{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-metric-space-01", "mko_id": "mko-metric-space", "title": { "zh-Hant": "度量空間：逐格與逐三元組核對有限距離表", "en": "Metric space: auditing a finite distance table cell by cell and triple by triple" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-metric-space-01", "source": { "mko_id": "mko-metric-space", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "50cc56a6caf28683227e2793c67e48b7091d583a514abf7fc895eb8fa92898a3", "english_sha256": "e9f4ba45ed025cc5ca926ed7f051b013a56283e0432cb0ed6334d1f0bb20b71e" }, "mode": "worked_example", "audience_prerequisites": [ "order_basics", "case_analysis", "contrapositive", "numeric_evaluation" ], "goal": { "zh-Hant": "給三張完整整數距離表。（甲）X={0,1,2,3} 上的表 d（四點環上的步數距離）：逐格核非負、對角線為零、非對角線不為零、對稱，並以涵蓋全部 4³=64 個三元組的分情形論證核三角不等式，得出 (X,d) 是度量空間。（乙）Y={0,1,2} 上的表 e 通過前四項卻在 (0,1,2) 違反三角不等式。（丙）Z={0,1,2} 上的表 f 非負、對稱、對角線為零，卻有非對角零值，違反分離性。", "en": "Three complete integer distance tables are given. (A) On X={0,1,2,3}, table d (step distance on a 4-cycle): check nonnegativity, zero diagonal, nonzero off-diagonal entries and symmetry cell by cell, and the triangle inequality by a case argument covering all 4³=64 triples, concluding that (X,d) is a metric space. (B) Table e on Y={0,1,2} passes the first four checks but violates the triangle inequality at (0,1,2). (C) Table f on Z={0,1,2} is nonnegative, symmetric and zero on the diagonal, yet has an off-diagonal zero and violates separation." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "候選甲：X={0,1,2,3}，表 d 的第 i 行（row）依序列出 d(i,0)、d(i,1)、d(i,2)、d(i,3)：第 0 行 0,1,2,1；第 1 行 1,0,1,2；第 2 行 2,1,0,1；第 3 行 1,2,1,0。", "en": "Candidate A: X={0,1,2,3}; row i of table d lists d(i,0), d(i,1), d(i,2), d(i,3) in order: row 0 is 0,1,2,1; row 1 is 1,0,1,2; row 2 is 2,1,0,1; row 3 is 1,2,1,0." } }, { "id": "G2", "statement": { "zh-Hant": "候選乙：Y={0,1,2}，表 e 的三行（row）為 0,1,3；1,0,1；3,1,0。", "en": "Candidate B: Y={0,1,2}; the three rows of table e are 0,1,3; 1,0,1; 3,1,0." } }, { "id": "G3", "statement": { "zh-Hant": "候選丙：Z={0,1,2}，表 f 的三行（row）為 0,0,2；0,0,2；2,2,0。", "en": "Candidate C: Z={0,1,2}; the three rows of table f are 0,0,2; 0,0,2; 2,2,0." } } ], "steps": [ { "id": "S01", "uses": [ "mko_metric_def" ], "before_tex": "d:X\\times X\\to\\mathbb{R}", "after_tex": "d\\ge 0;\\ d(x,y)=0\\Leftrightarrow x=y;\\ d(x,y)=d(y,x);\\ d(x,z)\\le d(x,y)+d(y,z)", "relation": "definition", "rule": "metric_def", "conditions": [], "explanation": { "zh-Hant": "依 MKO，度量要滿足四項，且每一項都對所有點（或點對、三元組）成立：非負；分離（距離為零當且僅當兩點相同）；對稱；三角不等式。分離包含兩個方向：對角線為零，以及非對角線不為零。", "en": "By the MKO a metric must satisfy four conditions, each for all points (or pairs, or triples): nonnegativity; separation (distance zero exactly when the points coincide); symmetry; the triangle inequality. Separation has two directions: a zero diagonal and nonzero off-diagonal entries." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_metric_def", "metric_def" ], "explanation": { "zh-Hant": "此步只列出要核的條件，不推出新事實；四項缺一不可。", "en": "This step only lists the conditions to check and derives no new fact; none of the four may be skipped." } }, "leaf_basis": "metric_def + mko_metric_def" }, { "id": "S02", "uses": [ "G1" ], "before_tex": "d(i,j),\\ i,j\\in X", "after_tex": "d(i,j)\\in\\{0,1,2\\},\\ d(i,j)\\ge 0", "relation": "implies", "rule": "finite_exhaustion", "conditions": [], "explanation": { "zh-Hant": "逐格查 16 個值：每個都是 0、1 或 2，因此都非負。同時記下兩個之後要用的界：所有值都 ≤2。", "en": "Inspect all 16 values: each is 0, 1 or 2, so all are nonnegative. Also note a bound used later: every value is at most 2." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "finite_exhaustion" ], "explanation": { "zh-Hant": "16 格逐一查完才算核完非負；這一步不涉及分離或三角不等式。", "en": "Nonnegativity is established only after all 16 cells are inspected; this step says nothing about separation or the triangle inequality." } }, "leaf_basis": "finite_exhaustion + order_basics" }, { "id": "S03", "uses": [ "G1" ], "before_tex": "d(i,i),\\ i\\in X", "after_tex": "x=y\\Rightarrow d(x,y)=0", "relation": "implies", "rule": "finite_exhaustion", "conditions": [], "explanation": { "zh-Hant": "對角線 d(0,0)、d(1,1)、d(2,2)、d(3,3) 都是 0，所以相同的點距離為零。這是分離的一個方向。", "en": "The diagonal entries d(0,0), d(1,1), d(2,2), d(3,3) are all 0, so a point has distance zero from itself. This is one direction of separation." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "finite_exhaustion" ], "explanation": { "zh-Hant": "只查了對角線四格；反向（距離為零推出同一點）不由此得到，留給 S04。", "en": "Only the four diagonal cells are inspected; the converse (zero distance forces the same point) does not follow and is left to S04." } }, "leaf_basis": "finite_exhaustion" }, { "id": "S04", "uses": [ "G1", "S02" ], "before_tex": "d(i,j),\\ i\\neq j", "after_tex": "d(x,y)=0\\Rightarrow x=y", "relation": "implies", "rule": "contrapositive", "conditions": [], "explanation": { "zh-Hant": "非對角線的 12 格都是 1 或 2，所以 x≠y 時 d(x,y)≥1，特別是 d(x,y)≠0。依逆否推理，這等價於「d(x,y)=0 推出 x=y」。", "en": "The 12 off-diagonal cells are all 1 or 2, so for x≠y we have d(x,y)≥1, in particular d(x,y)≠0. By contraposition this is equivalent to 'd(x,y)=0 implies x=y'." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "finite_exhaustion", "contrapositive" ], "explanation": { "zh-Hant": "必須查遍 12 個非對角格；若有一格為 0，分離就失敗（見候選丙）。此步同時記下 x≠y 時 d(x,y)≥1，供 S07.3 使用。", "en": "All 12 off-diagonal cells must be inspected; a single zero there would break separation (see candidate C). The step also records d(x,y)≥1 for x≠y, used in S07.3." } }, "leaf_basis": "finite_exhaustion + contrapositive" }, { "id": "S05", "uses": [ "S03", "S04" ], "before_tex": "x=y\\Rightarrow d(x,y)=0,\\ d(x,y)=0\\Rightarrow x=y", "after_tex": "d(x,y)=0\\Leftrightarrow x=y", "relation": "implies", "rule": "iff_intro", "conditions": [], "explanation": { "zh-Hant": "S03 與 S04 兩個方向合起來，就是分離性。MKO 第一段正是說：對角線須為零、非對角線不得為零，二者共同構成分離性。", "en": "S03 and S04 together give both directions, which is separation. This is the MKO's first paragraph: the diagonal must be zero and off-diagonal entries must not be zero, and together they make up separation." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "S04", "iff_intro" ], "explanation": { "zh-Hant": "雙條件需要兩個方向各自成立，兩者分別由 S03、S04 提供。", "en": "A biconditional needs both directions, supplied by S03 and S04 respectively." } }, "leaf_basis": "iff_intro" }, { "id": "S06", "uses": [ "G1" ], "before_tex": "d(i,j)\\ \\text{vs}\\ d(j,i)", "after_tex": "\\forall i,j\\in X:\\ d(i,j)=d(j,i)", "relation": "implies", "rule": "finite_exhaustion", "conditions": [], "explanation": { "zh-Hant": "比對 6 組對稱位置：d(0,1)=d(1,0)=1，d(0,2)=d(2,0)=2，d(0,3)=d(3,0)=1，d(1,2)=d(2,1)=1，d(1,3)=d(3,1)=2，d(2,3)=d(3,2)=1；對角線自然對稱。表等於它的轉置。", "en": "Compare the 6 mirror pairs: d(0,1)=d(1,0)=1, d(0,2)=d(2,0)=2, d(0,3)=d(3,0)=1, d(1,2)=d(2,1)=1, d(1,3)=d(3,1)=2, d(2,3)=d(3,2)=1; the diagonal is trivially symmetric. The table equals its transpose." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "finite_exhaustion" ], "explanation": { "zh-Hant": "6 組非對角對都查過，加上 4 個對角格，涵蓋全部 16 個有序對。", "en": "All 6 off-diagonal pairs are compared, and with the 4 diagonal cells all 16 ordered pairs are covered." } }, "leaf_basis": "finite_exhaustion" }, { "id": "S07", "uses": [ "S02", "S03", "S04" ], "before_tex": "(i,j,k)\\in X^3", "after_tex": "i=k\\ \\lor\\ (i\\neq k\\land(j=i\\lor j=k))\\ \\lor\\ i,j,k\\ \\text{pairwise distinct}", "relation": "case_split", "rule": "case_analysis", "conditions": [], "explanation": { "zh-Hant": "三角不等式 d(i,k)≤d(i,j)+d(j,k) 要對全部 64 個有序三元組成立。把三元組分成三種情形：A：i=k；B：i≠k 且 j 等於 i 或 k；C：i、j、k 兩兩相異。子步 S07.1 到 S07.3 逐一處理，S07.4 確認三種情形恰好涵蓋全部 64 個三元組。", "en": "The triangle inequality d(i,k)≤d(i,j)+d(j,k) must hold for all 64 ordered triples. Split the triples into three cases: A: i=k; B: i≠k and j equals i or k; C: i, j, k pairwise distinct. Sub-steps S07.1 to S07.3 treat each case, and S07.4 confirms that the three cases cover exactly the 64 triples." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "case_analysis", "S02", "S03", "S04" ], "explanation": { "zh-Hant": "分情形證明要求情形涵蓋所有可能；這由 S07.4 的計數核對。每個情形只用到 S02 到 S04 已核的表性質。", "en": "Proof by cases needs the cases to cover every possibility, which S07.4 checks by counting. Each case uses only table properties already established in S02 to S04." } }, "leaf_basis": "case_analysis" }, { "id": "S07.1", "uses": [ "S07", "S02", "S03" ], "before_tex": "i=k", "after_tex": "d(i,k)=0\\le d(i,j)+d(j,i)", "relation": "implies", "rule": "order_basics", "conditions": [ "i=k" ], "explanation": { "zh-Hant": "情形 A：i=k，左邊 d(i,i)=0（S03）；右邊是兩個非負數之和（S02），所以 ≥0。0≤右邊，不等式成立。", "en": "Case A: i=k, so the left side is d(i,i)=0 (S03); the right side is a sum of two nonnegative numbers (S02), hence ≥0. Since 0≤right side, the inequality holds." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "S03", "order_basics" ], "explanation": { "zh-Hant": "需要對角線為零與非負；不需要查任何具體的 j。", "en": "It needs the zero diagonal and nonnegativity; no particular j has to be inspected." } }, "leaf_basis": "order_basics + add_zero" }, { "id": "S07.2", "uses": [ "S07", "S03" ], "before_tex": "i\\neq k,\\ j\\in\\{i,k\\}", "after_tex": "d(i,j)+d(j,k)=d(i,k)", "relation": "implies", "rule": "eq_replace", "conditions": [ "i\\neq k" ], "explanation": { "zh-Hant": "情形 B：若 j=i，右邊為 d(i,i)+d(i,k)=0+d(i,k)=d(i,k)；若 j=k，右邊為 d(i,k)+d(k,k)=d(i,k)+0=d(i,k)。兩種子情形中右邊都恰好等於左邊，所以 d(i,k)≤右邊以等號成立。", "en": "Case B: if j=i the right side is d(i,i)+d(i,k)=0+d(i,k)=d(i,k); if j=k it is d(i,k)+d(k,k)=d(i,k)+0=d(i,k). In both sub-cases the right side equals the left, so d(i,k)≤right side holds with equality." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "eq_replace", "add_zero" ], "explanation": { "zh-Hant": "只用對角線為零與 0 的加法性質；因 i≠k，j 不能同時等於 i 與 k，兩個子情形互斥。", "en": "Only the zero diagonal and the additive identity are used; since i≠k, j cannot equal both i and k, so the two sub-cases are exclusive." } }, "leaf_basis": "eq_replace + add_zero + order_basics" }, { "id": "S07.3", "uses": [ "S07", "S02", "S04" ], "before_tex": "i,j,k\\ \\text{pairwise distinct}", "after_tex": "d(i,k)\\le 2\\le d(i,j)+d(j,k)", "relation": "implies", "rule": "order_basics", "conditions": [ "i\\neq j,\\ j\\neq k,\\ i\\neq k" ], "explanation": { "zh-Hant": "情形 C：i≠j 與 j≠k，由 S04，d(i,j)≥1 且 d(j,k)≥1，所以右邊 ≥1+1=2；由 S02，每個距離 ≤2，所以 d(i,k)≤2。依序的傳遞，d(i,k)≤右邊。", "en": "Case C: since i≠j and j≠k, S04 gives d(i,j)≥1 and d(j,k)≥1, so the right side is at least 1+1=2; by S02 every distance is at most 2, so d(i,k)≤2. By transitivity of order, d(i,k)≤right side." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "S04", "order_basics" ], "explanation": { "zh-Hant": "這個論證依賴表 d 的特性：非對角值的最小值 1 的兩倍不小於最大值 2。對候選乙不成立（最大值 3 大於 1+1），所以不能照搬。", "en": "The argument depends on a feature of table d: twice the smallest off-diagonal value, 1, is at least the largest value, 2. This fails for candidate B (largest value 3 exceeds 1+1), so it cannot be transferred." } }, "leaf_basis": "order_basics + numeric_evaluation" }, { "id": "S07.4", "uses": [ "S07" ], "before_tex": "|A|+|B|+|C|", "after_tex": "16+24+24=64=4^3", "relation": "implies", "rule": "finite_counting_basics", "conditions": [], "explanation": { "zh-Hant": "三種情形互斥且窮盡：若 i=k 屬 A；否則 i≠k，此時 j 等於 i 或 k 屬 B，與兩者都不同屬 C。計數：A 有 4·4=16 個（選 i 與 j）；B 有 2·(4·3)=24 個（選相異的 i、k，再讓 j 等於其中之一）；C 有 4·3·2=24 個。合計 64=4³，正是全部有序三元組。", "en": "The three cases are exclusive and exhaustive: if i=k the triple is in A; otherwise i≠k, and it is in B if j equals i or k, and in C if j differs from both. Counting: A has 4·4=16 triples (choose i and j); B has 2·(4·3)=24 (choose distinct i, k, then let j equal one of them); C has 4·3·2=24. The total 64=4³ is exactly the number of ordered triples." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "finite_counting_basics", "numeric_evaluation" ], "explanation": { "zh-Hant": "互斥性由情形定義保證，總數相符確認沒有漏掉任何三元組（N6、N7）。", "en": "Exclusivity follows from the definitions of the cases, and the matching total confirms that no triple is missed (N6, N7)." } }, "leaf_basis": "finite_counting_basics + numeric_evaluation" }, { "id": "S08", "uses": [ "S07", "S07.1", "S07.2", "S07.3", "S07.4" ], "before_tex": "\\text{cases A, B, C}", "after_tex": "\\forall i,j,k\\in X:\\ d(i,k)\\le d(i,j)+d(j,k)", "relation": "implies", "rule": "case_analysis", "conditions": [], "explanation": { "zh-Hant": "三種情形都已證明不等式成立，且涵蓋全部 64 個三元組，依分情形證明，三角不等式對 X 的每個有序三元組成立。N1 另以公式 d(i,j)=min(|i−j|, 4−|i−j|) 窮舉 64 個三元組作交叉核對。", "en": "The inequality is proved in all three cases, which cover all 64 triples, so by proof by cases the triangle inequality holds for every ordered triple of X. N1 additionally enumerates all 64 triples with the formula d(i,j)=min(|i−j|, 4−|i−j|) as a cross-check." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S07.1", "S07.2", "S07.3", "S07.4", "case_analysis" ], "explanation": { "zh-Hant": "分情形證明需要每個情形都有證明，且情形窮盡；兩者分別由 S07.1 到 S07.3 與 S07.4 提供。", "en": "Proof by cases needs a proof in every case and exhaustive cases; S07.1 to S07.3 and S07.4 supply these." } }, "leaf_basis": "case_analysis + finite_exhaustion" }, { "id": "S09", "uses": [ "S02", "S05", "S06", "S08" ], "before_tex": "\\text{nonnegativity, separation, symmetry, triangle inequality}", "after_tex": "(X,d)\\ \\text{is a metric space}", "relation": "implies", "rule": "mko_metric_def", "conditions": [], "explanation": { "zh-Hant": "四項條件分別由 S02、S05、S06、S08 給出，所以 (X,d) 是度量空間。結論只屬於這個指定的有限模型。", "en": "The four conditions come from S02, S05, S06 and S08, so (X,d) is a metric space. The conclusion concerns this specified finite model only." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S02", "S05", "S06", "S08", "mko_metric_def" ], "explanation": { "zh-Hant": "目標 G：(X,d) 是度量空間。足夠條件 H：四項條件同時成立；H⇒G 由定義。H 由表 d 正向證成，沒有把 G 當作前提。", "en": "Goal G: (X,d) is a metric space. Sufficient condition H: the four conditions hold together; H⇒G by the definition. H is proved forwards from table d; G is never used as a premise." } }, "leaf_basis": "mko_metric_def + and_intro_elim" }, { "id": "S10", "uses": [ "G2" ], "before_tex": "e\\ \\text{on } Y=\\{0,1,2\\}", "after_tex": "e\\ge 0,\\ e(x,y)=0\\Leftrightarrow x=y,\\ e(x,y)=e(y,x)", "relation": "implies", "rule": "finite_exhaustion", "conditions": [], "explanation": { "zh-Hant": "逐格查候選乙：9 個值都非負；對角線三格為 0；非對角線六格是 1、3、1、1、3、1，都不為 0；e(0,1)=e(1,0)=1，e(0,2)=e(2,0)=3，e(1,2)=e(2,1)=1，表對稱。前三項全部通過。", "en": "Inspect candidate B cell by cell: all 9 values are nonnegative; the three diagonal cells are 0; the six off-diagonal cells are 1, 3, 1, 1, 3, 1, none of them 0; e(0,1)=e(1,0)=1, e(0,2)=e(2,0)=3, e(1,2)=e(2,1)=1, so the table is symmetric. The first three conditions all pass." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G2", "finite_exhaustion" ], "explanation": { "zh-Hant": "這些都是逐格檢查，不涉及三元組，所以對三角不等式沒有任何保證。", "en": "These are cell-by-cell checks involving no triples, so they guarantee nothing about the triangle inequality." } }, "leaf_basis": "finite_exhaustion" }, { "id": "S11", "uses": [ "G2" ], "before_tex": "e(0,2)\\ \\text{vs}\\ e(0,1)+e(1,2)", "after_tex": "3>1+1=2", "relation": "implies", "rule": "order_basics", "conditions": [], "explanation": { "zh-Hant": "取三元組 (0,1,2)：e(0,2)=3，而 e(0,1)+e(1,2)=1+1=2。3>2，所以三角不等式在這個三元組失敗：經過點 1 的路比直接距離更短。", "en": "Take the triple (0,1,2): e(0,2)=3 while e(0,1)+e(1,2)=1+1=2. Since 3>2, the triangle inequality fails at this triple: going through point 1 is shorter than the direct distance." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G2", "order_basics", "numeric_evaluation" ], "explanation": { "zh-Hant": "一個反例就足以否定對所有三元組的全稱命題。", "en": "One counterexample suffices to refute the universal statement over all triples." } }, "leaf_basis": "order_basics + numeric_evaluation" }, { "id": "S12", "uses": [ "S10", "S11" ], "before_tex": "\\exists(i,j,k):\\ e(i,k)>e(i,j)+e(j,k)", "after_tex": "(Y,e)\\ \\text{is not a metric space}", "relation": "implies", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "三角不等式是全稱命題；S11 給出反例，依量詞否定它為假，所以 (Y,e) 不是度量空間，儘管 S10 的非負、分離、對稱全部通過。", "en": "The triangle inequality is a universal statement; S11 gives a counterexample, so by quantifier negation it is false and (Y,e) is not a metric space, although nonnegativity, separation and symmetry in S10 all pass." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S11", "quantifier_negation", "mko_metric_def" ], "explanation": { "zh-Hant": "目標 G：e 不是度量。足夠條件 H：某個三元組違反三角不等式；H⇒G 由定義。H 由 S11 直接計算，沒有把 G 當作前提。", "en": "Goal G: e is not a metric. Sufficient condition H: some triple violates the triangle inequality; H⇒G by the definition. H is computed directly in S11; G is never used as a premise." } }, "leaf_basis": "quantifier_negation + mko_metric_def" }, { "id": "S13", "uses": [ "G3" ], "before_tex": "f(0,1)=0,\\ 0\\neq 1", "after_tex": "(Z,f)\\ \\text{is not a metric space}", "relation": "implies", "rule": "mko_metric_def", "conditions": [], "explanation": { "zh-Hant": "候選丙的 9 個值都非負、表對稱、對角線為零；但 f(0,1)=0 而 0≠1，違反「d(x,y)=0 推出 x=y」，分離失敗，所以不是度量（這種表在文獻中稱為偽度量）。只看非負、對稱與對角線會漏掉它。", "en": "Candidate C has all 9 values nonnegative, a symmetric table and a zero diagonal; but f(0,1)=0 although 0≠1, violating 'd(x,y)=0 implies x=y'. Separation fails, so f is not a metric (such a table is usually called a pseudometric). Checking only nonnegativity, symmetry and the diagonal would miss this." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "G3", "mko_metric_def", "metric_def" ], "explanation": { "zh-Hant": "目標 G：f 不是度量。足夠條件 H：一個非對角零值（H⇒G 由分離的定義）。H 直接由 G3 的表讀出，沒有把 G 當作前提。", "en": "Goal G: f is not a metric. Sufficient condition H: one off-diagonal zero (H⇒G by the definition of separation). H is read directly from the table in G3; G is never used as a premise." } }, "leaf_basis": "mko_metric_def + metric_def + quantifier_negation" } ], "result": { "zh-Hant": "(X,d) 是度量空間：16 格非負、對角線為零且非對角線為 1 或 2（分離）、表對稱，三角不等式在全部 64 個三元組成立（分三種情形證明）。(Y,e) 通過非負、分離與對稱，但 e(0,2)=3>e(0,1)+e(1,2)=2，不是度量。(Z,f) 非負、對稱、對角線為零，但 f(0,1)=0 而 0≠1，分離失敗，不是度量。三個結論都只屬於指定的有限模型。", "en": "(X,d) is a metric space: all 16 cells are nonnegative, the diagonal is zero and off-diagonal entries are 1 or 2 (separation), the table is symmetric, and the triangle inequality holds for all 64 triples (proved in three cases). (Y,e) passes nonnegativity, separation and symmetry, but e(0,2)=3>e(0,1)+e(1,2)=2, so it is not a metric. (Z,f) is nonnegative, symmetric and zero on the diagonal, but f(0,1)=0 with 0≠1, so separation fails and it is not a metric. All three conclusions concern only the specified finite models." }, "negative_controls": [ { "claim": { "zh-Hant": "e 非負、對稱、對角線為零且非對角線不為零，所以 e 是度量", "en": "e is nonnegative, symmetric, zero on the diagonal and nonzero off it, so e is a metric" }, "expected": "reject", "reason": { "zh-Hant": "三角不等式在 (0,1,2) 失敗：3>1+1。", "en": "The triangle inequality fails at (0,1,2): 3>1+1." } }, { "claim": { "zh-Hant": "f 的值都非負、表對稱、對角線為零，所以 f 是度量", "en": "f has nonnegative values, a symmetric table and a zero diagonal, so f is a metric" }, "expected": "reject", "reason": { "zh-Hant": "f(0,1)=0 而 0≠1，分離失敗。", "en": "f(0,1)=0 although 0≠1, so separation fails." } }, { "claim": { "zh-Hant": "既然 (X,d) 的 64 個三元組都通過，任何集合上同類型的距離表都滿足三角不等式", "en": "Since all 64 triples of (X,d) pass, every distance table of this kind on any set satisfies the triangle inequality" }, "expected": "reject", "reason": { "zh-Hant": "有限窮舉只證指定模型（MKO 假設 a2）；候選乙就是失敗的例子。", "en": "A finite enumeration proves only the specified model (MKO assumption a2); candidate B is a failing example." } }, { "claim": { "zh-Hant": "三點載體上，只要核 ih(1,0)+h(0,2)=2。", "en": "Let h(0,1)=1, h(0,2)=1, h(1,2)=3 (symmetric, zero diagonal): (0,1,2) gives 1≤1+3 and passes, but (1,0,2) gives h(1,2)=3>h(1,0)+h(0,2)=2." } } ], "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": "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": "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": "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": "metric_def", "kind": "prerequisite", "statement": { "zh-Hant": "度量：d:X×X→ℝ 滿足 d(x,y)≥0 且 d(x,y)=0 ⇔ x=y；d(x,y)=d(y,x)；三角不等式 d(x,z)≤d(x,y)+d(y,z)。", "en": "Metric: d:X×X→ℝ with d(x,y)≥0 and d(x,y)=0 ⇔ x=y; d(x,y)=d(y,x); triangle inequality d(x,z)≤d(x,y)+d(y,z)." } }, { "id": "iff_intro", "kind": "prerequisite", "statement": { "zh-Hant": "雙條件：P⇒Q 與 Q⇒P 都成立時，P⇔Q 成立；反之亦然。", "en": "Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold." } }, { "id": "add_zero", "kind": "prerequisite", "statement": { "zh-Hant": "加法單位元：u+0=u。", "en": "Additive identity: u+0=u." } }, { "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": "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": "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": "quantifier_negation", "kind": "prerequisite", "statement": { "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。", "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)." } }, { "id": "mko_metric_def", "kind": "cited", "source": { "mko_id": "mko-metric-space", "sha256": "50cc56a6caf28683227e2793c67e48b7091d583a514abf7fc895eb8fa92898a3", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "集合 X 上的度量 d:X×X→R 滿足非負與分離、對稱及三角不等式。完整有限距離表可逐格及逐三元組核對指定候選。", "en": "A metric d:X×X→R on a set X satisfies nonnegativity and separation, symmetry and the triangle inequality. A complete finite distance table permits cell-by-cell and triple-by-triple auditing of a specified candidate." } }, { "id": "finite_exhaustion", "kind": "prerequisite", "statement": { "zh-Hant": "在有限集合上，對全部元素（或全部有序對、三元組）的命題成立，當且僅當它在列出的每一個情形都成立；逐一核完每個情形是該有限模型的完整證明，但不推及其他集合。", "en": "On a finite set, a statement about all elements (or all ordered pairs or triples) holds exactly when it holds in each of the finitely many listed cases; checking every case is a complete proof for that finite model, but says nothing about other sets." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S08", "vars": [ "i", "j", "k" ], "range": [ 0, 3 ], "lhs": "min(abs(i-k),4-abs(i-k))", "relation": "<=", "rhs": "min(abs(i-j),4-abs(i-j))+min(abs(j-k),4-abs(j-k))" }, { "id": "N2", "step": "S02", "lhs": "min(abs(0-3),4-abs(0-3))", "rhs": "1" }, { "id": "N3", "step": "S02", "lhs": "min(abs(1-3),4-abs(1-3))", "rhs": "2" }, { "id": "N4", "step": "S02", "lhs": "min(abs(0-2),4-abs(0-2))", "rhs": "2" }, { "id": "N5", "step": "S07.3", "lhs": "1+1", "relation": ">=", "rhs": "2" }, { "id": "N6", "step": "S07.4", "lhs": "4*4+2*(4*3)+4*3*2", "rhs": "4**3" }, { "id": "N7", "step": "S07.4", "lhs": "4*3*2", "rhs": "24" }, { "id": "N8", "step": "S11", "lhs": "3", "relation": ">", "rhs": "1+1" }, { "id": "N9", "lhs": "3", "relation": ">", "rhs": "1+1" }, { "id": "N10", "lhs": "1", "relation": "<=", "rhs": "1+3" } ] }, "presentation": { "md_scope": { "zh-Hant": "只處理載體標為 0 到 n−1、距離表完整且為整數的有限模型（MKO 假設 a1）；結論只屬於這三個指定模型，不外推到任意或無限度量空間（MKO 假設 a2）。不處理開球、收斂或完備性；不改動 canonical MKO 或英文 edition。", "en": "Treats only finite carriers labeled 0 to n−1 with complete integer distance tables (MKO assumption a1); the conclusions concern only these three specified models and are not extended to arbitrary or infinite metric spaces (MKO assumption a2). Open balls, convergence and completeness are not treated; the canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "X=\\{0,1,2,3\\}", "候選甲的載體", "the carrier of candidate A" ], [ "d(i,j)", "表 d 第 i 行（row）第 j 列（column）的距離", "the distance in row i, column j of table d" ], [ "e,\\ f", "候選乙、丙的距離表", "the distance tables of candidates B and C" ], [ "(i,j,k)", "三角不等式要核的有序三元組", "an ordered triple checked in the triangle inequality" ] ], "md_error_controls": [ { "title": "只核非負與對稱不夠 / Nonnegativity and symmetry are not enough", "zh": "MKO 指出的誤解是只核非負與對稱就叫度量。候選乙通過非負與對稱（甚至分離），卻違反三角不等式；候選丙通過非負與對稱，卻違反分離。四項都要核。", "en": "The MKO names the mistake of calling a table a metric after checking only nonnegativity and symmetry. Candidate B passes nonnegativity and symmetry (even separation) yet violates the triangle inequality; candidate C passes nonnegativity and symmetry yet violates separation. All four conditions must be checked." }, { "title": "分離包含非對角線不為零 / Separation includes nonzero off-diagonal entries", "zh": "對角線為零只是分離的一半（S03）。MKO 提醒漏掉非對角零值的錯誤：候選丙的 f(0,1)=0 使兩個不同的點距離為零。S04 必須查遍全部非對角格。", "en": "A zero diagonal is only half of separation (S03). The MKO warns against overlooking off-diagonal zeros: in candidate C, f(0,1)=0 gives two distinct points distance zero. S04 must inspect every off-diagonal cell." }, { "title": "三角不等式要涵蓋全部有序三元組 / The triangle inequality must cover every ordered triple", "zh": "在候選乙中，含重複點的三元組都會通過（理由同 S07.1、S07.2），失敗只出現在相異三元組 (0,1,2) 與 (2,1,0)。抽查或只查 i