{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-discrete-logistic-growth-01",
  "mko_id": "mko-discrete-logistic-growth",
  "title": {
    "zh-Hant": "離散 logistic 增量模型：[0,K] 的不變性、固定點與一條短軌道",
    "en": "Discrete logistic increment model: invariance of [0,K], fixed points and a short orbit"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-discrete-logistic-growth-01",
    "source": {
      "mko_id": "mko-discrete-logistic-growth",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "463931fbaa17a793076cfbf7af76f4490ba8c752e31fe52a9dc774d188ac3e4c",
      "english_sha256": "259717d8d2ce308dfd0854ed5a1909b22939cbe7fcc701286022adb367180cb9"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "order_basics",
      "distributive",
      "frac_rules",
      "zero_product",
      "nat_induction",
      "eq_congruence",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "對模型 G(x)=x+r x(1−x/K)（K>0、0≤r≤1），證明：若 0≤x≤K，則 0≤G(x)≤K，因此從 [0,K] 出發的整條軌道都留在 [0,K]；0 與 K 是固定點，且 r>0 時它們是僅有的固定點。再以 K=8、r=1、x0=2 精確算出 x1、x2、x3，並以 r=3 的例子說明條件 r≤1 不可省。",
      "en": "For the model G(x)=x+r x(1−x/K) with K>0 and 0≤r≤1, prove: if 0≤x≤K then 0≤G(x)≤K, so the whole orbit starting in [0,K] stays in [0,K]; 0 and K are fixed points, and for r>0 they are the only ones. Then compute x1, x2, x3 exactly for K=8, r=1, x0=2, and use an example with r=3 to show that the condition r≤1 cannot be dropped."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "模型（MKO）：固定 K>0 與 0≤r≤1，G(x)=x+r x(1−x/K)，x_(n+1)=G(x_n)，初值 x0∈[0,K]，所有計算為精確有理數。",
          "en": "Model (MKO): fix K>0 and 0≤r≤1, G(x)=x+r x(1−x/K), x_(n+1)=G(x_n), initial state x0∈[0,K], all computations in exact rationals."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "例子：K=8、r=1、x0=2。",
          "en": "Example: K=8, r=1, x0=2."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "對照（不滿足 r≤1）：K=8、r=3、x=6。",
          "en": "Comparison (violating r≤1): K=8, r=3, x=6."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "mko_logistic"
        ],
        "before_tex": "\\text{discrete logistic increment model}",
        "after_tex": "G(x)=x+rx\\left(1-\\tfrac{x}{K}\\right),\\ K>0,\\ 0\\le r\\le 1",
        "relation": "definition",
        "rule": "mko_logistic",
        "conditions": [
          "K>0",
          "0\\le r\\le 1"
        ],
        "explanation": {
          "zh-Hant": "固定 MKO 指定的模型：下一步等於現在的值加上增量 r x(1−x/K)。這是一個特定的離散模型；同名的其他「logistic map」是不同的模型，本稿的結論不轉移到它們（見錯誤控制）。",
          "en": "Fix the model specified by the MKO: the next state is the current value plus the increment r x(1−x/K). This is one particular discrete model; other maps also called 'logistic' are different models, and the conclusions here do not transfer to them (see the error controls)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_logistic"
          ],
          "explanation": {
            "zh-Hant": "此步只陳述模型與參數條件，不推出新事實。",
            "en": "This step only states the model and the parameter conditions and derives no new fact."
          }
        },
        "leaf_basis": "mko_logistic + def_intro"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "frac_rules",
          "distributive"
        ],
        "before_tex": "G(x)-x",
        "after_tex": "\\tfrac{rx(K-x)}{K}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "K\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "增量 G(x)−x=r x(1−x/K)。因 K≠0，1−x/K=K/K−x/K=(K−x)/K，所以增量 = r x(K−x)/K。這個形式把增量寫成三個因子 r、x、K−x 的乘積除以正數 K，方便判斷符號。",
          "en": "The increment is G(x)−x=r x(1−x/K). Since K≠0, 1−x/K=K/K−x/K=(K−x)/K, so the increment equals r x(K−x)/K. This writes the increment as the product of the three factors r, x, K−x divided by the positive number K, which makes its sign easy to read."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "同分母合併與拆分互為反運算，前提 K≠0（由 K>0）；N1 以本例數字交叉核對。",
            "en": "Combining and splitting over a common denominator undo each other given K≠0 (from K>0); N1 cross-checks with the example's numbers."
          }
        },
        "leaf_basis": "frac_rules + distributive + mul_assoc"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "G1"
        ],
        "before_tex": "0\\le x\\le K,\\ 0\\le r,\\ K>0",
        "after_tex": "G(x)-x=\\tfrac{rx(K-x)}{K}\\ge 0",
        "relation": "implies",
        "rule": "L_nonneg_prod",
        "conditions": [
          "0\\le x\\le K",
          "r\\ge 0",
          "K>0"
        ],
        "explanation": {
          "zh-Hant": "在 0≤x≤K 時，三個因子 r≥0、x≥0、K−x≥0 都非負。依引理 L_nonneg_prod 兩次，r x(K−x)≥0；再乘正數 1/K，不等號方向不變，所以增量 ≥0。",
          "en": "For 0≤x≤K the three factors r≥0, x≥0 and K−x≥0 are all nonnegative. Applying Lemma L_nonneg_prod twice gives r x(K−x)≥0; multiplying by the positive number 1/K keeps the direction, so the increment is ≥0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_nonneg_prod",
            "order_basics",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "需要 x≤K（使 K−x≥0）、x≥0、r≥0 與 K>0；不宣稱反向。x>K 且 r>0 時，K−x<0，所以增量為負；r=0 時增量為0。",
            "en": "It needs x≤K (so K−x≥0), x≥0, r≥0 and K>0; no converse is claimed. For x>K and r>0, K−x<0 makes the increment negative; when r=0, the increment is 0."
          }
        },
        "leaf_basis": "L_nonneg_prod + order_basics"
      },
      {
        "id": "S03.1",
        "uses": [
          "order_basics",
          "mul_zero"
        ],
        "before_tex": "u\\ge 0,\\ v\\ge 0",
        "after_tex": "uv\\ge 0",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "分兩種情況：若 v=0，uv=u·0=0≥0；若 v>0，由 u≥0，即 0≤u，兩邊乘正數 v 得 0·v≤uv，即 0≤uv（若 u=0 兩邊相等）。兩種情況涵蓋 v≥0 的全部可能。這證得引理 L_nonneg_prod。",
          "en": "Two cases: if v=0, uv=u·0=0≥0; if v>0, from u≥0, that is 0≤u, multiplying by the positive v gives 0·v≤uv, that is 0≤uv (with equality if u=0). The two cases cover every possibility for v≥0. This proves Lemma L_nonneg_prod."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "mul_zero",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "只用乘以正數保序與乘以零；不宣稱 uv≥0 推回 u、v 都非負（兩負數之積也非負）。",
            "en": "Only that multiplying by a positive number preserves order and multiplying by zero gives zero; it is not claimed that uv≥0 forces u, v≥0 (a product of two negatives is also nonnegative)."
          }
        },
        "leaf_basis": "order_basics + mul_zero + case_analysis"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "G1"
        ],
        "before_tex": "G(x)-x\\ge 0,\\ x\\ge 0",
        "after_tex": "G(x)\\ge x\\ge 0",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "G(x)−x≥0 兩邊加 x 得 G(x)≥x；再與 x≥0 用傳遞性得 G(x)≥0。下界證完，而且同時得知在 [0,K] 上更新不會使值減少。",
          "en": "Adding x to both sides of G(x)−x≥0 gives G(x)≥x; with x≥0, transitivity gives G(x)≥0. The lower bound is proved, and it also shows that on [0,K] an update never decreases the value."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "用加法保序與傳遞性；不宣稱 G(x)>x（例如 x=0 或 x=K 時相等）。",
            "en": "It uses that addition preserves order and transitivity; it does not claim G(x)>x (equality holds, for example, at x=0 and x=K)."
          }
        },
        "leaf_basis": "order_basics"
      },
      {
        "id": "S05",
        "uses": [
          "S01"
        ],
        "before_tex": "K\\cdot(K-G(x))",
        "after_tex": "(K-x)(K-rx)",
        "relation": "equivalent",
        "rule": "L_gap",
        "conditions": [
          "K>0"
        ],
        "explanation": {
          "zh-Hant": "上界的關鍵恆等式（引理 L_gap）：容量與下一步之差乘以 K，恰好分解成 (K−x)(K−r x)。這把「G(x)≤K」轉成兩個因子的符號問題。",
          "en": "The key identity for the upper bound (Lemma L_gap): the gap between capacity and the next state, multiplied by K, factors exactly as (K−x)(K−r x). This turns 'G(x)≤K' into a question about the signs of two factors."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_gap"
          ],
          "explanation": {
            "zh-Hant": "恆等式兩邊是同一個多項式表示式，雙向成立；證明在 S05.1，N2 以多項式恆等式核對。",
            "en": "Both sides are the same polynomial expression, so the step holds both ways; the proof is in S05.1, and N2 checks it as a polynomial identity."
          }
        },
        "leaf_basis": "distributive + frac_rules"
      },
      {
        "id": "S05.1",
        "uses": [
          "S01",
          "distributive",
          "frac_rules"
        ],
        "before_tex": "K\\cdot G(x)=Kx+rx(K-x)",
        "after_tex": "K(K-G(x))=K^2-Kx-rKx+rx^2=(K-x)(K-rx)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [
          "K\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "由 S02，K·G(x)=Kx+K·r x(K−x)/K=Kx+r x(K−x)。所以 K(K−G(x))=K²−Kx−rKx+r x²。另一方面 (K−x)(K−r x)=K²−rKx−Kx+r x²，兩者相同。這證得引理 L_gap。",
          "en": "By S02, K·G(x)=Kx+K·r x(K−x)/K=Kx+r x(K−x). So K(K−G(x))=K²−Kx−rKx+r x². On the other hand (K−x)(K−r x)=K²−rKx−Kx+r x², the same expression. This proves Lemma L_gap."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "兩邊展開成同一多項式（N2 在每變數次數 ≤2 的網格上核對），雙向成立；K/K=1 需要 K≠0。",
            "en": "Both sides expand to the same polynomial (N2 checks it on the grid for degree ≤2 per variable), so the step holds both ways; K/K=1 needs K≠0."
          }
        },
        "leaf_basis": "distributive + add_comm + mul_comm + frac_rules"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "order_basics"
        ],
        "before_tex": "0\\le r\\le 1,\\ 0\\le x\\le K",
        "after_tex": "rx\\le x\\le K",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "r\\le 1",
          "x\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "x−r x=(1−r)x，其中 1−r≥0（因 r≤1）且 x≥0，依 L_nonneg_prod，(1−r)x≥0，即 r x≤x；再由 x≤K 得 r x≤K。這一步是整個證明中唯一用到 r≤1 的地方。",
          "en": "x−r x=(1−r)x, where 1−r≥0 (since r≤1) and x≥0, so by L_nonneg_prod (1−r)x≥0, that is r x≤x; with x≤K this gives r x≤K. This is the only place in the proof where r≤1 is used."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_nonneg_prod",
            "order_basics",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "需要 r≤1 與 x≥0；r>1 時 r x 可能超過 K（S14 的例子 r x=18>8）。",
            "en": "It needs r≤1 and x≥0; for r>1, r x can exceed K (the example in S14 has r x=18>8)."
          }
        },
        "leaf_basis": "order_basics + distributive + L_nonneg_prod"
      },
      {
        "id": "S07",
        "uses": [
          "S05",
          "S06",
          "L_nonneg_prod"
        ],
        "before_tex": "K-x\\ge 0,\\ K-rx\\ge 0,\\ K>0",
        "after_tex": "G(x)\\le K",
        "relation": "implies",
        "rule": "L_nonneg_prod",
        "conditions": [
          "K>0"
        ],
        "explanation": {
          "zh-Hant": "K−x≥0（x≤K）且 K−r x≥0（S06），依 L_nonneg_prod，(K−x)(K−r x)≥0；由 S05，K(K−G(x))≥0。K>0，兩邊乘 1/K 得 K−G(x)≥0，即 G(x)≤K。上界證完。",
          "en": "K−x≥0 (since x≤K) and K−r x≥0 (S06), so by L_nonneg_prod (K−x)(K−r x)≥0; by S05, K(K−G(x))≥0. Since K>0, multiplying by 1/K gives K−G(x)≥0, that is G(x)≤K. The upper bound is proved."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S06",
            "L_nonneg_prod",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要 K>0 才能除以 K 而不改變不等號方向；兩個因子的非負性分別來自 x≤K 與 S06。",
            "en": "K>0 is needed to divide by K without reversing the inequality; the nonnegativity of the two factors comes from x≤K and S06."
          }
        },
        "leaf_basis": "L_nonneg_prod + order_basics + L_gap"
      },
      {
        "id": "S08",
        "uses": [
          "S04",
          "S07"
        ],
        "before_tex": "0\\le x\\le K",
        "after_tex": "0\\le G(x)\\le K",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [
          "K>0",
          "0\\le r\\le 1"
        ],
        "explanation": {
          "zh-Hant": "合併下界（S04）與上界（S07）：一步不變性成立。這正是 MKO 敘述的「若 0≤x_n≤K，下一步仍在 [0,K]」，且證明對區間中的每個 x 一致成立，不是有限核對。",
          "en": "Combine the lower bound (S04) and the upper bound (S07): one-step invariance holds. This is exactly the MKO's statement 'if 0≤x_n≤K, the next state remains in [0,K]', and the proof holds uniformly for every x in the interval; it is not a finite check."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "需要兩個方向的界；只證其中一個不能得到區間不變。",
            "en": "Both bounds are needed; proving only one does not give invariance of the interval."
          }
        },
        "leaf_basis": "and_intro_elim"
      },
      {
        "id": "S09",
        "uses": [
          "S08",
          "G1",
          "nat_induction"
        ],
        "before_tex": "x_0\\in[0,K]",
        "after_tex": "\\forall n\\in\\mathbb{N}:\\ x_n\\in[0,K]",
        "relation": "implies",
        "rule": "nat_induction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對 n 歸納：基底 x0∈[0,K] 由 G1 給定；若 x_n∈[0,K]，由 S08，x_(n+1)=G(x_n)∈[0,K]。所以整條軌道留在 [0,K]。",
          "en": "Induction on n: the base x0∈[0,K] is given by G1; if x_n∈[0,K], then by S08 x_(n+1)=G(x_n)∈[0,K]. So the whole orbit stays in [0,K]."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "nat_induction",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "歸納需要基底與步驟兩者；這只說明軌道有界，不說明它收斂或穩定。",
            "en": "The induction needs both base and step; it shows only that the orbit is bounded, not that it converges or is stable."
          }
        },
        "leaf_basis": "nat_induction + modus_ponens"
      },
      {
        "id": "S10",
        "uses": [
          "S02",
          "G1",
          "zero_product"
        ],
        "before_tex": "G(x)=x",
        "after_tex": "rx(K-x)=0;\\ G(0)=0,\\ G(K)=K;\\ r>0\\Rightarrow x\\in\\{0,K\\}",
        "relation": "equivalent",
        "rule": "zero_product",
        "conditions": [
          "K\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "由 S02，G(x)=x ⇔ r x(K−x)/K=0 ⇔ r x(K−x)=0（乘或除以 K≠0）。x=0 與 x=K 使乘積為零，所以對每個 r，0 與 K 都是固定點。若 r>0，依零乘積性質，r x(K−x)=0 ⇔ x=0 或 x=K，所以它們是僅有的固定點。若 r=0，G(x)=x 對所有 x 成立，每個點都是固定點。",
          "en": "By S02, G(x)=x ⇔ r x(K−x)/K=0 ⇔ r x(K−x)=0 (multiplying or dividing by K≠0). x=0 and x=K make the product zero, so 0 and K are fixed points for every r. If r>0, the zero-product property gives r x(K−x)=0 ⇔ x=0 or x=K, so they are the only fixed points. If r=0, G(x)=x for every x and every point is fixed."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "zero_product",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "乘以與除以 K≠0 互為反運算；零乘積性質是雙向的。r>0 的條件只在「僅有」這部分使用。",
            "en": "Multiplying and dividing by K≠0 undo each other; the zero-product property is two-way. The condition r>0 is used only for the 'only' part."
          }
        },
        "leaf_basis": "zero_product + mul_inverse_nonzero + case_analysis"
      },
      {
        "id": "S11",
        "uses": [
          "G2",
          "S01"
        ],
        "before_tex": "x_0=2,\\ K=8,\\ r=1",
        "after_tex": "x_1=2+2\\left(1-\\tfrac{2}{8}\\right)=\\tfrac{7}{2}",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "r=1 時 G(x)=x+x(1−x/8)=x(2−x/8)。x1=2·(2−1/4)=2·7/4=7/2。全部用精確分數（N3）。",
          "en": "With r=1, G(x)=x+x(1−x/8)=x(2−x/8). x1=2·(2−1/4)=2·7/4=7/2, all in exact fractions (N3)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "精確有理數計算，兩邊同值。",
            "en": "Exact rational computation; both sides have the same value."
          }
        },
        "leaf_basis": "numeric_evaluation + frac_rules"
      },
      {
        "id": "S12",
        "uses": [
          "S11"
        ],
        "before_tex": "x_1=\\tfrac{7}{2}",
        "after_tex": "x_2=\\tfrac{7}{2}\\left(2-\\tfrac{7}{16}\\right)=\\tfrac{7}{2}\\cdot\\tfrac{25}{16}=\\tfrac{175}{32}",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x1/8=7/16，2−7/16=25/16，x2=(7/2)(25/16)=175/32（N4）。",
          "en": "x1/8=7/16, 2−7/16=25/16, and x2=(7/2)(25/16)=175/32 (N4)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "精確分數乘法，兩邊同值。",
            "en": "Exact fraction multiplication; both sides have the same value."
          }
        },
        "leaf_basis": "numeric_evaluation + frac_rules"
      },
      {
        "id": "S13",
        "uses": [
          "S12",
          "S09",
          "S04"
        ],
        "before_tex": "x_2=\\tfrac{175}{32}",
        "after_tex": "x_3=\\tfrac{175}{32}\\cdot\\tfrac{337}{256}=\\tfrac{58975}{8192};\\ 2<\\tfrac{7}{2}<\\tfrac{175}{32}<\\tfrac{58975}{8192}<8",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x2/8=175/256，2−175/256=337/256，x3=(175/32)(337/256)=58975/8192（N5、N6），約 7.199，小於 8（N7）。四個值都在 [0,8] 且遞增，與 S09（留在區間）及 S04（G(x)≥x）一致。這三步只是一致性的例子，不證明軌道收斂到 8，也不證明 8 穩定（MKO 假設 a2）。",
          "en": "x2/8=175/256, 2−175/256=337/256, and x3=(175/32)(337/256)=58975/8192 (N5, N6), about 7.199 and below 8 (N7). All four values lie in [0,8] and increase, consistent with S09 (staying in the interval) and S04 (G(x)≥x). These three steps are only an example of consistency; they prove neither that the orbit converges to 8 nor that 8 is stable (MKO assumption a2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "numeric_evaluation",
            "S09",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "計算精確可重算；與定理一致不構成對定理的證明，定理的證明是 S03 到 S09。",
            "en": "The computation is exact and can be redone; agreement with the theorem is not a proof of it, which is S03 to S09."
          }
        },
        "leaf_basis": "numeric_evaluation + frac_rules + order_basics"
      },
      {
        "id": "S14",
        "uses": [
          "G3",
          "S05"
        ],
        "before_tex": "K=8,\\ r=3,\\ x=6",
        "after_tex": "G(6)=6+3\\cdot 6\\cdot\\tfrac{1}{4}=\\tfrac{21}{2}>8",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [
          "r=3>1"
        ],
        "explanation": {
          "zh-Hant": "條件 r≤1 不可省：r=3 時 r x=18>8=K，S06 失敗；L_gap 給出 8(8−G(6))=(8−6)(8−18)=−20，所以 G(6)=8+20/8=21/2>8，離開 [0,8]。這是一個具體反例，說明 S08 的結論依賴 r≤1。",
          "en": "The condition r≤1 cannot be dropped: with r=3, r x=18>8=K and S06 fails; L_gap gives 8(8−G(6))=(8−6)(8−18)=−20, so G(6)=8+20/8=21/2>8, which leaves [0,8]. This concrete counterexample shows that S08 depends on r≤1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "L_gap",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一個反例足以說明假設不可刪；不宣稱 r>1 時每個 x 都會離開區間（例如 x=0、x=K 仍固定）。",
            "en": "One counterexample suffices to show the hypothesis cannot be removed; it is not claimed that every x leaves the interval when r>1 (for example x=0 and x=K stay fixed)."
          }
        },
        "leaf_basis": "numeric_evaluation + L_gap"
      },
      {
        "id": "S15",
        "uses": [
          "S08",
          "S09",
          "S10",
          "S13",
          "S14"
        ],
        "before_tex": "\\text{model with}\\ K>0,\\ 0\\le r\\le 1",
        "after_tex": "G([0,K])\\subseteq[0,K],\\ \\mathrm{Fix}\\supseteq\\{0,K\\}\\ (=\\{0,K\\}\\ \\text{if}\\ r>0)",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整：[0,K] 在 G 下不變（S08），所以從 [0,K] 出發的軌道永遠留在 [0,K]（S09）；0 與 K 是固定點，r>0 時是僅有的固定點（S10）；例子軌道 2、7/2、175/32、58975/8192 符合這些結論（S13）；r≤1 不可省（S14）。",
          "en": "Summary: [0,K] is invariant under G (S08), so an orbit starting in [0,K] stays there forever (S09); 0 and K are fixed points, and the only ones when r>0 (S10); the example orbit 2, 7/2, 175/32, 58975/8192 agrees with these conclusions (S13); and r≤1 cannot be dropped (S14)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S08",
            "S09",
            "S10",
            "S14"
          ],
          "explanation": {
            "zh-Hant": "目標 G：不變性與固定點定理及其假設的必要性。足夠條件 H：S08、S09、S10 與反例 S14；H⇒G 由合取。H 由代數與歸納正向證成，沒有把 G 當作前提；S13 只是例子。",
            "en": "Goal G: the invariance and fixed-point theorem and the necessity of its hypothesis. Sufficient condition H: S08, S09, S10 and the counterexample S14; H⇒G by conjunction. H is proved forwards by algebra and induction; G is never used as a premise, and S13 is only an example."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "對 K>0、0≤r≤1 的模型 G(x)=x+r x(1−x/K)：若 0≤x≤K，則 x≤G(x)≤K，因此從 [0,K] 出發的軌道永遠留在 [0,K]；0 與 K 是固定點，r>0 時沒有其他固定點。K=8、r=1、x0=2 時 x1=7/2、x2=175/32、x3=58975/8192。條件 r≤1 不可省：r=3、x=6 時 G(6)=21/2>8。",
      "en": "For the model G(x)=x+r x(1−x/K) with K>0 and 0≤r≤1: if 0≤x≤K then x≤G(x)≤K, so an orbit starting in [0,K] stays in [0,K] forever; 0 and K are fixed points, and for r>0 there are no others. For K=8, r=1, x0=2: x1=7/2, x2=175/32, x3=58975/8192. The condition r≤1 cannot be dropped: with r=3 and x=6, G(6)=21/2>8."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "x_n 就是連續 logistic 方程 P'=rP(1−P/K) 的解在 t=n 的值",
          "en": "x_n is the value at t=n of the solution of the continuous logistic equation P'=rP(1−P/K)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "K=8、r=1、P(0)=2 時，所列連續解在 t=1 給 P(1)=8/(1+3/e)。若它等於離散的 x1=7/2，交叉相乘便需 e=7/3。此處另明示分析前提 e=Σ_(k≥0)1/k!（同離散指數成長稿的級數前提）；因各項非負，e≥1+1+1/2=5/2，而精確比較 5/2>7/3 排除該等式。這不是由 e 的無誤差界近似值推出，也不在此推導連續解。MKO 明言本模型不是連續方程的解析解。",
          "en": "For K=8, r=1 and P(0)=2, the stated continuous solution gives P(1)=8/(1+3/e). Equality with the discrete x1=7/2 would require e=7/3 by cross-multiplication. An analytic premise is made explicit here: e=Σ_(k≥0)1/k!, as in the exponential-growth guide's series premise. Nonnegative terms give e≥1+1+1/2=5/2, and the exact comparison 5/2>7/3 rules out equality. This is not an inference from an approximate decimal with no error bound, and the continuous solution is not derived here. The MKO states that this model is not an analytic solution of the continuous equation."
        }
      },
      {
        "claim": {
          "zh-Hant": "對任何 r>0，[0,K] 都在 G 下不變",
          "en": "For every r>0, [0,K] is invariant under G"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "K=8、r=3、x=6 時 G(6)=21/2>8（S14）。",
          "en": "With K=8, r=3, x=6, G(6)=21/2>8 (S14)."
        }
      },
      {
        "claim": {
          "zh-Hant": "x0 到 x3 遞增且小於 8，所以軌道收斂到 8 且 8 是穩定的",
          "en": "x0 to x3 increase and stay below 8, so the orbit converges to 8 and 8 is stable"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "有限軌道不能證明收斂或穩定（MKO 假設 a2）；這需要另外的論證，本稿沒有給。",
          "en": "A finite orbit cannot prove convergence or stability (MKO assumption a2); that needs a separate argument, which this guide does not give."
        }
      },
      {
        "claim": {
          "zh-Hant": "對所有 0≤r≤1，0 與 K 是僅有的固定點",
          "en": "For every 0≤r≤1, 0 and K are the only fixed points"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "r=0 時 G(x)=x，每個點都是固定點；「僅有」需要 r>0（S10）。",
          "en": "With r=0, G(x)=x and every point is fixed; 'only' needs r>0 (S10)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "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": "zero_product",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "零乘積性質：在 ℤ、ℚ、ℝ、ℂ（或任何整域）中，uv=0 ⇔ u=0 或 v=0。",
          "en": "Zero-product property: in ℤ, ℚ, ℝ, ℂ (or any integral domain), uv=0 ⇔ u=0 or v=0."
        }
      },
      {
        "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": "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_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_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": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "add_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法交換律：u+v=v+u。",
          "en": "Commutativity of addition: u+v=v+u."
        }
      },
      {
        "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": "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": "modus_ponens",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "肯定前件：由 P 與 P⇒Q 得 Q。",
          "en": "Modus ponens: from P and P⇒Q infer Q."
        }
      },
      {
        "id": "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": "mko_logistic",
        "kind": "cited",
        "source": {
          "mko_id": "mko-discrete-logistic-growth",
          "sha256": "463931fbaa17a793076cfbf7af76f4490ba8c752e31fe52a9dc774d188ac3e4c",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "本條目的離散 logistic 增量模型固定 K>0 與 0≤r≤1，以 x_(n+1)=x_n+r x_n(1−x_n/K) 更新。若 0≤x_n≤K，則下一步仍在 [0,K]；0 與 K 是固定點。這是指定離散模型，不是連續 logistic 方程的解析解。",
          "en": "This discrete logistic increment model fixes K>0 and 0≤r≤1 and updates x_(n+1)=x_n+r*x_n*(1−x_n/K). If 0≤x_n≤K, the next state remains in [0,K]; 0 and K are fixed points. This is the specified discrete model, not an analytic solution of the continuous logistic equation."
        }
      },
      {
        "id": "L_gap",
        "kind": "lemma",
        "proved_in": [
          "S05.1"
        ],
        "statement": {
          "zh-Hant": "對所有實數 x 與 K>0：K·(K−G(x))=(K−x)(K−r x)。",
          "en": "For all real x and K>0: K·(K−G(x))=(K−x)(K−r x)."
        }
      },
      {
        "id": "L_nonneg_prod",
        "kind": "lemma",
        "proved_in": [
          "S03.1"
        ],
        "statement": {
          "zh-Hant": "若 u≥0 且 v≥0，則 uv≥0。",
          "en": "If u≥0 and v≥0, then uv≥0."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "1*2*(1-2/8)",
        "rhs": "1*2*(8-2)/8"
      },
      {
        "id": "N2",
        "step": "S05.1",
        "vars": [
          "x",
          "K",
          "r"
        ],
        "degree": 2,
        "lhs": "K*K-(K*x+r*x*(K-x))",
        "rhs": "(K-x)*(K-r*x)"
      },
      {
        "id": "N3",
        "step": "S11",
        "lhs": "2+1*2*(1-2/8)",
        "rhs": "7/2"
      },
      {
        "id": "N4",
        "step": "S12",
        "lhs": "7/2+1*(7/2)*(1-(7/2)/8)",
        "rhs": "175/32"
      },
      {
        "id": "N5",
        "step": "S13",
        "lhs": "175/32+1*(175/32)*(1-(175/32)/8)",
        "rhs": "58975/8192"
      },
      {
        "id": "N6",
        "step": "S13",
        "lhs": "175*337",
        "rhs": "58975"
      },
      {
        "id": "N7",
        "step": "S13",
        "lhs": "58975/8192",
        "relation": "<",
        "rhs": "8"
      },
      {
        "id": "N8",
        "step": "S14",
        "lhs": "6+3*6*(1-6/8)",
        "rhs": "21/2"
      },
      {
        "id": "N9",
        "lhs": "4*1*(1-1)",
        "rhs": "0"
      },
      {
        "id": "N10",
        "step": "S10",
        "lhs": "1+1*1*(1-1/1)",
        "rhs": "1"
      },
      {
        "id": "N11",
        "step": "S10",
        "lhs": "8+1*8*(1-8/8)",
        "rhs": "8"
      },
      {
        "id": "N12",
        "step": "S14",
        "lhs": "(8-6)*(8-3*6)",
        "rhs": "8*(8-21/2)"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：K、r 固定，初值在 [0,K]，全部計算採精確有理數（假設 a1）。證明一步不變性與固定點，並只展開三步短軌道；不由有限結果宣稱穩定、收斂或真實族群適配（假設 a2）。不討論標準 logistic map 的動力學或連續 logistic 方程的解；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: K and r fixed, the initial state in [0,K], all computations in exact rationals (assumption a1). Proves one-step invariance and the fixed points and expands only a three-step orbit; infers no stability, convergence or fit to real populations from finite results (assumption a2). Neither the dynamics of the standard logistic map nor solutions of the continuous logistic equation are treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "K",
        "容量，K>0",
        "the capacity, K>0"
      ],
      [
        "r",
        "增量比率，0≤r≤1",
        "the increment rate, 0≤r≤1"
      ],
      [
        "G(x)=x+rx(1-\\tfrac{x}{K})",
        "一步更新映射",
        "the one-step update map"
      ],
      [
        "x_n",
        "第 n 步狀態，x_{n+1}=G(x_n)",
        "the state at step n, x_{n+1}=G(x_n)"
      ],
      [
        "[0,K]",
        "被證明為不變的區間",
        "the interval proved to be invariant"
      ]
    ],
    "md_error_controls": [
      {
        "title": "不同的 logistic 形式是不同的模型 / Different logistic forms are different models",
        "zh": "MKO 指出的誤解之一是把不同形式的 logistic map 混為一談。標準 logistic map f(x)=μx(1−x) 在 μ=4 時 f(1)=0，所以 1 不是它的固定點；而本模型取 K=1、r=1 時 G(1)=1（N9、N10）。參數範圍也不同（本模型 0≤r≤1）。本稿的不變性與固定點結論只屬於 MKO 指定的模型。",
        "en": "One mistake named by the MKO is to treat different logistic maps as one model. The standard logistic map f(x)=μx(1−x) with μ=4 has f(1)=0, so 1 is not a fixed point of it; whereas this model with K=1, r=1 has G(1)=1 (N9, N10). The parameter ranges differ too (here 0≤r≤1). The invariance and fixed-point results of this guide belong only to the model specified by the MKO."
      },
      {
        "title": "短軌道不證穩定或收斂 / A short orbit proves no stability or convergence",
        "zh": "MKO 指出的另一誤解。x0 到 x3 為 2、7/2、175/32、58975/8192，遞增且接近 8，但三步計算不證明收斂到 8 或 8 的穩定性，更不是對真實族群的預測（S13）。本稿證明的只有有界性（S09）與單調不減（S04）。",
        "en": "The other mistake named by the MKO. x0 to x3 are 2, 7/2, 175/32 and 58975/8192, increasing and moving closer to 8 over these three steps, but three computed steps prove neither convergence to 8 nor stability of 8, and they are no forecast for a real population (S13). What this guide proves is boundedness (S09) and that updates do not decrease the value (S04)."
      },
      {
        "title": "r≤1 用在哪一步 / Where r≤1 is used",
        "zh": "最容易忽略的是上界需要 r≤1：S06 由 (1−r)x≥0 得 r x≤K，才使 K−r x≥0。r=3、x=6 時 K−r x=−10，G(6)=21/2 超出 [0,8]（S14）。下界則只需要 r≥0。",
        "en": "The easiest thing to overlook is that the upper bound needs r≤1: S06 uses (1−r)x≥0 to get r x≤K, which makes K−r x≥0. With r=3 and x=6, K−r x=−10 and G(6)=21/2 lies outside [0,8] (S14). The lower bound needs only r≥0."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "有界且單調不減的軌道會收斂（單調收斂定理），但本稿不引用也不證明這一點，也不討論極限是否為 K 或其穩定性；這屬於 MKO 的 formal next。",
        "en": "A bounded nondecreasing orbit converges (monotone convergence theorem), but this guide neither cites nor proves that, nor discusses whether the limit is K or its stability; this belongs to the MKO's formal next step."
      },
      {
        "zh-Hant": "N2 是每變數次數 ≤2 的多項式恆等式網格核對。連續解的負例比較另明示 e 的級數表示為分析前提，並用 e≥5/2>7/3 的精確有理數界；這不等於原有數值 AST 已驗證指數級數或連續解公式，兩項仍屬分析前提與另行審閱範圍。",
        "en": "N2 is a polynomial-identity grid check with degree ≤2 per variable. The continuous-solution contrast separately declares the exponential-series representation as an analytic premise and uses the exact rational bound e≥5/2>7/3. The original numeric AST does not thereby verify that series representation or the continuous solution formula; those remain analytic premises subject to separate review."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "5ca9487593862c9c09d680b93f87e15ce9161f326f0796bb9f158056f6b57ecf",
    "original_source_sha256": "352e71a8886bf875d0af919e31e0a2fba9471fb99e640462f4b4b2996e07fd83",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M049"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}