{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-natural-number-01",
  "mko_id": "mko-natural-number",
  "title": {
    "zh-Hant": "自然數：以最小歸納集判定成員，並分辨數與數字符號",
    "en": "Natural number: deciding membership through the smallest inductive set, and telling numbers from numerals"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-natural-number-01",
    "source": {
      "mko_id": "mko-natural-number",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "222b72502fb661f3372012f602fd2f1eeef754b9e2069498c6b24234778730c4",
      "english_sha256": "f7444a7bfebcfed594478ed5e7c0f36cb1985114de58b5b473beddd1961c0bbc"
    },
    "mode": "definition_walkthrough",
    "audience_prerequisites": [
      "order_basics",
      "subset_def",
      "set_operations_def",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "依 OCME 慣例（0∈N），把 MKO 的「0 以及由 0 經有限次後繼所得的數」讀成「實數中包含 0 且對 x↦x+1 封閉的最小集合」。據此證明 0∈N、3∈N；說明二進位符號「11」與十進位符號「3」表示同一個自然數；證明 −1∉N 與 1/2∉N；並與「自然數從 1 開始」的慣例 N⁺=N∖{0} 比較最小元素。",
      "en": "Under the OCME convention (0∈N), read the MKO's '0 and the numbers obtained from 0 by finitely many successor steps' as 'the smallest set of reals that contains 0 and is closed under x↦x+1'. On this basis prove 0∈N and 3∈N; show that the binary numeral '11' and the decimal numeral '3' denote the same natural number; prove −1∉N and 1/2∉N; and compare least elements with the convention N⁺=N∖{0}, in which natural numbers start at 1."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "討論的數都是實數；後繼運算是 S(x)=x+1。n 表示任意一個待判定是否屬於 N 的數（MKO 假設 a2）。",
          "en": "The numbers discussed are real numbers; the successor operation is S(x)=x+1. n denotes an arbitrary number whose membership in N is under consideration (MKO assumption a2)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "MKO 的形式讀法：子集 T⊆ℝ 稱為歸納集，若 0∈T，且對每個 x∈T 都有 x+1∈T。N 定義為 ℝ 的所有歸納子集的交集，即包含 0 且對後繼封閉的最小集合。依本條目慣例，N 包含 0（MKO 假設 a1）。",
          "en": "Formal reading of the MKO: a subset T⊆ℝ is inductive if 0∈T and x+1∈T for every x∈T. N is defined as the intersection of all inductive subsets of ℝ, that is, the smallest set containing 0 and closed under successor. By this entry's convention N contains 0 (MKO assumption a1)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "位值記數：以 b 為底、由右至左數字依序為 d₀、d₁ 等直到 d_k 的數字符號，表示的數是 Σ_{i=0}^{k} d_i b^i。",
          "en": "Positional notation: a base-b numeral whose digits, read from right to left, are d₀, d₁ and so on up to d_k denotes the number Σ_{i=0}^{k} d_i b^i."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "要判定的數：0、3、−1、1/2；數字符號：十進位「3」、二進位「11」。",
          "en": "Numbers to decide: 0, 3, −1, 1/2; numerals: the decimal '3' and the binary '11'."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_nat_def"
        ],
        "before_tex": "N",
        "after_tex": "N=\\bigcap\\{T\\subseteq\\mathbb{R}\\mid 0\\in T,\\ \\forall x\\in T:\\ x+1\\in T\\}",
        "relation": "definition",
        "rule": "nat_as_intersection",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 說 N 包含 0 與由 0 經有限次後繼得到的數。若把「有限次」直接當定義會循環（必須先知道什麼是有限，也就是自然數），所以採 G2 的讀法：N 是所有歸納集的交集。這個集合族不空，因為 ℝ 本身是歸納集（0∈ℝ，x∈ℝ ⇒ x+1∈ℝ），所以交集有定義且 N⊆ℝ。",
          "en": "The MKO says N contains 0 and the numbers obtained from 0 by finitely many successor steps. Taking 'finitely many' literally as the definition would be circular (one would already need to know what finite, that is natural, means), so the reading of G2 is used: N is the intersection of all inductive sets. The family is nonempty because ℝ itself is inductive (0∈ℝ, and x∈ℝ ⇒ x+1∈ℝ), so the intersection is defined and N⊆ℝ."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "nat_as_intersection",
            "mko_nat_def",
            "big_intersection_def"
          ],
          "explanation": {
            "zh-Hant": "此步只採用定義；它與 MKO 文字讀法的等價列為開放項目，不在此證明。",
            "en": "This step only adopts the definition; its equivalence with the MKO's wording is listed as an open item and not proved here."
          }
        },
        "leaf_basis": "nat_as_intersection + inductive_def + big_intersection_def"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "L_N_inductive"
        ],
        "before_tex": "N=\\bigcap\\{T\\ \\text{inductive}\\}",
        "after_tex": "0\\in N,\\ \\forall x\\in N:\\ x+1\\in N",
        "relation": "implies",
        "rule": "L_N_inductive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_N_inductive，N 本身含 0 並對後繼封閉；這兩點正是 MKO 文字中的「包含 0」與「自然數的後繼仍是自然數」。0∈N 也就是 OCME 的慣例（假設 a1）。引理在 S02.1、S02.2 證明。",
          "en": "By Lemma L_N_inductive, N itself contains 0 and is closed under successor; these are exactly 'contains 0' and 'the successor of a natural number is natural' in the MKO's wording. 0∈N is the OCME convention (assumption a1). The lemma is proved in S02.1 and S02.2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_N_inductive",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "此步只引用引理；兩個部分各自在子步證明。",
            "en": "This step only invokes the lemma; its two parts are proved in the sub-steps."
          }
        },
        "leaf_basis": "L_N_inductive"
      },
      {
        "id": "S02.1",
        "uses": [
          "S01",
          "big_intersection_def"
        ],
        "before_tex": "T\\ \\text{inductive}\\Rightarrow 0\\in T",
        "after_tex": "0\\in N",
        "relation": "implies",
        "rule": "big_intersection_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "每個歸納集都含 0（歸納集定義的第一條），所以 0 屬於族中每一個集合；依交集的定義，0∈N。",
          "en": "Every inductive set contains 0 (the first clause of the definition), so 0 belongs to every set of the family; by the definition of intersection, 0∈N."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "inductive_def",
            "big_intersection_def"
          ],
          "explanation": {
            "zh-Hant": "需要「每一個」歸納集都含 0；只找到一個含 0 的集合不夠。",
            "en": "It needs every inductive set to contain 0; one set containing 0 would not be enough."
          }
        },
        "leaf_basis": "inductive_def + big_intersection_def + forall_intro"
      },
      {
        "id": "S02.2",
        "uses": [
          "S01",
          "big_intersection_def"
        ],
        "before_tex": "x\\in N",
        "after_tex": "x+1\\in N",
        "relation": "implies",
        "rule": "big_intersection_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "設 x∈N，任取一個歸納集 T。因 x 屬於每個歸納集，x∈T；T 對後繼封閉，所以 x+1∈T。T 是任意取的（沒有額外假設），所以 x+1 屬於每個歸納集，即 x+1∈N。S02.1 與 S02.2 合起來證得引理 L_N_inductive。",
          "en": "Let x∈N and take any inductive set T. Since x belongs to every inductive set, x∈T; T is closed under successor, so x+1∈T. T was arbitrary (no extra assumption), so x+1 belongs to every inductive set, that is x+1∈N. S02.1 and S02.2 together prove Lemma L_N_inductive."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "inductive_def",
            "big_intersection_def",
            "forall_intro",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "T 必須是任意的；若只對某一個歸納集驗證，不能推出 x+1 屬於交集。",
            "en": "T must be arbitrary; checking a single inductive set would not put x+1 into the intersection."
          }
        },
        "leaf_basis": "forall_elim + forall_intro + inductive_def + big_intersection_def"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "big_intersection_def"
        ],
        "before_tex": "T\\subseteq\\mathbb{R}\\ \\text{inductive}",
        "after_tex": "N\\subseteq T",
        "relation": "implies",
        "rule": "big_intersection_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "交集包含於族中的每一個集合：若 x∈N，x 屬於每個歸納集，特別屬於 T。這就是本稿用的歸納原理：要證每個自然數都在某個集合 T 中，只需證明 T 是歸納集。與 S02 合看，N 是最小的歸納集。",
          "en": "An intersection is contained in each set of the family: if x∈N, x belongs to every inductive set, in particular to T. This is the induction principle used in this guide: to show every natural number lies in a set T, it suffices to show that T is inductive. Together with S02, N is the smallest inductive set."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "big_intersection_def",
            "subset_def",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "只需要 T 是歸納集；不需要也不宣稱 T⊆N。",
            "en": "It needs only that T is inductive; T⊆N is neither needed nor claimed."
          }
        },
        "leaf_basis": "big_intersection_def + subset_def + forall_elim"
      },
      {
        "id": "S04",
        "uses": [
          "G4"
        ],
        "before_tex": "3",
        "after_tex": "((0+1)+1)+1",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 3 寫成從 0 出發的三次後繼：0+1=1，1+1=2，2+1=3，所以 3=S(S(S(0)))。",
          "en": "Write 3 as three successor steps from 0: 0+1=1, 1+1=2, 2+1=3, so 3=S(S(S(0)))."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確整數運算，雙向成立（N1）。",
            "en": "Exact integer arithmetic, valid both ways (N1)."
          }
        },
        "leaf_basis": "numeric_evaluation + add_assoc"
      },
      {
        "id": "S05",
        "uses": [
          "S02",
          "S04"
        ],
        "before_tex": "0\\in N",
        "after_tex": "1\\in N,\\ 2\\in N,\\ 3\\in N",
        "relation": "implies",
        "rule": "L_N_inductive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由 0∈N 與後繼封閉（S02）連用三次：0∈N ⇒ 1∈N ⇒ 2∈N ⇒ 3∈N。這正是 MKO 所說「經有限次（這裡是三次）後繼所得」。",
          "en": "Using 0∈N and closure under successor (S02) three times: 0∈N ⇒ 1∈N ⇒ 2∈N ⇒ 3∈N. This is exactly what the MKO means by 'obtained by finitely many (here three) successor steps'."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S04",
            "modus_ponens"
          ],
          "explanation": {
            "zh-Hant": "每一步都是肯定前件；反向（由 3∈N 推回 2∈N）並未宣稱，也不需要。",
            "en": "Each step is modus ponens; the backward direction (from 3∈N back to 2∈N) is neither claimed nor needed."
          }
        },
        "leaf_basis": "L_N_inductive + modus_ponens + eq_replace"
      },
      {
        "id": "S06",
        "uses": [
          "G3",
          "S05"
        ],
        "before_tex": "\\text{binary numeral }11,\\ \\text{decimal numeral }3",
        "after_tex": "1\\cdot2^1+1\\cdot2^0=3=3\\cdot10^0",
        "relation": "equivalent",
        "rule": "positional_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依位值記數，二進位符號「11」表示 1·2¹+1·2⁰=2+1=3，十進位符號「3」表示 3·10⁰=3。兩個不同的符號表示同一個數 3，而 3∈N（S05）。是否屬於 N 是數的性質，不是符號的性質。",
          "en": "By positional notation, the binary numeral '11' denotes 1·2¹+1·2⁰=2+1=3, and the decimal numeral '3' denotes 3·10⁰=3. Two different numerals denote the same number 3, and 3∈N (S05). Membership in N is a property of the number, not of the numeral."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "positional_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "兩個求值都是精確計算（N2、N3），等式雙向成立。",
            "en": "Both evaluations are exact computations (N2, N3), so the equality holds both ways."
          }
        },
        "leaf_basis": "positional_def + power_laws + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "inductive_def"
        ],
        "before_tex": "T_1=\\{x\\in\\mathbb{R}\\mid x\\ge 0\\}",
        "after_tex": "T_1\\ \\text{inductive}",
        "relation": "implies",
        "rule": "inductive_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "0≥0，所以 0∈T₁。若 x≥0，則 x+1≥0+1=1>0（序的加法性質與 0<1），所以 x+1∈T₁。兩條都成立，T₁ 是歸納集。",
          "en": "0≥0, so 0∈T₁. If x≥0 then x+1≥0+1=1>0 (order and addition, and 0<1), so x+1∈T₁. Both clauses hold, so T₁ is inductive."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "inductive_def",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要 0<1；若「後繼」是 x↦x−1，T₁ 就不再封閉。",
            "en": "It needs 0<1; if the 'successor' were x↦x−1, T₁ would no longer be closed."
          }
        },
        "leaf_basis": "inductive_def + order_basics + add_zero"
      },
      {
        "id": "S08",
        "uses": [
          "S03",
          "S07",
          "S02"
        ],
        "before_tex": "T_1\\ \\text{inductive}",
        "after_tex": "N\\subseteq T_1:\\ \\forall n\\in N,\\ n\\ge 0",
        "relation": "implies",
        "rule": "big_intersection_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 S03 的歸納原理，N⊆T₁：每個自然數都 ≥0。配合 0∈N（S02），0 是 N 的最小元素。",
          "en": "By the induction principle S03, N⊆T₁: every natural number is ≥0. Together with 0∈N (S02), 0 is the least element of N."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S07",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "只用到 T₁ 是歸納集；「最小元素」還需要 0 本身在 N 中。",
            "en": "It uses only that T₁ is inductive; 'least element' also needs 0 itself to be in N."
          }
        },
        "leaf_basis": "big_intersection_def + subset_def + order_basics"
      },
      {
        "id": "S09",
        "uses": [
          "S08",
          "G4"
        ],
        "before_tex": "-1<0",
        "after_tex": "-1\\notin N",
        "relation": "implies",
        "rule": "contrapositive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "−1<0，所以 −1∉T₁；因 N⊆T₁，依包含關係的逆否（x∈N ⇒ x∈T₁ 等價於 x∉T₁ ⇒ x∉N），−1∉N。注意 −1+1=0∈N 並不使 −1 成為自然數：後繼封閉只往前推，不往回推。",
          "en": "−1<0, so −1∉T₁; since N⊆T₁, the contrapositive of inclusion (x∈N ⇒ x∈T₁ is equivalent to x∉T₁ ⇒ x∉N) gives −1∉N. Note that −1+1=0∈N does not make −1 natural: closure under successor pushes forwards, not backwards."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "contrapositive",
            "subset_def"
          ],
          "explanation": {
            "zh-Hant": "逆否命題與原命題等價；不宣稱逆命題（x∈T₁ ⇒ x∈N），事實上 1/2∈T₁ 卻不是自然數。",
            "en": "The contrapositive is equivalent to the original; the converse (x∈T₁ ⇒ x∈N) is not claimed, and indeed 1/2∈T₁ is not natural."
          }
        },
        "leaf_basis": "contrapositive + subset_def + order_basics"
      },
      {
        "id": "S10",
        "uses": [
          "G1",
          "inductive_def"
        ],
        "before_tex": "T_2=\\{0\\}\\cup\\{x\\in\\mathbb{R}\\mid x\\ge 1\\}",
        "after_tex": "T_2\\ \\text{inductive}",
        "relation": "implies",
        "rule": "inductive_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "要排除 1/2，T₁ 太粗（1/2∈T₁），需要更細的歸納集 T₂。T₂ 含 0；對後繼的封閉性分兩種情況在 S10.1、S10.2 證明，因 T₂ 的元素只有 x=0 與 x≥1 兩類。",
          "en": "To exclude 1/2, T₁ is too coarse (1/2∈T₁), so a finer inductive set T₂ is needed. T₂ contains 0; closure under successor is proved by two cases in S10.1 and S10.2, since the elements of T₂ are of two kinds, x=0 and x≥1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "inductive_def",
            "case_analysis",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "x∈T₂ 只有兩種可能，兩種情況都必須處理，否則封閉性未證完。",
            "en": "x∈T₂ has only two possibilities, and both cases must be handled, otherwise closure is not fully proved."
          }
        },
        "leaf_basis": "inductive_def + case_analysis + set_operations_def"
      },
      {
        "id": "S10.1",
        "uses": [
          "S10"
        ],
        "before_tex": "x=0",
        "after_tex": "x+1=1\\in T_2",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x=0 時 x+1=1，而 1≥1，所以 1∈T₂。",
          "en": "If x=0 then x+1=1, and 1≥1, so 1∈T₂."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "add_zero"
          ],
          "explanation": {
            "zh-Hant": "直接計算，沒有附加條件。",
            "en": "Direct computation with no side condition."
          }
        },
        "leaf_basis": "order_basics + add_zero + set_operations_def"
      },
      {
        "id": "S10.2",
        "uses": [
          "S10",
          "S10.1"
        ],
        "before_tex": "x\\ge 1",
        "after_tex": "x+1\\ge 2\\ge 1,\\ x+1\\in T_2",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x≥1 時 x+1≥1+1=2≥1，所以 x+1∈T₂。兩種情況涵蓋 T₂ 的全部元素，依分情況論證 T₂ 對後繼封閉；加上 0∈T₂，T₂ 是歸納集。",
          "en": "If x≥1 then x+1≥1+1=2≥1, so x+1∈T₂. The two cases cover all elements of T₂, so by proof by cases T₂ is closed under successor; with 0∈T₂, T₂ is inductive."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "case_analysis",
            "S10.1"
          ],
          "explanation": {
            "zh-Hant": "用到序的加法性質 u<v ⇒ u+w<v+w（等號情況同理）；兩種情況合起來才得到封閉性。",
            "en": "It uses that adding the same number preserves order (and equality); only both cases together give closure."
          }
        },
        "leaf_basis": "order_basics + case_analysis"
      },
      {
        "id": "S11",
        "uses": [
          "S03",
          "S10",
          "S10.2"
        ],
        "before_tex": "T_2\\ \\text{inductive}",
        "after_tex": "N\\subseteq T_2",
        "relation": "implies",
        "rule": "big_intersection_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依歸納原理 S03，N⊆T₂：每個自然數要不是 0，就是 ≥1。所以 0 與 1 之間沒有自然數。",
          "en": "By the induction principle S03, N⊆T₂: every natural number is either 0 or ≥1. So there is no natural number strictly between 0 and 1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S10.2"
          ],
          "explanation": {
            "zh-Hant": "只需 T₂ 是歸納集；不宣稱 T₂⊆N。",
            "en": "Only that T₂ is inductive is needed; T₂⊆N is not claimed."
          }
        },
        "leaf_basis": "big_intersection_def + subset_def"
      },
      {
        "id": "S12",
        "uses": [
          "S11",
          "G4"
        ],
        "before_tex": "\\tfrac{1}{2}\\neq 0,\\ \\tfrac{1}{2}<1",
        "after_tex": "\\tfrac{1}{2}\\notin N",
        "relation": "implies",
        "rule": "contrapositive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "1/2≠0，且 1/2<1，所以 1/2∉T₂；因 N⊆T₂，依逆否 1/2∉N。這不能靠列出 0、1、2、3 再觀察 1/2 不在其中來證：列出有限個元素永遠列不完 N，必須用歸納集的論證。",
          "en": "1/2≠0 and 1/2<1, so 1/2∉T₂; since N⊆T₂, the contrapositive gives 1/2∉N. This cannot be proved by listing 0, 1, 2, 3 and observing that 1/2 is not among them: listing finitely many elements never exhausts N, so an inductive-set argument is required."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "contrapositive",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "1/2 必須同時不等於 0 且不 ≥1，才不在 T₂ 的兩個部分之中。",
            "en": "1/2 must both differ from 0 and fail to be ≥1 in order to lie in neither part of T₂."
          }
        },
        "leaf_basis": "contrapositive + subset_def + order_basics + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S02",
          "S05",
          "S08",
          "S11"
        ],
        "before_tex": "N^{+}=N\\setminus\\{0\\}",
        "after_tex": "0\\notin N^{+},\\ 3\\in N^{+},\\ \\min N^{+}=1,\\ \\min N=0",
        "relation": "definition",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "有些教材所說的自然數是 N⁺=N∖{0}（從 1 開始）。依集合差的定義 0∉N⁺；3∈N 且 3≠0，所以 3∈N⁺。由 N⊆T₂，N⁺ 的元素都 ≥1，而 1∈N（S05），所以 N⁺ 的最小元素是 1；在 OCME 的 N 中最小元素是 0（S08）。同一句「最小的自然數」在兩個慣例下答案不同。",
          "en": "Some textbooks mean N⁺=N∖{0} (starting at 1) by natural numbers. By the definition of set difference 0∉N⁺; 3∈N and 3≠0, so 3∈N⁺. Since N⊆T₂, every element of N⁺ is ≥1, and 1∈N (S05), so the least element of N⁺ is 1; in OCME's N the least element is 0 (S08). The same phrase 'the least natural number' has different answers under the two conventions."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "set_operations_def",
            "S11",
            "S08",
            "S05"
          ],
          "explanation": {
            "zh-Hant": "兩個最小元素的結論各自需要「下界」與「下界本身屬於集合」兩件事。",
            "en": "Each least-element conclusion needs two things: a lower bound, and the bound itself belonging to the set."
          }
        },
        "leaf_basis": "set_operations_def + def_intro + order_basics"
      },
      {
        "id": "S14",
        "uses": [
          "S05",
          "S06",
          "S09",
          "S12",
          "S13"
        ],
        "before_tex": "\\text{S05, S06, S09, S12, S13}",
        "after_tex": "0,3\\in N;\\ 11_{2}=3_{10};\\ -1,\\tfrac{1}{2}\\notin N;\\ \\min N=0,\\ \\min N^{+}=1",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 OCME 慣例下 0、3 是自然數；「11」（二進位）與「3」（十進位）是同一個自然數的兩種寫法；−1 與 1/2 不是自然數；N 的最小元素是 0，而 N⁺ 慣例下是 1。",
          "en": "Under the OCME convention 0 and 3 are natural numbers; '11' (binary) and '3' (decimal) are two ways of writing the same natural number; −1 and 1/2 are not natural numbers; the least element of N is 0, while under the N⁺ convention it is 1."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S03",
            "S05",
            "S06",
            "S09",
            "S12",
            "S13"
          ],
          "explanation": {
            "zh-Hant": "目標是 goal 所列各項。足夠條件 H：S05、S06、S09、S12、S13 各自成立；H 推出目標只用合取。排除 −1、1/2 的部分依賴 S03 的歸納原理與兩個歸納集，沒有把目標當作前提。",
            "en": "The goal is the list in the goal field. Sufficient condition H: S05, S06, S09, S12 and S13 each hold; H gives the goal by conjunction only. Excluding −1 and 1/2 relies on the induction principle S03 and the two inductive sets; the goal is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "把 N 讀成實數中含 0 且對 x↦x+1 封閉的最小集合：0∈N，3=S(S(S(0)))∈N；二進位「11」與十進位「3」都表示 3；因 N⊆{x ∣ x≥0}，−1∉N；因 N⊆{0}∪{x ∣ x≥1}，1/2∉N。OCME 的 N 最小元素是 0，而 N⁺=N∖{0} 的最小元素是 1。",
      "en": "Reading N as the smallest set of reals containing 0 and closed under x↦x+1: 0∈N and 3=S(S(S(0)))∈N; the binary '11' and the decimal '3' both denote 3; since N⊆{x ∣ x≥0}, −1∉N; since N⊆{0}∪{x ∣ x≥1}, 1/2∉N. The least element of OCME's N is 0, while that of N⁺=N∖{0} is 1."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "所有教材都把 0 算作自然數",
          "en": "Every textbook counts 0 as a natural number"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "慣例不同；許多教材的自然數從 1 開始（N⁺），MKO 明確提醒要先確認定義。",
          "en": "Conventions differ; many textbooks start natural numbers at 1 (N⁺), and the MKO explicitly says to check the definition first."
        }
      },
      {
        "claim": {
          "zh-Hant": "字串「10」永遠表示十",
          "en": "The string '10' always denotes ten"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "在二進位中「10」表示 1·2+0=2（N11）。",
          "en": "In binary '10' denotes 1·2+0=2 (N11)."
        }
      },
      {
        "claim": {
          "zh-Hant": "1/2 是自然數，因為它可由 0 加上 1/2 得到",
          "en": "1/2 is natural because it is obtained from 0 by adding 1/2"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "後繼只允許加 1；T₂ 是不含 1/2 的歸納集，所以 1/2∉N（S12）。",
          "en": "Successor allows adding 1 only; T₂ is an inductive set missing 1/2, so 1/2∉N (S12)."
        }
      },
      {
        "claim": {
          "zh-Hant": "−1 是自然數，因為 −1+1=0∈N",
          "en": "−1 is natural because −1+1=0∈N"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "封閉性是 x∈N ⇒ x+1∈N，不是反向；N⊆{x ∣ x≥0} 排除 −1（S09）。",
          "en": "Closure says x∈N ⇒ x+1∈N, not the reverse; N⊆{x ∣ x≥0} excludes −1 (S09)."
        }
      }
    ],
    "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": "subset_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "子集：A⊆B 表示 A 的每個元素都是 B 的元素，即 ∀x (x∈A⇒x∈B)。",
          "en": "Subset: A⊆B means every element of A is an element of B, i.e. ∀x (x∈A⇒x∈B)."
        }
      },
      {
        "id": "set_operations_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合運算：x∈A∪B ⇔ x∈A 或 x∈B；x∈A∩B ⇔ x∈A 且 x∈B；x∈A∖B ⇔ x∈A 且 x∉B；在全集 U 中 Aᶜ=U∖A。",
          "en": "Set operations: x∈A∪B ⇔ x∈A or x∈B; x∈A∩B ⇔ x∈A and x∈B; x∈A∖B ⇔ x∈A and x∉B; in a universe U, Aᶜ=U∖A."
        }
      },
      {
        "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": "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": "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": "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": "modus_ponens",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "肯定前件：由 P 與 P⇒Q 得 Q。",
          "en": "Modus ponens: from P and P⇒Q infer Q."
        }
      },
      {
        "id": "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": "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": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "id": "contrapositive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逆否命題：P⇒Q 與 ¬Q⇒¬P 等價。逆命題 Q⇒P 一般不與 P⇒Q 等價。",
          "en": "Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q."
        }
      },
      {
        "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": "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": "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_nat_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-natural-number",
          "sha256": "222b72502fb661f3372012f602fd2f1eeef754b9e2069498c6b24234778730c4",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在本條目的慣例中，自然數集合 N 包含 0，並包含 0 經有限次後繼所得的 1、2、3 等數。",
          "en": "Under this entry's convention, the natural-number set N contains 0 and the numbers 1, 2, 3, and so on obtained by finitely many successor steps from 0."
        }
      },
      {
        "id": "inductive_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "歸納集：T⊆ℝ 若 0∈T 且對每個 x∈T 都有 x+1∈T，則稱 T 為歸納集。",
          "en": "Inductive set: T⊆ℝ is inductive if 0∈T and x+1∈T for every x∈T."
        }
      },
      {
        "id": "nat_as_intersection",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "N=⋂{T⊆ℝ ∣ T 是歸納集}：x∈N 當且僅當 x 屬於每一個歸納集。",
          "en": "N=⋂{T⊆ℝ ∣ T is inductive}: x∈N if and only if x belongs to every inductive set."
        }
      },
      {
        "id": "big_intersection_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合族的交集：對非空集合族 𝓕，x∈⋂𝓕 ⇔ 對每個 T∈𝓕 都有 x∈T；因此對每個 T∈𝓕 都有 ⋂𝓕⊆T。",
          "en": "Intersection of a family: for a nonempty family 𝓕, x∈⋂𝓕 ⇔ x∈T for every T∈𝓕; hence ⋂𝓕⊆T for every T∈𝓕."
        }
      },
      {
        "id": "positional_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "位值記數：以 b 為底、數字由右至左為 d₀、d₁ 等到 d_k 的符號表示 Σ_{i=0}^{k} d_i b^i。",
          "en": "Positional notation: a base-b numeral with digits d₀, d₁ and so on up to d_k from right to left denotes Σ_{i=0}^{k} d_i b^i."
        }
      },
      {
        "id": "L_N_inductive",
        "kind": "lemma",
        "proved_in": [
          "S02.1",
          "S02.2"
        ],
        "statement": {
          "zh-Hant": "N 本身是歸納集：0∈N，且對每個 x∈N 都有 x+1∈N。",
          "en": "N itself is inductive: 0∈N, and x+1∈N for every x∈N."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S04",
        "lhs": "((0+1)+1)+1",
        "rhs": "3"
      },
      {
        "id": "N2",
        "step": "S06",
        "lhs": "1*2**1+1*2**0",
        "rhs": "3"
      },
      {
        "id": "N3",
        "step": "S06",
        "lhs": "3*10**0",
        "rhs": "3"
      },
      {
        "id": "N4",
        "step": "S07",
        "lhs": "0+1",
        "relation": ">",
        "rhs": "0"
      },
      {
        "id": "N5",
        "step": "S09",
        "lhs": "-1",
        "relation": "<",
        "rhs": "0"
      },
      {
        "id": "N6",
        "step": "S09",
        "lhs": "-1+1",
        "rhs": "0"
      },
      {
        "id": "N7",
        "step": "S10.1",
        "lhs": "0+1",
        "relation": ">=",
        "rhs": "1"
      },
      {
        "id": "N8",
        "step": "S10.2",
        "lhs": "1+1",
        "rhs": "2"
      },
      {
        "id": "N9",
        "step": "S12",
        "lhs": "1/2",
        "relation": "<",
        "rhs": "1"
      },
      {
        "id": "N10",
        "step": "S12",
        "lhs": "1/2",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N11",
        "lhs": "1*2**1+0*2**0",
        "rhs": "2"
      },
      {
        "id": "N12",
        "lhs": "1*10**1+0*10**0",
        "rhs": "10"
      },
      {
        "id": "N13",
        "lhs": "1*2**2+1*2**1+0*2**0",
        "rhs": "6"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在實數 ℝ 中討論（只用到加法與序的性質），以便判定 −1、1/2 這類數是否屬於 N；這是 N 在 ℝ 中的刻畫，不是 N 的基礎建構（在 OCME 的建構鏈中 ℝ 反而由 ℚ、ℤ、N 建成）。不處理 Peano 公理、與 Lean Nat 的語義橋，也不證明「最小歸納集」與「有限次後繼」兩種說法的等價；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the reals ℝ (using only properties of addition and order) so that membership of numbers such as −1 and 1/2 in N can be decided; this is a characterization of N inside ℝ, not a foundational construction of N (in OCME's construction chain ℝ is in fact built from ℚ, ℤ and N). It does not treat the Peano axioms or the semantic bridge to Lean's Nat, and does not prove that 'smallest inductive set' and 'finitely many successor steps' are equivalent readings. It does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "N",
        "OCME 慣例下的自然數集合（含 0）",
        "the set of natural numbers under the OCME convention (containing 0)"
      ],
      [
        "N^{+}",
        "N∖{0}，從 1 開始的慣例",
        "N∖{0}, the convention that starts at 1"
      ],
      [
        "S(x)=x+1",
        "後繼",
        "successor"
      ],
      [
        "T_1,\\ T_2",
        "兩個用來排除非自然數的歸納集",
        "two inductive sets used to exclude non-natural numbers"
      ],
      [
        "\\bigcap",
        "集合族的交集",
        "intersection of a family of sets"
      ]
    ],
    "md_error_controls": [
      {
        "title": "0 是否屬於自然數取決於慣例 / Whether 0 is natural depends on the convention",
        "zh": "MKO 指出的誤解是以為所有教材對自然數是否包含 0 都相同。OCME 固定 0∈N（與 Lean 的 Nat 一致），但有些教材的自然數從 1 開始（本稿的 N⁺）。同一句「最小的自然數」在 OCME 是 0，在 N⁺ 慣例是 1（S13）。閱讀外部資料時要先確認它的定義。",
        "en": "The misconception identified in the MKO is thinking that all textbooks agree on whether 0 is natural. OCME fixes 0∈N (matching Lean's Nat), but some textbooks start natural numbers at 1 (N⁺ here). The same phrase 'the least natural number' means 0 in OCME and 1 under the N⁺ convention (S13). Check an external source's definition before reading it."
      },
      {
        "title": "數與數字符號 / Numbers versus numerals",
        "zh": "MKO 的另一個誤解是把數字的書寫方式與數本身混為一談。字串「10」在十進位表示十（1·10+0=10，N12），在二進位表示二（1·2+0=2，N11）；「11」（二進位）與「3」（十進位）表示同一個數，二進位「110」與十進位「6」也表示同一個數（N13）。是否屬於 N 是數的性質，判定前要先確定符號用哪一個底。",
        "en": "Another misconception identified in the MKO is confusing how a number is written with the number itself. The string '10' denotes ten in decimal (1·10+0=10, N12) and two in binary (1·2+0=2, N11); '11' (binary) and '3' (decimal) denote the same number, and so do the binary '110' and the decimal '6' (N13). Membership in N is a property of the number, so the base of a numeral must be fixed before deciding."
      },
      {
        "title": "排除一個數需要歸納集，不是列舉 / Excluding a number needs an inductive set, not a listing",
        "zh": "要證 1/2∉N，不能列出 0、1、2、3 後說「1/2 不在表上」：表永遠是有限的，N 不是。也不能說「1/2 不是由 0 經有限次後繼得到」而不加證明，那正是要證的事。正確做法是找一個歸納集 T₂ 不含 1/2（S10 到 S12）。同理，−1+1=0∈N 不推出 −1∈N，因為封閉性只往前推。",
        "en": "To prove 1/2∉N one cannot list 0, 1, 2, 3 and say '1/2 is not on the list': the list is always finite and N is not. Nor can one assert without proof that '1/2 is not obtained from 0 by finitely many successor steps'; that is exactly what must be shown. The correct method is to find an inductive set T₂ that misses 1/2 (S10 to S12). Likewise −1+1=0∈N does not give −1∈N, because closure only pushes forwards."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿採用「最小歸納集」作為 MKO「由 0 經有限次後繼所得」的形式讀法；兩種說法等價的證明，以及與 Peano 公理或 Lean Nat 的語義橋（MKO formal next），未在此處理。",
        "en": "This guide adopts 'smallest inductive set' as the formal reading of the MKO's 'obtained from 0 by finitely many successor steps'; a proof that the two readings are equivalent, and the semantic bridge to the Peano axioms or Lean's Nat (MKO formal next), are not treated here."
      },
      {
        "zh-Hant": "本稿把 ℝ 當作討論成員問題的環境有序域；在 OCME 的建構鏈中 ℝ 由 ℚ、ℤ、N 建成，所以這是 N 在 ℝ 中的刻畫，不是 N 的基礎定義。",
        "en": "This guide uses ℝ as an ambient ordered field in which to ask membership questions; in OCME's construction chain ℝ is built from ℚ, ℤ and N, so this is a characterization of N inside ℝ, not a foundational definition of N."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "c89486e40b6bff801da1282ac05e77b9efe181f87c86c6b2e9a752974b116bf2",
    "original_source_sha256": "c89486e40b6bff801da1282ac05e77b9efe181f87c86c6b2e9a752974b116bf2",
    "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."
  }
}