{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-fundamental-theorem-calculus-01",
  "mko_id": "mko-fundamental-theorem-calculus",
  "title": {
    "zh-Hant": "微積分基本定理：由連續性證明 G′=f，再推出 ∫_a^b f=F(b)−F(a)",
    "en": "Fundamental theorem of calculus: proving G′=f from continuity, then ∫_a^b f=F(b)−F(a)"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-fundamental-theorem-calculus-01",
    "source": {
      "mko_id": "mko-fundamental-theorem-calculus",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "6b22aa2d4edd9541d84e10a77e3f555318aaef766fcb9ac34f8008c0e2b43b73",
      "english_sha256": "1b79750a283c6f448f4b0896689d9c71ff9688828879b8c62e6829d099913492"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "continuity_def",
      "func_limit_def",
      "abs_rules",
      "order_basics",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "設 f:ℝ→ℝ 在全 ℝ 連續，a∈ℝ，G(x)=∫_a^x f(t)dt 為有向積分。(1) 證明 G 在每一點可微且 G′(x)=f(x)；(2) 若 F:ℝ→ℝ 可微且 F′=f，證明對任意實數 a、b 有 ∫_a^b f(t)dt=F(b)−F(a)；(3) 以 f(t)=6t²−2t+1 與候選 F(x)=2x³−x²+x+1/2 算出 ∫_2^{−1} f(t)dt，並說明 G(2)=0 不推出 G′(2)=0、上下限在 x=0 重合也不推出導數為 0。",
      "en": "Let f:ℝ→ℝ be continuous on all of ℝ, a∈ℝ, and G(x)=∫_a^x f(t)dt an oriented integral. (1) Prove that G is differentiable at every point with G′(x)=f(x); (2) if F:ℝ→ℝ is differentiable with F′=f, prove ∫_a^b f(t)dt=F(b)−F(a) for arbitrary real a, b; (3) with f(t)=6t²−2t+1 and the candidate F(x)=2x³−x²+x+1/2, compute ∫_2^{−1} f(t)dt, and explain why G(2)=0 does not give G′(2)=0 and why bounds coinciding at x=0 do not give a zero derivative."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "f:ℝ→ℝ 在全 ℝ 連續；a∈ℝ；G(x)=∫_a^x f(t)dt。積分是有向的：x<a 時 ∫_a^x f=−∫_x^a f，且 ∫_a^a f=0（MKO 假設 a1、a2）。",
          "en": "f:ℝ→ℝ is continuous on all of ℝ; a∈ℝ; G(x)=∫_a^x f(t)dt. Integrals are oriented: for x<a, ∫_a^x f=−∫_x^a f, and ∫_a^a f=0 (MKO assumptions a1 and a2)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "F:ℝ→ℝ 在全 ℝ 可微，且對每個實數 x 有 F′(x)=f(x)。",
          "en": "F:ℝ→ℝ is differentiable on all of ℝ and F′(x)=f(x) for every real x."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "例題：f(t)=6t²−2t+1（升冪整數係數 [1,−2,6]，次數 2）；候選 F(x)=2x³−x²+x+1/2（升冪有理係數 [1/2,1,−1,2]，分母為正）；端點 a=2、b=−1。",
          "en": "Example: f(t)=6t²−2t+1 (ascending integer coefficients [1,−2,6], degree 2); candidate F(x)=2x³−x²+x+1/2 (ascending rational coefficients [1/2,1,−1,2] with positive denominators); endpoints a=2, b=−1."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "deriv_def"
        ],
        "before_tex": "G'(x_0)=f(x_0)",
        "after_tex": "\\lim_{h\\to 0,\\ h\\neq 0}\\frac{G(x_0+h)-G(x_0)}{h}=f(x_0)",
        "relation": "definition",
        "rule": "deriv_def",
        "conditions": [
          "x_0\\in\\mathbb{R}"
        ],
        "explanation": {
          "zh-Hant": "取定任意一點 x_0∈ℝ，除了 x_0∈ℝ 不加任何假設。依導數定義，第 (1) 小題在 x_0 的意思是：差商 Q(h)=(G(x_0+h)−G(x_0))/h（h≠0）在 h→0 時有有限極限，且極限等於 f(x_0)。",
          "en": "Fix an arbitrary point x_0∈ℝ, assuming nothing about it beyond x_0∈ℝ. By the definition of the derivative, part (1) at x_0 means: the difference quotient Q(h)=(G(x_0+h)−G(x_0))/h (h≠0) has a finite limit as h→0, and that limit is f(x_0)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "deriv_def"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出目標，不推出新事實。x_0 是任意的，S10 才能做全稱推廣；特別地不排除 x_0=a。",
            "en": "This step only states the goal and derives nothing new. x_0 is arbitrary, which allows the generalization in S10; in particular x_0=a is not excluded."
          }
        },
        "leaf_basis": "deriv_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "int_additive"
        ],
        "before_tex": "G(x_0+h)-G(x_0)",
        "after_tex": "\\int_{x_0}^{x_0+h} f(t)\\,dt",
        "relation": "equivalent",
        "rule": "int_additive",
        "conditions": [
          "h\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "取 p=a、q=x_0、r=x_0+h，可加性給 ∫_a^{x_0+h} f=∫_a^{x_0} f+∫_{x_0}^{x_0+h} f，即 G(x_0+h)=G(x_0)+∫_{x_0}^{x_0+h} f；兩邊同減 G(x_0)。因為採有向約定，a、x_0、x_0+h 的大小次序不限，h<0 也適用。",
          "en": "With p=a, q=x_0, r=x_0+h, additivity gives ∫_a^{x_0+h} f=∫_a^{x_0} f+∫_{x_0}^{x_0+h} f, that is G(x_0+h)=G(x_0)+∫_{x_0}^{x_0+h} f; subtract G(x_0) from both sides. Because the integrals are oriented, a, x_0 and x_0+h may come in any order, so h<0 is covered too."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "int_additive",
            "eq_congruence",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "反向：兩邊同加 G(x_0) 回到可加性的等式；兩式表示同一個數。",
            "en": "Backward: add G(x_0) to both sides to return to the additivity equation; both expressions name the same number."
          }
        },
        "leaf_basis": "int_additive + eq_congruence + add_inverse"
      },
      {
        "id": "S03",
        "uses": [
          "L_const_int"
        ],
        "before_tex": "\\int_{x_0}^{x_0+h} f(x_0)\\,dt",
        "after_tex": "f(x_0)\\,h",
        "relation": "equivalent",
        "rule": "L_const_int",
        "conditions": [
          "h\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "x_0 已取定，被積函數 t↦f(x_0) 是常數 c=f(x_0)；依引理 L_const_int（S03.1、S03.2 證明），其積分是 f(x_0)h。",
          "en": "With x_0 fixed, the integrand t↦f(x_0) is the constant c=f(x_0); by Lemma L_const_int (proved in S03.1 and S03.2) its integral is f(x_0)h."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_const_int"
          ],
          "explanation": {
            "zh-Hant": "兩式是同一個數，雙向成立；引理的兩種情況都已證。",
            "en": "Both expressions are the same number, so the step is two-way; both cases of the lemma are proved."
          }
        },
        "leaf_basis": "mko_moment + mko_reverse + case_analysis"
      },
      {
        "id": "S03.1",
        "uses": [
          "mko_moment"
        ],
        "before_tex": "\\int_{x_0}^{x_0+h} c\\,dt\\quad(h>0)",
        "after_tex": "c\\cdot\\frac{(x_0+h)^1-x_0^1}{1}=ch",
        "relation": "theorem_application",
        "rule": "mko_moment",
        "conditions": [
          "h>0"
        ],
        "explanation": {
          "zh-Hant": "h>0 時 x_0<x_0+h。常數 c 是多項式 c·x⁰，套用矩公式 n=0：c((x_0+h)−x_0)/1=ch。",
          "en": "For h>0, x_0<x_0+h. The constant c is the polynomial c·x⁰, and the moment formula with n=0 gives c((x_0+h)−x_0)/1=ch."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_moment"
          ],
          "explanation": {
            "zh-Hant": "矩公式要求端點 a≤b，此處由 h>0 保證；這一步不宣稱 h<0 的情形。",
            "en": "The moment formula needs endpoints a≤b, which h>0 guarantees; the case h<0 is not claimed here."
          }
        },
        "leaf_basis": "mko_moment + add_inverse"
      },
      {
        "id": "S03.2",
        "uses": [
          "mko_reverse",
          "mko_moment"
        ],
        "before_tex": "\\int_{x_0}^{x_0+h} c\\,dt\\quad(h<0)",
        "after_tex": "-\\int_{x_0+h}^{x_0} c\\,dt=-c\\bigl(x_0-(x_0+h)\\bigr)=ch",
        "relation": "theorem_application",
        "rule": "mko_reverse",
        "conditions": [
          "h<0"
        ],
        "explanation": {
          "zh-Hant": "h<0 時 x_0+h<x_0。先用反向規則 ∫_{x_0}^{x_0+h} c dt=−∫_{x_0+h}^{x_0} c dt，再對正向區間用矩公式：−c(x_0−(x_0+h))=−c(−h)=ch。依三分律，h≠0 表示 h>0 或 h<0，所以 S03.1 與 S03.2 合起來證得引理 L_const_int。",
          "en": "For h<0, x_0+h<x_0. First reverse: ∫_{x_0}^{x_0+h} c dt=−∫_{x_0+h}^{x_0} c dt; then the moment formula on the forward interval gives −c(x_0−(x_0+h))=−c(−h)=ch. By trichotomy, h≠0 means h>0 or h<0, so S03.1 and S03.2 together prove Lemma L_const_int."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_reverse",
            "mko_moment",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要反向規則與矩公式，兩者都來自 MKO 定積分；兩種情況覆蓋所有 h≠0。",
            "en": "It needs the reversal rule and the moment formula, both from the definite-integral MKO; the two cases cover every h≠0."
          }
        },
        "leaf_basis": "mko_reverse + mko_moment + neg_mul + order_basics"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "S03",
          "int_linear"
        ],
        "before_tex": "\\frac{G(x_0+h)-G(x_0)}{h}-f(x_0)",
        "after_tex": "\\frac{1}{h}\\int_{x_0}^{x_0+h}\\bigl(f(t)-f(x_0)\\bigr)\\,dt",
        "relation": "equivalent",
        "rule": "int_linear",
        "conditions": [
          "h\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "由 S02，分子是 ∫_{x_0}^{x_0+h} f；由 S03，f(x_0)=f(x_0)h/h=(1/h)∫_{x_0}^{x_0+h} f(x_0)dt（h≠0 才能除）。兩項都有因子 1/h，再依積分線性把兩個積分合成 ∫_{x_0}^{x_0+h}(f(t)−f(x_0))dt。",
          "en": "By S02 the numerator is ∫_{x_0}^{x_0+h} f; by S03, f(x_0)=f(x_0)h/h=(1/h)∫_{x_0}^{x_0+h} f(x_0)dt (division needs h≠0). Both terms carry the factor 1/h, and linearity of the integral merges the two integrals into ∫_{x_0}^{x_0+h}(f(t)−f(x_0))dt."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S02",
            "S03",
            "int_linear",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "反向：依線性把積分拆回兩項，再用 S03 與 S02 還原；每一步都是等式，前提 h≠0 不變。",
            "en": "Backward: split the integral by linearity and restore with S03 and S02; every move is an equality and the premise h≠0 is unchanged."
          }
        },
        "leaf_basis": "int_linear + mul_inverse_nonzero + distributive + eq_replace"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "continuity_def"
        ],
        "before_tex": "f\\ \\text{continuous at}\\ x_0,\\ \\varepsilon>0",
        "after_tex": "\\exists\\delta>0\\ \\forall t:\\ |t-x_0|<\\delta\\Rightarrow|f(t)-f(x_0)|<\\tfrac{\\varepsilon}{2}",
        "relation": "implies",
        "rule": "continuity_def",
        "conditions": [
          "\\varepsilon>0"
        ],
        "explanation": {
          "zh-Hant": "任取 ε>0。f 在 x_0 連續（G1），把連續的定義用在正數 ε/2 上，得到一個 δ>0 滿足上式。這是證明中唯一直接展開點連續 ε–δ 定義的地方；積分可加性、線性與上界估計的引用也依賴各自的連續性前提，S07 另核對誤差函數連續。",
          "en": "Take any ε>0. Since f is continuous at x_0 (G1), apply the definition of continuity to the positive number ε/2 to obtain δ>0 as stated. This is the only direct invocation of pointwise ε–δ continuity in the proof. The cited integral additivity, linearity and bound also use their continuity hypotheses, and S07 checks continuity of the error integrand."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "continuity_def"
          ],
          "explanation": {
            "zh-Hant": "需要 f 在 x_0 連續；若只有可積而無連續，這樣的 δ 可能不存在（見負面對照的階梯函數）。不宣稱反向。",
            "en": "It needs continuity of f at x_0; with integrability alone such a δ may fail to exist (see the step-function negative control). No converse is claimed."
          }
        },
        "leaf_basis": "continuity_def + forall_elim"
      },
      {
        "id": "S06",
        "uses": [
          "S05"
        ],
        "before_tex": "0<|h|<\\delta,\\ t\\ \\text{between}\\ x_0\\ \\text{and}\\ x_0+h",
        "after_tex": "|t-x_0|\\le|h|<\\delta,\\quad |f(t)-f(x_0)|<\\tfrac{\\varepsilon}{2}",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [
          "0<|h|<\\delta"
        ],
        "explanation": {
          "zh-Hant": "分兩種情況：h>0 時 x_0≤t≤x_0+h，所以 0≤t−x_0≤h=|h|；h<0 時 x_0+h≤t≤x_0，所以 0≤x_0−t≤−h=|h|。兩種情況都有 |t−x_0|≤|h|<δ，再由 S05 得 |f(t)−f(x_0)|<ε/2。",
          "en": "Two cases: if h>0 then x_0≤t≤x_0+h, so 0≤t−x_0≤h=|h|; if h<0 then x_0+h≤t≤x_0, so 0≤x_0−t≤−h=|h|. In both cases |t−x_0|≤|h|<δ, and S05 gives |f(t)−f(x_0)|<ε/2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "abs_rules",
            "order_basics",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "需要 |h|<δ；兩種情況涵蓋所有 h≠0。這是單向估計。",
            "en": "It needs |h|<δ; the two cases cover every h≠0. This is a one-way estimate."
          }
        },
        "leaf_basis": "abs_rules + order_basics + case_analysis"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "int_bound",
          "G1"
        ],
        "before_tex": "\\left|\\int_{x_0}^{x_0+h}\\bigl(f(t)-f(x_0)\\bigr)\\,dt\\right|",
        "after_tex": "\\le\\tfrac{\\varepsilon}{2}\\,|h|",
        "relation": "implies",
        "rule": "int_bound",
        "conditions": [
          "0<|h|<\\delta"
        ],
        "explanation": {
          "zh-Hant": "令 g(t)=f(t)−f(x_0)。g 連續，因為 |g(t)−g(s)|=|f(t)−f(s)|。由 S06，在 x_0 與 x_0+h 之間的閉區間上 |g|≤ε/2。h>0 時以 p=x_0、q=x_0+h 套用估計，得 |∫|≤(ε/2)h；h<0 時有向積分等於 −∫_{x_0+h}^{x_0} g，絕對值不變，估計給 (ε/2)(−h)。兩者都是 (ε/2)|h|。",
          "en": "Let g(t)=f(t)−f(x_0). g is continuous because |g(t)−g(s)|=|f(t)−f(s)|. By S06, |g|≤ε/2 on the closed interval between x_0 and x_0+h. For h>0 apply the bound with p=x_0, q=x_0+h to get |∫|≤(ε/2)h; for h<0 the oriented integral equals −∫_{x_0+h}^{x_0} g, whose absolute value is the same, and the bound gives (ε/2)(−h). Both are (ε/2)|h|."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "int_bound",
            "G1",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "估計需要 p≤q、g 連續、|g|≤M；h<0 時先用 G1 的有向約定換成正向區間。只得到上界，不宣稱相等。",
            "en": "The bound needs p≤q, g continuous and |g|≤M; for h<0 the orientation convention of G1 first turns it into a forward interval. Only an upper bound is obtained; equality is not claimed."
          }
        },
        "leaf_basis": "int_bound + abs_rules + case_analysis"
      },
      {
        "id": "S08",
        "uses": [
          "S04",
          "S07"
        ],
        "before_tex": "\\left|\\frac{G(x_0+h)-G(x_0)}{h}-f(x_0)\\right|",
        "after_tex": "\\le\\tfrac{\\varepsilon}{2}<\\varepsilon\\quad(0<|h|<\\delta)",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [
          "0<|h|<\\delta"
        ],
        "explanation": {
          "zh-Hant": "由 S04，左邊等於 |1/h|·|∫_{x_0}^{x_0+h}(f(t)−f(x_0))dt|=(1/|h|)·|∫|；由 S07，它 ≤(1/|h|)(ε/2)|h|=ε/2<ε。用到 |uv|=|u||v| 與 |h|>0。",
          "en": "By S04 the left side is |1/h|·|∫_{x_0}^{x_0+h}(f(t)−f(x_0))dt|=(1/|h|)·|∫|; by S07 this is ≤(1/|h|)(ε/2)|h|=ε/2<ε. It uses |uv|=|u||v| and |h|>0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S07",
            "abs_rules",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "需要 h≠0 才能除以 |h|；結論對每個 0<|h|<δ 成立。這是單向不等式。",
            "en": "h≠0 is needed to divide by |h|; the conclusion holds for every 0<|h|<δ. This is a one-way inequality."
          }
        },
        "leaf_basis": "abs_rules + frac_rules + order_basics"
      },
      {
        "id": "S09",
        "uses": [
          "S05",
          "S08",
          "func_limit_def"
        ],
        "before_tex": "\\forall\\varepsilon>0\\ \\exists\\delta>0\\ \\forall h:\\ 0<|h|<\\delta\\Rightarrow|Q(h)-f(x_0)|<\\varepsilon",
        "after_tex": "G'(x_0)=f(x_0)",
        "relation": "implies",
        "rule": "func_limit_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "ε>0 是任取的，S05 到 S08 對它找到了合用的 δ；這正是 lim_{h→0} Q(h)=f(x_0) 的定義（0 是 Q 的定義域 ℝ∖{0} 的極限點）。依 S01，G 在 x_0 可微且 G′(x_0)=f(x_0)。",
          "en": "ε>0 was arbitrary, and S05 to S08 found a δ that works for it; this is exactly the definition of lim_{h→0} Q(h)=f(x_0) (0 is a limit point of the domain ℝ∖{0} of Q). By S01, G is differentiable at x_0 with G′(x_0)=f(x_0)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S05",
            "S08",
            "func_limit_def",
            "deriv_def"
          ],
          "explanation": {
            "zh-Hant": "目標 G：G′(x_0)=f(x_0)。足夠條件 H：上面的 ε–δ 敘述；H⇒G 由函數極限與導數的定義。H 由 S05 到 S08 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: G′(x_0)=f(x_0). Sufficient condition H: the ε–δ statement above; H⇒G by the definitions of a function limit and of the derivative. H is proved forwards in S05 to S08; G is never used as a premise."
          }
        },
        "leaf_basis": "func_limit_def + deriv_def + forall_intro"
      },
      {
        "id": "S10",
        "uses": [
          "S09"
        ],
        "before_tex": "G'(x_0)=f(x_0)\\ \\text{for an arbitrary}\\ x_0\\in\\mathbb{R}",
        "after_tex": "\\forall x\\in\\mathbb{R}:\\ G'(x)=f(x)",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S01 到 S09 除了 x_0∈ℝ 沒有用到 x_0 的任何性質，基點 a 也是任意的，所以全稱推廣：對每個 a 與每個 x，G 在 x 可微且 G′(x)=f(x)。第 (1) 小題在所引用的積分性質下得證。",
          "en": "S01 to S09 used no property of x_0 beyond x_0∈ℝ, and the base point a was also arbitrary, so we generalize: for every a and every x, G is differentiable at x with G′(x)=f(x). Part (1) is proved, given the cited integral properties."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "推廣需要 x_0 未被特殊化；已檢查：沒有假設 x_0≠a，也沒有假設 h 的符號。",
            "en": "Generalization needs x_0 not to have been specialized; checked: neither x_0≠a nor a sign of h was assumed."
          }
        },
        "leaf_basis": "forall_intro"
      },
      {
        "id": "S11",
        "uses": [
          "G2",
          "S10",
          "deriv_rules"
        ],
        "before_tex": "D(x)=F(x)-G(x)",
        "after_tex": "D'(x)=F'(x)-G'(x)=f(x)-f(x)=0\\quad\\forall x\\in\\mathbb{R}",
        "relation": "implies",
        "rule": "deriv_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "F 在每點可微（G2），G 在每點可微（S10）。把 D 寫成 F+(−1)·G，由和與常數倍的導數規則，D 在每點可微，且 D′=F′−G′=f−f=0。",
          "en": "F is differentiable everywhere (G2) and so is G (S10). Writing D=F+(−1)·G, the sum and constant-multiple rules show D is differentiable everywhere with D′=F′−G′=f−f=0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "S10",
            "deriv_rules"
          ],
          "explanation": {
            "zh-Hant": "導數規則要求兩函數在該點都可微，由 G2 與 S10 提供。不宣稱反向。",
            "en": "The rules need both functions differentiable at the point, provided by G2 and S10. No converse is claimed."
          }
        },
        "leaf_basis": "deriv_rules + add_inverse"
      },
      {
        "id": "S12",
        "uses": [
          "S11",
          "thm_zero_deriv"
        ],
        "before_tex": "D'(x)=0\\quad\\forall x\\in\\mathbb{R}",
        "after_tex": "\\exists C\\in\\mathbb{R}\\ \\forall x\\in\\mathbb{R}:\\ F(x)=G(x)+C",
        "relation": "theorem_application",
        "rule": "thm_zero_deriv",
        "conditions": [
          "D\\ \\text{differentiable on}\\ \\mathbb{R}"
        ],
        "explanation": {
          "zh-Hant": "D 在整個 ℝ（一個區間）上可微且導數處處為 0；依引用的均值定理推論，D 是常數 C，即 F=G+C。",
          "en": "D is differentiable on all of ℝ (an interval) with derivative zero everywhere; by the cited corollary of the Mean Value Theorem, D is a constant C, that is F=G+C."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "thm_zero_deriv"
          ],
          "explanation": {
            "zh-Hant": "推論要求在一個區間上導數處處為 0；若定義域是兩段分離的區間，結論只能是「在每一段上為常數」。MKO 採全 ℝ 前提，正好避開這個問題。",
            "en": "The corollary needs zero derivative throughout an interval; on two separated intervals one could only conclude 'constant on each piece'. The MKO's premise that the domain is all of ℝ avoids this."
          }
        },
        "leaf_basis": "thm_zero_deriv"
      },
      {
        "id": "S13",
        "uses": [
          "S12",
          "G1"
        ],
        "before_tex": "F(b)-F(a)",
        "after_tex": "\\int_a^b f(t)\\,dt",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [
          "a,b\\in\\mathbb{R}"
        ],
        "explanation": {
          "zh-Hant": "由 S12，F(b)−F(a)=(G(b)+C)−(G(a)+C)=G(b)−G(a)（S13.2）；G(b)=∫_a^b f，且 G(a)=0（S13.1），所以等於 ∫_a^b f。b 可以小於、等於或大於 a，因為 G 用的是有向積分。",
          "en": "By S12, F(b)−F(a)=(G(b)+C)−(G(a)+C)=G(b)−G(a) (S13.2); G(b)=∫_a^b f and G(a)=0 (S13.1), so this equals ∫_a^b f. b may be less than, equal to or greater than a, because G uses oriented integrals."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S12",
            "G1",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "每一步都是等值代換，從右讀到左同樣成立。",
            "en": "Every move replaces equals by equals, so reading right to left works as well."
          }
        },
        "leaf_basis": "eq_replace + add_inverse + def_unfold"
      },
      {
        "id": "S13.1",
        "uses": [
          "G1"
        ],
        "before_tex": "G(a)",
        "after_tex": "\\int_a^a f(t)\\,dt=0",
        "relation": "equivalent",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 G 的定義 G(a)=∫_a^a f，再依 G1 的有向約定（相同端點的積分為 0）得 0。",
          "en": "By the definition of G, G(a)=∫_a^a f, and by the orientation convention of G1 (equal endpoints give 0) this is 0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "G1",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "展開定義與約定都是等式，兩向成立。",
            "en": "Unfolding the definition and the convention are equalities, valid both ways."
          }
        },
        "leaf_basis": "def_unfold"
      },
      {
        "id": "S13.2",
        "uses": [
          "S12"
        ],
        "before_tex": "(G(b)+C)-(G(a)+C)",
        "after_tex": "G(b)-G(a)",
        "relation": "equivalent",
        "rule": "add_inverse",
        "conditions": [],
        "explanation": {
          "zh-Hant": "C 與 −C 相消（加法結合律、交換律與反元素）。這也說明為什麼原函數加常數不改端點差。",
          "en": "C and −C cancel (associativity, commutativity and inverses for addition). This is also why adding a constant to an antiderivative does not change the endpoint difference."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "add_assoc",
            "add_comm",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "對所有實數成立的恆等式，雙向成立。",
            "en": "An identity for all reals, so it holds both ways."
          }
        },
        "leaf_basis": "add_assoc + add_comm + add_inverse + add_zero"
      },
      {
        "id": "S14",
        "uses": [
          "G3",
          "poly_deriv"
        ],
        "before_tex": "F(x)=\\tfrac{1}{2}+x-x^2+2x^3",
        "after_tex": "F'(x)=1-2x+6x^2=f(x)",
        "relation": "implies",
        "rule": "poly_deriv",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐係數求導：常數 1/2 的導數是 0；x 給 1；−x² 給 −2x；2x³ 給 6x²。F′ 的升冪係數 [1,−2,6] 與 f 的 [1,−2,6] 逐項相同，所以 F′=f 作為多項式相等，於每個實數 x 都成立。F 是多項式，在全 ℝ 可微，所以 G2 的前提對這個 F 成立。",
          "en": "Differentiate coefficientwise: the constant 1/2 gives 0; x gives 1; −x² gives −2x; 2x³ gives 6x². The ascending coefficients [1,−2,6] of F′ agree term by term with those of f, so F′=f as polynomials, hence at every real x. F is a polynomial and differentiable on all of ℝ, so the premise G2 holds for this F."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "poly_deriv"
          ],
          "explanation": {
            "zh-Hant": "係數表相等推出函數相等；這一步沒有用有限個點取樣（MKO 假設 a3）。不宣稱原函數唯一：F+7 等其他原函數也可用。",
            "en": "Equal coefficient lists give equal functions; no finite sampling is used (MKO assumption a3). Uniqueness of the antiderivative is not claimed: F+7 and others also work."
          }
        },
        "leaf_basis": "poly_deriv + numeric_evaluation"
      },
      {
        "id": "S15",
        "uses": [
          "S13",
          "S14",
          "G3"
        ],
        "before_tex": "\\int_2^{-1} f(t)\\,dt",
        "after_tex": "F(-1)-F(2)=-\\tfrac{7}{2}-\\tfrac{29}{2}=-18",
        "relation": "theorem_application",
        "rule": "forall_elim",
        "conditions": [
          "F'=f\\ \\text{on}\\ \\mathbb{R}"
        ],
        "explanation": {
          "zh-Hant": "S13 對任意實數端點成立，取 a=2、b=−1（端點反向）。F(−1)=2(−1)³−(−1)²+(−1)+1/2=−2−1−1+1/2=−7/2；F(2)=16−4+2+1/2=29/2；差為 −36/2=−18。",
          "en": "S13 holds for arbitrary real endpoints; take a=2, b=−1 (reversed endpoints). F(−1)=2(−1)³−(−1)²+(−1)+1/2=−2−1−1+1/2=−7/2; F(2)=16−4+2+1/2=29/2; the difference is −36/2=−18."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13",
            "S14",
            "forall_elim",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "需要 S14 先確認 F′=f；端點反向不需另外處理，因為 S13 的 a、b 是任意的。",
            "en": "It needs S14's check F′=f first; reversed endpoints need no special treatment because a and b in S13 are arbitrary."
          }
        },
        "leaf_basis": "forall_elim + numeric_evaluation"
      },
      {
        "id": "S15.1",
        "uses": [
          "G3",
          "mko_moment",
          "mko_reverse"
        ],
        "before_tex": "\\int_{-1}^{2} f(t)\\,dt",
        "after_tex": "6\\cdot\\tfrac{2^3-(-1)^3}{3}-2\\cdot\\tfrac{2^2-(-1)^2}{2}+\\bigl(2-(-1)\\bigr)=18",
        "relation": "theorem_application",
        "rule": "mko_moment",
        "conditions": [
          "-1\\le 2"
        ],
        "explanation": {
          "zh-Hant": "獨立核對：用定積分 MKO 的矩公式（不用 FTC）：6·9/3−2·3/2+3=18−3+3=18；再依反向規則 ∫_2^{−1} f=−18，與 S15 一致。",
          "en": "Independent check with the definite-integral MKO's moment formula (no FTC): 6·9/3−2·3/2+3=18−3+3=18; the reversal rule then gives ∫_2^{−1} f=−18, agreeing with S15."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_moment",
            "mko_reverse",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "這是另一條路的數值核對（N4），不是 FTC 的證明。",
            "en": "A numeric cross-check by another route (N4), not a proof of the FTC."
          }
        },
        "leaf_basis": "mko_moment + mko_reverse + numeric_evaluation"
      },
      {
        "id": "S15.2",
        "uses": [
          "S15"
        ],
        "before_tex": "\\bigl(F(-1)+7\\bigr)-\\bigl(F(2)+7\\bigr)",
        "after_tex": "F(-1)-F(2)=-18",
        "relation": "equivalent",
        "rule": "add_inverse",
        "conditions": [],
        "explanation": {
          "zh-Hant": "換用原函數 F+7：(F(−1)+7)−(F(2)+7)=F(−1)−F(2)=−18。加常數不改端點差，與 S13.2 消去 C 是同一件事。",
          "en": "Use the antiderivative F+7 instead: (F(−1)+7)−(F(2)+7)=F(−1)−F(2)=−18. Adding a constant does not change the endpoint difference, which is the same cancellation as C in S13.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "add_inverse",
            "add_assoc"
          ],
          "explanation": {
            "zh-Hant": "恆等式，雙向成立（N9）。",
            "en": "An identity, valid both ways (N9)."
          }
        },
        "leaf_basis": "add_inverse + add_assoc + add_comm"
      },
      {
        "id": "S16",
        "uses": [
          "S10",
          "S13",
          "G3"
        ],
        "before_tex": "G(x)=\\int_2^x f(t)\\,dt",
        "after_tex": "G(2)=0,\\quad G'(2)=f(2)=21\\neq 0",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取基點 a=2。G(2)=∫_2^2 f=0，但由 S10，G′(2)=f(2)=6·4−2·2+1=21≠0。函數值為 0 只說明累積量從 0 開始；導數是此刻的變化率。由 S13 也可寫 G(x)=F(x)−F(2)，所以 G′=F′=f，一致。",
          "en": "Take base point a=2. G(2)=∫_2^2 f=0, yet by S10 G′(2)=f(2)=6·4−2·2+1=21≠0. A zero value only says the accumulation starts at 0; the derivative is the rate of change at that moment. By S13 one may also write G(x)=F(x)−F(2), so G′=F′=f, consistently."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "G3",
            "forall_elim",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "只需 S10 在 x=2 的特例與 f(2) 的計算（N5）。",
            "en": "It needs only S10 at x=2 and the value of f(2) (N5)."
          }
        },
        "leaf_basis": "forall_elim + numeric_evaluation"
      },
      {
        "id": "S17",
        "uses": [
          "S13",
          "S14",
          "G3"
        ],
        "before_tex": "K(x)=\\int_x^{2x} f(t)\\,dt",
        "after_tex": "K(x)=F(2x)-F(x)=14x^3-3x^2+x",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對每個 x，以 a=x、b=2x 套用 S13：K(x)=F(2x)−F(x)=(16x³−4x²+2x+1/2)−(2x³−x²+x+1/2)=14x³−3x²+x（N6 為次數 3 的恆等式核對）。在 x=0，上下限都是 0，K(0)=0。",
          "en": "For each x apply S13 with a=x and b=2x: K(x)=F(2x)−F(x)=(16x³−4x²+2x+1/2)−(2x³−x²+x+1/2)=14x³−3x²+x (N6 checks this degree-3 identity). At x=0 both bounds are 0 and K(0)=0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13",
            "S14",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "S13 對任意實數端點成立，所以對每個 x 都可用；x<0 時端點是反向的，也被涵蓋。",
            "en": "S13 holds for arbitrary real endpoints, so it applies for every x; for x<0 the endpoints are reversed, and that is covered too."
          }
        },
        "leaf_basis": "forall_elim + distributive + power_laws"
      },
      {
        "id": "S17.1",
        "uses": [
          "S17",
          "poly_deriv"
        ],
        "before_tex": "K(x)=14x^3-3x^2+x",
        "after_tex": "K'(x)=42x^2-6x+1,\\quad K'(0)=1\\neq 0",
        "relation": "implies",
        "rule": "poly_deriv",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐項求導得 K′(x)=42x²−6x+1，所以 K′(0)=1≠0：上下限在 x=0 重合並不使導數為 0，因為上限以速度 2、下限以速度 1 移動。",
          "en": "Differentiating term by term gives K′(x)=42x²−6x+1, so K′(0)=1≠0: bounds coinciding at x=0 do not make the derivative zero, because the upper bound moves at speed 2 and the lower bound at speed 1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "poly_deriv",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "只用多項式求導（N8）。",
            "en": "Only polynomial differentiation is used (N8)."
          }
        },
        "leaf_basis": "poly_deriv + numeric_evaluation"
      },
      {
        "id": "S17.2",
        "uses": [
          "S17.1",
          "S14",
          "deriv_rules"
        ],
        "before_tex": "f(u(x))\\,u'(x)-f(l(x))\\,l'(x),\\quad u(x)=2x,\\ l(x)=x",
        "after_tex": "2f(2x)-f(x)=42x^2-6x+1",
        "relation": "implies",
        "rule": "deriv_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 假設 a2 的移動上下限公式給 f(2x)·2−f(x)·1=2(24x²−4x+1)−(6x²−2x+1)=42x²−6x+1，與 S17.1 相同（N7 為次數 2 的恆等式核對）。這也就是對 K(x)=F(2x)−F(x) 用合成規則：K′(x)=2F′(2x)−F′(x)。",
          "en": "The moving-bounds formula of MKO assumption a2 gives f(2x)·2−f(x)·1=2(24x²−4x+1)−(6x²−2x+1)=42x²−6x+1, the same as S17.1 (N7 checks this degree-2 identity). This is the chain rule applied to K(x)=F(2x)−F(x): K′(x)=2F′(2x)−F′(x)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S17.1",
            "deriv_rules",
            "S14"
          ],
          "explanation": {
            "zh-Hant": "這是兩種算法的一致性核對；合成規則的前提（F、u、l 可微）都成立。",
            "en": "A consistency check of two computations; the chain rule's premises (F, u and l differentiable) all hold."
          }
        },
        "leaf_basis": "deriv_rules + poly_deriv + numeric_evaluation"
      }
    ],
    "result": {
      "zh-Hant": "在 f 於 ℝ 連續的前提下（並引用積分的可加、線性與估計性質），G(x)=∫_a^x f 處處可微且 G′=f；若 F′=f 於 ℝ，則對任意實數 a、b，∫_a^b f=F(b)−F(a)（另引用「導數處處為 0 則為常數」）。例：∫_2^{−1}(6t²−2t+1)dt=−18，矩公式獨立核對一致；以 a=2 的累積函數有 G(2)=0 但 G′(2)=21；K(x)=∫_x^{2x} f 有 K(0)=0 但 K′(0)=1。",
      "en": "With f continuous on ℝ (and citing additivity, linearity and the bound for integrals), G(x)=∫_a^x f is differentiable everywhere with G′=f; if F′=f on ℝ, then ∫_a^b f=F(b)−F(a) for arbitrary real a, b (also citing 'zero derivative everywhere implies constant'). Example: ∫_2^{−1}(6t²−2t+1)dt=−18, confirmed independently by the moment formula; the accumulation function with a=2 has G(2)=0 but G′(2)=21; K(x)=∫_x^{2x} f has K(0)=0 but K′(0)=1."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "G(x)=∫_2^x f(t)dt 滿足 G(2)=0，所以 G′(2)=0",
          "en": "G(x)=∫_2^x f(t)dt has G(2)=0, so G′(2)=0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "由第一部分，G′(2)=f(2)=21。",
          "en": "By part (1), G′(2)=f(2)=21."
        }
      },
      {
        "claim": {
          "zh-Hant": "K(x)=∫_x^{2x} f(t)dt 在 x=0 上下限重合，所以 K′(0)=0",
          "en": "K(x)=∫_x^{2x} f(t)dt has coinciding bounds at x=0, so K′(0)=0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "K(x)=14x³−3x²+x，K′(0)=1。",
          "en": "K(x)=14x³−3x²+x, so K′(0)=1."
        }
      },
      {
        "claim": {
          "zh-Hant": "F₃(x)=1/2+x−(3/2)x²+2x³+(1/4)x⁴ 在 x=−1、0、1 都有 F₃′=f，所以 ∫_2^{−1} f=F₃(−1)−F₃(2)",
          "en": "F₃(x)=1/2+x−(3/2)x²+2x³+(1/4)x⁴ has F₃′=f at x=−1, 0, 1, so ∫_2^{−1} f=F₃(−1)−F₃(2)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "F₃′−f=x³−x，在 x=2 為 6≠0，前提 F₃′=f 不成立；F₃(−1)−F₃(2)=−81/4≠−18。",
          "en": "F₃′−f=x³−x is 6≠0 at x=2, so the premise F₃′=f fails; F₃(−1)−F₃(2)=−81/4≠−18."
        }
      },
      {
        "claim": {
          "zh-Hant": "對任何可積函數 f（不必連續），G(x)=∫_0^x f 處處可微且 G′=f",
          "en": "For every integrable f (not necessarily continuous), G(x)=∫_0^x f is differentiable everywhere with G′=f"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "f(t)=0（t<0）、1（t≥0）給 G(x)=max(x,0)，G′(0) 不存在。",
          "en": "f(t)=0 for t<0 and 1 for t≥0 gives G(x)=max(x,0), and G′(0) does not exist."
        }
      },
      {
        "claim": {
          "zh-Hant": "∫_a^b f=F(b)−F(a) 只在 a<b 時成立",
          "en": "∫_a^b f=F(b)−F(a) holds only when a<b"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "S13 對任意實數 a、b 成立；S15 以 a=2>b=−1 算得 −18，並經矩公式核對。",
          "en": "S13 holds for arbitrary real a, b; S15 computes −18 with a=2>b=−1, confirmed by the moment formula."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "continuity_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "連續（ε-δ）：f 在 a 連續表示對每個 ε>0，存在 δ>0，使所有滿足 |x−a|<δ 的定義域點 x 都有 |f(x)−f(a)|<ε。",
          "en": "Continuity (ε-δ): f is continuous at a if for every ε>0 there is δ>0 such that |f(x)−f(a)|<ε for every domain point x with |x−a|<δ."
        }
      },
      {
        "id": "func_limit_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "函數極限：lim_{x→a} f(x)=L 表示對每個 ε>0，存在 δ>0，使所有滿足 0<|x−a|<δ 的定義域點 x 都有 |f(x)−L|<ε（a 須為定義域的極限點）。",
          "en": "Limit of a function: lim_{x→a} f(x)=L means that for every ε>0 there is δ>0 such that |f(x)−L|<ε for every domain point x with 0<|x−a|<δ (a must be a limit point of the domain)."
        }
      },
      {
        "id": "abs_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "絕對值（實數）：|u|=u（u≥0）、|u|=−u（u<0）；|u|≥0；|uv|=|u||v|；三角不等式 |u+v|≤|u|+|v|；|u|<r ⇔ −r<u<r。",
          "en": "Absolute value (reals): |u|=u if u≥0 and |u|=−u if u<0; |u|≥0; |uv|=|u||v|; triangle inequality |u+v|≤|u|+|v|; |u|<r ⇔ −r<u<r."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "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": "add_inverse",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。",
          "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)."
        }
      },
      {
        "id": "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": "neg_mul",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。",
          "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv."
        }
      },
      {
        "id": "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": "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": "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": "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": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "id": "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": "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": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "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": "deriv_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-derivative",
          "sha256": "b523a35b9e5c20d66a7346f84ac5bece2c4167b17d46208fdb9b40115deeaf64",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對 f:ℝ→ℝ，f 在 x 的導數是 lim_{h→0,h≠0}(f(x+h)−f(x))/h，若此兩側有限極限存在；有限步 h 須非零。",
          "en": "For f:ℝ→ℝ, its derivative at x is lim_{h→0,h≠0}(f(x+h)−f(x))/h when this finite two-sided limit exists; a finite step h must be nonzero."
        }
      },
      {
        "id": "deriv_rules",
        "kind": "cited",
        "source": {
          "mko_id": "mko-derivative-rules",
          "sha256": "4756c1f6226c8cecc330975377ab5d42ed0e461b0fe039c05959fc60ab7c628f",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在相應點可微的實函數有 (f+g)′=f′+g′、(cf)′=cf′、(fg)′=f′g+fg′；若 g 在 x 可微且 f 在 g(x) 可微，則 (f∘g)′(x)=f′(g(x))g′(x)。",
          "en": "For real functions differentiable at the relevant points, (f+g)′=f′+g′, (cf)′=cf′ and (fg)′=f′g+fg′; if g is differentiable at x and f at g(x), then (f∘g)′(x)=f′(g(x))g′(x)."
        }
      },
      {
        "id": "mko_moment",
        "kind": "cited",
        "source": {
          "mko_id": "mko-definite-integral",
          "sha256": "f1fc56fc2ac48870dcaa53e6269bb2e365c1529c5540683ba6b6cd103ddcc5f7",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對連續多項式與有限區間 a≤b，I(x^n,a,b)=(b^{n+1}−a^{n+1})/(n+1)；對 f=Σc_k x^k 逐係數相加。",
          "en": "For continuous polynomials on a finite interval a≤b, I(x^n,a,b)=(b^{n+1}−a^{n+1})/(n+1); for f=Σc_k x^k the moments are summed coefficientwise."
        }
      },
      {
        "id": "mko_reverse",
        "kind": "cited",
        "source": {
          "mko_id": "mko-definite-integral",
          "sha256": "f1fc56fc2ac48870dcaa53e6269bb2e365c1529c5540683ba6b6cd103ddcc5f7",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "反向端點換號：I(f,a,b)=−I(f,b,a)；相同端點 I(f,a,a)=0。",
          "en": "Reversing endpoints changes the sign: I(f,a,b)=−I(f,b,a); equal endpoints give I(f,a,a)=0."
        }
      },
      {
        "id": "int_additive",
        "kind": "cited",
        "source": {
          "reference": "OpenStax, Calculus Volume 1, Section 5.2, Properties of the Definite Integral, rules 1, 2 and 6",
          "original_reference": "M. Spivak, Calculus, 4th edition (Publish or Perish, 2008), Chapter 13 'Integrals': additivity of the integral over adjacent intervals, extended to any order of endpoints by the convention that reversing the endpoints changes the sign",
          "url": "https://openstax.org/books/calculus-volume-1/pages/5-2-the-definite-integral",
          "locator": "Equations 5.9, 5.10, 5.14",
          "checked_date": "2026-10-19",
          "verification": "verified_replacement",
          "verification_note": "Additivity plus reversal handles all endpoint orders."
        },
        "needs_citation": false,
        "statement": {
          "zh-Hant": "若 f 在 ℝ 上連續，則對任意實數 p、q、r（大小次序不限，採有向約定）有 ∫_p^r f=∫_p^q f+∫_q^r f。",
          "en": "If f is continuous on ℝ, then for any reals p, q, r (in any order, with the orientation convention) ∫_p^r f=∫_p^q f+∫_q^r f."
        }
      },
      {
        "id": "int_linear",
        "kind": "cited",
        "source": {
          "reference": "OpenStax, Calculus Volume 1, Section 5.2, Properties of the Definite Integral, rule 4",
          "original_reference": "M. Spivak, Calculus, 4th edition (Publish or Perish, 2008), Chapter 13 'Integrals': linearity of the integral",
          "url": "https://openstax.org/books/calculus-volume-1/pages/5-2-the-definite-integral",
          "locator": "Equation 5.12, difference of integrals",
          "checked_date": "2026-10-19",
          "verification": "verified_replacement",
          "verification_note": "Continuous integrands; oriented endpoint convention retained."
        },
        "needs_citation": false,
        "statement": {
          "zh-Hant": "若 f、g 在 ℝ 上連續，則對任意實數 p、q（有向約定）有 ∫_p^q (f(t)−g(t))dt=∫_p^q f(t)dt−∫_p^q g(t)dt。",
          "en": "If f and g are continuous on ℝ, then for any reals p, q (orientation convention) ∫_p^q (f(t)−g(t))dt=∫_p^q f(t)dt−∫_p^q g(t)dt."
        }
      },
      {
        "id": "int_bound",
        "kind": "cited",
        "source": {
          "reference": "OpenStax, Calculus Volume 1, Section 5.2, Comparison Properties of Integrals",
          "original_reference": "M. Spivak, Calculus, 4th edition (Publish or Perish, 2008), Chapter 13 'Integrals': if m≤f(x)≤M on [a,b] then m(b−a)≤∫_a^b f≤M(b−a)",
          "url": "https://openstax.org/books/calculus-volume-1/pages/5-2-the-definite-integral",
          "locator": "Comparison bounds between constant functions",
          "checked_date": "2026-10-19",
          "verification": "verified_replacement",
          "verification_note": "Apply comparison to -M<=g<=M, then use the two-sided integral bound."
        },
        "needs_citation": false,
        "statement": {
          "zh-Hant": "若 p≤q，g 在 [p,q] 上連續，且對 [p,q] 中每個 t 有 |g(t)|≤M，則 |∫_p^q g(t)dt|≤M(q−p)。",
          "en": "If p≤q, g is continuous on [p,q], and |g(t)|≤M for every t in [p,q], then |∫_p^q g(t)dt|≤M(q−p)."
        }
      },
      {
        "id": "thm_zero_deriv",
        "kind": "cited",
        "source": {
          "reference": "OpenStax, Calculus Volume 1, Section 4.4, Theorem 4.6, Corollary 1",
          "original_reference": "M. Spivak, Calculus, 4th edition (Publish or Perish, 2008), Chapter 11, corollary of the Mean Value Theorem: a function whose derivative is zero at every point of an interval is constant on that interval",
          "url": "https://openstax.org/books/calculus-volume-1/pages/4-4-the-mean-value-theorem",
          "locator": "Functions with a Derivative of Zero",
          "checked_date": "2026-10-19",
          "verification": "verified_replacement",
          "verification_note": "Differentiability on an interval is required; the guide uses all real numbers."
        },
        "needs_citation": false,
        "statement": {
          "zh-Hant": "若 H:ℝ→ℝ 在全 ℝ 可微，且對每個 x 有 H′(x)=0，則 H 在 ℝ 上是常數。",
          "en": "If H:ℝ→ℝ is differentiable on all of ℝ and H′(x)=0 for every x, then H is constant on ℝ."
        }
      },
      {
        "id": "poly_deriv",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "多項式逐項求導：(Σ_{k=0}^{m} c_k x^k)′=Σ_{k=1}^{m} k c_k x^{k−1}，在全 ℝ 成立；它由 MKO 導數規則的和、常數倍與乘積規則逐次得到。",
          "en": "Polynomials differentiate term by term: (Σ_{k=0}^{m} c_k x^k)′=Σ_{k=1}^{m} k c_k x^{k−1} on all of ℝ; this follows by repeated use of the sum, constant-multiple and product rules of the derivative-rules MKO."
        }
      },
      {
        "id": "L_const_int",
        "kind": "lemma",
        "proved_in": [
          "S03.1",
          "S03.2"
        ],
        "statement": {
          "zh-Hant": "對任意實數 c、x_0 與 h≠0：∫_{x_0}^{x_0+h} c dt=ch。",
          "en": "For any reals c, x_0 and h≠0: ∫_{x_0}^{x_0+h} c dt=ch."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S15",
        "lhs": "2*(-1)**3 - (-1)**2 + (-1) + 1/2",
        "rhs": "-7/2"
      },
      {
        "id": "N2",
        "step": "S15",
        "lhs": "2*2**3 - 2**2 + 2 + 1/2",
        "rhs": "29/2"
      },
      {
        "id": "N3",
        "step": "S15",
        "lhs": "-7/2 - 29/2",
        "rhs": "-18"
      },
      {
        "id": "N4",
        "step": "S15.1",
        "lhs": "6*(2**3-(-1)**3)/3 - 2*(2**2-(-1)**2)/2 + (2-(-1))",
        "rhs": "18"
      },
      {
        "id": "N5",
        "step": "S16",
        "lhs": "6*2**2 - 2*2 + 1",
        "rhs": "21"
      },
      {
        "id": "N6",
        "step": "S17",
        "vars": [
          "x"
        ],
        "degree": 3,
        "lhs": "(2*(2*x)**3 - (2*x)**2 + 2*x + 1/2) - (2*x**3 - x**2 + x + 1/2)",
        "rhs": "14*x**3 - 3*x**2 + x"
      },
      {
        "id": "N7",
        "step": "S17.2",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "2*(6*(2*x)**2 - 2*(2*x) + 1) - (6*x**2 - 2*x + 1)",
        "rhs": "42*x**2 - 6*x + 1"
      },
      {
        "id": "N8",
        "step": "S17.1",
        "lhs": "42*0**2 - 6*0 + 1",
        "rhs": "1"
      },
      {
        "id": "N9",
        "step": "S15.2",
        "lhs": "((2*(-1)**3 - (-1)**2 + (-1) + 1/2) + 7) - ((2*2**3 - 2**2 + 2 + 1/2) + 7)",
        "rhs": "-18"
      },
      {
        "id": "N10",
        "step": "S14",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "3*2*x**2 + 2*(-1)*x + 1",
        "rhs": "6*x**2 - 2*x + 1"
      },
      {
        "id": "N11",
        "vars": [
          "x"
        ],
        "range": [
          -1,
          1
        ],
        "lhs": "x**3 - x",
        "rhs": "0"
      },
      {
        "id": "N12",
        "lhs": "2**3 - 2",
        "rhs": "6"
      },
      {
        "id": "N13",
        "lhs": "(-7/2 + 1/4 - 1/2) - (29/2 + 16/4 - 4/2)",
        "rhs": "-81/4"
      },
      {
        "id": "N14",
        "lhs": "-81/4",
        "relation": "!=",
        "rhs": "-18"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的充分前提下工作：f 在全 ℝ 連續、F 在全 ℝ 可微。第一部分的 ε–δ 論證完整寫出，但依賴三條一般積分性質（有向區間可加、線性、以上界估計積分）；這些性質對一般連續函數以外部文獻引用（citation_verified），不在本稿證明。第二部分另引用「導數處處為 0 的函數為常數」（均值定理推論，citation_verified）。例題中的多項式積分另以 MKO 定積分的矩公式交叉核對。不處理只可積而不連續的函數、端點單側導數或 Lebesgue 積分；不改動 canonical MKO 或英文 edition。",
      "en": "Works under the MKO's sufficient premises: f continuous on all of ℝ and F differentiable on all of ℝ. The ε–δ argument for part (1) is written out in full, but it rests on three general properties of the integral (additivity over oriented intervals, linearity, and an estimate from an upper bound on the integrand's absolute value); for general continuous functions these are cited from the literature (citation_verified) and not proved here. Part (2) also cites 'a function with zero derivative everywhere is constant' (a corollary of the Mean Value Theorem, citation_verified). The polynomial integral in the example is cross-checked with the moment formula of the definite-integral MKO. Merely integrable discontinuous functions, one-sided endpoint derivatives and Lebesgue integration are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "G(x)=\\int_a^x f(t)\\,dt",
        "累積函數（有向積分）",
        "the accumulation function (oriented integral)"
      ],
      [
        "F",
        "f 的原函數：F′=f",
        "an antiderivative of f: F′=f"
      ],
      [
        "x_0",
        "任意取定的一點",
        "an arbitrary fixed point"
      ],
      [
        "h",
        "非零增量",
        "a nonzero increment"
      ],
      [
        "Q(h)",
        "差商 (G(x_0+h)−G(x_0))/h",
        "the difference quotient (G(x_0+h)−G(x_0))/h"
      ],
      [
        "\\varepsilon,\\ \\delta",
        "誤差界與鄰域半徑",
        "error bound and neighbourhood radius"
      ],
      [
        "D=F-G",
        "兩個原函數之差",
        "the difference of the two antiderivatives"
      ],
      [
        "K(x)=\\int_x^{2x} f(t)\\,dt",
        "上下限都移動的積分",
        "an integral with both bounds moving"
      ]
    ],
    "md_error_controls": [
      {
        "title": "函數值為 0 不等於導數為 0 / A zero value is not a zero derivative",
        "zh": "MKO 指出的誤解：起點積分值為 0 就以為導數為 0。G(2)=∫_2^2 f=0 只說明累積從 0 開始；G′(2)=f(2)=21 是此刻的變化率（S16）。",
        "en": "The misconception the MKO names: since the integral is 0 at the starting point, the derivative must be 0. G(2)=∫_2^2 f=0 only says the accumulation starts at 0; G′(2)=f(2)=21 is the rate of change at that moment (S16)."
      },
      {
        "title": "上下限重合不等於導數為 0 / Coinciding bounds do not force a zero derivative",
        "zh": "K(x)=∫_x^{2x} f 在 x=0 時上下限都是 0，K(0)=0；但上限移動速度 2、下限速度 1，K′(0)=2f(0)−f(0)=f(0)=1（S17）。",
        "en": "K(x)=∫_x^{2x} f has both bounds equal to 0 at x=0, so K(0)=0; but the upper bound moves at speed 2 and the lower at speed 1, so K′(0)=2f(0)−f(0)=f(0)=1 (S17)."
      },
      {
        "title": "先核對 F′=f，而且要逐係數 / Check F′=f first, and coefficientwise",
        "zh": "第二部分需要 F′=f 這個前提。候選 F₃(x)=1/2+x−(3/2)x²+2x³+(1/4)x⁴ 在 x=−1、0、1 三點有 F₃′=f，但 F₃′−f=x³−x 在 x=2 等於 6；用它算端點差會得 F₃(−1)−F₃(2)=−81/4，而不是 −18。有限取樣不能代替逐係數核對（MKO 假設 a3）。",
        "en": "Part (2) needs the premise F′=f. The candidate F₃(x)=1/2+x−(3/2)x²+2x³+(1/4)x⁴ satisfies F₃′=f at x=−1, 0, 1, but F₃′−f=x³−x equals 6 at x=2; using it gives the endpoint difference F₃(−1)−F₃(2)=−81/4 instead of −18. Finite sampling cannot replace the coefficientwise check (MKO assumption a3)."
      },
      {
        "title": "連續性定義直接用一次，積分規則也依賴此假設 / Continuity is invoked directly once and also supports the integral rules",
        "zh": "點連續的定義只在 S05 直接展開，以控制 f(t)−f(x_0)；G1 的連續性亦供所引用的積分性質與 S07 使用。若 f 只可積不連續，例如 f(t)=0（t<0）、f(t)=1（t≥0），a=0，則 G(x)=max(x,0)，在 0 的右差商為 1、左差商為 0，G′(0) 不存在。",
        "en": "The definition of pointwise continuity is used directly only in S05 to control f(t)−f(x_0); G1's continuity hypothesis also supports the cited integral properties and S07. If f is merely integrable, for example f(t)=0 for t<0 and f(t)=1 for t≥0 with a=0, then G(x)=max(x,0), whose right quotient at 0 is 1 and left quotient is 0, so G′(0) does not exist."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "第一部分由三條引用的積分性質（int_additive、int_linear、int_bound）推出；這些性質對一般連續函數未在 OCME 中物化，本稿也未證明。被積函數連續時積分存在，同樣是前提。",
        "en": "Part (1) is derived from three cited integral properties (int_additive, int_linear, int_bound); for general continuous functions they are not materialized in OCME and are not proved here. Existence of the integral for a continuous integrand is likewise a premise."
      },
      {
        "zh-Hant": "第二部分引用均值定理推論 thm_zero_deriv，本稿未證明。",
        "en": "Part (2) cites the Mean Value Theorem corollary thm_zero_deriv, which is not proved here."
      },
      {
        "zh-Hant": "Spivak 的章節位置依記憶標示，尚未對原書核對，章節或定理編號可能需要修正。",
        "en": "The Spivak chapter locations are given from memory and have not been checked against the book; chapter or theorem numbers may need correction."
      },
      {
        "zh-Hant": "例題數字只對這個 f 與 F 有效，不能取代一般證明。N11 只是 x=−1、0、1 三點的有限核對，用來顯示取樣不足。",
        "en": "The example's numbers hold for this f and F only and cannot replace the general proof. N11 is only a finite check at x=−1, 0, 1, used to show that sampling is insufficient."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "f34b050fd0264c3575761dfd055ea69e5a8e7f1e68756823ca33ed236d0e4f17",
    "original_source_sha256": "ddfe7d1b1540889b94b73bb5fa4bb929138d1e74fc22af8ec30063b704076861",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M029"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}