{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-integer-01",
  "mko_id": "mko-integer",
  "title": {
    "zh-Hant": "整數：判定成員、負號的意義與對負號的封閉性",
    "en": "Integer: deciding membership, the meaning of the minus sign, and closure under negation"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-integer-01",
    "source": {
      "mko_id": "mko-integer",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "e1e3776eaf56e003f2eb07a66a39c7e221d26759df8fbc4db2652a2733318d66",
      "english_sha256": "24d4c8918b430c9dea2f942bb60a89a295d2386bf3bd9c14731ad56e0fb49034"
    },
    "mode": "definition_walkthrough",
    "audience_prerequisites": [
      "add_assoc",
      "add_comm",
      "add_zero",
      "add_inverse",
      "order_basics",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在實數中，取 Z={0}∪N⁺∪{−n ∣ n∈N⁺}（N 依 OCME 慣例含 0，N⁺=N∖{0}，−n 是 n 的加法反元素）。證明 −7∈Z；說明十進位符號「3.0」表示整數 3；證明 −2.5∉Z；由加法反元素的唯一性證明 −(−a)=a，特別是 −(−4)=4；對所有 z∈Z 證明 −z∈Z；並以 5−8=5+(−8)=−3 分辨二元減法與一元負號。",
      "en": "Inside the reals, take Z={0}∪N⁺∪{−n ∣ n∈N⁺} (N contains 0 by the OCME convention, N⁺=N∖{0}, and −n is the additive inverse of n). Prove −7∈Z; show that the decimal numeral '3.0' denotes the integer 3; prove −2.5∉Z; prove −(−a)=a from uniqueness of additive inverses, in particular −(−4)=4; prove −z∈Z for every z∈Z; and use 5−8=5+(−8)=−3 to tell binary subtraction from the unary minus sign."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "在實數 ℝ 中討論（只用到加法公理與序）。N⊆ℝ 是 OCME 慣例下的自然數（含 0，MKO 假設 a1），它是 ℝ 中最小的歸納集：0∈N、x∈N ⇒ x+1∈N，且任何含 0 並對 x↦x+1 封閉的 T⊆ℝ 都包含 N。",
          "en": "Work inside the reals ℝ (using only the addition axioms and order). N⊆ℝ is the natural numbers under the OCME convention (containing 0, MKO assumption a1), the smallest inductive subset of ℝ: 0∈N, x∈N ⇒ x+1∈N, and every T⊆ℝ that contains 0 and is closed under x↦x+1 contains N."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "N⁺=N∖{0}；Z={0}∪N⁺∪{−n ∣ n∈N⁺}。",
          "en": "N⁺=N∖{0}; Z={0}∪N⁺∪{−n ∣ n∈N⁺}."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "−a 表示 a 的加法反元素（a+(−a)=0）；二元減法 a−b 定義為 a+(−b)。",
          "en": "−a denotes the additive inverse of a (a+(−a)=0); binary subtraction a−b is defined as a+(−b)."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "要判定或計算的對象：−7、十進位符號「3.0」、−2.5、−(−4)、5−8。",
          "en": "Objects to decide or compute: −7, the decimal numeral '3.0', −2.5, −(−4), 5−8."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_int_def"
        ],
        "before_tex": "z\\in\\mathbb{Z}",
        "after_tex": "z=0\\ \\lor\\ z\\in N^{+}\\ \\lor\\ \\exists n\\in N^{+}:\\ z=-n",
        "relation": "definition",
        "rule": "mko_int_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，Z 由三部分組成：0、正自然數、正自然數的負數。所以判定 z∈Z，就是判定三個析取項中是否至少一個成立。這裡 −n 是 n 的加法反元素，不是「減掉 n」這個動作（G3）。",
          "en": "By the MKO, Z consists of three parts: 0, the positive natural numbers, and the negatives of positive natural numbers. So deciding z∈Z means deciding whether at least one of three disjuncts holds. Here −n is the additive inverse of n, not the action 'subtract n' (G3)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_int_def",
            "G2",
            "G3",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "此步只展開定義；三部分涵蓋 MKO 所列的全部成員。",
            "en": "This step only unfolds the definition; the three parts cover every member the MKO lists."
          }
        },
        "leaf_basis": "mko_int_def + def_unfold + set_operations_def"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "G1"
        ],
        "before_tex": "7=0+1+1+1+1+1+1+1",
        "after_tex": "7\\in N^{+},\\ -7\\in\\mathbb{Z}",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "7 由 0 經七次後繼得到（N1），G1 的封閉性連用七次給出 7∈N；7≠0，故 7∈N⁺。取 n=7，則 −7=−n，第三個析取項成立，依 ∃引入 −7∈Z。",
          "en": "7 is obtained from 0 by seven successor steps (N1), so applying the closure of G1 seven times gives 7∈N; 7≠0, so 7∈N⁺. Taking n=7, −7=−n, so the third disjunct holds and by ∃-introduction −7∈Z."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "exists_intro",
            "nat_least_inductive"
          ],
          "explanation": {
            "zh-Hant": "只需找到一個 n∈N⁺；不需要也不宣稱唯一性。",
            "en": "Only one n∈N⁺ needs to be found; uniqueness is neither needed nor claimed."
          }
        },
        "leaf_basis": "exists_intro + nat_least_inductive + set_operations_def + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "G4",
          "positional_def"
        ],
        "before_tex": "\\text{decimal numeral }3.0",
        "after_tex": "3\\cdot10^0+0\\cdot10^{-1}=3\\in N^{+}\\subseteq\\mathbb{Z}",
        "relation": "implies",
        "rule": "positional_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "十進位符號「3.0」表示 3·10⁰+0·10⁻¹=3。3 由 0 經三次後繼得到，3∈N 且 3≠0，所以 3∈N⁺，第二個析取項成立，3∈Z。符號中的小數點不改變數值；判定整數看的是數，不是符號的外觀。",
          "en": "The decimal numeral '3.0' denotes 3·10⁰+0·10⁻¹=3. 3 is obtained from 0 by three successor steps, so 3∈N and 3≠0, hence 3∈N⁺; the second disjunct holds and 3∈Z. The decimal point in the numeral does not change the value; integrality is decided by the number, not by how the numeral looks."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "positional_def",
            "S01",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "需要的是符號所表示的數；只看外觀（有沒有小數點）不能判定。",
            "en": "What is needed is the number the numeral denotes; its appearance (whether it has a decimal point) cannot decide."
          }
        },
        "leaf_basis": "positional_def + power_laws + numeric_evaluation + set_operations_def"
      },
      {
        "id": "S04",
        "uses": [
          "S01",
          "G4"
        ],
        "before_tex": "-2.5\\in\\mathbb{Z}",
        "after_tex": "-2.5=0\\ \\lor\\ -2.5\\in N^{+}\\ \\lor\\ \\exists n\\in N^{+}:\\ -2.5=-n",
        "relation": "equivalent",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對 −2.5（即 −5/2，N3）展開 S01 的成員條件。要證 −2.5∉Z，三個析取項都必須排除。",
          "en": "Unfold the membership condition of S01 for −2.5 (that is −5/2, N3). To prove −2.5∉Z, all three disjuncts must be excluded."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S01",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "定義展開，雙向成立。",
            "en": "Unfolding a definition holds both ways."
          }
        },
        "leaf_basis": "def_unfold + mko_int_def"
      },
      {
        "id": "S05",
        "uses": [
          "G3",
          "eq_congruence"
        ],
        "before_tex": "-2.5=-n",
        "after_tex": "n=2.5",
        "relation": "equivalent",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊同加 n+2.5：左邊 −2.5+(n+2.5)=n，右邊 −n+(n+2.5)=2.5（加法交換、結合與反元素），得 n=2.5。反向：n=2.5 時，兩邊同取加法反元素得 −n=−2.5。所以「−2.5 等於某個 −n」恰好等於「n=2.5」。",
          "en": "Add n+2.5 to both sides: the left side −2.5+(n+2.5)=n and the right side −n+(n+2.5)=2.5 (commutativity, associativity and inverses for addition), so n=2.5. Backward: if n=2.5, taking additive inverses of both sides gives −n=−2.5. So '−2.5 equals some −n' is exactly 'n=2.5'."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "add_inverse",
            "add_comm",
            "add_assoc"
          ],
          "explanation": {
            "zh-Hant": "兩個方向都是對等式兩邊做同一個有定義的運算。",
            "en": "Both directions apply the same defined operation to both sides of an equation."
          }
        },
        "leaf_basis": "eq_congruence + add_inverse + add_comm + add_assoc + add_zero"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "L_N_T3"
        ],
        "before_tex": "T_3=\\{0,1,2\\}\\cup\\{x\\in\\mathbb{R}\\mid x\\ge 3\\}",
        "after_tex": "N\\subseteq T_3",
        "relation": "implies",
        "rule": "L_N_T3",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_N_T3，每個自然數要不是 0、1、2，就是 ≥3。所以 2 與 3 之間沒有自然數，而且每個自然數都 ≥0。引理在 S06.1、S06.2 證明。",
          "en": "By Lemma L_N_T3, every natural number is 0, 1 or 2, or else ≥3. So there is no natural number strictly between 2 and 3, and every natural number is ≥0. The lemma is proved in S06.1 and S06.2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_N_T3"
          ],
          "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_N_T3"
      },
      {
        "id": "S06.1",
        "uses": [
          "G1",
          "nat_least_inductive"
        ],
        "before_tex": "0\\in T_3;\\ x\\in\\{0,1,2\\}",
        "after_tex": "x+1\\in\\{1,2,3\\}\\subseteq T_3",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "0∈T₃。x=0、1、2 時，x+1 分別是 1、2、3；1、2 在 {0,1,2} 中，3≥3，所以都在 T₃ 中。",
          "en": "0∈T₃. For x=0, 1, 2, x+1 is 1, 2, 3 respectively; 1 and 2 are in {0,1,2} and 3≥3, so all lie in T₃."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "case_analysis",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "三個情況逐一檢查。",
            "en": "The three cases are checked one by one."
          }
        },
        "leaf_basis": "case_analysis + numeric_evaluation + order_basics"
      },
      {
        "id": "S06.2",
        "uses": [
          "S06.1",
          "nat_least_inductive"
        ],
        "before_tex": "x\\ge 3",
        "after_tex": "x+1\\ge 4\\ge 3;\\ T_3\\ \\text{inductive},\\ N\\subseteq T_3",
        "relation": "implies",
        "rule": "nat_least_inductive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x≥3 時 x+1≥4≥3，所以 x+1∈T₃。T₃ 的元素不是在 {0,1,2} 中就是 ≥3，兩種情況（S06.1 與此處）涵蓋全部，所以 T₃ 對 x↦x+1 封閉；加上 0∈T₃，T₃ 是歸納集。依 G1 的最小性，N⊆T₃。S06.1 與 S06.2 合起來證得引理 L_N_T3。",
          "en": "If x≥3 then x+1≥4≥3, so x+1∈T₃. Every element of T₃ is in {0,1,2} or is ≥3, and the two cases (S06.1 and here) cover them all, so T₃ is closed under x↦x+1; with 0∈T₃, T₃ is inductive. By the minimality in G1, N⊆T₃. S06.1 and S06.2 together prove Lemma L_N_T3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06.1",
            "nat_least_inductive",
            "case_analysis",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "最小性只要求 T₃ 含 0 且對後繼封閉；不需要 T₃⊆N（事實上 3.5∈T₃ 不是自然數）。",
            "en": "Minimality only needs T₃ to contain 0 and be closed under successor; T₃⊆N is not needed (indeed 3.5∈T₃ is not natural)."
          }
        },
        "leaf_basis": "nat_least_inductive + case_analysis + order_basics"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "G4"
        ],
        "before_tex": "N\\subseteq T_3",
        "after_tex": "2.5\\notin N,\\ -2.5\\notin N",
        "relation": "implies",
        "rule": "contrapositive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "2.5 不等於 0、1、2，且 2.5<3，所以 2.5∉T₃；−2.5<0，而 T₃ 的元素都 ≥0，所以 −2.5∉T₃。因 N⊆T₃，依逆否 2.5∉N、−2.5∉N。",
          "en": "2.5 differs from 0, 1 and 2, and 2.5<3, so 2.5∉T₃; −2.5<0 while every element of T₃ is ≥0, so −2.5∉T₃. Since N⊆T₃, the contrapositive gives 2.5∉N and −2.5∉N."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "contrapositive",
            "subset_def",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "逆否與包含關係等價；不宣稱 T₃ 的每個元素都是自然數。",
            "en": "The contrapositive is equivalent to the inclusion; it is not claimed that every element of T₃ is natural."
          }
        },
        "leaf_basis": "contrapositive + subset_def + order_basics + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S04",
          "S05",
          "S07"
        ],
        "before_tex": "\\text{the three disjuncts of S04}",
        "after_tex": "-2.5\\notin\\mathbb{Z}",
        "relation": "implies",
        "rule": "de_morgan_logic",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第一項：−2.5≠0。第二項：−2.5∉N，所以也不在 N⁺⊆N 中。第三項：依 S05，它等於「2.5∈N⁺」，但 2.5∉N（S07）。三項都不成立，依德摩根律析取為假，依 S04 的等價 −2.5∉Z。",
          "en": "First disjunct: −2.5≠0. Second: −2.5∉N, so it is not in N⁺⊆N either. Third: by S05 it amounts to '2.5∈N⁺', but 2.5∉N (S07). All three fail, so by De Morgan's law the disjunction is false, and by the equivalence S04, −2.5∉Z."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S05",
            "S07",
            "de_morgan_logic"
          ],
          "explanation": {
            "zh-Hant": "三項都須排除；只說「−2.5 是負數」不夠，因為 −7 也是負數卻是整數。",
            "en": "All three must be excluded; saying '−2.5 is negative' is not enough, since −7 is negative too and is an integer."
          }
        },
        "leaf_basis": "de_morgan_logic + set_operations_def + quantifier_negation"
      },
      {
        "id": "S09",
        "uses": [
          "G3",
          "L_inv_unique"
        ],
        "before_tex": "(-a)+a=0,\\ (-a)+(-(-a))=0",
        "after_tex": "-(-a)=a\\quad(a\\in\\mathbb{R})",
        "relation": "implies",
        "rule": "L_inv_unique",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對任意實數 a：依加法交換律 (−a)+a=a+(−a)=0，所以 a 是 −a 的一個加法反元素；依定義 −(−a) 也是 −a 的加法反元素。依引理 L_inv_unique（取 L_inv_unique 中的 a 為 −a），−(−a)=a。a 是任意的，所以對所有實數成立。引理在 S09.1、S09.2 證明。",
          "en": "For any real a: by commutativity (−a)+a=a+(−a)=0, so a is an additive inverse of −a; by definition −(−a) is also an additive inverse of −a. By Lemma L_inv_unique (with the lemma's a taken to be −a), −(−a)=a. a was arbitrary, so this holds for every real. The lemma is proved in S09.1 and S09.2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_inv_unique",
            "add_comm",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "結論依賴反元素唯一；它不是背誦「負負得正」，而是唯一性的推論。",
            "en": "The conclusion rests on uniqueness of inverses; it is not a memorized 'two minuses make a plus' but a consequence of uniqueness."
          }
        },
        "leaf_basis": "L_inv_unique + add_comm + add_inverse + forall_intro"
      },
      {
        "id": "S09.1",
        "uses": [
          "add_zero",
          "add_assoc"
        ],
        "before_tex": "a+x=0,\\ a+y=0",
        "after_tex": "x=x+0=x+(a+y)=(x+a)+y",
        "relation": "implies",
        "rule": "add_assoc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x=x+0（加法單位元）=x+(a+y)（以假設 0=a+y 代換）=(x+a)+y（結合律）。",
          "en": "x=x+0 (additive identity) =x+(a+y) (replace 0 using the hypothesis a+y=0) =(x+a)+y (associativity)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "add_zero",
            "eq_replace",
            "add_assoc"
          ],
          "explanation": {
            "zh-Hant": "用到假設 a+y=0 作一次代換。",
            "en": "The hypothesis a+y=0 is used once, as a replacement."
          }
        },
        "leaf_basis": "add_zero + eq_replace + add_assoc"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "add_comm"
        ],
        "before_tex": "(x+a)+y",
        "after_tex": "y,\\ \\text{hence}\\ x=y",
        "relation": "implies",
        "rule": "add_comm",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(x+a)+y=(a+x)+y（交換律）=0+y（以假設 a+x=0 代換）=y。與 S09.1 串接得 x=y。S09.1 與 S09.2 合起來證得引理 L_inv_unique。",
          "en": "(x+a)+y=(a+x)+y (commutativity) =0+y (replace using the hypothesis a+x=0) =y. Chaining with S09.1 gives x=y. S09.1 and S09.2 together prove Lemma L_inv_unique."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "add_comm",
            "eq_replace",
            "add_zero",
            "eq_refl_symm_trans"
          ],
          "explanation": {
            "zh-Hant": "兩個假設 a+x=0、a+y=0 各用一次；等號傳遞把兩段串起來。",
            "en": "Each hypothesis, a+x=0 and a+y=0, is used once; transitivity of equality joins the two parts."
          }
        },
        "leaf_basis": "add_comm + eq_replace + add_zero + eq_refl_symm_trans"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "S01"
        ],
        "before_tex": "-(-4)",
        "after_tex": "4\\in N^{+}\\subseteq\\mathbb{Z}",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 S09 中取 a=4：−(−4)=4。驗算 (−4)+4=0（N8）。4 由 0 經四次後繼得到且 4≠0，所以 4∈N⁺⊆Z；−4 本身也是整數（第三部分，n=4）。",
          "en": "Take a=4 in S09: −(−4)=4. Check: (−4)+4=0 (N8). 4 is obtained from 0 by four successor steps and 4≠0, so 4∈N⁺⊆Z; −4 itself is also an integer (third part, n=4)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "這是一般結論的特例；反過來，這個例子不能證明一般結論。",
            "en": "This is an instance of the general result; conversely, the instance could not prove the general result."
          }
        },
        "leaf_basis": "forall_elim + numeric_evaluation + set_operations_def"
      },
      {
        "id": "S11",
        "uses": [
          "S01",
          "S09"
        ],
        "before_tex": "z\\in\\mathbb{Z}",
        "after_tex": "-z\\in\\mathbb{Z}",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 S01 分三種情況。(i) z=0：0+0=0，所以 0 是 0 的反元素，依唯一性 −0=0∈Z。(ii) z∈N⁺：取 n=z，−z 屬於第三部分。(iii) z=−n（n∈N⁺）：依 S09，−z=−(−n)=n∈N⁺⊆Z。三種情況涵蓋 Z 的全部元素，所以對每個整數 z，−z∈Z。這是對任意 z 的證明，不是由幾個例子歸納。",
          "en": "Split into the three cases of S01. (i) z=0: 0+0=0, so 0 is an inverse of 0 and by uniqueness −0=0∈Z. (ii) z∈N⁺: with n=z, −z lies in the third part. (iii) z=−n with n∈N⁺: by S09, −z=−(−n)=n∈N⁺⊆Z. The three cases cover every element of Z, so −z∈Z for every integer z. This is a proof for an arbitrary z, not a generalization from a few examples."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S09",
            "case_analysis",
            "L_inv_unique",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "三種情況必須對應 S01 的三個析取項，缺一不可。",
            "en": "The three cases must match the three disjuncts of S01; none may be left out."
          }
        },
        "leaf_basis": "case_analysis + L_inv_unique + forall_intro + exists_intro"
      },
      {
        "id": "S12",
        "uses": [
          "G3",
          "L_inv_unique"
        ],
        "before_tex": "5-8=5+(-8)",
        "after_tex": "5+(-8)=-3",
        "relation": "implies",
        "rule": "L_inv_unique",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 G3，二元減法 5−8 定義為 5+(−8)。計算 3+(5+(−8))=(3+5)+(−8)=8+(−8)=0（結合律與反元素），又 3+(−3)=0；依引理 L_inv_unique（a=3），5+(−8)=−3。這裡出現兩種「−」：5−8 中的是二元運算（減法），−3 中的是一元記號（3 的反元素）；運算的結果是 −3 這個數，而 −3∈Z（3∈N⁺）。",
          "en": "By G3, binary subtraction 5−8 is defined as 5+(−8). Compute 3+(5+(−8))=(3+5)+(−8)=8+(−8)=0 (associativity and inverses), and also 3+(−3)=0; by Lemma L_inv_unique (a=3), 5+(−8)=−3. Two kinds of '−' appear: in 5−8 it is a binary operation (subtraction), in −3 a unary sign (the inverse of 3); the result of the operation is the number −3, and −3∈Z (3∈N⁺)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_inv_unique",
            "add_assoc",
            "add_inverse",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "套用唯一性需要兩個等式 3+x=0 與 3+y=0，x=5+(−8)、y=−3；兩者都已驗證（N9、N11）。",
            "en": "Applying uniqueness needs the two equations 3+x=0 and 3+y=0 with x=5+(−8) and y=−3; both are verified (N9, N11)."
          }
        },
        "leaf_basis": "L_inv_unique + add_comm + add_assoc + add_inverse + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S02",
          "S03",
          "S08",
          "S10",
          "S11",
          "S12"
        ],
        "before_tex": "\\text{S02, S03, S08, S10, S11, S12}",
        "after_tex": "-7,\\ 3,\\ 4,\\ -3\\in\\mathbb{Z};\\ -2.5\\notin\\mathbb{Z};\\ \\forall z\\in\\mathbb{Z}:\\ -z\\in\\mathbb{Z}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "−7、3（寫成 3.0）、−(−4)=4、5−8=−3 都是整數；−2.5 不是；而且 Z 對負號封閉。",
          "en": "−7, 3 (written 3.0), −(−4)=4 and 5−8=−3 are integers; −2.5 is not; and Z is closed under negation."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S02",
            "S03",
            "S08",
            "S09",
            "S10",
            "S11",
            "S12"
          ],
          "explanation": {
            "zh-Hant": "目標是 goal 所列各項。足夠條件 H：S02、S03、S08、S10、S11、S12 各自成立；H 推出目標只用合取。普遍敘述（S09、S11）由任意元素的論證得到，沒有把目標當作前提。",
            "en": "The goal is the list in the goal field. Sufficient condition H: S02, S03, S08, S10, S11 and S12 each hold; H gives the goal by conjunction only. The universal statements (S09, S11) come from arguments about an arbitrary element; the goal is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "依 Z={0}∪N⁺∪{−n ∣ n∈N⁺}：−7∈Z；「3.0」表示 3∈Z；因 N⊆{0,1,2}∪{x ∣ x≥3}，2.5 與 −2.5 都不是自然數，所以 −2.5∉Z。加法反元素唯一，故 −(−a)=a，特別是 −(−4)=4；對每個 z∈Z 都有 −z∈Z。5−8 是運算 5+(−8)，其值是數 −3∈Z。",
      "en": "With Z={0}∪N⁺∪{−n ∣ n∈N⁺}: −7∈Z; '3.0' denotes 3∈Z; since N⊆{0,1,2}∪{x ∣ x≥3}, neither 2.5 nor −2.5 is natural, so −2.5∉Z. Additive inverses are unique, so −(−a)=a, in particular −(−4)=4; −z∈Z for every z∈Z. 5−8 is the operation 5+(−8), whose value is the number −3∈Z."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "−2.5 是整數，因為它是負數，而整數包含負數",
          "en": "−2.5 is an integer, because it is negative and the integers include negatives"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "Z 中的負數只能是正自然數的負數；2.5 不是自然數（S07），所以 −2.5∉Z（S08）。",
          "en": "The negative elements of Z are precisely the negatives of positive naturals; 2.5 is not natural (S07), so −2.5∉Z (S08)."
        }
      },
      {
        "claim": {
          "zh-Hant": "3.0 不是整數，因為它寫有小數點",
          "en": "3.0 is not an integer, because it is written with a decimal point"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "「3.0」表示 3，而 3∈N⁺⊆Z（S03）。",
          "en": "'3.0' denotes 3, and 3∈N⁺⊆Z (S03)."
        }
      },
      {
        "claim": {
          "zh-Hant": "−3 表示「從某數減去 3」這個運算",
          "en": "−3 denotes the operation 'subtract 3 from something'"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "−3 是滿足 3+(−3)=0 的數；5−8 才是運算 5+(−8)，其值為 −3（S12）。",
          "en": "−3 is the number with 3+(−3)=0; it is 5−8 that is an operation, 5+(−8), whose value is −3 (S12)."
        }
      },
      {
        "claim": {
          "zh-Hant": "因為 −(−4)=4、−(−7)=7，所以 −(−z)=z 已對所有整數證明",
          "en": "Since −(−4)=4 and −(−7)=7, −(−z)=z is proved for all integers"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "有限個例子不能證明普遍命題；一般證明在 S09（反元素唯一性）。",
          "en": "Finitely many examples cannot prove a universal statement; the general proof is S09 (uniqueness of inverses)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "add_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。",
          "en": "Associativity of addition: (u+v)+w=u+(v+w)."
        }
      },
      {
        "id": "add_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法交換律：u+v=v+u。",
          "en": "Commutativity of addition: u+v=v+u."
        }
      },
      {
        "id": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "id": "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": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "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": "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": "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": "eq_congruence",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。",
          "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule."
        }
      },
      {
        "id": "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": "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": "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": "de_morgan_logic",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "邏輯的德摩根律：¬(P∧Q) ⇔ ¬P∨¬Q；¬(P∨Q) ⇔ ¬P∧¬Q。",
          "en": "De Morgan's laws for logic: ¬(P∧Q) ⇔ ¬P∨¬Q and ¬(P∨Q) ⇔ ¬P∧¬Q."
        }
      },
      {
        "id": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬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": "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": "eq_refl_symm_trans",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。",
          "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w."
        }
      },
      {
        "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_int_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-integer",
          "sha256": "e1e3776eaf56e003f2eb07a66a39c7e221d26759df8fbc4db2652a2733318d66",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "整數集合 Z 包含 0、每個正自然數及其負數；符號 n∈Z 表示 n 是整數。",
          "en": "The integer set Z contains 0, every positive natural number, and its negative. The notation n∈Z states that n is an integer."
        }
      },
      {
        "id": "nat_least_inductive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "N 是 ℝ 中最小的歸納集：0∈N，x∈N ⇒ x+1∈N；若 T⊆ℝ 含 0 且對 x↦x+1 封閉，則 N⊆T。",
          "en": "N is the smallest inductive subset of ℝ: 0∈N and x∈N ⇒ x+1∈N; if T⊆ℝ contains 0 and is closed under x↦x+1, then N⊆T."
        }
      },
      {
        "id": "positional_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "十進位小數符號：整數部分各位數 d_i 表示 Σ d_i 10^i，小數點後第 j 位數 e_j 表示 e_j 10^{−j}，兩者相加；例如「3.0」表示 3·10⁰+0·10⁻¹。",
          "en": "Decimal numerals with a point: the digits d_i before the point contribute Σ d_i 10^i and the j-th digit e_j after the point contributes e_j 10^{−j}, added together; for example '3.0' denotes 3·10⁰+0·10⁻¹."
        }
      },
      {
        "id": "L_N_T3",
        "kind": "lemma",
        "proved_in": [
          "S06.1",
          "S06.2"
        ],
        "statement": {
          "zh-Hant": "令 T₃={0,1,2}∪{x∈ℝ ∣ x≥3}，則 N⊆T₃。",
          "en": "Let T₃={0,1,2}∪{x∈ℝ ∣ x≥3}; then N⊆T₃."
        }
      },
      {
        "id": "L_inv_unique",
        "kind": "lemma",
        "proved_in": [
          "S09.1",
          "S09.2"
        ],
        "statement": {
          "zh-Hant": "對實數 a、x、y：若 a+x=0 且 a+y=0，則 x=y（加法反元素唯一）。",
          "en": "For reals a, x, y: if a+x=0 and a+y=0, then x=y (additive inverses are unique)."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "0+1+1+1+1+1+1+1",
        "rhs": "7"
      },
      {
        "id": "N2",
        "step": "S03",
        "lhs": "3*10**0+0*10**(-1)",
        "rhs": "3"
      },
      {
        "id": "N3",
        "step": "S04",
        "lhs": "-2.5",
        "rhs": "-5/2"
      },
      {
        "id": "N4",
        "step": "S06.1",
        "lhs": "2+1",
        "rhs": "3"
      },
      {
        "id": "N5",
        "step": "S07",
        "lhs": "2.5",
        "relation": "<",
        "rhs": "3"
      },
      {
        "id": "N6",
        "step": "S07",
        "lhs": "-2.5",
        "relation": "<",
        "rhs": "0"
      },
      {
        "id": "N7",
        "step": "S07",
        "lhs": "2.5",
        "relation": "!=",
        "rhs": "2"
      },
      {
        "id": "N8",
        "step": "S10",
        "lhs": "(-4)+4",
        "rhs": "0"
      },
      {
        "id": "N9",
        "step": "S12",
        "lhs": "3+(5+(-8))",
        "rhs": "0"
      },
      {
        "id": "N10",
        "step": "S12",
        "lhs": "5-8",
        "rhs": "-3"
      },
      {
        "id": "N11",
        "step": "S12",
        "lhs": "3+(-3)",
        "rhs": "0"
      },
      {
        "id": "N12",
        "lhs": "-(-7)",
        "rhs": "7"
      },
      {
        "id": "N13",
        "step": "S06.2",
        "lhs": "3+1",
        "rhs": "4"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理 MKO 建立的成員與符號邊界：Z 的三部分、負號作為加法反元素、對負號的封閉性。在 ℝ 中描述 Z（只用到加法公理與序），N 以「最小歸納集」刻畫並作為前提。不證明 Z 對加法、乘法封閉，也不處理序與整除（MKO 說這些需要各自的聲明）；不處理與 Lean Int 的語義對應；不改動 canonical MKO 或英文 edition。",
      "en": "Treats only the membership and notation boundary the MKO sets up: the three parts of Z, the minus sign as an additive inverse, and closure under negation. Z is described inside ℝ (using only the addition axioms and order), with N characterized as the smallest inductive set and taken as a premise. It does not prove closure of Z under addition or multiplication and does not treat order or divisibility (the MKO says these need their own claims); it does not treat the semantic correspondence with Lean's Int; and it does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "\\mathbb{Z}",
        "整數集合",
        "the set of integers"
      ],
      [
        "N,\\ N^{+}",
        "自然數（含 0）與正自然數 N∖{0}",
        "natural numbers (with 0) and positive naturals N∖{0}"
      ],
      [
        "-a",
        "a 的加法反元素（一元負號）",
        "the additive inverse of a (unary minus)"
      ],
      [
        "a-b",
        "二元減法，定義為 a+(−b)",
        "binary subtraction, defined as a+(−b)"
      ],
      [
        "T_3",
        "歸納集 {0,1,2}∪{x ∣ x≥3}",
        "the inductive set {0,1,2}∪{x ∣ x≥3}"
      ]
    ],
    "md_error_controls": [
      {
        "title": "負號不是減法本身 / The minus sign is not subtraction itself",
        "zh": "MKO 指出的誤解之一是把負號當成減法操作本身。−3 是一個數：3 的加法反元素，滿足 3+(−3)=0；5−8 是一個運算，定義為 5+(−8)，它的值才是 −3（S12）。−(−4)=4 也不是背誦的規則，而是反元素唯一性的結果（S09）。",
        "en": "One misconception the MKO names is treating the minus sign as the subtraction operation itself. −3 is a number: the additive inverse of 3, with 3+(−3)=0; 5−8 is an operation, defined as 5+(−8), whose value is −3 (S12). −(−4)=4 is not a memorized rule either but a consequence of the uniqueness of inverses (S09)."
      },
      {
        "title": "小數點不決定是否為整數 / A decimal point does not decide integrality",
        "zh": "MKO 指出的另一誤解與帶小數點的寫法有關。「3.0」帶有小數點，表示的卻是整數 3（S03）；「−2.5」表示的數不是整數（S08）。判定看的是符號所表示的數，以及它是否落在 Z 的三部分之一，不看寫法。",
        "en": "Another misconception the MKO names concerns numerals with a decimal point. '3.0' has a decimal point yet denotes the integer 3 (S03); '−2.5' denotes a number that is not an integer (S08). What decides is the number a numeral denotes and whether it falls in one of the three parts of Z, not how it is written."
      },
      {
        "title": "例子不能推出普遍代數律 / Examples do not prove general laws",
        "zh": "MKO 提醒不要由幾個例子推得全部整數代數律。驗算 −(−4)=4、−(−7)=7（N12）不證明 −(−z)=z 對所有 z 成立；S09 對任意實數 a 證明，S11 以分情況證明對負號的封閉性。至於 Z 對加法、乘法的封閉性，本稿沒有證明，需要各自的聲明。",
        "en": "The MKO warns against deriving all integer algebra laws from a few examples. Checking −(−4)=4 and −(−7)=7 (N12) does not prove −(−z)=z for all z; S09 proves it for an arbitrary real a, and S11 proves closure under negation by cases. Closure of Z under addition and multiplication is not proved here and needs its own claims."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "Z 對加法、乘法的封閉性，以及序與整除的性質需要各自的聲明，本稿不證明（MKO 說明）。",
        "en": "Closure of Z under addition and multiplication, and properties of order and divisibility, need their own claims and are not proved here (as the MKO says)."
      },
      {
        "zh-Hant": "nat_least_inductive 是前提（自然數 MKO 的形式讀法）；在 OCME 的建構鏈中 ℝ 由 ℚ、ℤ、N 建成，本稿只把 ℝ 當作討論成員問題的環境。",
        "en": "nat_least_inductive is a premise (the formal reading of the natural-number MKO); in OCME's construction chain ℝ is built from ℚ, ℤ and N, and this guide uses ℝ only as the setting for membership questions."
      },
      {
        "zh-Hant": "與 Lean Int 的構造、自然數嵌入與負數的語義對應（MKO formal next）未在此處理。",
        "en": "The semantic correspondence with Lean's Int construction, the embedding of naturals and negatives (MKO formal next) is not treated here."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "02e167b67a0275433f107a2e6696fec332105d67efeeb4872a2ed6e36226c082",
    "original_source_sha256": "02e167b67a0275433f107a2e6696fec332105d67efeeb4872a2ed6e36226c082",
    "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."
  }
}