{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-order-relation-01",
  "mko_id": "mko-order-relation",
  "title": {
    "zh-Hant": "實數的通常順序：加法保序、正分母交叉比較、有限極值與閉區間",
    "en": "The usual order on ℝ: translation, cross-multiplication with positive denominators, finite extrema and closed intervals"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-order-relation-01",
    "source": {
      "mko_id": "mko-order-relation",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "95554189da3a451f8a69edd139e154fd9254524119837ea1643487a5646a2dc3",
      "english_sha256": "3bb53f0ce66755124a9e09f8d185cdebfbb4e6f8cc93a7733a5c08ceba6bdcd6"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "order_basics",
      "frac_rules",
      "add_inverse",
      "nat_induction",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "證明對所有實數 a、b、c：a<b ⇔ a+c<b+c；q、s>0 時 p/q<r/s ⇔ ps<rq；非空有限實數集合有最小元素，且最小元素唯一。據此比較 M={−5/9, 1/3, −7/12, −1/2}（照此次序列出）的元素，求 M 的最小與最大元素，並判定兩端點與 1/2 是否屬於閉區間 [min M, max M]。",
      "en": "Prove for all reals a, b, c that a<b ⇔ a+c<b+c; that p/q<r/s ⇔ ps<rq when q, s>0; and that a finite nonempty set of reals has a minimum, which is unique. Use these to compare the elements of M={−5/9, 1/3, −7/12, −1/2} (listed in this order), find the minimum and maximum of M, and decide whether the two endpoints and 1/2 belong to the closed interval [min M, max M]."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "ℝ 上的通常順序 ≤ 自反、反對稱、傳遞，且任兩數可比較；a<b 表示 a≤b 且 a≠b。",
          "en": "The usual order ≤ on ℝ is reflexive, antisymmetric, transitive and total; a<b means a≤b and a≠b."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "M={−5/9, 1/3, −7/12, −1/2}，依此次序列出；寫成正分母形式：(−5)/9、1/3、(−7)/12、(−1)/2。",
          "en": "M={−5/9, 1/3, −7/12, −1/2}, listed in this order; in positive-denominator form: (−5)/9, 1/3, (−7)/12, (−1)/2."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "定義：m 是集合 M 的最小元素，指 m∈M 且對所有 x∈M 有 m≤x；最大元素同理以 ≥。對 l≤u，閉區間 [l,u]={x∈ℝ ∣ l≤x 且 x≤u}。",
          "en": "Definitions: m is a minimum of a set M when m∈M and m≤x for all x∈M; a maximum likewise with ≥. For l≤u, the closed interval [l,u]={x∈ℝ ∣ l≤x and x≤u}."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "mko_order"
        ],
        "before_tex": "a,b\\in\\mathbb{R}",
        "after_tex": "a<b\\ :\\iff\\ a\\le b\\land a\\neq b;\\ \\text{exactly one of}\\ a<b,\\ a=b,\\ b<a",
        "relation": "definition",
        "rule": "mko_order",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 定義嚴格順序。由全序性（a≤b 或 b≤a）與反對稱（兩者同時成立則 a=b）可得三分律：a<b、a=b、b<a 恰有一個成立。這就是 MKO 所說的三分比較輸出 <、= 或 >。",
          "en": "Define the strict order as in the MKO. Totality (a≤b or b≤a) and antisymmetry (both together force a=b) give trichotomy: exactly one of a<b, a=b, b<a holds. This is the MKO's three-way comparison with outputs <, = or >."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_order",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "只是定義與公理的重述；三分律在 order_basics 中也列為前提。",
            "en": "This only restates a definition and the axioms; trichotomy is also listed in order_basics as a prerequisite."
          }
        },
        "leaf_basis": "mko_order + order_basics"
      },
      {
        "id": "S02",
        "uses": [
          "S01"
        ],
        "before_tex": "a<b",
        "after_tex": "a+c<b+c",
        "relation": "equivalent",
        "rule": "L_translate",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_translate，兩數同加 c 不改變比較。⇒ 在 S02.1，⇐ 在 S02.2。",
          "en": "By Lemma L_translate, adding c to both numbers does not change the comparison. ⇒ is S02.1 and ⇐ is S02.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_translate"
          ],
          "explanation": {
            "zh-Hant": "兩個方向都有證明，所以是等價。",
            "en": "Both directions are proved, so this is an equivalence."
          }
        },
        "leaf_basis": "order_basics + add_assoc + add_inverse + add_zero"
      },
      {
        "id": "S02.1",
        "uses": [
          "order_basics"
        ],
        "before_tex": "a<b",
        "after_tex": "a+c<b+c",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "（⇒）這正是順序公理「u<v ⇒ u+w<v+w」取 u=a、v=b、w=c。",
          "en": "(⇒) This is exactly the order axiom 'u<v ⇒ u+w<v+w' with u=a, v=b, w=c."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "直接套用公理，對任何實數 c 成立。",
            "en": "A direct application of the axiom, valid for every real c."
          }
        },
        "leaf_basis": "order_basics"
      },
      {
        "id": "S02.2",
        "uses": [
          "order_basics",
          "add_inverse"
        ],
        "before_tex": "a+c<b+c",
        "after_tex": "(a+c)+(-c)<(b+c)+(-c),\\ \\text{i.e.}\\ a<b",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "（⇐）對 a+c<b+c 再用同一公理，取 w=−c，得 (a+c)+(−c)<(b+c)+(−c)；化簡兩邊（結合律、c+(−c)=0、加 0 不變）得 a<b。S02.1 與 S02.2 合起來證得引理 L_translate。",
          "en": "(⇐) Apply the same axiom to a+c<b+c with w=−c to get (a+c)+(−c)<(b+c)+(−c); simplifying both sides (associativity, c+(−c)=0, adding 0) gives a<b. S02.1 and S02.2 together prove Lemma L_translate."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "add_assoc",
            "add_inverse",
            "add_zero"
          ],
          "explanation": {
            "zh-Hant": "每個實數 c 都有加法反元素 −c，所以這個方向不需附加條件。",
            "en": "Every real c has an additive inverse −c, so this direction needs no extra condition."
          }
        },
        "leaf_basis": "order_basics + add_assoc + add_inverse + add_zero"
      },
      {
        "id": "S03",
        "uses": [
          "S01"
        ],
        "before_tex": "\\frac{p}{q}<\\frac{r}{s}",
        "after_tex": "ps<rq",
        "relation": "equivalent",
        "rule": "L_cross",
        "conditions": [
          "q>0",
          "s>0"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_cross：分母都為正時，比較兩個分數等於比較交叉乘積。證明在 S03.1 至 S03.3；正分母條件不可省略。",
          "en": "By Lemma L_cross: when both denominators are positive, comparing two fractions is the same as comparing the cross products. The proof is in S03.1 to S03.3; the positive-denominator condition cannot be dropped."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_cross"
          ],
          "explanation": {
            "zh-Hant": "⇒ 與 ⇐ 分別在子步中證明，條件 q、s>0 兩個方向都要用。",
            "en": "⇒ and ⇐ are proved in the sub-steps, and both use the condition q, s>0."
          }
        },
        "leaf_basis": "order_basics + frac_rules + mul_inverse_nonzero"
      },
      {
        "id": "S03.1",
        "uses": [
          "order_basics",
          "mul_inverse_nonzero"
        ],
        "before_tex": "q>0,\\ s>0",
        "after_tex": "qs>0,\\ \\tfrac{1}{qs}>0",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "q>0",
          "s>0"
        ],
        "explanation": {
          "zh-Hant": "把 0<q 兩邊乘以 s>0 得 0·s<q·s，即 qs>0。qs≠0，所以 1/(qs) 存在；若 1/(qs)<0，乘以 qs>0 得 1<0，矛盾，而 1/(qs)≠0（因 (1/(qs))·qs=1），故 1/(qs)>0。",
          "en": "Multiply 0<q by s>0 to get 0·s<q·s, i.e. qs>0. Then qs≠0, so 1/(qs) exists; if 1/(qs)<0, multiplying by qs>0 would give 1<0, a contradiction, and 1/(qs)≠0 since (1/(qs))·qs=1, so 1/(qs)>0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "mul_inverse_nonzero",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "兩個正數的積為正，正數的倒數為正；兩者都依賴 q、s>0。",
            "en": "A product of two positives is positive, and the reciprocal of a positive is positive; both depend on q, s>0."
          }
        },
        "leaf_basis": "order_basics + mul_zero + mul_inverse_nonzero + contradiction"
      },
      {
        "id": "S03.2",
        "uses": [
          "S03.1",
          "frac_rules"
        ],
        "before_tex": "\\frac{p}{q}<\\frac{r}{s}",
        "after_tex": "\\frac{p}{q}\\cdot qs<\\frac{r}{s}\\cdot qs,\\ \\text{i.e.}\\ ps<rq",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "qs>0"
        ],
        "explanation": {
          "zh-Hant": "（⇒）兩邊乘以正數 qs，順序保持；化簡 (p/q)·qs=ps、(r/s)·qs=rq（分式規則，q、s≠0）。",
          "en": "(⇒) Multiply both sides by the positive number qs, which keeps the order; simplify (p/q)·qs=ps and (r/s)·qs=rq (fraction rules, q, s≠0)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03.1",
            "order_basics",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "需要 qs>0；若 qs<0，乘法會使不等號反向。",
            "en": "It needs qs>0; if qs<0, multiplication would reverse the inequality."
          }
        },
        "leaf_basis": "order_basics + frac_rules + mul_comm + mul_assoc"
      },
      {
        "id": "S03.3",
        "uses": [
          "S03.1",
          "frac_rules"
        ],
        "before_tex": "ps<rq",
        "after_tex": "\\frac{ps}{qs}<\\frac{rq}{qs},\\ \\text{i.e.}\\ \\frac{p}{q}<\\frac{r}{s}",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "\\tfrac{1}{qs}>0"
        ],
        "explanation": {
          "zh-Hant": "（⇐）兩邊乘以正數 1/(qs)（S03.1），順序保持；約分 ps/(qs)=p/q、rq/(qs)=r/s。S03.1 至 S03.3 合起來證得引理 L_cross。",
          "en": "(⇐) Multiply both sides by the positive number 1/(qs) (S03.1), which keeps the order; cancel ps/(qs)=p/q and rq/(qs)=r/s. S03.1 to S03.3 together prove Lemma L_cross."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03.1",
            "order_basics",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "需要 1/(qs)>0，由 S03.1 提供。",
            "en": "It needs 1/(qs)>0, supplied by S03.1."
          }
        },
        "leaf_basis": "order_basics + frac_rules"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "G2"
        ],
        "before_tex": "-\\tfrac{7}{12}\\ ?\\ -\\tfrac{5}{9}",
        "after_tex": "(-7)\\cdot9=-63<-60=(-5)\\cdot12\\ \\Rightarrow\\ -\\tfrac{7}{12}<-\\tfrac{5}{9}",
        "relation": "equivalent",
        "rule": "L_cross",
        "conditions": [
          "12>0",
          "9>0"
        ],
        "explanation": {
          "zh-Hant": "取 p=−7、q=12、r=−5、s=9，分母都為正。ps=−63、rq=−60，而 −60−(−63)=3>0，所以 −63<−60；依 L_cross，−7/12<−5/9。",
          "en": "Take p=−7, q=12, r=−5, s=9; both denominators are positive. ps=−63 and rq=−60, and −60−(−63)=3>0, so −63<−60; by L_cross, −7/12<−5/9."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_cross",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "L_cross 是等價，條件 12、9>0 成立，所以分數比較與整數比較同真同假。",
            "en": "L_cross is an equivalence and its conditions 12, 9>0 hold, so the fraction comparison and the integer comparison are true or false together."
          }
        },
        "leaf_basis": "order_basics + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "S03",
          "G2"
        ],
        "before_tex": "-\\tfrac{5}{9}\\ ?\\ -\\tfrac{1}{2}",
        "after_tex": "(-5)\\cdot2=-10<-9=(-1)\\cdot9\\ \\Rightarrow\\ -\\tfrac{5}{9}<-\\tfrac{1}{2}",
        "relation": "equivalent",
        "rule": "L_cross",
        "conditions": [
          "9>0",
          "2>0"
        ],
        "explanation": {
          "zh-Hant": "取 p=−5、q=9、r=−1、s=2：ps=−10、rq=−9，−10<−9，所以 −5/9<−1/2。",
          "en": "Take p=−5, q=9, r=−1, s=2: ps=−10, rq=−9 and −10<−9, so −5/9<−1/2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_cross",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "同 S04，分母為正，等價成立。",
            "en": "As in S04, the denominators are positive and the equivalence applies."
          }
        },
        "leaf_basis": "order_basics + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S03",
          "G2"
        ],
        "before_tex": "-\\tfrac{1}{2}\\ ?\\ \\tfrac{1}{3}",
        "after_tex": "(-1)\\cdot3=-3<2=1\\cdot2\\ \\Rightarrow\\ -\\tfrac{1}{2}<\\tfrac{1}{3}",
        "relation": "equivalent",
        "rule": "L_cross",
        "conditions": [
          "2>0",
          "3>0"
        ],
        "explanation": {
          "zh-Hant": "取 p=−1、q=2、r=1、s=3：ps=−3、rq=2，−3<2，所以 −1/2<1/3。",
          "en": "Take p=−1, q=2, r=1, s=3: ps=−3, rq=2 and −3<2, so −1/2<1/3."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_cross",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "同 S04，分母為正，等價成立。",
            "en": "As in S04, the denominators are positive and the equivalence applies."
          }
        },
        "leaf_basis": "order_basics + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S04",
          "S05",
          "S06"
        ],
        "before_tex": "-\\tfrac{7}{12}<-\\tfrac{5}{9},\\ -\\tfrac{5}{9}<-\\tfrac{1}{2},\\ -\\tfrac{1}{2}<\\tfrac{1}{3}",
        "after_tex": "-\\tfrac{7}{12}<-\\tfrac{5}{9}<-\\tfrac{1}{2}<\\tfrac{1}{3}",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "三個比較串成一條鏈；依 < 的傳遞性，鏈上任一數小於它右邊的每一個數，例如 −7/12<−1/2、−7/12<1/3、−5/9<1/3。M 的四個元素就此全部排好。",
          "en": "The three comparisons form a chain; by transitivity of <, each number in the chain is less than every number to its right, for example −7/12<−1/2, −7/12<1/3 and −5/9<1/3. All four elements of M are now ordered."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S05",
            "S06",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要的是三個相鄰比較與傳遞性；不需要再做其餘三對的交叉相乘。",
            "en": "What is needed is the three adjacent comparisons and transitivity; the other three pairs need no further cross-multiplication."
          }
        },
        "leaf_basis": "order_basics + logic_chain"
      },
      {
        "id": "S08",
        "uses": [
          "S01"
        ],
        "before_tex": "M\\subseteq\\mathbb{R}\\ \\text{finite},\\ M\\neq\\varnothing",
        "after_tex": "M\\ \\text{has a minimum}",
        "relation": "implies",
        "rule": "L_finite_min",
        "conditions": [
          "M\\neq\\varnothing"
        ],
        "explanation": {
          "zh-Hant": "MKO 斷言非空有限實數集合有最小與最大元素。依引理 L_finite_min，對元素個數 n≥1 歸納證明最小元素存在（S08.1、S08.2）；最大元素的論證相同，只需把 ≤ 換成 ≥。",
          "en": "The MKO asserts that a nonempty finite set of reals has a minimum and a maximum. By Lemma L_finite_min, the minimum exists, proved by induction on the number of elements n≥1 (S08.1, S08.2); the argument for the maximum is the same with ≥ in place of ≤."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_finite_min"
          ],
          "explanation": {
            "zh-Hant": "非空是必要的：歸納從 n=1 開始，空集合不在範圍內，也沒有最小元素。",
            "en": "Nonemptiness is essential: the induction starts at n=1, the empty set is outside its range, and it has no minimum."
          }
        },
        "leaf_basis": "nat_induction + order_basics + case_analysis"
      },
      {
        "id": "S08.1",
        "uses": [
          "G3",
          "min_max_def"
        ],
        "before_tex": "M=\\{a\\}",
        "after_tex": "\\min M=a",
        "relation": "induction_base",
        "rule": "nat_induction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "基礎 n=1：M={a}。a∈M，且 M 唯一的元素 x=a 滿足 a≤a（自反），所以 a 是最小元素（也是最大元素）。",
          "en": "Base n=1: M={a}. a∈M, and the only element x=a satisfies a≤a (reflexivity), so a is the minimum (and also the maximum)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "min_max_def",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "只用到自反性；單元素集合的兩個極值相同。",
            "en": "Only reflexivity is used; the two extrema of a singleton coincide."
          }
        },
        "leaf_basis": "min_max_def + order_basics"
      },
      {
        "id": "S08.2",
        "uses": [
          "S08.1",
          "min_max_def",
          "case_analysis"
        ],
        "before_tex": "|M|=n+1,\\ a\\in M,\\ m'=\\min(M\\setminus\\{a\\})",
        "after_tex": "\\min M=m'\\ \\text{if}\\ m'\\le a,\\ \\ \\min M=a\\ \\text{if}\\ a<m'",
        "relation": "induction_step",
        "rule": "nat_induction",
        "conditions": [
          "n\\ge1"
        ],
        "explanation": {
          "zh-Hant": "歸納步：設所有 n 個元素的非空集合都有最小元素（n≥1）。取 |M|=n+1 與 a∈M，M′=M∖{a} 有 n 個元素，依歸納假設有最小元素 m′。依三分律分兩種情形：若 m′≤a，則 m′ 不大於 M′ 的每個元素，也不大於 a，所以 m′ 是 M 的最小元素；若 a<m′，則對 x∈M′ 有 a<m′≤x，傳遞得 a≤x，又 a≤a，所以 a 是最小元素。S08.1 與 S08.2 合起來證得引理 L_finite_min。",
          "en": "Inductive step: assume every nonempty set with n elements has a minimum (n≥1). Take |M|=n+1 and a∈M; M′=M∖{a} has n elements and, by the induction hypothesis, a minimum m′. By trichotomy there are two cases: if m′≤a, then m′ is no greater than every element of M′ and than a, so m′ is the minimum of M; if a<m′, then for x∈M′ we have a<m′≤x, so a≤x by transitivity, and a≤a, so a is the minimum. S08.1 and S08.2 together prove Lemma L_finite_min."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "nat_induction",
            "case_analysis",
            "order_basics",
            "min_max_def"
          ],
          "explanation": {
            "zh-Hant": "需要 n≥1 使 M′ 非空，歸納假設才適用；兩個情形由全序性窮盡。",
            "en": "It needs n≥1 so that M′ is nonempty and the induction hypothesis applies; the two cases are exhaustive by totality."
          }
        },
        "leaf_basis": "nat_induction + case_analysis + order_basics + min_max_def"
      },
      {
        "id": "S09",
        "uses": [
          "S07",
          "G2",
          "min_max_def"
        ],
        "before_tex": "-\\tfrac{7}{12}\\in M,\\ -\\tfrac{7}{12}\\le x\\ \\forall x\\in M",
        "after_tex": "\\min M=-\\tfrac{7}{12}",
        "relation": "implies",
        "rule": "min_max_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "−7/12 屬於 M（列在第三個），由 S07 它小於其餘三個元素，對自己則 −7/12≤−7/12，所以它是最小元素。M 列出時 −5/9 排在第一，但極值由值決定，不由列出次序決定。",
          "en": "−7/12 belongs to M (listed third), by S07 it is less than the other three elements, and −7/12≤−7/12 for itself, so it is the minimum. M lists −5/9 first, but extrema are decided by value, not by the order of listing."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "min_max_def",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "兩個條件（屬於 M、不大於每個元素）都已檢查；由 x<y 得 x≤y 是嚴格順序的定義。",
            "en": "Both conditions (membership in M and being no greater than every element) are checked; x<y gives x≤y by the definition of the strict order."
          }
        },
        "leaf_basis": "min_max_def + order_basics"
      },
      {
        "id": "S10",
        "uses": [
          "S07",
          "G2",
          "min_max_def"
        ],
        "before_tex": "\\tfrac{1}{3}\\in M,\\ \\tfrac{1}{3}\\ge x\\ \\forall x\\in M",
        "after_tex": "\\max M=\\tfrac{1}{3}",
        "relation": "implies",
        "rule": "min_max_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "1/3 屬於 M，由 S07 其餘三個元素都小於 1/3，所以 1/3 是最大元素。",
          "en": "1/3 belongs to M and by S07 the other three elements are all less than 1/3, so 1/3 is the maximum."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "min_max_def"
          ],
          "explanation": {
            "zh-Hant": "與 S09 對稱。",
            "en": "Symmetric to S09."
          }
        },
        "leaf_basis": "min_max_def + order_basics"
      },
      {
        "id": "S11",
        "uses": [
          "S09",
          "S10"
        ],
        "before_tex": "m,m'\\ \\text{both minima of}\\ M",
        "after_tex": "m=m'",
        "relation": "implies",
        "rule": "L_min_unique",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_min_unique，最小元素唯一，所以可以說「這個」最小元素 −7/12；最大元素 1/3 同理唯一。證明在 S11.1。",
          "en": "By Lemma L_min_unique the minimum is unique, so one may speak of 'the' minimum −7/12; the maximum 1/3 is likewise unique. The proof is in S11.1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_min_unique"
          ],
          "explanation": {
            "zh-Hant": "唯一性依賴反對稱性。",
            "en": "Uniqueness relies on antisymmetry."
          }
        },
        "leaf_basis": "min_max_def + order_basics"
      },
      {
        "id": "S11.1",
        "uses": [
          "min_max_def",
          "G1"
        ],
        "before_tex": "m,m'\\in M,\\ m\\le x,\\ m'\\le x\\ \\forall x\\in M",
        "after_tex": "m\\le m',\\ m'\\le m\\ \\Rightarrow\\ m=m'",
        "relation": "implies",
        "rule": "mko_order",
        "conditions": [],
        "explanation": {
          "zh-Hant": "m 是最小元素且 m′∈M，所以 m≤m′；同理 m′≤m。依反對稱性 m=m′。最大元素以 ≥ 同樣證明。此步證得引理 L_min_unique。",
          "en": "m is a minimum and m′∈M, so m≤m′; likewise m′≤m. By antisymmetry m=m′. The maximum is handled the same way with ≥. This step proves Lemma L_min_unique."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "min_max_def",
            "mko_order"
          ],
          "explanation": {
            "zh-Hant": "只用到兩個最小元素都屬於 M 以及反對稱性。",
            "en": "Only that both minima belong to M and antisymmetry are used."
          }
        },
        "leaf_basis": "min_max_def + mko_order"
      },
      {
        "id": "S12",
        "uses": [
          "S07",
          "S09",
          "S10",
          "G3"
        ],
        "before_tex": "l=-\\tfrac{7}{12},\\ u=\\tfrac{1}{3}",
        "after_tex": "l\\le u",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 假設 a3 要求先驗閉區間的下界不大於上界。由 S07，l=−7/12<1/3=u，故 l≤u，[l,u] 是合法的閉區間。",
          "en": "MKO assumption a3 asks first to check that the lower endpoint of a closed interval is no greater than the upper one. By S07, l=−7/12<1/3=u, so l≤u and [l,u] is a valid closed interval."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "由 l<u 得 l≤u 是嚴格順序的定義。",
            "en": "From l<u, l≤u follows by the definition of the strict order."
          }
        },
        "leaf_basis": "order_basics + mko_order"
      },
      {
        "id": "S13",
        "uses": [
          "S12",
          "S09",
          "S10",
          "interval_def"
        ],
        "before_tex": "l\\le l\\le u,\\ l\\le u\\le u",
        "after_tex": "l\\in[l,u],\\ u\\in[l,u],\\ M\\subseteq[l,u]",
        "relation": "implies",
        "rule": "interval_def",
        "conditions": [
          "l\\le u"
        ],
        "explanation": {
          "zh-Hant": "閉區間的成員條件是 l≤x≤u。對 x=l：l≤l（自反）且 l≤u（S12），所以左端點屬於區間；右端點同理。M 的每個元素 x 滿足 min M≤x≤max M（S09、S10），所以 M⊆[l,u]。",
          "en": "Membership in a closed interval is l≤x≤u. For x=l: l≤l (reflexivity) and l≤u (S12), so the left endpoint belongs to the interval; the right endpoint likewise. Every element x of M satisfies min M≤x≤max M (S09, S10), so M⊆[l,u]."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "interval_def",
            "S12",
            "mko_order"
          ],
          "explanation": {
            "zh-Hant": "端點屬於閉區間只用到自反性與 l≤u；開區間是另一個物件，不在此討論。",
            "en": "Endpoint membership uses only reflexivity and l≤u; open intervals are a different object and are not discussed here."
          }
        },
        "leaf_basis": "interval_def + mko_order"
      },
      {
        "id": "S14",
        "uses": [
          "S03",
          "S13",
          "interval_def"
        ],
        "before_tex": "\\tfrac{1}{3}<\\tfrac{1}{2}\\ \\ (1\\cdot2<1\\cdot3)",
        "after_tex": "\\tfrac12\\notin[l,u]",
        "relation": "implies",
        "rule": "interval_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 L_cross（p=1、q=3、r=1、s=2）：1·2=2<3=1·3，所以 1/3<1/2。若 1/2≤1/3，與 1/3≤1/2 合用反對稱性得 1/2=1/3，與 1/3≠1/2 矛盾；所以 1/2≤u 不成立，1/2 不在區間內。",
          "en": "By L_cross (p=1, q=3, r=1, s=2): 1·2=2<3=1·3, so 1/3<1/2. If 1/2≤1/3, antisymmetry together with 1/3≤1/2 would give 1/2=1/3, contradicting 1/3≠1/2; so 1/2≤u fails and 1/2 is not in the interval."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_cross",
            "interval_def",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "成員條件是兩個不等式的合取，否定其中一個就足以排除。",
            "en": "Membership is a conjunction of two inequalities; refuting one of them suffices to exclude the point."
          }
        },
        "leaf_basis": "interval_def + contradiction + order_basics"
      },
      {
        "id": "S15",
        "uses": [
          "S02",
          "S04"
        ],
        "before_tex": "-\\tfrac{7}{12}<-\\tfrac{5}{9}",
        "after_tex": "-\\tfrac{7}{12}+\\tfrac{7}{12}<-\\tfrac{5}{9}+\\tfrac{7}{12},\\ \\text{i.e.}\\ 0<\\tfrac{1}{36}",
        "relation": "equivalent",
        "rule": "L_translate",
        "conditions": [],
        "explanation": {
          "zh-Hant": "用 L_translate 取 c=7/12：−7/12<−5/9 ⇔ 0<−5/9+7/12。計算 −5/9+7/12=−20/36+21/36=1/36>0，與 S04 一致。平移不改變比較，所以它只是重寫，不是新的題型。",
          "en": "Apply L_translate with c=7/12: −7/12<−5/9 ⇔ 0<−5/9+7/12. Compute −5/9+7/12=−20/36+21/36=1/36>0, consistent with S04. Translation does not change the comparison, so this is a rewriting, not a new kind of exercise."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_translate",
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "L_translate 是等價，兩邊同加 7/12 或同減 7/12 互為反向。",
            "en": "L_translate is an equivalence; adding 7/12 and subtracting 7/12 undo each other."
          }
        },
        "leaf_basis": "order_basics + frac_rules + numeric_evaluation"
      },
      {
        "id": "S16",
        "uses": [
          "S02",
          "S03",
          "S08",
          "S09",
          "S10",
          "S11",
          "S13",
          "S14"
        ],
        "before_tex": "\\text{S02 to S15}",
        "after_tex": "\\min M=-\\tfrac{7}{12},\\ \\max M=\\tfrac13,\\ l,u\\in[l,u],\\ \\tfrac12\\notin[l,u]",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "整理：加法保序與正分母交叉比較已證；非空有限集合的最小元素存在且唯一；M 的次序為 −7/12<−5/9<−1/2<1/3，最小 −7/12、最大 1/3；閉區間 [−7/12,1/3] 含兩端點與 M 全部元素，不含 1/2。",
          "en": "Summary: translation invariance and positive-denominator cross-comparison are proved; a nonempty finite set has a unique minimum; M is ordered −7/12<−5/9<−1/2<1/3 with minimum −7/12 and maximum 1/3; the closed interval [−7/12,1/3] contains both endpoints and all of M, but not 1/2."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S02",
            "S03",
            "S08",
            "S09",
            "S10",
            "S11",
            "S13",
            "S14",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G 為 goal 中的證明與判定；足夠條件 H 為各步結論的合取，皆已正向證成，H⇒G 只用合取引入。",
            "en": "Goal G is the proofs and decisions in the goal; the sufficient condition H is the conjunction of the step results, all proved forwards, and H⇒G uses only conjunction introduction."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "對所有實數 a、b、c，a<b ⇔ a+c<b+c；q、s>0 時 p/q<r/s ⇔ ps<rq；非空有限實數集合有唯一的最小元素（最大元素同理）。M 的元素排成 −7/12<−5/9<−1/2<1/3，所以 min M=−7/12、max M=1/3（與列出次序無關）；[−7/12,1/3] 是合法閉區間，包含兩端點與 M 的全部元素，不包含 1/2。",
      "en": "For all real a, b and c, a<b ⇔ a+c<b+c; for q, s>0, p/q<r/s ⇔ ps<rq; a nonempty finite set of reals has a unique minimum (and likewise maximum). The elements of M are ordered −7/12<−5/9<−1/2<1/3, so min M=−7/12 and max M=1/3 (regardless of the listing order); [−7/12,1/3] is a valid closed interval containing both endpoints and all of M, but not 1/2."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "1/(−2) 與 1/3 交叉相乘得 1·3=3>−2=1·(−2)，所以 1/(−2)>1/3",
          "en": "Cross-multiplying 1/(−2) and 1/3 gives 1·3=3>−2=1·(−2), so 1/(−2)>1/3"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "分母 −2 為負，L_cross 不適用；1/(−2)=−1/2<1/3。",
          "en": "The denominator −2 is negative, so L_cross does not apply; 1/(−2)=−1/2<1/3."
        }
      },
      {
        "claim": {
          "zh-Hant": "−7/12 是端點，所以不屬於 [−7/12,1/3]",
          "en": "−7/12 is an endpoint, so it is not in [−7/12,1/3]"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "−7/12≤−7/12≤1/3 成立，閉區間包含端點。",
          "en": "−7/12≤−7/12≤1/3 holds; a closed interval contains its endpoints."
        }
      },
      {
        "claim": {
          "zh-Hant": "M 的最小元素是 −5/9，因為它列在第一個",
          "en": "The minimum of M is −5/9 because it is listed first"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "−7/12<−5/9（S04），最小元素由值決定，是 −7/12。",
          "en": "−7/12<−5/9 (S04); the minimum is decided by value and is −7/12."
        }
      },
      {
        "claim": {
          "zh-Hant": "空集合的最小元素是 0",
          "en": "The minimum of the empty set is 0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "最小元素必須屬於該集合，而 0∉∅；空集合沒有最小元素。",
          "en": "A minimum must belong to the set, and 0∉∅; the empty set has no minimum."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "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": "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": "nat_induction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "數學歸納原理：對從 n₀ 起的命題族 P(n)，若 P(n₀) 成立，且對每個 n≥n₀ 都有 P(n)⇒P(n+1)，則所有 n≥n₀ 都有 P(n)。",
          "en": "Principle of induction: for a family P(n) starting at n₀, if P(n₀) holds and P(n)⇒P(n+1) for every n≥n₀, then P(n) holds for every n≥n₀."
        }
      },
      {
        "id": "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": "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_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "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": "mul_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "id": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "mul_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。",
          "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)."
        }
      },
      {
        "id": "mul_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法結合律：(uv)w=u(vw)。",
          "en": "Associativity of multiplication: (uv)w=u(vw)."
        }
      },
      {
        "id": "logic_chain",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "推論的傳遞：由 P⇒Q 與 Q⇒R 得 P⇒R；⇔ 同理。",
          "en": "Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔."
        }
      },
      {
        "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": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "mko_order",
        "kind": "cited",
        "source": {
          "mko_id": "mko-order-relation",
          "sha256": "95554189da3a451f8a69edd139e154fd9254524119837ea1643487a5646a2dc3",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "實數的通常順序 ≤ 滿足自反、反對稱、傳遞及任兩數可比較；a<b 表示 a≤b 且 a≠b。任意實數 a、b、c 都有 a<b 當且僅當 a+c<b+c。非空有限實數集合有最小與最大元素；閉區間 [l,u]（l≤u）包含兩端點。",
          "en": "On R, the usual total order ≤ is reflexive, antisymmetric and transitive; any two real numbers are comparable. Write a<b when a≤b and a≠b. For real a,b,c, a<b exactly when a+c<b+c. A finite nonempty real set has a minimum and a maximum; a closed interval [l,u] with l≤u includes both endpoints."
        }
      },
      {
        "id": "min_max_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "最小元素：m∈M 且 ∀x∈M, m≤x；最大元素：m∈M 且 ∀x∈M, m≥x。",
          "en": "Minimum: m∈M and m≤x for all x∈M; maximum: m∈M and m≥x for all x∈M."
        }
      },
      {
        "id": "interval_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "閉區間：對 l≤u，x∈[l,u] ⇔ l≤x 且 x≤u。",
          "en": "Closed interval: for l≤u, x∈[l,u] ⇔ l≤x and x≤u."
        }
      },
      {
        "id": "L_translate",
        "kind": "lemma",
        "proved_in": [
          "S02.1",
          "S02.2"
        ],
        "statement": {
          "zh-Hant": "對所有實數 a、b、c：a<b ⇔ a+c<b+c。",
          "en": "For all reals a, b, c: a<b ⇔ a+c<b+c."
        }
      },
      {
        "id": "L_cross",
        "kind": "lemma",
        "proved_in": [
          "S03.1",
          "S03.2",
          "S03.3"
        ],
        "statement": {
          "zh-Hant": "對實數 p、r 與 q>0、s>0：p/q<r/s ⇔ ps<rq。",
          "en": "For reals p, r and q>0, s>0: p/q<r/s ⇔ ps<rq."
        }
      },
      {
        "id": "L_finite_min",
        "kind": "lemma",
        "proved_in": [
          "S08.1",
          "S08.2"
        ],
        "statement": {
          "zh-Hant": "每個非空有限實數集合都有最小元素。",
          "en": "Every nonempty finite set of reals has a minimum."
        }
      },
      {
        "id": "L_min_unique",
        "kind": "lemma",
        "proved_in": [
          "S11.1"
        ],
        "statement": {
          "zh-Hant": "一個實數集合至多有一個最小元素（最大元素同理）。",
          "en": "A set of reals has at most one minimum (and likewise at most one maximum)."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S04",
        "lhs": "-7*9",
        "rhs": "-63"
      },
      {
        "id": "N2",
        "step": "S04",
        "lhs": "-5*12",
        "rhs": "-60"
      },
      {
        "id": "N3",
        "step": "S04",
        "lhs": "-63",
        "relation": "<",
        "rhs": "-60"
      },
      {
        "id": "N4",
        "step": "S05",
        "lhs": "-5*2",
        "relation": "<",
        "rhs": "-1*9"
      },
      {
        "id": "N5",
        "step": "S06",
        "lhs": "-1*3",
        "relation": "<",
        "rhs": "1*2"
      },
      {
        "id": "N6",
        "step": "S07",
        "lhs": "-7/12",
        "relation": "<",
        "rhs": "-5/9"
      },
      {
        "id": "N7",
        "step": "S07",
        "lhs": "-5/9",
        "relation": "<",
        "rhs": "-1/2"
      },
      {
        "id": "N8",
        "step": "S07",
        "lhs": "-1/2",
        "relation": "<",
        "rhs": "1/3"
      },
      {
        "id": "N9",
        "step": "S14",
        "lhs": "1*2",
        "relation": "<",
        "rhs": "1*3"
      },
      {
        "id": "N10",
        "step": "S14",
        "lhs": "1/3",
        "relation": "<",
        "rhs": "1/2"
      },
      {
        "id": "N11",
        "step": "S15",
        "lhs": "-5/9+7/12",
        "rhs": "1/36"
      },
      {
        "id": "N12",
        "step": "S15",
        "lhs": "-7/12+7/12",
        "rhs": "0"
      },
      {
        "id": "N13",
        "step": "S03",
        "lhs": "1*3",
        "relation": ">",
        "rhs": "1*(-2)"
      },
      {
        "id": "N14",
        "step": "S03",
        "lhs": "1/(-2)",
        "relation": "<",
        "rhs": "1/3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只談實數的通常全序（MKO a1），以 order_basics 中的順序公理為前提，不構造實數。比較題使用正分母有理數（a2）；極值只對非空有限集合，閉區間要求下界不大於上界（a3）。最大元素的存在以「同一論證把 ≤ 換成 ≥」帶過。不處理複數、一般偏序或無限集合的上確界；不改動 canonical MKO 或英文 edition。",
      "en": "Only the usual total order on the reals is considered (MKO a1), with the order axioms of order_basics as prerequisites; the reals are not constructed. Comparisons use rationals with positive denominators (a2); extrema are for nonempty finite sets, and a closed interval needs its lower endpoint no greater than its upper endpoint (a3). Existence of a maximum is covered by 'the same argument with ≥ in place of ≤'. Complex numbers, general partial orders and suprema of infinite sets are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "a\\le b,\\ a<b",
        "通常順序與嚴格順序",
        "the usual order and the strict order"
      ],
      [
        "\\frac{p}{q}",
        "有理數，分母 q>0",
        "a rational with denominator q>0"
      ],
      [
        "M",
        "待比較的有限集合",
        "the finite set being compared"
      ],
      [
        "\\min M,\\ \\max M",
        "最小與最大元素",
        "the minimum and maximum elements"
      ],
      [
        "[l,u]",
        "閉區間 {x ∣ l≤x≤u}",
        "the closed interval {x ∣ l≤x≤u}"
      ]
    ],
    "md_error_controls": [
      {
        "title": "交叉相乘前分母須為正 / Denominators must be positive before cross-multiplying",
        "zh": "MKO 的誤解之一是忽略分母正負。比較 1/(−2) 與 1/3 時直接交叉相乘得 1·3=3>−2=1·(−2)，會誤判 1/(−2)>1/3；實際上 1/(−2)=−1/2<1/3。原因是乘上負數 (−2)·3 使不等號反向。先寫成正分母 (−1)/2 再比較：(−1)·3=−3<2，結論正確。",
        "en": "One misconception identified in the MKO is ignoring the sign of denominators. Cross-multiplying 1/(−2) and 1/3 directly gives 1·3=3>−2=1·(−2) and wrongly suggests 1/(−2)>1/3; in fact 1/(−2)=−1/2<1/3. The cause is that multiplying by the negative (−2)·3 reverses the inequality. Rewriting with a positive denominator (−1)/2 first gives (−1)·3=−3<2 and the correct conclusion."
      },
      {
        "title": "閉區間包含端點 / A closed interval contains its endpoints",
        "zh": "另一個誤解是漏掉閉區間的端點。l≤l≤u 成立，所以 l∈[l,u]（S13）。l=u 時 [l,u] 只含一個值，例如 [2,2]={2}；不含端點的是開區間，那是另一個物件。",
        "en": "The other misconception is to leave out the endpoints of a closed interval. l≤l≤u holds, so l∈[l,u] (S13). When l=u the interval has a single value, for example [2,2]={2}; it is the open interval that excludes endpoints, and that is a different object."
      },
      {
        "title": "極值看值不看次序，空集合沒有極值 / Extrema go by value, not order, and the empty set has none",
        "zh": "M 列出時 −5/9 在第一個，但最小元素是 −7/12（S09）。空集合沒有最小元素，因為不存在 m∈∅；S08 的歸納從 n=1 開始，不涵蓋空集合，不能替空集合指定任意極值。",
        "en": "M lists −5/9 first, but the minimum is −7/12 (S09). The empty set has no minimum because there is no m∈∅; the induction in S08 starts at n=1 and does not cover the empty set, so no extremum may be assigned to it."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "順序公理（order_basics）作為前提使用，本稿不從實數的構造推導它們。",
        "en": "The order axioms (order_basics) are used as prerequisites; they are not derived from a construction of the reals."
      },
      {
        "zh-Hant": "最大元素的存在只以「同一論證把 ≤ 換成 ≥」說明，未逐步寫出。",
        "en": "Existence of a maximum is covered only by 'the same argument with ≥ in place of ≤' and is not written out step by step."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "2cbe9fb81fea4fdafc3a0280d691e7cbfb470c6ed78948d6657b4f589caa62a8",
    "original_source_sha256": "2cbe9fb81fea4fdafc3a0280d691e7cbfb470c6ed78948d6657b4f589caa62a8",
    "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."
  }
}