{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-convex-optimization-problem-01",
  "mko_id": "mko-convex-optimization-problem",
  "title": {
    "zh-Hant": "凸最佳化問題：資格核對、可行性、單點求值與兩候選比較，以及它們不能推出什麼",
    "en": "Convex optimization problem: qualification, feasibility, single-point evaluation and a two-candidate comparison, and what they do not imply"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-convex-optimization-problem-01",
    "source": {
      "mko_id": "mko-convex-optimization-problem",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "0484bac345126653c2ee68f9057b0c7bf124c996ee7e7cfd461d32ad4e78747b",
      "english_sha256": "30f46a6833f7f53681e66621ee7ede3cfaa0afe76de419c007c9b77b842434ad"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "order_basics",
      "square_nonneg",
      "distributive",
      "numeric_evaluation",
      "quantifier_negation",
      "forall_intro"
    ],
    "goal": {
      "zh-Hant": "問題 P：在可行域 X=[0,5] 上最小化 f(x)=x²−6x+10。（一）分別核對可行域凸與目標凸，判定 P 具凸問題資格；（二）判定候選 x₁=1、x₂=4、x₃=−1 的可行性；（三）求 f(1)、f(4) 並比較兩個可行候選；（四）說明兩候選中較低者不因此是 P 的最優解，並以 P′：在 X′=(0,1] 上最小化 x² 說明凸問題不一定有最小值點。",
      "en": "Problem P: minimize f(x)=x²−6x+10 over the feasible domain X=[0,5]. (1) Check separately that the domain and the objective are convex, so that P qualifies as a convex problem; (2) decide feasibility of the candidates x₁=1, x₂=4, x₃=−1; (3) evaluate f(1) and f(4) and compare the two feasible candidates; (4) explain why the candidate with the lower objective value is not thereby an optimal solution of P, and use P′: minimize x² over X′=(0,1], to show that a convex problem need not have a minimizer."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "問題 P：在 X=[0,5] 上最小化 f(x)=x²−6x+10（整係數 a=1、b=−6、c=10）。",
          "en": "Problem P: minimize f(x)=x²−6x+10 over X=[0,5] (integer coefficients a=1, b=−6, c=10)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "提出的候選：x₁=1、x₂=4、x₃=−1。",
          "en": "Proposed candidates: x₁=1, x₂=4, x₃=−1."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "問題 P′：在 X′=(0,1]={x ∣ 0<x≤1} 上最小化 f₀(x)=x²。",
          "en": "Problem P′: minimize f₀(x)=x² over X′=(0,1]={x ∣ 0<x≤1}."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "mko_cvxopt"
        ],
        "before_tex": "P\\ \\text{qualifies as convex}",
        "after_tex": "X\\ \\text{convex}\\ \\land\\ f\\ \\text{convex on}\\ X",
        "relation": "equivalent",
        "rule": "mko_cvxopt",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，凸問題資格由兩個獨立義務組成：可行域 X 是凸集合，目標 f 在 X 上是凸函數。兩者要分別核對（MKO para0）；任何一個失敗，P 就不具資格。",
          "en": "By the MKO, qualification as a convex problem consists of two independent obligations: the feasible domain X is convex, and the objective f is convex on X. They must be checked separately (MKO para0); if either fails, P does not qualify."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_cvxopt",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "展開與收回定義互逆，雙向成立。",
            "en": "Unfolding and folding the definition undo each other, so the step holds both ways."
          }
        },
        "leaf_basis": "mko_cvxopt + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "L_interval",
          "mko_cvxset"
        ],
        "before_tex": "x,y\\in[0,5],\\ \\lambda\\in[0,1]",
        "after_tex": "\\lambda x+(1-\\lambda)y\\in[0,5];\\ X\\ \\text{convex}",
        "relation": "implies",
        "rule": "L_interval",
        "conditions": [
          "\\lambda\\in[0,1]"
        ],
        "explanation": {
          "zh-Hant": "用引理 L_interval（α=0、β=5）：任取 x、y∈[0,5] 與 λ∈[0,1]，凸組合仍在 [0,5]。依凸集合定義與 ∀ 引入，X 是凸的。引理在 S02.1、S02.2 中證明。",
          "en": "Apply Lemma L_interval (α=0, β=5): for arbitrary x, y∈[0,5] and λ∈[0,1], the convex combination stays in [0,5]. By the definition of a convex set and ∀-introduction, X is convex. The lemma is proved in S02.1 and S02.2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_interval",
            "mko_cvxset",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要 x、y 在區間內與 λ∈[0,1]；這只是義務之一，不涉及 f。",
            "en": "It needs x, y in the interval and λ∈[0,1]; this is only one obligation and says nothing about f."
          }
        },
        "leaf_basis": "L_interval + forall_intro"
      },
      {
        "id": "S02.1",
        "uses": [
          "mono_nonneg",
          "ineq_add",
          "distributive"
        ],
        "before_tex": "\\alpha\\le s,\\ \\alpha\\le t,\\ 0\\le\\lambda\\le 1",
        "after_tex": "\\alpha=\\lambda\\alpha+(1-\\lambda)\\alpha\\le\\lambda s+(1-\\lambda)t",
        "relation": "implies",
        "rule": "mono_nonneg",
        "conditions": [
          "\\lambda\\ge 0",
          "1-\\lambda\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "λ≥0 且 1−λ≥0。α≤s 乘以 λ、α≤t 乘以 1−λ，不等號方向不變；相加，左邊依分配律等於 (λ+1−λ)α=α。",
          "en": "λ≥0 and 1−λ≥0. Multiplying α≤s by λ and α≤t by 1−λ keeps the direction; adding, the left side equals (λ+1−λ)α=α by distributivity."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mono_nonneg",
            "ineq_add",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "證明 L_interval 的下界，沒有使用 L_interval 本身。",
            "en": "It proves the lower bound of L_interval without using L_interval itself."
          }
        },
        "leaf_basis": "mono_nonneg + ineq_add + distributive"
      },
      {
        "id": "S02.2",
        "uses": [
          "S02.1",
          "mono_nonneg",
          "ineq_add",
          "distributive"
        ],
        "before_tex": "s\\le\\beta,\\ t\\le\\beta",
        "after_tex": "\\lambda s+(1-\\lambda)t\\le\\lambda\\beta+(1-\\lambda)\\beta=\\beta",
        "relation": "implies",
        "rule": "mono_nonneg",
        "conditions": [
          "\\lambda\\ge 0",
          "1-\\lambda\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "上界對稱：s≤β 乘以 λ、t≤β 乘以 1−λ，相加並化簡為 β。S02.1 與 S02.2 證得 L_interval。",
          "en": "The upper bound is symmetric: multiply s≤β by λ and t≤β by 1−λ, add and simplify to β. S02.1 and S02.2 prove L_interval."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mono_nonneg",
            "ineq_add",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "與 S02.1 對稱；不使用 L_interval 本身。",
            "en": "Symmetric to S02.1; L_interval itself is not used."
          }
        },
        "leaf_basis": "mono_nonneg + ineq_add + distributive"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "L_gap1"
        ],
        "before_tex": "G_f(x,y,\\lambda)=\\lambda f(x)+(1-\\lambda)f(y)-f(\\lambda x+(1-\\lambda)y)",
        "after_tex": "G_f(x,y,\\lambda)=\\lambda(1-\\lambda)(x-y)^2",
        "relation": "equivalent",
        "rule": "L_gap1",
        "conditions": [],
        "explanation": {
          "zh-Hant": "f 的首項係數是 1，用引理 L_gap1（b=−6、c=10）：f 的 Jensen gap 等於 λ(1−λ)(x−y)²，對所有實數 x、y、λ 成立。引理在 S03.1（一次與常數部分抵消）與 S03.2（二次部分化成完全平方）中證明；N1 核對整個恆等式。",
          "en": "The leading coefficient of f is 1, so Lemma L_gap1 (b=−6, c=10) applies: the Jensen gap of f equals λ(1−λ)(x−y)² for all reals x, y, λ. The lemma is proved in S03.1 (the linear and constant parts cancel) and S03.2 (the quadratic part becomes a perfect square); N1 checks the whole identity."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_gap1"
          ],
          "explanation": {
            "zh-Hant": "多項式恆等式，兩邊相同，雙向成立。",
            "en": "A polynomial identity; both sides are the same, valid both ways."
          }
        },
        "leaf_basis": "distributive + power_laws + add_inverse"
      },
      {
        "id": "S03.1",
        "uses": [
          "distributive"
        ],
        "before_tex": "\\lambda(bx+c)+(1-\\lambda)(by+c)-\\big(b(\\lambda x+(1-\\lambda)y)+c\\big)",
        "after_tex": "0",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "一次部分 λbx+(1−λ)by−b(λx+(1−λ)y)=0；常數部分 λc+(1−λ)c−c=0。兩者都由分配律得到，對任何 b、c 成立。",
          "en": "The linear part λbx+(1−λ)by−b(λx+(1−λ)y)=0 and the constant part λc+(1−λ)c−c=0, both by distributivity, for any b, c."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "多項式恆等式，雙向成立。",
            "en": "A polynomial identity, valid both ways."
          }
        },
        "leaf_basis": "distributive + add_inverse + mul_one"
      },
      {
        "id": "S03.2",
        "uses": [
          "S03.1",
          "distributive",
          "power_laws"
        ],
        "before_tex": "\\lambda x^2+(1-\\lambda)y^2-(\\lambda x+(1-\\lambda)y)^2",
        "after_tex": "\\lambda(1-\\lambda)(x-y)^2",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "展開 (λx+(1−λ)y)²=λ²x²+2λ(1−λ)xy+(1−λ)²y²。x² 的係數 λ−λ²=λ(1−λ)，y² 的係數 (1−λ)−(1−λ)²=λ(1−λ)，xy 的係數 −2λ(1−λ)，合起來是 λ(1−λ)(x−y)²。加上 S03.1 的 0，證得 L_gap1。",
          "en": "Expand (λx+(1−λ)y)²=λ²x²+2λ(1−λ)xy+(1−λ)²y². The coefficient of x² is λ−λ²=λ(1−λ), that of y² is (1−λ)−(1−λ)²=λ(1−λ), and that of xy is −2λ(1−λ); together this is λ(1−λ)(x−y)². Adding the 0 of S03.1 proves L_gap1."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "power_laws"
          ],
          "explanation": {
            "zh-Hant": "展開右邊即回到左邊；不使用 L_gap1 本身。",
            "en": "Expanding the right side returns the left; L_gap1 itself is not used."
          }
        },
        "leaf_basis": "distributive + power_laws + mul_comm"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "S03",
          "mono_nonneg",
          "square_nonneg",
          "mko_cvxf"
        ],
        "before_tex": "G_f=\\lambda(1-\\lambda)(x-y)^2,\\ \\lambda\\in[0,1]",
        "after_tex": "G_f\\ge 0;\\ f\\ \\text{convex on}\\ X",
        "relation": "implies",
        "rule": "mko_cvxf",
        "conditions": [
          "\\lambda\\in[0,1]"
        ],
        "explanation": {
          "zh-Hant": "λ≥0、1−λ≥0 給出 λ(1−λ)≥0；(x−y)²≥0；非負數相乘仍非負，所以 G_f≥0，即 f(λx+(1−λ)y)≤λf(x)+(1−λ)f(y)。x、y∈X、λ 任意，X 已在 S02 證為凸，依凸函數定義 f 在 X 上凸。",
          "en": "λ≥0 and 1−λ≥0 give λ(1−λ)≥0; (x−y)²≥0; a product of nonnegative numbers is nonnegative, so G_f≥0, that is f(λx+(1−λ)y)≤λf(x)+(1−λ)f(y). With x, y∈X and λ arbitrary, and X convex by S02, the definition gives that f is convex on X."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S03",
            "mono_nonneg",
            "square_nonneg",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要 λ∈[0,1] 與首項係數 1>0。這是第二個義務；不宣稱 f 有最小值點。",
            "en": "It needs λ∈[0,1] and the leading coefficient 1>0. This is the second obligation; it does not claim that f has a minimizer."
          }
        },
        "leaf_basis": "mono_nonneg + square_nonneg + mul_zero + forall_intro"
      },
      {
        "id": "S05",
        "uses": [
          "S01",
          "S02",
          "S04"
        ],
        "before_tex": "X\\ \\text{convex},\\ f\\ \\text{convex on}\\ X",
        "after_tex": "P\\ \\text{qualifies as a convex problem}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個義務都成立，依 S01，P 具凸問題資格。這只是資格判定：它沒有說哪一點是最優解，也沒有說最優解存在（MKO a2）。",
          "en": "Both obligations hold, so by S01 P qualifies as a convex problem. This is only qualification: it says neither which point is optimal nor that an optimal point exists (MKO assumption a2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S02",
            "S04",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "只合取兩個已證義務。",
            "en": "It only conjoins the two proved obligations."
          }
        },
        "leaf_basis": "and_intro_elim"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "G2",
          "set_builder"
        ],
        "before_tex": "x_1=1",
        "after_tex": "0\\le 1\\le 5,\\ x_1\\in X",
        "relation": "implies",
        "rule": "set_builder",
        "conditions": [],
        "explanation": {
          "zh-Hant": "可行性就是屬於 X=[0,5]={x ∣ 0≤x≤5}。0≤1 且 1≤5，所以 x₁ 可行。",
          "en": "Feasibility means membership in X=[0,5]={x ∣ 0≤x≤5}. 0≤1 and 1≤5, so x₁ is feasible."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "set_builder",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "單點可行性只說 x₁∈X，不說它的目標值高低。",
            "en": "Single-point feasibility says only that x₁∈X, not anything about how low its objective value is."
          }
        },
        "leaf_basis": "set_builder + order_basics"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "G2",
          "set_builder"
        ],
        "before_tex": "x_2=4,\\ x_3=-1",
        "after_tex": "0\\le 4\\le 5\\ (x_2\\in X);\\ -1<0\\ (x_3\\notin X)",
        "relation": "implies",
        "rule": "set_builder",
        "conditions": [],
        "explanation": {
          "zh-Hant": "0≤4≤5，所以 x₂ 可行。−1<0，違反下界，所以 x₃ 不可行；不論 f(−1)=17 是多少，x₃ 都不是 P 的候選，不能參加比較。",
          "en": "0≤4≤5, so x₂ is feasible. −1<0 violates the lower bound, so x₃ is infeasible; whatever f(−1)=17 is, x₃ is not a candidate for P and cannot enter the comparison."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "set_builder",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "兩個單點判定；x₃ 只要違反一個邊界條件就不可行。",
            "en": "Two single-point decisions; violating one bound is enough for x₃ to be infeasible."
          }
        },
        "leaf_basis": "set_builder + order_basics"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "S06"
        ],
        "before_tex": "f(1)",
        "after_tex": "f(1)=1-6+10=5",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "單點求值：f(1)=1²−6·1+10=1−6+10=5。",
          "en": "Single-point evaluation: f(1)=1²−6·1+10=1−6+10=5."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確整數計算，兩邊是同一個數。",
            "en": "Exact integer arithmetic; both sides are the same number."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "G1",
          "S07"
        ],
        "before_tex": "f(4)",
        "after_tex": "f(4)=16-24+10=2",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "單點求值：f(4)=4²−6·4+10=16−24+10=2。",
          "en": "Single-point evaluation: f(4)=4²−6·4+10=16−24+10=2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確整數計算，兩邊是同一個數。",
            "en": "Exact integer arithmetic; both sides are the same number."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "S06",
          "S07",
          "S08",
          "S09"
        ],
        "before_tex": "f(x_1)=5,\\ f(x_2)=2",
        "after_tex": "f(x_2)<f(x_1)\\ \\text{(smaller objective value at }x_2\\text{)}",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個可行候選的比較：2<5，所以在 {x₁,x₂} 這兩點之間，x₂ 的目標值較低。這個結論只關於這兩點（MKO para2），不涉及 X 中的其他點。",
          "en": "Comparison of the two feasible candidates: 2<5, so between the two points {x₁,x₂}, x₂ has the lower objective value. This conclusion concerns these two points only (MKO para2) and says nothing about the other points of X."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S07",
            "S08",
            "S09",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要兩點都可行與兩個精確值。不宣稱 x₂ 是 P 的最優解；S11 顯示它不是。",
            "en": "It needs both points feasible and the two exact values. It does not claim that x₂ is optimal for P; S11 shows it is not."
          }
        },
        "leaf_basis": "order_basics + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S10",
          "minimizer_def",
          "quantifier_negation"
        ],
        "before_tex": "3\\in X,\\ f(3)=9-18+10=1",
        "after_tex": "f(3)=1<2=f(x_2);\\ x_2\\ \\text{is not a minimizer of}\\ P",
        "relation": "implies",
        "rule": "minimizer_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "3 可行（0≤3≤5），單點求值 f(3)=9−18+10=1<2=f(4)。依最小值點的定義，x₂ 是最小值點需要 f(4)≤f(x) 對每個可行 x 成立；x=3 違反這一點，所以 x₂ 不是 P 的最優解。兩候選比較的勝者並不是整個問題的最優解。本稿也不宣稱 3 是最小值點：那需要本切片以外的最優性證書。",
          "en": "3 is feasible (0≤3≤5), and single-point evaluation gives f(3)=9−18+10=1<2=f(4). By the definition of a minimizer, x₂ would need f(4)≤f(x) for every feasible x; x=3 violates this, so x₂ is not optimal for P. The winner of a two-candidate comparison is not the optimum of the whole problem. Nor does this guide claim that 3 is a minimizer: that would need an optimality certificate outside this slice."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "minimizer_def",
            "quantifier_negation",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "否證只需一個可行且更低的點；正面宣稱任何點最優則需要對所有可行點的論證，本稿不提供。",
            "en": "A refutation needs only one feasible point with a lower value; a positive claim that some point is optimal would need an argument over all feasible points, which this guide does not provide."
          }
        },
        "leaf_basis": "minimizer_def + quantifier_negation + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "G3",
          "L_halfopen",
          "mko_cvxset"
        ],
        "before_tex": "s,t\\in(0,1],\\ \\lambda\\in[0,1]",
        "after_tex": "\\lambda s+(1-\\lambda)t\\in(0,1];\\ X'\\ \\text{convex}",
        "relation": "implies",
        "rule": "L_halfopen",
        "conditions": [
          "\\lambda\\in[0,1]"
        ],
        "explanation": {
          "zh-Hant": "轉向 P′。用引理 L_halfopen：(0,1] 中兩點的凸組合仍在 (0,1]，所以 X′ 是凸集合（半開區間也是明示實區間）。引理在 S12.1（嚴格正）與 S12.2（不超過 1）中證明。",
          "en": "Turn to P′. By Lemma L_halfopen, a convex combination of two points of (0,1] stays in (0,1], so X′ is convex (a half-open interval is also an explicit real interval). The lemma is proved in S12.1 (strictly positive) and S12.2 (at most 1)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_halfopen",
            "mko_cvxset",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要兩點都在 (0,1] 與 λ∈[0,1]；不宣稱反向。",
            "en": "It needs both points in (0,1] and λ∈[0,1]; no converse is claimed."
          }
        },
        "leaf_basis": "L_halfopen + forall_intro"
      },
      {
        "id": "S12.1",
        "uses": [
          "mono_nonneg",
          "order_basics",
          "case_analysis"
        ],
        "before_tex": "0<s,\\ 0<t,\\ 0\\le\\lambda\\le 1",
        "after_tex": "\\lambda s+(1-\\lambda)t>0",
        "relation": "case_split",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "分兩種情況。λ>0：λs>0（兩個正數相乘），而 (1−λ)t≥0（0≤t 乘以 1−λ≥0），正數加非負數仍為正。λ=0：和等於 t>0。λ∈[0,1] 只有這兩種情況，所以 λs+(1−λ)t>0。",
          "en": "Two cases. λ>0: λs>0 (a product of two positive numbers), and (1−λ)t≥0 (multiply 0≤t by 1−λ≥0); a positive number plus a nonnegative one is positive. λ=0: the sum equals t>0. For λ∈[0,1] these are the only two cases, so λs+(1−λ)t>0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mono_nonneg",
            "order_basics",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "兩個情況涵蓋 λ≥0 的全部可能；證明 L_halfopen 的下界，不使用 L_halfopen 本身。",
            "en": "The two cases cover every λ≥0; it proves the lower bound of L_halfopen without using L_halfopen itself."
          }
        },
        "leaf_basis": "case_analysis + order_basics + mono_nonneg"
      },
      {
        "id": "S12.2",
        "uses": [
          "S12.1",
          "mono_nonneg",
          "ineq_add",
          "distributive"
        ],
        "before_tex": "s\\le 1,\\ t\\le 1",
        "after_tex": "\\lambda s+(1-\\lambda)t\\le\\lambda+(1-\\lambda)=1",
        "relation": "implies",
        "rule": "mono_nonneg",
        "conditions": [
          "\\lambda\\ge 0",
          "1-\\lambda\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "s≤1 乘以 λ、t≤1 乘以 1−λ，相加得 λs+(1−λ)t≤λ+(1−λ)=1。S12.1 與 S12.2 證得 L_halfopen。",
          "en": "Multiply s≤1 by λ and t≤1 by 1−λ and add: λs+(1−λ)t≤λ+(1−λ)=1. S12.1 and S12.2 prove L_halfopen."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mono_nonneg",
            "ineq_add",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "與 L_interval 的上界相同的論證；不使用 L_halfopen 本身。",
            "en": "The same argument as the upper bound of L_interval; L_halfopen itself is not used."
          }
        },
        "leaf_basis": "mono_nonneg + ineq_add + distributive"
      },
      {
        "id": "S13",
        "uses": [
          "S12",
          "L_gap1",
          "mono_nonneg",
          "square_nonneg",
          "mko_cvxopt"
        ],
        "before_tex": "f_0(x)=x^2\\ \\text{on}\\ X'=(0,1]",
        "after_tex": "G_{f_0}=\\lambda(1-\\lambda)(x-y)^2\\ge 0;\\ P'\\ \\text{qualifies as convex}",
        "relation": "implies",
        "rule": "mko_cvxopt",
        "conditions": [
          "\\lambda\\in[0,1]"
        ],
        "explanation": {
          "zh-Hant": "f₀ 是 L_gap1 中 b=c=0 的情形，Jensen gap 為 λ(1−λ)(x−y)²≥0（同 S04 的符號論證），所以 f₀ 在凸集合 X′ 上是凸函數。兩個義務都成立，P′ 具凸問題資格。",
          "en": "f₀ is the case b=c=0 of L_gap1, with Jensen gap λ(1−λ)(x−y)²≥0 (by the same sign argument as S04), so f₀ is convex on the convex set X′. Both obligations hold, so P′ qualifies as a convex problem."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12",
            "L_gap1",
            "mono_nonneg",
            "square_nonneg"
          ],
          "explanation": {
            "zh-Hant": "與 S02 至 S05 相同的兩個義務；資格不說最小值點存在。",
            "en": "The same two obligations as S02 to S05; qualification does not say that a minimizer exists."
          }
        },
        "leaf_basis": "L_gap1 + mono_nonneg + square_nonneg + and_intro_elim"
      },
      {
        "id": "S14",
        "uses": [
          "G3",
          "S13",
          "minimizer_def"
        ],
        "before_tex": "x\\in(0,1]",
        "after_tex": "\\tfrac{x}{2}\\in(0,1],\\ f_0(\\tfrac{x}{2})=\\tfrac{x^2}{4}<x^2=f_0(x)",
        "relation": "implies",
        "rule": "minimizer_def",
        "conditions": [
          "x>0"
        ],
        "explanation": {
          "zh-Hant": "任取可行的 x∈(0,1]。x/2>0，且 x/2<x≤1，所以 x/2 也可行。x²−(x/2)²=3x²/4，而 x≠0 給出 x²>0，因此 3x²/4>0，即 f₀(x/2)<f₀(x)。依最小值點的定義，x 不是最小值點。",
          "en": "Take any feasible x∈(0,1]. x/2>0 and x/2<x≤1, so x/2 is feasible too. x²−(x/2)²=3x²/4, and x≠0 gives x²>0, hence 3x²/4>0, that is f₀(x/2)<f₀(x). By the definition of a minimizer, x is not a minimizer."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "minimizer_def",
            "square_nonneg",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要 x>0（使 x/2 在 X′ 中且 x²>0）。這對每個可行 x 都成立，下一步做全稱推廣。",
            "en": "It needs x>0 (so that x/2 lies in X′ and x²>0). It holds for every feasible x, which the next step generalizes."
          }
        },
        "leaf_basis": "minimizer_def + square_nonneg + order_basics + frac_rules"
      },
      {
        "id": "S15",
        "uses": [
          "S13",
          "S14",
          "forall_intro",
          "quantifier_negation"
        ],
        "before_tex": "\\forall x\\in X':\\ x\\ \\text{is not a minimizer}",
        "after_tex": "P'\\ \\text{is convex but has no minimizer}",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S14 中的 x 任意，依 ∀ 引入，X′ 中沒有任何一點是最小值點，即 P′ 沒有最小值點（f₀ 在 X′ 上的下確界 0 取不到，因為 0∉X′）。而 P′ 已具凸問題資格（S13）。所以「凸問題就有（唯一）全域解」是錯的：凸資格不保證解存在。",
          "en": "The x in S14 was arbitrary, so by ∀-introduction no point of X′ is a minimizer, i.e. P′ has no minimizer (the infimum 0 of f₀ on X′ is not attained, since 0∉X′). Yet P′ qualifies as convex (S13). So 'a convex problem has a (unique) global solution' is false: qualification does not guarantee that a solution exists."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13",
            "S14",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要 S14 對每個可行點成立。括號中關於下確界的說明只是補充，結論不依賴它。",
            "en": "It needs S14 for every feasible point. The remark on the infimum in parentheses is a side note; the conclusion does not depend on it."
          }
        },
        "leaf_basis": "forall_intro + quantifier_negation"
      }
    ],
    "result": {
      "zh-Hant": "P（在 [0,5] 上最小化 x²−6x+10）具凸問題資格：[0,5] 是凸集合，f 的 Jensen gap 為 λ(1−λ)(x−y)²≥0。x₁=1、x₂=4 可行，x₃=−1 不可行；f(1)=5、f(4)=2，所以兩候選中 x₂ 較低。但 f(3)=1<2，所以 x₂ 不是 P 的最優解。P′（在 (0,1] 上最小化 x²）也具凸問題資格，卻沒有最小值點。",
      "en": "P (minimize x²−6x+10 over [0,5]) qualifies as a convex problem: [0,5] is convex and the Jensen gap of f is λ(1−λ)(x−y)²≥0. x₁=1 and x₂=4 are feasible, x₃=−1 is not; f(1)=5 and f(4)=2, so x₂ has the lower objective value of the two candidates. But f(3)=1<2, so x₂ is not optimal for P. P′ (minimize x² over (0,1]) also qualifies as convex, yet it has no minimizer."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "P 是凸問題，且 f(4)<f(1)，所以 x₂=4 是 P 的最優解",
          "en": "P is convex and f(4)<f(1), so x₂=4 is the optimal solution of P"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "3 可行且 f(3)=1<2=f(4)。",
          "en": "3 is feasible and f(3)=1<2=f(4)."
        }
      },
      {
        "claim": {
          "zh-Hant": "每個凸最佳化問題都有唯一的全域最優解",
          "en": "Every convex optimization problem has a unique global optimal solution"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "P′：在 (0,1] 上最小化 x² 是凸問題，但沒有最小值點。",
          "en": "P′: minimizing x² over (0,1] is a convex problem without a minimizer."
        }
      },
      {
        "claim": {
          "zh-Hant": "f 是凸函數，所以在 [0,1]∪[4,5] 上最小化 f 也具本條目的凸問題資格",
          "en": "f is convex, so minimizing f over [0,1]∪[4,5] also qualifies as a convex problem here"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "可行域不凸：1、4 可行，但 5/2 不可行。",
          "en": "The feasible domain is not convex: 1 and 4 are feasible but 5/2 is not."
        }
      },
      {
        "claim": {
          "zh-Hant": "x₁=1 可行且 f(1)=5，所以 P 的最優值是 5",
          "en": "x₁=1 is feasible with f(1)=5, so the optimal value of P is 5"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "單點求值不是最優性；f(4)=2<5。",
          "en": "Single-point evaluation is not optimality; f(4)=2<5."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "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": "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": "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": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "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_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "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": "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": "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": "set_builder",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。",
          "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x)."
        }
      },
      {
        "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": "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": "mko_cvxopt",
        "kind": "cited",
        "source": {
          "mko_id": "mko-convex-optimization-problem",
          "sha256": "0484bac345126653c2ee68f9057b0c7bf124c996ee7e7cfd461d32ad4e78747b",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "最小化問題只有在目標函數凸且可行域凸時，才具本條目的凸問題資格。本輪只核一元區間與二次目標的資格、單點可行性、單點求值及兩個可行候選的局部比較；不由資格、單點或兩點比較宣稱已找到全域解。",
          "en": "A minimization problem qualifies here as convex only when its objective function and feasible domain are convex. This slice checks qualification for a univariate interval and quadratic objective, single-point feasibility, single-point evaluation and a comparison between two feasible candidates; qualification or one- or two-point checks do not establish a global solution."
        }
      },
      {
        "id": "mko_cvxf",
        "kind": "cited",
        "source": {
          "mko_id": "mko-convex-function",
          "sha256": "d5af460bbe41c8e967949f9398bde368d76471cd2b4a24e4f49f14ad90a05fed",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在凸定義域 C 上，f 為凸若對所有 x,y∈C 及 λ∈[0,1]，f(λx+(1−λ)y)≤λf(x)+(1−λ)f(y)。",
          "en": "On a convex domain C, f is convex if f(λx+(1−λ)y)≤λf(x)+(1−λ)f(y) for every x,y∈C and every λ∈[0,1]."
        }
      },
      {
        "id": "mko_cvxset",
        "kind": "cited",
        "source": {
          "mko_id": "mko-convex-set",
          "sha256": "0d52be71dca290f856018d176bbb26dcd877c887031f690bd1d656db017ce02d",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "實向量空間的子集 C 為凸，若對所有 x,y∈C 及所有 λ∈[0,1]，λx+(1−λ)y 仍屬於 C。",
          "en": "A subset C of a real vector space is convex if λx+(1−λ)y belongs to C for every x,y∈C and every λ∈[0,1]."
        }
      },
      {
        "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（w>0 時由實數的序，w=0 時兩邊都是 0）。",
          "en": "Multiplying by a nonnegative number keeps the inequality: if u≤v and w≥0, then wu≤wv (for w>0 by the order of the reals; for w=0 both sides are 0)."
        }
      },
      {
        "id": "ineq_add",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "不等式相加：若 u₁≤v₁ 且 u₂≤v₂，則 u₁+u₂≤v₁+v₂；若其中一個是嚴格不等，和也是嚴格不等。",
          "en": "Adding inequalities: if u₁≤v₁ and u₂≤v₂, then u₁+u₂≤v₁+v₂; if one of them is strict, so is the sum."
        }
      },
      {
        "id": "minimizer_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "最小值點：x*∈X 是 f 在 X 上的（全域）最小值點，當且僅當 x* 可行且對每個 x∈X 都有 f(x*)≤f(x)。因此只要找到可行的 x 使 f(x)<f(x*)，x* 就不是最小值點。",
          "en": "Minimizer: x*∈X is a (global) minimizer of f over X exactly when x* is feasible and f(x*)≤f(x) for every x∈X. Hence finding a feasible x with f(x)<f(x*) shows that x* is not a minimizer."
        }
      },
      {
        "id": "L_interval",
        "kind": "lemma",
        "proved_in": [
          "S02.1",
          "S02.2"
        ],
        "statement": {
          "zh-Hant": "若 α≤s≤β、α≤t≤β 且 λ∈[0,1]，則 α≤λs+(1−λ)t≤β；因此閉區間 [α,β] 是凸集合。",
          "en": "If α≤s≤β, α≤t≤β and λ∈[0,1], then α≤λs+(1−λ)t≤β; hence the closed interval [α,β] is convex."
        }
      },
      {
        "id": "L_gap1",
        "kind": "lemma",
        "proved_in": [
          "S03.1",
          "S03.2"
        ],
        "statement": {
          "zh-Hant": "對 q(x)=x²+bx+c 與任意實數 x、y、λ：λq(x)+(1−λ)q(y)−q(λx+(1−λ)y)=λ(1−λ)(x−y)²。",
          "en": "For q(x)=x²+bx+c and any reals x, y, λ: λq(x)+(1−λ)q(y)−q(λx+(1−λ)y)=λ(1−λ)(x−y)²."
        }
      },
      {
        "id": "L_halfopen",
        "kind": "lemma",
        "proved_in": [
          "S12.1",
          "S12.2"
        ],
        "statement": {
          "zh-Hant": "若 0<s≤1、0<t≤1 且 λ∈[0,1]，則 0<λs+(1−λ)t≤1；因此 (0,1] 是凸集合。",
          "en": "If 0<s≤1, 0<t≤1 and λ∈[0,1], then 0<λs+(1−λ)t≤1; hence (0,1] is convex."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "vars": [
          "l",
          "x",
          "y"
        ],
        "degree": 2,
        "lhs": "l*(x**2-6*x+10)+(1-l)*(y**2-6*y+10)-((l*x+(1-l)*y)**2-6*(l*x+(1-l)*y)+10)",
        "rhs": "l*(1-l)*(x-y)**2"
      },
      {
        "id": "N2",
        "step": "S03.1",
        "vars": [
          "b",
          "c",
          "l",
          "x",
          "y"
        ],
        "degree": 2,
        "lhs": "l*(b*x+c)+(1-l)*(b*y+c)-(b*(l*x+(1-l)*y)+c)",
        "rhs": "0"
      },
      {
        "id": "N3",
        "step": "S03.2",
        "vars": [
          "l",
          "x",
          "y"
        ],
        "degree": 2,
        "lhs": "l*x**2+(1-l)*y**2-(l*x+(1-l)*y)**2",
        "rhs": "l*(1-l)*(x-y)**2"
      },
      {
        "id": "N4",
        "step": "S02.2",
        "vars": [
          "l"
        ],
        "degree": 1,
        "lhs": "5*l+5*(1-l)",
        "rhs": "5"
      },
      {
        "id": "N5",
        "step": "S08",
        "lhs": "1**2-6*1+10",
        "rhs": "5"
      },
      {
        "id": "N6",
        "step": "S09",
        "lhs": "4**2-6*4+10",
        "rhs": "2"
      },
      {
        "id": "N7",
        "step": "S10",
        "lhs": "4**2-6*4+10",
        "relation": "<",
        "rhs": "1**2-6*1+10"
      },
      {
        "id": "N8",
        "step": "S11",
        "lhs": "3**2-6*3+10",
        "rhs": "1"
      },
      {
        "id": "N9",
        "step": "S11",
        "lhs": "3**2-6*3+10",
        "relation": "<",
        "rhs": "4**2-6*4+10"
      },
      {
        "id": "N10",
        "step": "S07",
        "lhs": "(-1)**2-6*(-1)+10",
        "rhs": "17"
      },
      {
        "id": "N11",
        "step": "S14",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "x**2-(x/2)**2",
        "rhs": "3*x**2/4"
      },
      {
        "id": "N12",
        "step": "S12.2",
        "vars": [
          "l"
        ],
        "degree": 1,
        "lhs": "l*1+(1-l)*1",
        "rhs": "1"
      },
      {
        "id": "N13",
        "lhs": "(1/2)*1+(1/2)*4",
        "rhs": "5/2"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "停在 MKO 的切片內：一元明示實區間作可行域、整係數二次目標；只做資格核對、單點可行性、單點求值與兩個可行候選的比較。為否證「某點最優」可再求一個可行點的值；不給任何點的全域最優證書，不引入 KKT、對偶、通用求解器或收斂結果；不改動 canonical MKO 或英文 edition。",
      "en": "Stays inside the MKO's slice: an explicit univariate real interval as feasible domain and an integer-coefficient quadratic objective; only qualification, single-point feasibility, single-point evaluation and a comparison of two feasible candidates. To refute 'this point is optimal' one more feasible point may be evaluated; no global optimality certificate is given for any point, and KKT, duality, general solvers and convergence results are not introduced; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "P:\\ \\min_{x\\in X} f(x)",
        "問題 P：在 X 上最小化 f",
        "problem P: minimize f over X"
      ],
      [
        "X=[0,5],\\ X'=(0,1]",
        "可行域",
        "the feasible domains"
      ],
      [
        "f(x)=x^2-6x+10",
        "P 的目標函數",
        "the objective of P"
      ],
      [
        "f_0(x)=x^2",
        "P′ 的目標函數",
        "the objective of P′"
      ],
      [
        "G_q(x,y,\\lambda)",
        "Jensen gap：λq(x)+(1−λ)q(y)−q(λx+(1−λ)y)",
        "Jensen gap: λq(x)+(1−λ)q(y)−q(λx+(1−λ)y)"
      ],
      [
        "x^\\ast",
        "最小值點（全域最優解）",
        "a minimizer (global optimal solution)"
      ]
    ],
    "md_error_controls": [
      {
        "title": "凸資格不保證有解，更不保證唯一 / Qualification guarantees neither a solution nor uniqueness",
        "zh": "MKO 指出的誤解是看到「凸」字樣就宣稱有唯一全域解。P′ 是凸問題，但對每個可行 x，x/2 都更低，所以沒有最小值點（S14、S15）。存在性與唯一性需要另外的條件與證書。",
        "en": "The mistake named by the MKO is to claim a unique global solution as soon as the word 'convex' appears. P′ is a convex problem, but for every feasible x the point x/2 has a lower objective value, so there is no minimizer (S14, S15). Existence and uniqueness need further conditions and certificates."
      },
      {
        "title": "兩候選中目標值較低者不一定是最優解 / The candidate with the lower objective value need not be optimal",
        "zh": "MKO 的另一個誤解是把兩個候選中較低者叫做整個問題的最優解。f(4)=2<5=f(1) 只說 4 比 1 好（S10）；可行點 3 的值 f(3)=1 更低（S11）。兩點比較只說兩點之間的高低（MKO para2）。",
        "en": "Another misconception identified in the MKO is to call the candidate with the lower objective value the optimum of the whole problem. f(4)=2<5=f(1) says only that 4 beats 1 (S10); the feasible point 3 has the even lower value f(3)=1 (S11). A two-point comparison says only which of the two is lower (MKO para2)."
      },
      {
        "title": "兩個義務分別核對，可行性先於比較 / Check both obligations separately; feasibility comes before comparison",
        "zh": "目標凸不代表問題凸：若可行域換成 [0,1]∪[4,5]，1、4 可行但 (1/2)·1+(1/2)·4=5/2 不可行，可行域不凸，問題不具資格。同樣，不可行的 x₃=−1 即使算得 f(−1)=17，也不能參加候選比較（S07）。",
        "en": "A convex objective does not make the problem convex: if the feasible domain were [0,1]∪[4,5], then 1 and 4 are feasible but (1/2)·1+(1/2)·4=5/2 is not, so the domain is not convex and the problem does not qualify. Likewise the infeasible x₃=−1 cannot enter the comparison even though f(−1)=17 can be computed (S07)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿不對 P 的任何點給出全域最優證書（包括 x=3）；最優性證書、KKT 與對偶屬於 MKO 的 formal next。",
        "en": "This guide gives no global optimality certificate for any point of P (including x=3); optimality certificates, KKT and duality belong to the MKO's formal next."
      },
      {
        "zh-Hant": "mono_nonneg、ineq_add 與 minimizer_def 是本稿的前提，不在本稿中推出。",
        "en": "mono_nonneg, ineq_add and minimizer_def are premises of this guide and are not derived here."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "41acd93a76eeb1cda8a7b0d56acf2deac4623d5ac9b30b940dad539544bc41e2",
    "original_source_sha256": "41acd93a76eeb1cda8a7b0d56acf2deac4623d5ac9b30b940dad539544bc41e2",
    "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."
  }
}