{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-sequence-01",
  "mko_id": "mko-sequence",
  "title": {
    "zh-Hant": "數列：由初始條件與遞迴確定通項，以及索引與有限前綴的陷阱",
    "en": "Sequences: fixing the general term from an initial condition and a recurrence, and the pitfalls of indexing and finite prefixes"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-sequence-01",
    "source": {
      "mko_id": "mko-sequence",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "61e652778c86fb6151fba32a0e61c914e4f99a1c496dd1c3d933c1f9297782f7",
      "english_sha256": "24722e5fa3e44eb82dbc3fccbd220761d9b4aa84bdf79ad9d04bad0d83621dde"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "nat_induction",
      "function_def",
      "distributive",
      "add_assoc",
      "eq_replace",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "設實數數列 a:ℕ→ℝ（ℕ 含 0）滿足 a₀=5、a_{n+1}=a_n+3。(1) 證明恰有一個這樣的數列，且 a_n=5+3n；(2) 計算 a₀ 到 a₄ 與 a₁₀，說明 a₀ 是起始項、a₁₀ 是從 1 數起的第 11 個值；(3) 寫出長度 L=4 的有限數列及其索引集合；(4) 說明前綴 5、8、11 不能唯一決定數列。",
      "en": "Let the real sequence a:ℕ→ℝ (with 0∈ℕ) satisfy a₀=5 and a_{n+1}=a_n+3. (1) Prove that exactly one such sequence exists and that a_n=5+3n; (2) compute a₀ to a₄ and a₁₀, explaining that a₀ is the initial term and a₁₀ is the 11th value when counting from 1; (3) write down the finite sequence of length L=4 and its index set; (4) show that the prefix 5, 8, 11 does not determine the sequence."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "ℕ 包含 0；實數數列是函數 a:ℕ→ℝ，a_n 表示 a(n)，a₀ 稱為起始項（MKO 假設 a1）。",
          "en": "ℕ includes 0; a real sequence is a function a:ℕ→ℝ, a_n denotes a(n), and a₀ is called the initial term (MKO assumption a1)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "遞迴與初始條件：a₀=5，且對每個 n∈ℕ，a_{n+1}=a_n+3。",
          "en": "Recurrence with initial condition: a₀=5 and a_{n+1}=a_n+3 for every n∈ℕ."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "有限數列：長度 L 是非負整數，索引集合為 {n∈ℕ ∣ n<L}，只有這些 n 可用（MKO 假設 a2）。",
          "en": "Finite sequence: the length L is a nonnegative integer, the index set is {n∈ℕ ∣ n<L}, and only these n are permitted (MKO assumption a2)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_seq_def"
        ],
        "before_tex": "(a_n)",
        "after_tex": "a:\\mathbb{N}\\to\\mathbb{R},\\ a_0=5,\\ a_{n+1}=a_n+3",
        "relation": "definition",
        "rule": "mko_seq_def",
        "conditions": [
          "0\\in\\mathbb{N}"
        ],
        "explanation": {
          "zh-Hant": "依 MKO，數列就是以 ℕ 為定義域的函數；索引 n 是輸入，a_n 是輸出。題目給的是遞迴式加初始條件，它約束 a 的值，但還沒有說明這樣的 a 存在或唯一，這正是 S02 到 S06 要證明的。",
          "en": "By the MKO a sequence is a function with domain ℕ: the index n is the input and a_n the output. The given information is a recurrence with an initial condition; it constrains the values of a but does not yet say that such an a exists or is unique, which is what S02 to S06 prove."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_seq_def"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出定義與條件，不推出新事實。",
            "en": "This step only states the definition and the conditions and derives no new fact."
          }
        },
        "leaf_basis": "mko_seq_def + function_def"
      },
      {
        "id": "S02",
        "uses": [
          "S01"
        ],
        "before_tex": "n\\in\\mathbb{N}",
        "after_tex": "s:\\mathbb{N}\\to\\mathbb{R},\\ s_n=5+3n",
        "relation": "definition",
        "rule": "def_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "提出候選數列 s_n=5+3n。對每個自然數 n，5+3n 是唯一確定的實數，所以 s 是 ℕ→ℝ 的函數，也就是一個數列。",
          "en": "Propose the candidate sequence s_n=5+3n. For each natural number n, 5+3n is a single well-defined real number, so s is a function ℕ→ℝ, that is, a sequence."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "def_intro",
            "function_def"
          ],
          "explanation": {
            "zh-Hant": "此步只引入名稱；s 是否滿足條件在 S03、S04 核對。",
            "en": "This step only introduces a name; whether s meets the conditions is checked in S03 and S04."
          }
        },
        "leaf_basis": "def_intro + function_def"
      },
      {
        "id": "S03",
        "uses": [
          "S02"
        ],
        "before_tex": "s_0",
        "after_tex": "s_0=5+3\\cdot 0=5",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代入 n=0：3·0=0（乘以零），5+0=5（加法單位元），所以 s₀=5，符合初始條件。",
          "en": "Substitute n=0: 3·0=0 (multiplication by zero) and 5+0=5 (additive identity), so s₀=5, as the initial condition requires."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_zero",
            "add_zero",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個數，等式雙向成立（N1）。",
            "en": "Both sides are the same number, so the equation holds both ways (N1)."
          }
        },
        "leaf_basis": "eq_replace + mul_zero + add_zero"
      },
      {
        "id": "S04",
        "uses": [
          "S02"
        ],
        "before_tex": "s_{n+1}=5+3(n+1)",
        "after_tex": "s_{n+1}=(5+3n)+3=s_n+3",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對任意 n：5+3(n+1)=5+(3n+3)（分配律）=(5+3n)+3（結合律）=s_n+3。所以 s 滿足遞迴，而且是對每個 n 都成立的恆等式。",
          "en": "For any n: 5+3(n+1)=5+(3n+3) by distributivity, =(5+3n)+3 by associativity, =s_n+3. So s satisfies the recurrence, as an identity valid for every n."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "add_assoc"
          ],
          "explanation": {
            "zh-Hant": "所用的是對所有 n 成立的恆等式，雙向成立；N2 以一次多項式恆等式核對。",
            "en": "Only identities valid for all n are used, so the step is two-way; N2 checks it as a degree-one polynomial identity."
          }
        },
        "leaf_basis": "distributive + add_assoc + mul_one"
      },
      {
        "id": "S05",
        "uses": [
          "S02",
          "S03",
          "S04"
        ],
        "before_tex": "s_0=5,\\ s_{n+1}=s_n+3",
        "after_tex": "\\exists\\ a:\\mathbb{N}\\to\\mathbb{R}\\ \\text{with}\\ a_0=5,\\ a_{n+1}=a_n+3",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "s 是一個數列（S02），並滿足初始條件（S03）與遞迴（S04），所以至少存在一個滿足 G2 的數列。存在性是直接寫出例子得到的。",
          "en": "s is a sequence (S02) satisfying the initial condition (S03) and the recurrence (S04), so at least one sequence satisfies G2. Existence is obtained by exhibiting an example."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S04",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "這只是存在；「只有一個」在 S06 證明。",
            "en": "This is existence only; that there is just one is proved in S06."
          }
        },
        "leaf_basis": "exists_intro + and_intro_elim"
      },
      {
        "id": "S06",
        "uses": [
          "S01",
          "S02",
          "L_unique",
          "function_def"
        ],
        "before_tex": "a_0=5,\\ a_{n+1}=a_n+3",
        "after_tex": "a_n=5+3n\\ \\text{for every}\\ n\\in\\mathbb{N},\\ \\text{so}\\ a=s",
        "relation": "theorem_application",
        "rule": "L_unique",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_unique（在 S06.1、S06.2 以歸納法證明），任何滿足 G2 的數列在每個索引都等於 s；兩者定義域都是 ℕ、對應域都是 ℝ，依 function_def 得 a=s。配合 S05，恰有一個這樣的數列。",
          "en": "By Lemma L_unique, proved by induction in S06.1 and S06.2, every sequence satisfying G2 agrees with s at every index; both have domain ℕ and codomain ℝ, so a=s by function_def. Together with S05, exactly one such sequence exists."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_unique",
            "function_def",
            "S05"
          ],
          "explanation": {
            "zh-Hant": "唯一性需要初始條件與遞迴兩者；只有遞迴而沒有 a₀，例如 a₀=0 也滿足遞迴，得到不同的數列。",
            "en": "Uniqueness needs both the initial condition and the recurrence; with the recurrence alone, a₀=0 would also fit and give a different sequence."
          }
        },
        "leaf_basis": "L_unique + function_def + nat_induction"
      },
      {
        "id": "S06.1",
        "uses": [
          "S02",
          "S03"
        ],
        "before_tex": "a_0=5",
        "after_tex": "a_0=s_0",
        "relation": "induction_base",
        "rule": "eq_refl_symm_trans",
        "conditions": [],
        "explanation": {
          "zh-Hant": "歸納基礎：a₀=5（G2 的初始條件），s₀=5（S03），所以 a₀=s₀。",
          "en": "Base case: a₀=5 by the initial condition and s₀=5 by S03, so a₀=s₀."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "eq_refl_symm_trans"
          ],
          "explanation": {
            "zh-Hant": "基礎只用到初始條件；它不能由遞迴本身推出。",
            "en": "The base case uses only the initial condition; it cannot be derived from the recurrence alone."
          }
        },
        "leaf_basis": "eq_refl_symm_trans"
      },
      {
        "id": "S06.2",
        "uses": [
          "S04",
          "nat_induction"
        ],
        "before_tex": "a_n=s_n",
        "after_tex": "a_{n+1}=a_n+3=s_n+3=s_{n+1}",
        "relation": "induction_step",
        "rule": "eq_replace",
        "conditions": [
          "n\\in\\mathbb{N}\\ \\text{arbitrary}"
        ],
        "explanation": {
          "zh-Hant": "設某個 n 有 a_n=s_n。則 a_{n+1}=a_n+3（a 的遞迴）=s_n+3（以歸納假設代換）=s_{n+1}（S04）。依歸納原理（n₀=0），S06.1 與 S06.2 證得引理 L_unique。",
          "en": "Suppose a_n=s_n for some n. Then a_{n+1}=a_n+3 by the recurrence for a, =s_n+3 by replacing with the induction hypothesis, =s_{n+1} by S04. By induction from n₀=0, S06.1 and S06.2 prove Lemma L_unique."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "eq_replace",
            "nat_induction"
          ],
          "explanation": {
            "zh-Hant": "只用同一個 n 的歸納假設；不宣稱反向。",
            "en": "Only the induction hypothesis for the same n is used; no converse is claimed."
          }
        },
        "leaf_basis": "eq_replace + nat_induction + forall_elim"
      },
      {
        "id": "S07",
        "uses": [
          "S06"
        ],
        "before_tex": "a_n=5+3n",
        "after_tex": "a_0=5,\\ a_1=8,\\ a_2=11,\\ a_3=14,\\ a_4=17,\\ a_{10}=35",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 n=0、1、2、3、4、10 代入通項：5、8、11、14、17、35。起始項是 a₀=5（索引 0），不是 a₁。",
          "en": "Substitute n=0, 1, 2, 3, 4 and 10 into the general term: 5, 8, 11, 14, 17 and 35. The initial term is a₀=5 (index 0), not a₁."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "forall_elim",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確整數計算，可獨立重算（N3 到 N5）。",
            "en": "Exact integer computation, independently recomputable (N3 to N5)."
          }
        },
        "leaf_basis": "forall_elim + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S07",
          "G1"
        ],
        "before_tex": "a_{10}=35",
        "after_tex": "a_{10}\\ \\text{is the 11th value};\\ b_k=a_{k-1}\\ (k\\geq 1)\\ \\text{gives}\\ b_{10}=a_9=32",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "索引 0 到 10 共有 11 個，所以 a₁₀ 是從 1 數起的第 11 個值。若有人改用從 1 起算的記號 b_k=a_{k−1}（k≥1，b₁=a₀=5），他的「第 10 項」b₁₀=a₉=5+27=32，與 a₁₀=35 不同。說「第 10 項」時必須說明索引從 0 還是從 1 起算。",
          "en": "The indices 0 to 10 number 11, so a₁₀ is the 11th value when counting from 1. If someone uses the 1-based labels b_k=a_{k−1} (k≥1, so b₁=a₀=5), their 'tenth term' is b₁₀=a₉=5+27=32, not a₁₀=35. Speaking of 'the tenth term' requires saying whether indexing starts at 0 or at 1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "G1",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "兩種慣例描述同一串數，只是標籤平移 1；本稿採 MKO 的從 0 起算。",
            "en": "Both conventions describe the same list of numbers with labels shifted by 1; this guide uses the MKO's convention of starting at 0."
          }
        },
        "leaf_basis": "def_unfold + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "G3",
          "S07",
          "mko_seq_def"
        ],
        "before_tex": "L=4",
        "after_tex": "(a_0,a_1,a_2,a_3)=(5,8,11,14),\\ \\ \\text{indices}\\ \\{0,1,2,3\\}",
        "relation": "definition",
        "rule": "mko_seq_def",
        "conditions": [
          "0\\leq n<4"
        ],
        "explanation": {
          "zh-Hant": "長度 L=4 的有限數列的索引集合是 {n∈ℕ ∣ n<4}={0,1,2,3}，它是函數 {0,1,2,3}→ℝ，值為 5、8、11、14。索引 4 不允許（4<4 不成立），所以這個有限數列沒有 a₄，雖然無限數列有 a₄=17。",
          "en": "The finite sequence of length L=4 has index set {n∈ℕ ∣ n<4}={0,1,2,3}; it is a function {0,1,2,3}→ℝ with values 5, 8, 11, 14. The index 4 is not permitted (4<4 fails), so this finite sequence has no a₄, even though the infinite sequence has a₄=17."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "mko_seq_def"
          ],
          "explanation": {
            "zh-Hant": "索引集合由 L 決定，從 0 到 L−1，共 L 個；常見錯誤是寫成 1 到 4 或 0 到 4。",
            "en": "The index set is fixed by L, from 0 to L−1, with L elements; common slips are writing 1 to 4 or 0 to 4."
          }
        },
        "leaf_basis": "mko_seq_def + set_builder"
      },
      {
        "id": "S10",
        "uses": [
          "S07"
        ],
        "before_tex": "c_n=5+3n+n(n-1)(n-2)",
        "after_tex": "c_0=5,\\ c_1=8,\\ c_2=11",
        "relation": "implies",
        "rule": "mul_zero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "考慮另一個數列 c_n=5+3n+n(n−1)(n−2)。當 n=0、1、2 時，乘積 n(n−1)(n−2) 中有一個因數是 0，所以乘積為 0，c_n=5+3n。因此 c 與 a 的前綴 5、8、11 完全相同。",
          "en": "Consider another sequence c_n=5+3n+n(n−1)(n−2). For n=0, 1, 2 one factor of the product n(n−1)(n−2) is 0, so the product is 0 and c_n=5+3n. Hence c has exactly the same prefix 5, 8, 11 as a."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mul_zero",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "只對 n=0、1、2 宣稱 c_n=a_n；N9 是這三點的有限核對，恰好涵蓋此步的全部主張。",
            "en": "c_n=a_n is claimed only for n=0, 1, 2; N9 is a finite check at these three points and covers exactly what this step claims."
          }
        },
        "leaf_basis": "mul_zero + add_zero + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S07",
          "S10",
          "function_def"
        ],
        "before_tex": "c_3",
        "after_tex": "c_3=5+9+3\\cdot 2\\cdot 1=20\\neq 14=a_3,\\ \\text{so}\\ c\\neq a",
        "relation": "implies",
        "rule": "function_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "c₃=5+3·3+3·2·1=14+6=20，而 a₃=14。兩個數列在索引 3 的值不同，依 function_def，c≠a。所以前綴 5、8、11 不能唯一決定數列：至少有兩個不同的數列以它開頭。",
          "en": "c₃=5+3·3+3·2·1=14+6=20, while a₃=14. The two sequences differ at index 3, so c≠a by function_def. Hence the prefix 5, 8, 11 does not determine the sequence: at least two different sequences begin with it."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "function_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一個不同的索引就足以說明兩個數列不同（N10、N11）。",
            "en": "One differing index suffices to show the sequences differ (N10, N11)."
          }
        },
        "leaf_basis": "function_def + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S06",
          "S10",
          "S11"
        ],
        "before_tex": "c_3-c_2",
        "after_tex": "c_3-c_2=20-11=9\\neq 3",
        "relation": "implies",
        "rule": "contrapositive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "c 滿足 c₀=5，但在 n=2 不滿足遞迴：c₃−c₂=9≠3。這與 S06 一致：滿足初始條件與遞迴的數列只有 a。能唯一決定數列的是規則加初始條件，而不是有限個前幾項。",
          "en": "c satisfies c₀=5 but fails the recurrence at n=2: c₃−c₂=9≠3. This agrees with S06: only a satisfies both the initial condition and the recurrence. What determines the sequence is the rule together with the initial condition, not finitely many first terms."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S11",
            "contrapositive"
          ],
          "explanation": {
            "zh-Hant": "由 S06 的逆否：c≠a 推出 c 不滿足 G2；這裡也直接找到失敗的位置 n=2。",
            "en": "By the contrapositive of S06, c≠a implies that c does not satisfy G2; here the failing index n=2 is also found directly."
          }
        },
        "leaf_basis": "contrapositive + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S05",
          "S06",
          "S07",
          "S08",
          "S09",
          "S11"
        ],
        "before_tex": "S05\\ \\text{to}\\ S09,\\ S11",
        "after_tex": "a_n=5+3n\\ \\text{unique};\\ a_{10}=35;\\ (5,8,11,14)\\ \\text{on}\\ \\{0,1,2,3\\};\\ \\text{prefix not determining}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "目標 (1) 由 S05、S06，(2) 由 S07、S08，(3) 由 S09，(4) 由 S10、S11 完成。",
          "en": "Goal (1) is completed by S05 and S06, (2) by S07 and S08, (3) by S09, and (4) by S10 and S11."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S05",
            "S06",
            "S07",
            "S08",
            "S09",
            "S11",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：(1) 到 (4)。足夠條件 H：所列各步成立；H⇒G 由合取引入。H 由定義、歸納與精確計算正向證成，沒有把 G 當作前提。",
            "en": "Goal G: (1) to (4). Sufficient condition H: the listed steps hold; H⇒G by conjunction introduction. H is proved forwards from the definitions, induction and exact computation; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "滿足 a₀=5、a_{n+1}=a_n+3 的數列恰有一個，即 a_n=5+3n；a₀ 到 a₄ 為 5、8、11、14、17，a₁₀=35（從 1 數起的第 11 個值；從 1 起算的記號下「第 10 項」是 32）。長度 4 的有限數列是 (5,8,11,14)，索引為 {0,1,2,3}。前綴 5、8、11 也是 c_n=5+3n+n(n−1)(n−2) 的前綴，而 c₃=20≠14，所以前綴不能唯一決定數列。",
      "en": "Exactly one sequence satisfies a₀=5 and a_{n+1}=a_n+3, namely a_n=5+3n; a₀ to a₄ are 5, 8, 11, 14, 17 and a₁₀=35 (the 11th value counting from 1; under 1-based labels the 'tenth term' is 32). The finite sequence of length 4 is (5,8,11,14) with indices {0,1,2,3}. The prefix 5, 8, 11 is also the prefix of c_n=5+3n+n(n−1)(n−2), but c₃=20≠14, so a prefix does not determine the sequence."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "由 5、8、11 可知下一項必定是 14",
          "en": "From 5, 8, 11 the next term must be 14"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "c_n=5+3n+n(n−1)(n−2) 有相同前綴，但 c₃=20。",
          "en": "c_n=5+3n+n(n−1)(n−2) has the same prefix but c₃=20."
        }
      },
      {
        "claim": {
          "zh-Hant": "在從 1 起算的記號下，第 10 項是 35",
          "en": "Under 1-based labels the tenth term is 35"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "b₁₀=a₉=32；35 是 a₁₀，是第 11 個值。",
          "en": "b₁₀=a₉=32; 35 is a₁₀, the 11th value."
        }
      },
      {
        "claim": {
          "zh-Hant": "長度 4 的有限數列 (5,8,11,14) 有一項 a₄=17",
          "en": "The finite sequence (5,8,11,14) of length 4 has a term a₄=17"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "索引只有 0、1、2、3；4<4 不成立。",
          "en": "Its indices are only 0, 1, 2, 3; 4<4 fails."
        }
      },
      {
        "claim": {
          "zh-Hant": "有限數列 (5,8) 與 (8,5) 相同，因為它們的值集合都是 {5,8}",
          "en": "The finite sequences (5,8) and (8,5) are equal because both have value set {5,8}"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "數列是函數；在索引 0 的值 5≠8。索引次序是資料的一部分。",
          "en": "A sequence is a function; the values at index 0 are 5≠8. The order of indices is part of the data."
        }
      },
      {
        "claim": {
          "zh-Hant": "c 也滿足 c₀=5 與 c_{n+1}=c_n+3",
          "en": "c also satisfies c₀=5 and c_{n+1}=c_n+3"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "c₃−c₂=9≠3。",
          "en": "c₃−c₂=9≠3."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "nat_induction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "數學歸納原理：對從 n₀ 起的命題族 P(n)，若 P(n₀) 成立，且對每個 n≥n₀ 都有 P(n)⇒P(n+1)，則所有 n≥n₀ 都有 P(n)。",
          "en": "Principle of induction: for a family P(n) starting at n₀, if P(n₀) holds and P(n)⇒P(n+1) for every n≥n₀, then P(n) holds for every n≥n₀."
        }
      },
      {
        "id": "function_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "函數：f:A→B 對每個 a∈A 指定恰好一個 f(a)∈B。兩函數相等 ⇔ 定義域、對應域相同且對每個 a 都有相同值。",
          "en": "Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "add_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。",
          "en": "Associativity of addition: (u+v)+w=u+(v+w)."
        }
      },
      {
        "id": "eq_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": "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_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_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "id": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "id": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "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": "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": "eq_refl_symm_trans",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。",
          "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w."
        }
      },
      {
        "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": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "set_builder",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。",
          "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x)."
        }
      },
      {
        "id": "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": "mko_seq_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-sequence",
          "sha256": "61e652778c86fb6151fba32a0e61c914e4f99a1c496dd1c3d933c1f9297782f7",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在 ℕ 包含 0 的慣例下，實數數列為函數 a:ℕ→ℝ，第 n 項記為 a(n) 或 a_n。有限數列則明示長度 L 及索引集合（從 0 到 L−1 的整數）。",
          "en": "Under the convention that ℕ includes 0, a real sequence is a function a:ℕ→ℝ whose nth term is written a(n) or a_n. A finite sequence instead states a length L and the index set (the integers from 0 to L−1)."
        }
      },
      {
        "id": "L_unique",
        "kind": "lemma",
        "proved_in": [
          "S06.1",
          "S06.2"
        ],
        "statement": {
          "zh-Hant": "若數列 a:ℕ→ℝ 滿足 a₀=5 且對每個 n 有 a_{n+1}=a_n+3，則對每個 n∈ℕ，a_n=5+3n。",
          "en": "If a sequence a:ℕ→ℝ satisfies a₀=5 and a_{n+1}=a_n+3 for every n, then a_n=5+3n for every n∈ℕ."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "5+3*0",
        "rhs": "5"
      },
      {
        "id": "N2",
        "step": "S04",
        "vars": [
          "n"
        ],
        "degree": 1,
        "lhs": "5+3*(n+1)",
        "rhs": "(5+3*n)+3"
      },
      {
        "id": "N3",
        "step": "S07",
        "lhs": "5+3*3",
        "rhs": "14"
      },
      {
        "id": "N4",
        "step": "S07",
        "lhs": "5+3*4",
        "rhs": "17"
      },
      {
        "id": "N5",
        "step": "S07",
        "lhs": "5+3*10",
        "rhs": "35"
      },
      {
        "id": "N6",
        "step": "S08",
        "lhs": "5+3*9",
        "rhs": "32"
      },
      {
        "id": "N7",
        "step": "S08",
        "lhs": "10-0+1",
        "rhs": "11"
      },
      {
        "id": "N8",
        "step": "S09",
        "lhs": "4",
        "relation": ">=",
        "rhs": "4"
      },
      {
        "id": "N9",
        "step": "S10",
        "vars": [
          "n"
        ],
        "range": [
          0,
          2
        ],
        "lhs": "5+3*n+n*(n-1)*(n-2)",
        "rhs": "5+3*n"
      },
      {
        "id": "N10",
        "step": "S11",
        "lhs": "5+3*3+3*2*1",
        "rhs": "20"
      },
      {
        "id": "N11",
        "step": "S11",
        "lhs": "20",
        "relation": "!=",
        "rhs": "5+3*3"
      },
      {
        "id": "N12",
        "step": "S12",
        "lhs": "20-11",
        "relation": "!=",
        "rhs": "3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內：實數數列是 ℕ→ℝ 的函數，ℕ 含 0（假設 a1）；有限數列明示長度 L 與索引集合，遞迴附初始條件（假設 a2），並採用 MKO 所用的形式 a₀=u、a_{n+1}=a_n+d（u=5、d=3）。存在性以寫出通項來證明，不需引用一般遞迴定理。不處理極限、單調性或級數；不改動 canonical MKO 或英文 edition。",
      "en": "Inside the MKO's scope: a real sequence is a function ℕ→ℝ with 0∈ℕ (assumption a1); a finite sequence states its length L and index set, and a recurrence comes with its initial condition (assumption a2), in the MKO's form a₀=u, a_{n+1}=a_n+d with u=5 and d=3. Existence is proved by exhibiting the general term, so no general recursion theorem is needed. It does not treat limits, monotonicity or series, and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "a_n",
        "數列 a 在索引 n 的值 a(n)",
        "the value a(n) of the sequence a at index n"
      ],
      [
        "s_n",
        "候選通項 5+3n",
        "the candidate general term 5+3n"
      ],
      [
        "c_n",
        "與 a 有相同前綴的另一個數列",
        "another sequence with the same prefix as a"
      ],
      [
        "L",
        "有限數列的長度",
        "the length of a finite sequence"
      ],
      [
        "\\mathbb{N}",
        "含 0 的自然數",
        "the natural numbers, including 0"
      ]
    ],
    "md_error_controls": [
      {
        "title": "索引從 0 還是從 1 起算要說清楚 / State whether indexing starts at 0 or at 1",
        "zh": "MKO 指出的第一個誤解是混淆從 0 或 1 起始的索引。本稿 a₀=5 是起始項，a₁₀=35；若改用 b_k=a_{k−1} 從 1 起算，「第 10 項」是 b₁₀=32。同一串數，兩種標籤給出不同的「第 10 項」，所以必須先說明慣例。",
        "en": "The first mistake named by the MKO is confusing indexing from 0 with indexing from 1. Here a₀=5 is the initial term and a₁₀=35; with 1-based labels b_k=a_{k−1}, the 'tenth term' is b₁₀=32. The same list of numbers gives different 'tenth terms' under the two labelings, so the convention must be stated first."
      },
      {
        "title": "有限前綴不能唯一決定數列 / A finite prefix does not determine a sequence",
        "zh": "MKO 指出的第二個誤解是僅憑一段前綴宣稱找出了唯一數列。5、8、11 之後，a 給 14，c 給 20；事實上任何有限前綴後面都可以接任意的值。能唯一決定數列的是明示的規則（通項或遞迴加初始條件），S06 正是用規則證明唯一性。",
        "en": "The second mistake named by the MKO is claiming that a prefix pins down a unique sequence. After 5, 8, 11, a continues with 14 and c with 20; in fact any finite prefix can be followed by any values. What determines a sequence is an explicit rule (a general term, or a recurrence with initial conditions), and S06 proves uniqueness from exactly such a rule."
      },
      {
        "title": "有限數列的索引只到 L−1 / A finite sequence has indices only up to L−1",
        "zh": "長度 L=4 的有限數列索引是 0、1、2、3，共 4 個。寫成 1 到 4 會丟掉 a₀ 並多出不存在的 a₄；寫成 0 到 4 則有 5 個索引，長度變成 5。a₄=17 屬於無限數列，不屬於這個有限數列。",
        "en": "The finite sequence of length L=4 has the 4 indices 0, 1, 2, 3. Writing 1 to 4 loses a₀ and adds a non-existent a₄; writing 0 to 4 gives 5 indices, so length 5. a₄=17 belongs to the infinite sequence, not to this finite one."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "N9 是 n=0、1、2 三點的有限核對，恰好是 S10 所宣稱的範圍；N2 是一次多項式恆等式的格點核對，對一次多項式是完整的判定。",
        "en": "N9 is a finite check at n=0, 1, 2, exactly the range claimed in S10; N2 is a grid check of a degree-one polynomial identity, which decides such identities completely."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "c85cb9781dafa2d70a69996e1317fce43de676cfa862621d70e5f98c2de30736",
    "original_source_sha256": "c85cb9781dafa2d70a69996e1317fce43de676cfa862621d70e5f98c2de30736",
    "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.",
    "english_binding_revision": {
      "id": "edition-symbol-bindings-20261101",
      "path": "content/edition-symbol-bindings-20261101/revision.json",
      "sha256": "e9b4c69f2c4732074c37abb624c8ab8a70afc3645c12bf9df2268ef650ad7b6c"
    }
  }
}