{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-inner-product-induced-norm-01",
  "mko_id": "mko-inner-product-induced-norm",
  "title": {
    "zh-Hant": "內積誘導範數：證明 √⟨x,x⟩ 是範數並在 ℝ² 正定矩陣下精確計算",
    "en": "Inner-product-induced norm: proving √⟨x,x⟩ is a norm and computing exactly for a positive-definite matrix on ℝ²"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-inner-product-induced-norm-01",
    "source": {
      "mko_id": "mko-inner-product-induced-norm",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "9b9498dd5a3aa838a1854d97c526a01a657f0f88bab64b00a78ef20fb1f80c97",
      "english_sha256": "ca3df586cdb1653a3796cb144948a8b5f7eb9a7d8b9950440e12d784e2757548"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "sqrt_def",
      "square_nonneg",
      "abs_rules",
      "order_basics",
      "vector_space_axioms",
      "case_analysis"
    ],
    "goal": {
      "zh-Hant": "在任一實內積空間中證明 N(v)=√⟨v,v⟩ 是範數：非負與分離、絕對齊次、三角不等式（經由完整證明的 Cauchy–Schwarz）。再在 ℝ² 上以 A=[[2,1],[1,2]] 定義的內積，對 x=(1,0)、y=(0,1) 精確算出平方範數、距離平方，並做 Cauchy–Schwarz、三角不等式與平行四邊形恆等式的實例核對。",
      "en": "In any real inner-product space, prove that N(v)=√⟨v,v⟩ is a norm: nonnegativity and separation, absolute homogeneity, and the triangle inequality (via a fully proved Cauchy–Schwarz inequality). Then, on ℝ² with the inner product given by A=[[2,1],[1,2]], compute exactly the squared norms and the squared distance for x=(1,0) and y=(0,1), and check Cauchy–Schwarz, the triangle inequality and the parallelogram identity on this instance."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "V 是實內積空間：⟨·,·⟩:V×V→ℝ 對每個變數線性（雙線性）、對稱 ⟨u,v⟩=⟨v,u⟩、正定：⟨v,v⟩≥0，且 ⟨v,v⟩=0 ⇔ v=0。",
          "en": "V is a real inner-product space: ⟨·,·⟩:V×V→ℝ is linear in each argument (bilinear), symmetric ⟨u,v⟩=⟨v,u⟩, and positive definite: ⟨v,v⟩≥0, with ⟨v,v⟩=0 ⇔ v=0."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "具體例子：V=ℝ²，⟨u,v⟩=uᵀAv=2u₁v₁+u₁v₂+u₂v₁+2u₂v₂，A=[[2,1],[1,2]] 已通過雙線性、對稱與正定（MKO 假設 a1）；x=(1,0)、y=(0,1) 位於同一 ℝ²。",
          "en": "Concrete example: V=ℝ², ⟨u,v⟩=uᵀAv=2u₁v₁+u₁v₂+u₂v₁+2u₂v₂, where A=[[2,1],[1,2]] already qualifies as bilinear, symmetric and positive definite (MKO assumption a1); x=(1,0) and y=(0,1) lie in the same ℝ²."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "定義 N(v)=√⟨v,v⟩（MKO 敘述）。",
          "en": "Define N(v)=√⟨v,v⟩ (MKO statement)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G3"
        ],
        "before_tex": "\\langle v,v\\rangle\\ge 0",
        "after_tex": "N(v)=\\sqrt{\\langle v,v\\rangle}\\ \\text{defined},\\ N(v)\\ge 0",
        "relation": "implies",
        "rule": "sqrt_def",
        "conditions": [
          "\\langle v,v\\rangle\\ge0"
        ],
        "explanation": {
          "zh-Hant": "正定性給 ⟨v,v⟩≥0，所以主平方根有定義；依定義主平方根非負，故 N(v)≥0。若 ⟨v,v⟩ 可能為負（例如不正定的對稱矩陣），N 根本無法定義。",
          "en": "Positive definiteness gives ⟨v,v⟩≥0, so the principal square root is defined, and by definition it is nonnegative, so N(v)≥0. If ⟨v,v⟩ could be negative (as for a symmetric matrix that is not positive definite), N would not even be defined."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "sqrt_def"
          ],
          "explanation": {
            "zh-Hant": "只用到 ⟨v,v⟩≥0 與主平方根的定義。",
            "en": "It uses only ⟨v,v⟩≥0 and the definition of the principal square root."
          }
        },
        "leaf_basis": "sqrt_def + ip_axioms"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "G1"
        ],
        "before_tex": "N(v)=0",
        "after_tex": "v=0",
        "relation": "equivalent",
        "rule": "ip_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "N(v)=0 ⇔ ⟨v,v⟩=0：√s=0 ⇔ s=0（0 是 0 的唯一非負平方根）。再由正定性 ⟨v,v⟩=0 ⇔ v=0。這就是分離性：正定性給分離（MKO）。",
          "en": "N(v)=0 ⇔ ⟨v,v⟩=0, because √s=0 ⇔ s=0 (0 is the only nonnegative square root of 0). Then positive definiteness gives ⟨v,v⟩=0 ⇔ v=0. This is separation: positive definiteness gives separation (MKO)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "sqrt_def",
            "ip_axioms"
          ],
          "explanation": {
            "zh-Hant": "兩個等價各自雙向：平方根的唯一性與正定性的「當且僅當」。",
            "en": "Both equivalences are two-way: uniqueness of the square root and the 'if and only if' of positive definiteness."
          }
        },
        "leaf_basis": "sqrt_def + ip_axioms + logic_chain"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "ip_axioms"
        ],
        "before_tex": "\\langle cv,cv\\rangle",
        "after_tex": "c^2\\langle v,v\\rangle",
        "relation": "equivalent",
        "rule": "ip_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第一個變數線性：⟨cv,cv⟩=c⟨v,cv⟩；第二個變數線性：c⟨v,cv⟩=c·c⟨v,v⟩=c²⟨v,v⟩。",
          "en": "Linearity in the first argument: ⟨cv,cv⟩=c⟨v,cv⟩; linearity in the second: c⟨v,cv⟩=c·c⟨v,v⟩=c²⟨v,v⟩."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "ip_axioms",
            "mul_assoc"
          ],
          "explanation": {
            "zh-Hant": "這是恆等式，兩邊對所有 c、v 相等，雙向成立。",
            "en": "This is an identity, equal for all c and v, so it holds both ways."
          }
        },
        "leaf_basis": "ip_axioms + mul_assoc + power_laws"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "S01"
        ],
        "before_tex": "N(cv)=\\sqrt{c^2\\langle v,v\\rangle}",
        "after_tex": "N(cv)=|c|\\,N(v)",
        "relation": "equivalent",
        "rule": "L_sqrt_scale",
        "conditions": [
          "\\langle v,v\\rangle\\ge0"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_sqrt_scale（在 S04.1 證明，s=⟨v,v⟩）得 √(c²⟨v,v⟩)=|c|√⟨v,v⟩=|c|N(v)。這是絕對齊次性：結果是 |c| 而不是 c；例如 N(−3x)=3N(x)。",
          "en": "By Lemma L_sqrt_scale (proved in S04.1, with s=⟨v,v⟩), √(c²⟨v,v⟩)=|c|√⟨v,v⟩=|c|N(v). This is absolute homogeneity: the factor is |c|, not c; for example N(−3x)=3N(x)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_sqrt_scale"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一實數的兩種寫法，雙向成立。",
            "en": "Both sides are two expressions for the same real number, so it holds both ways."
          }
        },
        "leaf_basis": "sqrt_def + abs_rules + power_laws"
      },
      {
        "id": "S04.1",
        "uses": [
          "sqrt_def",
          "abs_rules"
        ],
        "before_tex": "\\sqrt{c^2s},\\ s\\ge0",
        "after_tex": "|c|\\sqrt{s}",
        "relation": "equivalent",
        "rule": "sqrt_def",
        "conditions": [
          "s\\ge0"
        ],
        "explanation": {
          "zh-Hant": "b=|c|√s 滿足 b≥0（兩個非負數之積），且 b²=|c|²(√s)²=c²s（|c|²=c²，(√s)²=s）。主平方根是唯一滿足 b≥0、b²=c²s 的數，所以 √(c²s)=|c|√s。S04.1 證得引理 L_sqrt_scale。",
          "en": "b=|c|√s satisfies b≥0 (a product of two nonnegative numbers) and b²=|c|²(√s)²=c²s (since |c|²=c² and (√s)²=s). The principal square root is the unique number with b≥0 and b²=c²s, so √(c²s)=|c|√s. S04.1 proves Lemma L_sqrt_scale."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "sqrt_def",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "依唯一性兩式指同一數。若把 |c| 換成 c，c<0 且 s>0 時 c√s<0，不可能是主平方根；s=0 時兩式都為 0，不能以嚴格負號排除。",
            "en": "By uniqueness the two expressions name the same number. Replacing |c| by c fails for c<0 and s>0, since c√s<0 cannot be a principal square root. For s=0 both expressions are zero, so no strict-negative argument applies."
          }
        },
        "leaf_basis": "sqrt_def + abs_rules + power_laws + order_basics"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "ip_axioms"
        ],
        "before_tex": "\\langle x\\pm y,x\\pm y\\rangle",
        "after_tex": "\\langle x,x\\rangle\\pm2\\langle x,y\\rangle+\\langle y,y\\rangle",
        "relation": "equivalent",
        "rule": "ip_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "雙線性展開成四項：⟨x,x⟩±⟨x,y⟩±⟨y,x⟩+⟨y,y⟩；再用對稱 ⟨y,x⟩=⟨x,y⟩ 合併兩個交叉項，得 ±2⟨x,y⟩。",
          "en": "Bilinearity expands into four terms, ⟨x,x⟩±⟨x,y⟩±⟨y,x⟩+⟨y,y⟩; symmetry ⟨y,x⟩=⟨x,y⟩ then merges the two cross terms into ±2⟨x,y⟩."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "ip_axioms",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "恆等式，對所有 x、y 成立，雙向。交叉項必須保留：只有 ⟨x,y⟩=0 時才是畢氏形式。",
            "en": "An identity for all x and y, so two-way. The cross term must be kept: the Pythagorean form appears only when ⟨x,y⟩=0."
          }
        },
        "leaf_basis": "ip_axioms + add_comm + distributive"
      },
      {
        "id": "S06",
        "uses": [
          "G1"
        ],
        "before_tex": "x,y\\in V",
        "after_tex": "\\langle x,y\\rangle^2\\le\\langle x,x\\rangle\\langle y,y\\rangle",
        "relation": "implies",
        "rule": "L_cs",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_cs（Cauchy–Schwarz，在 S06.1 到 S06.3 完整證明），對所有 x、y 成立。證明分 y=0 與 y≠0 兩種情況，涵蓋全部可能。這是全稱證明，不是由有限個向量的核對得到（MKO 假設 a2）。",
          "en": "By Lemma L_cs (Cauchy–Schwarz, fully proved in S06.1 to S06.3), it holds for all x and y. The proof splits into the cases y=0 and y≠0, which cover every possibility. This is a universal proof, not obtained by checking finitely many vectors (MKO assumption a2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "L_cs"
          ],
          "explanation": {
            "zh-Hant": "需要雙線性、對稱與正定；不宣稱等號何時成立。",
            "en": "It needs bilinearity, symmetry and positive definiteness; it does not claim when equality holds."
          }
        },
        "leaf_basis": "case_analysis + ip_axioms + order_basics"
      },
      {
        "id": "S06.1",
        "uses": [
          "G1",
          "ip_axioms"
        ],
        "before_tex": "y=0",
        "after_tex": "\\langle x,0\\rangle^2=0=\\langle x,x\\rangle\\langle 0,0\\rangle",
        "relation": "implies",
        "rule": "ip_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "情況 y=0：由第二變數線性 ⟨x,0⟩=⟨x,0·0⟩=0·⟨x,0⟩=0，同理 ⟨0,0⟩=0。兩邊都是 0，不等式成立。",
          "en": "Case y=0: linearity in the second argument gives ⟨x,0⟩=⟨x,0·0⟩=0·⟨x,0⟩=0, and likewise ⟨0,0⟩=0. Both sides are 0, so the inequality holds."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "ip_axioms",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "只用線性；此情況必須單獨處理，因為 S06.3 要除以 ⟨y,y⟩。",
            "en": "Only linearity is used; this case must be handled separately because S06.3 divides by ⟨y,y⟩."
          }
        },
        "leaf_basis": "ip_axioms + mul_zero"
      },
      {
        "id": "S06.2",
        "uses": [
          "G1",
          "S05",
          "ip_axioms"
        ],
        "before_tex": "y\\neq0,\\ t\\in\\mathbb{R}",
        "after_tex": "0\\le\\langle x,x\\rangle-2t\\langle x,y\\rangle+t^2\\langle y,y\\rangle",
        "relation": "implies",
        "rule": "ip_axioms",
        "conditions": [
          "y\\neq0"
        ],
        "explanation": {
          "zh-Hant": "情況 y≠0：對每個實數 t，正定性給 ⟨x−ty,x−ty⟩≥0。依 S05 的展開方式（並用 S03 處理係數 t）：⟨x−ty,x−ty⟩=⟨x,x⟩−2t⟨x,y⟩+t²⟨y,y⟩。",
          "en": "Case y≠0: for every real t, positive definiteness gives ⟨x−ty,x−ty⟩≥0. Expanding as in S05 (with S03 handling the factor t): ⟨x−ty,x−ty⟩=⟨x,x⟩−2t⟨x,y⟩+t²⟨y,y⟩."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "ip_axioms",
            "S05"
          ],
          "explanation": {
            "zh-Hant": "不等式對每個 t 都成立，下一步才能選特定的 t。",
            "en": "The inequality holds for every t, which is what lets the next step choose a particular t."
          }
        },
        "leaf_basis": "ip_axioms + distributive + forall_intro"
      },
      {
        "id": "S06.3",
        "uses": [
          "S06.2",
          "G1"
        ],
        "before_tex": "t=\\frac{\\langle x,y\\rangle}{\\langle y,y\\rangle}",
        "after_tex": "\\langle x,y\\rangle^2\\le\\langle x,x\\rangle\\langle y,y\\rangle",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [
          "\\langle y,y\\rangle>0"
        ],
        "explanation": {
          "zh-Hant": "y≠0 時 ⟨y,y⟩>0，所以可取 t=⟨x,y⟩/⟨y,y⟩。記 a=⟨x,x⟩、b=⟨x,y⟩、c=⟨y,y⟩，代入 S06.2：0≤a−2b²/c+b²/c=a−b²/c。兩邊乘以 c>0（不等號方向不變）得 0≤ac−b²，即 b²≤ac。代數恆等式 c²(a−2(b/c)b+(b/c)²c)=c(ac−b²) 由 N2 核對。S06.1 到 S06.3 合起來（兩種情況涵蓋全部 y）證得引理 L_cs。",
          "en": "For y≠0, ⟨y,y⟩>0, so t=⟨x,y⟩/⟨y,y⟩ may be chosen. Write a=⟨x,x⟩, b=⟨x,y⟩, c=⟨y,y⟩ and substitute into S06.2: 0≤a−2b²/c+b²/c=a−b²/c. Multiplying by c>0 (which keeps the direction) gives 0≤ac−b², that is b²≤ac. The algebraic identity c²(a−2(b/c)b+(b/c)²c)=c(ac−b²) is checked by N2. S06.1 to S06.3 together (the two cases cover every y) prove Lemma L_cs."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06.2",
            "forall_elim",
            "order_basics",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "需要 c>0：除以 c 需要 c≠0，乘以 c 保持方向需要 c>0；兩者都由 y≠0 與正定性提供。",
            "en": "It needs c>0: dividing by c needs c≠0, and multiplying by c without reversing the inequality needs c>0; both come from y≠0 and positive definiteness."
          }
        },
        "leaf_basis": "forall_elim + frac_rules + order_basics + mul_inverse_nonzero"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "S01"
        ],
        "before_tex": "\\langle x,y\\rangle^2\\le N(x)^2N(y)^2",
        "after_tex": "|\\langle x,y\\rangle|\\le N(x)N(y)",
        "relation": "implies",
        "rule": "L_sq_order",
        "conditions": [],
        "explanation": {
          "zh-Hant": "令 u=|⟨x,y⟩|≥0、w=N(x)N(y)≥0。u²=⟨x,y⟩²（|a|²=a²），w²=N(x)²N(y)²=⟨x,x⟩⟨y,y⟩（(√s)²=s）。由 S06，u²≤w²；依引理 L_sq_order（在 S07.1 證明）得 u≤w。",
          "en": "Let u=|⟨x,y⟩|≥0 and w=N(x)N(y)≥0. Then u²=⟨x,y⟩² (|a|²=a²) and w²=N(x)²N(y)²=⟨x,x⟩⟨y,y⟩ ((√s)²=s). By S06, u²≤w², and Lemma L_sq_order (proved in S07.1) gives u≤w."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S01",
            "L_sq_order",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "本次 u=|⟨x,y⟩|、w=N(x)N(y) 都非負，足以套用所列引理；不把 u≥0 說成這個單向推論的必要條件。若取消 w≥0 則一般會失敗，例如 1²≤(−2)² 卻有 1>−2。",
            "en": "Here u=|⟨x,y⟩| and w=N(x)N(y) are both nonnegative, so the stated lemma applies. This does not make u≥0 necessary for the one-way implication. Removing w≥0 can fail: 1²≤(−2)² yet 1>−2."
          }
        },
        "leaf_basis": "abs_rules + sqrt_def + power_laws"
      },
      {
        "id": "S07.1",
        "uses": [
          "order_basics"
        ],
        "before_tex": "u,w\\ge0,\\ u^2\\le w^2",
        "after_tex": "u\\le w",
        "relation": "implies",
        "rule": "contrapositive",
        "conditions": [
          "u\\ge0",
          "w\\ge0"
        ],
        "explanation": {
          "zh-Hant": "證逆否命題：設 u>w≥0。則 u>0，把 u>w 乘以 u>0 得 u²>wu；又 u>w 且 w≥0 給 wu≥w²（w=0 時兩邊都是 0，w>0 時乘以 w）。所以 u²>w²。因此 u²≤w² 推出 u≤w。S07.1 證得引理 L_sq_order。",
          "en": "Prove the contrapositive: suppose u>w≥0. Then u>0, and multiplying u>w by u>0 gives u²>wu; also u>w with w≥0 gives wu≥w² (both sides are 0 when w=0, and multiply by w when w>0). So u²>w². Hence u²≤w² implies u≤w. S07.1 proves Lemma L_sq_order."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "contrapositive"
          ],
          "explanation": {
            "zh-Hant": "引理保留 u≥0、w≥0 作充分前提，但此單向證明不需另用 u≥0：逆否假設 u>w≥0 已推出 u>0；若 u<0≤w，結論也直接成立。w≥0 則保證可由平方比較推出原數上界。反向在兩者非負時成立，但本稿不需要。",
            "en": "The lemma keeps u≥0 and w≥0 as sufficient hypotheses, but this one-way proof does not separately need u≥0: its contrapositive assumption u>w≥0 already yields u>0. If u<0≤w the conclusion holds immediately. The condition w≥0 supplies the upper-bound sign needed for the implication. The converse holds when both are nonnegative, but is not needed here."
          }
        },
        "leaf_basis": "order_basics + contrapositive"
      },
      {
        "id": "S08",
        "uses": [
          "S05",
          "S07",
          "S01"
        ],
        "before_tex": "N(x+y)^2=\\langle x,x\\rangle+2\\langle x,y\\rangle+\\langle y,y\\rangle",
        "after_tex": "N(x+y)^2\\le\\big(N(x)+N(y)\\big)^2",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由 S05，N(x+y)²=N(x)²+2⟨x,y⟩+N(y)²。因 ⟨x,y⟩≤|⟨x,y⟩|≤N(x)N(y)（S07），兩邊同加相同的數保持不等式：N(x+y)²≤N(x)²+2N(x)N(y)+N(y)²=(N(x)+N(y))²。",
          "en": "By S05, N(x+y)²=N(x)²+2⟨x,y⟩+N(y)². Since ⟨x,y⟩≤|⟨x,y⟩|≤N(x)N(y) (S07), adding the same quantity to both sides keeps the inequality: N(x+y)²≤N(x)²+2N(x)N(y)+N(y)²=(N(x)+N(y))²."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S07",
            "order_basics",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "這裡正是 Cauchy–Schwarz 推出三角不等式之處（MKO）；不宣稱反向。",
            "en": "This is exactly where Cauchy–Schwarz yields the triangle inequality (MKO); no converse is claimed."
          }
        },
        "leaf_basis": "order_basics + abs_rules + distributive + sqrt_def"
      },
      {
        "id": "S09",
        "uses": [
          "S08",
          "S01",
          "S07"
        ],
        "before_tex": "N(x+y)^2\\le\\big(N(x)+N(y)\\big)^2",
        "after_tex": "N(x+y)\\le N(x)+N(y)",
        "relation": "implies",
        "rule": "L_sq_order",
        "conditions": [],
        "explanation": {
          "zh-Hant": "N(x+y)≥0 且 N(x)+N(y)≥0（S01），所以依引理 L_sq_order（u=N(x+y)、w=N(x)+N(y)）由 S08 得三角不等式。",
          "en": "N(x+y)≥0 and N(x)+N(y)≥0 (S01), so Lemma L_sq_order with u=N(x+y) and w=N(x)+N(y) turns S08 into the triangle inequality."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "S01",
            "L_sq_order"
          ],
          "explanation": {
            "zh-Hant": "需要兩邊非負，由 S01 提供。",
            "en": "Both sides must be nonnegative, which S01 provides."
          }
        },
        "leaf_basis": "order_basics + sqrt_def"
      },
      {
        "id": "S10",
        "uses": [
          "S01",
          "S02",
          "S04",
          "S09"
        ],
        "before_tex": "\\text{S01, S02, S04, S09}",
        "after_tex": "N\\ \\text{is a norm on}\\ V",
        "relation": "implies",
        "rule": "norm_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "非負與分離（S01、S02）、絕對齊次（S04）、三角不等式（S09）都對任意向量與純量證明，所以 N 是 V 上的範數。這是全稱證明，不依賴任何具體向量。",
          "en": "Nonnegativity and separation (S01, S02), absolute homogeneity (S04) and the triangle inequality (S09) are proved for arbitrary vectors and scalars, so N is a norm on V. This is a universal proof that depends on no particular vectors."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S01",
            "S02",
            "S04",
            "S09",
            "norm_def"
          ],
          "explanation": {
            "zh-Hant": "目標 G：N 是範數。足夠條件 H：範數定義的三項；H⇒G 由定義。H 由 G1 正向證成。",
            "en": "Goal G: N is a norm. Sufficient condition H: the three clauses of the definition; H⇒G by definition. H is proved forwards from G1."
          }
        },
        "leaf_basis": "norm_def + and_intro_elim + forall_intro"
      },
      {
        "id": "S11",
        "uses": [
          "G2",
          "square_nonneg"
        ],
        "before_tex": "v^{\\mathsf T}Av=2v_1^2+2v_1v_2+2v_2^2",
        "after_tex": "v_1^2+v_2^2+(v_1+v_2)^2",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "審核 G2 的前提：展開右邊 v₁²+v₂²+v₁²+2v₁v₂+v₂²=2v₁²+2v₁v₂+2v₂²（N1 作為多項式恆等式核對）。三個平方都非負，所以 vᵀAv≥0；若為 0，則 v₁=0、v₂=0，即 v=0。A 對稱且矩陣形式雙線性，所以 G2 確實定義內積。",
          "en": "Audit the premise of G2: expanding the right side gives v₁²+v₂²+v₁²+2v₁v₂+v₂²=2v₁²+2v₁v₂+2v₂² (N1 checks this as a polynomial identity). All three squares are nonnegative, so vᵀAv≥0; if it is 0, then v₁=0 and v₂=0, so v=0. A is symmetric and the matrix form is bilinear, so G2 really defines an inner product."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "square_nonneg"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一多項式，雙向成立；N1 在 3×3 格點上核對次數至多 2 的恆等式，足以判定。",
            "en": "Both sides are the same polynomial, so the step is two-way; N1 checks the identity of degree at most 2 on a 3×3 grid, which decides it."
          }
        },
        "leaf_basis": "distributive + square_nonneg + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "G2",
          "G3",
          "S11"
        ],
        "before_tex": "x=(1,0),\\ y=(0,1)",
        "after_tex": "\\langle x,x\\rangle=2,\\ \\langle y,y\\rangle=2,\\ \\langle x,y\\rangle=1,\\ N(x)=N(y)=\\sqrt2",
        "relation": "equivalent",
        "rule": "matrix_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "標準基底取出 A 的元素：⟨x,x⟩=A₁₁=2、⟨y,y⟩=A₂₂=2、⟨x,y⟩=A₁₂=1。平方範數是精確整數；範數保持精確符號 N(x)=N(y)=√2（MKO 假設 a2）。齊次性的實例：N(−3x)=√(9·2)=√18=3√2，與 |−3|N(x) 相同（N3）。",
          "en": "Standard basis vectors pick out entries of A: ⟨x,x⟩=A₁₁=2, ⟨y,y⟩=A₂₂=2, ⟨x,y⟩=A₁₂=1. The squared norms are exact integers; the norms stay exact symbols N(x)=N(y)=√2 (MKO assumption a2). An instance of homogeneity: N(−3x)=√(9·2)=√18=3√2, equal to |−3|N(x) (N3)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "matrix_product_def",
            "sqrt_def"
          ],
          "explanation": {
            "zh-Hant": "每個值都可由 G2 的公式直接重算，兩邊是同一數。",
            "en": "Each value can be recomputed directly from the formula in G2; both sides are the same number."
          }
        },
        "leaf_basis": "matrix_product_def + numeric_evaluation + sqrt_def"
      },
      {
        "id": "S13",
        "uses": [
          "S05",
          "S12"
        ],
        "before_tex": "\\langle x\\mp y,x\\mp y\\rangle",
        "after_tex": "\\langle x-y,x-y\\rangle=2,\\ \\langle x+y,x+y\\rangle=6",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代入 S05：⟨x−y,x−y⟩=2−2·1+2=2（N4），所以距離 N(x−y)=√2；⟨x+y,x+y⟩=2+2·1+2=6（N5），N(x+y)=√6。直接核對 x−y=(1,−1)：2·1·1+1·(−1)+(−1)·1+2·(−1)·(−1)=2（N6），與展開一致。交叉項 −2 不可省略：畢氏式 2+2=4 是錯的。",
          "en": "Substitute into S05: ⟨x−y,x−y⟩=2−2·1+2=2 (N4), so the distance is N(x−y)=√2; ⟨x+y,x+y⟩=2+2·1+2=6 (N5), so N(x+y)=√6. Direct check with x−y=(1,−1): 2·1·1+1·(−1)+(−1)·1+2·(−1)·(−1)=2 (N6), agreeing with the expansion. The cross term −2 cannot be dropped: the Pythagorean 2+2=4 is wrong."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S05",
            "S12",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "以等值代換代入恆等式，兩邊同值；兩種算法（展開與直接）得到同一結果。",
            "en": "Replacement of equals in an identity, so both sides have the same value; two methods (expansion and direct) give the same result."
          }
        },
        "leaf_basis": "eq_replace + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S05",
          "S06",
          "S09",
          "S12",
          "S13"
        ],
        "before_tex": "\\text{values of S12, S13}",
        "after_tex": "1\\le4,\\ \\ 6\\le8,\\ \\ 6+2=2(2+2)",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "實例核對：Cauchy–Schwarz 1²=1≤2·2=4（N7）；三角不等式的平方形式 N(x+y)²=6≤(√2+√2)²=2+2·2+2=8（N8，用 √2·√2=2）；平行四邊形恆等式：S05 的兩式相加得一般恆等式 ⟨x+y,x+y⟩+⟨x−y,x−y⟩=2⟨x,x⟩+2⟨y,y⟩，此例 6+2=8=2(2+2)（N9）。這些只是此實例與 S06、S09 及一般恆等式一致的核對，不是它們的證明。",
          "en": "Instance checks: Cauchy–Schwarz 1²=1≤2·2=4 (N7); the squared triangle inequality N(x+y)²=6≤(√2+√2)²=2+2·2+2=8 (N8, using √2·√2=2); the parallelogram identity: adding the two formulas of S05 gives the general identity ⟨x+y,x+y⟩+⟨x−y,x−y⟩=2⟨x,x⟩+2⟨y,y⟩, here 6+2=8=2(2+2) (N9). These only check that this instance agrees with S06, S09 and the general identity; they are not proofs of them."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12",
            "S13",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "數值可獨立重算；但有限個向量通過核對不能建立全稱命題，全稱部分由 S06 到 S10 證明。",
            "en": "The numbers can be recomputed independently; but finitely many vectors passing a check cannot establish a universal statement, and the universal part is proved in S06 to S10."
          }
        },
        "leaf_basis": "numeric_evaluation + sqrt_def"
      },
      {
        "id": "S15",
        "uses": [
          "S10",
          "S11",
          "S12",
          "S13",
          "S14"
        ],
        "before_tex": "\\text{S10 to S14}",
        "after_tex": "N\\ \\text{norm};\\ N(x)=N(y)=N(x-y)=\\sqrt2,\\ N(x+y)=\\sqrt6",
        "relation": "implies",
        "rule": "mko_ip_norm_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "N 在任一實內積空間是範數（S10）；在 G2 的例子中精確值為 N(x)=N(y)=√2、距離 N(x−y)=√2、N(x+y)=√6，實例核對與一般結論一致（S14）。本稿不由「範數來自內積」推出完備性或 Hilbert 空間。",
          "en": "N is a norm on every real inner-product space (S10); in the example of G2 the exact values are N(x)=N(y)=√2, distance N(x−y)=√2 and N(x+y)=√6, and the instance checks agree with the general results (S14). This guide does not infer completeness, or a Hilbert space, from the norm coming from an inner product."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S10",
            "S12",
            "S13",
            "S14"
          ],
          "explanation": {
            "zh-Hant": "目標 G：範數定理與實例的精確值。足夠條件 H：S10 與 S12 到 S14；H⇒G 由合取。H 由 G1 到 G3 正向證成，沒有用有限樣本代替全稱證明。",
            "en": "Goal G: the norm theorem and the exact values of the instance. Sufficient condition H: S10 and S12 to S14; H⇒G by conjunction. H is proved forwards from G1 to G3, and no finite sample replaces a universal proof."
          }
        },
        "leaf_basis": "and_intro_elim + mko_ip_norm_def"
      }
    ],
    "result": {
      "zh-Hant": "在任一實內積空間中，N(v)=√⟨v,v⟩ 是範數：正定性給非負與分離，雙線性給 N(cv)=|c|N(v)，完整證明的 Cauchy–Schwarz ⟨x,y⟩²≤⟨x,x⟩⟨y,y⟩ 給三角不等式。在 ℝ² 以 A=[[2,1],[1,2]] 定義的內積下，⟨x,x⟩=⟨y,y⟩=2、⟨x,y⟩=1，N(x)=N(y)=√2，距離平方 ⟨x−y,x−y⟩=2，⟨x+y,x+y⟩=6，實例滿足 1≤4、6≤8 與 6+2=2(2+2)。",
      "en": "In every real inner-product space, N(v)=√⟨v,v⟩ is a norm: positive definiteness gives nonnegativity and separation, bilinearity gives N(cv)=|c|N(v), and the fully proved Cauchy–Schwarz inequality ⟨x,y⟩²≤⟨x,x⟩⟨y,y⟩ gives the triangle inequality. On ℝ² with the inner product given by A=[[2,1],[1,2]], ⟨x,x⟩=⟨y,y⟩=2 and ⟨x,y⟩=1, so N(x)=N(y)=√2, the squared distance is ⟨x−y,x−y⟩=2 and ⟨x+y,x+y⟩=6; the instance satisfies 1≤4, 6≤8 and 6+2=2(2+2)."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "x、y、x+y 這幾個向量都滿足三角不等式，所以 N 在每個內積空間都是範數",
          "en": "x, y and x+y all satisfy the triangle inequality, so N is a norm on every inner-product space"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "有限個實例不能證明全稱命題；全稱證明在 S06 到 S09。",
          "en": "Finitely many instances cannot prove a universal statement; the universal proof is S06 to S09."
        }
      },
      {
        "claim": {
          "zh-Hant": "N(x+y)=N(x)+N(y)=2√2",
          "en": "N(x+y)=N(x)+N(y)=2√2"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "N(x+y)²=6，而 (2√2)²=8，6≠8；三角不等式一般是不等式。",
          "en": "N(x+y)²=6 while (2√2)²=8, and 6≠8; the triangle inequality need not hold with equality."
        }
      },
      {
        "claim": {
          "zh-Hant": "由畢氏定理，N(x−y)²=N(x)²+N(y)²=4",
          "en": "By Pythagoras, N(x−y)²=N(x)²+N(y)²=4"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "此內積下 ⟨x,y⟩=1≠0，交叉項 −2 使 N(x−y)²=2（S13）。",
          "en": "Under this inner product ⟨x,y⟩=1≠0, and the cross term −2 makes N(x−y)²=2 (S13)."
        }
      },
      {
        "claim": {
          "zh-Hant": "任何對稱矩陣都定義內積，例如 B=[[1,2],[2,1]]",
          "en": "Every symmetric matrix defines an inner product, for example B=[[1,2],[2,1]]"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "v=(1,−1) 時 vᵀBv=1−2−2+1=−2<0（N10），不正定，√ 無定義。",
          "en": "For v=(1,−1), vᵀBv=1−2−2+1=−2<0 (N10), so B is not positive definite and √ is undefined."
        }
      },
      {
        "claim": {
          "zh-Hant": "範數來自內積，所以 (V,N) 必是 Hilbert 空間",
          "en": "The norm comes from an inner product, so (V,N) must be a Hilbert space"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "Hilbert 空間另需完備性，本稿與 MKO 都不由內積誘導範數推出它。",
          "en": "A Hilbert space additionally needs completeness, which neither this guide nor the MKO derives from the induced norm."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "sqrt_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "主平方根：對 a≥0，√a 是唯一滿足 b≥0 且 b²=a 的實數 b。負數在實數中沒有平方根。",
          "en": "Principal square root: for a≥0, √a is the unique real b with b≥0 and b²=a. Negative numbers have no real square root."
        }
      },
      {
        "id": "square_nonneg",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數平方非負：對所有實數 u，u²≥0，且 u²=0 ⇔ u=0。",
          "en": "Squares are nonnegative: for every real u, u²≥0, and u²=0 ⇔ u=0."
        }
      },
      {
        "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": "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": "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": "logic_chain",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "推論的傳遞：由 P⇒Q 與 Q⇒R 得 P⇒R；⇔ 同理。",
          "en": "Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔."
        }
      },
      {
        "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": "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": "add_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法交換律：u+v=v+u。",
          "en": "Commutativity of addition: u+v=v+u."
        }
      },
      {
        "id": "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_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "id": "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": "forall_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。",
          "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds."
        }
      },
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "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": "contrapositive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逆否命題：P⇒Q 與 ¬Q⇒¬P 等價。逆命題 Q⇒P 一般不與 P⇒Q 等價。",
          "en": "Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q."
        }
      },
      {
        "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": "matrix_product_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "矩陣乘法：若 A 為 m×n、B 為 n×p，則 (AB)_{ij}=Σ_{k=1}^{n} A_{ik}B_{kj}。矩陣乘法一般不可交換。",
          "en": "Matrix product: for A of size m×n and B of size n×p, (AB)_{ij}=Σ_{k=1}^{n} A_{ik}B_{kj}. Matrix multiplication is in general not commutative."
        }
      },
      {
        "id": "eq_replace",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。",
          "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true."
        }
      },
      {
        "id": "mko_ip_norm_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-inner-product-induced-norm",
          "sha256": "9b9498dd5a3aa838a1854d97c526a01a657f0f88bab64b00a78ef20fb1f80c97",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在同一實內積空間中定義 norm(x)=sqrt(<x,x>)。正定性給分離，齊次性給絕對齊次，Cauchy–Schwarz 推出三角不等式。本輪以 R² 正定矩陣精確計算平方範數與恆等式，不以樣本取代全稱證書。",
          "en": "In one real inner-product space define norm(x)=sqrt(<x,x>). Positive definiteness gives separation, homogeneity gives absolute homogeneity, and Cauchy–Schwarz gives the triangle inequality. This slice calculates exact squared norms and identities for positive-definite matrices on R² without replacing universal certificates with vector samples."
        }
      },
      {
        "id": "ip_axioms",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "內積公理：⟨au+bw,v⟩=a⟨u,v⟩+b⟨w,v⟩（第二個變數同理）；⟨u,v⟩=⟨v,u⟩；⟨v,v⟩≥0 且 ⟨v,v⟩=0 ⇔ v=0。",
          "en": "Inner-product axioms: ⟨au+bw,v⟩=a⟨u,v⟩+b⟨w,v⟩ (likewise in the second argument); ⟨u,v⟩=⟨v,u⟩; ⟨v,v⟩≥0 and ⟨v,v⟩=0 ⇔ v=0."
        }
      },
      {
        "id": "norm_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實向量空間上的範數 N：N(v)≥0 且 N(v)=0 ⇔ v=0；N(cv)=|c|N(v)；N(u+v)≤N(u)+N(v)。",
          "en": "A norm N on a real vector space: N(v)≥0 with N(v)=0 ⇔ v=0; N(cv)=|c|N(v); N(u+v)≤N(u)+N(v)."
        }
      },
      {
        "id": "L_sqrt_scale",
        "kind": "lemma",
        "proved_in": [
          "S04.1"
        ],
        "statement": {
          "zh-Hant": "對實數 c 與 s≥0：√(c²s)=|c|√s。",
          "en": "For a real c and s≥0: √(c²s)=|c|√s."
        }
      },
      {
        "id": "L_cs",
        "kind": "lemma",
        "proved_in": [
          "S06.1",
          "S06.2",
          "S06.3"
        ],
        "statement": {
          "zh-Hant": "Cauchy–Schwarz：在實內積空間中，對所有 x、y，⟨x,y⟩²≤⟨x,x⟩⟨y,y⟩。",
          "en": "Cauchy–Schwarz: in a real inner-product space, ⟨x,y⟩²≤⟨x,x⟩⟨y,y⟩ for all x, y."
        }
      },
      {
        "id": "L_sq_order",
        "kind": "lemma",
        "proved_in": [
          "S07.1"
        ],
        "statement": {
          "zh-Hant": "對實數 u≥0、w≥0：若 u²≤w²，則 u≤w。",
          "en": "For reals u≥0 and w≥0: if u²≤w², then u≤w."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S11",
        "vars": [
          "a",
          "b"
        ],
        "degree": 2,
        "lhs": "2*a*a+2*a*b+2*b*b",
        "rhs": "a*a+b*b+(a+b)*(a+b)"
      },
      {
        "id": "N2",
        "step": "S06.3",
        "vars": [
          "a",
          "b",
          "c"
        ],
        "degree": 2,
        "lhs": "c*c*a-2*b*b*c+b*b*c",
        "rhs": "c*(a*c-b*b)"
      },
      {
        "id": "N3",
        "step": "S12",
        "lhs": "(-3)**2*2",
        "rhs": "3**2*2"
      },
      {
        "id": "N4",
        "step": "S13",
        "lhs": "2-2*1+2",
        "rhs": "2"
      },
      {
        "id": "N5",
        "step": "S13",
        "lhs": "2+2*1+2",
        "rhs": "6"
      },
      {
        "id": "N6",
        "step": "S13",
        "lhs": "2*1*1+1*(-1)+(-1)*1+2*(-1)*(-1)",
        "rhs": "2"
      },
      {
        "id": "N7",
        "step": "S14",
        "lhs": "1**2",
        "relation": "<=",
        "rhs": "2*2"
      },
      {
        "id": "N8",
        "step": "S14",
        "lhs": "6",
        "relation": "<=",
        "rhs": "2+2*2+2"
      },
      {
        "id": "N9",
        "step": "S14",
        "lhs": "6+2",
        "rhs": "2*(2+2)"
      },
      {
        "id": "N10",
        "lhs": "1*1*1+2*1*(-1)+2*(-1)*1+1*(-1)*(-1)",
        "relation": "<",
        "rhs": "0"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只在 MKO 範圍內工作：同一實內積空間；具體例子用已通過雙線性、對稱與正定的 ℝ² 矩陣內積（假設 a1），平方根保持精確符號（假設 a2）。範數公理對一般實內積空間完整證明；具體向量的核對只是實例，不取代全稱證明。不討論完備性或 Hilbert 空間；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: one real inner-product space; the concrete example uses a matrix inner product on ℝ² that already qualifies as bilinear, symmetric and positive definite (assumption a1), with square roots kept as exact symbols (assumption a2). The norm axioms are proved in full for a general real inner-product space; checks on particular vectors are instances only and do not replace the universal proofs. Completeness and Hilbert spaces are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "\\langle u,v\\rangle",
        "內積",
        "the inner product"
      ],
      [
        "N(v)=\\sqrt{\\langle v,v\\rangle}",
        "內積誘導範數",
        "the inner-product-induced norm"
      ],
      [
        "A=\\begin{pmatrix}2&1\\\\1&2\\end{pmatrix}",
        "ℝ² 上定義內積的矩陣",
        "the matrix defining the inner product on ℝ²"
      ],
      [
        "x=(1,0),\\ y=(0,1)",
        "兩個例子向量",
        "the two example vectors"
      ],
      [
        "|c|",
        "實數 c 的絕對值",
        "the absolute value of the real c"
      ]
    ],
    "md_error_controls": [
      {
        "title": "樣本通過不是全稱證明 / Passing samples is not a universal proof",
        "zh": "MKO 指出的誤解：看到若干向量的三角不等式成立就宣稱一般範數。S14 中 6≤8 只說明 x、y 這一對沒有違反；三角不等式對所有向量成立是由 S06 到 S09 的證明給出，那裡沒有用到任何具體向量。",
        "en": "The MKO's misconception: declaring a general norm after seeing the triangle inequality hold for some vectors. In S14, 6≤8 only says that the pair x, y does not violate it; that the triangle inequality holds for all vectors comes from the proof in S06 to S09, which uses no particular vectors."
      },
      {
        "title": "內積誘導範數不等於 Hilbert 空間 / An induced norm is not yet a Hilbert space",
        "zh": "MKO 的第二個誤解是由內積誘導範數直接稱 Hilbert 空間。Hilbert 空間還要求對此範數完備，這是另一個性質，Cauchy–Schwarz 與平行四邊形恆等式都不推出它。本稿沒有證明也沒有使用完備性。",
        "en": "The MKO's second misconception is to call a space Hilbert directly because its norm comes from an inner product. A Hilbert space also requires completeness with respect to this norm, a separate property that neither Cauchy–Schwarz nor the parallelogram identity implies. This guide neither proves nor uses completeness."
      },
      {
        "title": "交叉項與絕對值 / The cross term and the absolute value",
        "zh": "最容易做錯的兩處：S05 的交叉項 2⟨x,y⟩ 不能省略（此例 ⟨x−y,x−y⟩=2，不是 4）；S04 的係數是 |c| 不是 c（N(−3x)=3√2，不是 −3√2）。另外 √ 不對加法分配：N(x+y)=√6≠√2+√2。",
        "en": "The two easiest slips: the cross term 2⟨x,y⟩ in S05 cannot be dropped (here ⟨x−y,x−y⟩=2, not 4), and the factor in S04 is |c|, not c (N(−3x)=3√2, not −3√2). Also √ does not distribute over addition: N(x+y)=√6≠√2+√2."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "完備性（因而 Hilbert 空間，MKO 的下一步）不在本稿範圍：本稿既不證明 ℝ² 在此範數下完備，也不給出不完備的例子。",
        "en": "Completeness (and hence Hilbert spaces, the MKO's next step) is outside this guide: it neither proves that ℝ² is complete under this norm nor gives an incomplete example."
      },
      {
        "zh-Hant": "N1 與 N2 是格點上的多項式恆等式核對；N3 到 N9 只對本例的數字有效。",
        "en": "N1 and N2 are polynomial-identity checks on a grid; N3 to N9 are valid for this example's numbers only."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "c4ddbc627d1b1a07ab027fee5efedc07a1bcf302d187aa0e3c6b47ad21ca3b2f",
    "original_source_sha256": "5c985ba54e11c7600e0ba5b543ac1e169154c8abd3f78b01cd3b3d654e479277",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M091",
      "BR-M092"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}