(-2-5)+abs(3-1)",
        "rhs": "10"
      },
      {
        "id": "N5",
        "step": "S13",
        "lhs": "max(abs(1-2),abs(-2-5),abs(3-1))",
        "rhs": "7"
      },
      {
        "id": "N6",
        "step": "S13",
        "lhs": "abs(4-2)+abs(0-5)+abs(-1-1)",
        "rhs": "9"
      },
      {
        "id": "N7",
        "step": "S13",
        "lhs": "max(abs(4-2),abs(0-5),abs(-1-1))",
        "rhs": "5"
      },
      {
        "id": "N8",
        "step": "S14",
        "lhs": "10",
        "relation": "<=",
        "rhs": "9+9"
      },
      {
        "id": "N9",
        "step": "S14",
        "lhs": "7",
        "relation": "<=",
        "rhs": "4+5"
      },
      {
        "id": "N10",
        "step": "S15",
        "lhs": "9",
        "relation": "!=",
        "rhs": "4"
      },
      {
        "id": "N11",
        "lhs": "1",
        "relation": "!=",
        "rhs": "2*1"
      },
      {
        "id": "N12",
        "lhs": "max(-1,-2*(-1))",
        "relation": "!=",
        "rhs": "max(1,-2*1)"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只在 MKO 的範圍內：已認證的實賦範向量空間，以及 ℝⁿ 上命名的 l1 與 l∞ 範數；一般定理中兩點只需在同一實賦範空間並使用該範數，不要求有限維；有限維座標與同一命名範數的要求只用於後面的 ℝⁿ 計算例子（假設 a1）。誘導性由全稱公式與範數公理證明，有限點的距離只作例子（假設 a2）。不處理完備性或 Banach 空間；不改動 canonical MKO 或英文 edition。",
      "en": "Stays inside the MKO: a certified real normed vector space, and the named l1 and l∞ norms on ℝⁿ; in the general theorem both points need only lie in the same real normed space and use its norm, with no finite-dimensional hypothesis. The finite-coordinate and named-norm requirements apply only to the later ℝⁿ examples (assumption a1). The metric property is proved from the universal formula and the norm axioms; distances between finitely many points are examples only (assumption a2). Completeness and Banach spaces are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "\\|v\\|",
        "向量 v 的範數",
        "norm of the vector v"
      ],
      [
        "d(x,y)",
        "誘導距離 ‖x−y‖",
        "induced distance ‖x−y‖"
      ],
      [
        "\\|v\\|_1",
        "l1 範數 Σ|v_i|",
        "l1 norm Σ|v_i|"
      ],
      [
        "\\|v\\|_\\infty",
        "l∞ 範數 max|v_i|",
        "l∞ norm max|v_i|"
      ],
      [
        "d_1,\\ d_\\infty",
        "l1 與 l∞ 誘導的度量",
        "metrics induced by l1 and l∞"
      ],
      [
        "x,\\ y,\\ z",
        "ℝ³ 中的三個整數座標點",
        "three integer-coordinate points of ℝ³"
      ]
    ],
    "md_error_controls": [
      {
        "title": "有限樣本不能建立誘導 / A finite sample does not establish the metric or its origin",
        "zh": "MKO 指出的常見誤解是核對有限幾對距離後就宣稱某個度量由範數誘導。S14 的 10≤18、7≤9 只是例子。反例：ℝ³ 上的離散度量（相異點距離 1）滿足度量公理，但範數誘導的度量必有 d(2v,0)=‖2v‖=2d(v,0)，離散度量對 v≠0 給出 1≠2，所以它不由任何範數誘導。",
        "en": "The common mistake named by the MKO is to claim, after checking finitely many distances, that a metric is induced by a norm. The 10≤18 and 7≤9 of S14 are only examples. Counterexample: the discrete metric on ℝ³ (distance 1 between distinct points) satisfies the metric axioms, but a norm-induced metric always has d(2v,0)=‖2v‖=2d(v,0), while the discrete metric gives 1≠2 for v≠0, so no norm induces it."
      },
      {
        "title": "N1 與 N3 不足以保證對稱性 / N1 and N3 alone do not ensure symmetry",
        "zh": "只有 N1 與 N3 不夠。在 ℝ 上 f(v)=max(v,−2v) 非負、只在 0 為零、且次可加（線性函數的最大值），但 f(1)=1、f(−1)=2，所以 f(x−y) 不對稱：從 1 到 0 是 1，從 0 到 1 是 2。S05.2 正是用 N2 的 t=−1 才得到對稱。",
        "en": "N1 and N3 alone are not enough. On ℝ, f(v)=max(v,−2v) is nonnegative, zero only at 0, and subadditive (a maximum of linear functions), but f(1)=1 and f(−1)=2, so f(x−y) is not symmetric: from 1 to 0 it is 1, from 0 to 1 it is 2. S05.2 obtains symmetry exactly through N2 with t=−1."
      },
      {
        "title": "同一命名範數、同一維度 / One named norm, one dimension",
        "zh": "d1(x,y)=9 而 d∞(x,y)=4（S15）。混合範數不能只引用同一度量的三角不等式，還須核比較估計；本例因 d∞(x,y)≤d1(x,y)，確有 d∞(x,z)≤d∞(x,y)+d∞(y,z)≤d1(x,y)+d∞(y,z)。兩點維度不同時，x−y 在本模型中沒有定義（S02）。",
        "en": "d1(x,y)=9 while d∞(x,y)=4 (S15). Mixing norms requires a comparison estimate in addition to the triangle inequality for one metric. In this example d∞(x,y)≤d1(x,y), so d∞(x,z)≤d∞(x,y)+d∞(y,z)≤d1(x,y)+d∞(y,z) does hold. When the two points have different dimensions, x−y is undefined in this model (S02)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "max_rules 是本稿的前提，不在本稿中由實數序公理推出。",
        "en": "max_rules is a premise of this guide and is not derived here from the order axioms of the reals."
      },
      {
        "zh-Hant": "錯誤控制中 f(v)=max(v,−2v) 的次可加性只作說明（線性函數最大值的標準性質），未列為編號步驟。",
        "en": "The subadditivity of f(v)=max(v,−2v) in the error controls is explanatory (a standard property of a maximum of linear functions) and is not a numbered step."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "49854f1c5f3eec27f6de9fba35448a82d0db140a86fbabd4f87f370e0e39ac1a",
    "original_source_sha256": "6d24d167b92a2af8ec2a7ab0a267082c9664a86cde10e8728624bf97a61aff02",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M071"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}