{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-linear-programming-01",
  "mko_id": "mko-linear-programming",
  "title": {
    "zh-Hant": "線性規劃：在明示候選中求最佳，並說明它不是連續最優",
    "en": "Linear programming: the best supplied candidate, and why it is not the continuous optimum"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-linear-programming-01",
    "source": {
      "mko_id": "mko-linear-programming",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "e954454f13d345f635417e689ccae76261e754c182baf343173c0279585e8ba4",
      "english_sha256": "b16ebfdf1e39849a91ad524e7052c80e4c5e21153a5faf6290fd437bb617a283"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "order_basics",
      "frac_rules",
      "distributive",
      "and_intro_elim",
      "de_morgan_logic",
      "set_builder",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "限制 3x+2y≤12、x+2y≤6、−x≤0、−y≤0，目標為最大化 L(x,y)=x+y。對明示的七個候選 P1=(0,0)、P2=(4,0)、P3=(3,1)、P4=(2,2)、P5=(0,3)、P6=(4,1)、P7=(3,2)，逐限制核對可行性，只在可行候選間比較目標值，求出候選中的最大值與全部最佳點；再以可行點 Q=(3,3/2) 與一個不等式證書證明連續可行域上的最大值是 9/2，而不是候選中的 4。",
      "en": "Constraints 3x+2y≤12, x+2y≤6, −x≤0, −y≤0; objective: maximize L(x,y)=x+y. For the seven supplied candidates P1=(0,0), P2=(4,0), P3=(3,1), P4=(2,2), P5=(0,3), P6=(4,1), P7=(3,2), check feasibility constraint by constraint, compare objective values only among feasible candidates, and find the best value among the candidates and all points attaining it; then use the feasible point Q=(3,3/2) and an inequality certificate to prove that the maximum over the continuous feasible region is 9/2, not the candidate value 4."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "限制（形式 ax+by≤c，整係數）：C1：3x+2y≤12；C2：x+2y≤6；C3：−x≤0；C4：−y≤0。目標：最大化 L(x,y)=x+y。",
          "en": "Constraints (form ax+by≤c, integer coefficients): C1: 3x+2y≤12; C2: x+2y≤6; C3: −x≤0; C4: −y≤0. Objective: maximize L(x,y)=x+y."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "完整明示的候選表 K：P1=(0,0)、P2=(4,0)、P3=(3,1)、P4=(2,2)、P5=(0,3)、P6=(4,1)、P7=(3,2)。",
          "en": "The complete supplied candidate list K: P1=(0,0), P2=(4,0), P3=(3,1), P4=(2,2), P5=(0,3), P6=(4,1), P7=(3,2)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "連續可行域 F_ℝ={(x,y)∈ℝ² ∣ C1、C2、C3、C4 都成立}。",
          "en": "The continuous feasible region F_ℝ={(x,y)∈ℝ² ∣ C1, C2, C3 and C4 all hold}."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "G3",
          "mko_lp"
        ],
        "before_tex": "\\text{linear program with supplied candidates}",
        "after_tex": "\\max L(x,y)=x+y\\ \\text{s.t.}\\ C_1,C_2,C_3,C_4;\\ K=\\{P_1,P_2,P_3,P_4,P_5,P_6,P_7\\}",
        "relation": "definition",
        "rule": "mko_lp",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，線性規劃由線性目標與線性限制組成。這裡有兩個不同的集合：明示的候選集合 K（七個點）與連續可行域 F_ℝ。S02 到 S11 只在 K 上工作，結論稱為「所給候選中的最佳」；S12 到 S15 才討論 F_ℝ。",
          "en": "By the MKO, a linear program consists of a linear objective and linear constraints. Two different sets appear here: the supplied candidate set K (seven points) and the continuous feasible region F_ℝ. S02 to S11 work on K only, and their conclusion is called 'best among the supplied candidates'; only S12 to S15 discuss F_ℝ."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "G3",
            "mko_lp"
          ],
          "explanation": {
            "zh-Hant": "此步只陳述問題與兩個集合，不推出新事實。",
            "en": "This step only states the problem and the two sets and derives no new fact."
          }
        },
        "leaf_basis": "mko_lp + def_intro"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "G1",
          "set_builder"
        ],
        "before_tex": "P\\in F_K",
        "after_tex": "P\\in K\\ \\land\\ C_1(P)\\land C_2(P)\\land C_3(P)\\land C_4(P)",
        "relation": "equivalent",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "可行候選的定義：候選 P 可行，當且僅當四條限制全部成立。這是合取：先逐限制核對，再只在可行候選間比較目標值（MKO 第一段）。限制是「≤」，所以等號成立也算滿足。",
          "en": "Definition of a feasible candidate: P is feasible exactly when all four constraints hold. This is a conjunction: check constraint by constraint first, then compare objective values only among feasible candidates (MKO, first paragraph). The constraints use '≤', so equality also satisfies them."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "def_unfold",
            "set_builder",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "集合建構式的展開，兩向成立。",
            "en": "Unfolding the set-builder definition holds both ways."
          }
        },
        "leaf_basis": "set_builder + def_unfold + and_intro_elim"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "C_1,C_2,C_3,C_4\\ \\text{at}\\ P_1,\\ P_2,\\ P_3,\\ P_4,\\ P_5,\\ P_6,\\ P_7",
        "after_tex": "\\text{table of left-hand sides}",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把每個候選代入每條限制的左邊，分三個子步：C1、C2、以及符號限制 C3、C4。",
          "en": "Substitute every candidate into the left side of every constraint, in three sub-steps: C1, C2, and the sign constraints C3 and C4."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "每個值都是精確整數計算，可逐一重算（N1 至 N7）。",
            "en": "Every value is an exact integer computation and can be recomputed one by one (N1 to N7)."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S03.1",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "3x+2y\\ \\text{at}\\ P_1,\\ P_2,\\ P_3,\\ P_4,\\ P_5,\\ P_6,\\ P_7",
        "after_tex": "0,\\ 12,\\ 11,\\ 10,\\ 6,\\ 14,\\ 13",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "3·0+2·0=0，3·4+2·0=12，3·3+2·1=11，3·2+2·2=10，3·0+2·3=6，3·4+2·1=14，3·3+2·2=13。與上界 12 比較：前五個 ≤12，P6 的 14 與 P7 的 13 >12。",
          "en": "3·0+2·0=0, 3·4+2·0=12, 3·3+2·1=11, 3·2+2·2=10, 3·0+2·3=6, 3·4+2·1=14, 3·3+2·2=13. Against the bound 12: the first five are ≤12, while 14 for P6 and 13 for P7 are >12."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "整數算術的值唯一。",
            "en": "Integer arithmetic has unique values."
          }
        },
        "leaf_basis": "numeric_evaluation + order_basics"
      },
      {
        "id": "S03.2",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "x+2y\\ \\text{at}\\ P_1,\\ P_2,\\ P_3,\\ P_4,\\ P_5,\\ P_6,\\ P_7",
        "after_tex": "0,\\ 4,\\ 5,\\ 6,\\ 6,\\ 6,\\ 7",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "0+0=0，4+0=4，3+2=5，2+4=6，0+6=6，4+2=6，3+4=7。與上界 6 比較：P1 至 P6 都 ≤6（P4、P5、P6 取等號），P7 的 7>6。",
          "en": "0+0=0, 4+0=4, 3+2=5, 2+4=6, 0+6=6, 4+2=6, 3+4=7. Against the bound 6: P1 to P6 are all ≤6 (P4, P5 and P6 with equality), while 7 for P7 is >6."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "整數算術的值唯一。",
            "en": "Integer arithmetic has unique values."
          }
        },
        "leaf_basis": "numeric_evaluation + order_basics"
      },
      {
        "id": "S03.3",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "-x\\le 0,\\ -y\\le 0\\ \\text{at all}\\ P_i",
        "after_tex": "\\text{true for all seven}",
        "relation": "equivalent",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "七個候選的坐標都是 0 到 4 的非負整數；對非負數 u，−u≤0。所以 C3、C4 對全部候選成立，可行性只由 C1、C2 決定。",
          "en": "All coordinates of the seven candidates are nonnegative integers from 0 to 4; for a nonnegative u, −u≤0. So C3 and C4 hold for every candidate, and feasibility is decided by C1 and C2 alone."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "u≥0 ⇔ −u≤0（兩邊乘 −1 反轉不等號），雙向成立。",
            "en": "u≥0 ⇔ −u≤0 (multiplying by −1 reverses the inequality), which holds both ways."
          }
        },
        "leaf_basis": "order_basics"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "S03.1",
          "S03.2",
          "S03.3"
        ],
        "before_tex": "P_1,\\ P_2,\\ P_3,\\ P_4,\\ P_5",
        "after_tex": "C_1\\land C_2\\land C_3\\land C_4\\ \\text{hold}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "P1 到 P5 的 C1 值 0、12、11、10、6 都 ≤12，C2 值 0、4、5、6、6 都 ≤6，C3、C4 也成立。四條全部成立，所以這五點可行。P2 使 C1 取等號，P4、P5 使 C2 取等號；等號仍屬可行。",
          "en": "For P1 to P5 the C1 values 0, 12, 11, 10, 6 are all ≤12, the C2 values 0, 4, 5, 6, 6 are all ≤6, and C3, C4 hold. All four constraints hold, so these five points are feasible. P2 makes C1 tight and P4, P5 make C2 tight; equality is still feasible."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S03.1",
            "S03.2",
            "S03.3"
          ],
          "explanation": {
            "zh-Hant": "合取需要四條都核過；不宣稱可行點有任何最優性。",
            "en": "The conjunction needs all four constraints checked; no optimality of these feasible points is claimed."
          }
        },
        "leaf_basis": "and_intro_elim + order_basics"
      },
      {
        "id": "S05",
        "uses": [
          "S02",
          "S03.1",
          "S03.2"
        ],
        "before_tex": "P_6=(4,1),\\ P_7=(3,2)",
        "after_tex": "P_6,P_7\\notin F_K",
        "relation": "implies",
        "rule": "de_morgan_logic",
        "conditions": [],
        "explanation": {
          "zh-Hant": "否定合取只需一項失敗：P6 的 C1 值 14>12；P7 的 C1 值 13>12（C2 值 7>6 也失敗）。所以 P6、P7 不可行，不參加目標值比較。",
          "en": "Negating a conjunction needs only one failing part: P6 has C1 value 14>12; P7 has C1 value 13>12 (and C2 value 7>6 fails too). So P6 and P7 are infeasible and take no part in comparing objective values."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "de_morgan_logic",
            "S03.1"
          ],
          "explanation": {
            "zh-Hant": "¬(A∧B∧C∧D) ⇔ ¬A∨¬B∨¬C∨¬D；一條失敗的限制就是充分的見證。",
            "en": "¬(A∧B∧C∧D) ⇔ ¬A∨¬B∨¬C∨¬D; one failing constraint is a sufficient witness."
          }
        },
        "leaf_basis": "de_morgan_logic + order_basics"
      },
      {
        "id": "S06",
        "uses": [
          "S04",
          "S05",
          "G2"
        ],
        "before_tex": "F_K=\\{P\\in K\\mid C_1\\land C_2\\land C_3\\land C_4\\}",
        "after_tex": "F_K=\\{P_1,P_2,P_3,P_4,P_5\\}",
        "relation": "equivalent",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "K 的七個元素都已判定：P1 到 P5 可行（S04），P6、P7 不可行（S05）。所以可行候選集合恰為 {P1,P2,P3,P4,P5}。",
          "en": "All seven elements of K have been decided: P1 to P5 are feasible (S04) and P6, P7 are not (S05). So the set of feasible candidates is exactly {P1,P2,P3,P4,P5}."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_extensionality",
            "S04",
            "S05"
          ],
          "explanation": {
            "zh-Hant": "逐元素判定涵蓋 K 的全部七點，兩個集合元素相同，依外延相等。",
            "en": "The element-by-element decision covers all seven points of K, so the two sets have the same elements and are equal by extensionality."
          }
        },
        "leaf_basis": "set_extensionality + case_analysis"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "G1"
        ],
        "before_tex": "L(P_1),\\ L(P_2),\\ L(P_3),\\ L(P_4),\\ L(P_5)",
        "after_tex": "0,\\ 4,\\ 4,\\ 4,\\ 3",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "只對可行候選計算目標值：L(0,0)=0，L(4,0)=4，L(3,1)=4，L(2,2)=4，L(0,3)=3。",
          "en": "Compute the objective only on feasible candidates: L(0,0)=0, L(4,0)=4, L(3,1)=4, L(2,2)=4, L(0,3)=3."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確整數計算。",
            "en": "Exact integer computation."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S07",
          "mko_opt"
        ],
        "before_tex": "\\max_{P\\in F_K}L(P)",
        "after_tex": "4",
        "relation": "implies",
        "rule": "mko_opt",
        "conditions": [
          "F_K\\neq\\varnothing"
        ],
        "explanation": {
          "zh-Hant": "依最大化的定義，要找可行點 P*，使每個可行 P 都有 L(P)≤L(P*)。值 4 被 P2 達成，且 0、4、4、4、3 每個都 ≤4；所以候選中的最大值為 4。F_K 是非空有限集，最大值必存在。",
          "en": "By the definition of maximization, find a feasible P* with L(P)≤L(P*) for every feasible P. The value 4 is attained by P2, and each of 0, 4, 4, 4, 3 is ≤4; so the maximum among the candidates is 4. F_K is a nonempty finite set, so a maximum exists."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_opt",
            "S07",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要兩件事：4 被某個可行候選達成，且每個可行候選的值都 ≤4。比較只在 F_K 內進行。",
            "en": "Two things are needed: 4 is attained by some feasible candidate, and every feasible candidate has value ≤4. The comparison is made within F_K only."
          }
        },
        "leaf_basis": "mko_opt + order_basics + finite_counting_basics"
      },
      {
        "id": "S09",
        "uses": [
          "S07",
          "S08",
          "mko_opt"
        ],
        "before_tex": "\\arg\\max_{P\\in F_K}L(P)",
        "after_tex": "\\{P_2,P_3,P_4\\}",
        "relation": "equivalent",
        "rule": "mko_opt",
        "conditions": [],
        "explanation": {
          "zh-Hant": "達到最大值 4 的可行候選是 P2、P3、P4，三者平手；P1（0）與 P5（3）沒有達到。依 MKO，argmax 應保留全部平手的點，不能只報其中一個。",
          "en": "The feasible candidates attaining the maximum 4 are P2, P3 and P4, a three-way tie; P1 (0) and P5 (3) do not. By the MKO, the argmax keeps all tied points rather than reporting just one."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_opt",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "argmax 是 {P∈F_K ∣ L(P)=4}，逐點核對五個可行候選，兩向成立。",
            "en": "The argmax is {P∈F_K ∣ L(P)=4}; checking the five feasible candidates one by one makes this hold both ways."
          }
        },
        "leaf_basis": "mko_opt + set_builder + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "S05",
          "S08",
          "G1"
        ],
        "before_tex": "L(P_6)=5,\\ L(P_7)=5",
        "after_tex": "5>4\\ \\text{but}\\ P_6,P_7\\notin F_K",
        "relation": "implies",
        "rule": "mko_opt",
        "conditions": [],
        "explanation": {
          "zh-Hant": "P6、P7 的目標值都是 5，比 4 大，但它們不可行（S05），所以不在比較範圍內，不影響最大值。若把它們算進來，就會錯把 5 當作最優值，這正是 MKO 指出的誤解之一。",
          "en": "P6 and P7 both have objective value 5, larger than 4, but they are infeasible (S05), so they are outside the comparison and do not affect the maximum. Counting them would wrongly report 5 as the optimum, which is one of the mistakes named by the MKO."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "mko_opt"
          ],
          "explanation": {
            "zh-Hant": "最大化的定義只量化可行點；不可行點的目標值無論多大都與結論無關。",
            "en": "The definition of maximization quantifies over feasible points only; the objective value of an infeasible point, however large, is irrelevant to the conclusion."
          }
        },
        "leaf_basis": "mko_opt + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S08",
          "S09",
          "mko_lp"
        ],
        "before_tex": "\\max_{F_K}L=4,\\ \\arg\\max=\\{P_2,P_3,P_4\\}",
        "after_tex": "\\text{best among the supplied candidates only}",
        "relation": "implies",
        "rule": "mko_lp",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，沒有證明候選表完整涵蓋可行域，結論只能稱為「所給候選中的最佳」。到目前為止，沒有任何步驟討論 F_ℝ 中不在 K 裡的點。",
          "en": "By the MKO, without a proof that the candidate list covers the feasible region, the conclusion can only be called 'best among the supplied candidates'. So far no step has considered points of F_ℝ that are not in K."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_lp",
            "S08"
          ],
          "explanation": {
            "zh-Hant": "此步只界定結論範圍；S12 會證明這個限制是實質的。",
            "en": "This step only fixes the scope of the conclusion; S12 shows that the restriction is substantive."
          }
        },
        "leaf_basis": "mko_lp"
      },
      {
        "id": "S12",
        "uses": [
          "G1",
          "G3",
          "S08"
        ],
        "before_tex": "Q=(3,\\tfrac{3}{2})",
        "after_tex": "Q\\in F_{\\mathbb{R}},\\ L(Q)=\\tfrac{9}{2}>4",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取候選表外的點 Q=(3,3/2)：C1：3·3+2·(3/2)=12≤12；C2：3+2·(3/2)=6≤6；坐標非負，C3、C4 成立。所以 Q∈F_ℝ，且 L(Q)=3+3/2=9/2>4。因此候選中的最佳值 4 不是連續可行域上的最大值。",
          "en": "Take the point Q=(3,3/2) outside the candidate list: C1: 3·3+2·(3/2)=12≤12; C2: 3+2·(3/2)=6≤6; the coordinates are nonnegative, so C3 and C4 hold. Hence Q∈F_ℝ and L(Q)=3+3/2=9/2>4. Therefore the candidate best value 4 is not the maximum over the continuous feasible region."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "numeric_evaluation",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "一個可行見證點就足以否定「4 是 F_ℝ 上的最大值」（N8、N9、N10）；這一步還沒有證明 9/2 是最大值。",
            "en": "One feasible witness suffices to refute '4 is the maximum on F_ℝ' (N8, N9, N10); this step does not yet prove that 9/2 is the maximum."
          }
        },
        "leaf_basis": "numeric_evaluation + order_basics + exists_intro"
      },
      {
        "id": "S13",
        "uses": [
          "G1",
          "G3"
        ],
        "before_tex": "(x,y)\\in F_{\\mathbb{R}}",
        "after_tex": "x+y\\le\\tfrac{9}{2}",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "不等式證書：以非負權重 1/4、1/4 合併 C1 與 C2，對每個可行點得到 x+y≤9/2。分兩個子步：一個恆等式與一次不等式相加。這是對 F_ℝ 中所有點的證明，不是取樣。",
          "en": "An inequality certificate: combining C1 and C2 with the nonnegative weights 1/4 and 1/4 gives x+y≤9/2 for every feasible point. Two sub-steps: one identity and one addition of inequalities. This is a proof for all points of F_ℝ, not a sample."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "G3"
          ],
          "explanation": {
            "zh-Hant": "只用到 C1、C2 與權重非負；C3、C4 在這個證書中沒有用到。",
            "en": "Only C1, C2 and the nonnegativity of the weights are used; C3 and C4 are not needed for this certificate."
          }
        },
        "leaf_basis": "order_basics + distributive + frac_rules"
      },
      {
        "id": "S13.1",
        "uses": [
          "distributive",
          "frac_rules"
        ],
        "before_tex": "x+y",
        "after_tex": "\\tfrac{1}{4}(3x+2y)+\\tfrac{1}{4}(x+2y)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "展開右邊：(1/4)(3x+2y)+(1/4)(x+2y)=(1/4)(4x+4y)=x+y。這是對所有實數 x、y 成立的恆等式（N11）。",
          "en": "Expand the right side: (1/4)(3x+2y)+(1/4)(x+2y)=(1/4)(4x+4y)=x+y. This identity holds for all real x and y (N11)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個一次式，雙向成立。",
            "en": "Both sides are the same linear expression, so the step holds both ways."
          }
        },
        "leaf_basis": "distributive + add_assoc + add_comm + frac_rules"
      },
      {
        "id": "S13.2",
        "uses": [
          "S13.1",
          "G1",
          "order_basics"
        ],
        "before_tex": "3x+2y\\le 12,\\ x+2y\\le 6",
        "after_tex": "\\tfrac{1}{4}(3x+2y)+\\tfrac{1}{4}(x+2y)\\le\\tfrac{12}{4}+\\tfrac{6}{4}=\\tfrac{9}{2}",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "\\tfrac{1}{4}>0"
        ],
        "explanation": {
          "zh-Hant": "C1 兩邊乘正數 1/4，不等號方向不變：(1/4)(3x+2y)≤3；C2 同理：(1/4)(x+2y)≤3/2。兩個不等式相加（u1≤v1 且 u2≤v2 推出 u1+u2≤v1+u2≤v1+v2）得和 ≤9/2；再由 S13.1，x+y≤9/2。若權重取負，乘法會反轉不等號，相加就不成立。",
          "en": "Multiply C1 by the positive number 1/4, keeping the direction: (1/4)(3x+2y)≤3; likewise C2: (1/4)(x+2y)≤3/2. Add the two inequalities (u1≤v1 and u2≤v2 give u1+u2≤v1+u2≤v1+v2) to get a sum ≤9/2; by S13.1, x+y≤9/2. With a negative weight, multiplication would reverse the inequality and the addition would fail."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "S13.1"
          ],
          "explanation": {
            "zh-Hant": "需要權重為正（或非負）；不宣稱反向，即不由 x+y≤9/2 推回 C1、C2。",
            "en": "It needs positive (or nonnegative) weights; no converse is claimed, that is, C1 and C2 are not recovered from x+y≤9/2."
          }
        },
        "leaf_basis": "order_basics + frac_rules + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S12",
          "S13"
        ],
        "before_tex": "Q\\in F_{\\mathbb{R}},\\ L(Q)=\\tfrac{9}{2},\\ \\forall P\\in F_{\\mathbb{R}}:\\ L(P)\\le\\tfrac{9}{2}",
        "after_tex": "\\max_{F_{\\mathbb{R}}}L=\\tfrac{9}{2}",
        "relation": "implies",
        "rule": "mko_opt",
        "conditions": [],
        "explanation": {
          "zh-Hant": "最大值的兩個條件都成立：9/2 被可行點 Q 達成（S12），且每個可行點的值都 ≤9/2（S13）。所以連續可行域上的最大值是 9/2，比候選中的最佳 4 大。",
          "en": "Both conditions for a maximum hold: 9/2 is attained by the feasible point Q (S12), and every feasible point has value ≤9/2 (S13). So the maximum over the continuous feasible region is 9/2, larger than the candidate best 4."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12",
            "S13",
            "mko_opt"
          ],
          "explanation": {
            "zh-Hant": "上界證書與達成點缺一不可：只有見證點只給下界，只有證書不保證上界被達成。",
            "en": "The upper-bound certificate and the attaining point are both needed: the witness alone gives only a lower bound, and the certificate alone does not show the bound is attained."
          }
        },
        "leaf_basis": "mko_opt + and_intro_elim"
      },
      {
        "id": "S15",
        "uses": [
          "S13.1",
          "S13.2",
          "S14",
          "G1"
        ],
        "before_tex": "x+y=\\tfrac{9}{2},\\ (x,y)\\in F_{\\mathbb{R}}",
        "after_tex": "(x,y)=(3,\\tfrac{3}{2})",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若 x+y=9/2，則 S13.2 中兩個不等式都必須取等號：若任一嚴格小於，和就嚴格小於 9/2。所以 3x+2y=12 且 x+2y=6；相減得 2x=6，x=3，再得 y=3/2。因此連續最大值只在 Q 達成；它不在候選表中，候選中的三個平手點都只有 4。",
          "en": "If x+y=9/2, both inequalities in S13.2 must be equalities: if either were strict, the sum would be strictly below 9/2. So 3x+2y=12 and x+2y=6; subtracting gives 2x=6, x=3, and then y=3/2. Hence the continuous maximum is attained only at Q; Q is not in the candidate list, where the three tied points reach only 4."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13.2",
            "order_basics",
            "eq_congruence"
          ],
          "explanation": {
            "zh-Hant": "需要「嚴格小於加上小於等於得到嚴格小於」與一個 2×2 線性方程組的消去；不宣稱這個唯一性對其他目標函數成立。",
            "en": "It needs 'a strict inequality plus a non-strict one gives a strict one' and elimination in a 2×2 linear system; uniqueness is not claimed for other objectives."
          }
        },
        "leaf_basis": "order_basics + eq_congruence + numeric_evaluation"
      }
    ],
    "result": {
      "zh-Hant": "在七個明示候選中，P6、P7 不可行（違反 3x+2y≤12）；可行候選中 x+y 的最大值是 4，由 P2=(4,0)、P3=(3,1)、P4=(2,2) 三點平手達成，這只是所給候選中的最佳。連續可行域上，可行點 Q=(3,3/2) 給出 9/2，而證書 x+y=(1/4)(3x+2y)+(1/4)(x+2y)≤9/2 證明 9/2 是最大值，且只在 Q 達成。",
      "en": "Among the seven supplied candidates, P6 and P7 are infeasible (they violate 3x+2y≤12); among the feasible ones the maximum of x+y is 4, attained in a three-way tie by P2=(4,0), P3=(3,1) and P4=(2,2), which is only the best among the supplied candidates. Over the continuous feasible region, the feasible point Q=(3,3/2) gives 9/2, and the certificate x+y=(1/4)(3x+2y)+(1/4)(x+2y)≤9/2 proves that 9/2 is the maximum, attained only at Q."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "P6 的目標值 5 最大，所以 P6 是最優解",
          "en": "P6 has the largest objective value 5, so P6 is optimal"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "P6 不可行：3·4+2·1=14>12。",
          "en": "P6 is infeasible: 3·4+2·1=14>12."
        }
      },
      {
        "claim": {
          "zh-Hant": "候選中的最優解唯一，是 P2=(4,0)",
          "en": "The best candidate is unique, namely P2=(4,0)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "P3=(3,1) 與 P4=(2,2) 也可行且目標值為 4，三點平手。",
          "en": "P3=(3,1) and P4=(2,2) are also feasible with objective value 4, a three-way tie."
        }
      },
      {
        "claim": {
          "zh-Hant": "候選中最大值是 4，所以這個線性規劃在連續可行域上的最大值是 4",
          "en": "The best candidate value is 4, so the maximum of this linear program over the continuous feasible region is 4"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "Q=(3,3/2) 可行且 x+y=9/2>4。",
          "en": "Q=(3,3/2) is feasible and has x+y=9/2>4."
        }
      },
      {
        "claim": {
          "zh-Hant": "以權重 1 與 −1 合併 C1、C2，得對所有可行點 2x≤6，即 x≤3",
          "en": "Combining C1 and C2 with weights 1 and −1 gives 2x≤6, that is x≤3, for every feasible point"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "乘以負數會反轉不等號，不能這樣相加；P2=(4,0) 可行而 x=4>3。",
          "en": "Multiplying by a negative number reverses the inequality, so the inequalities cannot be added this way; P2=(4,0) is feasible with x=4>3."
        }
      }
    ],
    "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": "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": "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": "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": "de_morgan_logic",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "邏輯的德摩根律：¬(P∧Q) ⇔ ¬P∨¬Q；¬(P∨Q) ⇔ ¬P∧¬Q。",
          "en": "De Morgan's laws for logic: ¬(P∧Q) ⇔ ¬P∨¬Q and ¬(P∨Q) ⇔ ¬P∧¬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": "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_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": "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": "set_extensionality",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合相等：A=B 當且僅當 A⊆B 且 B⊆A，即兩者有相同元素。",
          "en": "Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements."
        }
      },
      {
        "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": "finite_counting_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限計數：互斥有限集合 A、B 有 |A∪B|=|A|+|B|；|A×B|=|A|·|B|；有限集之間存在雙射 ⇔ 元素個數相同。",
          "en": "Finite counting: for disjoint finite sets A, B, |A∪B|=|A|+|B|; |A×B|=|A|·|B|; there is a bijection between finite sets ⇔ they have the same number of elements."
        }
      },
      {
        "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": "add_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。",
          "en": "Associativity of addition: (u+v)+w=u+(v+w)."
        }
      },
      {
        "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": "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": "mko_lp",
        "kind": "cited",
        "source": {
          "mko_id": "mko-linear-programming",
          "sha256": "e954454f13d345f635417e689ccae76261e754c182baf343173c0279585e8ba4",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "線性規劃由線性目標與線性限制構成，尋找可行域上的最小或最大值。有限候選網格可以完整核對列出的點，但除非已證候選完整涵蓋一般可行域，結論只能稱為所給候選中的最佳。",
          "en": "A linear program consists of a linear objective and linear constraints and seeks a minimum or maximum on the feasible domain. A finite candidate grid permits complete checking of its listed points, but unless coverage of the general feasible domain has been proved, its conclusion is only best among the supplied candidates."
        }
      },
      {
        "id": "mko_opt",
        "kind": "cited",
        "source": {
          "mko_id": "mko-optimization-problem",
          "sha256": "55c8accf83a0b3a16cb86f438bdc6575f8dcec45e670f324e863e79a1a233514",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "給可行集合 S 與目標函數 f:S→R，最小化問題尋找 x*∈S，使每個 x∈S 都有 f(x*)≤f(x)；最大化反向。最優值可能由多個最優點同時達成。",
          "en": "For a feasible set S and objective f:S→R, minimization seeks x*∈S with f(x*)≤f(x) for every x∈S; maximization reverses the inequality. Several optimal points may attain the same value."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03.1",
        "lhs": "3*4+2*0",
        "rhs": "12"
      },
      {
        "id": "N2",
        "step": "S03.1",
        "lhs": "3*3+2*1",
        "rhs": "11"
      },
      {
        "id": "N3",
        "step": "S03.1",
        "lhs": "3*2+2*2",
        "rhs": "10"
      },
      {
        "id": "N4",
        "step": "S03.1",
        "lhs": "3*4+2*1",
        "relation": ">",
        "rhs": "12"
      },
      {
        "id": "N5",
        "step": "S03.1",
        "lhs": "3*3+2*2",
        "rhs": "13"
      },
      {
        "id": "N6",
        "step": "S03.2",
        "lhs": "4+2*1",
        "rhs": "6"
      },
      {
        "id": "N7",
        "step": "S03.2",
        "lhs": "3+2*2",
        "relation": ">",
        "rhs": "6"
      },
      {
        "id": "N8",
        "step": "S12",
        "lhs": "3*3+2*(3/2)",
        "rhs": "12"
      },
      {
        "id": "N9",
        "step": "S12",
        "lhs": "3+2*(3/2)",
        "rhs": "6"
      },
      {
        "id": "N10",
        "step": "S12",
        "lhs": "3+3/2",
        "rhs": "9/2"
      },
      {
        "id": "N11",
        "step": "S13.1",
        "vars": [
          "x",
          "y"
        ],
        "degree": 1,
        "lhs": "(3*x+2*y)/4+(x+2*y)/4",
        "rhs": "x+y"
      },
      {
        "id": "N12",
        "step": "S13.2",
        "lhs": "12/4+6/4",
        "rhs": "9/2"
      },
      {
        "id": "N13",
        "step": "S14",
        "lhs": "9/2",
        "relation": ">",
        "rhs": "4"
      },
      {
        "id": "N14",
        "step": "S15",
        "lhs": "(12-6)/2",
        "rhs": "3"
      },
      {
        "id": "N15",
        "step": "S07",
        "lhs": "max(0+0,4+0,3+1,2+2,0+3)",
        "rhs": "4"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：二變數整係數限制 ax+by≤c、整係數目標與完整明示的有限候選（假設 a1）。候選枚舉只得出「所給候選中的最佳」（假設 a2）。連續可行域上的最大值另以一個可行見證點與一個非負權重組合的不等式證書證明；這只針對本題，不證明一般的頂點定理或對偶定理。不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: integer-coefficient constraints ax+by≤c in two variables, an integer-coefficient objective and a complete explicitly supplied finite candidate list (assumption a1). Enumerating candidates yields only 'best among the supplied candidates' (assumption a2). The maximum over the continuous feasible region is proved separately with one feasible witness point and one inequality certificate built from nonnegative weights; this is specific to this problem and proves no general vertex or duality theorem. The canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "C_1,C_2,C_3,C_4",
        "四條線性限制",
        "the four linear constraints"
      ],
      [
        "L(x,y)=x+y",
        "線性目標",
        "the linear objective"
      ],
      [
        "K=\\{P_1,P_2,P_3,P_4,P_5,P_6,P_7\\}",
        "明示候選集合",
        "the supplied candidate set"
      ],
      [
        "F_K",
        "可行候選集合",
        "the set of feasible candidates"
      ],
      [
        "F_{\\mathbb{R}}",
        "連續可行域",
        "the continuous feasible region"
      ],
      [
        "Q=(3,\\tfrac{3}{2})",
        "候選表之外的可行見證點",
        "a feasible witness point outside the candidate list"
      ]
    ],
    "md_error_controls": [
      {
        "title": "不可行的點不能當最優 / An infeasible point cannot be optimal",
        "zh": "MKO 指出的第一個誤解。P6、P7 的目標值 5 比所有可行候選都大，但 P6 的 3x+2y=14>12、P7 的 13>12，它們不可行（S05、S10）。正確順序是先逐限制核對可行性，再只在可行點間比較目標值。",
        "en": "The first mistake named by the MKO. P6 and P7 have objective value 5, larger than every feasible candidate, but P6 has 3x+2y=14>12 and P7 has 13>12, so they are infeasible (S05, S10). The correct order is to check feasibility constraint by constraint first, and then compare objective values among feasible points only."
      },
      {
        "title": "有限候選最佳不一定是一般 LP 最優 / The best candidate need not attain the LP optimum",
        "zh": "MKO 指出的第二個誤解。候選中的最佳值 4 只涵蓋所給的七點；候選表外的可行點 Q=(3,3/2) 給出 9/2（S12）。要宣稱連續最優，需要涵蓋整個可行域的論證，例如 S13 的不等式證書。",
        "en": "The second mistake named by the MKO. The candidate best value 4 covers only the seven supplied points; the feasible point Q=(3,3/2) outside the list gives 9/2 (S12). Claiming a continuous optimum needs an argument covering the whole feasible region, such as the inequality certificate of S13."
      },
      {
        "title": "平手要全部保留 / Keep every tied point",
        "zh": "P2、P3、P4 的目標值都是 4。只回報其中一個，會讓讀者以為最佳點唯一；依 MKO，argmax 應保留全部平手的索引（S09）。",
        "en": "P2, P3 and P4 all have objective value 4. Reporting only one of them suggests a unique best point; by the MKO the argmax keeps every tied index (S09)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "S13 的證書只針對本題；一般線性規劃的頂點定理、標準形與對偶定理屬於 MKO 的 formal next，本稿沒有證明。",
        "en": "The certificate of S13 is specific to this problem; the vertex theorem, standard form and duality theorem for general linear programs belong to the MKO's formal next step and are not proved here."
      },
      {
        "zh-Hant": "N11 是一次多項式恆等式的網格核對；其餘數值核對都是精確的單點計算。",
        "en": "N11 is a grid check of a degree-one polynomial identity; the other numeric checks are exact single computations."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "3314594ea08c425708a9014997e20b768f85d871eee2977d20033d7ea5f171e0",
    "original_source_sha256": "3314594ea08c425708a9014997e20b768f85d871eee2977d20033d7ea5f171e0",
    "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."
  }
}