{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-real-number-01",
  "mko_id": "mko-real-number",
  "title": {
    "zh-Hant": "實數：在 Cauchy 完備化中證明「循環的 9」與有理數 1 是同一個實數",
    "en": "Real number: proving in the Cauchy completion that 'repeating nines' and the rational 1 are the same real"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-real-number-01",
    "source": {
      "mko_id": "mko-real-number",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "2bc8c0a1defae2276ce50808a0761bf4d62f26892e443fcab79fc8090113d880",
      "english_sha256": "5dde700ff98e1ae0b35a5924253f8ad22515aa33c9a17deab8b7fa9a35e1ef5c"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "nat_induction",
      "frac_rules",
      "power_laws",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在以有理數 Cauchy 序列的等價類建構的實數中（有理數 q 嵌入為常數列的類 ι(q)），令 a_n=1−10^{−n}（n≥1），即有限小數 0.9、0.99、0.999 等。證明 (a_n) 是 Cauchy 序列，且與常數列 1 等價，所以 [(a_n)]=ι(1)；證明 ι 是單射，因此有限小數 0.999=999/1000 所表示的實數不等於 1；並說明有限小數 0.375=3/8 是有理數，在 ℝ 中就是 ι(3/8)。",
      "en": "In the reals constructed as equivalence classes of Cauchy sequences of rationals (a rational q embedded as the class ι(q) of its constant sequence), let a_n=1−10^{−n} (n≥1), the finite decimals 0.9, 0.99, 0.999 and so on. Prove that (a_n) is a Cauchy sequence equivalent to the constant sequence 1, so [(a_n)]=ι(1); prove that ι is injective, so the real denoted by the finite decimal 0.999=999/1000 is not 1; and show that the finite decimal 0.375=3/8 is rational and is ι(3/8) in ℝ."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "ℚ 是有序域：有理數的加、乘、序與絕對值遵守有序域的規則（含三角不等式 |x+y|≤|x|+|y|）。本稿的 ε、δ 都是正有理數。",
          "en": "ℚ is an ordered field: addition, multiplication, order and absolute value of rationals obey the ordered-field rules (including the triangle inequality |x+y|≤|x|+|y|). Every ε and δ in this guide is a positive rational."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "Cauchy 序列：有理數列 (a_n)_{n≥1} 是 Cauchy 序列，若對每個有理數 ε>0，存在 N，使所有 m,n≥N 都有 |a_m−a_n|<ε。",
          "en": "Cauchy sequence: a sequence of rationals (a_n)_{n≥1} is Cauchy if for every rational ε>0 there is N such that |a_m−a_n|<ε for all m,n≥N."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "等價與實數（MKO 假設 a1）：Cauchy 序列 (a_n)、(b_n) 等價，記作 (a_n)∼(b_n)，若對每個有理數 ε>0，存在 N，使所有 n≥N 都有 |a_n−b_n|<ε。ℝ 是 Cauchy 序列等價類 [(a_n)] 的集合；有理數 q 的嵌入是常數列的類 ι(q)=[(q)_{n≥1}]。",
          "en": "Equivalence and reals (MKO assumption a1): Cauchy sequences (a_n), (b_n) are equivalent, written (a_n)∼(b_n), if for every rational ε>0 there is N such that |a_n−b_n|<ε for all n≥N. ℝ is the set of equivalence classes [(a_n)] of Cauchy sequences; the embedding of a rational q is the class of its constant sequence, ι(q)=[(q)_{n≥1}]."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "資料：a_n=1−10^{−n}（n≥1）；有限小數 0.999 與 0.375。",
          "en": "Data: a_n=1−10^{−n} (n≥1); the finite decimals 0.999 and 0.375."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G2",
          "G3",
          "mko_real_def"
        ],
        "before_tex": "\\mathbb{R}",
        "after_tex": "\\{[(a_n)]\\mid (a_n)\\ \\text{Cauchy in}\\ \\mathbb{Q}\\},\\ \\iota(q)=[(q)_{n\\ge1}]",
        "relation": "definition",
        "rule": "equiv_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 假設 a1，在 Cauchy 完備化的語境中工作：實數是有理數 Cauchy 序列的等價類，有理數 q 經 ι 嵌入為常數列（每一項都是 q）的類。常數列是 Cauchy 序列，因為 |q−q|=0<ε 對所有 m、n 成立。",
          "en": "By MKO assumption a1, work in the Cauchy-completion setting: a real number is an equivalence class of Cauchy sequences of rationals, and a rational q is embedded by ι as the class of its constant sequence (every term equal to q). A constant sequence is Cauchy, since |q−q|=0<ε for all m, n."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "G3",
            "equiv_def",
            "mko_real_def"
          ],
          "explanation": {
            "zh-Hant": "此步只採用建構；不同建構（例如 Dedekind 分割）的等價不在此處理。",
            "en": "This step only adopts the construction; equivalence with other constructions (such as Dedekind cuts) is not treated here."
          }
        },
        "leaf_basis": "equiv_def + cauchy_def + mko_real_def"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "thm_equiv_relation"
        ],
        "before_tex": "[(a_n)]=[(b_n)]",
        "after_tex": "(a_n)\\sim(b_n)",
        "relation": "equivalent",
        "rule": "thm_equiv_relation",
        "conditions": [
          "(a_n),(b_n)\\ \\text{Cauchy}"
        ],
        "explanation": {
          "zh-Hant": "∼ 是等價關係（外部引用 thm_equiv_relation），所以等價類有定義，而且兩個類相等恰好當兩個代表序列等價。之後要證兩個實數相等，只需證兩個序列等價；要證不相等，只需證不等價。",
          "en": "∼ is an equivalence relation (cited as thm_equiv_relation), so the classes are well defined, and two classes are equal exactly when their representative sequences are equivalent. From now on, to prove two reals equal it suffices to prove the sequences equivalent, and to prove them different, that the sequences are not equivalent."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "thm_equiv_relation"
          ],
          "explanation": {
            "zh-Hant": "這是等價類的基本性質，需要 ∼ 是等價關係；本稿不重證，標記 citation_verified。",
            "en": "This is the basic property of equivalence classes and needs ∼ to be an equivalence relation; it is not reproved here and is marked citation_verified."
          }
        },
        "leaf_basis": "thm_equiv_relation + def_unfold"
      },
      {
        "id": "S03",
        "uses": [
          "G4"
        ],
        "before_tex": "a_n=1-10^{-n}\\ \\ (n\\ge1)",
        "after_tex": "a_1=\\tfrac{9}{10},\\ a_2=\\tfrac{99}{100},\\ a_3=\\tfrac{999}{1000}",
        "relation": "definition",
        "rule": "def_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "定義 a_n=1−10^{−n}。每一項都是有理數（10^{−n}=1/10^n，且 10^n≠0）。前三項是 9/10、99/100、999/1000（N1 到 N3），也就是有限小數 0.9、0.99、0.999；一般地，a_n 是小數點後有 n 個 9 的有限小數。",
          "en": "Define a_n=1−10^{−n}. Every term is rational (10^{−n}=1/10^n with 10^n≠0). The first three terms are 9/10, 99/100, 999/1000 (N1 to N3), the finite decimals 0.9, 0.99, 0.999; in general a_n is the finite decimal with n nines after the point."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "def_intro",
            "q_ordered_field",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "此步只命名序列，並核對前三項。",
            "en": "This step only names the sequence and checks its first three terms."
          }
        },
        "leaf_basis": "def_intro + q_ordered_field + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "q_ordered_field"
        ],
        "before_tex": "a_n-1",
        "after_tex": "-10^{-n},\\quad |a_n-1|=10^{-n}",
        "relation": "equivalent",
        "rule": "q_ordered_field",
        "conditions": [],
        "explanation": {
          "zh-Hant": "a_n−1=(1−10^{−n})−1=−10^{−n}。10^n>0，所以 10^{−n}=1/10^n>0，|−10^{−n}|=10^{−n}。例：a_3−1=999/1000−1=−1/1000（N4）。",
          "en": "a_n−1=(1−10^{−n})−1=−10^{−n}. 10^n>0, so 10^{−n}=1/10^n>0 and |−10^{−n}|=10^{−n}. Example: a_3−1=999/1000−1=−1/1000 (N4)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "q_ordered_field",
            "add_inverse",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "代數恆等式與絕對值的定義，雙向成立。",
            "en": "An algebraic identity and the definition of absolute value; it holds both ways."
          }
        },
        "leaf_basis": "q_ordered_field + add_inverse + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "L_pow_ten"
        ],
        "before_tex": "n\\ge 1\\ \\text{an integer}",
        "after_tex": "10^n>n",
        "relation": "implies",
        "rule": "L_pow_ten",
        "conditions": [
          "n\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "引理 L_pow_ten：對每個整數 n≥1，10^n>n。以數學歸納法在 S05.1、S05.2 證明。N8 檢查 n=1 到 40，只是有限檢查，不是證明；證明是歸納法。",
          "en": "Lemma L_pow_ten: for every integer n≥1, 10^n>n. It is proved by induction in S05.1 and S05.2. N8 checks n=1 to 40; that is a finite check only, not a proof; the proof is the induction."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_pow_ten"
          ],
          "explanation": {
            "zh-Hant": "此步只引用引理；證明在子步。",
            "en": "This step only invokes the lemma; the proof is in the sub-steps."
          }
        },
        "leaf_basis": "L_pow_ten"
      },
      {
        "id": "S05.1",
        "uses": [
          "q_ordered_field"
        ],
        "before_tex": "n=1",
        "after_tex": "10^1=10>1",
        "relation": "induction_base",
        "rule": "nat_induction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "基礎情形 n=1：10¹=10>1（N5）。",
          "en": "Base case n=1: 10¹=10>1 (N5)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "nat_induction",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "歸納法需要基礎情形成立。",
            "en": "Induction needs the base case to hold."
          }
        },
        "leaf_basis": "nat_induction + numeric_evaluation"
      },
      {
        "id": "S05.2",
        "uses": [
          "S05.1",
          "q_ordered_field"
        ],
        "before_tex": "10^n>n\\ \\ (n\\ge1)",
        "after_tex": "10^{n+1}>n+1",
        "relation": "induction_step",
        "rule": "nat_induction",
        "conditions": [
          "n\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "設 10^n>n（n≥1）。則 10^{n+1}=10·10^n>10n（乘以正數 10 保持不等號）。又 10n−(n+1)=9n−1（N6 是恆等式），n≥1 時 9n−1≥9·1−1=8>0（N7），所以 10n>n+1。合起來 10^{n+1}>n+1。依歸納原理（S05.1 與此步），對所有 n≥1 有 10^n>n，證得引理 L_pow_ten。",
          "en": "Assume 10^n>n (n≥1). Then 10^{n+1}=10·10^n>10n (multiplying by the positive 10 preserves the inequality). Also 10n−(n+1)=9n−1 (N6 is an identity), and for n≥1, 9n−1≥9·1−1=8>0 (N7), so 10n>n+1. Together 10^{n+1}>n+1. By the induction principle (S05.1 and this step), 10^n>n for all n≥1, which proves Lemma L_pow_ten."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "nat_induction",
            "q_ordered_field",
            "power_laws"
          ],
          "explanation": {
            "zh-Hant": "歸納步需要 n≥1 才有 9n−1>0；n=0 時 9·0−1<0，但那不在範圍內。",
            "en": "The step needs n≥1 for 9n−1>0; at n=0, 9·0−1<0, but that case is outside the range."
          }
        },
        "leaf_basis": "nat_induction + power_laws + q_ordered_field"
      },
      {
        "id": "S06",
        "uses": [
          "L_small"
        ],
        "before_tex": "\\varepsilon\\in\\mathbb{Q},\\ \\varepsilon>0",
        "after_tex": "\\exists N\\ \\forall n\\ge N:\\ 10^{-n}<\\varepsilon",
        "relation": "implies",
        "rule": "L_small",
        "conditions": [
          "\\varepsilon>0"
        ],
        "explanation": {
          "zh-Hant": "引理 L_small：每個正有理數 ε 都有 N，使 n≥N 時 10^{−n}<ε。在 S06.1、S06.2 證明。這裡只用有理數的運算與序，沒有用到正在建構的實數，避免循環。",
          "en": "Lemma L_small: every positive rational ε has an N with 10^{−n}<ε for n≥N. It is proved in S06.1 and S06.2. Only rational arithmetic and order are used, not the reals being constructed, which avoids circularity."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_small"
          ],
          "explanation": {
            "zh-Hant": "此步只引用引理；證明在子步。",
            "en": "This step only invokes the lemma; the proof is in the sub-steps."
          }
        },
        "leaf_basis": "L_small"
      },
      {
        "id": "S06.1",
        "uses": [
          "q_ordered_field",
          "pos_int_ge_one"
        ],
        "before_tex": "\\varepsilon=\\frac{p}{q}>0,\\ p,q\\in\\mathbb{Z},\\ q\\neq0",
        "after_tex": "\\varepsilon=\\frac{r}{s},\\ r,s\\in\\mathbb{Z},\\ r\\ge1,\\ s\\ge1",
        "relation": "implies",
        "rule": "frac_rules",
        "conditions": [
          "\\varepsilon>0"
        ],
        "explanation": {
          "zh-Hant": "ε 是有理數，可寫成 p/q（q≠0）。若 q<0，依分數規則 p/q=(−p)/(−q)，所以可設 q>0。此時 p=εq 是兩個正數的積，p>0。令 r=p、s=q，兩者都是正整數，所以 r≥1、s≥1。",
          "en": "ε is rational, so ε=p/q with q≠0. If q<0, the fraction rule gives p/q=(−p)/(−q), so we may assume q>0. Then p=εq is a product of two positives, so p>0. Let r=p and s=q; both are positive integers, so r≥1 and s≥1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "frac_rules",
            "q_ordered_field",
            "pos_int_ge_one"
          ],
          "explanation": {
            "zh-Hant": "需要 ε>0 才能得到 r>0；ε 是有理數才能寫成整數之比。",
            "en": "ε>0 is needed to get r>0; ε being rational is what lets it be written as a ratio of integers."
          }
        },
        "leaf_basis": "frac_rules + q_ordered_field + pos_int_ge_one + exists_elim"
      },
      {
        "id": "S06.2",
        "uses": [
          "S06.1",
          "L_pow_ten"
        ],
        "before_tex": "N:=s,\\ n\\ge N",
        "after_tex": "10^{-n}<\\tfrac{1}{n}\\le\\tfrac{1}{s}\\le\\tfrac{r}{s}=\\varepsilon",
        "relation": "implies",
        "rule": "q_ordered_field",
        "conditions": [
          "n\\ge s\\ge 1",
          "r\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "取 N=s。對 n≥s：依 L_pow_ten，10^n>n>0，正數取倒數使不等號反向，10^{−n}=1/10^n<1/n；n≥s>0 推出 1/n≤1/s；r≥1 且 1/s>0 推出 1/s≤r/s=ε。串起來 10^{−n}<ε。例：ε=1/100 時 r=1、s=100，N=100 可用（N9）；引理只需要某個 N，不要求最小，例如 n=3 時已有 10^{−3}<1/100（N10）。S06.1 與 S06.2 合起來證得引理 L_small。",
          "en": "Take N=s. For n≥s: by L_pow_ten, 10^n>n>0, and taking reciprocals of positives reverses the inequality, so 10^{−n}=1/10^n<1/n; n≥s>0 gives 1/n≤1/s; r≥1 and 1/s>0 give 1/s≤r/s=ε. Chaining, 10^{−n}<ε. Example: for ε=1/100, r=1 and s=100, and N=100 works (N9); the lemma needs some N, not the least one; for instance already 10^{−3}<1/100 at n=3 (N10). S06.1 and S06.2 together prove Lemma L_small."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_pow_ten",
            "q_ordered_field",
            "S06.1",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "需要正數取倒數反轉不等號；若 n 可以為 0，1/n 沒有定義，所以 n≥s≥1 很重要。",
            "en": "It needs that reciprocals of positives reverse inequalities; if n could be 0, 1/n would be undefined, so n≥s≥1 matters."
          }
        },
        "leaf_basis": "q_ordered_field + L_pow_ten + exists_intro + forall_intro"
      },
      {
        "id": "S07",
        "uses": [
          "S04",
          "S06",
          "G2"
        ],
        "before_tex": "(a_n)_{n\\ge1}",
        "after_tex": "(a_n)\\ \\text{is Cauchy}",
        "relation": "implies",
        "rule": "cauchy_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "任取有理數 ε>0。對 ε/2（仍是正有理數）使用 L_small，得 N 使 n≥N 時 10^{−n}<ε/2。對 m,n≥N：|a_m−a_n|=|(a_m−1)+(1−a_n)|≤|a_m−1|+|a_n−1|=10^{−m}+10^{−n}<ε/2+ε/2=ε（三角不等式與 S04）。ε 是任意的，所以 (a_n) 是 Cauchy 序列。",
          "en": "Take any rational ε>0. Apply L_small to ε/2 (still a positive rational) to get N with 10^{−n}<ε/2 for n≥N. For m,n≥N: |a_m−a_n|=|(a_m−1)+(1−a_n)|≤|a_m−1|+|a_n−1|=10^{−m}+10^{−n}<ε/2+ε/2=ε (triangle inequality and S04). ε was arbitrary, so (a_n) is a Cauchy sequence."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S06",
            "q_ordered_field",
            "forall_intro",
            "cauchy_def"
          ],
          "explanation": {
            "zh-Hant": "需要對 ε/2 使用引理，使兩項各小於 ε/2；N 可以依賴 ε，這是定義允許的。",
            "en": "The lemma is applied to ε/2 so that each of the two terms is below ε/2; N may depend on ε, as the definition allows."
          }
        },
        "leaf_basis": "cauchy_def + q_ordered_field + forall_intro + exists_intro"
      },
      {
        "id": "S08",
        "uses": [
          "S01",
          "S04",
          "S06",
          "S07",
          "G3"
        ],
        "before_tex": "(a_n),\\ (1)_{n\\ge1}",
        "after_tex": "(a_n)\\sim(1)_{n\\ge1}",
        "relation": "implies",
        "rule": "equiv_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個序列都是 Cauchy 序列（S07 與 S01）。任取有理數 ε>0，L_small 給出 N，使 n≥N 時 |a_n−1|=10^{−n}<ε（S04）。ε 是任意的，依 G3，(a_n) 與常數列 1 等價。",
          "en": "Both sequences are Cauchy (S07 and S01). Take any rational ε>0; L_small gives N with |a_n−1|=10^{−n}<ε for n≥N (S04). ε was arbitrary, so by G3, (a_n) is equivalent to the constant sequence 1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S06",
            "equiv_def",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "等價要求對每個 ε 都找到 N；只看幾項（例如 a_3 與 1 差 1/1000）不夠。",
            "en": "Equivalence requires an N for every ε; looking at a few terms (for example a_3 differs from 1 by 1/1000) is not enough."
          }
        },
        "leaf_basis": "equiv_def + forall_intro + exists_intro"
      },
      {
        "id": "S09",
        "uses": [
          "S02",
          "S08"
        ],
        "before_tex": "(a_n)\\sim(1)_{n\\ge1}",
        "after_tex": "[(a_n)]=\\iota(1)",
        "relation": "equivalent",
        "rule": "thm_equiv_relation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 S02，序列等價恰好表示類相等，所以 [(a_n)]=[(1)_{n≥1}]=ι(1)。也就是說，0.9、0.99、0.999 等截斷組成的整個序列所定的實數，正是有理數 1 的嵌入；常說的「0.9 循環等於 1」在這個建構中就是這個等式。",
          "en": "By S02, equivalence of sequences is exactly equality of classes, so [(a_n)]=[(1)_{n≥1}]=ι(1). That is, the real number determined by the whole sequence of truncations 0.9, 0.99, 0.999 and so on is exactly the embedding of the rational 1; the usual statement '0.9 with the digit 9 repeating indefinitely equals 1' is, in this construction, this equality."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S02",
            "thm_equiv_relation"
          ],
          "explanation": {
            "zh-Hant": "S02 是雙條件，雙向成立。",
            "en": "S02 is a biconditional, so the step holds both ways."
          }
        },
        "leaf_basis": "thm_equiv_relation + def_unfold"
      },
      {
        "id": "S10",
        "uses": [
          "S02",
          "G3",
          "q_ordered_field"
        ],
        "before_tex": "p\\neq q\\ \\text{in}\\ \\mathbb{Q}",
        "after_tex": "\\iota(p)\\neq\\iota(q)",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "設 p≠q，令 δ=|p−q|>0（有理數）。若常數列 (p) 與 (q) 等價，取 ε=δ，存在 N 使 n≥N 時 |p−q|<δ，即 δ<δ，矛盾。所以兩個常數列不等價，依 S02，ι(p)≠ι(q)：ι 是單射，不同的有理數嵌入成不同的實數。",
          "en": "Let p≠q and δ=|p−q|>0 (a rational). If the constant sequences (p) and (q) were equivalent, taking ε=δ would give N with |p−q|<δ for n≥N, that is δ<δ, a contradiction. So the constant sequences are not equivalent, and by S02, ι(p)≠ι(q): ι is injective, and different rationals embed as different reals."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "contradiction",
            "q_ordered_field"
          ],
          "explanation": {
            "zh-Hant": "關鍵是常數列的差永遠是 δ，不會變小；ε 取 δ 就排除等價。",
            "en": "The key is that the absolute difference between corresponding terms of the constant sequences is always δ and never shrinks; choosing ε=δ rules out equivalence."
          }
        },
        "leaf_basis": "contradiction + q_ordered_field + equiv_def"
      },
      {
        "id": "S11",
        "uses": [
          "S10",
          "G4"
        ],
        "before_tex": "\\tfrac{999}{1000}\\neq 1",
        "after_tex": "\\iota(\\tfrac{999}{1000})\\neq\\iota(1)",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "999/1000≠1（兩者差 1/1000，N12、N13），依 S10，有限小數 0.999 所表示的實數 ι(999/1000) 不等於 1。等於 1 的是整個截斷序列的類（S09），不是任何一個有限截斷；有限截斷只是近似值。",
          "en": "999/1000≠1 (they differ by 1/1000, N12, N13), so by S10 the real ι(999/1000) denoted by the finite decimal 0.999 is not 1. What equals 1 is the class of the whole sequence of truncations (S09), not any single finite truncation; a finite truncation is only an approximation."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "forall_elim",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "S09 與 S11 不矛盾：S09 比較的是整個序列，S11 比較的是單一個常數。",
            "en": "S09 and S11 do not conflict: S09 compares a whole sequence, S11 a single constant."
          }
        },
        "leaf_basis": "forall_elim + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S01",
          "G4"
        ],
        "before_tex": "0.375",
        "after_tex": "\\tfrac{375}{1000}=\\tfrac{3}{8},\\quad \\iota(\\tfrac{3}{8})\\in\\mathbb{R}",
        "relation": "implies",
        "rule": "mko_real_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "有限小數 0.375=3/10+7/100+5/1000=375/1000=3/8（N14、N15），是有理數；它在 ℝ 中是 ι(3/8)，即常數列 3/8 的類。有限小數不是另一種獨立數系，更不是無理數。",
          "en": "The finite decimal 0.375=3/10+7/100+5/1000=375/1000=3/8 (N14, N15) is rational; in ℝ it is ι(3/8), the class of the constant sequence 3/8. A finite decimal is not a separate number system, and certainly not irrational."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_real_def",
            "equiv_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "只用到有限小數的意義與嵌入 ι。",
            "en": "Only the meaning of a finite decimal and the embedding ι are used."
          }
        },
        "leaf_basis": "mko_real_def + equiv_def + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S07",
          "S09",
          "S10",
          "S11",
          "S12"
        ],
        "before_tex": "\\text{S07, S09, S10, S11, S12}",
        "after_tex": "[(a_n)]=\\iota(1);\\ \\iota\\ \\text{injective};\\ \\iota(\\tfrac{999}{1000})\\neq\\iota(1);\\ 0.375=\\tfrac{3}{8}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "截斷序列 (a_n) 是 Cauchy 序列且與常數列 1 等價，所以它所定的實數就是 ι(1)；ι 是單射，所以單一的有限截斷 0.999 不等於 1；有限小數 0.375 是有理數 3/8 的嵌入。",
          "en": "The truncation sequence (a_n) is Cauchy and equivalent to the constant sequence 1, so the real it determines is ι(1); ι is injective, so the single finite truncation 0.999 is not 1; and the finite decimal 0.375 is the embedding of the rational 3/8."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S07",
            "S08",
            "S09",
            "S10",
            "S11",
            "S12"
          ],
          "explanation": {
            "zh-Hant": "目標是 goal 所列各項。足夠條件 H：S07、S09、S10、S11、S12 各自成立；H 推出目標只用合取。普遍敘述由任意 ε 的論證與歸納法得到，沒有把目標當作前提；唯一外部依賴是 thm_equiv_relation。",
            "en": "The goal is the list in the goal field. Sufficient condition H: S07, S09, S10, S11 and S12 each hold; H gives the goal by conjunction only. The universal statements come from arguments for arbitrary ε and from induction; the goal is never used as a premise, and the only external dependency is thm_equiv_relation."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在 Cauchy 完備化中，a_n=1−10^{−n} 是 Cauchy 序列，且對每個 ε>0 都有 N 使 |a_n−1|=10^{−n}<ε，所以 (a_n)∼(1)，[(a_n)]=ι(1)。ι 是單射，所以有限截斷 0.999=999/1000 所表示的實數不是 1。有限小數 0.375=3/8 是有理數，在 ℝ 中是 ι(3/8)。",
      "en": "In the Cauchy completion, a_n=1−10^{−n} is a Cauchy sequence, and for every ε>0 there is N with |a_n−1|=10^{−n}<ε, so (a_n)∼(1) and [(a_n)]=ι(1). ι is injective, so the real denoted by the finite truncation 0.999=999/1000 is not 1. The finite decimal 0.375=3/8 is rational and is ι(3/8) in ℝ."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "0.375 是無理數，因為它是小數",
          "en": "0.375 is irrational because it is a decimal"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "0.375=375/1000=3/8 是有理數（S12）。",
          "en": "0.375=375/1000=3/8 is rational (S12)."
        }
      },
      {
        "claim": {
          "zh-Hant": "有限小數 0.999（三個 9）等於 1",
          "en": "The finite decimal 0.999 (three nines) equals 1"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "999/1000≠1，而 ι 是單射，所以 ι(999/1000)≠ι(1)（S10、S11）。",
          "en": "999/1000≠1 and ι is injective, so ι(999/1000)≠ι(1) (S10, S11)."
        }
      },
      {
        "claim": {
          "zh-Hant": "截斷序列 0.9、0.99、0.999 等所定的實數比 1 小一點點",
          "en": "The real determined by the truncations 0.9, 0.99, 0.999 and so on is slightly less than 1"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(a_n)∼(1)，所以 [(a_n)]=ι(1)，兩者是同一個實數（S08、S09）。",
          "en": "(a_n)∼(1), so [(a_n)]=ι(1): they are the same real number (S08, S09)."
        }
      },
      {
        "claim": {
          "zh-Hant": "實數線沒有空隙，所以完備性已經證明",
          "en": "The real line has no gaps, so completeness is proved"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "「沒有空隙」是直覺，不是證明；本稿沒有證明完備性（MKO）。",
          "en": "'No gaps' is intuition, not proof; completeness is not proved in this guide (MKO)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "nat_induction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "數學歸納原理：對從 n₀ 起的命題族 P(n)，若 P(n₀) 成立，且對每個 n≥n₀ 都有 P(n)⇒P(n+1)，則所有 n≥n₀ 都有 P(n)。",
          "en": "Principle of induction: for a family P(n) starting at n₀, if P(n₀) holds and P(n)⇒P(n+1) for every n≥n₀, then P(n) holds for every n≥n₀."
        }
      },
      {
        "id": "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": "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": "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": "def_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。",
          "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable."
        }
      },
      {
        "id": "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": "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": "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": "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": "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": "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": "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_real_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-real-number",
          "sha256": "2bc8c0a1defae2276ce50808a0761bf4d62f26892e443fcab79fc8090113d880",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "實數系 R 可由有理數的 Cauchy 完備化建構；每個有理數都可嵌入 R，而不同建構仍須證明彼此等價。",
          "en": "The real-number system R can be constructed as a Cauchy completion of the rationals. Every rational number embeds into R, while different constructions still require proofs of equivalence."
        }
      },
      {
        "id": "q_ordered_field",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "ℚ 是有序域：域的運算規則、序的規則（三一律、遞移、加同數保序、乘正數保序、正數取倒數使不等號反向）與絕對值規則（|x|≥0、|−x|=|x|、三角不等式）都在 ℚ 中成立。",
          "en": "ℚ is an ordered field: the field rules, the order rules (trichotomy, transitivity, adding preserves order, multiplying by a positive preserves order, reciprocals of positives reverse order) and the absolute-value rules (|x|≥0, |−x|=|x|, the triangle inequality) all hold in ℚ."
        }
      },
      {
        "id": "pos_int_ge_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "每個正整數都 ≥1。",
          "en": "Every positive integer is ≥1."
        }
      },
      {
        "id": "cauchy_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "Cauchy 序列：對每個有理數 ε>0，存在 N，使所有 m,n≥N 都有 |a_m−a_n|<ε。",
          "en": "Cauchy sequence: for every rational ε>0 there is N with |a_m−a_n|<ε for all m,n≥N."
        }
      },
      {
        "id": "equiv_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "(a_n)∼(b_n) 表示對每個有理數 ε>0，存在 N，使所有 n≥N 都有 |a_n−b_n|<ε；ℝ 是等價類的集合，ι(q)=[(q)_{n≥1}]。",
          "en": "(a_n)∼(b_n) means that for every rational ε>0 there is N with |a_n−b_n|<ε for all n≥N; ℝ is the set of equivalence classes, and ι(q)=[(q)_{n≥1}]."
        }
      },
      {
        "id": "thm_equiv_relation",
        "kind": "cited",
        "source": {
          "reference": "Keith Conrad, Metric Spaces, Section 8 (completions), p. 39",
          "original_reference": "T. Tao, Analysis I, 3rd edition (Hindustan Book Agency and Springer, 2016), Section 5.3, Proposition 5.3.3 (formal limits are well defined)",
          "url": "https://kconrad.math.uconn.edu/blurbs/analysis/metricspaces.pdf#page=39",
          "locator": "Definition of equivalence on Cauchy sequences and quotient construction",
          "checked_date": "2026-10-19",
          "verification": "verified_replacement",
          "verification_note": "Specialize the stated metric construction to the rational metric |x-y|; class equality is the equivalence relation."
        },
        "needs_citation": false,
        "statement": {
          "zh-Hant": "有理數 Cauchy 序列之間的等價 ∼ 是等價關係（自反、對稱、遞移）；因此等價類有定義，且 [(a_n)]=[(b_n)] ⇔ (a_n)∼(b_n)。",
          "en": "Equivalence ∼ of Cauchy sequences of rationals is an equivalence relation (reflexive, symmetric, transitive); hence equivalence classes are well defined, and [(a_n)]=[(b_n)] ⇔ (a_n)∼(b_n)."
        }
      },
      {
        "id": "L_pow_ten",
        "kind": "lemma",
        "proved_in": [
          "S05.1",
          "S05.2"
        ],
        "statement": {
          "zh-Hant": "對每個整數 n≥1：10^n>n。",
          "en": "For every integer n≥1: 10^n>n."
        }
      },
      {
        "id": "L_small",
        "kind": "lemma",
        "proved_in": [
          "S06.1",
          "S06.2"
        ],
        "statement": {
          "zh-Hant": "對每個有理數 ε>0，存在正整數 N，使所有整數 n≥N 都有 10^{−n}<ε。",
          "en": "For every rational ε>0 there is a positive integer N such that 10^{−n}<ε for every integer n≥N."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "1-10**(-1)",
        "rhs": "9/10"
      },
      {
        "id": "N2",
        "step": "S03",
        "lhs": "1-10**(-2)",
        "rhs": "99/100"
      },
      {
        "id": "N3",
        "step": "S03",
        "lhs": "1-10**(-3)",
        "rhs": "999/1000"
      },
      {
        "id": "N4",
        "step": "S04",
        "lhs": "999/1000-1",
        "rhs": "-1/1000"
      },
      {
        "id": "N5",
        "step": "S05.1",
        "lhs": "10**1",
        "relation": ">",
        "rhs": "1"
      },
      {
        "id": "N6",
        "step": "S05.2",
        "vars": [
          "n"
        ],
        "degree": 1,
        "lhs": "10*n-(n+1)",
        "rhs": "9*n-1"
      },
      {
        "id": "N7",
        "step": "S05.2",
        "lhs": "9*1-1",
        "rhs": "8"
      },
      {
        "id": "N8",
        "step": "S05",
        "vars": [
          "n"
        ],
        "range": [
          1,
          40
        ],
        "lhs": "10**n",
        "relation": ">",
        "rhs": "n"
      },
      {
        "id": "N9",
        "step": "S06.2",
        "lhs": "10**(-100)",
        "relation": "<",
        "rhs": "1/100"
      },
      {
        "id": "N10",
        "step": "S06.2",
        "lhs": "10**(-3)",
        "relation": "<",
        "rhs": "1/100"
      },
      {
        "id": "N11",
        "step": "S07",
        "lhs": "1/200+1/200",
        "rhs": "1/100"
      },
      {
        "id": "N12",
        "step": "S11",
        "lhs": "abs(1-999/1000)",
        "rhs": "1/1000"
      },
      {
        "id": "N13",
        "step": "S11",
        "lhs": "999/1000",
        "relation": "!=",
        "rhs": "1"
      },
      {
        "id": "N14",
        "step": "S12",
        "lhs": "3/10+7/100+5/1000",
        "rhs": "375/1000"
      },
      {
        "id": "N15",
        "step": "S12",
        "lhs": "375/1000",
        "rhs": "3/8"
      },
      {
        "id": "N16",
        "lhs": "1.4142**2",
        "rhs": "1.99996164"
      },
      {
        "id": "N17",
        "lhs": "1.4142**2",
        "relation": "!=",
        "rhs": "2"
      },
      {
        "id": "N18",
        "lhs": "9/10",
        "relation": "!=",
        "rhs": "1"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 假設 a1 的 Cauchy 完備化語境中工作，只用有理數的運算、序與絕對值（不使用正在建構的實數的性質，避免循環）。「∼ 是等價關係」以外部文獻引用（citation_verified）。不定義 ℝ 上的加法、乘法與序，不證明完備性，也不證明與 Dedekind 分割等其他建構的等價（MKO 指出這需要另外證明）；不改動 canonical MKO 或英文 edition。",
      "en": "Works in the Cauchy-completion setting of MKO assumption a1, using only arithmetic, order and absolute value of rationals (no property of the reals being constructed, to avoid circularity). That ∼ is an equivalence relation is cited from outside (citation_verified). It does not define addition, multiplication or order on ℝ, does not prove completeness, and does not prove equivalence with other constructions such as Dedekind cuts (which the MKO says needs a separate proof); it does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "(a_n)_{n\\ge1}",
        "有理數列",
        "a sequence of rationals"
      ],
      [
        "(a_n)\\sim(b_n)",
        "兩個 Cauchy 序列等價",
        "two Cauchy sequences are equivalent"
      ],
      [
        "[(a_n)]",
        "(a_n) 的等價類，即一個實數",
        "the equivalence class of (a_n), that is, a real number"
      ],
      [
        "\\iota(q)",
        "有理數 q 的嵌入：常數列 q 的類",
        "the embedding of a rational q: the class of the constant sequence q"
      ],
      [
        "\\varepsilon,\\ \\delta",
        "正有理數",
        "positive rationals"
      ]
    ],
    "md_error_controls": [
      {
        "title": "有限小數是有理數 / Finite decimals are rational",
        "zh": "MKO 指出的誤解之一是把有限小數視為無理數。0.375=375/1000=3/8（S12），任何有限小數都是「整數除以 10 的乘冪」，因此是有理數；它在 ℝ 中就是某個 ι(q)。",
        "en": "One misconception the MKO names is treating finite decimals as irrational. 0.375=375/1000=3/8 (S12); every finite decimal is an integer divided by a power of 10 and so is rational; in ℝ it is some ι(q)."
      },
      {
        "title": "近似小數不是精確值 / An approximate decimal is not the exact value",
        "zh": "MKO 的另一誤解是把近似小數當成精確值。0.999 與 1 只差 1/1000，但 ι(999/1000)≠ι(1)（S11）；等於 1 的是整個截斷序列的類（S09）。同理，有理數 1.4142 的平方是 1.99996164≠2（N16、N17），所以它不是 2 的平方根，只是近似。",
        "en": "Another misconception the MKO names is taking an approximate decimal for the exact value. 0.999 and 1 differ by only 1/1000, yet ι(999/1000)≠ι(1) (S11); what equals 1 is the class of the whole sequence of truncations (S09). Likewise the rational 1.4142 has square 1.99996164≠2 (N16, N17), so it is not a square root of 2, only an approximation."
      },
      {
        "title": "「沒有空隙」不是完備性的證明 / 'No gaps' is not a proof of completeness",
        "zh": "MKO 提醒：只說實數線「沒有空隙」便宣稱完備性已證明，是誤解；『補入極限點』只是建構直覺。本稿建構了具體的實數並證明一個等式與 ι 的單射性，但沒有定義 ℝ 上的序與運算，也沒有證明完備性（例如每個有上界的非空集合有最小上界），更沒有證明與 Dedekind 分割等其他建構的等價。這些都需要另外的定義與證明。",
        "en": "The MKO warns that claiming completeness is proved merely by saying the real line 'has no gaps' is a misconception; 'adding limit points' is only construction intuition. This guide constructs specific reals and proves one equality and the injectivity of ι, but it defines neither order nor arithmetic on ℝ, does not prove completeness (for instance that every nonempty set bounded above has a least upper bound), and does not prove equivalence with other constructions such as Dedekind cuts. Each of these needs its own definitions and proofs."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "ℝ 上的加法、乘法、序與完備性，以及與 Dedekind 分割等其他建構的等價，未在本稿處理（MKO 指出需另外證明）。",
        "en": "Addition, multiplication, order and completeness of ℝ, and equivalence with other constructions such as Dedekind cuts, are not treated here (the MKO says they need separate proofs)."
      },
      {
        "zh-Hant": "與 Mathlib Real 的 Cauchy completion、有理數嵌入、線性序與完備性聲明的對接（MKO formal next）未在此處理。",
        "en": "The link to Mathlib's Real (Cauchy completion, rational embedding, linear order and completeness statements; MKO formal next) is not treated here."
      },
      {
        "zh-Hant": "N8 只是 n=1 到 40 的有限檢查；10^n>n 的證明是 S05.1、S05.2 的歸納法。",
        "en": "N8 is a finite check for n=1 to 40 only; the proof of 10^n>n is the induction in S05.1 and S05.2."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "cf5c66c9854af5d95d9c38472102f8e076b1a986cf8bbfbef4f0f539680fda81",
    "original_source_sha256": "cf5c66c9854af5d95d9c38472102f8e076b1a986cf8bbfbef4f0f539680fda81",
    "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."
  }
}