{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-metric-convergence-01",
  "mko_id": "mko-metric-convergence",
  "title": {
    "zh-Hant": "度量空間中的收斂：有限度量表上最終常值序列的極限與唯一性",
    "en": "Convergence in a metric space: limit and uniqueness for an ultimately constant sequence on a finite metric table"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-metric-convergence-01",
    "source": {
      "mko_id": "mko-metric-convergence",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "185db7b6b678014d6791d0448e2eccc1efc643b288d9b95c2b7b91f3ea710969",
      "english_sha256": "8970fc590fd89152780c85bc3235042b210a5536388a57ef8b5ea934bc5eb1d5"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "metric_def",
      "order_basics",
      "quantifier_negation",
      "forall_intro",
      "exists_intro"
    ],
    "goal": {
      "zh-Hant": "在三點集合 X={a,b,c} 上給定完整度量表 d(a,b)=1、d(a,c)=2、d(b,c)=3/2。序列 x_0=a、x_1=c、x_2=a，並明示對所有 n≥3 有 x_n=b。(1) 核對 d 是度量；(2) 以 ε-N 定義證明 x_n→b，並找出各 ε 下最小的有效 N；(3) 證明度量空間中極限唯一，並直接給出 x_n 不收斂到 a、也不收斂到 c 的見證；(4) 用一個在有限尾段上與 x_n 完全一致、極限卻不同的序列，說明只核有限尾段不足以判定極限是否為指定的 b。",
      "en": "On the three-point set X={a,b,c} a complete metric table is given: d(a,b)=1, d(a,c)=2, d(b,c)=3/2. The sequence has x_0=a, x_1=c, x_2=a, and it is declared that x_n=b for every n≥3. (1) Check that d is a metric; (2) prove x_n→b from the ε-N definition and find the least valid N for each ε; (3) prove that limits in a metric space are unique, and give direct witnesses that x_n converges neither to a nor to c; (4) using a sequence that agrees with x_n on a finite tail but has a different limit, show that checking a finite tail cannot decide whether the limit is the specified b."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "X={a,b,c}，a、b、c 是三個相異標籤。度量表：對每個 p∈X 有 d(p,p)=0；d(a,b)=d(b,a)=1；d(a,c)=d(c,a)=2；d(b,c)=d(c,b)=3/2（MKO 假設 a1：完整有限度量表）。",
          "en": "X={a,b,c} with three distinct labels a, b, c. Metric table: d(p,p)=0 for every p∈X; d(a,b)=d(b,a)=1; d(a,c)=d(c,a)=2; d(b,c)=d(c,b)=3/2 (MKO assumption a1: a complete finite metric table)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "序列 (x_n)_{n∈ℕ0}：x_0=a、x_1=c、x_2=a，並明示對每一個 n≥3 有 x_n=b（MKO 假設 a2：最終常值模型，前綴長度 3，尾值 b）。",
          "en": "The sequence (x_n)_{n∈ℕ0}: x_0=a, x_1=c, x_2=a, and it is declared that x_n=b for every n≥3 (MKO assumption a2: ultimately constant model with prefix length 3 and tail value b)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "比較序列 (y_n)_{n∈ℕ0}：對 0≤n≤10 有 y_n=x_n，並明示對每一個 n≥11 有 y_n=c。",
          "en": "Comparison sequence (y_n)_{n∈ℕ0}: y_n=x_n for 0≤n≤10, and it is declared that y_n=c for every n≥11."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "mko_metric_conv"
        ],
        "before_tex": "x_n\\to p\\ \\text{in}\\ (X,d)",
        "after_tex": "\\forall\\varepsilon>0\\ \\exists N\\in\\mathbb{N}_0\\ \\forall n\\ge N:\\ d(x_n,p)<\\varepsilon",
        "relation": "definition",
        "rule": "mko_metric_conv",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 的定義展開「x_n 收斂到 p」：先給任意 ε>0，然後才找 N，再要求所有 n≥N 都有 d(x_n,p)<ε。量詞順序是 ∀ε ∃N ∀n，不等號是嚴格的 <（開球）。之後每一個收斂或不收斂的判斷都回到這個式子。",
          "en": "Unfold 'x_n converges to p' by the MKO's definition: an arbitrary ε>0 is given first, then N is sought, and every n≥N must satisfy d(x_n,p)<ε. The quantifier order is ∀ε ∃N ∀n and the inequality is strict (an open ball). Every later claim of convergence or non-convergence returns to this formula."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_metric_conv"
          ],
          "explanation": {
            "zh-Hant": "這一步只展開定義，不產生新事實；定義預設 (X,d) 是度量空間，這在 S02 核對。",
            "en": "This step only unfolds the definition and produces no new fact; the definition presupposes that (X,d) is a metric space, which S02 checks."
          }
        },
        "leaf_basis": "mko_metric_conv + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "L_metric_table"
        ],
        "before_tex": "d:\\ \\text{the table of G1}",
        "after_tex": "(X,d)\\ \\text{is a metric space}",
        "relation": "implies",
        "rule": "L_metric_table",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 假設 a1 使用完整有限度量表；但表是否真的是度量必須先核對，否則後面的唯一性證明（用到三角不等式與分離性）沒有根據。依引理 L_metric_table（在 S02.1 到 S02.3 證明），G1 的表滿足度量的全部公理。",
          "en": "MKO assumption a1 uses complete finite metric tables, but whether a table really is a metric must be checked first, or the later uniqueness proof (which uses the triangle inequality and separation) has no basis. By Lemma L_metric_table (proved in S02.1 to S02.3), the table of G1 satisfies every metric axiom."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_metric_table",
            "metric_def"
          ],
          "explanation": {
            "zh-Hant": "只宣稱這一張表是度量，不宣稱其他表也是。三角不等式在這裡是有限多個不等式，全部列舉之後才得到結論。",
            "en": "Only this table is claimed to be a metric, not any other table. Here the triangle inequality amounts to finitely many inequalities, and the conclusion follows only after all of them are enumerated."
          }
        },
        "leaf_basis": "L_metric_table + metric_def"
      },
      {
        "id": "S02.1",
        "uses": [
          "G1",
          "metric_def"
        ],
        "before_tex": "d(p,p)=0,\\ d(a,b)=1,\\ d(a,c)=2,\\ d(b,c)=\\tfrac32\\ \\text{(with mirror entries)}",
        "after_tex": "d\\ge 0,\\ \\ (d(p,q)=0\\Leftrightarrow p=q),\\ \\ d(p,q)=d(q,p)",
        "relation": "implies",
        "rule": "metric_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐格讀表：對角線三格都是 0；非對角線六格是 1、2、3/2 及它們的對稱位置，都大於 0。所以 d≥0，而且 d(p,q)=0 恰好發生在 p=q。三對對稱格 d(a,b)=d(b,a)、d(a,c)=d(c,a)、d(b,c)=d(c,b) 相同，所以對稱性成立。這是對全部 9 格的完整列舉。",
          "en": "Read the table cell by cell: the three diagonal cells are 0; the six off-diagonal cells are 1, 2, 3/2 and their mirror entries, all greater than 0. So d≥0, and d(p,q)=0 happens exactly when p=q. The three mirror pairs d(a,b)=d(b,a), d(a,c)=d(c,a), d(b,c)=d(c,b) agree, so symmetry holds. This is a complete enumeration of all 9 cells."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "metric_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "只有完整列舉 9 格才能宣稱這三條公理；少看一格就只是部分核對。",
            "en": "Only the enumeration of all 9 cells supports these three axioms; leaving out a cell would make it a partial check."
          }
        },
        "leaf_basis": "metric_def + numeric_evaluation + order_basics"
      },
      {
        "id": "S02.2",
        "uses": [
          "S02.1"
        ],
        "before_tex": "x,y,z\\ \\text{pairwise distinct}",
        "after_tex": "2\\le 1+\\tfrac32,\\quad 1\\le 2+\\tfrac32,\\quad \\tfrac32\\le 1+2",
        "relation": "implies",
        "rule": "metric_def",
        "conditions": [
          "x\\neq y,\\ y\\neq z,\\ x\\neq z"
        ],
        "explanation": {
          "zh-Hant": "三點兩兩相異時，三角不等式 d(x,z)≤d(x,y)+d(y,z) 由中間點 y 決定：y=b 給 d(a,c)=2≤d(a,b)+d(b,c)=5/2；y=c 給 d(a,b)=1≤d(a,c)+d(c,b)=7/2；y=a 給 d(b,c)=3/2≤d(b,a)+d(a,c)=3。交換 x 與 z 只把各項換成對稱格（S02.1 已證相等），所以 6 個相異有序三元組化為這 3 個不等式，全部成立。",
          "en": "For pairwise distinct points, the triangle inequality d(x,z)≤d(x,y)+d(y,z) is determined by the middle point y: y=b gives d(a,c)=2≤d(a,b)+d(b,c)=5/2; y=c gives d(a,b)=1≤d(a,c)+d(c,b)=7/2; y=a gives d(b,c)=3/2≤d(b,a)+d(a,c)=3. Swapping x and z only replaces each entry by its mirror entry (equal by S02.1), so the 6 ordered triples of distinct points reduce to these 3 inequalities, all of which hold."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02.1",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "每個不等式都由 N1、N2、N3 獨立重算；化約到 3 個不等式用到 S02.1 的對稱性。",
            "en": "Each inequality is recomputed independently in N1, N2 and N3; the reduction to 3 inequalities uses the symmetry from S02.1."
          }
        },
        "leaf_basis": "metric_def + numeric_evaluation + order_basics"
      },
      {
        "id": "S02.3",
        "uses": [
          "S02.1",
          "case_analysis"
        ],
        "before_tex": "x=z\\ \\lor\\ y=x\\ \\lor\\ y=z",
        "after_tex": "d(x,z)\\le d(x,y)+d(y,z)",
        "relation": "case_split",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "其餘 27−6=21 個有序三元組至少有兩點相同，分三種情況：若 x=z，左邊 d(x,x)=0，而右邊是兩個非負數之和；若 y=x，右邊是 0+d(x,z)；若 y=z，右邊是 d(x,z)+0。三種情況都成立（情況可以重疊，但涵蓋所有含重複點的三元組）。連同 S02.2，三角不等式對全部 27 個三元組成立，S02.1 到 S02.3 證完引理 L_metric_table。",
          "en": "The remaining 27−6=21 ordered triples have at least two equal points, in three cases: if x=z the left side is d(x,x)=0 and the right side is a sum of two nonnegative numbers; if y=x the right side is 0+d(x,z); if y=z it is d(x,z)+0. All three cases hold (they may overlap, but together they cover every triple with a repeated point). With S02.2, the triangle inequality holds for all 27 triples, and S02.1 to S02.3 complete Lemma L_metric_table."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02.1",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "情況必須涵蓋所有含重複點的三元組：三點不全相異時必有 x=y、y=z 或 x=z 之一，正是這三種情況。",
            "en": "The cases must cover every triple with a repeated point: if the three points are not all distinct, then x=y, y=z or x=z, which are exactly these cases."
          }
        },
        "leaf_basis": "case_analysis + metric_def + order_basics"
      },
      {
        "id": "S03",
        "uses": [
          "G2"
        ],
        "before_tex": "x_0=a,\\ x_1=c,\\ x_2=a",
        "after_tex": "x_n=b\\ \\text{for every}\\ n\\ge 3\\ \\text{(declared tail)}",
        "relation": "definition",
        "rule": "def_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 假設 a2，最終常值模型明示前綴（長度 3：a、c、a）之後的所有項都等於尾值 b。這是一條涵蓋無限多個 n 的規則，不是有限列表；正因為它說的是「每一個 n≥3」，後面才能給出完整見證。",
          "en": "Per MKO assumption a2, the ultimately constant model declares that every term after the prefix (length 3: a, c, a) equals the tail value b. This is a rule covering infinitely many n, not a finite list; precisely because it speaks of every n≥3, a complete witness can be given later."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "def_intro"
          ],
          "explanation": {
            "zh-Hant": "此步只寫下 G2 的宣告；若沒有這個對所有 n≥3 的宣告，任何有限列表都無法決定尾段行為。",
            "en": "This step only records the declaration of G2; without a declaration for all n≥3, no finite list could determine the behaviour of the tail."
          }
        },
        "leaf_basis": "def_intro"
      },
      {
        "id": "S04",
        "uses": [
          "S01",
          "S03",
          "L_tail_const"
        ],
        "before_tex": "x_n=b\\ \\ (n\\ge 3)",
        "after_tex": "x_n\\to b",
        "relation": "theorem_application",
        "rule": "L_tail_const",
        "conditions": [
          "M=3",
          "p=b"
        ],
        "explanation": {
          "zh-Hant": "引理 L_tail_const：若一個序列對所有 n≥M 都等於 p，則它收斂到 p。取 M=3、p=b，由 S03 得 x_n→b。引理在 S04.1 到 S04.3 對任意度量空間與任意 M 證明。",
          "en": "Lemma L_tail_const: if a sequence equals p for every n≥M, it converges to p. With M=3 and p=b, S03 gives x_n→b. The lemma is proved in S04.1 to S04.3 for any metric space and any M."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "L_tail_const"
          ],
          "explanation": {
            "zh-Hant": "前提只有「對每個 n≥3 有 x_n=b」與 d(b,b)=0。反向（收斂推出最終常值）在這裡不需要；它在有限度量空間中其實成立，但在一般度量空間如 ℝ 中不成立（另取正整數索引 n≥1 的序列 1/n→0，卻不是最終常值）。",
            "en": "The only premises are 'x_n=b for every n≥3' and d(b,b)=0. The converse (convergence implies ultimately constant) is not needed here; it does hold in finite metric spaces, but fails in general metric spaces such as ℝ (the separately positive-indexed sequence 1/n, n≥1, tends to 0 without being ultimately constant)."
          }
        },
        "leaf_basis": "L_tail_const + forall_elim"
      },
      {
        "id": "S04.1",
        "uses": [
          "metric_def"
        ],
        "before_tex": "s_n=p\\ \\text{for all}\\ n\\ge M",
        "after_tex": "d(s_n,p)=d(p,p)=0\\ \\text{for all}\\ n\\ge M",
        "relation": "implies",
        "rule": "eq_replace",
        "conditions": [
          "n\\ge M"
        ],
        "explanation": {
          "zh-Hant": "設 (s_n) 是任意度量空間中的序列，並且對所有 n≥M 有 s_n=p。對這樣的 n，把 s_n 換成 p（等量代換）得 d(s_n,p)=d(p,p)；度量公理 d(x,y)=0 ⇔ x=y 給 d(p,p)=0。",
          "en": "Let (s_n) be a sequence in any metric space with s_n=p for every n≥M. For such n, replacing s_n by p (substitution of equals) gives d(s_n,p)=d(p,p), and the metric axiom d(x,y)=0 ⇔ x=y gives d(p,p)=0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "eq_replace",
            "metric_def"
          ],
          "explanation": {
            "zh-Hant": "只對 n≥M 宣稱；n<M 的項不受任何限制，也不影響結論。",
            "en": "This is claimed only for n≥M; the terms with n<M are unrestricted and do not affect the conclusion."
          }
        },
        "leaf_basis": "eq_replace + metric_def"
      },
      {
        "id": "S04.2",
        "uses": [
          "S04.1",
          "exists_intro"
        ],
        "before_tex": "\\varepsilon>0\\ \\text{arbitrary};\\ d(s_n,p)=0\\ (n\\ge M)",
        "after_tex": "\\exists N\\ (N=M)\\ \\forall n\\ge N:\\ d(s_n,p)<\\varepsilon",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [
          "\\varepsilon>0"
        ],
        "explanation": {
          "zh-Hant": "任取 ε>0。因為 0<ε，S04.1 給所有 n≥M 都有 d(s_n,p)=0<ε。所以 N=M 是滿足要求的見證，依 ∃引入得到 ∃N。這裡 N 與 ε 無關，這是最終常值序列的特點；一般收斂序列可能需要隨 ε 調整 N，但定義不要求每次都改變，也不要求任選見證嚴格單調。",
          "en": "Take any ε>0. Since 0<ε, S04.1 gives d(s_n,p)=0<ε for every n≥M. So N=M is a valid witness, and ∃-introduction gives ∃N. Here N does not depend on ε, which is special to ultimately constant sequences; a general convergent sequence may require N to be adjusted with ε, but the definition requires neither a change every time nor strict monotonicity of arbitrarily chosen witnesses."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04.1",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要 ε>0 才有 0<ε；若允許 ε=0，嚴格不等式 0<0 不成立，定義也不會這樣要求。",
            "en": "ε>0 is needed for 0<ε; if ε=0 were allowed, the strict inequality 0<0 would fail, and the definition does not ask for it."
          }
        },
        "leaf_basis": "exists_intro + order_basics"
      },
      {
        "id": "S04.3",
        "uses": [
          "S04.2",
          "forall_intro",
          "S01"
        ],
        "before_tex": "\\varepsilon\\ \\text{arbitrary}:\\ \\exists N\\ \\forall n\\ge N:\\ d(s_n,p)<\\varepsilon",
        "after_tex": "\\forall\\varepsilon>0\\ \\exists N\\ \\forall n\\ge N:\\ d(s_n,p)<\\varepsilon,\\ \\text{i.e.}\\ s_n\\to p",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S04.2 除了 ε>0 之外對 ε 沒有任何假設，所以依 ∀引入對每個 ε>0 成立；收回 S01 的定義即得 s_n→p。S04.1 到 S04.3 合起來證得引理 L_tail_const。",
          "en": "S04.2 made no assumption on ε beyond ε>0, so ∀-introduction gives the statement for every ε>0; folding back the definition of S01 gives s_n→p. S04.1 to S04.3 together prove Lemma L_tail_const."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04.2",
            "forall_intro",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "推廣的前提是 ε 為任意；證明中沒有選定特定的 ε。",
            "en": "The generalization needs ε to be arbitrary; no particular ε was chosen in the proof."
          }
        },
        "leaf_basis": "forall_intro + def_unfold"
      },
      {
        "id": "S05",
        "uses": [
          "S01",
          "S03",
          "G1"
        ],
        "before_tex": "\\varepsilon>0",
        "after_tex": "N_{\\min}(\\varepsilon)=0\\ (\\varepsilon>\\tfrac32),\\quad 2\\ (1<\\varepsilon\\le\\tfrac32),\\quad 3\\ (0<\\varepsilon\\le 1)",
        "relation": "case_split",
        "rule": "case_analysis",
        "conditions": [
          "\\varepsilon>0"
        ],
        "explanation": {
          "zh-Hant": "雖然 N=3 對每個 ε 都有效，最小的有效 N 仍隨 ε 改變。到 b 的距離依序是 d(x_0,b)=d(a,b)=1、d(x_1,b)=d(c,b)=3/2、d(x_2,b)=1，之後全為 0。三種情況互斥並涵蓋所有 ε>0，分別在 S05.1 到 S05.3 處理。這具體顯示最小有效 N 可以依賴 ε，與同一個 N=3 對所有 ε 有效並不矛盾。",
          "en": "Although N=3 works for every ε, the least valid N still changes with ε. The distances to b are d(x_0,b)=d(a,b)=1, d(x_1,b)=d(c,b)=3/2, d(x_2,b)=1, and 0 from then on. The three cases are mutually exclusive and cover every ε>0; they are treated in S05.1 to S05.3. This shows that the least valid N may depend on ε, consistently with the single valid cutoff N=3 working for every ε."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "三個區間 (3/2,∞)、(1,3/2]、(0,1] 兩兩不交且聯集為 (0,∞)，所以情況完整。",
            "en": "The intervals (3/2,∞), (1,3/2] and (0,1] are pairwise disjoint with union (0,∞), so the cases are exhaustive."
          }
        },
        "leaf_basis": "case_analysis + order_basics"
      },
      {
        "id": "S05.1",
        "uses": [
          "S05"
        ],
        "before_tex": "\\varepsilon>\\tfrac32",
        "after_tex": "N=0\\ \\text{valid, hence}\\ N_{\\min}=0",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "\\varepsilon>\\tfrac32"
        ],
        "explanation": {
          "zh-Hant": "所有 d(x_n,b) 都屬於 {0,1,3/2}，最大值是 3/2（N5），而 3/2<ε，所以 N=0 時每一項都滿足 d(x_n,b)<ε。N 不能小於 0，所以最小有效 N 是 0。",
          "en": "Every d(x_n,b) lies in {0,1,3/2}, whose maximum is 3/2 (N5), and 3/2<ε, so with N=0 every term satisfies d(x_n,b)<ε. N cannot be below 0, so the least valid N is 0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "用到距離值的完整清單（前綴三項加上 S03 的常值尾段）與遞移性 d≤3/2<ε。",
            "en": "Uses the complete list of distance values (three prefix terms plus the constant tail of S03) and transitivity d≤3/2<ε."
          }
        },
        "leaf_basis": "order_basics + numeric_evaluation"
      },
      {
        "id": "S05.2",
        "uses": [
          "S05"
        ],
        "before_tex": "1<\\varepsilon\\le\\tfrac32",
        "after_tex": "N\\in\\{0,1\\}\\ \\text{invalid},\\ N=2\\ \\text{valid, hence}\\ N_{\\min}=2",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "1<\\varepsilon\\le\\tfrac32"
        ],
        "explanation": {
          "zh-Hant": "N=0 或 N=1 時，n=1 在尾段裡，而 d(x_1,b)=3/2≥ε，嚴格不等式失敗，所以這兩個 N 無效。N=2 時尾段是 x_2=a（距離 1<ε，N10）與之後的 b（距離 0），全部 <ε。所以最小有效 N 是 2。",
          "en": "With N=0 or N=1 the index n=1 is in the tail, and d(x_1,b)=3/2≥ε, so the strict inequality fails and these N are invalid. With N=2 the tail is x_2=a (distance 1<ε, N10) followed by b (distance 0), all below ε. So the least valid N is 2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "否定一個 N 只需一個反例項（n=1）；肯定 N=2 需要尾段每一項，這由前綴值與 S03 的宣告提供。",
            "en": "Rejecting an N needs one bad index (n=1); accepting N=2 needs every tail term, which the prefix values and the declaration of S03 supply."
          }
        },
        "leaf_basis": "order_basics + numeric_evaluation"
      },
      {
        "id": "S05.3",
        "uses": [
          "S05"
        ],
        "before_tex": "0<\\varepsilon\\le 1",
        "after_tex": "N\\le 2\\ \\text{invalid},\\ N=3\\ \\text{valid, hence}\\ N_{\\min}=3",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "0<\\varepsilon\\le 1"
        ],
        "explanation": {
          "zh-Hant": "任何 N≤2 的尾段都含 n=2，而 d(x_2,b)=d(a,b)=1≥ε，所以無效。N=3 時尾段全是 b，距離 0<ε。所以最小有效 N 是 3；例如 ε=1/1000 時 N 必須至少是 3。",
          "en": "Every tail with N≤2 contains n=2, and d(x_2,b)=d(a,b)=1≥ε, so such N are invalid. With N=3 the whole tail is b, at distance 0<ε. So the least valid N is 3; for example with ε=1/1000, N must be at least 3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S03",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "否定 N≤2 用到 n=2 這一項；肯定 N=3 用到 S03 對所有 n≥3 的宣告。",
            "en": "Rejecting N≤2 uses the term n=2; accepting N=3 uses the declaration of S03 for all n≥3."
          }
        },
        "leaf_basis": "order_basics + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S02",
          "S04",
          "L_limit_unique"
        ],
        "before_tex": "x_n\\to b\\ \\land\\ x_n\\to q",
        "after_tex": "q=b",
        "relation": "theorem_application",
        "rule": "L_limit_unique",
        "conditions": [
          "(X,d)\\ \\text{metric space}"
        ],
        "explanation": {
          "zh-Hant": "引理 L_limit_unique：在度量空間中，若同一序列收斂到 p 也收斂到 q，則 p=q。S02 確認 (X,d) 是度量空間，S04 給 x_n→b，所以 x_n 的任何極限 q 都等於 b。引理在 S06.1 到 S06.4 證明，用到三角不等式與分離性（MKO 段落 2：度量分離性給極限唯一）。",
          "en": "Lemma L_limit_unique: in a metric space, if one sequence converges to p and to q, then p=q. S02 confirms that (X,d) is a metric space and S04 gives x_n→b, so every limit q of x_n equals b. The lemma is proved in S06.1 to S06.4 using the triangle inequality and separation (MKO paragraph 2: metric separation gives uniqueness of the limit)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S04",
            "L_limit_unique"
          ],
          "explanation": {
            "zh-Hant": "需要 d 是真正的度量（S02）；若表違反三角不等式，S06.2 就不能做。",
            "en": "Requires d to be a genuine metric (S02); if the table violated the triangle inequality, S06.2 could not be taken."
          }
        },
        "leaf_basis": "L_limit_unique + modus_ponens"
      },
      {
        "id": "S06.1",
        "uses": [
          "S01",
          "forall_elim",
          "exists_elim"
        ],
        "before_tex": "s_n\\to p,\\ s_n\\to q,\\ \\varepsilon>0",
        "after_tex": "\\exists N:\\ d(s_N,p)<\\tfrac{\\varepsilon}{2}\\ \\land\\ d(s_N,q)<\\tfrac{\\varepsilon}{2}",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [
          "\\tfrac{\\varepsilon}{2}>0"
        ],
        "explanation": {
          "zh-Hant": "任取 ε>0，則 ε/2>0。把 s_n→p 的定義用在 ε/2 上，得到 N_1 使 n≥N_1 時 d(s_n,p)<ε/2；把 s_n→q 用在 ε/2 上，得到 N_2。取 N=max(N_1,N_2)，則 N≥N_1 且 N≥N_2，所以第 N 項同時滿足兩個不等式。",
          "en": "Take any ε>0; then ε/2>0. Applying the definition of s_n→p at ε/2 gives N_1 with d(s_n,p)<ε/2 for n≥N_1; applying s_n→q at ε/2 gives N_2. Let N=max(N_1,N_2); then N≥N_1 and N≥N_2, so the term s_N satisfies both inequalities."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "forall_elim",
            "exists_elim",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "代入特例需要 ε/2>0；取 max 讓同一個指標同時落在兩個尾段中。N_1、N_2 是 ∃消去取的新名字，最後結論 p=q 不含它們。",
            "en": "Instantiation needs ε/2>0; taking the max puts one index in both tails at once. N_1 and N_2 are fresh names from ∃-elimination, and the final conclusion p=q does not mention them."
          }
        },
        "leaf_basis": "forall_elim + exists_elim + order_basics"
      },
      {
        "id": "S06.2",
        "uses": [
          "S06.1",
          "metric_def"
        ],
        "before_tex": "d(s_N,p)<\\tfrac{\\varepsilon}{2},\\ d(s_N,q)<\\tfrac{\\varepsilon}{2}",
        "after_tex": "d(p,q)\\le d(p,s_N)+d(s_N,q)<\\tfrac{\\varepsilon}{2}+\\tfrac{\\varepsilon}{2}=\\varepsilon",
        "relation": "implies",
        "rule": "metric_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "三角不等式給 d(p,q)≤d(p,s_N)+d(s_N,q)；對稱性把 d(p,s_N) 換成 d(s_N,p)。兩個嚴格不等式相加得到小於 ε/2+ε/2=ε（N6 驗證 ε/2+ε/2=ε 是恆等式）。因為 ε 在 S06.1 是任意取的，所以對每個 ε>0 都有 d(p,q)<ε。",
          "en": "The triangle inequality gives d(p,q)≤d(p,s_N)+d(s_N,q); symmetry replaces d(p,s_N) by d(s_N,p). Adding the two strict inequalities gives less than ε/2+ε/2=ε (N6 checks that ε/2+ε/2=ε is an identity). Since ε was arbitrary in S06.1, d(p,q)<ε holds for every ε>0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "metric_def",
            "order_basics",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "用到三角不等式、對稱性與不等式相加；只得到上界，不宣稱等號。",
            "en": "Uses the triangle inequality, symmetry and the addition of inequalities; only an upper bound is obtained, no equality is claimed."
          }
        },
        "leaf_basis": "metric_def + order_basics + forall_intro"
      },
      {
        "id": "S06.3",
        "uses": [
          "S06.2",
          "contradiction"
        ],
        "before_tex": "\\forall\\varepsilon>0:\\ d(p,q)<\\varepsilon",
        "after_tex": "d(p,q)=0",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [
          "d(p,q)\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "因為 d≥0，只需排除 d(p,q)>0。假設 d(p,q)>0，則 ε=d(p,q) 是合法的正數，S06.2 給 d(p,q)<d(p,q)，矛盾。所以 d(p,q)=0。",
          "en": "Since d≥0, it suffices to rule out d(p,q)>0. Suppose d(p,q)>0; then ε=d(p,q) is an admissible positive number and S06.2 gives d(p,q)<d(p,q), a contradiction. Hence d(p,q)=0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06.2",
            "contradiction",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "關鍵是 S06.2 對每個 ε>0 都成立，所以可以取 ε=d(p,q)；若只對某些 ε 成立，就不能這樣取。",
            "en": "The key point is that S06.2 holds for every ε>0, so ε=d(p,q) may be used; if it held only for some ε, this choice would not be allowed."
          }
        },
        "leaf_basis": "contradiction + order_basics"
      },
      {
        "id": "S06.4",
        "uses": [
          "S06.3",
          "metric_def"
        ],
        "before_tex": "d(p,q)=0",
        "after_tex": "p=q",
        "relation": "implies",
        "rule": "metric_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "度量的分離性 d(p,q)=0 ⇔ p=q 給出 p=q。S06.1 到 S06.4 合起來證得引理 L_limit_unique；證明對任意度量空間有效，不依賴 X 有限。",
          "en": "Separation, d(p,q)=0 ⇔ p=q, gives p=q. S06.1 to S06.4 together prove Lemma L_limit_unique; the proof works in any metric space and does not use that X is finite."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "metric_def"
          ],
          "explanation": {
            "zh-Hant": "這裡用到分離性的 ⇒ 方向；若只有偽度量（允許 d(p,q)=0 而 p≠q），極限就可能不唯一。",
            "en": "This uses the ⇒ direction of separation; for a pseudometric (allowing d(p,q)=0 with p≠q) limits need not be unique."
          }
        },
        "leaf_basis": "metric_def"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "G1"
        ],
        "before_tex": "a\\neq b,\\ c\\neq b",
        "after_tex": "x_n\\not\\to a,\\ \\ x_n\\not\\to c",
        "relation": "implies",
        "rule": "contrapositive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S06 說：若 x_n→q 則 q=b。逆否命題：若 q≠b 則 x_n 不收斂到 q。a、c 都與 b 相異（G1），所以 x_n 既不收斂到 a，也不收斂到 c。",
          "en": "S06 says: if x_n→q then q=b. Its contrapositive: if q≠b then x_n does not converge to q. Both a and c differ from b (G1), so x_n converges neither to a nor to c."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "contrapositive"
          ],
          "explanation": {
            "zh-Hant": "逆否命題與原命題等價，所以這是合法的單向推論；S08 到 S10 另外給出直接見證作交叉核對。",
            "en": "The contrapositive is equivalent to the original statement, so this is a valid one-way inference; S08 to S10 give direct witnesses as a cross-check."
          }
        },
        "leaf_basis": "contrapositive + L_limit_unique"
      },
      {
        "id": "S08",
        "uses": [
          "S01",
          "quantifier_negation"
        ],
        "before_tex": "\\neg\\,\\forall\\varepsilon>0\\ \\exists N\\ \\forall n\\ge N:\\ d(x_n,q)<\\varepsilon",
        "after_tex": "\\exists\\varepsilon>0\\ \\forall N\\ \\exists n\\ge N:\\ d(x_n,q)\\ge\\varepsilon",
        "relation": "equivalent",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把「x_n 不收斂到 q」改寫成正面形式：逐一翻轉量詞（∀ 變 ∃、∃ 變 ∀），最後把 d<ε 否定成 d≥ε。所以要證不收斂，只要找一個 ε>0，使每一個 N 之後都還有某項距離 q 至少 ε。",
          "en": "Rewrite 'x_n does not converge to q' in positive form: flip each quantifier (∀ to ∃, ∃ to ∀) and negate d<ε to d≥ε. So to prove non-convergence it suffices to find one ε>0 such that after every N some term is at distance at least ε from q."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "quantifier_negation",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "量詞否定律雙向成立；實數的三分律保證 ¬(d<ε) 與 d≥ε 等價。",
            "en": "Quantifier negation holds in both directions; trichotomy of the reals makes ¬(d<ε) equivalent to d≥ε."
          }
        },
        "leaf_basis": "quantifier_negation + order_basics"
      },
      {
        "id": "S09",
        "uses": [
          "S08",
          "S03",
          "G1"
        ],
        "before_tex": "q=a,\\ \\varepsilon=1,\\ N\\ \\text{arbitrary}",
        "after_tex": "n=\\max(N,3)\\ge N:\\ d(x_n,a)=d(b,a)=1\\ge 1",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [
          "\\varepsilon=1>0"
        ],
        "explanation": {
          "zh-Hant": "取 ε=1。對任意 N，令 n=max(N,3)；則 n≥N 且 n≥3，所以 x_n=b，d(x_n,a)=d(b,a)=1≥1（例如 N=7 時 n=7，見 N7）。依 S08，x_n 不收斂到 a。注意這裡正好用到嚴格不等式：d=1 不小於 ε=1；任何 0<ε≤1 也都可以當見證。",
          "en": "Take ε=1. For any N, let n=max(N,3); then n≥N and n≥3, so x_n=b and d(x_n,a)=d(b,a)=1≥1 (for example N=7 gives n=7, see N7). By S08, x_n does not converge to a. Note that the strict inequality is used exactly here: d=1 is not less than ε=1; any 0<ε≤1 would also serve as a witness."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "S03",
            "exists_intro",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "ε 只需一個；n 必須對每個 N 都找得到，max(N,3) 的構造對任意 N 有效。前綴裡的 x_0=a、x_2=a 對結論毫無幫助。",
            "en": "One ε is enough; an n must be found for every N, and the construction max(N,3) works for arbitrary N. The prefix terms x_0=a and x_2=a do not help the opposite conclusion at all."
          }
        },
        "leaf_basis": "exists_intro + forall_intro + order_basics"
      },
      {
        "id": "S10",
        "uses": [
          "S08",
          "S03",
          "G1"
        ],
        "before_tex": "q=c,\\ \\varepsilon=\\tfrac32,\\ N\\ \\text{arbitrary}",
        "after_tex": "n=\\max(N,3)\\ge N:\\ d(x_n,c)=d(b,c)=\\tfrac32\\ge\\tfrac32",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [
          "\\varepsilon=\\tfrac32>0"
        ],
        "explanation": {
          "zh-Hant": "同樣的構造：取 ε=3/2，對任意 N 令 n=max(N,3)，則 x_n=b 而 d(b,c)=3/2≥3/2（N9）。依 S08，x_n 不收斂到 c。這與 S07 由唯一性得到的結論一致。",
          "en": "The same construction: take ε=3/2, and for any N let n=max(N,3); then x_n=b and d(b,c)=3/2≥3/2 (N9). By S08, x_n does not converge to c. This agrees with the conclusion S07 drew from uniqueness."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "S03",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "與 S09 相同的前提；只換了目標點與 ε。",
            "en": "The same premises as S09; only the target point and ε change."
          }
        },
        "leaf_basis": "exists_intro + forall_intro + order_basics"
      },
      {
        "id": "S11",
        "uses": [
          "S04",
          "S09",
          "S10"
        ],
        "before_tex": "x_n\\to b,\\ x_n\\not\\to a,\\ x_n\\not\\to c",
        "after_tex": "\\{q\\in X\\mid x_n\\to q\\}=\\{b\\}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "X 只有 a、b、c 三點：S04 說 b 是極限，S09 與 S10 排除 a 與 c。合起來，x_n 的極限集合恰為 {b}。S06 從一般唯一性得到同一結論。",
          "en": "X has only the three points a, b, c: S04 shows b is a limit, and S09 and S10 exclude a and c. Together, the set of limits of x_n is exactly {b}. S06 reaches the same conclusion from general uniqueness."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S09",
            "S10",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "需要 X 的完整清單（G1）才能說「恰為」；只排除部分點不夠。",
            "en": "Saying 'exactly' needs the complete list of X (G1); excluding only some points would not suffice."
          }
        },
        "leaf_basis": "and_intro_elim + case_analysis"
      },
      {
        "id": "S12",
        "uses": [
          "G3",
          "L_tail_const",
          "L_limit_unique",
          "G1"
        ],
        "before_tex": "y_n=c\\ \\ (n\\ge 11)",
        "after_tex": "y_n\\to c\\ \\land\\ y_n\\not\\to b",
        "relation": "theorem_application",
        "rule": "L_tail_const",
        "conditions": [
          "M=11",
          "p=c",
          "b\\neq c"
        ],
        "explanation": {
          "zh-Hant": "G3 明示 n≥11 時 y_n=c，引理 L_tail_const（M=11、p=c）給 y_n→c。再用 L_limit_unique：y_n 的極限唯一，而 b≠c，所以 y_n 不收斂到 b。",
          "en": "G3 declares y_n=c for n≥11, and Lemma L_tail_const (M=11, p=c) gives y_n→c. By L_limit_unique the limit of y_n is unique, and b≠c, so y_n does not converge to b."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "L_tail_const",
            "L_limit_unique"
          ],
          "explanation": {
            "zh-Hant": "兩個引理都已在任意度量空間中證明；這裡只做代入。",
            "en": "Both lemmas were proved for any metric space; this step only instantiates them."
          }
        },
        "leaf_basis": "L_tail_const + L_limit_unique + contrapositive"
      },
      {
        "id": "S13",
        "uses": [
          "S03",
          "S04",
          "S12",
          "G3"
        ],
        "before_tex": "x_n=y_n=b\\ \\ (3\\le n\\le 10)",
        "after_tex": "\\lim x_n=b\\neq c=\\lim y_n",
        "relation": "implies",
        "rule": "L_tail_const",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x 與 y 在 n=0 到 10 完全一致，特別是有限尾段 x_3 到 x_10 全是 b。若只核這段有限尾段，兩者看起來一樣，但 x_n→b 而 y_n→c。所以任何只列出有限多項的資料都無法決定極限。一般地，對任意 K，在 K 之前與 x 相同、之後恆為 c 的序列由 L_tail_const（M=K+1）收斂到 c；這正是 MKO 所說「有限樣本尾段通過不足以證一般收斂」。",
          "en": "x and y agree for n=0 to 10, and in particular the finite tail x_3 to x_10 is all b. Checking only this finite tail, the two look identical, yet x_n→b while y_n→c. So no data listing finitely many terms can decide the limit. In general, for any K the sequence equal to x up to K and constantly c afterwards converges to c by L_tail_const (M=K+1); this is the MKO's point that passing a sampled finite tail is insufficient for general convergence."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S12",
            "L_tail_const"
          ],
          "explanation": {
            "zh-Hant": "結論只是「有限資料不決定極限」；它不否認 x_n→b，因為 x 的收斂來自 S03 對所有 n≥3 的宣告，而不是來自有限尾段。",
            "en": "The conclusion is only that finite data do not decide the limit; it does not deny x_n→b, because the convergence of x comes from the declaration of S03 for all n≥3, not from the finite tail."
          }
        },
        "leaf_basis": "L_tail_const + L_limit_unique + forall_intro"
      },
      {
        "id": "S14",
        "uses": [
          "S02",
          "S04",
          "S05",
          "S11",
          "S13"
        ],
        "before_tex": "\\text{S02},\\ \\text{S04},\\ \\text{S05},\\ \\text{S11},\\ \\text{S13}",
        "after_tex": "x_n\\to b\\ \\text{with witness}\\ N=3\\ \\text{for every}\\ \\varepsilon;\\ \\text{limit unique}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "總結：d 是度量（S02）；x_n→b，N=3 對每個 ε 都有效，最小有效 N 為 0、2、3 依 ε 而定（S04、S05）；極限恰為 b（S11）；這個完整見證依賴 G2 對所有 n≥3 的宣告，有限尾段本身不夠（S13）。",
          "en": "Summary: d is a metric (S02); x_n→b, with N=3 valid for every ε and the least valid N equal to 0, 2 or 3 depending on ε (S04, S05); the limit is exactly b (S11); this complete witness rests on the declaration of G2 for all n≥3, and a finite tail by itself is not enough (S13)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S04",
            "S05",
            "S11",
            "S13"
          ],
          "explanation": {
            "zh-Hant": "只是把已證結論合取，沒有新推論。",
            "en": "This only conjoins conclusions already proved; there is no new inference."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "G1 的表是度量；序列 x_n（前綴 a、c、a，n≥3 時恆為 b）收斂到 b，N=3 對每個 ε>0 都有效，最小有效 N 在 ε>3/2、1<ε≤3/2、0<ε≤1 時分別為 0、2、3；度量空間中極限唯一，x_n 不收斂到 a（見證 ε=1）也不收斂到 c（見證 ε=3/2）。在有限尾段上與 x_n 一致的 y_n 卻收斂到 c，兩者都收斂但極限不同，所以有限尾段核對不足以判定極限是否為指定的 b；這一對例子本身不區分「有極限」與「沒有極限」。",
      "en": "The table of G1 is a metric; the sequence x_n (prefix a, c, a, then constantly b for n≥3) converges to b, with N=3 valid for every ε>0 and least valid N equal to 0, 2, 3 for ε>3/2, 1<ε≤3/2, 0<ε≤1 respectively; limits in a metric space are unique, and x_n converges neither to a (witness ε=1) nor to c (witness ε=3/2). The sequence y_n, which agrees with x_n on a finite tail, converges to c instead, both sequences converge but have different limits, so a finite tail check cannot decide whether the limit is the specified b. This pair by itself does not distinguish existence from nonexistence of a limit."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "x_3 到 x_10 都等於 b，所以任何具有這些項的序列都收斂到 b",
          "en": "x_3 to x_10 all equal b, so every sequence with these terms converges to b"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "y_n 具有完全相同的這些項，卻收斂到 c（S12、S13）。",
          "en": "y_n has exactly these terms but converges to c (S12, S13)."
        }
      },
      {
        "claim": {
          "zh-Hant": "因為 x_0=a 且 x_2=a，所以 x_n→a",
          "en": "Since x_0=a and x_2=a, x_n→a"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "收斂只看尾段：取 ε=1，對每個 N，n=max(N,3) 給 d(x_n,a)=1≥1（S09）。",
          "en": "Convergence depends only on the tail: with ε=1, for every N the index n=max(N,3) gives d(x_n,a)=1≥1 (S09)."
        }
      },
      {
        "claim": {
          "zh-Hant": "n≥3 時 d(x_n,a)=1≤1，所以取 ε=1 就說明 x_n 收斂到 a",
          "en": "For n≥3, d(x_n,a)=1≤1, so taking ε=1 shows that x_n converges to a"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "定義要求對每一個 ε>0 都成立，而且是嚴格的 <；ε=1 時 1<1 已不成立，ε=1/2 更不成立。",
          "en": "The definition requires every ε>0 and a strict <; at ε=1 already 1<1 fails, and at ε=1/2 it fails as well."
        }
      },
      {
        "claim": {
          "zh-Hant": "在某個度量空間中，一個序列可以同時收斂到 b 與 c",
          "en": "In some metric space a sequence can converge to both b and c"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "L_limit_unique 在任意度量空間成立：兩個極限的距離小於每個 ε>0，因此為 0，分離性給 b=c；這裡 d(b,c)=3/2>0。",
          "en": "L_limit_unique holds in every metric space: the distance between two limits is below every ε>0, hence 0, and separation gives b=c; here d(b,c)=3/2>0."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "id": "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": "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": "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": "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": "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": "forall_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。",
          "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds."
        }
      },
      {
        "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": "modus_ponens",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "肯定前件：由 P 與 P⇒Q 得 Q。",
          "en": "Modus ponens: from P and P⇒Q infer Q."
        }
      },
      {
        "id": "exists_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃消去：由 ∃x∈D, P(x)，可取一個新名字 a 表示某個滿足 P(a) 的元素；只要最後結論不含 a，就可用 a 推理。",
          "en": "∃-elimination: from ∃x∈D, P(x) we may name some element a with P(a) by a fresh name; reasoning with a is allowed as long as the final conclusion does not mention a."
        }
      },
      {
        "id": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "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": "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": "mko_metric_conv",
        "kind": "cited",
        "source": {
          "mko_id": "mko-metric-convergence",
          "sha256": "185db7b6b678014d6791d0448e2eccc1efc643b288d9b95c2b7b91f3ea710969",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "序列 x_n 在度量空間中收斂到 x，若對每個 ε>0 存在 N，使所有 n≥N 都有 d(x_n,x)<ε。有限樣本尾段通過不足以證一般收斂；明示最終常值的無限序列則可給完整見證。",
          "en": "A sequence x_n in a metric space converges to x if, for every ε>0, there exists N such that all n≥N satisfy d(x_n,x)<ε. Passing a sampled finite tail is insufficient for general convergence; an explicitly ultimately constant infinite sequence can instead provide a complete witness."
        }
      },
      {
        "id": "L_metric_table",
        "kind": "lemma",
        "proved_in": [
          "S02.1",
          "S02.2",
          "S02.3"
        ],
        "statement": {
          "zh-Hant": "G1 的表 d 是 X 上的度量：d≥0 且 d(p,q)=0 ⇔ p=q；d 對稱；對所有 27 個有序三元組成立三角不等式。",
          "en": "The table d of G1 is a metric on X: d≥0 and d(p,q)=0 ⇔ p=q; d is symmetric; the triangle inequality holds for all 27 ordered triples."
        }
      },
      {
        "id": "L_tail_const",
        "kind": "lemma",
        "proved_in": [
          "S04.1",
          "S04.2",
          "S04.3"
        ],
        "statement": {
          "zh-Hant": "在任意度量空間中，若序列 (s_n) 對所有 n≥M 都有 s_n=p，則 s_n→p，而且 N=M 對每個 ε>0 都是有效截斷。",
          "en": "In any metric space, if a sequence (s_n) has s_n=p for every n≥M, then s_n→p, and N=M is a valid cutoff for every ε>0."
        }
      },
      {
        "id": "L_limit_unique",
        "kind": "lemma",
        "proved_in": [
          "S06.1",
          "S06.2",
          "S06.3",
          "S06.4"
        ],
        "statement": {
          "zh-Hant": "在任意度量空間中，若 s_n→p 且 s_n→q，則 p=q。",
          "en": "In any metric space, if s_n→p and s_n→q, then p=q."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02.2",
        "lhs": "2",
        "relation": "<=",
        "rhs": "1+3/2"
      },
      {
        "id": "N2",
        "step": "S02.2",
        "lhs": "1",
        "relation": "<=",
        "rhs": "2+3/2"
      },
      {
        "id": "N3",
        "step": "S02.2",
        "lhs": "3/2",
        "relation": "<=",
        "rhs": "1+2"
      },
      {
        "id": "N4",
        "step": "S02.3",
        "lhs": "3**3-factorial(3)",
        "rhs": "21"
      },
      {
        "id": "N5",
        "step": "S05.1",
        "lhs": "max(0,1,3/2)",
        "rhs": "3/2"
      },
      {
        "id": "N6",
        "step": "S06.2",
        "vars": [
          "e"
        ],
        "degree": 1,
        "lhs": "e/2+e/2",
        "rhs": "e"
      },
      {
        "id": "N7",
        "step": "S09",
        "lhs": "max(7,3)",
        "rhs": "7"
      },
      {
        "id": "N8",
        "step": "S09",
        "lhs": "1",
        "relation": ">=",
        "rhs": "1"
      },
      {
        "id": "N9",
        "step": "S10",
        "lhs": "3/2",
        "relation": ">=",
        "rhs": "3/2"
      },
      {
        "id": "N10",
        "step": "S05.2",
        "lhs": "1",
        "relation": "<",
        "rhs": "3/2"
      },
      {
        "id": "N11",
        "step": "S02.2",
        "lhs": "3",
        "relation": ">",
        "rhs": "1+3/2"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理 G1 的有限度量空間與 G2 的最終常值序列模型（MKO 假設 a1、a2）。極限唯一性引理對任意度量空間完整證明；最終常值引理也對任意度量空間證明。不處理 Cauchy 序列、完備性或連續映射，也不把有限樣本的尾段核對當作收斂的證明。不改動 canonical MKO 或其英文 edition。",
      "en": "Treats only the finite metric space of G1 and the ultimately constant sequence model of G2 (MKO assumptions a1, a2). The uniqueness-of-limits lemma is proved in full for every metric space, and so is the ultimately-constant lemma. Cauchy sequences, completeness and continuous maps are not treated, and a check of a finite sampled tail is never taken as a proof of convergence. The canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "X=\\{a,b,c\\}",
        "三點載體",
        "the three-point carrier"
      ],
      [
        "d",
        "G1 的完整度量表",
        "the complete metric table of G1"
      ],
      [
        "(x_n)_{n\\in\\mathbb{N}_0}",
        "最終常值序列：前綴 a、c、a，n≥3 時 x_n=b",
        "the ultimately constant sequence: prefix a, c, a and x_n=b for n≥3"
      ],
      [
        "(y_n)_{n\\in\\mathbb{N}_0}",
        "n≤10 時與 x_n 相同、n≥11 時等於 c 的比較序列",
        "the comparison sequence equal to x_n for n≤10 and to c for n≥11"
      ],
      [
        "\\varepsilon,\\ N",
        "容許誤差（ε>0）與截斷指標",
        "tolerance (ε>0) and cutoff index"
      ],
      [
        "(s_n),\\ p,\\ q,\\ M",
        "引理中的一般序列、候選極限與前綴長度",
        "the generic sequence, candidate limits and prefix length in the lemmas"
      ]
    ],
    "md_error_controls": [
      {
        "title": "有限尾段不是收斂證明 / A finite tail is not a proof of convergence",
        "zh": "MKO 指出的誤解是：看到一段很長但有限的尾段都很接近，就斷言收斂。S13 的 y_n 在 n=3 到 10 與 x_n 完全相同（都等於 b），卻收斂到 c。x_n→b 之所以成立，是因為 G2 明示「每一個 n≥3」都有 x_n=b；這是一條涵蓋無限多項的規則，而不是有限的核對結果。",
        "en": "The misconception named by the MKO is to assert convergence because a long but finite tail looks close. The sequence y_n of S13 equals x_n for n=3 to 10 (all equal to b), yet it converges to c. x_n→b holds because G2 declares x_n=b for every n≥3; that is a rule covering infinitely many terms, not the result of a finite check."
      },
      {
        "title": "嚴格不等式與量詞順序 / Strict inequality and quantifier order",
        "zh": "定義用 d(x_n,x)<ε，是嚴格不等式。S09 中 d(x_n,a)=1 對 ε=1 不滿足 <，所以 ε=1 就足以排除 a；若誤用 ≤，這個特定見證就失效（雖然改用 ε=1/2 仍可排除）。量詞順序允許 N 依賴 ε，但不要求一定依賴：本題 N=3 對所有 ε>0 都有效；S05 顯示的是最小有效 N 隨 ε 由 0 變到 2 再變到 3。",
        "en": "The definition uses d(x_n,x)<ε, a strict inequality. In S09, d(x_n,a)=1 does not satisfy < for ε=1, so ε=1 already excludes a; with ≤ by mistake this particular witness would fail (although ε=1/2 would still exclude a). The quantifier order allows N to depend on ε but does not require dependence: N=3 works for every ε>0 here, while S05 shows that the least valid N changes from 0 to 2 to 3 as ε shrinks."
      },
      {
        "title": "先核度量表 / Check the metric table first",
        "zh": "唯一性證明 S06.2 用到三角不等式。若表被誤寫成 d(a,c)=3（其他不變），則 d(a,c)=3>d(a,b)+d(b,c)=5/2，三角不等式失敗，這張表就不是度量，S06 的論證也不能用。所以 S02 必須在任何收斂推理之前完成。",
        "en": "The uniqueness proof S06.2 uses the triangle inequality. If the table were miswritten with d(a,c)=3 (everything else unchanged), then d(a,c)=3>d(a,b)+d(b,c)=5/2, the triangle inequality fails, the table is not a metric, and the argument of S06 cannot be used. So S02 must be completed before any convergence reasoning."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿不處理 Cauchy 序列、完備性與連續映射（MKO 的 formal next）。",
        "en": "This guide does not treat Cauchy sequences, completeness or continuous maps (the MKO's formal next)."
      },
      {
        "zh-Hant": "S04 審查中提到「有限度量空間中收斂推出最終常值」只是附註（取 ε 為最小正距離即可證），本稿的論證不依賴它。",
        "en": "The remark in the audit of S04 that convergence implies ultimately constant in a finite metric space (provable by taking ε to be the least positive distance) is only a note; the argument of this guide does not depend on it."
      },
      {
        "zh-Hant": "S05 的最小有效 N 只對這個序列與這張表成立。",
        "en": "The least valid N of S05 holds only for this sequence and this table."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "9f78ba61eeae5e730d7f5c30614aceca18c228eb2343bf50bee946868c588393",
    "original_source_sha256": "c050f16c9639e75c444249ed4ba0351bd36dc14296bbf9c4278170c501246d93",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M055",
      "BR-M056",
      "BR-M057"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}