{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-sequence-limit-01",
  "mko_id": "mko-sequence-limit",
  "title": {
    "zh-Hant": "數列極限：倒數模型的精確尾界與 ε-N 證明",
    "en": "Sequence limits: the exact tail cutoff and an ε-N proof for the reciprocal model"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-sequence-limit-01",
    "source": {
      "mko_id": "mko-sequence-limit",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "3018ee44ad1a41db719f4c5275fd215d436e1dc4e73fdc44993604a211486fff",
      "english_sha256": "ee964b7eba696e940cf9793cbba93912244a2bd9ed11ec3f4e647fafa8fd38ae"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "abs_rules",
      "order_basics",
      "frac_rules",
      "archimedean",
      "seq_limit_def",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "對 u(n)=2−3/(n+1)（n∈ℕ0，即 MKO 模型 L=2、c=−3、d=1）：(1) 證明候選尾界 N 對 ε 合法當且僅當 3/(N+1)<ε；(2) 求 ε=1/2 的最小合法 N，並說明 N=5 為何不合法；(3) 以 ε-N 定義證明 u(n)→2；(4) 說明 N 必須依 ε 選取。",
      "en": "For u(n)=2−3/(n+1) with n∈ℕ0 (the MKO model with L=2, c=−3, d=1): (1) prove that a candidate cutoff N is valid for ε exactly when 3/(N+1)<ε; (2) find the least valid N for ε=1/2 and explain why N=5 fails; (3) prove u(n)→2 from the ε-N definition; (4) show that N must be chosen after ε."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "ℕ0 包含 0；對每個 n∈ℕ0，n+1≥1>0。",
          "en": "ℕ0 includes 0; for every n∈ℕ0, n+1≥1>0."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "u(n)=2+(−3)/(n+1)，即 MKO 模型 u(n)=L+c/(n+d) 取 L=2、c=−3、d=1。",
          "en": "u(n)=2+(−3)/(n+1), that is the MKO model u(n)=L+c/(n+d) with L=2, c=−3, d=1."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "尾界題的容差 ε=1/2（p=1、q=2）；比較用容差 ε=1/10。",
          "en": "The tolerance for the cutoff question is ε=1/2 (p=1, q=2); the comparison tolerance is ε=1/10."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "seq_limit_def",
          "mko_seq_limit"
        ],
        "before_tex": "u(n)\\to 2",
        "after_tex": "\\forall\\varepsilon>0\\ \\exists N\\in\\mathbb{N}_0\\ \\forall n\\ge N:\\ |u(n)-2|<\\varepsilon",
        "relation": "definition",
        "rule": "mko_seq_limit",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 的定義展開 u(n)→2。量詞順序是「對每個 ε，存在 N，對所有 n≥N」：N 在 ε 之後選，可以依 ε 改變；選定後要管住整個尾段，而且誤差條件是嚴格的「<」。",
          "en": "Unfold u(n)→2 by the MKO's definition. The quantifier order is 'for every ε there is N such that for all n≥N': N is chosen after ε and may depend on it; once chosen it must control the whole tail, and the error condition is the strict '<'."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "seq_limit_def",
            "mko_seq_limit"
          ],
          "explanation": {
            "zh-Hant": "此步只展開定義，不推出新事實。",
            "en": "This step only unfolds the definition and derives no new fact."
          }
        },
        "leaf_basis": "seq_limit_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "u(n)=L+\\frac{c}{n+d}",
        "after_tex": "L=2,\\ c=-3,\\ d=1,\\ n+1\\ge 1>0\\ (n\\in\\mathbb{N}_0)",
        "relation": "definition",
        "rule": "mko_seq_limit",
        "conditions": [
          "d\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "把本例放進 MKO 模型：L=2、c=−3、d=1。因 n≥0，分母 n+1≥1>0，每一項都有定義，而且分母為正（後面翻轉不等號時要用）。",
          "en": "Place the example in the MKO model: L=2, c=−3, d=1. Since n≥0, the denominator n+1≥1>0, so every term is defined and the denominator is positive (needed later when inequalities are inverted)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_seq_limit"
          ],
          "explanation": {
            "zh-Hant": "只是對照參數；d=1 滿足 MKO 的 d≥1。",
            "en": "It only matches parameters; d=1 meets the MKO's d≥1."
          }
        },
        "leaf_basis": "def_intro + order_basics"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "abs_rules"
        ],
        "before_tex": "e(n)=|u(n)-2|",
        "after_tex": "\\left|\\frac{-3}{n+1}\\right|=\\frac{3}{n+1}",
        "relation": "equivalent",
        "rule": "abs_rules",
        "conditions": [
          "n+1>0"
        ],
        "explanation": {
          "zh-Hant": "u(n)−2=−3/(n+1)。因 n+1>0，3/(n+1)>0，所以 −3/(n+1)<0，依絕對值定義 |−3/(n+1)|=3/(n+1)。誤差只依 |c|=3 與 n，與 L=2 無關（N5 對 n=0 到 60 抽查）。",
          "en": "u(n)−2=−3/(n+1). Since n+1>0, 3/(n+1)>0, so −3/(n+1)<0 and by the definition of absolute value |−3/(n+1)|=3/(n+1). The error depends only on |c|=3 and n, not on L=2 (N5 spot-checks n=0 to 60)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "abs_rules",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個數；需要 n+1>0 來決定 −3/(n+1) 的正負。",
            "en": "Both sides name the same number; n+1>0 is needed to know the sign of −3/(n+1)."
          }
        },
        "leaf_basis": "abs_rules + order_basics + frac_rules"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "S03"
        ],
        "before_tex": "n\\ge N\\ (n,N\\in\\mathbb{N}_0)",
        "after_tex": "\\frac{3}{n+1}\\le\\frac{3}{N+1}",
        "relation": "implies",
        "rule": "L_recip_mono",
        "conditions": [
          "0<N+1\\le n+1"
        ],
        "explanation": {
          "zh-Hant": "n≥N 推出 0<N+1≤n+1。依引理 L_recip_mono（a=N+1、b=n+1），1/(n+1)≤1/(N+1)；再乘以正數 3 保持不等號。所以本模型的誤差隨 n 不增：尾段中最大的誤差在 n=N。",
          "en": "n≥N gives 0<N+1≤n+1. By Lemma L_recip_mono (a=N+1, b=n+1), 1/(n+1)≤1/(N+1); multiplying by the positive 3 keeps the inequality. So in this model the error does not increase with n: the largest error in the tail is at n=N."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_recip_mono",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "只宣稱單向不等；需要分母為正（S02）。這個單調性是本模型特有的，不是一般數列的性質。",
            "en": "Only a one-way inequality is claimed; positive denominators (S02) are needed. This monotonicity belongs to this model, not to sequences in general."
          }
        },
        "leaf_basis": "L_recip_mono + order_basics"
      },
      {
        "id": "S04.1",
        "uses": [
          "order_basics",
          "frac_rules"
        ],
        "before_tex": "0<a\\le b",
        "after_tex": "\\frac{1}{b}\\le\\frac{1}{a}",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "a>0",
          "b>0"
        ],
        "explanation": {
          "zh-Hant": "a、b>0，所以 ab>0，1/(ab)>0。把 a≤b 兩邊乘正數 1/(ab)，不等號方向不變：a/(ab)≤b/(ab)，即 1/b≤1/a。這證得引理 L_recip_mono。",
          "en": "a, b>0, so ab>0 and 1/(ab)>0. Multiplying a≤b by the positive 1/(ab) keeps the direction: a/(ab)≤b/(ab), that is 1/b≤1/a. This proves Lemma L_recip_mono."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "需要 a>0；若 a<0<b，例如 a=−1、b=1，則 1/b=1>−1=1/a，結論不成立。",
            "en": "It needs a>0; with a<0<b, for example a=−1 and b=1, 1/b=1>−1=1/a and the conclusion fails."
          }
        },
        "leaf_basis": "order_basics + frac_rules + mul_inverse_nonzero"
      },
      {
        "id": "S05",
        "uses": [
          "S03",
          "S04"
        ],
        "before_tex": "\\forall n\\ge N:\\ \\frac{3}{n+1}<\\varepsilon",
        "after_tex": "\\frac{3}{N+1}<\\varepsilon",
        "relation": "equivalent",
        "rule": "L_cutoff",
        "conditions": [
          "\\varepsilon>0"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_cutoff，候選 N 管住整個尾段，當且僅當第 N 項本身的誤差小於 ε。這把 MKO 假設 a2「均勻尾界等價於一個精確不等式」落實為證明。引理在 S05.1、S05.2 證明。",
          "en": "By Lemma L_cutoff, a candidate N controls the whole tail exactly when the N-th error itself is below ε. This turns MKO assumption a2, 'a uniform cutoff reduces to one exact inequality', into a proof. The lemma is proved in S05.1 and S05.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_cutoff"
          ],
          "explanation": {
            "zh-Hant": "兩個方向分別在 S05.1（⇒）與 S05.2（⇐）證明；⇐ 方向依賴 S04 的單調性。",
            "en": "The two directions are proved in S05.1 (⇒) and S05.2 (⇐); the ⇐ direction relies on the monotonicity of S04."
          }
        },
        "leaf_basis": "forall_elim + forall_intro + order_basics + iff_intro"
      },
      {
        "id": "S05.1",
        "uses": [
          "forall_elim"
        ],
        "before_tex": "\\forall n\\ge N:\\ \\frac{3}{n+1}<\\varepsilon",
        "after_tex": "\\frac{3}{N+1}<\\varepsilon",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "⇒：全稱命題對尾段每一項成立，特別取 n=N（N≥N）即得。",
          "en": "⇒: the universal statement holds for every term of the tail, in particular for n=N (as N≥N)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "全稱消去，不需其他前提。",
            "en": "Universal elimination; no other premise is needed."
          }
        },
        "leaf_basis": "forall_elim"
      },
      {
        "id": "S05.2",
        "uses": [
          "S04",
          "order_basics",
          "forall_intro"
        ],
        "before_tex": "\\frac{3}{N+1}<\\varepsilon",
        "after_tex": "\\forall n\\ge N:\\ \\frac{3}{n+1}\\le\\frac{3}{N+1}<\\varepsilon",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "⇐：取任意 n≥N。由 S04，3/(n+1)≤3/(N+1)，再由假設 3/(N+1)<ε，依序的傳遞得 3/(n+1)<ε。n 任意，依全稱推廣對所有 n≥N 成立。S05.1、S05.2 合起來證得引理 L_cutoff。",
          "en": "⇐: take an arbitrary n≥N. By S04, 3/(n+1)≤3/(N+1), and by hypothesis 3/(N+1)<ε, so by transitivity of order 3/(n+1)<ε. As n was arbitrary, generalization gives it for every n≥N. S05.1 and S05.2 together prove Lemma L_cutoff."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "order_basics",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "關鍵依賴是 S04 的誤差不增；對誤差可能回升的一般數列，只檢查第 N 項並不足夠。",
            "en": "The key dependence is the nonincreasing error of S04; for a general sequence whose errors can rise again, checking the N-th term alone is not enough."
          }
        },
        "leaf_basis": "order_basics + forall_intro"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "G3",
          "frac_rules"
        ],
        "before_tex": "\\frac{3}{N+1}<\\frac{p}{q}",
        "after_tex": "3q<p(N+1)",
        "relation": "equivalent",
        "rule": "order_basics",
        "conditions": [
          "q>0",
          "N+1>0"
        ],
        "explanation": {
          "zh-Hant": "兩邊同乘正數 q(N+1)，不等號方向不變：3q<p(N+1)。這就是 MKO 的精確條件 |c|q<p(N+d)，只含整數。嚴格的「<」保留下來：等號不合格。",
          "en": "Multiply both sides by the positive q(N+1); the direction is unchanged: 3q<p(N+1). This is the MKO's exact condition |c|q<p(N+d), involving integers only. The strict '<' is kept: equality does not qualify."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "order_basics",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "反向是兩邊同除以正數 q(N+1)；乘除正數都保持嚴格不等號。",
            "en": "Backward, divide both sides by the positive q(N+1); multiplying or dividing by a positive number keeps a strict inequality."
          }
        },
        "leaf_basis": "order_basics + frac_rules + mul_inverse_nonzero"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "G3"
        ],
        "before_tex": "3\\cdot 2<1\\cdot(N+1)",
        "after_tex": "N\\ge 6\\ (N\\in\\mathbb{N}_0)",
        "relation": "equivalent",
        "rule": "order_basics",
        "conditions": [
          "N\\ \\text{integer}"
        ],
        "explanation": {
          "zh-Hant": "代入 p=1、q=2：6<N+1，即 N>5。N 是整數，5 與 6 之間沒有整數，所以 N>5 ⇔ N≥6。",
          "en": "Substitute p=1, q=2: 6<N+1, that is N>5. N is an integer and there is no integer strictly between 5 and 6, so N>5 ⇔ N≥6."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "order_basics",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "對整數 N 兩邊等價；若 N 可以是任意實數，N>5 不能寫成 N≥6。",
            "en": "For integer N the two are equivalent; if N could be any real, N>5 could not be rewritten as N≥6."
          }
        },
        "leaf_basis": "order_basics + add_inverse + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S03",
          "S05",
          "G3"
        ],
        "before_tex": "N=5",
        "after_tex": "e(5)=\\frac{3}{6}=\\frac12\\not<\\frac12:\\ N=5\\ \\text{invalid}",
        "relation": "implies",
        "rule": "L_cutoff",
        "conditions": [],
        "explanation": {
          "zh-Hant": "N=5 時第 5 項的誤差 3/6=1/2，恰好等於 ε，不小於 ε；依 S05 的 ⇒ 方向的逆否，N=5 不是合法尾界。這正是 MKO 警告的「誤差等於 ε 不算小於 ε」。",
          "en": "For N=5 the error of term 5 is 3/6=1/2, exactly ε, not below ε; by the contrapositive of the ⇒ direction of S05, N=5 is not a valid cutoff. This is the MKO's warning that an error equal to ε is not less than ε."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S05",
            "contrapositive"
          ],
          "explanation": {
            "zh-Hant": "一項不滿足就足以否定 N=5；計算見 N2。",
            "en": "One failing term is enough to reject N=5; the computation is N2."
          }
        },
        "leaf_basis": "L_cutoff + contrapositive + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S05",
          "S07",
          "S08"
        ],
        "before_tex": "N=6:\\ \\frac{3}{7}<\\frac12",
        "after_tex": "\\forall n\\ge 6:\\ e(n)<\\frac12;\\ \\text{least valid}\\ N=6",
        "relation": "implies",
        "rule": "L_cutoff",
        "conditions": [],
        "explanation": {
          "zh-Hant": "3·2=6<7=1·(6+1)，所以 3/7<1/2；依 S05（⇐），所有 n≥6 的誤差都小於 1/2。由 S07，合法的 N 恰是 N≥6，所以最小合法 N=6；與 MKO 公式 max(0, floor(3·2/1)−1+1)=6 一致（N4）。N6 是 n=6 到 200 的有限抽查，不是證明，證明是 S05。",
          "en": "3·2=6<7=1·(6+1), so 3/7<1/2; by S05 (⇐), every n≥6 has error below 1/2. By S07 the valid cutoffs are exactly N≥6, so the least valid N is 6, agreeing with the MKO formula max(0, floor(3·2/1)−1+1)=6 (N4). N6 is a finite spot check for n=6 to 200, not a proof; the proof is S05."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S05",
            "S07",
            "S08"
          ],
          "explanation": {
            "zh-Hant": "目標 G：最小合法 N 是 6。足夠條件 H：6 合法且每個 N≤5 不合法；H⇒G 由最小值定義。H 由 S07（合法 ⇔ N≥6）正向證成，S08 是其中 N=5 的具體核對。",
            "en": "Goal G: the least valid N is 6. Sufficient condition H: 6 is valid and every N≤5 is not; H⇒G by the definition of least. H follows forwards from S07 (valid ⇔ N≥6), with S08 as the concrete check at N=5."
          }
        },
        "leaf_basis": "L_cutoff + order_basics + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "archimedean",
          "S01"
        ],
        "before_tex": "\\varepsilon>0\\ \\text{arbitrary}",
        "after_tex": "N\\in\\mathbb{N}_0,\\ N>\\frac{3}{\\varepsilon},\\ \\text{hence}\\ \\frac{3}{N+1}<\\varepsilon",
        "relation": "implies",
        "rule": "archimedean",
        "conditions": [
          "\\varepsilon>0"
        ],
        "explanation": {
          "zh-Hant": "取任意 ε>0，不加其他條件。3/ε 是實數，依阿基米德性質存在自然數 N>3/ε，於是 N+1>3/ε>0。兩邊乘正數 ε/(N+1)：ε>3/(N+1)。例如 ε=1/10 時可取 N=31（N12）；不必取最小的 N。",
          "en": "Take an arbitrary ε>0 with no other condition. 3/ε is a real number, so by the Archimedean property there is a natural number N>3/ε, and then N+1>3/ε>0. Multiplying both sides by the positive ε/(N+1) gives ε>3/(N+1). For example, for ε=1/10 one may take N=31 (N12); N need not be the least one."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "archimedean",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "N 依 ε 而選，存在性來自阿基米德性質；乘的是正數，不等號方向不變。",
            "en": "N is chosen depending on ε, and its existence comes from the Archimedean property; the multiplier is positive, so the inequality keeps its direction."
          }
        },
        "leaf_basis": "archimedean + order_basics + frac_rules"
      },
      {
        "id": "S11",
        "uses": [
          "S10",
          "S05",
          "S03"
        ],
        "before_tex": "\\frac{3}{N+1}<\\varepsilon",
        "after_tex": "\\forall n\\ge N:\\ |u(n)-2|=\\frac{3}{n+1}<\\varepsilon",
        "relation": "implies",
        "rule": "L_cutoff",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 S05 的 ⇐ 方向，S10 選出的 N 管住整個尾段；再由 S03，3/(n+1) 正是誤差 |u(n)−2|。",
          "en": "By the ⇐ direction of S05, the N chosen in S10 controls the whole tail; by S03, 3/(n+1) is exactly the error |u(n)−2|."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "S05",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "只用引理的 ⇐ 方向與誤差公式。",
            "en": "Only the ⇐ direction of the lemma and the error formula are used."
          }
        },
        "leaf_basis": "L_cutoff + eq_replace"
      },
      {
        "id": "S12",
        "uses": [
          "S10",
          "S11",
          "S01"
        ],
        "before_tex": "\\varepsilon\\ \\text{arbitrary},\\ \\exists N\\ \\forall n\\ge N:\\ |u(n)-2|<\\varepsilon",
        "after_tex": "u(n)\\to 2",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S10、S11 對任意 ε>0 構造出合法的 N，沒有對 ε 加額外條件；依全稱推廣，對每個 ε>0 都存在合法 N。依 S01 的定義，u(n)→2。證明對所有 ε 有效，不是只檢查了幾個 ε 或幾項。",
          "en": "S10 and S11 construct a valid N for an arbitrary ε>0, with no extra condition on ε; by generalization, every ε>0 has a valid N. By the definition in S01, u(n)→2. The proof covers every ε; it does not merely test a few values of ε or a few terms."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S01",
            "S10",
            "S11",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "目標 G：u(n)→2。足夠條件 H：對任意 ε>0 有合法 N；H⇒G 由定義。H 由 S10、S11 正向證成，未把 G 當作前提。",
            "en": "Goal G: u(n)→2. Sufficient condition H: every ε>0 has a valid N; H⇒G by the definition. H is proved forwards in S10 and S11, without assuming G."
          }
        },
        "leaf_basis": "forall_intro + exists_intro + seq_limit_def"
      },
      {
        "id": "S13",
        "uses": [
          "S03",
          "S05",
          "S06",
          "G3"
        ],
        "before_tex": "\\exists N\\ \\forall\\varepsilon>0\\ \\forall n\\ge N:\\ e(n)<\\varepsilon\\ ?",
        "after_tex": "\\text{false: for each}\\ N,\\ \\varepsilon=\\tfrac{3}{N+1}\\ \\text{gives}\\ e(N)=\\varepsilon\\not<\\varepsilon",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若把 N 放在 ε 之前固定，命題變成「有一個 N 對所有 ε 都合法」。它是假的：對任一 N，取 ε=3/(N+1)>0，第 N 項誤差恰等於 ε，不小於 ε。具體例子：ε=1/2 的最小 N 是 6，但 ε=1/10 時 3/7>1/10，N=6 不合法；由 S06，ε=1/10 需要 3·10<N+1，即 N≥30（N=29 時誤差恰為 1/10）。N 必須依 ε 改變。",
          "en": "Fixing N before ε turns the statement into 'one N is valid for every ε'. That is false: for any N take ε=3/(N+1)>0; the N-th error equals ε exactly and is not below it. Concretely, the least N for ε=1/2 is 6, but for ε=1/10, 3/7>1/10, so N=6 fails; by S06, ε=1/10 needs 3·10<N+1, that is N≥30 (at N=29 the error is exactly 1/10). N must change with ε."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S05",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "對每個 N 給出一個使其失效的 ε，即否定了「∃N∀ε」；這不影響 S12 的「∀ε∃N」。",
            "en": "Giving, for each N, an ε that defeats it refutes '∃N∀ε'; this does not affect the '∀ε∃N' of S12."
          }
        },
        "leaf_basis": "quantifier_negation + L_cutoff + numeric_evaluation"
      }
    ],
    "result": {
      "zh-Hant": "對 u(n)=2−3/(n+1)，誤差為 3/(n+1)，且隨 n 不增；候選 N 對 ε 合法當且僅當 3/(N+1)<ε，對 ε=p/q 即 3q<p(N+1)。ε=1/2 時合法的 N 恰為 N≥6，最小 N=6；N=5 的誤差恰為 1/2，因嚴格不等而不合法。對任意 ε>0 由阿基米德性質取 N>3/ε 即合法，所以 u(n)→2。沒有一個 N 能對所有 ε 合法：N 必須在 ε 之後選。",
      "en": "For u(n)=2−3/(n+1) the error is 3/(n+1) and does not increase with n; a candidate N is valid for ε exactly when 3/(N+1)<ε, which for ε=p/q reads 3q<p(N+1). For ε=1/2 the valid cutoffs are exactly N≥6, so the least is 6; at N=5 the error equals 1/2 exactly and fails the strict inequality. For any ε>0, any natural N>3/ε (which exists by the Archimedean property) is valid, so u(n)→2. No single N is valid for every ε: N must be chosen after ε."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "ε=1/2 時 N=5 合法，因為第 5 項誤差等於 1/2",
          "en": "For ε=1/2, N=5 is valid because the error of term 5 equals 1/2"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "條件是嚴格的 |u(n)−2|<ε；3/6=1/2 不小於 1/2。",
          "en": "The condition is the strict |u(n)−2|<ε; 3/6=1/2 is not below 1/2."
        }
      },
      {
        "claim": {
          "zh-Hant": "N=6 對每個 ε>0 都合法",
          "en": "N=6 is valid for every ε>0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "ε=1/10 時第 6 項誤差 3/7>1/10。",
          "en": "For ε=1/10 the error of term 6 is 3/7>1/10."
        }
      },
      {
        "claim": {
          "zh-Hant": "v(n) 的前 101 項與 u(n) 相同，所以 v(n)→2",
          "en": "The first 101 terms of v(n) equal those of u(n), so v(n)→2"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "v(n)=5（n>100），取 ε=1，任何 N 都在 n=max(N,101) 失敗。",
          "en": "v(n)=5 for n>100; with ε=1 every N fails at n=max(N,101)."
        }
      },
      {
        "claim": {
          "zh-Hant": "把 L=2 改成 L=100 會改變 ε=1/2 的最小尾界",
          "en": "Changing L=2 to L=100 changes the least cutoff for ε=1/2"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "誤差 |c|/(n+d) 不含 L（S03），最小 N 仍是 6。",
          "en": "The error |c|/(n+d) does not involve L (S03); the least N is still 6."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "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": "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": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_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": "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": "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": "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": "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": "iff_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "雙條件：P⇒Q 與 Q⇒P 都成立時，P⇔Q 成立；反之亦然。",
          "en": "Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold."
        }
      },
      {
        "id": "add_inverse",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。",
          "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)."
        }
      },
      {
        "id": "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": "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": "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": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "mko_seq_limit",
        "kind": "cited",
        "source": {
          "mko_id": "mko-sequence-limit",
          "sha256": "3018ee44ad1a41db719f4c5275fd215d436e1dc4e73fdc44993604a211486fff",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "實數列 u:ℕ0→ℝ 趨向 L，指對每個 ε>0，都存在自然數 N≥0，使所有 n≥N 有 |u(n)−L|<ε。本輪具體模型 u(n)=L+c/(n+d)，d≥1，誤差為 |c|/(n+d)；最小合法 N 為 max(0, floor(|c|q/p)−d+1)。",
          "en": "A real sequence u:ℕ0→ℝ tends to L when, for every ε>0, there is a natural N≥0 such that |u(n)−L|<ε for all n≥N. The model is u(n)=L+c/(n+d) with d≥1; its error is |c|/(n+d), and the least permitted cutoff is max(0, floor(|c|q/p)−d+1)."
        }
      },
      {
        "id": "L_recip_mono",
        "kind": "lemma",
        "proved_in": [
          "S04.1"
        ],
        "statement": {
          "zh-Hant": "對實數 0<a≤b：1/b≤1/a。",
          "en": "For reals 0<a≤b: 1/b≤1/a."
        }
      },
      {
        "id": "L_cutoff",
        "kind": "lemma",
        "proved_in": [
          "S05.1",
          "S05.2"
        ],
        "statement": {
          "zh-Hant": "對 N∈ℕ0 與 ε>0：(對所有 n≥N 有 3/(n+1)<ε) ⇔ 3/(N+1)<ε。",
          "en": "For N∈ℕ0 and ε>0: (3/(n+1)<ε for all n≥N) ⇔ 3/(N+1)<ε."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S07",
        "lhs": "3*2",
        "relation": "<",
        "rhs": "1*(6+1)"
      },
      {
        "id": "N2",
        "step": "S08",
        "lhs": "3/(5+1)",
        "rhs": "1/2"
      },
      {
        "id": "N3",
        "step": "S09",
        "lhs": "3/(6+1)",
        "relation": "<",
        "rhs": "1/2"
      },
      {
        "id": "N4",
        "step": "S09",
        "lhs": "max(0, (3*2)//1 - 1 + 1)",
        "rhs": "6"
      },
      {
        "id": "N5",
        "step": "S03",
        "vars": [
          "n"
        ],
        "range": [
          0,
          60
        ],
        "lhs": "abs((2-3/(n+1))-2)",
        "rhs": "3/(n+1)"
      },
      {
        "id": "N6",
        "step": "S09",
        "vars": [
          "n"
        ],
        "range": [
          6,
          200
        ],
        "lhs": "3/(n+1)",
        "relation": "<",
        "rhs": "1/2"
      },
      {
        "id": "N7",
        "step": "S13",
        "lhs": "3*10",
        "relation": "<",
        "rhs": "30+1"
      },
      {
        "id": "N8",
        "step": "S13",
        "lhs": "3/(29+1)",
        "rhs": "1/10"
      },
      {
        "id": "N9",
        "step": "S13",
        "lhs": "3/(6+1)",
        "relation": ">",
        "rhs": "1/10"
      },
      {
        "id": "N10",
        "step": "S13",
        "lhs": "max(0, (3*10)//1 - 1 + 1)",
        "rhs": "30"
      },
      {
        "id": "N11",
        "step": "S04.1",
        "lhs": "1/7",
        "relation": "<=",
        "rhs": "1/6"
      },
      {
        "id": "N12",
        "step": "S10",
        "lhs": "3/(31+1)",
        "relation": "<",
        "rhs": "1/10"
      },
      {
        "id": "N13",
        "lhs": "max(0, (1*2)//1 - 1 + 1)",
        "rhs": "2"
      },
      {
        "id": "N14",
        "lhs": "1/(1+1)",
        "rhs": "1/2"
      },
      {
        "id": "N15",
        "lhs": "abs(5-2)",
        "relation": ">=",
        "rhs": "1"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理 MKO 的固定倒數模型 u(n)=L+c/(n+d) 的一個實例；「只看第 N 項」的化約依賴本模型誤差單調不增，不適用於一般數列。不處理一般收斂定理或形式橋。不改動 canonical MKO 或英文 edition。",
      "en": "Treats one instance of the MKO's fixed reciprocal model u(n)=L+c/(n+d); the reduction to the N-th term relies on this model's nonincreasing errors and does not apply to sequences in general. General convergence theorems and the formal bridge are not treated. It does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "\\mathbb{N}_0",
        "包含 0 的自然數",
        "the natural numbers including 0"
      ],
      [
        "u(n)",
        "本稿的數列 2−3/(n+1)",
        "the sequence 2−3/(n+1) of this guide"
      ],
      [
        "\\varepsilon=\\tfrac{p}{q}",
        "正的容差，p、q>0",
        "a positive tolerance, p, q>0"
      ],
      [
        "N",
        "尾界：要求所有 n≥N 都滿足誤差條件",
        "the cutoff: every n≥N must meet the error bound"
      ],
      [
        "e(n)",
        "誤差 |u(n)−2|",
        "the error |u(n)−2|"
      ]
    ],
    "md_error_controls": [
      {
        "title": "誤差等於 ε 不算小於 ε / An error equal to ε is not below ε",
        "zh": "MKO 的誤解之一是把誤差等於 ε 當作小於 ε。ε=1/2 時 N=5 的第 5 項誤差 3/6 恰為 1/2，不合法（S08），所以最小 N 是 6 而不是 5。MKO 的另一例 c=1、d=1、ε=1/2 也一樣：N=1 的誤差 1/2 等於容差，最小 N 是 2（N13、N14）。",
        "en": "One misconception the MKO lists is treating an error equal to ε as below ε. For ε=1/2 the error of term 5 is 3/6=1/2 exactly, so N=5 is invalid (S08), and the least N is 6, not 5. The MKO's own example c=1, d=1, ε=1/2 is the same: at N=1 the error 1/2 equals the tolerance and the least N is 2 (N13, N14)."
      },
      {
        "title": "N 在 ε 之後選 / N is chosen after ε",
        "zh": "把 ∀ε 後的 N 固定死是另一個誤解。S13 證明沒有單一 N 對所有 ε 合法：ε=1/2 用 N=6 即可，ε=1/10 卻需要 N≥30。正確的量詞順序是 ∀ε∃N（S12），不是 ∃N∀ε。",
        "en": "Freezing the N that comes after ∀ε is another misconception. S13 proves that no single N works for every ε: N=6 suffices for ε=1/2, but ε=1/10 needs N≥30. The correct quantifier order is ∀ε∃N (S12), not ∃N∀ε."
      },
      {
        "title": "有限項不能決定極限 / Finitely many terms do not determine a limit",
        "zh": "只看這些有限項不能判定整個序列是否趨於指定的 2；以下 v 雖然收斂到 5，卻不趨於 2。設 v(n)=u(n)（n≤100）、v(n)=5（n>100）：前 101 項與 u 相同，但取 ε=1，對任何 N，令 n=max(N,101)，則 |v(n)−2|=3≥1，所以 v 不趨向 2。本稿只看第 N 項就足夠，是因為 S04 證明了本模型誤差不增；一般數列沒有這個保證。",
        "en": "These finitely many terms do not determine convergence of the whole sequence to the specified value 2: the extension v below converges to 5 but not to 2. Let v(n)=u(n) for n≤100 and v(n)=5 for n>100: its first 101 terms equal those of u, but with ε=1 and any N, taking n=max(N,101) gives |v(n)−2|=3≥1, so v does not tend to 2. Here checking the N-th term suffices only because S04 proved this model's errors do not increase; a general sequence has no such guarantee."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "「合法 ⇔ 第 N 項合格」的化約（S05）只對本模型成立，依賴 S04 的誤差單調性。",
        "en": "The reduction 'valid ⇔ the N-th term qualifies' (S05) holds for this model only and relies on the monotone errors of S04."
      },
      {
        "zh-Hant": "N5、N6 是有限抽查，只對所列的 n 有效；一般結論由 S03、S05 證明。",
        "en": "N5 and N6 are finite spot checks valid only for the listed n; the general statements are proved by S03 and S05."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "38a147bff4e8a2b6b50d3919ef8225a220e20fc8deac81e0442527c1086c28d0",
    "original_source_sha256": "7c7efb8f2ac366108474ea59eedcd5bdb8bed79fdc71a3419059aa6ab5b196f7",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M025"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}