{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-discrete-exponential-growth-01",
  "mko_id": "mko-discrete-exponential-growth",
  "title": {
    "zh-Hant": "離散指數成長模型：由 x_(n+1)=r·x_n 歸納證明 x_n=x_0·r^n",
    "en": "Discrete exponential growth: proving x_n=x_0·r^n from x_(n+1)=r·x_n by induction"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-discrete-exponential-growth-01",
    "source": {
      "mko_id": "mko-discrete-exponential-growth",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "81d7d953b56dc3b4f26d1051ae74ae30b298a2c6085c59c5f06439b68ea8e780",
      "english_sha256": "e3b7db219c9df5e1a4a5757566489aba9a73916a7d659377948d93c20b4c77b0"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "nat_induction",
      "eq_congruence",
      "mul_assoc",
      "mul_comm",
      "forall_elim"
    ],
    "goal": {
      "zh-Hant": "在固定時間步、常數因子 r≥0 的模型 x_(n+1)=r·x_n 中，以數學歸納法證明對每個自然數 n 有 x_n=x_0·r^n，並說明遞迴式與閉式描述同一數列；以 x_0=8、r=3/2 計算 x_1 到 x_4 並與閉式比對；說明 r>1、r=1、0≤r<1 與 x_0=0 的行為；判定資料 8、12、18、26 不符合任何以 x_0=8 開始的固定因子模型；並說明每步因子 3/2 不是連續成長率 1/2。",
      "en": "In the model x_(n+1)=r·x_n with a fixed time step and constant factor r≥0, prove by induction that x_n=x_0·r^n for every natural number n, and explain that the recurrence and the closed form describe the same sequence; with x_0=8 and r=3/2 compute x_1 to x_4 and compare with the closed form; describe the behaviour for r>1, r=1, 0≤r<1 and x_0=0; decide that the data 8, 12, 18, 26 fit no fixed-factor model starting at x_0=8; and explain why the per-step factor 3/2 is not the continuous growth rate 1/2."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "時間步固定，n 從 0 起算，因子 r 在整段期間保持不變（MKO 假設 a1）。",
          "en": "Time steps are fixed, n starts at 0, and the factor r stays constant throughout (MKO assumption a1)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "模型：對每個自然數 n，x_(n+1)=r·x_n；x_0 與 r 為精確有理數，r≥0（假設 a2）。",
          "en": "Model: x_(n+1)=r·x_n for every natural number n; x_0 and r are exact rationals with r≥0 (assumption a2)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "主例：x_0=8，r=3/2。對照例：x_0=8，r=1/2；以及 x_0=0。",
          "en": "Main instance: x_0=8, r=3/2. Comparison instances: x_0=8, r=1/2; and x_0=0."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "給定的觀測資料：d_0=8，d_1=12，d_2=18，d_3=26。",
          "en": "Given observed data: d_0=8, d_1=12, d_2=18, d_3=26."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "\\text{fixed step, constant factor }r\\ge 0",
        "after_tex": "x_{n+1}=r\\,x_n\\ (n\\in\\mathbb{N}),\\ x_0\\ \\text{given}",
        "relation": "definition",
        "rule": "mko_exp_model",
        "conditions": [
          "r\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "依 MKO 寫出模型：每一步把目前的量乘以同一個非負因子 r。G1 要求時間步固定、r 不隨時間改變；這是模型的假設，不是由任何資料推出的。",
          "en": "State the model as in the MKO: each step multiplies the current quantity by the same nonnegative factor r. G1 requires fixed steps and an r that does not change over time; these are assumptions of the model, not conclusions from any data."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_exp_model"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出模型，不推出新事實；之後所有結論都只在此模型中成立。",
            "en": "This step only states the model and derives no new fact; all later conclusions hold within this model only."
          }
        },
        "leaf_basis": "mko_exp_model + def_intro"
      },
      {
        "id": "S02",
        "uses": [
          "nat_power_def"
        ],
        "before_tex": "r^n",
        "after_tex": "r^0=1,\\ r^{n+1}=r^n\\cdot r",
        "relation": "definition",
        "rule": "nat_power_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把次冪的意義說清楚：r^n 是 n 個 r 相乘，正式地以遞迴定義。這個定義對 r=0 也適用，所以 0^0=1，閉式在 r=0 時也有意義。",
          "en": "Make the meaning of powers precise: r^n is the product of n copies of r, defined formally by recursion. The definition also applies to r=0, so 0^0=1 and the closed form makes sense when r=0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "nat_power_def"
          ],
          "explanation": {
            "zh-Hant": "只引入定義，不推出新事實。",
            "en": "Only a definition is introduced; no new fact is derived."
          }
        },
        "leaf_basis": "nat_power_def + def_intro"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "S02",
          "L_closed_form"
        ],
        "before_tex": "x_{n+1}=r\\,x_n,\\ x_0",
        "after_tex": "\\forall n\\in\\mathbb{N}:\\ x_n=x_0\\,r^n",
        "relation": "implies",
        "rule": "L_closed_form",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_closed_form（在 S03.1 到 S03.3 以數學歸納法證明），遞迴式推出閉式，對所有 n 成立，而不是只對算過的幾項成立。",
          "en": "By Lemma L_closed_form (proved by induction in S03.1 to S03.3), the recurrence implies the closed form for all n, not just for the few terms that were computed."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_closed_form"
          ],
          "explanation": {
            "zh-Hant": "依賴該引理；引理只用到 G2 的遞迴式與次冪定義。",
            "en": "It depends on the lemma, which uses only the recurrence of G2 and the definition of powers."
          }
        },
        "leaf_basis": "nat_induction + nat_power_def + mul_assoc + mul_comm"
      },
      {
        "id": "S03.1",
        "uses": [
          "S02"
        ],
        "before_tex": "n=0",
        "after_tex": "x_0=x_0\\cdot 1=x_0\\,r^0",
        "relation": "induction_base",
        "rule": "nat_induction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "基底：令 P(n) 為 x_n=x_0·r^n。因 r^0=1 且 x_0·1=x_0，P(0) 成立。",
          "en": "Base: let P(n) be x_n=x_0·r^n. Since r^0=1 and x_0·1=x_0, P(0) holds."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "nat_power_def",
            "mul_one"
          ],
          "explanation": {
            "zh-Hant": "只用到 r^0=1 與乘法單位元；對 r=0 也成立。",
            "en": "It uses only r^0=1 and the multiplicative identity; it also holds for r=0."
          }
        },
        "leaf_basis": "nat_power_def + mul_one"
      },
      {
        "id": "S03.2",
        "uses": [
          "S01",
          "S02"
        ],
        "before_tex": "x_n=x_0\\,r^n",
        "after_tex": "x_{n+1}=r\\,x_n=r\\,(x_0\\,r^n)=x_0\\,(r^n\\cdot r)=x_0\\,r^{n+1}",
        "relation": "induction_step",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "歸納步：假設 P(n)。由遞迴式 x_(n+1)=r·x_n，把 x_n 換成 x_0·r^n，再用乘法交換律與結合律重排為 x_0·(r^n·r)，依次冪定義即 x_0·r^(n+1)，所以 P(n+1) 成立。",
          "en": "Inductive step: assume P(n). By the recurrence x_(n+1)=r·x_n; replace x_n by x_0·r^n, rearrange by commutativity and associativity of multiplication to x_0·(r^n·r), which by the definition of powers is x_0·r^(n+1); so P(n+1) holds."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "eq_replace",
            "mul_assoc",
            "mul_comm",
            "nat_power_def"
          ],
          "explanation": {
            "zh-Hant": "需要遞迴式對這個 n 成立、歸納假設，以及乘法的交換結合律；r 不能隨 n 改變，否則第一個等號不成立。",
            "en": "It needs the recurrence at this n, the inductive hypothesis and commutativity and associativity of multiplication; r must not depend on n, or the first equality would fail."
          }
        },
        "leaf_basis": "eq_replace + mul_assoc + mul_comm + nat_power_def"
      },
      {
        "id": "S03.3",
        "uses": [
          "S03.1",
          "S03.2"
        ],
        "before_tex": "P(0),\\ \\forall n\\,(P(n)\\Rightarrow P(n+1))",
        "after_tex": "\\forall n\\in\\mathbb{N}:\\ x_n=x_0\\,r^n",
        "relation": "implies",
        "rule": "nat_induction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依數學歸納原理，P(n) 對所有自然數 n 成立。S03.1 到 S03.3 合起來證得引理 L_closed_form。",
          "en": "By the principle of induction, P(n) holds for every natural number n. S03.1 to S03.3 together prove Lemma L_closed_form."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03.1",
            "S03.2",
            "nat_induction"
          ],
          "explanation": {
            "zh-Hant": "兩個前提由 S03.1、S03.2 提供，n 在歸納步中任意。",
            "en": "The two premises come from S03.1 and S03.2, with n arbitrary in the inductive step."
          }
        },
        "leaf_basis": "nat_induction"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "G3"
        ],
        "before_tex": "\\forall n:\\ x_n=x_0\\,r^n",
        "after_tex": "x_n=8\\left(\\tfrac{3}{2}\\right)^n",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [
          "x_0=8,\\ r=\\tfrac{3}{2}\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "引理對任意 x_0 與 r≥0 成立；代入主例 x_0=8、r=3/2 得 x_n=8·(3/2)^n。",
          "en": "The lemma holds for any x_0 and r≥0; substituting the main instance x_0=8, r=3/2 gives x_n=8·(3/2)^n."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "全稱消去只需要代入的值符合條件，r=3/2≥0。",
            "en": "Universal instantiation needs only that the substituted values meet the conditions, r=3/2≥0."
          }
        },
        "leaf_basis": "forall_elim"
      },
      {
        "id": "S05",
        "uses": [
          "G2",
          "G3"
        ],
        "before_tex": "x_0=8,\\ x_{n+1}=\\tfrac{3}{2}x_n",
        "after_tex": "x_1=12,\\ x_2=18,\\ x_3=27,\\ x_4=\\tfrac{81}{2}",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依遞迴式逐步計算：8·3/2=12，12·3/2=18，18·3/2=27，27·3/2=81/2。",
          "en": "Compute step by step from the recurrence: 8·3/2=12, 12·3/2=18, 18·3/2=27, 27·3/2=81/2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "每項只由前一項決定，精確有理數計算；不宣稱這四項本身證明了閉式。",
            "en": "Each term is determined by the previous one, with exact rational arithmetic; these four terms are not claimed to prove the closed form by themselves."
          }
        },
        "leaf_basis": "numeric_evaluation + eq_replace"
      },
      {
        "id": "S06",
        "uses": [
          "S04",
          "S05"
        ],
        "before_tex": "8\\left(\\tfrac{3}{2}\\right)^4",
        "after_tex": "8\\cdot\\tfrac{81}{16}=\\tfrac{81}{2}=x_4",
        "relation": "equivalent",
        "rule": "power_laws",
        "conditions": [],
        "explanation": {
          "zh-Hant": "閉式在 n=4 的值：(3/2)^4=81/16，8·81/16=81/2，與 S05 的遞迴計算一致。這是對 S03 的數值交叉核對，不是它的證明。",
          "en": "The closed form at n=4: (3/2)^4=81/16 and 8·81/16=81/2, agreeing with the recursive computation of S05. This is a numerical cross-check of S03, not its proof."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "power_laws",
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確有理數計算，雙向是同一個數。",
            "en": "Exact rational arithmetic; both sides are the same number."
          }
        },
        "leaf_basis": "power_laws + frac_rules + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S03",
          "S02"
        ],
        "before_tex": "y_n=x_0\\,r^n",
        "after_tex": "y_0=x_0,\\ y_{n+1}=r\\,y_n",
        "relation": "equivalent",
        "rule": "nat_power_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "反方向：閉式數列 y_n=x_0·r^n 滿足遞迴式與同一初值，因為 y_0=x_0·1=x_0，且 y_(n+1)=x_0·r^n·r=r·y_n。配合 S03（滿足遞迴與初值的數列必等於閉式），遞迴加初值與閉式描述的是同一個數列，雙向成立，這正是 MKO 第二段說的等價。",
          "en": "The other direction: the closed-form sequence y_n=x_0·r^n satisfies the recurrence with the same initial value, since y_0=x_0·1=x_0 and y_(n+1)=x_0·r^n·r=r·y_n. Together with S03 (any sequence satisfying the recurrence and initial value equals the closed form), the recurrence with initial value and the closed form describe the same sequence, both ways, which is the equivalence in the MKO's second paragraph."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "nat_power_def",
            "S03",
            "mul_assoc",
            "mul_comm"
          ],
          "explanation": {
            "zh-Hant": "一個方向是 S03，另一個方向是此步的直接代入；兩個方向都有理由。",
            "en": "One direction is S03 and the other is the direct substitution in this step; both directions have reasons."
          }
        },
        "leaf_basis": "nat_power_def + mul_assoc + mul_comm + iff_intro"
      },
      {
        "id": "S08",
        "uses": [
          "S04",
          "mko_exp_cases"
        ],
        "before_tex": "x_{n+1}-x_n=(r-1)x_n,\\ r=\\tfrac{3}{2}",
        "after_tex": "x_{n+1}-x_n=\\tfrac{1}{2}x_n>0",
        "relation": "implies",
        "rule": "mko_exp_cases",
        "conditions": [
          "x_n=8(\\tfrac{3}{2})^n>0"
        ],
        "explanation": {
          "zh-Hant": "x_(n+1)−x_n=r·x_n−x_n=(r−1)x_n。主例 r−1=1/2，而 x_n=8·(3/2)^n 是正數的乘積，所以 x_n>0，於是每步嚴格增加，且增量本身與 x_n 成正比：這就是 r>1 的「每步按因子成長」。",
          "en": "x_(n+1)−x_n=r·x_n−x_n=(r−1)x_n. In the main instance r−1=1/2, and x_n=8·(3/2)^n is a product of positive numbers, so x_n>0; hence each step strictly increases, and the increment itself is proportional to x_n: this is the 'growth by a factor' of r>1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "distributive",
            "order_basics",
            "mko_exp_cases"
          ],
          "explanation": {
            "zh-Hant": "需要 x_n>0；若 x_0=0，即使 r>1 也不會成長（S10）。",
            "en": "It needs x_n>0; with x_0=0 there is no growth even when r>1 (S10)."
          }
        },
        "leaf_basis": "distributive + order_basics + mko_exp_cases"
      },
      {
        "id": "S09",
        "uses": [
          "S03",
          "G3",
          "mko_exp_cases"
        ],
        "before_tex": "x_0=8,\\ r=\\tfrac{1}{2}",
        "after_tex": "x_n=8\\left(\\tfrac{1}{2}\\right)^n:\\ 8,4,2,1,\\tfrac{1}{2};\\ x_{n+1}-x_n=-\\tfrac{1}{2}x_n<0",
        "relation": "implies",
        "rule": "mko_exp_cases",
        "conditions": [
          "0\\le r<1"
        ],
        "explanation": {
          "zh-Hant": "對照例 r=1/2：依 L_closed_form，x_n=8·(1/2)^n，前五項為 8、4、2、1、1/2；x_(n+1)−x_n=(1/2−1)x_n=−x_n/2<0，每步縮為一半。這是 0≤r<1 的縮放情形。 一般分類不另加 x0≥0 假設：由閉式，r>0 時 x_n 與 x0 同號，且 x_(n+1)−x_n=(r−1)x_n。故 r>1 且 x0>0 時嚴格遞增、x0<0 時嚴格遞減，非零初值的絕對值每步乘 r 而增大；0<r<1 時絕對值縮小，正初值遞減、負初值遞增；r=0 時 x1 及以後全為0。r=1 與 x0=0 的常數情形見 S10。",
          "en": "Comparison instance r=1/2: by L_closed_form, x_n=8·(1/2)^n, with first five terms 8, 4, 2, 1, 1/2; x_(n+1)−x_n=(1/2−1)x_n=−x_n/2<0, halving at each step. This is the shrinking case 0≤r<1. The general classification does not add an assumption x0≥0: by the closed form, for r>0, x_n has the sign of x0 and x_(n+1)−x_n=(r−1)x_n. Thus for r>1 the sequence strictly increases when x0>0 and strictly decreases when x0<0; its magnitude grows by the factor r for nonzero x0. For 0<r<1 the magnitude shrinks, with positive initial values decreasing and negative initial values increasing. For r=0, x1 and all later values are 0. The constant cases r=1 and x0=0 are handled in S10."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "mko_exp_cases",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "同一引理的正初值對照例 x0=8、r=1/2 對每個 n 都為正，不會有限步變0；一般符號分類另外由閉式和差分恆等式得出，r=0 則在第一步到0。",
            "en": "In the positive-initial-value instance x0=8, r=1/2, every value is positive and the sequence never reaches 0 in finitely many steps. The general sign classification follows separately from the closed form and the difference identity; r=0 reaches 0 at the first step."
          }
        },
        "leaf_basis": "L_closed_form + order_basics + mko_exp_cases"
      },
      {
        "id": "S10",
        "uses": [
          "S03",
          "mko_exp_cases"
        ],
        "before_tex": "r=1\\ \\text{or}\\ x_0=0",
        "after_tex": "r=1:\\ x_n=x_0;\\ \\ x_0=0:\\ x_n=0\\cdot r^n=0",
        "relation": "implies",
        "rule": "mko_exp_cases",
        "conditions": [],
        "explanation": {
          "zh-Hant": "r=1 時 1^n=1，所以 x_n=x_0，數列不變。x_0=0 時，對任何 r≥0（包括 r=0，依 0^0=1 的定義）x_n=0·r^n=0，序列全為零。",
          "en": "For r=1, 1^n=1, so x_n=x_0 and the sequence is constant. For x_0=0 and any r≥0 (including r=0, with 0^0=1 by definition), x_n=0·r^n=0, so the sequence is all zeros."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "mul_zero",
            "mko_exp_cases"
          ],
          "explanation": {
            "zh-Hant": "兩個特例都是 L_closed_form 的代入；用到 1 的冪為 1 與乘以零為零。",
            "en": "Both special cases are substitutions into L_closed_form; they use that powers of 1 are 1 and that multiplying by zero gives zero."
          }
        },
        "leaf_basis": "L_closed_form + mul_zero + mul_one"
      },
      {
        "id": "S11",
        "uses": [
          "G4",
          "S03"
        ],
        "before_tex": "d=(8,12,18,26),\\ \\text{fixed-}r\\ \\text{model with }x_0=8",
        "after_tex": "r=\\tfrac{3}{2}\\ \\text{forced},\\ x_3=27\\neq 26:\\ \\text{no fixed-}r\\ \\text{model fits}",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [
          "d_0=8\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "假設某個固定因子模型 x_(n+1)=r·x_n、x_0=8 與四個資料都相符。由 x_1=12=r·8 且 8≠0 得 r=12/8=3/2（S11.1）。依 L_closed_form，模型給出 x_3=8·(3/2)^3=27，但 d_3=26，矛盾（S11.2）。所以沒有任何固定 r 的模型同時符合四個資料。前三個資料的比值都是 3/2，看起來像固定因子，但不足以推出第四期。",
          "en": "Assume some fixed-factor model x_(n+1)=r·x_n with x_0=8 matches all four data. From x_1=12=r·8 and 8≠0, r=12/8=3/2 (S11.1). By L_closed_form the model gives x_3=8·(3/2)^3=27, but d_3=26, a contradiction (S11.2). So no fixed-r model matches all four data. The first three data have ratios 3/2, which looks like a fixed factor, but that does not determine the fourth period."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "contradiction",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "需要 d_0≠0 才能解出 r；結論只否定以 x_0=8 開始、r 固定的模型，不否定其他模型（例如 r 隨時間改變）。",
            "en": "Solving for r needs d_0≠0; the conclusion rules out only fixed-r models starting at x_0=8, not other models (such as a time-varying r)."
          }
        },
        "leaf_basis": "contradiction + mul_inverse_nonzero + L_closed_form"
      },
      {
        "id": "S11.1",
        "uses": [
          "G4",
          "G2"
        ],
        "before_tex": "12=r\\cdot 8",
        "after_tex": "r=\\tfrac{12}{8}=\\tfrac{3}{2}",
        "relation": "equivalent",
        "rule": "mul_inverse_nonzero",
        "conditions": [
          "8\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "兩邊同乘 1/8（因 8≠0）：r=12/8=3/2。",
          "en": "Multiply both sides by 1/8 (since 8≠0): r=12/8=3/2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_inverse_nonzero",
            "eq_congruence"
          ],
          "explanation": {
            "zh-Hant": "乘 1/8 與乘 8 互為反運算，雙向成立。",
            "en": "Multiplying by 1/8 and by 8 undo each other, so both directions hold."
          }
        },
        "leaf_basis": "mul_inverse_nonzero + eq_congruence + frac_rules"
      },
      {
        "id": "S11.2",
        "uses": [
          "S11.1",
          "S03",
          "G4"
        ],
        "before_tex": "x_3=8\\left(\\tfrac{3}{2}\\right)^3",
        "after_tex": "x_3=27\\neq 26=d_3",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "8·27/8=27，而 d_3=26；兩數不等，與「模型符合資料」的假設矛盾。另外 d_3/d_2=26/18=13/9≠3/2，同樣顯示第三步的比值不同。",
          "en": "8·27/8=27, while d_3=26; the numbers differ, contradicting the assumption that the model fits the data. Also d_3/d_2=26/18=13/9≠3/2, showing that the ratio at the third step differs."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "numeric_evaluation",
            "S11.1"
          ],
          "explanation": {
            "zh-Hant": "精確計算；一個不符的資料點就足以否定固定因子模型。",
            "en": "Exact computation; one mismatching data point suffices to rule out the fixed-factor model."
          }
        },
        "leaf_basis": "numeric_evaluation + power_laws"
      },
      {
        "id": "S12",
        "uses": [
          "G1",
          "exp_lower_bound"
        ],
        "before_tex": "r=\\tfrac{3}{2}\\ \\text{per step},\\ \\text{continuous }x(t)=x_0e^{kt}",
        "after_tex": "e^{k}=\\tfrac{3}{2}\\ \\text{needed};\\ e^{1/2}\\ge\\tfrac{13}{8}>\\tfrac{3}{2},\\ \\text{so }k\\neq\\tfrac{1}{2}",
        "relation": "implies",
        "rule": "exp_lower_bound",
        "conditions": [
          "u=\\tfrac{1}{2}\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "每步因子 3/2 表示每步增加 50%。對本題非零初值 x_0=8，若連續時間的 x(t)=x_0·e^(kt) 在整數時間與模型一致，由第 1 步可消去 x_0，得到 e^k=3/2。若 x_0=0，兩者都是零軌跡，不能由軌跡一致推定 k；以下比較的是非零軌跡所需的因子一致。把 50% 直接當作連續成長率 k=1/2 是錯的：依級數下界 e^(1/2)≥1+1/2+1/8=13/8，而 13/8>3/2，所以 e^(1/2)≠3/2。離散因子與連續成長率是不同的量，本稿不建立連續模型。",
          "en": "A per-step factor 3/2 means a 50% increase per step. For the worked nonzero initial state x_0=8, agreement of x(t)=x_0·e^(kt) with the model at integer times permits cancellation of x_0 at time 1, giving e^k=3/2. If x_0=0 both trajectories are zero and their agreement does not determine k; the following factor comparison concerns the nonzero trajectory. Taking the 50% directly as the continuous rate k=1/2 is wrong: by the series bound e^(1/2)≥1+1/2+1/8=13/8, and 13/8>3/2, so e^(1/2)≠3/2. A discrete factor and a continuous growth rate are different quantities; no continuous model is built here."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "exp_lower_bound",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "只需要 e^u 的下界與有理數比較；此步只證明 k≠1/2，不計算 k 的值。",
            "en": "It needs only the lower bound for e^u and a rational comparison; this step proves only k≠1/2 and does not compute k."
          }
        },
        "leaf_basis": "exp_lower_bound + order_basics"
      },
      {
        "id": "S13",
        "uses": [
          "S03",
          "S06",
          "S07",
          "S08",
          "S09",
          "S10",
          "S11",
          "S12"
        ],
        "before_tex": "\\text{proved facts}",
        "after_tex": "x_n=x_0r^n\\ \\forall n;\\ x_4=\\tfrac{81}{2};\\ \\text{data }(8,12,18,26)\\ \\text{fit no fixed }r",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "合併結論：模型中 x_n=x_0·r^n 對所有 n 成立且與遞迴式等價；主例 x_4=81/2；非零初值在 r>1 時絕對值增大、0<r<1 時絕對值縮小，數值單調方向另依初值正負；r=0 時從第一步起為0，r=1 不變，x_0=0 全為零；資料 8、12、18、26 不符合固定因子模型；每步因子不是連續率。",
          "en": "Combine the conclusions: in the model x_n=x_0·r^n for all n, equivalent to the recurrence; in the main instance x_4=81/2; for nonzero initial values the magnitude grows when r>1 and shrinks when 0<r<1, with the numerical monotonicity depending on the initial sign; r=0 gives zero from the first step, r=1 is constant, and x_0=0 stays zero; the data 8, 12, 18, 26 fit no fixed-factor model; the per-step factor is not a continuous rate."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S03",
            "S07",
            "S08",
            "S09",
            "S10",
            "S11",
            "S12",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：本稿目標的各項判定。足夠條件 H：S03、S07 到 S12 各自已證（H⇒G 由合取引入）。H 由給定條件正向推出，沒有把 G 當作前提。",
            "en": "Goal G: the determinations in the goal. Sufficient condition H: S03 and S07 to S12, each already proved (H⇒G by conjunction introduction). H is derived forwards from the givens; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在模型 x_(n+1)=r·x_n（r≥0 固定）中，數學歸納法證明對所有 n 有 x_n=x_0·r^n，且閉式也滿足遞迴式，兩者描述同一數列。x_0=8、r=3/2 時 x_1 到 x_4 為 12、18、27、81/2，與閉式一致。非零初值在 r>1 時絕對值增大、0<r<1 時絕對值縮小；正初值的數值分別遞增、遞減，負初值的數值方向相反。r=0 時從第一步起為0，r=1 時不變，x_0=0 時全為零。資料 8、12、18、26 迫使 r=3/2，卻與預測 x_3=27 不符，所以不符合任何以 8 開始的固定因子模型。每步因子 3/2 不等於連續成長率 1/2，因為 e^(1/2)≥13/8>3/2。",
      "en": "In the model x_(n+1)=r·x_n with fixed r≥0, induction proves x_n=x_0·r^n for all n, and the closed form also satisfies the recurrence, so both describe the same sequence. With x_0=8 and r=3/2, x_1 to x_4 are 12, 18, 27, 81/2, agreeing with the closed form. For nonzero initial values the magnitude grows when r>1 and shrinks when 0<r<1; the numerical sequence respectively increases or decreases for a positive initial value, with opposite directions for a negative one. For r=0 it is zero from the first step, for r=1 it is constant, and for x_0=0 it stays zero. The data 8, 12, 18, 26 force r=3/2 but contradict the prediction x_3=27, so they fit no fixed-factor model starting at 8. The per-step factor 3/2 is not the continuous growth rate 1/2, since e^(1/2)≥13/8>3/2."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "8、12、18 三期都乘 3/2，所以第四期必定是 27",
          "en": "8, 12, 18 each multiply by 3/2, so the fourth period must be 27"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "給定資料的第四期是 26；固定因子只是模型假設，資料不必遵守（S11）。",
          "en": "The given fourth value is 26; a fixed factor is a model assumption that data need not obey (S11)."
        }
      },
      {
        "claim": {
          "zh-Hant": "每步因子 3/2 等於連續成長率 1/2",
          "en": "A per-step factor 3/2 equals a continuous growth rate 1/2"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "e^(1/2)≥13/8>3/2，所以 k=1/2 給出的每步因子大於 3/2（S12）。",
          "en": "e^(1/2)≥13/8>3/2, so k=1/2 gives a per-step factor larger than 3/2 (S12)."
        }
      },
      {
        "claim": {
          "zh-Hant": "因為 n=0 到 4 時閉式都與遞迴相符，所以 x_n=8·(3/2)^n 對所有 n 成立已被證明",
          "en": "Since the closed form matches the recurrence for n=0 to 4, x_n=8·(3/2)^n has been proved for all n"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "有限個數值核對不能證明全稱命題；證明來自 S03 的歸納。",
          "en": "Finitely many numerical checks cannot prove a universal statement; the proof is the induction of S03."
        }
      },
      {
        "claim": {
          "zh-Hant": "r=3/2>1，所以不論初值為何數列都會成長",
          "en": "r=3/2>1, so the sequence grows whatever the initial value"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x_0=0 時 x_n=0 對所有 n 成立（S10）。",
          "en": "With x_0=0, x_n=0 for every n (S10)."
        }
      }
    ],
    "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": "eq_congruence",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。",
          "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule."
        }
      },
      {
        "id": "mul_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法結合律：(uv)w=u(vw)。",
          "en": "Associativity of multiplication: (uv)w=u(vw)."
        }
      },
      {
        "id": "mul_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。",
          "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)."
        }
      },
      {
        "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_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_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "id": "eq_replace",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。",
          "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true."
        }
      },
      {
        "id": "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": "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": "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": "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": "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": "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": "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": "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": "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": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "mko_exp_model",
        "kind": "cited",
        "source": {
          "mko_id": "mko-discrete-exponential-growth",
          "sha256": "81d7d953b56dc3b4f26d1051ae74ae30b298a2c6085c59c5f06439b68ea8e780",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "固定一個時間步與非負常數因子 r，離散指數模型由 x_(n+1)=r x_n 定義。給初值 x_0，反覆代入或歸納得到 x_n=x_0 r^n；這是指定模型的代數結果，不是資料自動符合模型的證明。",
          "en": "Fix a time step and a nonnegative constant r. The discrete exponential model is x_(n+1)=r*x_n. From initial x_0, repeated substitution or induction gives x_n=x_0*r^n. This is an algebraic result within the specified model, not proof that observed data fit it."
        }
      },
      {
        "id": "mko_exp_cases",
        "kind": "cited",
        "source": {
          "mko_id": "mko-discrete-exponential-growth",
          "sha256": "81d7d953b56dc3b4f26d1051ae74ae30b298a2c6085c59c5f06439b68ea8e780",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "獨立來源追加式勘誤（使用者已採用於本地修訂；未合併或發布）：r>1 是每步乘以大於 1 的固定因子，不一定是加倍；非零初值的絕對值增加，正初值的值增加而負初值的值減少。r=1 時數值不變；0<r<1 時非零數值的絕對值縮小；r=0 時第一步起為零。初值為零時全序列為零。原始引文與 canonical MKO 保留不動。",
          "en": "Independent append-only source erratum (owner-adopted in this local revision; not merged or published): r>1 multiplies each step by a fixed factor greater than 1, not necessarily by 2. Nonzero magnitudes increase; positive state values increase and negative state values decrease. For r=1 values stay fixed; for 0<r<1 nonzero magnitudes shrink; for r=0 states are zero from the first step onward. A zero initial state stays zero. The original citation and canonical MKO are preserved unchanged."
        }
      },
      {
        "id": "nat_power_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "自然數次冪的遞迴定義：對任何有理數 r，r^0=1，r^(n+1)=r^n·r（因此 0^0=1 是此定義的一部分）。",
          "en": "Recursive definition of natural-number powers: for every rational r, r^0=1 and r^(n+1)=r^n·r (so 0^0=1 is part of this definition)."
        }
      },
      {
        "id": "exp_lower_bound",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "對 u≥0，eᵘ=Σ_(k≥0) uᵏ/k! 的每一項都非負，所以 eᵘ≥1+u+u²/2。",
          "en": "For u≥0 every term of eᵘ=Σ_(k≥0) uᵏ/k! is nonnegative, so eᵘ≥1+u+u²/2."
        }
      },
      {
        "id": "L_closed_form",
        "kind": "lemma",
        "proved_in": [
          "S03.1",
          "S03.2",
          "S03.3"
        ],
        "statement": {
          "zh-Hant": "在模型 G2 中，對每個自然數 n，x_n=x_0·r^n。",
          "en": "In the model G2, x_n=x_0·r^n for every natural number n."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S05",
        "lhs": "8*3/2",
        "rhs": "12"
      },
      {
        "id": "N2",
        "step": "S05",
        "lhs": "12*3/2",
        "rhs": "18"
      },
      {
        "id": "N3",
        "step": "S05",
        "lhs": "18*3/2",
        "rhs": "27"
      },
      {
        "id": "N4",
        "step": "S05",
        "lhs": "27*3/2",
        "rhs": "81/2"
      },
      {
        "id": "N5",
        "step": "S06",
        "lhs": "8*(3/2)**4",
        "rhs": "81/2"
      },
      {
        "id": "N6",
        "step": "S07",
        "vars": [
          "n"
        ],
        "range": [
          0,
          20
        ],
        "lhs": "8*(3/2)**(n+1)",
        "rhs": "(3/2)*(8*(3/2)**n)"
      },
      {
        "id": "N7",
        "step": "S09",
        "lhs": "8*(1/2)**4",
        "rhs": "1/2"
      },
      {
        "id": "N8",
        "step": "S10",
        "lhs": "0*(3/2)**5",
        "rhs": "0"
      },
      {
        "id": "N9",
        "step": "S11.1",
        "lhs": "12/8",
        "rhs": "3/2"
      },
      {
        "id": "N10",
        "step": "S11.2",
        "lhs": "8*(3/2)**3",
        "rhs": "27"
      },
      {
        "id": "N11",
        "step": "S11.2",
        "lhs": "26/18",
        "relation": "!=",
        "rhs": "3/2"
      },
      {
        "id": "N12",
        "step": "S12",
        "lhs": "1+1/2+(1/2)**2/2",
        "rhs": "13/8"
      },
      {
        "id": "N13",
        "step": "S12",
        "lhs": "13/8",
        "relation": ">",
        "rhs": "3/2"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只在指定模型中作代數推導（MKO 假設 a1、a2）：初值與因子為精確有理數，r 在整段期間不變。結論是模型內的代數結果，不是資料符合模型的證明，也不是生物族群的實測預測。S12 只用指數函數的級數下界來說明離散因子與連續率不同，不建立連續時間模型。不改動 canonical MKO 或英文 edition。",
      "en": "Algebra within the specified model only (MKO assumptions a1 and a2): initial values and factors are exact rationals, and r stays constant throughout. The conclusions are algebraic results within the model, not proof that data fit the model and not empirical forecasts of a biological population. S12 uses only a series lower bound for the exponential function to show that a discrete factor differs from a continuous rate; no continuous-time model is built. The canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "x_n",
        "第 n 個時間步的量",
        "the quantity at time step n"
      ],
      [
        "r",
        "每步的常數因子，r≥0",
        "the constant factor per step, r≥0"
      ],
      [
        "x_0",
        "初值",
        "the initial value"
      ],
      [
        "r^n",
        "自然數次冪，r^0=1，r^(n+1)=r^n·r",
        "natural-number power, r^0=1, r^(n+1)=r^n·r"
      ],
      [
        "d_0,d_1,d_2,d_3",
        "給定的觀測資料",
        "the given observed data"
      ]
    ],
    "md_error_controls": [
      {
        "title": "幾期成長不代表因子永遠固定 / A few periods of growth do not make the factor fixed forever",
        "zh": "MKO 指出的誤解之一：資料 8、12、18 的比值都是 3/2，但第四期是 26 而不是 27（S11）。L_closed_form 是「假設 r 固定」之下的代數結果；資料是否符合模型要另外核對，模型本身不證明適配。",
        "en": "One misconception named by the MKO: the data 8, 12, 18 have ratios 3/2, but the fourth value is 26, not 27 (S11). L_closed_form is an algebraic result under the assumption that r is fixed; whether data fit the model must be checked separately, and the model itself proves no fit."
      },
      {
        "title": "離散因子不是連續成長率 / A discrete factor is not a continuous rate",
        "zh": "MKO 指出的另一個誤解：每步因子 3/2（增加 50%）若直接當作連續率 k=1/2，會得到每步因子 e^(1/2)≥13/8，比 3/2 大（S12）。兩者之間需要明確的換算，不能直接互換。",
        "en": "Another misconception named by the MKO: treating the per-step factor 3/2 (a 50% increase) as a continuous rate k=1/2 gives a per-step factor e^(1/2)≥13/8, larger than 3/2 (S12). The two need an explicit conversion and cannot be swapped directly."
      },
      {
        "title": "歸納步要用同一個 r / The inductive step needs the same r",
        "zh": "S03.2 的第一個等號用的是 x_(n+1)=r·x_n，其中 r 與 n 無關。若因子隨時間改變（r_n），乘積是 r_0·r_1 一直乘到 r_(n−1)，不再是 r^n；把幾個數值代入閉式成立也不能取代歸納。",
        "en": "The first equality in S03.2 uses x_(n+1)=r·x_n with r independent of n. If the factor changes with time (r_n), the product is r_0·r_1 and so on up to r_(n−1), no longer r^n; checking the closed form at a few values cannot replace the induction."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "N6 是 n=0 到 20 的有限核對，只核對這些點；閉式對所有 n 的成立由 S03 的歸納證明。",
        "en": "N6 is a finite check for n=0 to 20 only; the closed form for all n is proved by the induction of S03."
      },
      {
        "zh-Hant": "logistic 模型、時間變動因子與模型校準屬於 MKO 的 formal next。",
        "en": "Logistic models, time-varying factors and model calibration belong to the MKO's formal next step."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "d241eac2e24aeedd4c822c1586a4997617fbe5ca988f7a3c508da4081980817f",
    "original_source_sha256": "31bb4abc1099f94cff294e388d438434f5734c98c44377638bc50ff98ef5d26e",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M042",
      "BR-M043"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}