{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-finite-convex-combination-01",
  "mko_id": "mko-finite-convex-combination",
  "title": {
    "zh-Hant": "有限凸組合：以精確權重核對見證，並區分負權重、近似權重與唯一性",
    "en": "Finite convex combination: checking a witness with exact weights, and separating negative weights, approximate weights and uniqueness"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-finite-convex-combination-01",
    "source": {
      "mko_id": "mko-finite-convex-combination",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "84206c94ef6c80e08ea032783fce4ef3d1e4b15b55088556463d360b07c78d80",
      "english_sha256": "cea2a4c5c0d854cbc55dac8f42fc0521d0afbfa6d70790f79fe1f6b1326a7214"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "vector_space_axioms",
      "finite_sum_rules",
      "order_basics",
      "cartesian_product_def",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在 ℝ² 中令 v_1=(0,0)、v_2=(6,0)、v_3=(0,6)。(1) 核對 λ=(1/6,1/3,1/2) 是凸權重並重建 w=(2,3)；(2) 說明零權重允許、兩點凸組合是線段點；(3) 證明 u=(4,4) 不是 v_1、v_2、v_3 的凸組合（唯一的仿射權重含 −1/3），並以線性泛函上界給第二個證明；(4) 說明 0.333 三個權重不合格；(5) 加入 v_4=(3,3) 後，同一點有兩組不同的凸權重。",
      "en": "In ℝ² let v_1=(0,0), v_2=(6,0), v_3=(0,6). (1) Check that λ=(1/6,1/3,1/2) are convex weights and reconstruct w=(2,3); (2) show that zero weights are allowed and that two-point convex combinations are segment points; (3) prove that u=(4,4) is not a convex combination of v_1, v_2, v_3 (its only affine weights include −1/3), with a second proof by a linear-functional bound; (4) show that three weights of 0.333 do not qualify; (5) after adding v_4=(3,3), exhibit two different convex weight vectors for the same point."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "ℝ² 中 v_1=(0,0)、v_2=(6,0)、v_3=(0,6)；權重為精確有理數，不接受浮點近似（MKO 假設 a1）。",
          "en": "In ℝ², v_1=(0,0), v_2=(6,0), v_3=(0,6); weights are exact rationals and floating-point approximations are not accepted (MKO assumption a1)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "候選見證：w=(2,3)，λ=(1/6,1/3,1/2)。",
          "en": "Candidate witness: w=(2,3) with λ=(1/6,1/3,1/2)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "其他資料：點 (3,0) 配權重 (1/2,1/2,0)；u=(4,4)；十進位權重 (0.333,0.333,0.333)；第四點 v_4=(3,3)。",
          "en": "Further data: the point (3,0) with weights (1/2,1/2,0); u=(4,4); decimal weights (0.333,0.333,0.333); a fourth point v_4=(3,3)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "mko_convex_def"
        ],
        "before_tex": "w\\ \\text{convex combination of}\\ v_1,v_2,v_3",
        "after_tex": "\\exists\\lambda:\\ \\lambda_i\\ge0,\\ \\sum_i\\lambda_i=1,\\ \\sum_i\\lambda_iv_i=w",
        "relation": "definition",
        "rule": "mko_convex_def",
        "conditions": [
          "n=3\\ge1"
        ],
        "explanation": {
          "zh-Hant": "依 MKO 展開：凸組合需要三件事分別成立：每個權重非負、權重和精確為 1、加權和等於目標（MKO 段落 0、2）。見證就是一組滿足三件事的權重。",
          "en": "Unfold by the MKO: a convex combination needs three separate things: every weight nonnegative, the weights summing exactly to 1, and the weighted sum equal to the target (MKO paragraphs 0 and 2). A witness is a weight vector meeting all three."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_convex_def"
          ],
          "explanation": {
            "zh-Hant": "只展開定義；三個條件缺一不可，之後逐一核對。",
            "en": "Only unfolds the definition; none of the three conditions may be skipped, and they are checked one by one below."
          }
        },
        "leaf_basis": "mko_convex_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G2",
          "S01"
        ],
        "before_tex": "\\lambda=(\\tfrac16,\\tfrac13,\\tfrac12)",
        "after_tex": "\\lambda_1,\\lambda_2,\\lambda_3\\ge0",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "三個權重都是正有理數，所以非負。這是第一個條件。",
          "en": "All three weights are positive rationals, hence nonnegative. This is the first condition."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "逐一核對三個權重；非負性與總和為 1 是分開的條件。",
            "en": "Each of the three weights is checked; nonnegativity and total 1 are separate conditions."
          }
        },
        "leaf_basis": "order_basics"
      },
      {
        "id": "S03",
        "uses": [
          "G2",
          "S01"
        ],
        "before_tex": "\\tfrac16+\\tfrac13+\\tfrac12",
        "after_tex": "1",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "通分：1/6+2/6+3/6=6/6=1（N1），是精確等式。這是第二個條件。",
          "en": "With a common denominator: 1/6+2/6+3/6=6/6=1 (N1), an exact equality. This is the second condition."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "精確有理數求和，可獨立重算。",
            "en": "An exact rational sum, independently recomputable."
          }
        },
        "leaf_basis": "numeric_evaluation + frac_rules"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "G2",
          "vector_space_axioms"
        ],
        "before_tex": "\\tfrac16(0,0)+\\tfrac13(6,0)+\\tfrac12(0,6)",
        "after_tex": "(2,3)=w",
        "relation": "equivalent",
        "rule": "vector_space_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐坐標重建（S04.1、S04.2）：第一坐標 2、第二坐標 3，正是 w。這是第三個條件，也是 MKO 段落 2 所說的「以精確重建目標核對」。",
          "en": "Reconstruct coordinatewise (S04.1, S04.2): first coordinate 2, second coordinate 3, which is exactly w. This is the third condition, the exact reconstruction of the target described in MKO paragraph 2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "vector_space_axioms",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "向量相等 ⇔ 兩個坐標都相等，雙向成立。",
            "en": "Vectors are equal ⇔ both coordinates agree, so it holds both ways."
          }
        },
        "leaf_basis": "vector_space_axioms + cartesian_product_def"
      },
      {
        "id": "S04.1",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "\\tfrac16\\cdot0+\\tfrac13\\cdot6+\\tfrac12\\cdot0",
        "after_tex": "2",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第一坐標：只有 v_2 的第一坐標非零，1/3·6=2（N2）。",
          "en": "First coordinate: only v_2 has a nonzero first coordinate, and 1/3·6=2 (N2)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確求值。",
            "en": "Exact evaluation."
          }
        },
        "leaf_basis": "numeric_evaluation + mul_zero"
      },
      {
        "id": "S04.2",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "\\tfrac16\\cdot0+\\tfrac13\\cdot0+\\tfrac12\\cdot6",
        "after_tex": "3",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第二坐標：只有 v_3 的第二坐標非零，1/2·6=3（N3）。",
          "en": "Second coordinate: only v_3 has a nonzero second coordinate, and 1/2·6=3 (N3)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確求值。",
            "en": "Exact evaluation."
          }
        },
        "leaf_basis": "numeric_evaluation + mul_zero"
      },
      {
        "id": "S05",
        "uses": [
          "S02",
          "S03",
          "S04",
          "S01"
        ],
        "before_tex": "\\lambda_i\\ge0,\\ \\sum\\lambda_i=1,\\ \\sum\\lambda_iv_i=w",
        "after_tex": "w=(2,3)\\ \\text{is a convex combination of}\\ v_1,v_2,v_3",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "三個條件都通過，λ=(1/6,1/3,1/2) 是見證，依 ∃引入得 w 是凸組合。見證只證明這個點是凸組合，不說明權重唯一（MKO 假設 a2）。",
          "en": "All three conditions pass, so λ=(1/6,1/3,1/2) is a witness, and ∃-introduction makes w a convex combination. The witness proves only that this point is a convex combination, not that the weights are unique (MKO assumption a2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S03",
            "S04",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "存在性只需一組見證；唯一性另議（S13）。",
            "en": "Existence needs one witness; uniqueness is a separate matter (S13)."
          }
        },
        "leaf_basis": "exists_intro + and_intro_elim"
      },
      {
        "id": "S06",
        "uses": [
          "G3",
          "S01"
        ],
        "before_tex": "\\tfrac12v_1+\\tfrac12v_2+0\\cdot v_3",
        "after_tex": "(3,0)\\ \\text{with weights}\\ (\\tfrac12,\\tfrac12,0)\\ \\text{convex}",
        "relation": "implies",
        "rule": "mko_convex_def",
        "conditions": [
          "\\lambda_3=0"
        ],
        "explanation": {
          "zh-Hant": "權重 1/2、1/2、0 都非負（0 也是非負），和為 1（N5）；重建：第一坐標 1/2·6=3（N4），第二坐標 0。所以 (3,0) 是凸組合；零權重是允許的，它只是讓 v_3 不起作用。",
          "en": "The weights 1/2, 1/2, 0 are all nonnegative (0 is nonnegative too) and sum to 1 (N5); reconstruction: first coordinate 1/2·6=3 (N4), second coordinate 0. So (3,0) is a convex combination; a zero weight is allowed and simply makes v_3 play no role."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_convex_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "定義要求 λ_i≥0，不是 λ_i>0；三個條件都已核。",
            "en": "The definition asks for λ_i≥0, not λ_i>0; all three conditions are checked."
          }
        },
        "leaf_basis": "mko_convex_def + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "S01",
          "vector_space_axioms"
        ],
        "before_tex": "\\lambda v_2+(1-\\lambda)v_3,\\ \\lambda\\in[0,1]",
        "after_tex": "(6\\lambda,\\ 6-6\\lambda);\\ \\lambda=\\tfrac13:\\ (2,4)",
        "relation": "implies",
        "rule": "vector_space_axioms",
        "conditions": [
          "0\\le\\lambda\\le1"
        ],
        "explanation": {
          "zh-Hant": "兩點凸組合用 λ 與 1−λ（MKO 段落 1）：兩者都非負若且唯若 0≤λ≤1，和自動為 1。坐標為 (6λ, 6−6λ)（N6、N7 為恆等式），λ 從 0 到 1 走過從 v_3 到 v_2 的線段。λ=1/3 給 (2,4)（N8、N9）。",
          "en": "A two-point convex combination uses λ and 1−λ (MKO paragraph 1): both are nonnegative exactly when 0≤λ≤1, and they sum to 1 automatically. The coordinates are (6λ, 6−6λ) (N6, N7 are identities), and as λ goes from 0 to 1 the point runs along the segment from v_3 to v_2. λ=1/3 gives (2,4) (N8, N9)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "vector_space_axioms",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "λ 在 [0,1] 外時 λ 或 1−λ 為負，得到的是直線上的仿射組合而不是線段上的凸組合。",
            "en": "For λ outside [0,1], λ or 1−λ is negative, giving an affine combination on the line rather than a convex combination on the segment."
          }
        },
        "leaf_basis": "vector_space_axioms + order_basics + distributive"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "G3",
          "p_affine_comb"
        ],
        "before_tex": "\\sum_i\\lambda_i=1,\\ \\sum_i\\lambda_iv_i=(4,4)",
        "after_tex": "\\lambda=(-\\tfrac13,\\ \\tfrac23,\\ \\tfrac23)\\ \\text{(the only solution)}",
        "relation": "equivalent",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "設任意權重和為 1 且加權和為 u=(4,4)。因 v_1=(0,0)，第一坐標給 6λ_2=4、第二坐標給 6λ_3=4（S08.1），所以 λ_2=λ_3=2/3；和為 1 給 λ_1=1−4/3=−1/3（S08.2）。每一步都是等價變形，所以這是唯一的仿射權重。",
          "en": "Take any weights summing to 1 whose weighted sum is u=(4,4). Since v_1=(0,0), the first coordinate gives 6λ_2=4 and the second gives 6λ_3=4 (S08.1), so λ_2=λ_3=2/3; the total 1 gives λ_1=1−4/3=−1/3 (S08.2). Every step is an equivalence, so these are the only affine weights."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "反向：代回 (−1/3,2/3,2/3) 得和 1、加權和 (4,4)；除以 6 需 6≠0。",
            "en": "Backward: substituting (−1/3,2/3,2/3) gives total 1 and weighted sum (4,4); dividing by 6 needs 6≠0."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero"
      },
      {
        "id": "S08.1",
        "uses": [
          "G1",
          "cartesian_product_def"
        ],
        "before_tex": "\\lambda_1(0,0)+\\lambda_2(6,0)+\\lambda_3(0,6)=(4,4)",
        "after_tex": "6\\lambda_2=4,\\ 6\\lambda_3=4\\ \\Longleftrightarrow\\ \\lambda_2=\\lambda_3=\\tfrac23",
        "relation": "equivalent",
        "rule": "cartesian_product_def",
        "conditions": [
          "6\\neq0"
        ],
        "explanation": {
          "zh-Hant": "加權和為 (6λ_2, 6λ_3)，與 (4,4) 逐坐標相等。兩邊乘 1/6 得 λ_2=λ_3=4/6=2/3（N10）。λ_1 不出現在坐標方程中，因為 v_1 是原點。",
          "en": "The weighted sum is (6λ_2, 6λ_3), equal coordinatewise to (4,4). Multiplying by 1/6 gives λ_2=λ_3=4/6=2/3 (N10). λ_1 does not appear in the coordinate equations because v_1 is the origin."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "cartesian_product_def",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "有序對相等與乘非零數都可逆。",
            "en": "Equality of ordered pairs and multiplication by a nonzero number are both reversible."
          }
        },
        "leaf_basis": "cartesian_product_def + mul_inverse_nonzero"
      },
      {
        "id": "S08.2",
        "uses": [
          "S08.1",
          "eq_congruence"
        ],
        "before_tex": "\\lambda_1+\\tfrac23+\\tfrac23=1",
        "after_tex": "\\lambda_1=-\\tfrac13",
        "relation": "equivalent",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊同減 4/3：λ_1=1−4/3=−1/3（N11）。",
          "en": "Subtract 4/3 from both sides: λ_1=1−4/3=−1/3 (N11)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "減與加互逆。",
            "en": "Subtraction and addition undo each other."
          }
        },
        "leaf_basis": "eq_congruence + add_inverse + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S08",
          "S01",
          "p_affine_comb"
        ],
        "before_tex": "\\lambda_1=-\\tfrac13<0",
        "after_tex": "u=(4,4)\\ \\text{affine but not a convex combination of}\\ v_1,v_2,v_3",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若 u 是凸組合，其凸權重也是和為 1 的權重，由 S08 必為 (−1/3,2/3,2/3)，但 −1/3<0（N12）違反非負性，矛盾。所以 u 不是凸組合；它只是仿射組合（MKO：負權重只形成一般仿射組合）。",
          "en": "If u were a convex combination, its convex weights would be weights summing to 1, hence equal to (−1/3,2/3,2/3) by S08; but −1/3<0 (N12) violates nonnegativity, a contradiction. So u is not a convex combination; it is only an affine combination (the MKO: negative weights give only a general affine combination)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "contradiction",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "關鍵是 S08 的唯一性：若仿射權重不唯一，一組負權重不足以排除凸組合。",
            "en": "The key is the uniqueness in S08: if affine weights were not unique, one negative set of weights would not rule out a convex combination."
          }
        },
        "leaf_basis": "contradiction + order_basics"
      },
      {
        "id": "S10",
        "uses": [
          "L_bound"
        ],
        "before_tex": "\\lambda_i\\ge0,\\ \\sum\\lambda_i=1,\\ \\ell(v_i)\\le M",
        "after_tex": "\\ell\\Big(\\sum_i\\lambda_iv_i\\Big)\\le M",
        "relation": "theorem_application",
        "rule": "L_bound",
        "conditions": [],
        "explanation": {
          "zh-Hant": "引理 L_bound（S10.1、S10.2 證明）：凸組合在線性泛函下的值不超過各點值的上界。這給出不依賴解方程的第二種排除方法，而且明顯用到權重非負。",
          "en": "Lemma L_bound (proved in S10.1, S10.2): under a linear functional, a convex combination does not exceed an upper bound of the values at the points. This gives a second exclusion method that does not need solving equations, and it visibly uses nonnegativity of the weights."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_bound"
          ],
          "explanation": {
            "zh-Hant": "需要凸權重；仿射權重（可負）不滿足這個上界（S11）。",
            "en": "Needs convex weights; affine weights (possibly negative) do not obey the bound (S11)."
          }
        },
        "leaf_basis": "L_bound + vector_space_axioms"
      },
      {
        "id": "S10.1",
        "uses": [
          "vector_space_axioms",
          "distributive"
        ],
        "before_tex": "\\ell\\Big(\\sum_i\\lambda_iv_i\\Big)",
        "after_tex": "\\sum_i\\lambda_i\\ell(v_i)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "寫 v_i=(x_i,y_i)。Σλ_iv_i=(Σλ_ix_i, Σλ_iy_i)，所以 ℓ 的值為 αΣλ_ix_i+βΣλ_iy_i=Σλ_i(αx_i+βy_i)=Σλ_iℓ(v_i)（分配律；N13 對三點、α=β=1 驗證恆等式）。",
          "en": "Write v_i=(x_i,y_i). Σλ_iv_i=(Σλ_ix_i, Σλ_iy_i), so ℓ takes the value αΣλ_ix_i+βΣλ_iy_i=Σλ_i(αx_i+βy_i)=Σλ_iℓ(v_i) (distributivity; N13 checks the identity for three points with α=β=1)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "finite_sum_rules"
          ],
          "explanation": {
            "zh-Hant": "恆等變形，雙向成立；這一步不需要權重非負。",
            "en": "An identity, valid both ways; this step does not need nonnegative weights."
          }
        },
        "leaf_basis": "distributive + finite_sum_rules"
      },
      {
        "id": "S10.2",
        "uses": [
          "S10.1",
          "order_basics",
          "finite_sum_rules"
        ],
        "before_tex": "\\sum_i\\lambda_i\\ell(v_i)",
        "after_tex": "\\le\\sum_i\\lambda_iM=M\\sum_i\\lambda_i=M",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "\\lambda_i\\ge0",
          "\\sum_i\\lambda_i=1"
        ],
        "explanation": {
          "zh-Hant": "對每個 i，ℓ(v_i)≤M 且 λ_i≥0，乘以非負數保持 ≤：λ_iℓ(v_i)≤λ_iM。相加得 Σλ_iℓ(v_i)≤MΣλ_i=M。S10.1 與 S10.2 證得引理 L_bound。若某個 λ_i<0，乘法會反轉不等號，這一步就失效。",
          "en": "For each i, ℓ(v_i)≤M and λ_i≥0, and multiplying by a nonnegative number preserves ≤: λ_iℓ(v_i)≤λ_iM. Adding gives Σλ_iℓ(v_i)≤MΣλ_i=M. S10.1 and S10.2 prove Lemma L_bound. If some λ_i<0, multiplication would reverse the inequality and the step would fail."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "finite_sum_rules"
          ],
          "explanation": {
            "zh-Hant": "只宣稱上界；兩個條件（非負、和為 1）都被用到。",
            "en": "Only an upper bound is claimed; both conditions (nonnegative, total 1) are used."
          }
        },
        "leaf_basis": "order_basics + finite_sum_rules + mul_one"
      },
      {
        "id": "S11",
        "uses": [
          "S10",
          "G1",
          "G3"
        ],
        "before_tex": "\\ell(x,y)=x+y:\\ \\ell(v_1)=0,\\ \\ell(v_2)=6,\\ \\ell(v_3)=6",
        "after_tex": "\\ell(u)=8>6\\ \\Rightarrow\\ u\\ \\text{not a convex combination}",
        "relation": "implies",
        "rule": "contrapositive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取 ℓ(x,y)=x+y，三點的值為 0、6、6，所以 M=6。若 u 是凸組合，L_bound 給 ℓ(u)≤6，但 ℓ(u)=4+4=8>6（N14、N15）。由逆否，u 不是凸組合，與 S09 一致。仿射權重 (−1/3,2/3,2/3) 給 −1/3·0+2/3·6+2/3·6=8，正是負權重讓值越過上界。",
          "en": "Take ℓ(x,y)=x+y; its values at the three points are 0, 6, 6, so M=6. If u were a convex combination, L_bound would give ℓ(u)≤6, but ℓ(u)=4+4=8>6 (N14, N15). By contraposition u is not a convex combination, agreeing with S09. The affine weights (−1/3,2/3,2/3) give −1/3·0+2/3·6+2/3·6=8: the negative weight is exactly what pushes the value past the bound."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_bound",
            "contrapositive",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "上界論證只能排除，不能證明某點是凸組合；例如 (−1,1) 滿足 x+y=0≤6，卻不是凸組合：它的第一坐標為負，而每個凸組合的第一坐標都至少為 0。所以反向不成立。",
            "en": "The bound argument can only exclude, not prove membership: (−1,1) satisfies x+y=0≤6 yet is not a convex combination (its first coordinate is negative, while every convex combination has first coordinate at least 0), so the converse fails."
          }
        },
        "leaf_basis": "contrapositive + L_bound + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "G3",
          "S01",
          "G1"
        ],
        "before_tex": "\\lambda=(0.333,\\ 0.333,\\ 0.333)",
        "after_tex": "\\sum\\lambda_i=0.999\\neq1:\\ \\text{not convex weights}",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "0.333 是精確的有理數 333/1000，三個相加為 0.999（N16），不等於 1（N17），總和條件失敗，所以不是凸權重；加權和 (1.998, 1.998) 也不是 (2,2)。精確權重 (1/3,1/3,1/3) 和為 1（N18），給出 (2,2)，才是合格見證。MKO 假設 a1 不接受浮點近似。",
          "en": "0.333 is the exact rational 333/1000, and three of them add to 0.999 (N16), which is not 1 (N17), so the total condition fails and these are not convex weights; the weighted sum (1.998, 1.998) is not (2,2) either. The exact weights (1/3,1/3,1/3) sum to 1 (N18) and give (2,2), a valid witness. MKO assumption a1 does not accept floating-point approximations."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "截斷小數是一個確定的有理數；無限循環的 0.9（9 循環）作為實數等於 1，但截斷後的 0.999 不等於 1。",
            "en": "A truncated decimal is a definite rational number; the infinitely repeating decimal with all 9s equals 1 as a real number, but the truncated 0.999 does not."
          }
        },
        "leaf_basis": "numeric_evaluation + frac_rules"
      },
      {
        "id": "S13",
        "uses": [
          "G3",
          "S01",
          "G1"
        ],
        "before_tex": "v_4=(3,3)",
        "after_tex": "(3,3)=1\\cdot v_4=\\tfrac12v_2+\\tfrac12v_3:\\ \\text{weights}\\ (0,0,0,1)\\ \\text{and}\\ (0,\\tfrac12,\\tfrac12,0)",
        "relation": "implies",
        "rule": "mko_convex_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在四點 v_1、v_2、v_3、v_4 上，(3,3) 有兩組凸權重：(0,0,0,1) 直接取 v_4；(0,1/2,1/2,0) 給 (1/2·6, 1/2·6)=(3,3)（N19、N20）。兩組都非負、和為 1，但不同。所以凸組合見證不帶來座標唯一性；仿射獨立可保證整個族上凸表示的唯一性，但不是個別點唯一表示的必要條件；此處 v_4 在 v_2、v_3 的線段上，已明示 (3,3) 不唯一，不能把這個結論外推成每個點都不唯一。",
          "en": "On the four points v_1, v_2, v_3, v_4, the point (3,3) has two convex weight vectors: (0,0,0,1) takes v_4 directly, and (0,1/2,1/2,0) gives (1/2·6, 1/2·6)=(3,3) (N19, N20). Both are nonnegative with total 1, yet they differ. So a convex-combination witness does not bring uniqueness of coordinates; affine independence guarantees uniqueness throughout a family, but is not necessary for uniqueness at an individual point. Here v_4 lies on the segment from v_2 to v_3 and the two witnesses show nonuniqueness at (3,3), not at every point."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_convex_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "兩組見證都完整核對三個條件；一個例子就否定「凸權重必唯一」。",
            "en": "Both witnesses pass all three conditions; one example refutes 'convex weights are always unique'."
          }
        },
        "leaf_basis": "mko_convex_def + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S05",
          "S06",
          "S09",
          "S11",
          "S12",
          "S13"
        ],
        "before_tex": "\\text{S05 to S13}",
        "after_tex": "\\text{convex}\\iff(\\lambda_i\\ge0)\\land(\\textstyle\\sum\\lambda_i=1)\\land(\\text{exact reconstruction})",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "總結：w=(2,3) 以 (1/6,1/3,1/2) 為見證；零權重允許；u=(4,4) 的唯一仿射權重含負數，且 x+y=8>6，不是凸組合；0.333 三個權重和為 0.999，不合格；凸權重可以不唯一。",
          "en": "Summary: w=(2,3) has the witness (1/6,1/3,1/2); zero weights are allowed; u=(4,4) has a negative entry in its only affine weights and x+y=8>6, so it is not a convex combination; three weights of 0.333 sum to 0.999 and do not qualify; convex weights need not be unique."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S09",
            "S12",
            "S13"
          ],
          "explanation": {
            "zh-Hant": "只合取已證結論。",
            "en": "Only conjoins conclusions already proved."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "λ=(1/6,1/3,1/2) 非負、和為 1、重建 (2,3)，所以 w=(2,3) 是 v_1、v_2、v_3 的凸組合；(3,0)=1/2v_1+1/2v_2+0v_3 也是（零權重允許）；λv_2+(1−λ)v_3（0≤λ≤1）走過線段，λ=1/3 給 (2,4)。u=(4,4) 唯一的仿射權重是 (−1/3,2/3,2/3)，且 x+y=8 超過上界 6，故不是凸組合。(0.333,0.333,0.333) 和為 0.999，不合格。加入 v_4=(3,3) 後，(3,3) 有兩組不同的凸權重。",
      "en": "λ=(1/6,1/3,1/2) is nonnegative, sums to 1 and reconstructs (2,3), so w=(2,3) is a convex combination of v_1, v_2, v_3; so is (3,0)=1/2v_1+1/2v_2+0v_3 (zero weights allowed); λv_2+(1−λ)v_3 (0≤λ≤1) runs along a segment, with λ=1/3 giving (2,4). The only affine weights of u=(4,4) are (−1/3,2/3,2/3), and x+y=8 exceeds the bound 6, so u is not a convex combination. (0.333,0.333,0.333) sums to 0.999 and does not qualify. After adding v_4=(3,3), the point (3,3) has two different convex weight vectors."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "權重 (−1/3,2/3,2/3) 和為 1，所以 u=(4,4) 是 v_1、v_2、v_3 的凸組合",
          "en": "The weights (−1/3,2/3,2/3) sum to 1, so u=(4,4) is a convex combination of v_1, v_2, v_3"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "−1/3<0；而且這是唯一的仿射權重（S08、S09），x+y=8>6（S11）。",
          "en": "−1/3<0; moreover these are the only affine weights (S08, S09), and x+y=8>6 (S11)."
        }
      },
      {
        "claim": {
          "zh-Hant": "(0.333,0.333,0.333) 是凸權重",
          "en": "(0.333,0.333,0.333) are convex weights"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "和為 0.999≠1（S12）。",
          "en": "Their total is 0.999≠1 (S12)."
        }
      },
      {
        "claim": {
          "zh-Hant": "(3,3)=1·v_4，所以它在 v_1 到 v_4 上的凸座標唯一為 (0,0,0,1)",
          "en": "(3,3)=1·v_4, so its convex coordinates over v_1 through v_4 are uniquely (0,0,0,1)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(0,1/2,1/2,0) 也給出 (3,3)（S13）。",
          "en": "(0,1/2,1/2,0) also gives (3,3) (S13)."
        }
      },
      {
        "claim": {
          "zh-Hant": "凸組合的權重必須全為正，所以 (3,0)=1/2v_1+1/2v_2+0v_3 不是凸組合",
          "en": "Convex weights must all be positive, so (3,0)=1/2v_1+1/2v_2+0v_3 is not a convex combination"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "定義只要求 λ_i≥0，零權重允許（S06）。",
          "en": "The definition only requires λ_i≥0; zero weights are allowed (S06)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "vector_space_axioms",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "向量空間公理（實數或其他域 F 上）：加法成交換群；純量乘法滿足 a(u+v)=au+av、(a+b)u=au+bu、(ab)u=a(bu)、1u=u。",
          "en": "Vector space axioms (over ℝ or another field F): vectors form a commutative group under addition; scalar multiplication satisfies a(u+v)=au+av, (a+b)u=au+bu, (ab)u=a(bu) and 1u=u."
        }
      },
      {
        "id": "finite_sum_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限和：Σ_{i=1}^{0} a_i=0、Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}；Σ(a_i+b_i)=Σa_i+Σb_i；Σ c·a_i=c·Σa_i。",
          "en": "Finite sums: Σ_{i=1}^{0} a_i=0 and Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}; Σ(a_i+b_i)=Σa_i+Σb_i; Σ c·a_i=c·Σa_i."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "cartesian_product_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "笛卡兒積：A×B 是所有有序對 (a,b)（a∈A、b∈B）的集合；(a,b)=(c,d) ⇔ a=c 且 b=d。",
          "en": "Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d."
        }
      },
      {
        "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": "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_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": "exists_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。",
          "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)."
        }
      },
      {
        "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": "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": "eq_congruence",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。",
          "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule."
        }
      },
      {
        "id": "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": "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": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "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": "mko_convex_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-finite-convex-combination",
          "sha256": "84206c94ef6c80e08ea032783fce4ef3d1e4b15b55088556463d360b07c78d80",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "向量 v_1 到 v_n 的有限凸組合為 Σ λ_i v_i，其中 n≥1、每個 λ_i≥0 且 Σ λ_i=1。零權重允許；負權重或只近似和為 1 只形成一般仿射或線性組合。",
          "en": "A finite convex combination of vectors v_1 through v_n is Σ_i λ_i v_i, where n≥1, each λ_i≥0 and Σ_i λ_i=1. Zero weights are allowed. Negative weights or a total only approximately one do not qualify as convex weights; the expression may instead be a general affine or linear combination."
        }
      },
      {
        "id": "p_affine_comb",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "仿射組合：Σ λ_i v_i 且 Σ λ_i=1，權重符號不限；線性組合：權重沒有任何限制。",
          "en": "Affine combination: Σ λ_i v_i with Σ λ_i=1 and weights of any sign; linear combination: no restriction on the weights."
        }
      },
      {
        "id": "L_bound",
        "kind": "lemma",
        "proved_in": [
          "S10.1",
          "S10.2"
        ],
        "statement": {
          "zh-Hant": "設 ℓ(x,y)=αx+βy，λ_i≥0、Σλ_i=1，且對每個 i 有 ℓ(v_i)≤M。則 ℓ(Σλ_i v_i)≤M。",
          "en": "Let ℓ(x,y)=αx+βy, λ_i≥0 with Σλ_i=1, and ℓ(v_i)≤M for every i. Then ℓ(Σλ_i v_i)≤M."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "1/6+1/3+1/2",
        "rhs": "1"
      },
      {
        "id": "N2",
        "step": "S04.1",
        "lhs": "1/6*0+1/3*6+1/2*0",
        "rhs": "2"
      },
      {
        "id": "N3",
        "step": "S04.2",
        "lhs": "1/6*0+1/3*0+1/2*6",
        "rhs": "3"
      },
      {
        "id": "N4",
        "step": "S06",
        "lhs": "1/2*0+1/2*6+0*0",
        "rhs": "3"
      },
      {
        "id": "N5",
        "step": "S06",
        "lhs": "1/2+1/2+0",
        "rhs": "1"
      },
      {
        "id": "N6",
        "step": "S07",
        "vars": [
          "l"
        ],
        "degree": 1,
        "lhs": "l*6+(1-l)*0",
        "rhs": "6*l"
      },
      {
        "id": "N7",
        "step": "S07",
        "vars": [
          "l"
        ],
        "degree": 1,
        "lhs": "l*0+(1-l)*6",
        "rhs": "6-6*l"
      },
      {
        "id": "N8",
        "step": "S07",
        "lhs": "6*(1/3)",
        "rhs": "2"
      },
      {
        "id": "N9",
        "step": "S07",
        "lhs": "6-6*(1/3)",
        "rhs": "4"
      },
      {
        "id": "N10",
        "step": "S08.1",
        "lhs": "4/6",
        "rhs": "2/3"
      },
      {
        "id": "N11",
        "step": "S08.2",
        "lhs": "1-2/3-2/3",
        "rhs": "-1/3"
      },
      {
        "id": "N12",
        "step": "S09",
        "lhs": "-1/3",
        "relation": "<",
        "rhs": "0"
      },
      {
        "id": "N13",
        "step": "S10.1",
        "vars": [
          "a",
          "b",
          "c",
          "x1",
          "x2",
          "x3",
          "y1",
          "y2",
          "y3"
        ],
        "degree": 1,
        "lhs": "(a*x1+b*x2+c*x3)+(a*y1+b*y2+c*y3)",
        "rhs": "a*(x1+y1)+b*(x2+y2)+c*(x3+y3)"
      },
      {
        "id": "N14",
        "step": "S11",
        "lhs": "4+4",
        "rhs": "8"
      },
      {
        "id": "N15",
        "step": "S11",
        "lhs": "-1/3*0+2/3*6+2/3*6",
        "relation": ">",
        "rhs": "6"
      },
      {
        "id": "N16",
        "step": "S12",
        "lhs": "0.333+0.333+0.333",
        "rhs": "0.999"
      },
      {
        "id": "N17",
        "step": "S12",
        "lhs": "0.999",
        "relation": "!=",
        "rhs": "1"
      },
      {
        "id": "N18",
        "step": "S12",
        "lhs": "1/3+1/3+1/3",
        "rhs": "1"
      },
      {
        "id": "N19",
        "step": "S13",
        "lhs": "1/2*6+1/2*0",
        "rhs": "3"
      },
      {
        "id": "N20",
        "step": "S13",
        "lhs": "1/2*0+1/2*6",
        "rhs": "3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理 G1 到 G3 的整數座標向量與精確有理權重（MKO 假設 a1）。凸組合見證只證給定加權點，不主張座標唯一（a2）。凸集合、凸包與仿射獨立的一般理論不在範圍內。不改動 canonical MKO 或其英文 edition。",
      "en": "Treats only the integer-coordinate vectors and exact rational weights of G1 to G3 (MKO assumption a1). A convex-combination witness establishes only the given weighted point, not uniqueness of coordinates (a2). The general theory of convex sets, convex hulls and affine independence is out of scope. The canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "v_1=(0,0),\\ v_2=(6,0),\\ v_3=(0,6)",
        "給定的三個點",
        "the three given points"
      ],
      [
        "v_4=(3,3)",
        "加入的第四個點",
        "the added fourth point"
      ],
      [
        "\\lambda=(\\lambda_1,\\lambda_2,\\lambda_3)",
        "權重",
        "the weights"
      ],
      [
        "w=(2,3),\\ u=(4,4)",
        "目標點與反例點",
        "the target point and the counterexample point"
      ],
      [
        "\\ell(x,y)=x+y",
        "用於上界論證的線性泛函",
        "the linear functional used for the bound"
      ]
    ],
    "md_error_controls": [
      {
        "title": "負權重只給仿射組合 / Negative weights give only affine combinations",
        "zh": "MKO 指出的誤解之一是允許負權重。u=(4,4) 可寫成 −1/3v_1+2/3v_2+2/3v_3，權重和為 1，但這只是仿射組合。S08 證明這是唯一的仿射權重，S09 因而排除凸組合；S11 用 x+y≤6 的上界獨立確認。",
        "en": "One misconception named by the MKO is to allow negative weights. u=(4,4) can be written −1/3v_1+2/3v_2+2/3v_3 with weights summing to 1, but that is only an affine combination. S08 proves these are the only affine weights, so S09 rules out a convex combination; S11 confirms this independently with the bound x+y≤6."
      },
      {
        "title": "近似和不等於精確 1 / An approximate total is not exactly 1",
        "zh": "MKO 的另一個誤解是把 0.999 這類近似值當成精確的 1。三個 0.333 的和是 0.999，不是 1，所以不合格；正確的見證要用精確有理數 1/3。（無限循環的 9 作為實數等於 1，但題庫只用有限、精確的有理權重。）",
        "en": "Another misconception named by the MKO is to treat an approximation such as 0.999 as exactly 1. Three weights of 0.333 add to 0.999, not 1, so they do not qualify; a correct witness uses the exact rational 1/3. (The infinitely repeating 9s equal 1 as a real number, but the exercises use only finite, exact rational weights.)"
      },
      {
        "title": "見證不帶來唯一座標 / A witness does not give unique coordinates",
        "zh": "MKO 的第三個誤解是由凸組合直接聲稱唯一座標。S13 中 (3,3) 在 v_1 到 v_4 上有 (0,0,0,1) 與 (0,1/2,1/2,0) 兩組凸權重。唯一性須另作論證；仿射獨立是保證全族唯一性的充分條件，不是每個個別點唯一表示的必要條件；本稿在 S08 對三點 v_1、v_2、v_3 的仿射權重唯一性是直接解方程得到的，不是由見證推出的。",
        "en": "A third misconception named by the MKO is to claim unique coordinates directly from a convex combination. In S13, (3,3) has the two convex weight vectors (0,0,0,1) and (0,1/2,1/2,0) over v_1 through v_4. Uniqueness requires its own argument; affine independence is sufficient for uniqueness throughout a family, not necessary for uniqueness at every individual point; the uniqueness of affine weights for the three points v_1, v_2, v_3 in S08 came from solving the equations directly, not from the witness."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "凸集合、凸包與仿射獨立的一般理論（MKO 的 formal next）不在範圍內；S08 的唯一性只對 v_1、v_2、v_3 直接解方程得到。",
        "en": "The general theory of convex sets, convex hulls and affine independence (the MKO's formal next) is out of scope; the uniqueness in S08 is obtained only for v_1, v_2, v_3 by solving the equations directly."
      },
      {
        "zh-Hant": "S11 的上界論證只能排除點，不能證明一點是凸組合。",
        "en": "The bound argument of S11 can only exclude points; it cannot prove that a point is a convex combination."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "883bd8ea9e8aa6424a1a57388935dbdcff2c47f965eba50f18d4f37366e19d03",
    "original_source_sha256": "0a525035a113a0483c2c6902ae703bbc991232a443c60a2cb924e27999cf2bad",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M065"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}