{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-real-inner-product-space-01", "mko_id": "mko-real-inner-product-space", "title": { "zh-Hant": "實內積空間：以 Sylvester 條件為 ℝ² 上的矩陣形式作全稱正定證書", "en": "Real inner-product space: Sylvester conditions as a universal positive-definiteness certificate for a matrix form on ℝ²" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-real-inner-product-space-01", "source": { "mko_id": "mko-real-inner-product-space", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "70d1a304624117e0e814b5b9ca816a5b851798e54cc213f1f2b3460e6450f963", "english_sha256": "37a8aec17484c89c814b26ce16792d6a4b33d8d0ef4fffb7e9e9e05e3ec4c42f" }, "mode": "worked_example", "audience_prerequisites": [ "matrix_product_def", "distributive", "square_nonneg", "order_basics", "numeric_evaluation" ], "goal": { "zh-Hant": "設 M 為第一行 (2,1)、第二行 (1,3) 的整數對稱矩陣，⟨x,y⟩_M=xᵀMy（x,y∈ℝ²）。（一）分別證明 ⟨·,·⟩_M 雙線性、對稱、正定，所以是 ℝ² 上的內積；正定性以 2×2 Sylvester 條件（完整證明其充分與必要方向）作全稱證書。（二）判定 N（行 (1,2),(2,1)）、P（行 (1,1),(1,1)）、R（行 (2,1),(0,3)）給出的形式為何不是內積。", "en": "Let M be the symmetric integer matrix with first row (2,1) and second row (1,3), and ⟨x,y⟩_M=xᵀMy for x,y∈ℝ². (1) Prove separately that ⟨·,·⟩_M is bilinear, symmetric and positive definite, hence an inner product on ℝ²; positive definiteness is certified universally by the 2-by-2 Sylvester conditions, with both sufficiency and necessity proved in full. (2) Decide why the forms given by N (rows (1,2),(2,1)), P (rows (1,1),(1,1)) and R (rows (2,1),(0,3)) are not inner products." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "V=ℝ²，向量 x=(x₁,x₂) 視為列向量，運算逐坐標進行。", "en": "V=ℝ², with vectors x=(x₁,x₂) treated as columns and operations coordinatewise." } }, { "id": "G2", "statement": { "zh-Hant": "M 的列為 (2,1) 與 (1,3)；⟨x,y⟩_M=xᵀMy。", "en": "M has rows (2,1) and (1,3); ⟨x,y⟩_M=xᵀMy." } }, { "id": "G3", "statement": { "zh-Hant": "對照矩陣：N 的列為 (1,2),(2,1)；P 的列為 (1,1),(1,1)；R 的列為 (2,1),(0,3)；各自定義 ⟨x,y⟩=xᵀ(矩陣)y。", "en": "Comparison matrices: N with rows (1,2),(2,1); P with rows (1,1),(1,1); R with rows (2,1),(0,3); each defines ⟨x,y⟩=xᵀ(matrix)y." } } ], "steps": [ { "id": "S01", "uses": [ "G1", "G2", "mko_ips_def" ], "before_tex": "\\langle\\cdot,\\cdot\\rangle_M\\ \\text{inner product on}\\ \\mathbb{R}^2", "after_tex": "\\text{bilinear}\\ \\land\\ \\text{symmetric}\\ \\land\\ \\forall x\\neq0:\\ \\langle x,x\\rangle_M>0", "relation": "definition", "rule": "mko_ips_def", "conditions": [], "explanation": { "zh-Hant": "依 MKO，內積有三個分開的責任：兩槽雙線性、對稱、正定（每個非零向量的 ⟨x,x⟩ 嚴格大於 0）。之後 S03 到 S07 分別證明，缺一不可；S12 到 S14 的對照矩陣各自缺一項。", "en": "By the MKO, an inner product has three separate obligations: bilinear in both slots, symmetric, and positive definite (⟨x,x⟩ strictly positive for every nonzero vector). S03 to S07 prove them one by one, and none can be skipped; each comparison matrix in S12 to S14 lacks one of them." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_ips_def" ], "explanation": { "zh-Hant": "只展開定義，不推出新事實。", "en": "It only unfolds the definition and derives no new fact." } }, "leaf_basis": "mko_ips_def + def_unfold" }, { "id": "S02", "uses": [ "G2", "matrix_product_def" ], "before_tex": "\\langle x,y\\rangle_M=x^{T}My", "after_tex": "\\langle x,y\\rangle_M=2x_1y_1+x_1y_2+x_2y_1+3x_2y_2", "relation": "equivalent", "rule": "matrix_product_def", "conditions": [], "explanation": { "zh-Hant": "先算 My=(2y₁+y₂, y₁+3y₂)，再與 xᵀ 相乘：x₁(2y₁+y₂)+x₂(y₁+3y₂)，展開得四項。係數 2、1、1、3 正是 M 的四個元素 M_{ij}，乘在 x_iy_j 上。", "en": "First My=(2y₁+y₂, y₁+3y₂); multiplying by xᵀ gives x₁(2y₁+y₂)+x₂(y₁+3y₂), which expands into four terms. The coefficients 2, 1, 1, 3 are exactly the entries M_{ij}, attached to x_iy_j." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "matrix_product_def", "distributive" ], "explanation": { "zh-Hant": "展開是恆等式；反向把四項重組回 xᵀMy，兩邊同值。", "en": "The expansion is an identity; backward, the four terms regroup into xᵀMy, so both sides have the same value." } }, "leaf_basis": "matrix_product_def + distributive" }, { "id": "S03", "uses": [ "S02", "mko_bilinear_def" ], "before_tex": "\\langle\\alpha x+\\beta z,\\ y\\rangle_M", "after_tex": "\\alpha\\langle x,y\\rangle_M+\\beta\\langle z,y\\rangle_M", "relation": "equivalent", "rule": "mko_bilinear_def", "conditions": [], "explanation": { "zh-Hant": "第一槽線性：αx+βz 的坐標是 αx_i+βz_i；代入 S02 的四項，每一項 M_{ij}(αx_i+βz_i)y_j 依分配律拆成 α M_{ij}x_iy_j+β M_{ij}z_iy_j，再分組。x、z、y、α、β 都是任意的，所以這是全稱證書（N1 以多項式恆等式核對）。", "en": "Linearity in the first slot: αx+βz has coordinates αx_i+βz_i; substituting into the four terms of S02, each term M_{ij}(αx_i+βz_i)y_j splits by distributivity into αM_{ij}x_iy_j+βM_{ij}z_iy_j, and the terms regroup. x, z, y, α, β are arbitrary, so this is a universal certificate (N1 checks it as a polynomial identity)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "mul_assoc", "mul_comm" ], "explanation": { "zh-Hant": "所用都是實數恆等式，兩邊同值，雙向成立。", "en": "Only real identities are used; both sides have the same value, so the step holds both ways." } }, "leaf_basis": "distributive + mul_assoc + mul_comm + finite_sum_rules" }, { "id": "S04", "uses": [ "S02", "mko_bilinear_def" ], "before_tex": "\\langle x,\\ \\alpha y+\\beta w\\rangle_M", "after_tex": "\\alpha\\langle x,y\\rangle_M+\\beta\\langle x,w\\rangle_M", "relation": "equivalent", "rule": "mko_bilinear_def", "conditions": [], "explanation": { "zh-Hant": "第二槽線性同理：每一項 M_{ij}x_i(αy_j+βw_j) 拆成 αM_{ij}x_iy_j+βM_{ij}x_iw_j。第二槽必須另外核對：雙線性是兩個槽各自的責任。", "en": "Linearity in the second slot works the same way: each term M_{ij}x_i(αy_j+βw_j) splits into αM_{ij}x_iy_j+βM_{ij}x_iw_j. The second slot must be checked separately: bilinearity is an obligation for each slot." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "mul_assoc", "mul_comm" ], "explanation": { "zh-Hant": "恆等式，雙向成立。", "en": "An identity, valid both ways." } }, "leaf_basis": "distributive + mul_assoc + mul_comm + finite_sum_rules" }, { "id": "S05", "uses": [ "S02", "G2" ], "before_tex": "\\langle y,x\\rangle_M=2y_1x_1+y_1x_2+y_2x_1+3y_2x_2", "after_tex": "\\langle y,x\\rangle_M=\\langle x,y\\rangle_M\\ \\ (\\forall x,y)", "relation": "implies", "rule": "mul_comm", "conditions": [ "M_{12}=M_{21}" ], "explanation": { "zh-Hant": "交換 x、y 後，項 y₁x₂ 對應原式的 x₂y₁、y₂x₁ 對應 x₁y₂；兩者係數都是 1，因為 M_{12}=M_{21}=1。再用乘法交換律，四項逐一相同。對稱性依賴 M 的對稱。", "en": "After swapping x and y, the term y₁x₂ matches x₂y₁ of the original and y₂x₁ matches x₁y₂; both carry coefficient 1 because M_{12}=M_{21}=1. Commutativity of multiplication then makes the four terms agree one by one. Symmetry relies on M being symmetric." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mul_comm", "G2" ], "explanation": { "zh-Hant": "條件 M_{12}=M_{21} 不可省略：S14 的 R 違反它，形式就不對稱（N2 以恆等式核對本步）。", "en": "The condition M_{12}=M_{21} cannot be dropped: R in S14 violates it and its form is not symmetric (N2 checks this step as an identity)." } }, "leaf_basis": "mul_comm + add_comm" }, { "id": "S06", "uses": [ "S02" ], "before_tex": "q(x)=\\langle x,x\\rangle_M", "after_tex": "q(x)=2x_1^2+2x_1x_2+3x_2^2=ax_1^2+2bx_1x_2+cx_2^2\\ \\ (a,b,c)=(2,1,3)", "relation": "definition", "rule": "def_intro", "conditions": [], "explanation": { "zh-Hant": "在 S02 中令 y=x，x₁x₂ 與 x₂x₁ 合併為 2x₁x₂。一般對稱矩陣（行 (a,b),(b,c)）給 q=ax₁²+2bx₁x₂+cx₂²；此處 a=2、b=1、c=3。", "en": "Put y=x in S02; the terms x₁x₂ and x₂x₁ combine into 2x₁x₂. A general symmetric matrix with rows (a,b),(b,c) gives q=ax₁²+2bx₁x₂+cx₂²; here a=2, b=1, c=3." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "def_intro" ], "explanation": { "zh-Hant": "只命名與合併同類項；之後的正定判斷都在 q 上進行。", "en": "It only names and collects like terms; the positive-definiteness check is made on q from here on." } }, "leaf_basis": "def_intro + mul_comm + distributive" }, { "id": "S07", "uses": [ "S06" ], "before_tex": "a=2>0,\\ \\ ac-b^2=6-1=5>0", "after_tex": "\\forall x\\neq0:\\ q(x)>0", "relation": "theorem_application", "rule": "L_sylvester_suff", "conditions": [ "a>0", "ac-b^2>0" ], "explanation": { "zh-Hant": "Sylvester 條件：首主子式 a=2>0，行行式 ac−b²=2·3−1²=5>0。依引理 L_sylvester_suff，q 在每個非零向量上為正。引理在子步 S07.1 到 S07.3 以配方完整證明，對所有 x 一次成立，不靠樣本。", "en": "Sylvester conditions: the leading principal minor a=2>0 and the determinant ac−b²=2·3−1²=5>0. By Lemma L_sylvester_suff, q is positive at every nonzero vector. The lemma is proved completely by completing the square in sub-steps S07.1 to S07.3, for all x at once, without samples." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S06", "L_sylvester_suff", "numeric_evaluation" ], "explanation": { "zh-Hant": "需要兩個嚴格不等式；若 ac−b²=0（見 S13）結論就不成立。", "en": "Both strict inequalities are needed; if ac−b²=0 (see S13) the conclusion fails." } }, "leaf_basis": "square_nonneg + order_basics + zero_product" }, { "id": "S07.1", "uses": [ "distributive" ], "before_tex": "a\\,q(x)=a(ax_1^2+2bx_1x_2+cx_2^2)", "after_tex": "a\\,q(x)=(ax_1+bx_2)^2+(ac-b^2)x_2^2", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "展開右邊：(ax₁+bx₂)²=a²x₁²+2abx₁x₂+b²x₂²，加上 (ac−b²)x₂² 得 a²x₁²+2abx₁x₂+acx₂²，正是左邊。乘以 a 而不除以 a，所以這個恆等式對所有實數 a 都成立（N3）。", "en": "Expand the right side: (ax₁+bx₂)²=a²x₁²+2abx₁x₂+b²x₂², and adding (ac−b²)x₂² gives a²x₁²+2abx₁x₂+acx₂², which is the left side. Multiplying by a instead of dividing keeps the identity valid for every real a (N3)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "power_laws" ], "explanation": { "zh-Hant": "多項式恆等式，兩邊同值。", "en": "A polynomial identity; both sides have the same value." } }, "leaf_basis": "distributive + power_laws + add_inverse" }, { "id": "S07.2", "uses": [ "S07.1", "square_nonneg", "order_basics" ], "before_tex": "a>0,\\ ac-b^2>0,\\ a\\,q(x)=(ax_1+bx_2)^2+(ac-b^2)x_2^2", "after_tex": "a\\,q(x)\\ge 0\\ \\Rightarrow\\ q(x)\\ge 0", "relation": "implies", "rule": "square_nonneg", "conditions": [ "a>0", "ac-b^2>0" ], "explanation": { "zh-Hant": "平方非負，(ac−b²)>0 乘非負的 x₂² 仍非負，所以 a·q(x)≥0；再乘以 1/a>0（順序公理保持不等號方向）得 q(x)≥0。", "en": "Squares are nonnegative, and (ac−b²)>0 times the nonnegative x₂² is nonnegative, so a·q(x)≥0; multiplying by 1/a>0 (the order axioms keep the direction) gives q(x)≥0." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "square_nonneg", "order_basics" ], "explanation": { "zh-Hant": "需要 a>0 才能除以 a 而不翻轉不等號；不宣稱反向。", "en": "a>0 is needed to divide by a without flipping the inequality; no converse is claimed." } }, "leaf_basis": "square_nonneg + order_basics + mul_inverse_nonzero" }, { "id": "S07.3", "uses": [ "S07.1", "S07.2", "zero_product" ], "before_tex": "q(x)=0", "after_tex": "x_2=0\\ \\land\\ x_1=0,\\ \\text{so}\\ x\\neq0\\Rightarrow q(x)>0", "relation": "implies", "rule": "zero_product", "conditions": [ "a>0", "ac-b^2>0" ], "explanation": { "zh-Hant": "若 q(x)=0，則兩個非負項之和為 0，所以兩項皆為 0：(ac−b²)x₂²=0 且 ac−b²≠0，得 x₂=0；再由 (ax₁+bx₂)²=0 得 ax₁=0，因 a≠0 得 x₁=0。逆否：x≠0 ⇒ q(x)≠0，配合 S07.2 得 q(x)>0。S07.1 到 S07.3 合起來證得引理 L_sylvester_suff。", "en": "If q(x)=0, the two nonnegative terms sum to 0, so both are 0: (ac−b²)x₂²=0 with ac−b²≠0 gives x₂=0; then (ax₁+bx₂)²=0 gives ax₁=0, and a≠0 gives x₁=0. Contrapositively x≠0 ⇒ q(x)≠0, and with S07.2, q(x)>0. S07.1 to S07.3 together prove Lemma L_sylvester_suff." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "zero_product", "square_nonneg", "contrapositive" ], "explanation": { "zh-Hant": "「非負數之和為 0 ⇒ 各項為 0」與零積性質；兩個嚴格條件都用到。", "en": "It uses 'a sum of nonnegatives is 0 ⇒ each is 0' and the zero-product property; both strict conditions are used." } }, "leaf_basis": "zero_product + square_nonneg + order_basics + contrapositive" }, { "id": "S08", "uses": [ "S06", "S07.1" ], "before_tex": "2\\,q(x)=2(2x_1^2+2x_1x_2+3x_2^2)", "after_tex": "2\\,q(x)=(2x_1+x_2)^2+5x_2^2", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "把 S07.1 的恆等式代入 a=2、b=1、c=3：2q(x)=(2x₁+x₂)²+5x₂²。這是 M 的具體配方證書：右邊兩項非負，且同時為 0 只在 x=0。", "en": "Substitute a=2, b=1, c=3 into the identity of S07.1: 2q(x)=(2x₁+x₂)²+5x₂². This is the concrete completing-the-square certificate for M: both terms on the right are nonnegative and vanish together only at x=0." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "S07.1", "numeric_evaluation" ], "explanation": { "zh-Hant": "多項式恆等式（N4），雙向成立。", "en": "A polynomial identity (N4), valid both ways." } }, "leaf_basis": "distributive + power_laws" }, { "id": "S09", "uses": [ "S01", "S03", "S04", "S05", "S07" ], "before_tex": "\\text{S03},\\ \\text{S04},\\ \\text{S05},\\ \\text{S07}", "after_tex": "\\langle\\cdot,\\cdot\\rangle_M\\ \\text{is an inner product};\\ (\\mathbb{R}^2,\\langle\\cdot,\\cdot\\rangle_M)\\ \\text{real inner-product space}", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "三項責任都已證明：雙線性（S03、S04）、對稱（S05）、正定（S07）。依 S01 的定義，⟨·,·⟩_M 是 ℝ² 上的內積。", "en": "All three obligations are proved: bilinearity (S03, S04), symmetry (S05) and positive definiteness (S07). By the definition in S01, ⟨·,·⟩_M is an inner product on ℝ²." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S01", "S03", "S04", "S05", "S07", "and_intro_elim" ], "explanation": { "zh-Hant": "目標 G：內積。足夠條件 H：三項責任；H⇒G 由 S01。H 由 G2 正向證得，沒有用到 G，也沒有用到向量樣本。", "en": "Goal G: an inner product. Sufficient condition H: the three obligations; H⇒G by S01. H is proved forwards from G2, using neither G nor sampled vectors." } }, "leaf_basis": "and_intro_elim + mko_ips_def" }, { "id": "S10", "uses": [ "S02", "S06" ], "before_tex": "e_1=(1,0),\\ e_2=(0,1),\\ x=(1,-1)", "after_tex": "\\langle e_1,e_2\\rangle_M=1,\\ q(e_1)=2,\\ q(e_2)=3,\\ q(1,-1)=2-2+3=3", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "精確求值：⟨e₁,e₂⟩_M=M₁₂=1；q(e₁)=a=2，q(e₂)=c=3；q(1,−1)=2·1+2·1·(−1)+3·1=3>0。與 S08 一致：2q(1,−1)=(2−1)²+5=6。這些數值只是例子，正定性的證明在 S07。", "en": "Exact evaluation: ⟨e₁,e₂⟩_M=M₁₂=1; q(e₁)=a=2 and q(e₂)=c=3; q(1,−1)=2·1+2·1·(−1)+3·1=3>0. Consistent with S08: 2q(1,−1)=(2−1)²+5=6. These values are examples only; positive definiteness is proved in S07." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "numeric_evaluation" ], "explanation": { "zh-Hant": "可由 N5 重算；有限個值不能取代 S07 的全稱證書（見錯誤控制）。", "en": "Recomputable by N5; finitely many values cannot replace the universal certificate of S07 (see the error controls)." } }, "leaf_basis": "numeric_evaluation" }, { "id": "S11", "uses": [ "S06" ], "before_tex": "\\forall x\\neq0:\\ ax_1^2+2bx_1x_2+cx_2^2>0", "after_tex": "a>0\\ \\land\\ ac-b^2>0", "relation": "implies", "rule": "L_sylvester_nec", "conditions": [], "explanation": { "zh-Hant": "Sylvester 條件也是必要的（引理 L_sylvester_nec），所以對 2×2 對稱矩陣它是精確判準：正定 ⇔ a>0 且 ac−b²>0。必要方向讓我們能用條件失敗直接否定正定（S12、S13）。證明在子步 S11.1、S11.2。", "en": "The Sylvester conditions are also necessary (Lemma L_sylvester_nec), so for 2-by-2 symmetric matrices they are an exact criterion: positive definite ⇔ a>0 and ac−b²>0. The necessary direction lets a failed condition refute positive definiteness directly (S12, S13). The proof is in sub-steps S11.1 and S11.2." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_sylvester_nec" ], "explanation": { "zh-Hant": "與 S07 合起來給出等價；本步只用必要方向。", "en": "Together with S07 it gives an equivalence; this step uses only the necessary direction." } }, "leaf_basis": "order_basics + mul_inverse_nonzero" }, { "id": "S11.1", "uses": [ "S06" ], "before_tex": "x=e_1=(1,0)\\neq0", "after_tex": "q(e_1)=a>0", "relation": "implies", "rule": "forall_elim", "conditions": [], "explanation": { "zh-Hant": "在正定假設中取 x=e₁≠0：q(e₁)=a·1+2b·0+c·0=a，所以 a>0。", "en": "Specialize the positivity hypothesis to x=e₁≠0: q(e₁)=a·1+2b·0+c·0=a, so a>0." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "forall_elim", "mul_zero" ], "explanation": { "zh-Hant": "只用正定假設的一個特例。", "en": "Only one instance of the positivity hypothesis is used." } }, "leaf_basis": "forall_elim + mul_zero + numeric_evaluation" }, { "id": "S11.2", "uses": [ "S11.1" ], "before_tex": "x=(-b,a)\\neq0", "after_tex": "q(-b,a)=ab^2-2ab^2+ca^2=a(ac-b^2)>0\\ \\Rightarrow\\ ac-b^2>0", "relation": "implies", "rule": "forall_elim", "conditions": [ "a>0" ], "explanation": { "zh-Hant": "因 a>0，向量 (−b,a) 的第二坐標非零，所以它非零。代入：a·b²+2b·(−b)·a+c·a²=a(ac−b²)（N6）。正定給出 a(ac−b²)>0，兩邊乘 1/a>0 得 ac−b²>0。S11.1 與 S11.2 合起來證得引理 L_sylvester_nec。", "en": "Since a>0, the vector (−b,a) has a nonzero second coordinate and is nonzero. Substituting gives a·b²+2b·(−b)·a+c·a²=a(ac−b²) (N6). Positivity gives a(ac−b²)>0, and multiplying by 1/a>0 gives ac−b²>0. S11.1 and S11.2 together prove Lemma L_sylvester_nec." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S11.1", "forall_elim", "order_basics" ], "explanation": { "zh-Hant": "需要 S11.1 的 a>0，才能保證 (−b,a)≠0 並除以 a。", "en": "It needs a>0 from S11.1, both to ensure (−b,a)≠0 and to divide by a." } }, "leaf_basis": "forall_elim + order_basics + distributive + mul_inverse_nonzero" }, { "id": "S12", "uses": [ "G3", "S11" ], "before_tex": "N:\\ a=1,\\ b=2,\\ c=1,\\ ac-b^2=1-4=-3<0", "after_tex": "q_N(-2,1)=4-8+1=-3<0\\ \\Rightarrow\\ \\langle\\cdot,\\cdot\\rangle_N\\ \\text{not an inner product}", "relation": "implies", "rule": "contrapositive", "conditions": [], "explanation": { "zh-Hant": "N 對稱、雙線性，對角元 1、1 都為正，但行行式 −3<0，依 L_sylvester_nec 的逆否不正定。S11.2 的見證向量 (−b,a)=(−2,1) 給出 q_N=1·4+4·(−2)·1+1·1=−3<0。另一個見證 (1,−1) 給 1−4+1=−2。", "en": "N is symmetric and bilinear and its diagonal entries 1, 1 are positive, but its determinant is −3<0, so by the contrapositive of L_sylvester_nec it is not positive definite. The witness vector (−b,a)=(−2,1) of S11.2 gives q_N=1·4+4·(−2)·1+1·1=−3<0. Another witness, (1,−1), gives 1−4+1=−2." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S11", "contrapositive", "numeric_evaluation" ], "explanation": { "zh-Hant": "一個負值見證就否定正定（N7）；不需要行舉其他向量。", "en": "One negative witness refutes positive definiteness (N7); no other vectors are needed." } }, "leaf_basis": "contrapositive + numeric_evaluation" }, { "id": "S13", "uses": [ "G3", "S11" ], "before_tex": "P:\\ a=b=c=1,\\ ac-b^2=0", "after_tex": "q_P(x)=(x_1+x_2)^2\\ge0,\\ q_P(1,-1)=0\\ \\Rightarrow\\ \\text{semi-definite, not an inner product}", "relation": "implies", "rule": "contrapositive", "conditions": [], "explanation": { "zh-Hant": "P 對稱雙線性，q_P=x₁²+2x₁x₂+x₂²=(x₁+x₂)²≥0，是半正定；但行行式 0 不是嚴格大於 0，非零向量 (1,−1) 給 q_P=0，違反「非零向量必須嚴格為正」。所以這個僅半正定、具有非零零向量的形式不是內積。廣義的半正定包含正定，不能把結論寫成所有半正定形式都不是內積。", "en": "P is symmetric and bilinear, with q_P=x₁²+2x₁x₂+x₂²=(x₁+x₂)²≥0, so it is positive semi-definite; but its determinant 0 is not strictly positive, and the nonzero vector (1,−1) gives q_P=0, violating 'strictly positive on nonzero vectors'. Thus this merely positive-semidefinite form, with a nonzero null vector, is not an inner product. Positive-semidefinite in the inclusive sense also includes positive-definite forms, so the conclusion is not that every semidefinite form fails to be an inner product." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S11", "contrapositive", "numeric_evaluation" ], "explanation": { "zh-Hant": "見證 (1,−1)≠0 而 q_P=0（N8）。", "en": "Witness: (1,−1)≠0 while q_P=0 (N8)." } }, "leaf_basis": "contrapositive + square_nonneg + numeric_evaluation" }, { "id": "S14", "uses": [ "G3", "S02" ], "before_tex": "R=\\begin{pmatrix}2&1\\\\0&3\\end{pmatrix}", "after_tex": "\\langle e_1,e_2\\rangle_R=R_{12}=1\\neq 0=R_{21}=\\langle e_2,e_1\\rangle_R\\ \\Rightarrow\\ \\text{not symmetric}", "relation": "implies", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "xᵀRy 仍是雙線性（S03、S04 的論證不用對稱），但 ⟨e₁,e₂⟩_R=R₁₂=1，⟨e₂,e₁⟩_R=R₂₁=0，對稱性失敗。所以 R 不定義內積，不論它的二次型是否為正；三項責任必須分開核對。", "en": "xᵀRy is still bilinear (the arguments of S03 and S04 do not use symmetry), but ⟨e₁,e₂⟩_R=R₁₂=1 and ⟨e₂,e₁⟩_R=R₂₁=0, so symmetry fails. Hence R does not define an inner product, whatever its quadratic form does; the three obligations must be checked separately." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "quantifier_negation", "matrix_product_def" ], "explanation": { "zh-Hant": "一對反例否定全稱對稱性（N9）。", "en": "One counterexample pair refutes universal symmetry (N9)." } }, "leaf_basis": "quantifier_negation + matrix_product_def" }, { "id": "S15", "uses": [ "S09", "S12", "S13", "S14" ], "before_tex": "\\text{S09},\\ \\text{S12},\\ \\text{S13},\\ \\text{S14}", "after_tex": "M\\ \\text{inner product};\\ N\\ \\text{indefinite};\\ P\\ \\text{semi-definite};\\ R\\ \\text{not symmetric}", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "合併：M 通過三項責任，是內積；N 不定（行行式 −3）；P 只是半正定（行行式 0）；R 不對稱。", "en": "Combine: M passes all three obligations and is an inner product; N is indefinite (determinant −3); P is only semi-definite (determinant 0); R is not symmetric." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S09", "S12", "S13", "S14", "and_intro_elim" ], "explanation": { "zh-Hant": "目標是四項判定的合取；各項由所行步驟正向證得，H⇒G 由合取引入。", "en": "The goal is the conjunction of the four decisions; each is proved forwards by the listed steps, and H⇒G by conjunction introduction." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "⟨x,y⟩_M=2x₁y₁+x₁y₂+x₂y₁+3x₂y₂ 對兩槽雙線性、對稱，且 2q(x)=(2x₁+x₂)²+5x₂² 只在 x=0 時為 0，Sylvester 條件 a=2>0、ac−b²=5>0 給出全稱正定證書，所以 (ℝ²,⟨·,·⟩_M) 是實內積空間。對 2×2 對稱矩陣，Sylvester 條件是正定的充要條件。N（行行式 −3，q_N(−2,1)=−3）、P（行行式 0，q_P(1,−1)=0）、R（不對稱）都不給出內積。", "en": "⟨x,y⟩_M=2x₁y₁+x₁y₂+x₂y₁+3x₂y₂ is bilinear in both slots and symmetric, and 2q(x)=(2x₁+x₂)²+5x₂² vanishes only at x=0; the Sylvester conditions a=2>0 and ac−b²=5>0 give a universal positive-definiteness certificate, so (ℝ²,⟨·,·⟩_M) is a real inner-product space. For 2-by-2 symmetric matrices the Sylvester conditions are necessary and sufficient for positive definiteness. N (determinant −3, q_N(−2,1)=−3), P (determinant 0, q_P(1,−1)=0) and R (not symmetric) give no inner product." }, "negative_controls": [ { "claim": { "zh-Hant": "N 的二次型在 (1,0)、(0,1)、(1,1) 都為正，所以 ⟨·,·⟩_N 是內積", "en": "The quadratic form of N is positive at (1,0), (0,1), (1,1), so ⟨·,·⟩_N is an inner product" }, "expected": "reject", "reason": { "zh-Hant": "q_N(1,−1)=1−4+1=−2<0。", "en": "q_N(1,−1)=1−4+1=−2<0." } }, { "claim": { "zh-Hant": "P 對稱且 q_P≥0，所以 ⟨·,·⟩_P 是內積", "en": "P is symmetric and q_P≥0, so ⟨·,·⟩_P is an inner product" }, "expected": "reject", "reason": { "zh-Hant": "q_P(1,−1)=0 而 (1,−1)≠0，只是半正定。", "en": "q_P(1,−1)=0 although (1,−1)≠0; it is only semi-definite." } }, { "claim": { "zh-Hant": "xᵀRy 雙線性且對角元為正，所以是內積", "en": "xᵀRy is bilinear with positive diagonal entries, so it is an inner product" }, "expected": "reject", "reason": { "zh-Hant": "⟨e₁,e₂⟩_R=1 但 ⟨e₂,e₁⟩_R=0，不對稱。", "en": "⟨e₁,e₂⟩_R=1 but ⟨e₂,e₁⟩_R=0; it is not symmetric." } }, { "claim": { "zh-Hant": "對角元都為正的 2×2 對稱矩陣必定正定", "en": "A 2-by-2 symmetric matrix with positive diagonal entries is always positive definite" }, "expected": "reject", "reason": { "zh-Hant": "N 的對角元為 1、1，但行行式 −3，q_N(−2,1)=−3。", "en": "N has diagonal entries 1, 1 but determinant −3, and q_N(−2,1)=−3." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "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": "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": "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": "order_basics", "kind": "prerequisite", "statement": { "zh-Hant": "實數的序：任兩數恰有 uv 之一；u0 推出 uwvw。", "en": "Order of the reals: exactly one of uv holds; u0 give uwvw." } }, { "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": "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": "mul_assoc", "kind": "prerequisite", "statement": { "zh-Hant": "乘法結合律：(uv)w=u(vw)。", "en": "Associativity of multiplication: (uv)w=u(vw)." } }, { "id": "mul_comm", "kind": "prerequisite", "statement": { "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。", "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)." } }, { "id": "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": "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": "def_intro", "kind": "prerequisite", "statement": { "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。", "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable." } }, { "id": "zero_product", "kind": "prerequisite", "statement": { "zh-Hant": "零乘積性質：在 ℤ、ℚ、ℝ、ℂ（或任何整域）中，uv=0 ⇔ u=0 或 v=0。", "en": "Zero-product property: in ℤ, ℚ, ℝ, ℂ (or any integral domain), uv=0 ⇔ u=0 or v=0." } }, { "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_inverse", "kind": "prerequisite", "statement": { "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。", "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)." } }, { "id": "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": "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": "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": "quantifier_negation", "kind": "prerequisite", "statement": { "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。", "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)." } }, { "id": "mko_ips_def", "kind": "cited", "source": { "mko_id": "mko-real-inner-product-space", "sha256": "70d1a304624117e0e814b5b9ca816a5b851798e54cc213f1f2b3460e6450f963", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "在同一實向量空間上，內積須對兩槽雙線性、對稱，且對每個非零向量 x 都有 ⟨x,x⟩>0。本輪對 ℝ² 使用整係數對稱矩陣及 Sylvester 條件作全稱正定證書。", "en": "On the same real vector space, an inner product must be bilinear in both slots, symmetric and satisfy ⟨x,x⟩>0 for every nonzero x. This slice uses integer-coefficient symmetric matrices on ℝ² and the Sylvester conditions as a universal positive-definiteness certificate." } }, { "id": "mko_bilinear_def", "kind": "cited", "source": { "mko_id": "mko-real-bilinear-form", "sha256": "fa2ccad4a26e9cf94b590f02594c9e1c0db0ba86fc50c60ee410e2b934d4c7d5", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "B:V×V→ℝ 為雙線性，若固定任一輸入後，另一輸入上的映射皆為線性；整係數矩陣公式 B(x,y)=xᵀAy 作全稱座標證書。", "en": "B:V×V→ℝ is bilinear when fixing either input leaves a linear map in the other; the integer-coefficient matrix formula B(x,y)=xᵀAy serves as a universal coordinate certificate." } }, { "id": "L_sylvester_suff", "kind": "lemma", "proved_in": [ "S07.1", "S07.2", "S07.3" ], "statement": { "zh-Hant": "對實數 a,b,c，若 a>0 且 ac−b²>0，則 q(x)=ax₁²+2bx₁x₂+cx₂² 對每個 x≠0 都大於 0。", "en": "For reals a,b,c, if a>0 and ac−b²>0, then q(x)=ax₁²+2bx₁x₂+cx₂² is positive for every x≠0." } }, { "id": "L_sylvester_nec", "kind": "lemma", "proved_in": [ "S11.1", "S11.2" ], "statement": { "zh-Hant": "對實數 a,b,c，若 q(x)=ax₁²+2bx₁x₂+cx₂² 對每個 x≠0 都大於 0，則 a>0 且 ac−b²>0。", "en": "For reals a,b,c, if q(x)=ax₁²+2bx₁x₂+cx₂² is positive for every x≠0, then a>0 and ac−b²>0." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S03", "vars": [ "al", "be", "x1", "x2", "z1", "z2", "y1", "y2" ], "degree": 1, "lhs": "2*(al*x1+be*z1)*y1+(al*x1+be*z1)*y2+(al*x2+be*z2)*y1+3*(al*x2+be*z2)*y2", "rhs": "al*(2*x1*y1+x1*y2+x2*y1+3*x2*y2)+be*(2*z1*y1+z1*y2+z2*y1+3*z2*y2)" }, { "id": "N2", "step": "S05", "vars": [ "x1", "x2", "y1", "y2" ], "degree": 1, "lhs": "2*y1*x1+y1*x2+y2*x1+3*y2*x2", "rhs": "2*x1*y1+x1*y2+x2*y1+3*x2*y2" }, { "id": "N3", "step": "S07.1", "vars": [ "a", "b", "c", "x1", "x2" ], "degree": 2, "lhs": "a*(a*x1**2+2*b*x1*x2+c*x2**2)", "rhs": "(a*x1+b*x2)**2+(a*c-b**2)*x2**2" }, { "id": "N4", "step": "S08", "vars": [ "x1", "x2" ], "degree": 2, "lhs": "2*(2*x1**2+2*x1*x2+3*x2**2)", "rhs": "(2*x1+x2)**2+5*x2**2" }, { "id": "N5", "step": "S10", "lhs": "2*1**2+2*1*(-1)+3*(-1)**2", "rhs": "3" }, { "id": "N6", "step": "S11.2", "vars": [ "a", "b", "c" ], "degree": 2, "lhs": "a*(-b)**2+2*b*(-b)*a+c*a**2", "rhs": "a*(a*c-b**2)" }, { "id": "N7", "step": "S12", "lhs": "1*(-2)**2+4*(-2)*1+1*1**2", "rhs": "1*1-2**2" }, { "id": "N8", "step": "S13", "lhs": "(1+(-1))**2", "rhs": "1*1-1**2" }, { "id": "N9", "step": "S14", "lhs": "1", "relation": "!=", "rhs": "0" }, { "id": "N10", "step": "S07", "lhs": "2*3-1**2", "rhs": "5" } ] }, "presentation": { "md_scope": { "zh-Hant": "在 MKO 的範圍內工作：矩陣與向量完整明示，候選矩陣為 2×2 整數矩陣（假設 a1）；正定性以首主子式與行行式的精確條件核對，不由有限個向量樣本建立（假設 a2）。雙線性、對稱、正定是三個分開的責任。不處理內積誘導範數（下一節點）或完備性，不處理 n>2 的 Sylvester 判準，不改動 canonical MKO 或英文 edition。", "en": "Works inside the MKO's scope: matrices and vectors are fully specified, with 2-by-2 integer candidate matrices (assumption a1); positive definiteness is checked by exact leading-principal-minor and determinant conditions, not established from finitely many sampled vectors (assumption a2). Bilinearity, symmetry and positive definiteness are three separate obligations. The induced norm (the next node), completeness and the Sylvester criterion for n>2 are not treated, and the canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "M=\\begin{pmatrix}2&1\\\\1&3\\end{pmatrix}", "候選對稱整數矩陣", "the candidate symmetric integer matrix" ], [ "\\langle x,y\\rangle_M=x^{T}My", "由 M 定義的雙線性形式", "the bilinear form defined by M" ], [ "q(x)=\\langle x,x\\rangle_M", "對應的二次型", "the associated quadratic form" ], [ "\\begin{pmatrix}a&b\\\\b&c\\end{pmatrix}", "一般 2×2 實對稱矩陣，q(x)=ax_1^2+2bx_1x_2+cx_2^2", "a general 2-by-2 real symmetric matrix, q(x)=ax_1^2+2bx_1x_2+cx_2^2" ], [ "a,\\ ac-b^2", "首主子式與行行式", "the leading principal minor and the determinant" ], [ "N,P,R", "三個對照矩陣", "the three comparison matrices" ] ], "md_error_controls": [ { "title": "對稱雙線性或半正定都不夠 / Symmetric bilinear or semi-definite is not enough", "zh": "MKO 指出的誤解是把對稱雙線性形式或半正定形式直接叫內積。N 與 P 都對稱雙線性；N 在 (1,−1) 取負值，P 在非零的 (1,−1) 取 0。內積要求每個非零向量嚴格為正。", "en": "The MKO's misconception is to call a symmetric bilinear form, or a semi-definite one, an inner product. N and P are both symmetric and bilinear; N is negative at (1,−1) and P is 0 at the nonzero vector (1,−1). An inner product must be strictly positive on every nonzero vector." }, { "title": "向量樣本不證正定 / Sampled vectors do not prove positive definiteness", "zh": "N 的二次型在 (1,0)、(0,1)、(1,1) 的值是 1、1、6，全為正，抽樣會誤判為正定；但 q_N(1,−1)=−2。正定必須由全稱證書建立：配方恆等式加 Sylvester 條件（S07）。", "en": "The quadratic form of N takes the values 1, 1, 6 at (1,0), (0,1), (1,1), all positive, so sampling would wrongly suggest positive definiteness; but q_N(1,−1)=−2. Positive definiteness must come from a universal certificate: the completing-the-square identity plus the Sylvester conditions (S07)." }, { "title": "對角元為正不夠 / Positive diagonal entries are not enough", "zh": "N 的對角元 1、1 都為正，只滿足第一個條件 a>0；行行式 ac−b²=−3 才揭露它不正定。2×2 時兩個條件缺一不可（S11 證明兩者都是必要的）。", "en": "The diagonal entries 1, 1 of N are positive, satisfying only the first condition a>0; the determinant ac−b²=−3 reveals that it is not positive definite. For 2-by-2 matrices both conditions are required (S11 proves both are necessary)." } ], "md_open_items": [ { "zh-Hant": "MKO 提醒內積空間不自動完備；ℝ² 本身是完備的，但本稿不處理完備性或 Hilbert 空間。", "en": "The MKO notes that an inner-product space is not automatically complete; ℝ² itself is complete, but this guide does not treat completeness or Hilbert spaces." }, { "zh-Hant": "Sylvester 判準只對 2×2 完整證明；n×n 的一般形式不在本稿範圍內。", "en": "The Sylvester criterion is proved completely only for 2-by-2 matrices; the general n-by-n form is outside this guide's scope." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "d02e546e2588353f6bbb496de2a5d6006b38b2adc1288620b1b7141c8175932f", "original_source_sha256": "b8b97eaec9185806e2d7a87378d14b180297a8ad4a22308761b5db72c6a6627c", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [ "BR-M081" ], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }