{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-real-normed-vector-space-01",
  "mko_id": "mko-real-normed-vector-space",
  "title": {
    "zh-Hant": "實賦範向量空間：ℓ¹ 與 ℓ^∞ 公式的全稱範數證書，以及兩個非範數公式",
    "en": "Real normed vector space: universal norm certificates for the ℓ¹ and ℓ^∞ formulas, and two formulas that are not norms"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-real-normed-vector-space-01",
    "source": {
      "mko_id": "mko-real-normed-vector-space",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "8f2547c8a3312254b34134e96a3083f313e6e7334ab0a87a1ba73cc147931964",
      "english_sha256": "2d1073f7489e9288fffcc9421a16d5080d3a63d7d3ae3f185810c959313006e9"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "abs_rules",
      "finite_sum_rules",
      "order_basics",
      "vector_space_axioms",
      "forall_intro",
      "quantifier_negation"
    ],
    "goal": {
      "zh-Hant": "在 ℝⁿ（n≥1 固定）上，對所有向量 u、v 與所有實數 a，證明 ‖v‖₁=Σ|v_i| 與 ‖v‖_∞=max|v_i| 都滿足範數的三條公理：非負且 ‖v‖=0 ⇔ v=0、‖av‖=|a|‖v‖、‖u+v‖≤‖u‖+‖v‖。再在 ℝ³ 中對整數座標向量 u=(3,−4,1)、v=(−1,2,5) 計算兩種範數與三角不等式、齊次性的實例；最後說明 φ(v)=|v₁| 與 ψ(v)=(‖v‖₁)² 不是範數。",
      "en": "On ℝⁿ (n≥1 fixed), for all vectors u, v and all reals a, prove that ‖v‖₁=Σ|v_i| and ‖v‖_∞=max|v_i| both satisfy the three norm axioms: nonnegativity with ‖v‖=0 ⇔ v=0, ‖av‖=|a|‖v‖, and ‖u+v‖≤‖u‖+‖v‖. Then, in ℝ³, compute both norms and instances of the triangle inequality and homogeneity for the integer-coordinate vectors u=(3,−4,1) and v=(−1,2,5); finally show that φ(v)=|v₁| and ψ(v)=(‖v‖₁)² are not norms."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "n≥1 固定；ℝⁿ 是實向量空間，加法與純量乘法逐座標進行；v=0 當且僅當每個 v_i=0。",
          "en": "n≥1 is fixed; ℝⁿ is a real vector space with coordinatewise addition and scalar multiplication; v=0 exactly when every v_i=0."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "命名公式：‖v‖₁=Σ_{i=1}^{n}|v_i|，‖v‖_∞=max_{1≤i≤n}|v_i|。",
          "en": "Named formulas: ‖v‖₁=Σ_{i=1}^{n}|v_i| and ‖v‖_∞=max_{1≤i≤n}|v_i|."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "範數公理（MKO）：(N1) ‖v‖≥0，且 ‖v‖=0 ⇔ v=0；(N2) ‖av‖=|a|‖v‖；(N3) ‖u+v‖≤‖u‖+‖v‖；三條都要對所有 u,v∈ℝⁿ、a∈ℝ 成立。",
          "en": "Norm axioms (MKO): (N1) ‖v‖≥0, and ‖v‖=0 ⇔ v=0; (N2) ‖av‖=|a|‖v‖; (N3) ‖u+v‖≤‖u‖+‖v‖; all three must hold for all u,v∈ℝⁿ and a∈ℝ."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "ℝ³ 中的具體資料：u=(3,−4,1)、v=(−1,2,5)；候選公式 φ(v)=|v₁| 與 ψ(v)=(‖v‖₁)²。",
          "en": "Concrete data in ℝ³: u=(3,−4,1) and v=(−1,2,5); candidate formulas φ(v)=|v₁| and ψ(v)=(‖v‖₁)²."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G2",
          "G3",
          "mko_norm"
        ],
        "before_tex": "\\|\\cdot\\|_1,\\ \\|\\cdot\\|_\\infty\\ \\text{are norms on}\\ \\mathbb{R}^n",
        "after_tex": "(N1)\\land(N2)\\land(N3)\\ \\text{for all } u,v\\in\\mathbb{R}^n,\\ a\\in\\mathbb{R},\\ \\text{for each formula}",
        "relation": "definition",
        "rule": "mko_norm",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，公式是範數，當且僅當三條公理對所有向量與所有純量成立。所以要給的是全稱證書：每一步對任意 u、v、a 論證，而不是代入幾個向量。",
          "en": "By the MKO, a formula is a norm exactly when the three axioms hold for all vectors and all scalars. What is needed is therefore a universal certificate: every step argues for arbitrary u, v, a, rather than substituting a few vectors."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "mko_norm"
          ],
          "explanation": {
            "zh-Hant": "此步只列出義務，不推出新事實。",
            "en": "This step only lists the obligations and derives no new fact."
          }
        },
        "leaf_basis": "mko_norm + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G2",
          "abs_rules",
          "sum_order"
        ],
        "before_tex": "|v_i|\\ge 0\\ \\text{for every } i",
        "after_tex": "\\|v\\|_1=\\sum_{i=1}^{n}|v_i|\\ge 0",
        "relation": "implies",
        "rule": "sum_order",
        "conditions": [],
        "explanation": {
          "zh-Hant": "絕對值非負：每個 |v_i|≥0。非負數的有限和非負，所以 ‖v‖₁≥0（N1 的前半）。",
          "en": "Absolute values are nonnegative: every |v_i|≥0. A finite sum of nonnegative numbers is nonnegative, so ‖v‖₁≥0 (the first half of N1)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "sum_order"
          ],
          "explanation": {
            "zh-Hant": "只需絕對值非負；不宣稱反向。",
            "en": "Only the nonnegativity of absolute values is needed; no converse is claimed."
          }
        },
        "leaf_basis": "abs_rules + sum_order"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "G1",
          "sum_order",
          "L_abs_zero"
        ],
        "before_tex": "\\|v\\|_1=0",
        "after_tex": "v=0",
        "relation": "equivalent",
        "rule": "sum_order",
        "conditions": [],
        "explanation": {
          "zh-Hant": "正向：若 ‖v‖₁=0，對每個 j，0≤|v_j|≤Σ|v_i|=0（非負項的和不小於任一項），所以 |v_j|=0；依引理 L_abs_zero（S03.1），v_j=0；每個座標為 0，即 v=0。反向：v=0 時每個 |v_i|=|0|=0，和為 0。這是 N1 的後半。",
          "en": "Forward: if ‖v‖₁=0, then for every j, 0≤|v_j|≤Σ|v_i|=0 (a sum of nonnegative terms is at least each term), so |v_j|=0; by Lemma L_abs_zero (S03.1), v_j=0; every coordinate is 0, that is v=0. Backward: if v=0 then every |v_i|=|0|=0 and the sum is 0. This is the second half of N1."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "sum_order",
            "L_abs_zero",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "兩個方向都有各自的理由：正向用「和不小於每一項」與 L_abs_zero，反向用 |0|=0。",
            "en": "Each direction has its own reason: forward uses 'the sum is at least each term' and L_abs_zero; backward uses |0|=0."
          }
        },
        "leaf_basis": "sum_order + L_abs_zero + coord_ops"
      },
      {
        "id": "S03.1",
        "uses": [
          "abs_rules",
          "case_analysis"
        ],
        "before_tex": "t\\in\\mathbb{R}",
        "after_tex": "|t|=0\\iff t=0",
        "relation": "equivalent",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若 t=0，|0|=0。若 t≠0，分兩種情況：t>0 時 |t|=t>0；t<0 時 |t|=−t>0。兩種情況 |t|≠0。所以 |t|=0 恰好在 t=0 時成立。這證明了引理 L_abs_zero。",
          "en": "If t=0, |0|=0. If t≠0 there are two cases: for t>0, |t|=t>0; for t<0, |t|=−t>0. In both cases |t|≠0. So |t|=0 holds exactly when t=0. This proves Lemma L_abs_zero."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "abs_rules",
            "case_analysis",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "t=0、t>0、t<0 三種情況涵蓋全部實數（三分律），兩個方向都已論證；不使用 L_abs_zero 本身。",
            "en": "The cases t=0, t>0, t<0 cover every real (trichotomy) and both directions are argued; L_abs_zero itself is not used."
          }
        },
        "leaf_basis": "abs_rules + case_analysis + order_basics"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "G2",
          "abs_rules",
          "finite_sum_rules"
        ],
        "before_tex": "\\|av\\|_1=\\sum_{i=1}^{n}|a v_i|",
        "after_tex": "\\|av\\|_1=\\sum_{i=1}^{n}|a||v_i|=|a|\\,\\|v\\|_1",
        "relation": "equivalent",
        "rule": "finite_sum_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(av)_i=a v_i，且 |a v_i|=|a||v_i|；常數 |a| 可從有限和中提出：Σ|a||v_i|=|a|Σ|v_i|。對任意 a、v 成立，這是 N2。",
          "en": "(av)_i=a v_i and |a v_i|=|a||v_i|; the constant |a| factors out of the finite sum: Σ|a||v_i|=|a|Σ|v_i|. This holds for arbitrary a and v, which is N2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "abs_rules",
            "finite_sum_rules",
            "coord_ops"
          ],
          "explanation": {
            "zh-Hant": "每個等號都是等值改寫，可逆向讀。",
            "en": "Every equality is a rewrite into an equal expression and can be read backwards."
          }
        },
        "leaf_basis": "coord_ops + abs_rules + finite_sum_rules"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "G2",
          "abs_rules",
          "sum_order",
          "finite_sum_rules"
        ],
        "before_tex": "\\|u+v\\|_1=\\sum_{i=1}^{n}|u_i+v_i|",
        "after_tex": "\\|u+v\\|_1\\le\\sum_{i=1}^{n}(|u_i|+|v_i|)=\\|u\\|_1+\\|v\\|_1",
        "relation": "implies",
        "rule": "sum_order",
        "conditions": [],
        "explanation": {
          "zh-Hant": "每個座標用實數的三角不等式 |u_i+v_i|≤|u_i|+|v_i|；逐項不等式相加得和的不等式；再把 Σ(|u_i|+|v_i|) 拆成 Σ|u_i|+Σ|v_i|。對任意 u、v 成立，這是 N3。",
          "en": "In each coordinate use the real triangle inequality |u_i+v_i|≤|u_i|+|v_i|; adding the termwise inequalities gives the inequality for the sums; then split Σ(|u_i|+|v_i|) into Σ|u_i|+Σ|v_i|. This holds for arbitrary u and v, which is N3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "sum_order",
            "finite_sum_rules"
          ],
          "explanation": {
            "zh-Hant": "需要逐座標的三角不等式；不宣稱等號成立。",
            "en": "It needs the coordinatewise triangle inequality; equality is not claimed."
          }
        },
        "leaf_basis": "abs_rules + sum_order + finite_sum_rules + coord_ops"
      },
      {
        "id": "S06",
        "uses": [
          "S02",
          "S03",
          "S04",
          "S05",
          "forall_intro"
        ],
        "before_tex": "(N1),(N2),(N3)\\ \\text{for}\\ \\|\\cdot\\|_1\\ \\text{with arbitrary } u,v,a",
        "after_tex": "\\|\\cdot\\|_1\\ \\text{is a norm on}\\ \\mathbb{R}^n",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S02 至 S05 對任意 u、v、a 證明三條公理，沒有用到具體數值；依 ∀ 引入與合取，ℓ¹ 公式是 ℝⁿ 上的範數。",
          "en": "S02 to S05 prove the three axioms for arbitrary u, v, a, without using particular values; by ∀-introduction and conjunction, the ℓ¹ formula is a norm on ℝⁿ."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S03",
            "S04",
            "S05",
            "forall_intro",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "三條公理缺一不可。",
            "en": "None of the three axioms may be missing."
          }
        },
        "leaf_basis": "forall_intro + and_intro_elim"
      },
      {
        "id": "S07",
        "uses": [
          "G2",
          "max_finite",
          "abs_rules"
        ],
        "before_tex": "\\|v\\|_\\infty=\\max_i|v_i|",
        "after_tex": "\\|v\\|_\\infty=|v_k|\\ge 0\\ \\text{for some } k",
        "relation": "implies",
        "rule": "max_finite",
        "conditions": [
          "n\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "n≥1，有限個數的最大值存在且等於其中某個 |v_k|，而絕對值非負，所以 ‖v‖_∞≥0（N1 的前半）。",
          "en": "Since n≥1, the maximum of finitely many numbers exists and equals some |v_k|, and absolute values are nonnegative, so ‖v‖_∞≥0 (the first half of N1)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "max_finite",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "需要 n≥1 使最大值存在；不宣稱反向。",
            "en": "It needs n≥1 so that the maximum exists; no converse is claimed."
          }
        },
        "leaf_basis": "max_finite + abs_rules"
      },
      {
        "id": "S08",
        "uses": [
          "S07",
          "G1",
          "max_finite",
          "L_abs_zero"
        ],
        "before_tex": "\\|v\\|_\\infty=0",
        "after_tex": "v=0",
        "relation": "equivalent",
        "rule": "max_finite",
        "conditions": [],
        "explanation": {
          "zh-Hant": "正向：若 max|v_i|=0，對每個 i 有 0≤|v_i|≤0，所以 |v_i|=0，依 L_abs_zero 得 v_i=0，即 v=0。反向：v=0 時每個 |v_i|=0，最大值為 0。",
          "en": "Forward: if max|v_i|=0, then 0≤|v_i|≤0 for every i, so |v_i|=0 and by L_abs_zero v_i=0, that is v=0. Backward: if v=0, every |v_i|=0 and the maximum is 0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "max_finite",
            "L_abs_zero",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "正向用「每項不超過最大值」與 L_abs_zero，反向用 |0|=0；兩個方向都有理由。",
            "en": "Forward uses 'each term is at most the maximum' and L_abs_zero, backward uses |0|=0; both directions have reasons."
          }
        },
        "leaf_basis": "max_finite + L_abs_zero + coord_ops"
      },
      {
        "id": "S09",
        "uses": [
          "G1",
          "G2",
          "abs_rules",
          "max_finite",
          "mono_nonneg"
        ],
        "before_tex": "\\|av\\|_\\infty=\\max_i|a||v_i|",
        "after_tex": "\\|av\\|_\\infty=|a|\\max_i|v_i|=|a|\\,\\|v\\|_\\infty",
        "relation": "equivalent",
        "rule": "max_finite",
        "conditions": [
          "|a|\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "|(av)_i|=|a||v_i|。記 M=max|v_i|=|v_k|。上界（S09.1）：每個 |a||v_i|≤|a|M，所以最大值 ≤|a|M；下界（S09.2）：|a||v_k|=|a|M 本身是其中一個值，所以最大值 ≥|a|M。兩者合起來相等，這是 N2。提出的是 |a|≥0，不是 a。",
          "en": "|(av)_i|=|a||v_i|. Let M=max|v_i|=|v_k|. Upper bound (S09.1): every |a||v_i|≤|a|M, so the maximum is ≤|a|M; lower bound (S09.2): |a||v_k|=|a|M is itself one of the values, so the maximum is ≥|a|M. Together they are equal, which is N2. What is factored out is |a|≥0, not a."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "abs_rules",
            "max_finite",
            "mono_nonneg"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個數（上下界相等），等式雙向成立；關鍵條件是 |a|≥0。",
            "en": "Both sides are the same number (the upper and lower bounds agree), so the equality holds both ways; the key condition is |a|≥0."
          }
        },
        "leaf_basis": "abs_rules + max_finite + mono_nonneg + order_basics"
      },
      {
        "id": "S09.1",
        "uses": [
          "max_finite",
          "mono_nonneg",
          "abs_rules"
        ],
        "before_tex": "|v_i|\\le M\\ \\text{for every } i,\\ |a|\\ge 0",
        "after_tex": "\\max_i|a||v_i|\\le|a|M",
        "relation": "implies",
        "rule": "mono_nonneg",
        "conditions": [
          "|a|\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "每個 |v_i|≤M，乘以 |a|≥0 不改變方向：|a||v_i|≤|a|M。所有值都不超過 |a|M，所以它們的最大值也不超過。",
          "en": "Every |v_i|≤M; multiplying by |a|≥0 keeps the direction: |a||v_i|≤|a|M. All values are at most |a|M, so their maximum is too."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "max_finite",
            "mono_nonneg"
          ],
          "explanation": {
            "zh-Hant": "需要乘數非負；若乘以負數，方向會反轉（見錯誤控制）。",
            "en": "It needs a nonnegative multiplier; multiplying by a negative number would reverse the direction (see the error controls)."
          }
        },
        "leaf_basis": "mono_nonneg + max_finite"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "max_finite"
        ],
        "before_tex": "M=|v_k|",
        "after_tex": "|a|M=|a||v_k|\\le\\max_i|a||v_i|",
        "relation": "implies",
        "rule": "max_finite",
        "conditions": [],
        "explanation": {
          "zh-Hant": "最大值 M 由某個 |v_k| 取到，所以 |a|M=|a||v_k| 是被取最大值的數之一，不超過它們的最大值。與 S09.1 合起來得等號。",
          "en": "The maximum M is attained by some |v_k|, so |a|M=|a||v_k| is one of the numbers whose maximum is taken and is at most that maximum. Together with S09.1 this gives equality."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "max_finite"
          ],
          "explanation": {
            "zh-Hant": "需要最大值被取到（有限個數）。",
            "en": "It needs the maximum to be attained (finitely many numbers)."
          }
        },
        "leaf_basis": "max_finite"
      },
      {
        "id": "S10",
        "uses": [
          "G1",
          "G2",
          "abs_rules",
          "max_finite",
          "ineq_add"
        ],
        "before_tex": "|u_i+v_i|\\ \\text{for each } i",
        "after_tex": "\\|u+v\\|_\\infty\\le\\|u\\|_\\infty+\\|v\\|_\\infty",
        "relation": "implies",
        "rule": "max_finite",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對每個 i：|u_i+v_i|≤|u_i|+|v_i|（三角不等式），而 |u_i|≤‖u‖_∞、|v_i|≤‖v‖_∞，相加得 |u_i+v_i|≤‖u‖_∞+‖v‖_∞。每個值都不超過同一個上界 K=‖u‖_∞+‖v‖_∞，所以最大值也不超過 K。這是 N3。",
          "en": "For each i: |u_i+v_i|≤|u_i|+|v_i| (triangle inequality), while |u_i|≤‖u‖_∞ and |v_i|≤‖v‖_∞; adding gives |u_i+v_i|≤‖u‖_∞+‖v‖_∞. Every value is at most the same bound K=‖u‖_∞+‖v‖_∞, so the maximum is at most K. This is N3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "max_finite",
            "ineq_add"
          ],
          "explanation": {
            "zh-Hant": "需要一個與 i 無關的共同上界；不宣稱等號。",
            "en": "It needs a common bound independent of i; equality is not claimed."
          }
        },
        "leaf_basis": "abs_rules + max_finite + ineq_add + coord_ops"
      },
      {
        "id": "S11",
        "uses": [
          "S07",
          "S08",
          "S09",
          "S10",
          "forall_intro"
        ],
        "before_tex": "(N1),(N2),(N3)\\ \\text{for}\\ \\|\\cdot\\|_\\infty\\ \\text{with arbitrary } u,v,a",
        "after_tex": "\\|\\cdot\\|_\\infty\\ \\text{is a norm on}\\ \\mathbb{R}^n",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S07 至 S10 對任意 u、v、a 證明三條公理；依 ∀ 引入與合取，ℓ^∞ 公式是 ℝⁿ 上的範數。",
          "en": "S07 to S10 prove the three axioms for arbitrary u, v, a; by ∀-introduction and conjunction, the ℓ^∞ formula is a norm on ℝⁿ."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "S08",
            "S09",
            "S10",
            "forall_intro",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "三條公理缺一不可。",
            "en": "None of the three axioms may be missing."
          }
        },
        "leaf_basis": "forall_intro + and_intro_elim"
      },
      {
        "id": "S12",
        "uses": [
          "G4",
          "G2"
        ],
        "before_tex": "u=(3,-4,1),\\ v=(-1,2,5)",
        "after_tex": "\\|u\\|_1=8,\\ \\|u\\|_\\infty=4,\\ \\|v\\|_1=8,\\ \\|v\\|_\\infty=5",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "直接用公式：‖u‖₁=3+4+1=8，‖u‖_∞=max(3,4,1)=4；‖v‖₁=1+2+5=8，‖v‖_∞=max(1,2,5)=5。兩個向量的 ℓ¹ 範數相同，但 ℓ^∞ 範數不同：不同範數衡量的大小不同。",
          "en": "Apply the formulas: ‖u‖₁=3+4+1=8 and ‖u‖_∞=max(3,4,1)=4; ‖v‖₁=1+2+5=8 and ‖v‖_∞=max(1,2,5)=5. The two vectors have the same ℓ¹ norm but different ℓ^∞ norms: different norms measure size differently."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G4",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確整數計算；只關於這兩個向量。",
            "en": "Exact integer arithmetic; it concerns these two vectors only."
          }
        },
        "leaf_basis": "numeric_evaluation + abs_rules"
      },
      {
        "id": "S13",
        "uses": [
          "S12",
          "S05",
          "S10"
        ],
        "before_tex": "u+v=(2,-2,6)",
        "after_tex": "\\|u+v\\|_1=10\\le 16,\\ \\|u+v\\|_\\infty=6\\le 9",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "u+v=(2,−2,6)。‖u+v‖₁=2+2+6=10≤8+8=16；‖u+v‖_∞=6≤4+5=9。這兩個數值與 S05、S10 的一般結論一致，但它們只是實例，不是證明。",
          "en": "u+v=(2,−2,6). ‖u+v‖₁=2+2+6=10≤8+8=16; ‖u+v‖_∞=6≤4+5=9. These values agree with the general results of S05 and S10, but they are instances only, not proofs."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確計算；不推出任何全稱結論。",
            "en": "Exact computation; it implies no universal conclusion."
          }
        },
        "leaf_basis": "numeric_evaluation + coord_ops"
      },
      {
        "id": "S14",
        "uses": [
          "S12",
          "S04",
          "S09"
        ],
        "before_tex": "-2u=(-6,8,-2)",
        "after_tex": "\\|-2u\\|_1=16=|-2|\\cdot 8,\\ \\|-2u\\|_\\infty=8=|-2|\\cdot 4",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "−2u=(−6,8,−2)。‖−2u‖₁=6+8+2=16=2·8，‖−2u‖_∞=8=2·4。乘的是 |−2|=2，不是 −2；範數不會是負的。",
          "en": "−2u=(−6,8,−2). ‖−2u‖₁=6+8+2=16=2·8 and ‖−2u‖_∞=8=2·4. The factor is |−2|=2, not −2; a norm is never negative."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "實例，與 S04、S09 一致；不作為證明。",
            "en": "An instance consistent with S04 and S09; not used as a proof."
          }
        },
        "leaf_basis": "numeric_evaluation + coord_ops"
      },
      {
        "id": "S15",
        "uses": [
          "G4",
          "quantifier_negation",
          "G3"
        ],
        "before_tex": "\\varphi(v)=|v_1|\\ \\text{on}\\ \\mathbb{R}^3",
        "after_tex": "\\varphi(0,1,0)=0,\\ (0,1,0)\\neq 0;\\ \\varphi\\ \\text{is not a norm}",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "φ 滿足非負、齊次（|a v₁|=|a||v₁|）與三角不等式（|u₁+v₁|≤|u₁|+|v₁|），而且都對所有向量成立。但 N1 要求 φ(v)=0 只在 v=0 時發生；v=(0,1,0)≠0 卻有 φ(v)=|0|=0。一個反例否定這條全稱公理，所以 φ 不是範數（它只是半範數）。",
          "en": "φ satisfies nonnegativity, homogeneity (|a v₁|=|a||v₁|) and the triangle inequality (|u₁+v₁|≤|u₁|+|v₁|), all for every vector. But N1 requires φ(v)=0 only when v=0; v=(0,1,0)≠0 has φ(v)=|0|=0. One counterexample refutes this universal axiom, so φ is not a norm (it is only a seminorm)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "quantifier_negation",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "否證只需一個向量；其餘兩條公理成立並不能補救。",
            "en": "A refutation needs one vector; the other two axioms holding cannot repair it."
          }
        },
        "leaf_basis": "quantifier_negation + abs_rules + numeric_evaluation"
      },
      {
        "id": "S16",
        "uses": [
          "G4",
          "S06",
          "quantifier_negation"
        ],
        "before_tex": "\\psi(v)=(\\|v\\|_1)^2,\\ e_1=(1,0,0),\\ e_2=(0,1,0)",
        "after_tex": "\\psi(e_1+e_2)=4>2=\\psi(e_1)+\\psi(e_2);\\ \\psi(2e_1)=4\\neq 2=|2|\\psi(e_1)",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "ψ≥0，且由 S03，ψ(v)=0 ⇔ ‖v‖₁=0 ⇔ v=0，所以 N1 成立。但 ψ(e₁+e₂)=(1+1)²=4，而 ψ(e₁)+ψ(e₂)=1+1=2，三角不等式失敗；ψ(2e₁)=2²=4，而 |2|ψ(e₁)=2，齊次性也失敗。所以 ψ 不是範數。",
          "en": "ψ≥0, and by S03 ψ(v)=0 ⇔ ‖v‖₁=0 ⇔ v=0, so N1 holds. But ψ(e₁+e₂)=(1+1)²=4 while ψ(e₁)+ψ(e₂)=1+1=2, so the triangle inequality fails; ψ(2e₁)=2²=4 while |2|ψ(e₁)=2, so homogeneity fails too. Hence ψ is not a norm."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "quantifier_negation",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "每條失敗的公理只需一個精確反例；ψ 通過 N1 不能補救。",
            "en": "Each failing axiom needs only one exact counterexample; ψ passing N1 cannot repair it."
          }
        },
        "leaf_basis": "quantifier_negation + numeric_evaluation + power_laws"
      }
    ],
    "result": {
      "zh-Hant": "對所有 n≥1、所有 u,v∈ℝⁿ 與 a∈ℝ，‖v‖₁=Σ|v_i| 與 ‖v‖_∞=max|v_i| 都滿足 (N1) 非負且只在 v=0 時為 0、(N2) ‖av‖=|a|‖v‖、(N3) 三角不等式，所以兩者都是 ℝⁿ 上的範數。例：u=(3,−4,1)、v=(−1,2,5) 有 ‖u‖₁=‖v‖₁=8、‖u‖_∞=4、‖v‖_∞=5、‖u+v‖₁=10≤16、‖u+v‖_∞=6≤9。φ(v)=|v₁| 因 φ(0,1,0)=0 不是範數；ψ(v)=(‖v‖₁)² 因 ψ(e₁+e₂)=4>2 不是範數。",
      "en": "For every n≥1, all u,v∈ℝⁿ and a∈ℝ, ‖v‖₁=Σ|v_i| and ‖v‖_∞=max|v_i| satisfy (N1) nonnegativity with value 0 only at v=0, (N2) ‖av‖=|a|‖v‖, and (N3) the triangle inequality, so both are norms on ℝⁿ. Example: u=(3,−4,1) and v=(−1,2,5) have ‖u‖₁=‖v‖₁=8, ‖u‖_∞=4, ‖v‖_∞=5, ‖u+v‖₁=10≤16 and ‖u+v‖_∞=6≤9. φ(v)=|v₁| is not a norm since φ(0,1,0)=0; ψ(v)=(‖v‖₁)² is not a norm since ψ(e₁+e₂)=4>2."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "ψ(v)=(‖v‖₁)² 處處非負，且在 e₁、−e₁ 上通過三角不等式，所以 ψ 是範數",
          "en": "ψ(v)=(‖v‖₁)² is nonnegative everywhere and passes the triangle inequality on e₁ and −e₁, so ψ is a norm"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "ψ(e₁+e₂)=4>2=ψ(e₁)+ψ(e₂)。",
          "en": "ψ(e₁+e₂)=4>2=ψ(e₁)+ψ(e₂)."
        }
      },
      {
        "claim": {
          "zh-Hant": "φ(v)=|v₁| 是 ℝ³ 上的範數",
          "en": "φ(v)=|v₁| is a norm on ℝ³"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "φ(0,1,0)=0，但 (0,1,0)≠0。",
          "en": "φ(0,1,0)=0 although (0,1,0)≠0."
        }
      },
      {
        "claim": {
          "zh-Hant": "‖u+v‖₁=10≤16 這個例子證明了 ℓ¹ 的三角不等式",
          "en": "The example ‖u+v‖₁=10≤16 proves the triangle inequality for ℓ¹"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "一個實例不是全稱證明；證明是 S05 的逐座標論證。",
          "en": "One instance is not a universal proof; the proof is the coordinatewise argument of S05."
        }
      },
      {
        "claim": {
          "zh-Hant": "對任意實數 c，max_i(c t_i)=c·max_i t_i",
          "en": "For every real c, max_i(c t_i)=c·max_i t_i"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "c=−1、t=(1,3)：左邊 −1，右邊 −3。",
          "en": "c=−1, t=(1,3): the left side is −1 and the right side is −3."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "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": "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": "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": "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": "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": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "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": "power_laws",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。",
          "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1."
        }
      },
      {
        "id": "mko_norm",
        "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(av)=|a| norm(v)、norm(u+v) 不超過 norm(u)+norm(v)。本輪只使用 ℝⁿ 上的 l1 與 linfinity 命名公式；資格來自公式的全稱代數證書，有限抽樣不冒充一般範數證明。",
          "en": "A norm sends vectors to nonnegative real numbers, with norm(v)=0 iff v=0, norm(av)=|a| norm(v), and norm(u+v)≤norm(u)+norm(v). This slice uses only the named l1 and linfinity formulas on ℝⁿ; qualification comes from universal algebraic certificates for the formulas, and finite samples do not stand in for general norm proofs."
        }
      },
      {
        "id": "coord_ops",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逐座標運算：(u+v)_i=u_i+v_i，(av)_i=a v_i；v=0 ⇔ 每個 v_i=0。",
          "en": "Coordinatewise operations: (u+v)_i=u_i+v_i and (av)_i=a v_i; v=0 ⇔ every v_i=0."
        }
      },
      {
        "id": "sum_order",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限和與序：若對每個 i 都有 s_i≤t_i，則 Σs_i≤Σt_i；若每個 s_i≥0，則 Σs_i≥0，且對每個 j 有 Σs_i≥s_j。",
          "en": "Finite sums and order: if s_i≤t_i for every i, then Σs_i≤Σt_i; if every s_i≥0, then Σs_i≥0 and Σs_i≥s_j for every j."
        }
      },
      {
        "id": "max_finite",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限個實數的最大值：對 t_i（1≤i≤n，n≥1），M=max_i t_i 存在且等於某個 t_k；對每個 i 有 t_i≤M；若對每個 i 有 t_i≤K，則 M≤K。",
          "en": "Maximum of finitely many reals: for t_i (1≤i≤n, n≥1), M=max_i t_i exists and equals some t_k; t_i≤M for every i; if t_i≤K for every i, then M≤K."
        }
      },
      {
        "id": "mono_nonneg",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以非負數保持不等號：若 u≤v 且 w≥0，則 wu≤wv。",
          "en": "Multiplying by a nonnegative number keeps the inequality: if u≤v and w≥0, then wu≤wv."
        }
      },
      {
        "id": "ineq_add",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "不等式相加：若 u₁≤v₁ 且 u₂≤v₂，則 u₁+u₂≤v₁+v₂。",
          "en": "Adding inequalities: if u₁≤v₁ and u₂≤v₂, then u₁+u₂≤v₁+v₂."
        }
      },
      {
        "id": "L_abs_zero",
        "kind": "lemma",
        "proved_in": [
          "S03.1"
        ],
        "statement": {
          "zh-Hant": "對實數 t：|t|=0 ⇔ t=0。",
          "en": "For a real t: |t|=0 ⇔ t=0."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S12",
        "lhs": "abs(3)+abs(-4)+abs(1)",
        "rhs": "8"
      },
      {
        "id": "N2",
        "step": "S12",
        "lhs": "max(abs(3),abs(-4),abs(1))",
        "rhs": "4"
      },
      {
        "id": "N3",
        "step": "S12",
        "lhs": "abs(-1)+abs(2)+abs(5)",
        "rhs": "8"
      },
      {
        "id": "N4",
        "step": "S12",
        "lhs": "max(abs(-1),abs(2),abs(5))",
        "rhs": "5"
      },
      {
        "id": "N5",
        "step": "S13",
        "lhs": "abs(3-1)+abs(-4+2)+abs(1+5)",
        "rhs": "10"
      },
      {
        "id": "N6",
        "step": "S13",
        "lhs": "abs(3-1)+abs(-4+2)+abs(1+5)",
        "relation": "<=",
        "rhs": "8+8"
      },
      {
        "id": "N7",
        "step": "S13",
        "lhs": "max(abs(3-1),abs(-4+2),abs(1+5))",
        "rhs": "6"
      },
      {
        "id": "N8",
        "step": "S13",
        "lhs": "max(abs(3-1),abs(-4+2),abs(1+5))",
        "relation": "<=",
        "rhs": "4+5"
      },
      {
        "id": "N9",
        "step": "S14",
        "lhs": "abs(-2*3)+abs(-2*(-4))+abs(-2*1)",
        "rhs": "abs(-2)*8"
      },
      {
        "id": "N10",
        "step": "S14",
        "lhs": "max(abs(-2*3),abs(-2*(-4)),abs(-2*1))",
        "rhs": "abs(-2)*4"
      },
      {
        "id": "N11",
        "step": "S15",
        "lhs": "abs(0)",
        "rhs": "0"
      },
      {
        "id": "N12",
        "step": "S16",
        "lhs": "(abs(1)+abs(1)+abs(0))**2",
        "relation": ">",
        "rhs": "(abs(1)+abs(0)+abs(0))**2+(abs(0)+abs(1)+abs(0))**2"
      },
      {
        "id": "N13",
        "step": "S16",
        "lhs": "(abs(2)+abs(0)+abs(0))**2",
        "relation": "!=",
        "rhs": "abs(2)*(abs(1)+abs(0)+abs(0))**2"
      },
      {
        "id": "N14",
        "lhs": "max(-1*1,-1*3)",
        "relation": "!=",
        "rhs": "-1*max(1,3)"
      },
      {
        "id": "N15",
        "lhs": "(abs(1-1)+abs(0)+abs(0))**2",
        "relation": "<=",
        "rhs": "1+1"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "停在 MKO 的切片內：只用 ℝⁿ 上命名的 ℓ¹ 與 ℓ^∞ 公式，資格來自對所有向量與純量成立的代數證書；具體例子用整數座標向量。不處理誘導度量、完備性（Banach 空間）、範數等價或有界算子；不改動 canonical MKO 或英文 edition。",
      "en": "Stays inside the MKO's slice: only the named ℓ¹ and ℓ^∞ formulas on ℝⁿ, with qualification coming from algebraic certificates valid for all vectors and scalars; concrete examples use integer-coordinate vectors. Induced metrics, completeness (Banach spaces), equivalence of norms and bounded operators are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "v=(v_i)_{i=1}^{n}",
        "ℝⁿ 中的向量及其座標",
        "a vector of ℝⁿ and its coordinates"
      ],
      [
        "\\|v\\|_1=\\sum_{i=1}^{n}|v_i|",
        "ℓ¹ 範數：座標絕對值總和",
        "the ℓ¹ norm: sum of absolute values of coordinates"
      ],
      [
        "\\|v\\|_\\infty=\\max_{1\\le i\\le n}|v_i|",
        "ℓ^∞ 範數：座標絕對值最大值",
        "the ℓ^∞ norm: largest absolute value of a coordinate"
      ],
      [
        "(N1),(N2),(N3)",
        "範數三公理：正定、齊次、三角不等式",
        "the three norm axioms: definiteness, homogeneity, triangle inequality"
      ],
      [
        "e_1,\\ e_2",
        "ℝ³ 的標準基向量 (1,0,0)、(0,1,0)",
        "the standard basis vectors (1,0,0) and (0,1,0) of ℝ³"
      ],
      [
        "\\varphi,\\ \\psi",
        "兩個待審的候選公式",
        "two candidate formulas under review"
      ]
    ],
    "md_error_controls": [
      {
        "title": "有限點通過測試不是範數證明 / Passing tests at finitely many points is not a norm proof",
        "zh": "MKO 指出的誤解是以有限點通過非負與三角測試就宣稱任意公式是範數。ψ 處處非負，且在 e₁ 與 −e₁ 上 ψ(0)=0≤1+1 通過三角測試，但在 e₁、e₂ 上失敗（S16）。S13、S14 對 ℓ¹、ℓ^∞ 的數值也只是實例；資格來自 S02 至 S11 的全稱證書。",
        "en": "The mistake named by the MKO is to declare an arbitrary formula a norm because it passes nonnegativity and triangle tests at finitely many points. ψ is nonnegative everywhere and passes the triangle test on e₁ and −e₁ (ψ(0)=0≤1+1), yet fails on e₁ and e₂ (S16). The values in S13 and S14 for ℓ¹ and ℓ^∞ are likewise only instances; qualification comes from the universal certificates of S02 to S11."
      },
      {
        "title": "三條公理都要核：半範數不一定是範數 / Check all three axioms: a seminorm need not be a norm",
        "zh": "φ(v)=|v₁| 的非負、齊次與三角不等式對所有向量都成立，最容易被誤認為範數；它失敗的是「只在 v=0 時為 0」（S15）。正定性與其他兩條同樣必須證明。",
        "en": "φ(v)=|v₁| satisfies nonnegativity, homogeneity and the triangle inequality for all vectors, which makes it easy to mistake for a norm; what fails is 'zero only at v=0' (S15). Definiteness must be proved just like the other two axioms."
      },
      {
        "title": "最大值的齊次性要提出 |a| / Homogeneity of the maximum factors out |a|",
        "zh": "max_i(c t_i)=c·max_i t_i 只對 c≥0 成立：t=(1,3)、c=−1 時，max(−1,−3)=−1，而 −1·max(1,3)=−3。S09 先把 |a v_i| 寫成 |a||v_i|，再提出非負的 |a|，才可用 S09.1 的單調性。",
        "en": "max_i(c t_i)=c·max_i t_i holds only for c≥0: with t=(1,3) and c=−1, max(−1,−3)=−1 while −1·max(1,3)=−3. S09 first writes |a v_i| as |a||v_i| and then factors out the nonnegative |a|, which is what allows the monotonicity of S09.1."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "sum_order、max_finite、mono_nonneg、ineq_add 與 coord_ops 是本稿的前提，不在本稿中由實數公理推出。",
        "en": "sum_order, max_finite, mono_nonneg, ineq_add and coord_ops are premises of this guide and are not derived here from the axioms of the reals."
      },
      {
        "zh-Hant": "N1 至 N10 只是 S12 至 S14 具體向量的數值；範數資格不依賴它們。誘導度量與 Banach 空間屬於 MKO 的 formal next。",
        "en": "N1 to N10 are only the values for the concrete vectors of S12 to S14; the norm qualification does not depend on them. Induced metrics and Banach spaces belong to the MKO's formal next."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "69969d495840bc7b4387481a8711aa588a4d19c47ffa88dc52585857ed099067",
    "original_source_sha256": "69969d495840bc7b4387481a8711aa588a4d19c47ffa88dc52585857ed099067",
    "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."
  }
}