{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-cauchy-sequence-01",
  "mko_id": "mko-cauchy-sequence",
  "title": {
    "zh-Hant": "Cauchy 序列：倒數序列的完整證書、序列在其中無極限的空間、交替反例與最終常值模型",
    "en": "Cauchy sequence: a complete certificate for the reciprocal sequence, a space in which it has no limit, an alternating counterexample and an ultimately constant model"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-cauchy-sequence-01",
    "source": {
      "mko_id": "mko-cauchy-sequence",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "f833bc771b9e0a259993b3560086625a99bfce9069edf8b08acd37a09b08c004",
      "english_sha256": "99682dd196d4595b954c7f7161c3deb478e8b66887dc9eba68ebb95204765b5f"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "metric_def",
      "seq_limit_def",
      "archimedean",
      "abs_rules",
      "order_basics",
      "quantifier_negation"
    ],
    "goal": {
      "zh-Hant": "（一）證明 x_n=1/n（n≥1）在 (ℝ, |s−t|) 中是 Cauchy 序列：對每個 ε>0 給出 N，使所有 m,n≥N 都有 |1/m−1/n|<ε。（二）證明同一序列在 X=(0,1]（限制度量）中仍是 Cauchy，但在 X 中沒有極限。（三）證明 y_n=(−1)^n 不是 Cauchy。（四）證明最終常值序列 z（z₁=0、z₂=7、n≥3 時 z_n=5）是 Cauchy。並說明有限尾段為何不能證明 Cauchy。",
      "en": "(1) Prove that x_n=1/n (n≥1) is a Cauchy sequence in (ℝ, |s−t|): for every ε>0 give N such that |1/m−1/n|<ε for all m,n≥N. (2) Prove that the same sequence is still Cauchy in X=(0,1] with the restricted metric, but has no limit in X. (3) Prove that y_n=(−1)^n is not Cauchy. (4) Prove that the ultimately constant sequence z (z₁=0, z₂=7, z_n=5 for n≥3) is Cauchy. Also explain why a finite tail cannot prove the Cauchy property."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "(ℝ,d)，d(s,t)=|s−t|；X=(0,1]={t ∣ 0<t≤1}，度量為同一公式 d(s,t)=|s−t| 限制到 X。",
          "en": "(ℝ,d) with d(s,t)=|s−t|; X=(0,1]={t ∣ 0<t≤1}, with the same formula d(s,t)=|s−t| restricted to X."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "x_n=1/n，n≥1。",
          "en": "x_n=1/n for n≥1."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "y_n=(−1)^n，n≥1。",
          "en": "y_n=(−1)^n for n≥1."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "z₁=0、z₂=7，且對所有 n≥3 有 z_n=5。",
          "en": "z₁=0, z₂=7, and z_n=5 for every n≥3."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_cauchy"
        ],
        "before_tex": "(x_n)\\ \\text{Cauchy in}\\ (\\mathbb{R},d)",
        "after_tex": "\\forall\\varepsilon>0\\ \\exists N\\ \\forall m,n\\ge N:\\ \\left|\\tfrac1m-\\tfrac1n\\right|<\\varepsilon",
        "relation": "equivalent",
        "rule": "mko_cauchy",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 的定義展開：比較的是尾段中任意兩項之間的距離，定義中沒有出現任何極限。N 可以依賴 ε，但不能依賴 m、n。",
          "en": "Unfold the MKO's definition: what is compared is the distance between any two terms of the tail, and no limit appears in the definition. N may depend on ε but not on m or n."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_cauchy",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "展開與收回定義互逆，雙向成立。",
            "en": "Unfolding and folding the definition undo each other, so the step holds both ways."
          }
        },
        "leaf_basis": "mko_cauchy + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "L_absmax"
        ],
        "before_tex": "s,t\\ge 0",
        "after_tex": "|s-t|\\le\\max(s,t)",
        "relation": "implies",
        "rule": "L_absmax",
        "conditions": [
          "s\\ge 0",
          "t\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "引理 L_absmax：兩個非負數之差的絕對值不超過其中較大者。證明在 S02.1、S02.2 分 s≥t 與 s<t 兩種情況。之後用 s=1/m、t=1/n。",
          "en": "Lemma L_absmax: the absolute difference of two nonnegative numbers is at most the larger of them. The proof in S02.1 and S02.2 splits into the cases s≥t and s<t. It will be used with s=1/m and t=1/n."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_absmax"
          ],
          "explanation": {
            "zh-Hant": "需要 s、t 非負；對負數不成立，例如 |(−1)−1|=2>1=max(−1,1)。",
            "en": "It needs s and t nonnegative; it fails for negatives, for example |(−1)−1|=2>1=max(−1,1)."
          }
        },
        "leaf_basis": "L_absmax"
      },
      {
        "id": "S02.1",
        "uses": [
          "abs_rules",
          "max_two",
          "order_basics"
        ],
        "before_tex": "s\\ge t\\ge 0",
        "after_tex": "|s-t|=s-t\\le s=\\max(s,t)",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [
          "s\\ge t",
          "t\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "s≥t 時 s−t≥0，所以 |s−t|=s−t；又 t≥0，所以 s−t≤s；而 s≥t 時 max(s,t)=s。",
          "en": "When s≥t, s−t≥0, so |s−t|=s−t; since t≥0, s−t≤s; and when s≥t, max(s,t)=s."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "max_two",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "用到 t≥0；不使用 L_absmax 本身。",
            "en": "It uses t≥0; L_absmax itself is not used."
          }
        },
        "leaf_basis": "abs_rules + max_two + order_basics"
      },
      {
        "id": "S02.2",
        "uses": [
          "S02.1",
          "abs_rules",
          "max_two",
          "case_analysis"
        ],
        "before_tex": "t>s\\ge 0",
        "after_tex": "|s-t|=t-s\\le t=\\max(s,t)",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [
          "t>s",
          "s\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "t>s 時 s−t<0，|s−t|=t−s；又 s≥0，所以 t−s≤t=max(s,t)。s≥t 與 t>s 涵蓋全部情況，S02.1、S02.2 合起來證得 L_absmax。",
          "en": "When t>s, s−t<0 and |s−t|=t−s; since s≥0, t−s≤t=max(s,t). The cases s≥t and t>s cover everything, so S02.1 and S02.2 together prove L_absmax."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "max_two",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "用到 s≥0；兩個情況涵蓋全部實數對；不使用 L_absmax 本身。",
            "en": "It uses s≥0; the two cases cover all pairs of reals; L_absmax itself is not used."
          }
        },
        "leaf_basis": "abs_rules + max_two + case_analysis"
      },
      {
        "id": "S03",
        "uses": [
          "L_recip"
        ],
        "before_tex": "m,n\\ge N\\ge 1",
        "after_tex": "0<\\tfrac1m\\le\\tfrac1N,\\ 0<\\tfrac1n\\le\\tfrac1N",
        "relation": "implies",
        "rule": "L_recip",
        "conditions": [
          "N\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_recip（S03.1 證明），m≥N>0 推出 0<1/m≤1/N，n 同理。尾段中每一項都在 (0,1/N] 內。",
          "en": "By Lemma L_recip (proved in S03.1), m≥N>0 gives 0<1/m≤1/N, and likewise for n. Every term of the tail lies in (0,1/N]."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_recip"
          ],
          "explanation": {
            "zh-Hant": "需要 N>0；不宣稱反向。",
            "en": "It needs N>0; no converse is claimed."
          }
        },
        "leaf_basis": "L_recip"
      },
      {
        "id": "S03.1",
        "uses": [
          "order_basics",
          "mul_inverse_nonzero",
          "frac_rules"
        ],
        "before_tex": "m\\ge N>0",
        "after_tex": "\\tfrac1N=m\\cdot\\tfrac{1}{mN}\\ge N\\cdot\\tfrac{1}{mN}=\\tfrac1m>0",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "N>0"
        ],
        "explanation": {
          "zh-Hant": "m≥N>0 推出 m>0、mN>0，所以 1/(mN)>0。把 m≥N 兩邊乘以正數 1/(mN)，方向不變：1/N≥1/m；若 m>N 則嚴格：1/N>1/m。又 1/m>0，因為正數的倒數為正。這證明了 L_recip。",
          "en": "m≥N>0 gives m>0 and mN>0, so 1/(mN)>0. Multiplying m≥N by the positive number 1/(mN) keeps the direction: 1/N≥1/m; if m>N it is strict: 1/N>1/m. Also 1/m>0, since the reciprocal of a positive number is positive. This proves L_recip."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "mul_inverse_nonzero",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "需要 N>0 使所有倒數存在且乘數為正；不使用 L_recip 本身。",
            "en": "It needs N>0 so that the reciprocals exist and the multiplier is positive; L_recip itself is not used."
          }
        },
        "leaf_basis": "order_basics + mul_inverse_nonzero + frac_rules"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "S03",
          "max_two"
        ],
        "before_tex": "m,n\\ge N\\ge 1",
        "after_tex": "\\left|\\tfrac1m-\\tfrac1n\\right|\\le\\max\\left(\\tfrac1m,\\tfrac1n\\right)\\le\\tfrac1N",
        "relation": "implies",
        "rule": "max_two",
        "conditions": [
          "N\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "用 L_absmax（s=1/m≥0、t=1/n≥0）得第一個不等號；由 S03，兩者都 ≤1/N，所以它們的最大值 ≤1/N。這是一個對尾段中所有 m、n 一致的界，與 m、n 的大小關係無關。",
          "en": "L_absmax (s=1/m≥0, t=1/n≥0) gives the first inequality; by S03 both are ≤1/N, so their maximum is ≤1/N. This bound is uniform over all m, n in the tail, whatever their relative size."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S03",
            "max_two"
          ],
          "explanation": {
            "zh-Hant": "需要兩項都非負且都 ≤1/N；不宣稱等號。",
            "en": "It needs both terms nonnegative and both ≤1/N; equality is not claimed."
          }
        },
        "leaf_basis": "L_absmax + max_two + order_basics"
      },
      {
        "id": "S05",
        "uses": [
          "archimedean"
        ],
        "before_tex": "\\varepsilon>0\\ \\text{arbitrary}",
        "after_tex": "\\exists N\\in\\mathbb{N},\\ N\\ge 1:\\ \\tfrac1N<\\varepsilon",
        "relation": "implies",
        "rule": "archimedean",
        "conditions": [
          "\\varepsilon>0"
        ],
        "explanation": {
          "zh-Hant": "任取 ε>0（不加其他假設）。依阿基米德性質，存在自然數 N 使 1/N<ε；可取 N≥1。例如 ε=1/10 時 N=11 可用。",
          "en": "Take an arbitrary ε>0, with no other assumption. By the Archimedean property there is a natural number N with 1/N<ε; N≥1 can be taken. For example, with ε=1/10, N=11 works."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "archimedean"
          ],
          "explanation": {
            "zh-Hant": "N 只依賴 ε；這正是 Cauchy 定義中 ∃N 的位置所允許的。",
            "en": "N depends only on ε, which is exactly what the position of ∃N in the Cauchy definition allows."
          }
        },
        "leaf_basis": "archimedean"
      },
      {
        "id": "S06",
        "uses": [
          "S04",
          "S05",
          "order_basics"
        ],
        "before_tex": "m,n\\ge N,\\ \\tfrac1N<\\varepsilon",
        "after_tex": "\\left|\\tfrac1m-\\tfrac1n\\right|\\le\\tfrac1N<\\varepsilon",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對 S05 選出的 N 與任意 m,n≥N，S04 給出 |1/m−1/n|≤1/N，再由 1/N<ε 與序的傳遞得 |1/m−1/n|<ε。",
          "en": "For the N chosen in S05 and any m,n≥N, S04 gives |1/m−1/n|≤1/N, and 1/N<ε with transitivity of order gives |1/m−1/n|<ε."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S05",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "≤ 與 < 串接得 <；不宣稱反向。",
            "en": "Chaining ≤ and < gives <; no converse is claimed."
          }
        },
        "leaf_basis": "order_basics + forall_intro"
      },
      {
        "id": "S07",
        "uses": [
          "S01",
          "S06",
          "forall_intro",
          "exists_intro"
        ],
        "before_tex": "\\text{for arbitrary } \\varepsilon>0:\\ N\\ \\text{found with S06}",
        "after_tex": "(x_n)=(\\tfrac1n)\\ \\text{is Cauchy in}\\ (\\mathbb{R},d)",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對每個 ε 都找到了 N（∃ 引入），而 ε 是任取的（∀ 引入）；依 S01，(1/n) 在 (ℝ,d) 中是 Cauchy。這是覆蓋所有 ε 的完整證書（MKO a2 允許的倒數序列界）。",
          "en": "For each ε an N was found (∃-introduction), and ε was arbitrary (∀-introduction); by S01, (1/n) is Cauchy in (ℝ,d). This is a complete certificate covering every ε (the reciprocal-sequence bound allowed by MKO assumption a2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S06",
            "forall_intro",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "需要 ε 的任意性與 N 的存在；論證沒有用到任何極限。",
            "en": "It needs the arbitrariness of ε and the existence of N; the argument uses no limit at all."
          }
        },
        "leaf_basis": "forall_intro + exists_intro"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "S03",
          "S07",
          "subspace_metric"
        ],
        "before_tex": "x_n\\in X?\\ \\text{and}\\ d_X(x_m,x_n)",
        "after_tex": "x_n\\in(0,1],\\ d_X(x_m,x_n)=\\left|\\tfrac1m-\\tfrac1n\\right|;\\ (x_n)\\ \\text{Cauchy in}\\ X",
        "relation": "implies",
        "rule": "subspace_metric",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由 L_recip（N=1），0<1/n≤1，所以每一項都在 X 中。X 上的距離是同一個公式，所以 S04 至 S07 的每個不等式不變，(x_n) 在 X 中也是 Cauchy。Cauchy 只依賴項與項之間的距離。",
          "en": "By L_recip with N=1, 0<1/n≤1, so every term lies in X. The distance on X is the same formula, so every inequality of S04 to S07 is unchanged and (x_n) is Cauchy in X as well. The Cauchy property depends only on distances between terms."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S07",
            "subspace_metric"
          ],
          "explanation": {
            "zh-Hant": "需要所有項都在 X 中與度量相同；不涉及任何極限。",
            "en": "It needs all terms in X and the same metric; no limit is involved."
          }
        },
        "leaf_basis": "subspace_metric + L_recip"
      },
      {
        "id": "S09",
        "uses": [
          "S08",
          "subspace_metric",
          "contradiction"
        ],
        "before_tex": "\\text{suppose } L\\in X\\ \\text{and}\\ x_n\\to L\\ \\text{in}\\ X",
        "after_tex": "\\text{contradiction; no } L\\in X\\ \\text{is a limit of}\\ (x_n)",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "反證：假設某個 L∈X（所以 L>0）是極限。S09.1 用 ε=L/2 取得 N₁；S09.2 用阿基米德性質找到 n≥N₁ 使 1/n<L/2；S09.3 顯示此時 |1/n−L|>L/2，與 S09.1 矛盾。所以 (x_n) 在 X 中沒有極限。",
          "en": "By contradiction: suppose some L∈X (so L>0) is a limit. S09.1 takes N₁ for ε=L/2; S09.2 uses the Archimedean property to find n≥N₁ with 1/n<L/2; S09.3 shows that then |1/n−L|>L/2, contradicting S09.1. So (x_n) has no limit in X."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "subspace_metric",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "L 是任意的 X 中的點，所以否定了所有可能的極限；只宣稱在 X 中沒有極限（在 ℝ 中極限 0 存在，但 0∉X）。",
            "en": "L is an arbitrary point of X, so every possible limit is ruled out; the claim is only that there is no limit in X (in ℝ the limit 0 exists, but 0∉X)."
          }
        },
        "leaf_basis": "contradiction + seq_limit_def"
      },
      {
        "id": "S09.1",
        "uses": [
          "subspace_metric",
          "seq_limit_def"
        ],
        "before_tex": "x_n\\to L\\in X,\\ \\varepsilon=\\tfrac L2>0",
        "after_tex": "\\exists N_1\\ \\forall n\\ge N_1:\\ \\left|\\tfrac1n-L\\right|<\\tfrac L2",
        "relation": "implies",
        "rule": "seq_limit_def",
        "conditions": [
          "L>0"
        ],
        "explanation": {
          "zh-Hant": "L∈X 表示 L>0，所以 ε=L/2>0 是合法的容許誤差；依極限定義得 N₁。",
          "en": "L∈X means L>0, so ε=L/2>0 is a legitimate tolerance; the definition of limit gives N₁."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "seq_limit_def",
            "subspace_metric"
          ],
          "explanation": {
            "zh-Hant": "需要 L>0；這正是 0∉X 的作用。",
            "en": "It needs L>0; this is exactly where 0∉X matters."
          }
        },
        "leaf_basis": "seq_limit_def + forall_elim"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "archimedean",
          "L_recip",
          "max_two"
        ],
        "before_tex": "n_0>\\tfrac2L\\ \\text{(Archimedean)},\\ n=\\max(n_0,N_1)",
        "after_tex": "n\\ge N_1,\\ \\tfrac1n\\le\\tfrac1{n_0}<\\tfrac L2",
        "relation": "implies",
        "rule": "archimedean",
        "conditions": [
          "L>0"
        ],
        "explanation": {
          "zh-Hant": "依阿基米德性質取自然數 n₀>2/L>0。令 n=max(n₀,N₁)，則 n≥N₁ 且 n≥n₀。依 L_recip，1/n≤1/n₀，且因 n₀>2/L，1/n₀<1/(2/L)=L/2。",
          "en": "By the Archimedean property take a natural number n₀>2/L>0. Let n=max(n₀,N₁); then n≥N₁ and n≥n₀. By L_recip, 1/n≤1/n₀, and since n₀>2/L, 1/n₀<1/(2/L)=L/2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "archimedean",
            "L_recip",
            "max_two"
          ],
          "explanation": {
            "zh-Hant": "同一個 n 同時滿足 n≥N₁ 與 1/n<L/2，這是取最大值的目的。",
            "en": "The same n satisfies both n≥N₁ and 1/n<L/2, which is the purpose of taking the maximum."
          }
        },
        "leaf_basis": "archimedean + L_recip + max_two"
      },
      {
        "id": "S09.3",
        "uses": [
          "S09.1",
          "S09.2",
          "abs_rules"
        ],
        "before_tex": "\\tfrac1n<\\tfrac L2,\\ n\\ge N_1",
        "after_tex": "\\left|\\tfrac1n-L\\right|\\ge L-\\tfrac1n>L-\\tfrac L2=\\tfrac L2",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "|w|≥−w 對任何實數成立；取 w=1/n−L 得 |1/n−L|≥L−1/n。由 1/n<L/2 得 L−1/n>L/2。所以 |1/n−L|>L/2，但 n≥N₁，S09.1 要求 |1/n−L|<L/2，矛盾。",
          "en": "|w|≥−w holds for every real w; with w=1/n−L this gives |1/n−L|≥L−1/n. From 1/n<L/2, L−1/n>L/2. So |1/n−L|>L/2, yet n≥N₁ and S09.1 requires |1/n−L|<L/2, a contradiction."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09.1",
            "S09.2",
            "abs_rules",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "同一個 n 同時滿足兩個互相矛盾的不等式；矛盾來自「L∈X 是極限」這個假設。",
            "en": "The same n satisfies two contradictory inequalities; the contradiction comes from the assumption that some L∈X is a limit."
          }
        },
        "leaf_basis": "abs_rules + order_basics + contradiction"
      },
      {
        "id": "S10",
        "uses": [
          "S08",
          "S09",
          "mko_cauchy"
        ],
        "before_tex": "(x_n)\\ \\text{Cauchy in}\\ X,\\ \\text{no limit in}\\ X",
        "after_tex": "\\text{Cauchy does not presuppose a limit in the space}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 X=(0,1] 中，(1/n) 是 Cauchy（S08）卻沒有極限（S09）。這正是 MKO 所說：定義不先假設極限存在或位於空間內。X 不是完備的度量空間，但本稿不展開完備性理論。",
          "en": "In X=(0,1], (1/n) is Cauchy (S08) yet has no limit (S09). This is exactly the MKO's point: the definition does not first assume that a limit exists or belongs to the space. X is not a complete metric space, but this guide does not develop the theory of completeness."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "S09",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "只合取兩個已證結論；關於完備性的一句只是說明。",
            "en": "It only conjoins two proved conclusions; the sentence on completeness is a remark."
          }
        },
        "leaf_basis": "and_intro_elim + mko_cauchy"
      },
      {
        "id": "S11",
        "uses": [
          "G3",
          "abs_rules",
          "power_laws"
        ],
        "before_tex": "y_{n+1}=(-1)^{n+1}=-y_n",
        "after_tex": "|y_n-y_{n+1}|=|2y_n|=2|y_n|=2\\ \\text{for every } n\\ge 1",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(−1)^(n+1)=(−1)^n·(−1)=−y_n，所以 y_n−y_(n+1)=2y_n，而 |y_n|=|±1|=1，得 |y_n−y_(n+1)|=2。這對每個 n 成立，不是只看列出的幾項。",
          "en": "(−1)^(n+1)=(−1)^n·(−1)=−y_n, so y_n−y_(n+1)=2y_n, and |y_n|=|±1|=1, giving |y_n−y_(n+1)|=2. This holds for every n, not just for some listed terms."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "power_laws"
          ],
          "explanation": {
            "zh-Hant": "用到指數律與 |uv|=|u||v|；對所有 n 成立。",
            "en": "It uses the exponent laws and |uv|=|u||v|; it holds for all n."
          }
        },
        "leaf_basis": "power_laws + abs_rules + neg_mul"
      },
      {
        "id": "S12",
        "uses": [
          "mko_cauchy",
          "quantifier_negation"
        ],
        "before_tex": "\\neg\\,(y_n)\\ \\text{Cauchy}",
        "after_tex": "\\exists\\varepsilon>0\\ \\forall N\\ \\exists m,n\\ge N:\\ |y_m-y_n|\\ge\\varepsilon",
        "relation": "equivalent",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "否定 ∀ε∃N∀m,n 的命題：依序把每個量詞翻轉，並把 <ε 換成 ≥ε。所以要否證 Cauchy，只需一個 ε，使每個 N 之後都找得到兩項相距至少 ε。",
          "en": "Negating the statement ∀ε∃N∀m,n flips each quantifier in turn and replaces <ε by ≥ε. So to refute the Cauchy property one ε is enough, such that beyond every N there are two terms at distance at least ε."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "quantifier_negation",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "量詞否定與 ¬(d<ε) ⇔ d≥ε 都是雙向的。",
            "en": "Quantifier negation and ¬(d<ε) ⇔ d≥ε are both two-way."
          }
        },
        "leaf_basis": "quantifier_negation + order_basics"
      },
      {
        "id": "S13",
        "uses": [
          "S11",
          "S12",
          "exists_intro"
        ],
        "before_tex": "\\varepsilon=1,\\ N\\ \\text{arbitrary},\\ m=N,\\ n=N+1",
        "after_tex": "|y_N-y_{N+1}|=2\\ge 1;\\ (y_n)\\ \\text{is not Cauchy}",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取 ε=1。對任意 N，取相鄰的 m=N、n=N+1（都 ≥N），由 S11 距離為 2≥1。依 S12 的等價，(y_n) 不是 Cauchy。這是 MKO 所說的交替反例族：對每個 N 都有相鄰反例。",
          "en": "Take ε=1. For any N take the adjacent m=N and n=N+1 (both ≥N); by S11 their distance is 2≥1. By the equivalence of S12, (y_n) is not Cauchy. This is the MKO's alternating counterexample family: for every N there is an adjacent counterexample."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "S12",
            "exists_intro",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要對每個 N 都有見證；S11 對所有 n 成立，所以見證對每個 N 都存在。",
            "en": "It needs a witness for every N; S11 holds for all n, so a witness exists for every N."
          }
        },
        "leaf_basis": "exists_intro + forall_intro + order_basics"
      },
      {
        "id": "S14",
        "uses": [
          "G4",
          "mko_cauchy",
          "forall_intro"
        ],
        "before_tex": "\\varepsilon>0\\ \\text{arbitrary},\\ N=3",
        "after_tex": "m,n\\ge 3\\Rightarrow|z_m-z_n|=|5-5|=0<\\varepsilon;\\ (z_n)\\ \\text{Cauchy}",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [
          "\\varepsilon>0"
        ],
        "explanation": {
          "zh-Hant": "最終常值模型：對任意 ε>0，取 N=3（與 ε 無關）。m,n≥3 時 z_m=z_n=5，距離 0<ε。ε 任意，所以 (z_n) 是 Cauchy。前兩項 0 與 7 不影響結論，因為定義只看尾段（MKO para1）。",
          "en": "Ultimately constant model: for any ε>0 take N=3, independent of ε. For m,n≥3, z_m=z_n=5 and the distance is 0<ε. Since ε is arbitrary, (z_n) is Cauchy. The first two terms 0 and 7 do not affect the conclusion, because the definition looks only at the tail (MKO para1)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G4",
            "mko_cauchy",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要 n≥3 時常值 5 對所有 n 成立（G4 明示），而不是只在列出的幾項成立。",
            "en": "It needs the constant value 5 for all n≥3 (stated in G4), not only for some listed terms."
          }
        },
        "leaf_basis": "forall_intro + exists_intro + numeric_evaluation"
      }
    ],
    "result": {
      "zh-Hant": "(1/n) 在 (ℝ,|·|) 中是 Cauchy：對每個 ε>0，取 N 使 1/N<ε，則所有 m,n≥N 有 |1/m−1/n|≤1/N<ε。它在 X=(0,1] 中仍是 Cauchy，卻在 X 中沒有極限。(−1)^n 不是 Cauchy：ε=1 時對每個 N，相鄰兩項相距 2。最終常值序列 z 以 N=3 對所有 ε 成立，是 Cauchy。",
      "en": "(1/n) is Cauchy in (ℝ,|·|): for every ε>0 take N with 1/N<ε; then |1/m−1/n|≤1/N<ε for all m,n≥N. It is still Cauchy in X=(0,1], yet has no limit in X. (−1)^n is not Cauchy: with ε=1, for every N two adjacent terms are 2 apart. The ultimately constant sequence z is Cauchy, with N=3 working for every ε."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "調和部分和 H_n 的相鄰差 1/(n+1) 趨於 0，所以 (H_n) 是 Cauchy",
          "en": "The adjacent differences 1/(n+1) of the harmonic partial sums H_n tend to 0, so (H_n) is Cauchy"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "對每個 N，H_{2N}−H_N≥1/2。",
          "en": "For every N, H_{2N}−H_N≥1/2."
        }
      },
      {
        "claim": {
          "zh-Hant": "(y_n)=((−1)^n) 只取 −1 與 1 兩個值，所以是 Cauchy",
          "en": "(y_n)=((−1)^n) takes only the two values −1 and 1, so it is Cauchy"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "ε=1 時，對每個 N 有 |y_N−y_(N+1)|=2≥1。",
          "en": "With ε=1, |y_N−y_(N+1)|=2≥1 for every N."
        }
      },
      {
        "claim": {
          "zh-Hant": "(1/n) 在 X=(0,1] 中是 Cauchy，所以它在 X 中收斂",
          "en": "(1/n) is Cauchy in X=(0,1], so it converges in X"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "S09：X 中沒有任何點是它的極限。",
          "en": "S09: no point of X is its limit."
        }
      },
      {
        "claim": {
          "zh-Hant": "列出 1/n 的前 1000 項並核對相互距離，就證明了 (1/n) 是 Cauchy",
          "en": "Listing the first 1000 terms of 1/n and checking their mutual distances proves that (1/n) is Cauchy"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "有限尾段只是有界核對（MKO a1）；證明是 S02 至 S07 對每個 ε 的論證。",
          "en": "A finite tail is only a bounded audit (MKO assumption a1); the proof is the argument of S02 to S07 for every ε."
        }
      }
    ],
    "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": "seq_limit_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "數列極限：a_n→L 表示對每個 ε>0，存在 N 使所有 n≥N 都有 |a_n−L|<ε。",
          "en": "Limit of a sequence: a_n→L means that for every ε>0 there is N such that |a_n−L|<ε for all n≥N."
        }
      },
      {
        "id": "archimedean",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "阿基米德性質：對任意實數 x，存在自然數 n>x；因此對任意 ε>0，存在 n 使 1/n<ε。",
          "en": "Archimedean property: for every real x there is a natural number n>x; hence for every ε>0 there is n with 1/n<ε."
        }
      },
      {
        "id": "abs_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "絕對值（實數）：|u|=u（u≥0）、|u|=−u（u<0）；|u|≥0；|uv|=|u||v|；三角不等式 |u+v|≤|u|+|v|；|u|<r ⇔ −r<u<r。",
          "en": "Absolute value (reals): |u|=u if u≥0 and |u|=−u if u<0; |u|≥0; |uv|=|u||v|; triangle inequality |u+v|≤|u|+|v|; |u|<r ⇔ −r<u<r."
        }
      },
      {
        "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": "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": "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_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "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": "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": "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": "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": "power_laws",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。",
          "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1."
        }
      },
      {
        "id": "neg_mul",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。",
          "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv."
        }
      },
      {
        "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": "mko_cauchy",
        "kind": "cited",
        "source": {
          "mko_id": "mko-cauchy-sequence",
          "sha256": "f833bc771b9e0a259993b3560086625a99bfce9069edf8b08acd37a09b08c004",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "序列 (x_n) 為 Cauchy，若對每個 ε>0 存在 N，使所有 m,n≥N 都有 d(x_m,x_n)<ε。定義不先假設極限存在或位於空間內。有限列出的尾段只作有界核對，不能證一般 Cauchy。",
          "en": "A sequence (x_n) is Cauchy if for every ε>0 there exists N such that all m,n≥N satisfy d(x_m,x_n)<ε. The definition does not first assume that a limit exists or belongs to the space. A finitely listed tail supplies only a bounded audit, not a proof of the general Cauchy property."
        }
      },
      {
        "id": "max_two",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "兩數的最大值：max(s,t) 是 s、t 中較大者；max(s,t)≥s、max(s,t)≥t；若 s≤K 且 t≤K，則 max(s,t)≤K。",
          "en": "Maximum of two numbers: max(s,t) is the larger of s and t; max(s,t)≥s and max(s,t)≥t; if s≤K and t≤K, then max(s,t)≤K."
        }
      },
      {
        "id": "subspace_metric",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "子空間度量：對 X⊆ℝ，d(s,t)=|s−t| 限制到 X 仍是度量。X 中的序列在 X 中收斂到 L，表示 L∈X 且對每個 ε>0 有 N 使 n≥N 時 |x_n−L|<ε。",
          "en": "Subspace metric: for X⊆ℝ, d(s,t)=|s−t| restricted to X is still a metric. A sequence in X converges in X to L when L∈X and for every ε>0 there is N with |x_n−L|<ε for n≥N."
        }
      },
      {
        "id": "L_absmax",
        "kind": "lemma",
        "proved_in": [
          "S02.1",
          "S02.2"
        ],
        "statement": {
          "zh-Hant": "若 s≥0 且 t≥0，則 |s−t|≤max(s,t)。",
          "en": "If s≥0 and t≥0, then |s−t|≤max(s,t)."
        }
      },
      {
        "id": "L_recip",
        "kind": "lemma",
        "proved_in": [
          "S03.1"
        ],
        "statement": {
          "zh-Hant": "對實數 m≥N>0：0<1/m≤1/N；若 m>N，則 1/m<1/N。",
          "en": "For reals m≥N>0: 0<1/m≤1/N; and if m>N, then 1/m<1/N."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S05",
        "lhs": "1/11",
        "relation": "<",
        "rhs": "1/10"
      },
      {
        "id": "N2",
        "step": "S06",
        "lhs": "abs(1/11-1/1000)",
        "relation": "<",
        "rhs": "1/10"
      },
      {
        "id": "N3",
        "step": "S02",
        "vars": [
          "m",
          "n"
        ],
        "range": [
          1,
          30
        ],
        "lhs": "abs(1/m-1/n)",
        "relation": "<=",
        "rhs": "max(1/m,1/n)"
      },
      {
        "id": "N4",
        "step": "S11",
        "lhs": "abs((-1)**5-(-1)**6)",
        "rhs": "2"
      },
      {
        "id": "N5",
        "step": "S11",
        "vars": [
          "n"
        ],
        "range": [
          1,
          40
        ],
        "lhs": "abs((-1)**n-(-1)**(n+1))",
        "rhs": "2"
      },
      {
        "id": "N6",
        "step": "S13",
        "lhs": "2",
        "relation": ">=",
        "rhs": "1"
      },
      {
        "id": "N7",
        "step": "S14",
        "lhs": "abs(5-5)",
        "rhs": "0"
      },
      {
        "id": "N8",
        "step": "S09.3",
        "lhs": "abs(1/5-1/2)",
        "relation": ">=",
        "rhs": "1/4"
      },
      {
        "id": "N9",
        "step": "S09.2",
        "lhs": "1/5",
        "relation": "<",
        "rhs": "(1/2)/2"
      },
      {
        "id": "N10",
        "lhs": "1/3+1/4",
        "relation": ">=",
        "rhs": "1/2"
      },
      {
        "id": "N11",
        "lhs": "1/4+1/5+1/6",
        "relation": ">=",
        "rhs": "1/2"
      },
      {
        "id": "N12",
        "step": "S03.1",
        "lhs": "1/7",
        "relation": "<=",
        "rhs": "1/3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "停在 MKO 的切片內：完整證書只用於明示的最終常值模型、倒數序列的界，以及一個精確反例族（交替序列）。度量是 ℝ 的 |s−t| 及其在 (0,1] 上的限制。不處理完備度量空間的一般理論或 Banach 空間；有限列出的尾段只作有界核對；不改動 canonical MKO 或英文 edition。",
      "en": "Stays inside the MKO's slice: complete certificates only for an explicit ultimately constant model, reciprocal-sequence bounds and one exact counterexample family (the alternating sequence). The metric is |s−t| on ℝ and its restriction to (0,1]. The general theory of complete metric spaces and Banach spaces is not treated; finitely listed tails serve only as bounded audits; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "d(s,t)=|s-t|",
        "ℝ 上的標準度量",
        "the standard metric on ℝ"
      ],
      [
        "X=(0,1]",
        "帶限制度量的子空間",
        "the subspace with the restricted metric"
      ],
      [
        "x_n=\\tfrac1n",
        "倒數序列",
        "the reciprocal sequence"
      ],
      [
        "y_n=(-1)^n",
        "交替序列",
        "the alternating sequence"
      ],
      [
        "z_n",
        "最終常值序列",
        "the ultimately constant sequence"
      ],
      [
        "\\varepsilon,\\ N",
        "容許誤差與尾段起點",
        "the tolerance and the start of the tail"
      ]
    ],
    "md_error_controls": [
      {
        "title": "有限尾段很近不證 Cauchy / A close finite tail does not prove Cauchy",
        "zh": "MKO 指出的誤解是長有限尾段很靠近就宣稱整個無限序列 Cauchy。調和部分和 H_n=Σ_{k=1}^{n} 1/k 的相鄰差 1/(n+1) 越來越小，任何列出的有限尾段都可能看起來很接近；但對每個 N，H_{2N}−H_N=Σ_{k=N+1}^{2N} 1/k≥N·1/(2N)=1/2（每一項 1/k≥1/(2N)），所以 ε=1/2 時 Cauchy 條件失敗。",
        "en": "The mistake named by the MKO is to call the whole infinite sequence Cauchy because a long finite tail is close together. For the harmonic partial sums H_n=Σ_{k=1}^{n} 1/k the adjacent differences 1/(n+1) keep shrinking, and any listed finite tail may look close; but for every N, H_{2N}−H_N=Σ_{k=N+1}^{2N} 1/k≥N·1/(2N)=1/2 (each term 1/k≥1/(2N)), so the Cauchy condition fails for ε=1/2."
      },
      {
        "title": "比較尾段任兩項，不只相鄰項 / Compare any two terms of the tail, not only adjacent ones",
        "zh": "Cauchy 要求所有 m,n≥N，而不是只有 n 與 n+1。調和部分和的相鄰差趨於 0 卻不是 Cauchy。另一方面，(−1)^n 的否證恰好可以用相鄰項（S13），因為否證只需一對見證。",
        "en": "Cauchy requires all m,n≥N, not only n and n+1. The harmonic partial sums have adjacent differences tending to 0 yet are not Cauchy. On the other hand, adjacent terms suffice to refute (−1)^n (S13), because a refutation needs only one pair of witnesses."
      },
      {
        "title": "不要先假設極限 / Do not presuppose a limit",
        "zh": "S01 至 S07 從未使用極限。若把「Cauchy」誤寫成「與某個候選極限很近」，在 X=(0,1] 中就找不到候選（S09），卻仍然是 Cauchy（S08）。",
        "en": "S01 to S07 never use a limit. If 'Cauchy' is misread as 'close to some candidate limit', then in X=(0,1] there is no candidate (S09), and yet the sequence is still Cauchy (S08)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "N3 與 N5 是有限範圍的核對（m,n 從 1 到 30；n 從 1 到 40），只支持這些點，不證明 L_absmax 或 S11 的一般結論；一般結論由 S02.1、S02.2 與 S11 的論證給出。",
        "en": "N3 and N5 are finite-range checks (m, n from 1 to 30; n from 1 to 40); they support only those points and do not prove L_absmax or the general claim of S11, which come from the arguments of S02.1, S02.2 and S11."
      },
      {
        "zh-Hant": "錯誤控制中的調和部分和不在 MKO 列出的完整證書類型內，只作為反例說明；其下界論證在錯誤控制中完整給出，N10、N11 只核 N=2、N=3。",
        "en": "The harmonic partial sums in the error controls are not among the certificate types listed by the MKO and serve only as a counterexample illustration; their lower-bound argument is given in full in the error control, and N10, N11 check only N=2 and N=3."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "3cf4cca5b06ba03d3f526ddbf2d5e9556d77fa989091a6210c0ad46c7169d2a5",
    "original_source_sha256": "3cf4cca5b06ba03d3f526ddbf2d5e9556d77fa989091a6210c0ad46c7169d2a5",
    "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."
  }
}