{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-convex-set-01", "mko_id": "mko-convex-set", "title": { "zh-Hant": "凸集合：半空間與軸對齊盒的完整凸性證書，及有限點集的中點反例", "en": "Convex set: complete convexity certificates for a halfspace and an axis-aligned box, and a midpoint counterexample for a finite point set" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-convex-set-01", "source": { "mko_id": "mko-convex-set", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "0d52be71dca290f856018d176bbb26dcd877c887031f690bd1d656db017ce02d", "english_sha256": "f81d0d2b42596c9a4b95349d75fc98970570af520e6edb9c1af988358ff84f0c" }, "mode": "theorem_proof", "audience_prerequisites": [ "order_basics", "distributive", "mul_zero", "forall_intro", "quantifier_negation", "cartesian_product_def" ], "goal": { "zh-Hant": "在實向量空間 ℝ² 中：（一）證明閉半空間 H={(x,y) ∣ 2x−3y≤6} 是凸集合；（二）證明非空軸對齊盒 B=[−1,2]×[0,3] 是凸集合；兩個證明都必須對所有 p、q 與所有 λ∈[0,1] 成立。（三）用一個精確的中點見證否證有限點集 F={(0,0),(2,0),(0,2)} 的凸性，並說明為何抽查有限個 λ、或找不到反例，都不能證明凸性。", "en": "In the real vector space ℝ²: (1) prove that the closed halfspace H={(x,y) ∣ 2x−3y≤6} is convex; (2) prove that the nonempty axis-aligned box B=[−1,2]×[0,3] is convex; both proofs must hold for all p, q and all λ∈[0,1]. (3) Refute the convexity of the finite point set F={(0,0),(2,0),(0,2)} with one exact midpoint witness, and explain why checking finitely many λ, or failing to find a counterexample, cannot prove convexity." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "ℝ² 是實向量空間，加法與純量乘法逐座標進行：(a,b)+(c,d)=(a+c,b+d)，t(a,b)=(ta,tb)。", "en": "ℝ² is a real vector space with coordinatewise addition and scalar multiplication: (a,b)+(c,d)=(a+c,b+d) and t(a,b)=(ta,tb)." } }, { "id": "G2", "statement": { "zh-Hant": "H={(x,y)∈ℝ² ∣ 2x−3y≤6}，法向量 (2,−3)≠(0,0) 的閉半空間。", "en": "H={(x,y)∈ℝ² ∣ 2x−3y≤6}, the closed halfspace with normal vector (2,−3)≠(0,0)." } }, { "id": "G3", "statement": { "zh-Hant": "B=[−1,2]×[0,3]={(x,y) ∣ −1≤x≤2 且 0≤y≤3}，非空軸對齊盒。", "en": "B=[−1,2]×[0,3]={(x,y) ∣ −1≤x≤2 and 0≤y≤3}, a nonempty axis-aligned box." } }, { "id": "G4", "statement": { "zh-Hant": "F={(0,0),(2,0),(0,2)}，恰有這三個元素的明示有限點集。", "en": "F={(0,0),(2,0),(0,2)}, an explicit finite point set with exactly these three elements." } } ], "steps": [ { "id": "S01", "uses": [ "G2", "mko_convex" ], "before_tex": "H\\ \\text{convex}", "after_tex": "\\forall p,q\\in H\\ \\forall\\lambda\\in[0,1]:\\ \\lambda p+(1-\\lambda)q\\in H", "relation": "equivalent", "rule": "mko_convex", "conditions": [], "explanation": { "zh-Hant": "依 MKO 的定義展開「H 是凸的」：對 H 中所有點 p、q 與 [0,1] 中所有 λ，凸組合仍在 H 中。這是一個對三個變數的全稱命題，證明必須覆蓋所有情況，不能只檢查幾個 λ。", "en": "Unfold 'H is convex' by the MKO's definition: for all points p, q in H and all λ in [0,1], the convex combination stays in H. This is a universal statement over three variables, so the proof must cover every case rather than check a few values of λ." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mko_convex", "def_unfold" ], "explanation": { "zh-Hant": "展開定義與收回定義互逆，雙向成立。", "en": "Unfolding and folding the definition undo each other, so the step holds both ways." } }, "leaf_basis": "mko_convex + def_unfold" }, { "id": "S02", "uses": [ "S01", "G2", "set_builder" ], "before_tex": "p=(x_1,y_1)\\in H,\\ q=(x_2,y_2)\\in H,\\ \\lambda\\in[0,1]", "after_tex": "2x_1-3y_1\\le 6,\\ 2x_2-3y_2\\le 6,\\ 0\\le\\lambda\\le 1", "relation": "equivalent", "rule": "set_builder", "conditions": [], "explanation": { "zh-Hant": "任取 H 中的 p、q 與 [0,1] 中的 λ，除此之外不加任何假設（以便最後做 ∀ 引入）。依集合建構式，p∈H 就是 2x₁−3y₁≤6，q∈H 就是 2x₂−3y₂≤6；λ∈[0,1] 就是 0≤λ≤1。", "en": "Take arbitrary p, q in H and λ in [0,1], with no further assumption (so that ∀-introduction applies at the end). By set-builder notation, p∈H means 2x₁−3y₁≤6, q∈H means 2x₂−3y₂≤6, and λ∈[0,1] means 0≤λ≤1." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "set_builder", "G2" ], "explanation": { "zh-Hant": "對 ℝ² 中的點，屬於 H 與滿足線性不等式是同一件事，雙向成立。", "en": "For points of ℝ², belonging to H and satisfying the linear inequality are the same thing, so the step holds both ways." } }, "leaf_basis": "set_builder + def_unfold" }, { "id": "S03", "uses": [ "G1", "S02", "coord_ops" ], "before_tex": "z=\\lambda p+(1-\\lambda)q", "after_tex": "z=(\\lambda x_1+(1-\\lambda)x_2,\\ \\lambda y_1+(1-\\lambda)y_2)", "relation": "equivalent", "rule": "coord_ops", "conditions": [], "explanation": { "zh-Hant": "依 G1 的逐座標運算：λp=(λx₁,λy₁)，(1−λ)q=((1−λ)x₂,(1−λ)y₂)，相加即得 z 的兩個座標。", "en": "By the coordinatewise operations of G1: λp=(λx₁,λy₁) and (1−λ)q=((1−λ)x₂,(1−λ)y₂); adding gives the two coordinates of z." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "coord_ops" ], "explanation": { "zh-Hant": "兩邊表示同一個有序對，等值代換雙向成立。", "en": "Both sides name the same ordered pair, so the replacement holds both ways." } }, "leaf_basis": "coord_ops + eq_replace" }, { "id": "S04", "uses": [ "S03" ], "before_tex": "2z_1-3z_2", "after_tex": "\\lambda(2x_1-3y_1)+(1-\\lambda)(2x_2-3y_2)", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "把 z₁=λx₁+(1−λ)x₂、z₂=λy₁+(1−λ)y₂ 代入 2z₁−3z₂，展開後按 λ 與 1−λ 重新分組：2λx₁+2(1−λ)x₂−3λy₁−3(1−λ)y₂=λ(2x₁−3y₁)+(1−λ)(2x₂−3y₂)。這說明線性式在凸組合上等於兩端值的同權重組合（N1 是這個多項式恆等式的核對）。", "en": "Substitute z₁=λx₁+(1−λ)x₂ and z₂=λy₁+(1−λ)y₂ into 2z₁−3z₂, expand and regroup by λ and 1−λ: 2λx₁+2(1−λ)x₂−3λy₁−3(1−λ)y₂=λ(2x₁−3y₁)+(1−λ)(2x₂−3y₂). So the linear form at a convex combination is the combination of its endpoint values with the same weights (N1 checks this polynomial identity)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "mul_comm", "add_comm" ], "explanation": { "zh-Hant": "兩式是同一個多項式，展開與分組可互相還原，對所有實數成立。", "en": "Both are the same polynomial; expanding and regrouping undo each other, valid for all reals." } }, "leaf_basis": "distributive + mul_comm + add_comm + add_assoc" }, { "id": "S05", "uses": [ "S02", "order_basics" ], "before_tex": "0\\le\\lambda\\le 1", "after_tex": "\\lambda\\ge 0,\\ 1-\\lambda\\ge 0", "relation": "implies", "rule": "order_basics", "conditions": [], "explanation": { "zh-Hant": "λ≥0 直接來自 S02。由 λ≤1，兩邊同加 −λ 得 0≤1−λ。兩個權重都非負，下一步乘不等式時方向不變就靠這一點。", "en": "λ≥0 comes directly from S02. From λ≤1, adding −λ to both sides gives 0≤1−λ. Both weights are nonnegative, which is what keeps the direction of the inequalities in the next step." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "order_basics" ], "explanation": { "zh-Hant": "只需 λ∈[0,1]。若 λ 在 [0,1] 之外，其中一個權重為負，S06 不能做；錯誤控制給出這種情形離開 H 的例子。", "en": "Only λ∈[0,1] is needed. If λ were outside [0,1], one weight would be negative and S06 could not be done; an error control gives an example that then leaves H." } }, "leaf_basis": "order_basics" }, { "id": "S06", "uses": [ "S02", "S05", "L_mono" ], "before_tex": "2x_1-3y_1\\le 6,\\ 2x_2-3y_2\\le 6", "after_tex": "\\lambda(2x_1-3y_1)\\le 6\\lambda,\\ (1-\\lambda)(2x_2-3y_2)\\le 6(1-\\lambda)", "relation": "implies", "rule": "L_mono", "conditions": [ "\\lambda\\ge 0", "1-\\lambda\\ge 0" ], "explanation": { "zh-Hant": "用引理 L_mono 兩次：第一式乘以 w=λ≥0，第二式乘以 w=1−λ≥0，不等號方向不變。引理在 S06.1、S06.2 中分 w>0 與 w=0 兩種情況證明。", "en": "Apply Lemma L_mono twice: multiply the first inequality by w=λ≥0 and the second by w=1−λ≥0; the direction stays the same. The lemma is proved in S06.1 and S06.2 by the cases w>0 and w=0." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S05", "L_mono" ], "explanation": { "zh-Hant": "需要兩個權重非負（S05）。不反向宣稱：λ=0 時第一式變成 0≤0，無法還原 2x₁−3y₁≤6。", "en": "It needs both weights nonnegative (S05). No converse is claimed: when λ=0 the first inequality becomes 0≤0, from which 2x₁−3y₁≤6 cannot be recovered." } }, "leaf_basis": "L_mono" }, { "id": "S06.1", "uses": [ "order_basics", "eq_congruence" ], "before_tex": "u\\le v,\\ w>0", "after_tex": "wu\\le wv", "relation": "implies", "rule": "order_basics", "conditions": [ "w>0" ], "explanation": { "zh-Hant": "u≤v 表示 u0，實數的序給出 wu0 the order of the reals gives wu0；u0; the cases u0（S06.1）與 w=0（本步）兩種可能，分情況論證完成。S06.1 與 S06.2 合起來證得引理 L_mono。", "en": "If w=0, multiplying by zero gives wu=0=wv, so wu≤wv also holds. w≥0 allows only w>0 (S06.1) and w=0 (this step), so the case analysis is complete. S06.1 and S06.2 together prove Lemma L_mono." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S06.1", "mul_zero", "case_analysis" ], "explanation": { "zh-Hant": "兩個情況涵蓋 w≥0 的全部可能；w<0 不在引理範圍內，且那時不等號會反向。", "en": "The two cases cover every possibility for w≥0; w<0 is outside the lemma, and there the inequality would reverse." } }, "leaf_basis": "mul_zero + case_analysis" }, { "id": "S07", "uses": [ "S04", "S06", "ineq_add" ], "before_tex": "\\lambda(2x_1-3y_1)\\le 6\\lambda,\\ (1-\\lambda)(2x_2-3y_2)\\le 6(1-\\lambda)", "after_tex": "2z_1-3z_2\\le 6\\lambda+6(1-\\lambda)=6", "relation": "implies", "rule": "ineq_add", "conditions": [], "explanation": { "zh-Hant": "兩個不等式相加，左邊依 S04 正是 2z₁−3z₂；右邊依分配律 6λ+6(1−λ)=6(λ+1−λ)=6。所以 2z₁−3z₂≤6。", "en": "Add the two inequalities; by S04 the left side is exactly 2z₁−3z₂, and by distributivity the right side is 6λ+6(1−λ)=6(λ+1−λ)=6. So 2z₁−3z₂≤6." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S04", "S06", "ineq_add", "distributive" ], "explanation": { "zh-Hant": "需要兩個同向不等式與 S04 的恆等式。相加不可逆：由和的不等式不能推回各自的不等式，所以不宣稱反向。", "en": "It needs two inequalities in the same direction and the identity of S04. Adding is not reversible: the inequality for the sum does not give back each inequality, so no converse is claimed." } }, "leaf_basis": "ineq_add + distributive + eq_replace" }, { "id": "S08", "uses": [ "S07", "S01", "S02", "forall_intro" ], "before_tex": "2z_1-3z_2\\le 6", "after_tex": "z\\in H\\ \\text{for all } p,q\\in H,\\ \\lambda\\in[0,1];\\ H\\ \\text{convex}", "relation": "implies", "rule": "forall_intro", "conditions": [], "explanation": { "zh-Hant": "依集合建構式，2z₁−3z₂≤6 表示 z∈H。p、q、λ 在 S02 中是任取的，沒有任何額外假設，所以依 ∀ 引入，S01 的全稱命題成立：H 是凸集合。這是覆蓋所有 λ∈[0,1] 的完整證書，而不是抽樣。", "en": "By set-builder notation, 2z₁−3z₂≤6 means z∈H. The p, q, λ of S02 were arbitrary, with no extra assumption, so by ∀-introduction the universal statement of S01 holds: H is convex. This is a complete certificate covering every λ∈[0,1], not a sample." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S07", "S02", "set_builder", "forall_intro" ], "explanation": { "zh-Hant": "∀ 引入要求 p、q、λ 任意；S02 至 S07 沒有用到它們的具體值，條件成立。", "en": "∀-introduction requires p, q, λ to be arbitrary; S02 to S07 used no particular values of them, so the condition holds." } }, "leaf_basis": "forall_intro + set_builder" }, { "id": "S09", "uses": [ "G3", "S03", "mko_convex", "cartesian_product_def" ], "before_tex": "B\\ \\text{convex}", "after_tex": "\\forall p,q\\in B\\ \\forall\\lambda\\in[0,1]:\\ -1\\le z_1\\le 2\\ \\land\\ 0\\le z_2\\le 3", "relation": "equivalent", "rule": "cartesian_product_def", "conditions": [], "explanation": { "zh-Hant": "依定義展開 B 的凸性，並用 B=[−1,2]×[0,3]：z∈B 等價於第一座標在 [−1,2]、第二座標在 [0,3]。依 S03（同樣的逐座標計算），z₁=λx₁+(1−λ)x₂、z₂=λy₁+(1−λ)y₂，而 p、q∈B 給出 −1≤x₁,x₂≤2 與 0≤y₁,y₂≤3。盒的條件逐座標分開，所以只需對每個座標證明區間條件。", "en": "Unfold the convexity of B by the definition and use B=[−1,2]×[0,3]: z∈B is equivalent to the first coordinate lying in [−1,2] and the second in [0,3]. By S03 (the same coordinatewise computation), z₁=λx₁+(1−λ)x₂ and z₂=λy₁+(1−λ)y₂, while p, q∈B give −1≤x₁,x₂≤2 and 0≤y₁,y₂≤3. The box conditions separate by coordinate, so it suffices to prove the interval condition in each coordinate." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "cartesian_product_def", "mko_convex", "G3" ], "explanation": { "zh-Hant": "屬於笛卡兒積等價於每個座標各自屬於對應區間，雙向成立。", "en": "Belonging to a Cartesian product is equivalent to each coordinate lying in its interval, so the step holds both ways." } }, "leaf_basis": "cartesian_product_def + def_unfold" }, { "id": "S10", "uses": [ "S09", "L_interval" ], "before_tex": "-1\\le x_1,x_2\\le 2,\\ 0\\le y_1,y_2\\le 3,\\ \\lambda\\in[0,1]", "after_tex": "-1\\le z_1\\le 2,\\ 0\\le z_2\\le 3", "relation": "implies", "rule": "L_interval", "conditions": [ "\\lambda\\in[0,1]" ], "explanation": { "zh-Hant": "對第一座標用引理 L_interval（α=−1、β=2、s=x₁、t=x₂），對第二座標用同一引理（α=0、β=3、s=y₁、t=y₂）。引理在 S10.1 至 S10.3 中對任意 α≤β 證明，所以一次涵蓋所有 λ∈[0,1]。", "en": "Apply Lemma L_interval to the first coordinate (α=−1, β=2, s=x₁, t=x₂) and to the second (α=0, β=3, s=y₁, t=y₂). The lemma is proved for arbitrary α≤β in S10.1 to S10.3, so it covers every λ∈[0,1] at once." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S09", "L_interval" ], "explanation": { "zh-Hant": "需要兩端點都在區間內與 λ∈[0,1]。不宣稱反向：z 在區間內不推出 p、q 都在區間內。", "en": "It needs both endpoints in the interval and λ∈[0,1]. No converse is claimed: z in the interval does not imply that p and q are." } }, "leaf_basis": "L_interval" }, { "id": "S10.1", "uses": [ "L_mono", "ineq_add", "S05" ], "before_tex": "\\alpha\\le s,\\ \\alpha\\le t", "after_tex": "\\lambda\\alpha+(1-\\lambda)\\alpha\\le\\lambda s+(1-\\lambda)t", "relation": "implies", "rule": "L_mono", "conditions": [ "\\lambda\\ge 0", "1-\\lambda\\ge 0" ], "explanation": { "zh-Hant": "依 L_mono（已在 S06.1、S06.2 證明），α≤s 乘以 λ≥0 得 λα≤λs；α≤t 乘以 1−λ≥0 得 (1−λ)α≤(1−λ)t（權重非負同 S05）。兩式相加得下界。", "en": "By L_mono (proved in S06.1 and S06.2), multiplying α≤s by λ≥0 gives λα≤λs, and multiplying α≤t by 1−λ≥0 gives (1−λ)α≤(1−λ)t (the weights are nonnegative as in S05). Adding the two gives the lower bound." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_mono", "ineq_add", "S05" ], "explanation": { "zh-Hant": "本步證明引理 L_interval 的一部分，只用 L_mono 與不等式相加，沒有用 L_interval 本身。", "en": "This step proves part of Lemma L_interval using only L_mono and the addition of inequalities, not L_interval itself." } }, "leaf_basis": "L_mono + ineq_add" }, { "id": "S10.2", "uses": [ "S10.1", "distributive" ], "before_tex": "\\lambda\\alpha+(1-\\lambda)\\alpha", "after_tex": "\\alpha", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "依分配律 λα+(1−λ)α=(λ+1−λ)α=1·α=α。代回 S10.1 得 α≤λs+(1−λ)t。", "en": "By distributivity λα+(1−λ)α=(λ+1−λ)α=1·α=α. Substituting into S10.1 gives α≤λs+(1−λ)t." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "mul_one" ], "explanation": { "zh-Hant": "反向：α=1·α=(λ+(1−λ))α=λα+(1−λ)α，對所有 λ 成立。", "en": "Backwards: α=1·α=(λ+(1−λ))α=λα+(1−λ)α, valid for every λ." } }, "leaf_basis": "distributive + mul_one + add_inverse" }, { "id": "S10.3", "uses": [ "S10.1", "S10.2", "L_mono", "ineq_add" ], "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": "L_mono", "conditions": [ "\\lambda\\ge 0", "1-\\lambda\\ge 0" ], "explanation": { "zh-Hant": "上界同理：s≤β 乘以 λ、t≤β 乘以 1−λ，相加，再用 S10.2 的同一計算（把 α 換成 β）得 β。S10.1 至 S10.3 合起來證得引理 L_interval。", "en": "The upper bound follows similarly: multiply s≤β by λ and t≤β by 1−λ, add, and use the computation of S10.2 with β in place of α to get β. S10.1 to S10.3 together prove Lemma L_interval." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_mono", "ineq_add", "distributive" ], "explanation": { "zh-Hant": "與 S10.1 對稱，只用 L_mono、不等式相加與分配律；沒有用 L_interval 本身。", "en": "Symmetric to S10.1, using only L_mono, the addition of inequalities and distributivity; L_interval itself is not used." } }, "leaf_basis": "L_mono + ineq_add + distributive" }, { "id": "S11", "uses": [ "S10", "S09", "forall_intro" ], "before_tex": "-1\\le z_1\\le 2,\\ 0\\le z_2\\le 3", "after_tex": "z\\in B\\ \\text{for all } p,q\\in B,\\ \\lambda\\in[0,1];\\ B\\ \\text{convex}", "relation": "implies", "rule": "forall_intro", "conditions": [], "explanation": { "zh-Hant": "兩個座標條件合起來，依 S09 的等價就是 z∈B。p、q、λ 是任取的，依 ∀ 引入，B 是凸集合。這正是 MKO 所說盒由逐座標上下界證書處理所有 λ。", "en": "The two coordinate conditions together are, by the equivalence of S09, z∈B. Since p, q, λ were arbitrary, ∀-introduction gives that B is convex. This is what the MKO means by handling all λ for a box with coordinatewise upper and lower bound certificates." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S09", "S10", "forall_intro" ], "explanation": { "zh-Hant": "需要兩個座標的結論與 p、q、λ 的任意性。", "en": "It needs both coordinate conclusions and the arbitrariness of p, q, λ." } }, "leaf_basis": "forall_intro + and_intro_elim" }, { "id": "S12", "uses": [ "G4", "mko_convex", "quantifier_negation" ], "before_tex": "\\neg(F\\ \\text{convex})", "after_tex": "\\exists p,q\\in F\\ \\exists\\lambda\\in[0,1]:\\ \\lambda p+(1-\\lambda)q\\notin F", "relation": "equivalent", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "否定凸性的定義：¬∀ 等價於 ∃¬。所以要否證 F 的凸性，只需找到一組 p、q∈F 與 λ∈[0,1]，使凸組合不在 F 中。", "en": "Negate the definition of convexity: ¬∀ is equivalent to ∃¬. So to refute the convexity of F it suffices to find one choice of p, q∈F and λ∈[0,1] whose convex combination is not in F." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "quantifier_negation", "mko_convex" ], "explanation": { "zh-Hant": "量詞否定是雙向等價。", "en": "Quantifier negation is a two-way equivalence." } }, "leaf_basis": "quantifier_negation + def_unfold" }, { "id": "S13", "uses": [ "G1", "G4", "coord_ops" ], "before_tex": "p=(0,0),\\ q=(2,0),\\ \\lambda=\\tfrac12", "after_tex": "\\tfrac12 p+\\tfrac12 q=(1,0)", "relation": "equivalent", "rule": "coord_ops", "conditions": [], "explanation": { "zh-Hant": "取 F 中的 p=(0,0)、q=(2,0) 與 λ=1/2∈[0,1]。逐座標計算：(1/2)·0+(1/2)·2=1，(1/2)·0+(1/2)·0=0，所以中點是 (1,0)。", "en": "Take p=(0,0) and q=(2,0) in F and λ=1/2∈[0,1]. Coordinatewise: (1/2)·0+(1/2)·2=1 and (1/2)·0+(1/2)·0=0, so the midpoint is (1,0)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "coord_ops", "numeric_evaluation" ], "explanation": { "zh-Hant": "精確的有理數計算，兩邊是同一個點。", "en": "An exact rational computation; both sides are the same point." } }, "leaf_basis": "coord_ops + numeric_evaluation" }, { "id": "S14", "uses": [ "S13", "G4", "cartesian_product_def" ], "before_tex": "(1,0)\\ \\text{vs.}\\ (0,0),(2,0),(0,2)", "after_tex": "(1,0)\\notin F", "relation": "implies", "rule": "cartesian_product_def", "conditions": [], "explanation": { "zh-Hant": "F 恰有三個元素。有序對相等需兩座標都相等：(1,0) 與 (0,0)、(2,0)、(0,2) 的第一座標分別是 1 對 0、1 對 2、1 對 0，都不相等。所以 (1,0) 不等於 F 的任何元素，(1,0)∉F。", "en": "F has exactly three elements. Equal ordered pairs need equal coordinates: the first coordinates of (1,0) against (0,0), (2,0), (0,2) are 1 vs 0, 1 vs 2 and 1 vs 0, all different. So (1,0) equals no element of F, and (1,0)∉F." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G4", "cartesian_product_def" ], "explanation": { "zh-Hant": "需要 F 的完整列舉（G4 明示只有三點）；若 F 只是「至少含」這三點，此步不成立。", "en": "It needs the complete listing of F (G4 states exactly three points); if F were only known to contain these three points, the step would fail." } }, "leaf_basis": "cartesian_product_def + numeric_evaluation" }, { "id": "S15", "uses": [ "S12", "S13", "S14", "exists_intro" ], "before_tex": "p=(0,0),\\ q=(2,0)\\in F,\\ \\lambda=\\tfrac12,\\ (1,0)\\notin F", "after_tex": "F\\ \\text{is not convex}", "relation": "implies", "rule": "exists_intro", "conditions": [], "explanation": { "zh-Hant": "S13、S14 給出具體的 p、q、λ 使凸組合不在 F 中；依 ∃ 引入，S12 的存在命題成立，所以 F 不是凸集合。一個精確反例就足以否證（MKO para2）。", "en": "S13 and S14 give specific p, q, λ whose convex combination is not in F; by ∃-introduction the existential statement of S12 holds, so F is not convex. One exact counterexample suffices to refute (MKO para2)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S12", "S13", "S14", "exists_intro" ], "explanation": { "zh-Hant": "需要見證中的 p、q 確實屬於 F、λ 確實在 [0,1]，且結果確實不在 F；三者都已核對。", "en": "It needs the witness points p, q to be in F, λ to be in [0,1], and the result to be outside F; all three were checked." } }, "leaf_basis": "exists_intro + quantifier_negation" }, { "id": "S16", "uses": [ "S15", "G1", "mul_zero" ], "before_tex": "\\lambda\\in\\{0,1\\}", "after_tex": "1\\cdot p+0\\cdot q=p\\in F,\\ 0\\cdot p+1\\cdot q=q\\in F", "relation": "implies", "rule": "mul_zero", "conditions": [], "explanation": { "zh-Hant": "對任何集合與其中任何 p、q，λ=1 給出 p、λ=0 給出 q，都在集合內。所以 F 通過所有 λ∈{0,1} 的抽查，但 S15 已證 F 不凸。由此可見：只抽查有限個 λ（例如端點）通過，不能推出凸性；正面結論需要 S02 至 S08 那樣覆蓋所有 λ 的證書。", "en": "For any set and any p, q in it, λ=1 gives p and λ=0 gives q, both in the set. So F passes every check with λ∈{0,1}, yet S15 proved F is not convex. Hence passing a check at finitely many λ (for example the endpoints) does not imply convexity; a positive conclusion needs a certificate covering every λ, as in S02 to S08." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S15", "mul_zero", "mul_one" ], "explanation": { "zh-Hant": "只用乘以 0 與乘以 1 的性質及 S15。結論是「抽查通過不足以證凸」，由 F 這個具體反例支持；不宣稱任何抽查都會通過。", "en": "It uses only multiplication by 0 and by 1, and S15. The conclusion, that passing samples is not enough for convexity, is supported by the concrete counterexample F; it does not claim that every sample passes." } }, "leaf_basis": "mul_zero + mul_one + add_zero" } ], "result": { "zh-Hant": "閉半空間 H={2x−3y≤6} 與盒 B=[−1,2]×[0,3] 都是凸集合：證明對所有 p、q 與所有 λ∈[0,1] 成立（線性式在凸組合上等於同權重組合，再乘以非負權重相加；盒逐座標用區間證書）。有限點集 F={(0,0),(2,0),(0,2)} 不是凸集合：(0,0) 與 (2,0) 的中點 (1,0) 不在 F 中。", "en": "The closed halfspace H={2x−3y≤6} and the box B=[−1,2]×[0,3] are convex: the proofs hold for all p, q and all λ∈[0,1] (the linear form at a convex combination equals the same-weight combination, then nonnegative weights are applied and the inequalities added; the box uses the interval certificate in each coordinate). The finite point set F={(0,0),(2,0),(0,2)} is not convex: the midpoint (1,0) of (0,0) and (2,0) is not in F." }, "negative_controls": [ { "claim": { "zh-Hant": "對 F 中所有 p、q，λ=0 與 λ=1 的凸組合都在 F 中，所以 F 是凸的", "en": "For all p, q in F, the convex combinations with λ=0 and λ=1 lie in F, so F is convex" }, "expected": "reject", "reason": { "zh-Hant": "端點抽查對任何集合都通過；λ=1/2 時 (0,0) 與 (2,0) 的中點 (1,0)∉F。", "en": "Endpoint samples pass for every set; with λ=1/2 the midpoint (1,0) of (0,0) and (2,0) is not in F." } }, { "claim": { "zh-Hant": "嘗試 p=q=(0,0)、λ=1/2 沒有得到 F 外的點，所以 F 是凸的", "en": "Trying p=q=(0,0) with λ=1/2 gives no point outside F, so F is convex" }, "expected": "reject", "reason": { "zh-Hant": "一個見證失敗不證明凸性；另一個見證 (0,0)、(2,0)、λ=1/2 成功否證。", "en": "One failed witness does not prove convexity; another witness, (0,0), (2,0) with λ=1/2, succeeds in refuting it." } }, { "claim": { "zh-Hant": "對所有實數 λ（不限於 [0,1]），H 都對 λp+(1−λ)q 封閉", "en": "H is closed under λp+(1−λ)q for every real λ, not only λ∈[0,1]" }, "expected": "reject", "reason": { "zh-Hant": "p=(3,0)、q=(0,0)、λ=2 給出 (6,0)，2·6−3·0=12>6。", "en": "p=(3,0), q=(0,0), λ=2 gives (6,0), and 2·6−3·0=12>6." } }, { "claim": { "zh-Hant": "B 的凸性必須對每個有理數 λ 分別檢查才算證明", "en": "The convexity of B is proved only by checking each rational λ separately" }, "expected": "reject", "reason": { "zh-Hant": "L_interval 對任意實數 λ∈[0,1] 一次證明，不需要也不能靠逐一檢查（λ 有無限多個）。", "en": "L_interval proves it for every real λ∈[0,1] at once; checking values one by one is neither needed nor possible, since there are infinitely many." } } ], "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 uv 之一；u0 推出 uwvw。", "en": "Order of the reals: exactly one of uv holds; u0 give uwvw." } }, { "id": "distributive", "kind": "prerequisite", "statement": { "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。", "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu." } }, { "id": "mul_zero", "kind": "prerequisite", "statement": { "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。", "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0." } }, { "id": "forall_intro", "kind": "prerequisite", "statement": { "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。", "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain." } }, { "id": "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": "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": "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_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": "eq_replace", "kind": "prerequisite", "statement": { "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。", "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true." } }, { "id": "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": "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": "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": "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": "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": "mul_one", "kind": "prerequisite", "statement": { "zh-Hant": "乘法單位元：1·u=u。", "en": "Multiplicative identity: 1·u=u." } }, { "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": "and_intro_elim", "kind": "prerequisite", "statement": { "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。", "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q." } }, { "id": "numeric_evaluation", "kind": "prerequisite", "statement": { "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。", "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable." } }, { "id": "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_zero", "kind": "prerequisite", "statement": { "zh-Hant": "加法單位元：u+0=u。", "en": "Additive identity: u+0=u." } }, { "id": "mko_convex", "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。正例用區間、盒與半空間的完整代數證書；單一精確反例可否證凸性。", "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]. Positive cases use complete algebraic certificates for intervals, boxes and halfspaces; a single exact counterexample can refute convexity." } }, { "id": "coord_ops", "kind": "prerequisite", "statement": { "zh-Hant": "逐座標運算：(a,b)+(c,d)=(a+c,b+d)，t(a,b)=(ta,tb)。", "en": "Coordinatewise operations: (a,b)+(c,d)=(a+c,b+d) and t(a,b)=(ta,tb)." } }, { "id": "ineq_add", "kind": "prerequisite", "statement": { "zh-Hant": "不等式相加：若 u₁≤v₁ 且 u₂≤v₂，則 u₁+u₂≤v₁+v₂。", "en": "Adding inequalities: if u₁≤v₁ and u₂≤v₂, then u₁+u₂≤v₁+v₂." } }, { "id": "L_mono", "kind": "lemma", "proved_in": [ "S06.1", "S06.2" ], "statement": { "zh-Hant": "乘以非負數保持不等號：若 u≤v 且 w≥0，則 wu≤wv。", "en": "Multiplying by a nonnegative number keeps the inequality: if u≤v and w≥0, then wu≤wv." } }, { "id": "L_interval", "kind": "lemma", "proved_in": [ "S10.1", "S10.2", "S10.3" ], "statement": { "zh-Hant": "區間的凸性證書：若 α≤s≤β、α≤t≤β 且 λ∈[0,1]，則 α≤λs+(1−λ)t≤β。", "en": "Convexity certificate for an interval: if α≤s≤β, α≤t≤β and λ∈[0,1], then α≤λs+(1−λ)t≤β." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S04", "vars": [ "l", "x1", "y1", "x2", "y2" ], "degree": 2, "lhs": "2*(l*x1+(1-l)*x2)-3*(l*y1+(1-l)*y2)", "rhs": "l*(2*x1-3*y1)+(1-l)*(2*x2-3*y2)" }, { "id": "N2", "step": "S07", "vars": [ "l" ], "degree": 1, "lhs": "6*l+6*(1-l)", "rhs": "6" }, { "id": "N3", "step": "S10.2", "vars": [ "a", "l" ], "degree": 2, "lhs": "l*a+(1-l)*a", "rhs": "a" }, { "id": "N4", "step": "S13", "lhs": "(1/2)*0+(1/2)*2", "rhs": "1" }, { "id": "N5", "step": "S13", "lhs": "(1/2)*0+(1/2)*0", "rhs": "0" }, { "id": "N6", "step": "S07", "lhs": "(1/3)*3+(2/3)*0", "rhs": "1" }, { "id": "N7", "step": "S07", "lhs": "(1/3)*0+(2/3)*(-2)", "rhs": "-4/3" }, { "id": "N8", "step": "S07", "lhs": "2*1-3*(-4/3)", "relation": "<=", "rhs": "6" }, { "id": "N9", "lhs": "2*6-3*0", "relation": ">", "rhs": "6" }, { "id": "N10", "step": "S02", "lhs": "2*3-3*0", "relation": "<=", "rhs": "6" }, { "id": "N11", "step": "S02", "lhs": "2*0-3*(-2)", "relation": "<=", "rhs": "6" } ] }, "presentation": { "md_scope": { "zh-Hant": "停在 MKO 的切片內：閉半空間、非空軸對齊盒（逐座標區間）與明示有限點集的反例。正面結論都用對所有 p、q、λ 成立的代數證書；不使用拓撲、凸包或分離定理；不改動 canonical MKO 或英文 edition。", "en": "Stays inside the MKO's slice: a closed halfspace, a nonempty axis-aligned box (coordinatewise intervals) and a counterexample for an explicit finite point set. Every positive conclusion uses an algebraic certificate valid for all p, q, λ; topology, convex hulls and separation theorems are not used; the canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "H", "閉半空間 2x−3y≤6", "the closed halfspace 2x−3y≤6" ], [ "B=[-1,2]\\times[0,3]", "軸對齊盒", "the axis-aligned box" ], [ "F", "三點的有限集合", "the three-point finite set" ], [ "p=(x_1,y_1),\\ q=(x_2,y_2)", "任取的兩點", "two arbitrary points" ], [ "\\lambda\\in[0,1]", "凸組合的權重", "the weight of the convex combination" ], [ "z=\\lambda p+(1-\\lambda)q", "p 與 q 的凸組合", "the convex combination of p and q" ] ], "md_error_controls": [ { "title": "抽查 λ 不能證凸 / Sampling λ does not prove convexity", "zh": "MKO 指出的誤解是檢查幾個 λ 就宣稱凸。S16 說明 λ=0 與 λ=1 對任何集合都會通過，F 通過了卻不凸。只檢查 λ=1/2 同樣是有限抽樣。H 與 B 的證明對任意 λ∈[0,1] 一次成立，這才是完整證書。", "en": "The mistake named by the MKO is to check a few λ and declare convexity. S16 shows that λ=0 and λ=1 pass for every set, and F passes them while not being convex. Checking only λ=1/2 is equally a finite sample. The proofs for H and B hold for every λ∈[0,1] at once, which is what makes them complete certificates." }, { "title": "找不到反例不等於凸 / Not finding a counterexample is not convexity", "zh": "若對 F 只嘗試 p=q=(0,0)、λ=1/2，得到 (0,0)∈F，這個見證失敗了，但 F 仍然不凸（S15）。MKO a2：提出的反例失敗，不能反向證明凸性。", "en": "If for F one only tries p=q=(0,0) with λ=1/2, the result (0,0)∈F, so this witness fails, yet F is still not convex (S15). MKO assumption a2: a failed proposed counterexample does not conversely establish convexity." }, { "title": "權重必須非負 / The weights must be nonnegative", "zh": "S06 乘不等式時方向不變，只因 λ 與 1−λ 都非負（S05）。若允許 λ=2：p=(3,0)、q=(0,0) 都在 H 中（6≤6、0≤6），但 2p+(1−2)q=(6,0)，2·6−3·0=12>6，離開了 H。所以定義中的 λ∈[0,1] 不能省略。", "en": "S06 keeps the direction of the inequalities only because λ and 1−λ are both nonnegative (S05). If λ=2 were allowed: p=(3,0) and q=(0,0) lie in H (6≤6 and 0≤6), but 2p+(1−2)q=(6,0) and 2·6−3·0=12>6, which leaves H. So the condition λ∈[0,1] in the definition cannot be dropped." } ], "md_open_items": [ { "zh-Hant": "N6 至 N8、N10、N11 是 S07 的一個具體實例（p=(3,0)、q=(0,−2)、λ=1/3，z=(1,−4/3)），只作示範，證明不依賴它們。", "en": "N6 to N8, N10 and N11 are one concrete instance of S07 (p=(3,0), q=(0,−2), λ=1/3, z=(1,−4/3)), for illustration only; the proof does not depend on them." }, { "zh-Hant": "ineq_add 與 coord_ops 是本稿的前提，不在本稿中由實數公理推出。", "en": "ineq_add and coord_ops are premises of this guide and are not derived here from the axioms of the reals." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "72519a9b66965202b14774efb4fde882285f42c4cd38475be7d32bffffbb565f", "original_source_sha256": "72519a9b66965202b14774efb4fde882285f42c4cd38475be7d32bffffbb565f", "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." } }