{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-continuity-01",
  "mko_id": "mko-continuity",
  "title": {
    "zh-Hant": "連續：分支函數的中心值與可去商的唯一補點",
    "en": "Continuity: centre values of branch functions and the unique continuous extension across a removable singularity"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-continuity-01",
    "source": {
      "mko_id": "mko-continuity",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "500bf8bf13b82d10311338ec1e1ec96d1b7b1d3fcff6113697be491180bdb9e6",
      "english_sha256": "3d9b3ecf4edf7bc99e0a106ce8d24abe7e3fe8a8423ef9ab92a33510611f8d61"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "abs_rules",
      "order_basics",
      "distributive",
      "frac_rules",
      "continuity_def",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "(1) 對 f_V(x)=2x+1（x<1）、f_V(1)=V、f_V(x)=5x−2（x>1），證明 V=3 時 f_V 在 1 連續，V=4 時不連續；(2) 對 g(x)=2x+1（x<1）、g(1)=3、g(x)=5x−1（x>1），證明 g 在 1 不連續；(3) 對 q(x)=(x²−4)/(x−2)（論域 ℝ∖{2}），證明全實數連續延拓在 2 的值必為 4，且 4 確實可行。",
      "en": "(1) For f_V(x)=2x+1 (x<1), f_V(1)=V, f_V(x)=5x−2 (x>1), prove that f_V is continuous at 1 when V=3 and not when V=4; (2) for g(x)=2x+1 (x<1), g(1)=3, g(x)=5x−1 (x>1), prove that g is not continuous at 1; (3) for q(x)=(x²−4)/(x−2) on ℝ∖{2}, prove that a continuous extension to all real numbers must take the value 4 at 2, and that 4 does work."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "函數 f:ℝ→ℝ 定義在整個實數上，中心點值明示。",
          "en": "Functions f:ℝ→ℝ are defined on all reals, with the centre value stated explicitly."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "f_V(x)=2x+1（x<1），f_V(1)=V，f_V(x)=5x−2（x>1）；比較 V=3 與 V=4。",
          "en": "f_V(x)=2x+1 (x<1), f_V(1)=V, f_V(x)=5x−2 (x>1); compare V=3 and V=4."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "g(x)=2x+1（x<1），g(1)=3，g(x)=5x−1（x>1）。",
          "en": "g(x)=2x+1 (x<1), g(1)=3, g(x)=5x−1 (x>1)."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "q(x)=(x²−4)/(x−2)，論域 ℝ∖{2}；延拓 F_W(x)=q(x)（x≠2）、F_W(2)=W。",
          "en": "q(x)=(x²−4)/(x−2) on ℝ∖{2}; extension F_W(x)=q(x) for x≠2 and F_W(2)=W."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "continuity_def"
        ],
        "before_tex": "f\\ \\text{continuous at}\\ a",
        "after_tex": "\\forall\\varepsilon>0\\ \\exists\\delta>0\\ \\forall x\\in\\mathbb{R}:\\ |x-a|<\\delta\\Rightarrow|f(x)-f(a)|<\\varepsilon",
        "relation": "definition",
        "rule": "mko_continuity",
        "conditions": [
          "f(a)\\ \\text{defined}"
        ],
        "explanation": {
          "zh-Hant": "依 MKO 展開在 a 的連續性。與去心極限不同，條件 |x−a|<δ 包含 x=a，而且比較對象是已定義的點值 f(a)；所以中心值本身是判定的一部分。",
          "en": "Unfold continuity at a by the MKO. Unlike the punctured limit, the condition |x−a|<δ includes x=a, and the comparison is with the defined value f(a); so the centre value itself is part of the decision."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_continuity",
            "continuity_def"
          ],
          "explanation": {
            "zh-Hant": "此步只展開定義。",
            "en": "This step only unfolds the definition."
          }
        },
        "leaf_basis": "continuity_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G2",
          "function_def"
        ],
        "before_tex": "f_V\\ \\text{by three cases}",
        "after_tex": "f_V:\\mathbb{R}\\to\\mathbb{R}\\ \\text{total},\\ f_V(1)=V",
        "relation": "definition",
        "rule": "function_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "三個情況 x<1、x=1、x>1 互斥且涵蓋全部實數，每個 x 恰得一個值，所以 f_V 是 ℝ 上的函數；中心值 V 是獨立給定的資料，不由分支公式決定。",
          "en": "The three cases x<1, x=1, x>1 are exclusive and cover all reals, so each x gets exactly one value and f_V is a function on ℝ; the centre value V is separate data, not determined by the branch formulas."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "function_def",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "用到實數三分律保證三個情況恰有其一。",
            "en": "Trichotomy of the reals guarantees exactly one case applies."
          }
        },
        "leaf_basis": "function_def + order_basics"
      },
      {
        "id": "S03",
        "uses": [
          "G2"
        ],
        "before_tex": "\\text{left branch at }1,\\ \\text{right branch at }1",
        "after_tex": "L^-=2\\cdot 1+1=3,\\ L^+=5\\cdot 1-2=3",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 假設 a2，仿射分支的單側極限是分支公式在 a 的值：左 2·1+1=3，右 5·1−2=3（N1、N2）。兩者都與 V 無關。",
          "en": "By MKO assumption a2, the one-sided limit of an affine branch is the branch formula evaluated at a: left 2·1+1=3, right 5·1−2=3 (N1, N2). Neither depends on V."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確求值，兩邊是同一個數。",
            "en": "Exact evaluation; both sides name the same number."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "mko_continuity"
        ],
        "before_tex": "L^-=L^+=3",
        "after_tex": "f_V\\ \\text{continuous at}\\ 1\\iff V=3",
        "relation": "theorem_application",
        "rule": "mko_continuity",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 的模型判準：連續當且僅當左極限=右極限=V。左右都是 3，所以條件化為 V=3。等價地，點缺陷對 (3−V, 3−V) 必須是 (0,0)。S05 至 S08 以 ε-δ 直接核對兩個方向的具體情況，不只依賴這個引用的判準。",
          "en": "By the MKO's model criterion: continuous exactly when left limit = right limit = V. Both are 3, so the condition becomes V=3. Equivalently, the point-defect pair (3−V, 3−V) must be (0,0). S05 to S08 check both concrete cases directly by ε-δ, rather than relying only on this cited criterion."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "mko_continuity"
          ],
          "explanation": {
            "zh-Hant": "判準引自 MKO；本稿在 S06 與 S08 對 V=3、V=4 各自獨立驗證。",
            "en": "The criterion is cited from the MKO; this guide independently verifies V=3 and V=4 in S06 and S08."
          }
        },
        "leaf_basis": "mko_continuity"
      },
      {
        "id": "S05",
        "uses": [
          "G2",
          "S02"
        ],
        "before_tex": "V=3",
        "after_tex": "\\forall x\\in\\mathbb{R}:\\ |f_3(x)-3|\\le 5|x-1|",
        "relation": "implies",
        "rule": "L_branch_bound",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_branch_bound，三個情況合起來，f_3 與 3 的距離不超過 5|x−1|。引理在 S05.1 至 S05.3 分情況證明。",
          "en": "By Lemma L_branch_bound, combining the three cases, the distance from f_3(x) to 3 is at most 5|x−1|. The lemma is proved by cases in S05.1 to S05.3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_branch_bound",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "三個情況互斥且涵蓋全部 x（S02）。",
            "en": "The three cases are exclusive and cover every x (S02)."
          }
        },
        "leaf_basis": "case_analysis + abs_rules + order_basics"
      },
      {
        "id": "S05.1",
        "uses": [
          "G2",
          "abs_rules"
        ],
        "before_tex": "x<1:\\ |f_3(x)-3|",
        "after_tex": "|2x-2|=2|x-1|\\le 5|x-1|",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [
          "x<1"
        ],
        "explanation": {
          "zh-Hant": "x<1 時 f_3(x)=2x+1，(2x+1)−3=2(x−1)（N3），絕對值 2|x−1|；又 |x−1|≥0，所以 2|x−1|≤5|x−1|。",
          "en": "For x<1, f_3(x)=2x+1 and (2x+1)−3=2(x−1) (N3), with absolute value 2|x−1|; as |x−1|≥0, 2|x−1|≤5|x−1|."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "只宣稱不超過；2 與 5 的比較需要 |x−1|≥0。",
            "en": "Only an upper bound is claimed; comparing 2 with 5 needs |x−1|≥0."
          }
        },
        "leaf_basis": "abs_rules + distributive + order_basics"
      },
      {
        "id": "S05.2",
        "uses": [
          "G2",
          "abs_rules"
        ],
        "before_tex": "x>1:\\ |f_3(x)-3|",
        "after_tex": "|5x-5|=5|x-1|",
        "relation": "implies",
        "rule": "abs_rules",
        "conditions": [
          "x>1"
        ],
        "explanation": {
          "zh-Hant": "x>1 時 f_3(x)=5x−2，(5x−2)−3=5(x−1)（N4），絕對值 5|x−1|。",
          "en": "For x>1, f_3(x)=5x−2 and (5x−2)−3=5(x−1) (N4), with absolute value 5|x−1|."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "abs_rules",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "等式自然也是 ≤ 的情形。",
            "en": "An equality is in particular a ≤ bound."
          }
        },
        "leaf_basis": "abs_rules + distributive"
      },
      {
        "id": "S05.3",
        "uses": [
          "G2"
        ],
        "before_tex": "x=1:\\ |f_3(1)-3|",
        "after_tex": "|3-3|=0=5|1-1|",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [
          "x=1"
        ],
        "explanation": {
          "zh-Hant": "x=1 時 f_3(1)=V=3，差為 0，等於 5|1−1|=0。中心點也滿足界；這一情況只有在 V=3 時成立。三個情況合起來證得引理 L_branch_bound。",
          "en": "At x=1, f_3(1)=V=3 and the difference is 0, equal to 5|1−1|=0. The centre also satisfies the bound; this case holds only because V=3. The three cases together prove Lemma L_branch_bound."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "若 V=4，此情況給出 1≤0，不成立，所以引理只對 V=3 宣稱。",
            "en": "With V=4 this case would give 1≤0, which fails; so the lemma is claimed for V=3 only."
          }
        },
        "leaf_basis": "numeric_evaluation + abs_rules"
      },
      {
        "id": "S06",
        "uses": [
          "S01",
          "S05"
        ],
        "before_tex": "\\varepsilon>0\\ \\text{arbitrary},\\ \\delta=\\tfrac{\\varepsilon}{5}",
        "after_tex": "|x-1|<\\delta\\Rightarrow|f_3(x)-f_3(1)|\\le 5|x-1|<\\varepsilon;\\ f_3\\ \\text{continuous at}\\ 1",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [
          "\\varepsilon>0"
        ],
        "explanation": {
          "zh-Hant": "取任意 ε>0，令 δ=ε/5>0。若 |x−1|<δ，由 S05 與 f_3(1)=3，|f_3(x)−f_3(1)|≤5|x−1|<5δ=ε。ε 任意，依 S01，f_3 在 1 連續，與 S04 的判準一致。例如 ε=1/10 時 δ=1/50（N11）。",
          "en": "Take an arbitrary ε>0 and set δ=ε/5>0. If |x−1|<δ, then by S05 and f_3(1)=3, |f_3(x)−f_3(1)|≤5|x−1|<5δ=ε. As ε is arbitrary, S01 gives continuity of f_3 at 1, agreeing with the criterion of S04. For example ε=1/10 gives δ=1/50 (N11)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S01",
            "S05",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "目標 G：f_3 在 1 連續。足夠條件 H：對任意 ε，δ=ε/5 有效；H⇒G 由定義。H 由 S05 正向證成，未把 G 當作前提。",
            "en": "Goal G: f_3 is continuous at 1. Sufficient condition H: for arbitrary ε, δ=ε/5 works; H⇒G by the definition. H is proved forwards from S05, without assuming G."
          }
        },
        "leaf_basis": "forall_intro + exists_intro + order_basics + continuity_def"
      },
      {
        "id": "S07",
        "uses": [
          "S03",
          "S04",
          "G2"
        ],
        "before_tex": "V=4",
        "after_tex": "(L^- - V,\\ L^+ - V)=(-1,-1)\\neq(0,0)",
        "relation": "implies",
        "rule": "mko_continuity",
        "conditions": [],
        "explanation": {
          "zh-Hant": "V=4 時點缺陷對為 (3−4, 3−4)=(−1,−1)（N5），不是 (0,0)。左右極限相等（都是 3），卻不等於中心值，依 S04 的判準不連續。S08 以 ε-δ 直接證明。",
          "en": "With V=4 the point-defect pair is (3−4, 3−4)=(−1,−1) (N5), not (0,0). The one-sided limits agree (both 3) but differ from the centre value, so by the criterion of S04 the function is not continuous. S08 proves this directly by ε-δ."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S04",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "缺陷對只是判準的讀法；獨立的反證在 S08。",
            "en": "The defect pair is only a reading of the criterion; the independent refutation is S08."
          }
        },
        "leaf_basis": "mko_continuity + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S01",
          "G2"
        ],
        "before_tex": "f_4\\ \\text{continuous at}\\ 1\\ \\text{(assumed)},\\ \\varepsilon=\\tfrac12,\\ x=1-\\tfrac{\\delta}{2}",
        "after_tex": "|f_4(x)-4|=|(3-\\delta)-4|=1+\\delta>\\tfrac12:\\ \\text{contradiction}",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [
          "\\delta>0"
        ],
        "explanation": {
          "zh-Hant": "反證：假設 f_4 在 1 連續。對 ε=1/2 取得某個 δ>0。令 x=1−δ/2，則 |x−1|=δ/2<δ 且 x<1，所以 f_4(x)=2(1−δ/2)+1=3−δ，|f_4(x)−f_4(1)|=|(3−δ)−4|=|−1−δ|=1+δ>1/2（N6）。這與連續性給出的 <1/2 矛盾，所以 f_4 在 1 不連續。",
          "en": "By contradiction: suppose f_4 is continuous at 1. For ε=1/2 obtain some δ>0. Let x=1−δ/2; then |x−1|=δ/2<δ and x<1, so f_4(x)=2(1−δ/2)+1=3−δ and |f_4(x)−f_4(1)|=|(3−δ)−4|=|−1−δ|=1+δ>1/2 (N6). This contradicts the <1/2 given by continuity, so f_4 is not continuous at 1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "exists_elim",
            "contradiction",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "δ 是由假設取出的任意正數；論證對每個可能的 δ 都成立，所以結論不依賴 δ。",
            "en": "δ is whatever positive number the assumption supplies; the argument works for every possible δ, so the conclusion does not depend on δ."
          }
        },
        "leaf_basis": "contradiction + exists_elim + abs_rules + order_basics"
      },
      {
        "id": "S09",
        "uses": [
          "G3",
          "mko_continuity"
        ],
        "before_tex": "g:\\ L^-=2\\cdot1+1=3,\\ L^+=5\\cdot1-1=4,\\ V=3",
        "after_tex": "(L^- - V,\\ L^+ - V)=(0,1)\\neq(0,0)",
        "relation": "implies",
        "rule": "mko_continuity",
        "conditions": [],
        "explanation": {
          "zh-Hant": "g 的左極限 3、右極限 4（N7），中心值 3。缺陷對 (0,1)：中心值只等於左側，右側不同，依判準不連續。",
          "en": "g has left limit 3 and right limit 4 (N7), with centre value 3. The defect pair is (0,1): the centre value matches only the left side, the right differs, so by the criterion g is not continuous."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "mko_continuity",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "缺陷對的讀法；獨立反證在 S10。",
            "en": "A reading of the defect pair; the independent refutation is S10."
          }
        },
        "leaf_basis": "mko_continuity + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "S01",
          "G3"
        ],
        "before_tex": "g\\ \\text{continuous at}\\ 1\\ \\text{(assumed)},\\ \\varepsilon=\\tfrac12,\\ x=1+\\tfrac{\\delta}{2}",
        "after_tex": "|g(x)-3|=1+\\tfrac{5\\delta}{2}>\\tfrac12:\\ \\text{contradiction}",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [
          "\\delta>0"
        ],
        "explanation": {
          "zh-Hant": "假設 g 在 1 連續，對 ε=1/2 取得 δ>0。令 x=1+δ/2（x>1、|x−1|<δ），g(x)=5(1+δ/2)−1=4+5δ/2，|g(x)−g(1)|=1+5δ/2>1/2（N9），矛盾。所以 g 在 1 不連續，雖然 g(1) 等於左極限。",
          "en": "Suppose g is continuous at 1 and obtain δ>0 for ε=1/2. Let x=1+δ/2 (x>1, |x−1|<δ); g(x)=5(1+δ/2)−1=4+5δ/2 and |g(x)−g(1)|=1+5δ/2>1/2 (N9), a contradiction. So g is not continuous at 1, although g(1) equals the left limit."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "exists_elim",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "與 S08 同理；只需右側一點。",
            "en": "As in S08; one point on the right suffices."
          }
        },
        "leaf_basis": "contradiction + exists_elim + abs_rules + order_basics"
      },
      {
        "id": "S11",
        "uses": [
          "G4",
          "distributive",
          "frac_rules"
        ],
        "before_tex": "q(x)=\\frac{x^2-4}{x-2},\\ x\\neq 2",
        "after_tex": "q(x)=x+2\\ (x\\neq 2);\\ 2\\notin\\mathrm{dom}(q)",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "x-2\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "x²−4=(x−2)(x+2)（N10），x≠2 時約去 x−2≠0，得 q(x)=x+2。q 的論域是 ℝ∖{2}，它不是 ℝ 上的函數，所以在本模型中「q 在 2 連續」沒有意義；有意義的問題是：怎樣的全實數延拓 F_W 在 2 連續。",
          "en": "x²−4=(x−2)(x+2) (N10); for x≠2 cancel x−2≠0 to get q(x)=x+2. The domain of q is ℝ∖{2}; q is not a function on ℝ, so in this model 'q is continuous at 2' has no meaning. The meaningful question is which extension F_W to all real numbers is continuous at 2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "在論域 x≠2 上雙向成立；不延伸到 x=2。",
            "en": "Valid both ways on the domain x≠2; it does not extend to x=2."
          }
        },
        "leaf_basis": "frac_rules + distributive + mul_inverse_nonzero"
      },
      {
        "id": "S12",
        "uses": [
          "S01",
          "S11",
          "G4"
        ],
        "before_tex": "W=4:\\ |F_4(x)-4|",
        "after_tex": "|x-2|\\ \\text{for every}\\ x;\\ \\delta=\\varepsilon\\ \\Rightarrow\\ F_4\\ \\text{continuous at}\\ 2",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "存在性：W=4 時，x≠2 有 F_4(x)−4=(x+2)−4=x−2，x=2 有 F_4(2)−4=0=2−2；所以對所有 x，|F_4(x)−F_4(2)|=|x−2|。對任意 ε>0 取 δ=ε 即得連續。",
          "en": "Existence: with W=4, for x≠2, F_4(x)−4=(x+2)−4=x−2, and at x=2, F_4(2)−4=0=2−2; so |F_4(x)−F_4(2)|=|x−2| for all x. Taking δ=ε for arbitrary ε>0 gives continuity."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S11",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "兩種 x 都檢查了；ε 任意。",
            "en": "Both kinds of x are checked; ε is arbitrary."
          }
        },
        "leaf_basis": "forall_intro + exists_intro + continuity_def + case_analysis"
      },
      {
        "id": "S13",
        "uses": [
          "S11",
          "S12"
        ],
        "before_tex": "F_W\\ \\text{continuous at}\\ 2",
        "after_tex": "W=4",
        "relation": "implies",
        "rule": "L_unique_ext",
        "conditions": [],
        "explanation": {
          "zh-Hant": "唯一性：依引理 L_unique_ext，任何在 2 連續的延拓都必須取 W=4。與 MKO 的「唯一所需值為 2a」一致（a=2）。引理在 S13.1 至 S13.3 以反證證明。",
          "en": "Uniqueness: by Lemma L_unique_ext, every extension continuous at 2 must take W=4, agreeing with the MKO's 'the unique value is 2a' (a=2). The lemma is proved by contradiction in S13.1 to S13.3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_unique_ext"
          ],
          "explanation": {
            "zh-Hant": "引理只用 x≠2 的值 x+2 與連續性定義，不引用極限唯一性的一般定理。",
            "en": "The lemma uses only the values x+2 at x≠2 and the definition of continuity, not a general theorem on uniqueness of limits."
          }
        },
        "leaf_basis": "contradiction + abs_rules + order_basics"
      },
      {
        "id": "S13.1",
        "uses": [
          "continuity_def",
          "exists_elim"
        ],
        "before_tex": "W\\neq 4\\ \\text{(assumed)},\\ F_W\\ \\text{continuous at}\\ 2",
        "after_tex": "\\varepsilon=\\tfrac{|W-4|}{2}>0,\\ \\delta>0:\\ |x-2|<\\delta\\Rightarrow|F_W(x)-W|<\\varepsilon",
        "relation": "implies",
        "rule": "exists_elim",
        "conditions": [
          "W\\neq 4"
        ],
        "explanation": {
          "zh-Hant": "假設 W≠4 且 F_W 在 2 連續。|W−4|>0，所以 ε=|W−4|/2>0；依連續性定義取得某個 δ>0。",
          "en": "Suppose W≠4 and F_W is continuous at 2. Then |W−4|>0, so ε=|W−4|/2>0; the definition of continuity supplies some δ>0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "continuity_def",
            "exists_elim",
            "abs_rules"
          ],
          "explanation": {
            "zh-Hant": "需要 W≠4 才能使 ε>0。",
            "en": "W≠4 is needed to make ε>0."
          }
        },
        "leaf_basis": "exists_elim + continuity_def + abs_rules"
      },
      {
        "id": "S13.2",
        "uses": [
          "S13.1",
          "S11"
        ],
        "before_tex": "r=\\tfrac12\\min(\\delta,\\varepsilon),\\ x=2+r",
        "after_tex": "|F_W(x)-W|<\\varepsilon,\\ |F_W(x)-4|=r<\\varepsilon",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [
          "r>0"
        ],
        "explanation": {
          "zh-Hant": "令 r=min(δ,ε)/2>0，x=2+r。則 |x−2|=r<δ，由 S13.1，|F_W(x)−W|<ε；又 x≠2，F_W(x)=x+2=4+r，所以 |F_W(x)−4|=r<ε。",
          "en": "Let r=min(δ,ε)/2>0 and x=2+r. Then |x−2|=r<δ, so by S13.1, |F_W(x)−W|<ε; also x≠2, so F_W(x)=x+2=4+r and |F_W(x)−4|=r<ε."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13.1",
            "S11",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "取 x≠2 是為了使用 q 的公式；r 同時小於 δ 與 ε。",
            "en": "x≠2 is chosen so that the formula of q applies; r is below both δ and ε."
          }
        },
        "leaf_basis": "forall_elim + order_basics + eq_replace"
      },
      {
        "id": "S13.3",
        "uses": [
          "S13.2",
          "abs_rules"
        ],
        "before_tex": "|W-4|\\le|W-F_W(x)|+|F_W(x)-4|",
        "after_tex": "|W-4|<2\\varepsilon=|W-4|:\\ \\text{contradiction, so}\\ W=4",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "三角不等式：|W−4|≤|W−F_W(x)|+|F_W(x)−4|<ε+ε=2ε=|W−4|，得 |W−4|<|W−4|，矛盾（例如 W=5 時 2·(1/2)=1=|5−4|，N12）。所以假設 W≠4 不成立，W=4。S13.1 至 S13.3 證得引理 L_unique_ext。",
          "en": "Triangle inequality: |W−4|≤|W−F_W(x)|+|F_W(x)−4|<ε+ε=2ε=|W−4|, giving |W−4|<|W−4|, a contradiction (for example, W=5 gives 2·(1/2)=1=|5−4|, N12). So the assumption W≠4 fails and W=4. S13.1 to S13.3 prove Lemma L_unique_ext."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13.2",
            "abs_rules",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "矛盾只否定 W≠4；存在性另由 S12 證明。",
            "en": "The contradiction refutes only W≠4; existence is proved separately in S12."
          }
        },
        "leaf_basis": "contradiction + abs_rules + order_basics"
      },
      {
        "id": "S14",
        "uses": [
          "S06",
          "S08",
          "S10",
          "S12",
          "S13"
        ],
        "before_tex": "\\text{S06, S08, S10, S12, S13}",
        "after_tex": "f_3\\ \\text{cont.},\\ f_4,\\ g\\ \\text{not};\\ F_W\\ \\text{cont. at 2}\\iff W=4",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "匯總：V=3 連續、V=4 不連續、g 不連續；可去商的延拓在 2 連續當且僅當 W=4（S12 給出 ⇐，S13 給出 ⇒）。原商 q 仍只在 ℝ∖{2} 上有定義。",
          "en": "Summary: V=3 continuous, V=4 not, g not; the extension of the removable quotient is continuous at 2 exactly when W=4 (S12 gives ⇐, S13 gives ⇒). The original quotient q is still defined only on ℝ∖{2}."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S06",
            "S08",
            "S10",
            "S12",
            "S13"
          ],
          "explanation": {
            "zh-Hant": "目標 G：本稿的四個結論。每一部分已正向證成；W=4 的雙向分別來自 S12 與 S13。",
            "en": "Goal G: the four conclusions of this guide. Each part is proved forwards; the two directions for W=4 come from S12 and S13."
          }
        },
        "leaf_basis": "and_intro_elim + iff_intro"
      }
    ],
    "result": {
      "zh-Hant": "f_V 的左右極限都是 3：V=3 時在 1 連續（δ=ε/5），V=4 時不連續（缺陷對 (−1,−1)）。g 的左右極限 3、4 不等，中心值 3 只配合左側，所以不連續（缺陷對 (0,1)）。q=(x²−4)/(x−2) 只在 ℝ∖{2} 有定義；全實數延拓在 2 連續當且僅當補值 W=4。",
      "en": "f_V has both one-sided limits equal to 3: with V=3 it is continuous at 1 (δ=ε/5); with V=4 it is not (defect pair (−1,−1)). g has one-sided limits 3 and 4, which differ, and its centre value 3 matches only the left side, so it is not continuous (defect pair (0,1)). q=(x²−4)/(x−2) is defined only on ℝ∖{2}; an extension to all real numbers is continuous at 2 exactly when the assigned value is W=4."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "f_4 的左右極限都是 3，所以 f_4 在 1 連續",
          "en": "Both one-sided limits of f_4 are 3, so f_4 is continuous at 1"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "f_4(1)=4≠3；x=1−δ/2 給出誤差 1+δ>1/2（S08）。",
          "en": "f_4(1)=4≠3; x=1−δ/2 gives error 1+δ>1/2 (S08)."
        }
      },
      {
        "claim": {
          "zh-Hant": "g(1)=3 等於左極限，所以 g 在 1 連續",
          "en": "g(1)=3 equals the left limit, so g is continuous at 1"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "右極限為 4；x=1+δ/2 給出誤差 1+5δ/2>1/2（S10）。",
          "en": "The right limit is 4; x=1+δ/2 gives error 1+5δ/2>1/2 (S10)."
        }
      },
      {
        "claim": {
          "zh-Hant": "q(x)=(x²−4)/(x−2) 在 2 的極限是 4，所以 q 在 2 連續",
          "en": "q(x)=(x²−4)/(x−2) has limit 4 at 2, so q is continuous at 2"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "q 在 2 無定義；只有補值 W=4 的延拓 F_4 在 2 連續。",
          "en": "q is undefined at 2; only the extension F_4 with W=4 is continuous at 2."
        }
      },
      {
        "claim": {
          "zh-Hant": "可去奇點補任何值都能得到連續延拓，例如 F(2)=0",
          "en": "Assigning any value at a removable singularity gives a continuous extension, for example F(2)=0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "S13 證明必須 W=4；W=0 時取 ε=2 即可否定連續。",
          "en": "S13 proves W=4 is forced; with W=0, ε=2 already refutes continuity."
        }
      },
      {
        "claim": {
          "zh-Hant": "檢查 f_3(0.9) 與 f_3(1.1) 都接近 3，就證明了 f_3 在 1 連續",
          "en": "Checking that f_3(0.9) and f_3(1.1) are close to 3 proves f_3 is continuous at 1"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "有限測點不是證明；證明是 S06 對任意 ε 的論證。",
          "en": "Finitely many test points are not a proof; the proof is S06's argument for arbitrary ε."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "abs_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "絕對值（實數）：|u|=u（u≥0）、|u|=−u（u<0）；|u|≥0；|uv|=|u||v|；三角不等式 |u+v|≤|u|+|v|；|u|<r ⇔ −r<u<r。",
          "en": "Absolute value (reals): |u|=u if u≥0 and |u|=−u if u<0; |u|≥0; |uv|=|u||v|; triangle inequality |u+v|≤|u|+|v|; |u|<r ⇔ −r<u<r."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "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": "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": "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": "function_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "函數：f:A→B 對每個 a∈A 指定恰好一個 f(a)∈B。兩函數相等 ⇔ 定義域、對應域相同且對每個 a 都有相同值。",
          "en": "Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a."
        }
      },
      {
        "id": "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": "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": "exists_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。",
          "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)."
        }
      },
      {
        "id": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "exists_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃消去：由 ∃x∈D, P(x)，可取一個新名字 a 表示某個滿足 P(a) 的元素；只要最後結論不含 a，就可用 a 推理。",
          "en": "∃-elimination: from ∃x∈D, P(x) we may name some element a with P(a) by a fresh name; reasoning with a is allowed as long as the final conclusion does not mention a."
        }
      },
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "forall_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。",
          "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds."
        }
      },
      {
        "id": "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": "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": "iff_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "雙條件：P⇒Q 與 Q⇒P 都成立時，P⇔Q 成立；反之亦然。",
          "en": "Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold."
        }
      },
      {
        "id": "mko_continuity",
        "kind": "cited",
        "source": {
          "mko_id": "mko-continuity",
          "sha256": "500bf8bf13b82d10311338ec1e1ec96d1b7b1d3fcff6113697be491180bdb9e6",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對 f:ℝ→ℝ，f 在 a 連續指每個 ε>0 都有 δ>0，使 |x−a|<δ 時 |f(x)−f(a)|<ε；等價地去心兩側極限等於已定義點值 f(a)。兩側仿射、中心明示值 V 的模型，在 a 連續當且僅當左極限=右極限=V。可去商的全 ℝ 連續延拓唯一所需中心值為 2a。",
          "en": "For f:ℝ→ℝ, continuity at a means every ε>0 admits δ>0 with |f(x)−f(a)|<ε whenever |x−a|<δ; equivalently the punctured two-sided limit equals the defined f(a). In the two-affine-branch model with explicit centre value V, continuity at a holds exactly when left limit = right limit = V. A continuous extension to all real numbers of the removable quotient needs the unique centre value 2a."
        }
      },
      {
        "id": "L_branch_bound",
        "kind": "lemma",
        "proved_in": [
          "S05.1",
          "S05.2",
          "S05.3"
        ],
        "statement": {
          "zh-Hant": "對所有實數 x：|f_3(x)−3|≤5|x−1|。",
          "en": "For every real x: |f_3(x)−3|≤5|x−1|."
        }
      },
      {
        "id": "L_unique_ext",
        "kind": "lemma",
        "proved_in": [
          "S13.1",
          "S13.2",
          "S13.3"
        ],
        "statement": {
          "zh-Hant": "若 F:ℝ→ℝ 在 x≠2 處 F(x)=x+2，且 F 在 2 連續，則 F(2)=4。",
          "en": "If F:ℝ→ℝ has F(x)=x+2 for x≠2 and F is continuous at 2, then F(2)=4."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "2*1+1",
        "rhs": "3"
      },
      {
        "id": "N2",
        "step": "S03",
        "lhs": "5*1-2",
        "rhs": "3"
      },
      {
        "id": "N3",
        "step": "S05.1",
        "vars": [
          "x"
        ],
        "degree": 1,
        "lhs": "(2*x+1)-3",
        "rhs": "2*(x-1)"
      },
      {
        "id": "N4",
        "step": "S05.2",
        "vars": [
          "x"
        ],
        "degree": 1,
        "lhs": "(5*x-2)-3",
        "rhs": "5*(x-1)"
      },
      {
        "id": "N5",
        "step": "S07",
        "lhs": "3-4",
        "rhs": "-1"
      },
      {
        "id": "N6",
        "step": "S08",
        "vars": [
          "d"
        ],
        "degree": 1,
        "lhs": "(2*(1-d/2)+1)-4",
        "rhs": "-1-d"
      },
      {
        "id": "N7",
        "step": "S09",
        "lhs": "5*1-1",
        "rhs": "4"
      },
      {
        "id": "N8",
        "step": "S09",
        "lhs": "4-3",
        "rhs": "1"
      },
      {
        "id": "N9",
        "step": "S10",
        "vars": [
          "d"
        ],
        "degree": 1,
        "lhs": "(5*(1+d/2)-1)-3",
        "rhs": "1+5*d/2"
      },
      {
        "id": "N10",
        "step": "S11",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "(x-2)*(x+2)",
        "rhs": "x**2-4"
      },
      {
        "id": "N11",
        "step": "S06",
        "lhs": "5*(1/50)",
        "rhs": "1/10"
      },
      {
        "id": "N12",
        "step": "S13.3",
        "lhs": "2*(abs(5-4)/2)",
        "rhs": "abs(5-4)"
      },
      {
        "id": "N13",
        "step": "S08",
        "lhs": "abs(2*(1-1/20)+1-4)",
        "rhs": "1+1/10"
      },
      {
        "id": "N14",
        "lhs": "abs(0-4)/2",
        "rhs": "2"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理 MKO 的兩側仿射分支模型（中心值明示）與可去商的補點延拓，全部以 ε-δ 直接論證，並與 MKO 的「左極限=右極限=V」判準對照。不證明一般的連續性代數或中間值定理。不改動 canonical MKO 或英文 edition。",
      "en": "Treats only the MKO's two-affine-branch model with an explicit centre value and extension of a quotient across a removable singularity, all argued directly by ε-δ and compared with the MKO's criterion 'left limit = right limit = V'. General continuity algebra and the intermediate value theorem are not proved. It does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "f_V",
        "中心值為 V 的兩側仿射分支函數",
        "the two-affine-branch function with centre value V"
      ],
      [
        "L^-,\\ L^+",
        "左、右極限",
        "the left and right limits"
      ],
      [
        "(L^- - V,\\ L^+ - V)",
        "點缺陷對",
        "the point-defect pair"
      ],
      [
        "q",
        "可去商，論域 ℝ∖{2}",
        "the removable quotient on ℝ∖{2}"
      ],
      [
        "F_W",
        "q 的全實數延拓，F_W(2)=W",
        "the extension of q to all real numbers with F_W(2)=W"
      ]
    ],
    "md_error_controls": [
      {
        "title": "左右極限相等仍不足 / Equal one-sided limits are not enough",
        "zh": "MKO 的誤解之一是左右極限相等就忽略點值。f_4 的左右極限都是 3，但 f_4(1)=4，S08 以 ε=1/2 直接證明它不連續。連續要求的是「兩側極限 = 中心值」，中心值是判定的一部分。",
        "en": "One misconception the MKO lists is ignoring the point value once the one-sided limits agree. Both one-sided limits of f_4 are 3, but f_4(1)=4, and S08 proves directly with ε=1/2 that it is not continuous. Continuity requires 'two-sided limit = centre value'; the centre value is part of the decision."
      },
      {
        "title": "中心值只配合一側也不夠 / A centre value matching one side is not enough",
        "zh": "g(1)=3 等於左極限，但右極限是 4（S09、S10），所以不連續。點缺陷對 (L⁻−V, L⁺−V) 兩個分量都必須為 0。",
        "en": "g(1)=3 equals the left limit, but the right limit is 4 (S09, S10), so g is not continuous. Both components of the defect pair (L⁻−V, L⁺−V) must be 0."
      },
      {
        "title": "可去不等於不需補定 / Removable does not mean no filling is needed",
        "zh": "另一個誤解是未補定便聲稱原商在中心連續。q 在 2 沒有定義（S11），談不上在 2 連續；要得到全實數連續函數必須補值，而且只有 W=4 可行（S12、S13）。補了其他值（例如 W=0）的延拓在 2 不連續。",
        "en": "Another misconception is claiming the original quotient is continuous at the centre without filling it. q is undefined at 2 (S11), so continuity at 2 does not even apply; a continuous function on all real numbers requires a value at the missing point, and only W=4 works (S12, S13). An extension filled with another value, for example W=0, is not continuous at 2."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "MKO 的判準「左極限=右極限=V ⇔ 連續」在本稿中被引用，並在兩個具體情況下以 ε-δ 獨立核對；一般等價的證明不在本稿範圍。",
        "en": "The MKO's criterion 'left limit = right limit = V ⇔ continuous' is cited here and independently checked by ε-δ in the two concrete cases; the general equivalence is not proved in this guide."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "47982f7ca8622e26e45615f6dc3a8550fa228ca562132006c98685246f8606e4",
    "original_source_sha256": "47982f7ca8622e26e45615f6dc3a8550fa228ca562132006c98685246f8606e4",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}