{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-arithmetic-operations-01",
  "mko_id": "mko-arithmetic-operations",
  "title": {
    "zh-Hant": "四則運算：同一組輸入在 ℕ、ℤ、ℚ 中是否有定義",
    "en": "Arithmetic operations: whether the same inputs give a defined result in ℕ, ℤ and ℚ"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-arithmetic-operations-01",
    "source": {
      "mko_id": "mko-arithmetic-operations",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "452d77028e2659b860c94200c1de159c4184dde86d6d73950e1998a0080e265a",
      "english_sha256": "cd55f6d985015e73e560232ec9e4f77f3559c06dc72f0cf006f7083206a54169"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "order_basics",
      "add_inverse",
      "mul_inverse_nonzero",
      "mul_zero",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "取 a=3、b=5。在 ℕ（含 0）、ℤ、ℚ 三個數系中，判定 a+b、a·b、a−b、a÷b 是否有定義；有定義時求出結果，並判定結果屬於哪些數系。再判定 3÷0 與 0÷0 在這三個數系中是否有定義。",
      "en": "Take a=3 and b=5. In each of the number systems ℕ (including 0), ℤ and ℚ, decide whether a+b, a·b, a−b and a÷b are defined; when they are, compute the result and decide which systems it belongs to. Then decide whether 3÷0 and 0÷0 are defined in these three systems."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "數系：ℕ={n∈ℤ ∣ n≥0}，且 ℕ⊂ℤ⊂ℚ；三者共用同一套加法、乘法與大小順序（ℚ 的運算與順序限制到 ℤ、ℕ 上）。",
          "en": "Number systems: ℕ={n∈ℤ ∣ n≥0} and ℕ⊂ℤ⊂ℚ; the three share one addition, one multiplication and one order (those of ℚ restricted to ℤ and ℕ)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "差與商的定義：設 S 是上述數系之一，a,b∈S。若恰有一個 c∈S 使 b+c=a，則稱 a−b 在 S 中有定義，且 a−b=c；若恰有一個 q∈S 使 b·q=a，則稱 a÷b 在 S 中有定義，且 a÷b=q。若沒有這樣的元素，或不只一個，該運算對這組輸入在 S 中沒有定義。",
          "en": "Difference and quotient: let S be one of these systems and a,b∈S. If exactly one c∈S satisfies b+c=a, then a−b is defined in S and a−b=c; if exactly one q∈S satisfies b·q=a, then a÷b is defined in S and a÷b=q. If there is no such element, or more than one, the operation is undefined in S for these inputs."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "輸入：a=3、b=5；另外檢查除數為零的兩組輸入 (3,0) 與 (0,0)。",
          "en": "Inputs: a=3 and b=5; in addition, the two inputs with divisor zero, (3,0) and (0,0)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "a,b\\ \\text{and an operation, system not yet stated}",
        "after_tex": "S\\in\\{\\mathbb{N},\\mathbb{Z},\\mathbb{Q}\\};\\ a-b=c\\iff b+c=a\\ (\\text{unique }c\\in S);\\ a\\div b=q\\iff bq=a\\ (\\text{unique }q\\in S)",
        "relation": "definition",
        "rule": "mko_arith",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 要求運算的定義連同數系與除零邊界一起指定，所以先固定：三個數系 ℕ⊂ℤ⊂ℚ 共用同一套加法與乘法（G1）；減法與除法定義為「解方程」：a−b 是使 b+c=a 的唯一 c，a÷b 是使 b·q=a 的唯一 q，而且解必須在所討論的數系 S 之中（G2）。之後說「有定義」一律指這個意義。",
          "en": "The MKO requires operations to be specified together with the number system and the division-by-zero boundary, so these are fixed first: the three systems ℕ⊂ℤ⊂ℚ share one addition and one multiplication (G1); subtraction and division are defined by solving equations: a−b is the unique c with b+c=a, and a÷b is the unique q with b·q=a, where the solution must lie in the system S under discussion (G2). From here on 'defined' always has this meaning."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_arith"
          ],
          "explanation": {
            "zh-Hant": "此步只固定數系與定義，不推出新事實；之後每個結論都相對於某個明示的 S。",
            "en": "This step only fixes the systems and the definitions and derives no new fact; every later conclusion is relative to a stated S."
          }
        },
        "leaf_basis": "mko_arith + def_intro"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "G3",
          "nat_closed"
        ],
        "before_tex": "3,5\\in\\mathbb{N}",
        "after_tex": "3+5=8\\in\\mathbb{N},\\quad 3\\cdot 5=15\\in\\mathbb{N}",
        "relation": "implies",
        "rule": "nat_closed",
        "conditions": [
          "3\\ge 0",
          "5\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "3 與 5 是非負整數，所以屬於 ℕ。依 nat_closed，它們的和與積仍在 ℕ 中；計算得 3+5=8、3·5=15，兩者都 ≥0。加法與乘法對這組輸入在 ℕ 中有定義，結果留在 ℕ。因三個數系共用同一套運算（G1），在 ℤ、ℚ 中的和與積也是 8 與 15。",
          "en": "3 and 5 are nonnegative integers, so they lie in ℕ. By nat_closed their sum and product stay in ℕ; computing gives 3+5=8 and 3·5=15, both ≥0. Addition and multiplication are defined in ℕ for these inputs and the results stay in ℕ. Since the three systems share one set of operations (G1), the sum and product in ℤ and ℚ are also 8 and 15."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G3",
            "nat_closed",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "需要 3,5∈ℕ 以及 ℕ 對加法、乘法封閉；封閉性只對這兩種運算宣稱，不延伸到減法與除法（見 S05、S09）。",
            "en": "It needs 3,5∈ℕ and the closure of ℕ under addition and multiplication; closure is claimed for these two operations only, not for subtraction or division (see S05 and S09)."
          }
        },
        "leaf_basis": "nat_closed + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "G1"
        ],
        "before_tex": "c\\in\\mathbb{N}",
        "after_tex": "5+c\\ge 5",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "c\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "為了判定 3−5 在 ℕ 中是否有定義，先看 5+c 在 ℕ 中能取哪些值。任取 c∈ℕ，則 c≥0。若 c>0，依 order_basics（u<v 推出 u+w<v+w，取 u=0、v=c、w=5）得 5+0<5+c；若 c=0，則 5+c=5。兩種情形都有 5+c≥5+0=5。",
          "en": "To decide whether 3−5 is defined in ℕ, first see which values 5+c can take for c in ℕ. Take any c∈ℕ, so c≥0. If c>0, order_basics (u<v gives u+w<v+w, with u=0, v=c, w=5) gives 5+0<5+c; if c=0 then 5+c=5. In both cases 5+c≥5+0=5."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "order_basics",
            "add_zero"
          ],
          "explanation": {
            "zh-Hant": "只用到 c≥0 與加法保持順序；c 是任取的，沒有其他假設。反向（由 5+c≥5 推回 c∈ℕ）不需要，也未宣稱。",
            "en": "Only c≥0 and the fact that adding preserves order are used; c is arbitrary with no other assumption. The converse (recovering c∈ℕ from 5+c≥5) is neither needed nor claimed."
          }
        },
        "leaf_basis": "order_basics + add_zero"
      },
      {
        "id": "S04",
        "uses": [
          "S03"
        ],
        "before_tex": "\\forall c\\in\\mathbb{N}:\\ 5+c\\ge 5>3",
        "after_tex": "\\neg\\exists c\\in\\mathbb{N}:\\ 5+c=3",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由 S03，每個 c∈ℕ 都有 5+c≥5，而 5>3，依順序的傳遞得 5+c>3；依三一律，5+c>3 排除 5+c=3。所以每個 c∈ℕ 都有 5+c≠3，依量詞否定，這等價於「不存在 c∈ℕ 使 5+c=3」。",
          "en": "By S03 every c∈ℕ has 5+c≥5, and 5>3, so by transitivity of order 5+c>3; by trichotomy 5+c>3 rules out 5+c=3. So 5+c≠3 for every c∈ℕ, which by negation of quantifiers is the same as 'there is no c∈ℕ with 5+c=3'."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "order_basics",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "需要的是對 ℕ 中每個 c 都成立的不等式，而不只是試了幾個 c；S03 對任意 c 證成，所以結論涵蓋全部 c∈ℕ。",
            "en": "What is needed is an inequality for every c in ℕ, not a few trial values; S03 was proved for an arbitrary c, so the conclusion covers all c∈ℕ."
          }
        },
        "leaf_basis": "quantifier_negation + order_basics"
      },
      {
        "id": "S05",
        "uses": [
          "S04",
          "G2"
        ],
        "before_tex": "\\neg\\exists c\\in\\mathbb{N}:\\ 5+c=3",
        "after_tex": "3-5\\ \\text{is undefined in}\\ \\mathbb{N}",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 G2，3−5 在 ℕ 中有定義需要恰有一個 c∈ℕ 使 5+c=3；S04 說連一個都沒有，所以 3−5 在 ℕ 中沒有定義。這表示 ℕ 對減法不封閉。",
          "en": "By G2, 3−5 is defined in ℕ only if exactly one c∈ℕ satisfies 5+c=3; S04 says there is none at all, so 3−5 is undefined in ℕ. This shows that ℕ is not closed under subtraction."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "G2",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "定義要求解存在且唯一；不存在就已足以判定沒有定義。此結論只對 ℕ 成立；ℤ 在 S06、S07 另行處理。",
            "en": "The definition requires a solution that exists and is unique; non-existence already settles that it is undefined. The conclusion is for ℕ only; ℤ is treated separately in S06 and S07."
          }
        },
        "leaf_basis": "def_unfold"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "G3"
        ],
        "before_tex": "5+c=3,\\ c\\in\\mathbb{Z}",
        "after_tex": "c=3+(-5)=-2",
        "relation": "equivalent",
        "rule": "L_sub",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 ℤ 中依引理 L_sub（a=3、b=5）：5+c=3 ⇔ c=3+(−5)，而 3+(−5)=−2。所以在 ℤ 中方程 5+c=3 恰有一個解 c=−2。同一論證在 ℚ 中也成立，ℚ 中的唯一解也是 −2。",
          "en": "In ℤ, by Lemma L_sub (a=3, b=5): 5+c=3 ⇔ c=3+(−5), and 3+(−5)=−2. So in ℤ the equation 5+c=3 has exactly one solution, c=−2. The same argument works in ℚ, where the only solution is also −2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_sub",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "反向：代回 5+(−2)=3。引理的兩個方向都在子步證明，所以「是解」與「等於 −2」等價，這同時給出存在與唯一。",
            "en": "Backward: substitute, 5+(−2)=3. Both directions of the lemma are proved in the sub-steps, so 'is a solution' and 'equals −2' are equivalent, which gives both existence and uniqueness."
          }
        },
        "leaf_basis": "eq_congruence + add_assoc + add_comm + add_inverse + add_zero + numeric_evaluation"
      },
      {
        "id": "S06.1",
        "uses": [
          "eq_congruence",
          "add_inverse"
        ],
        "before_tex": "b+c=a",
        "after_tex": "(-b)+(b+c)=(-b)+a",
        "relation": "equivalent",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 ℤ（或 ℚ）中每個 b 都有 −b；兩邊同時在左側加上 −b。反向是兩邊同加 b，再用 S06.2 的化簡。",
          "en": "In ℤ (or ℚ) every b has a −b; add −b on the left of both sides. The backward move is to add b to both sides and simplify as in S06.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "加 −b 與加 b 互為反運算，在 ℤ、ℚ 中沒有附加條件。在 ℕ 中 −b 一般不在 ℕ 內，這一步無法在 ℕ 中進行，這正是 S05 的情形。",
            "en": "Adding −b and adding b undo each other, with no side condition in ℤ or ℚ. In ℕ, −b is in general not in ℕ, so the step cannot be carried out inside ℕ, which is exactly the situation of S05."
          }
        },
        "leaf_basis": "eq_congruence + add_inverse"
      },
      {
        "id": "S06.2",
        "uses": [
          "S06.1"
        ],
        "before_tex": "(-b)+(b+c)=(-b)+a",
        "after_tex": "c=a+(-b)",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "化簡左邊：(−b)+(b+c)=((−b)+b)+c=0+c=c（加法結合律、反元素、交換律與單位元）；右邊依交換律 (−b)+a=a+(−b)。S06.1 與 S06.2 合起來證得引理 L_sub。",
          "en": "Simplify the left side: (−b)+(b+c)=((−b)+b)+c=0+c=c (associativity, inverses, commutativity and identity for addition); on the right, commutativity gives (−b)+a=a+(−b). S06.1 and S06.2 together prove Lemma L_sub."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "add_assoc",
            "add_comm",
            "add_inverse",
            "add_zero"
          ],
          "explanation": {
            "zh-Hant": "只把式子換成相等的式子，等式的真假不變，雙向成立。",
            "en": "Expressions are only replaced by equal ones, which does not change whether the equation holds, so the step is two-way."
          }
        },
        "leaf_basis": "add_assoc + add_comm + add_inverse + add_zero + eq_replace"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "G1",
          "G2"
        ],
        "before_tex": "c=-2\\ \\text{is the only solution in}\\ \\mathbb{Z}",
        "after_tex": "3-5=-2\\ \\text{in}\\ \\mathbb{Z},\\quad -2\\notin\\mathbb{N}",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [
          "-2<0"
        ],
        "explanation": {
          "zh-Hant": "S06 給出 ℤ 中恰有一個解，依 G2，3−5 在 ℤ（以及 ℚ）中有定義且等於 −2。又 −2<0，依 G1 的 ℕ={n∈ℤ ∣ n≥0}，−2∉ℕ。這與 S05 一致：ℤ 中唯一的解不在 ℕ 中，所以 ℕ 中沒有解。同一組輸入，換了數系，判定就不同。",
          "en": "S06 gives exactly one solution in ℤ, so by G2, 3−5 is defined in ℤ (and in ℚ) and equals −2. Also −2<0, so by ℕ={n∈ℤ ∣ n≥0} in G1, −2∉ℕ. This agrees with S05: the only solution in ℤ is not in ℕ, so ℕ has none. The same inputs get a different verdict in a different system."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "G1",
            "G2",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要 S06 的存在與唯一，以及 ℕ 的定義；並不宣稱 ℤ 對所有運算都封閉（S09 的除法就不封閉）。",
            "en": "It needs existence and uniqueness from S06 and the definition of ℕ; it does not claim that ℤ is closed under every operation (division in S09 is not)."
          }
        },
        "leaf_basis": "def_unfold + order_basics"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "q\\in\\mathbb{Z}\\ \\text{with}\\ 5q=3\\ ?",
        "after_tex": "q\\le 0\\ \\lor\\ q\\ge 1",
        "relation": "case_split",
        "rule": "int_no_gap",
        "conditions": [
          "q\\in\\mathbb{Z}"
        ],
        "explanation": {
          "zh-Hant": "要判定 3÷5 在 ℤ 中是否有定義，需問有沒有整數 q 使 5q=3。依 int_no_gap（m=0），每個整數 q 都滿足 q≤0 或 q≥1；兩種情形涵蓋全部整數，分別在 S08.1、S08.2 處理。",
          "en": "To decide whether 3÷5 is defined in ℤ, ask whether some integer q has 5q=3. By int_no_gap (m=0), every integer q satisfies q≤0 or q≥1; the two cases cover all integers and are handled in S08.1 and S08.2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "int_no_gap"
          ],
          "explanation": {
            "zh-Hant": "分情況需要兩支涵蓋所有整數；這依賴 q 是整數，對有理數 q（例如 3/5）不成立。",
            "en": "The case split needs the two branches to cover every integer; this depends on q being an integer and fails for a rational q such as 3/5."
          }
        },
        "leaf_basis": "int_no_gap + case_analysis"
      },
      {
        "id": "S08.1",
        "uses": [
          "S08"
        ],
        "before_tex": "q\\le 0",
        "after_tex": "5q\\le 0<3,\\ \\text{so}\\ 5q\\neq 3",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "5>0"
        ],
        "explanation": {
          "zh-Hant": "若 q<0，乘以正數 5 保持順序，得 5q<5·0=0；若 q=0，則 5q=0。所以 5q≤0<3，於是 5q≠3。",
          "en": "If q<0, multiplying by the positive number 5 preserves order, so 5q<5·0=0; if q=0 then 5q=0. So 5q≤0<3, and therefore 5q≠3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "用到 5>0 才能保持不等號方向；只推出 5q≠3，不宣稱反向。",
            "en": "5>0 is what keeps the direction of the inequality; only 5q≠3 is concluded and no converse is claimed."
          }
        },
        "leaf_basis": "order_basics + mul_zero"
      },
      {
        "id": "S08.2",
        "uses": [
          "S08"
        ],
        "before_tex": "q\\ge 1",
        "after_tex": "5q\\ge 5>3,\\ \\text{so}\\ 5q\\neq 3",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "5>0"
        ],
        "explanation": {
          "zh-Hant": "若 q>1，乘以 5>0 得 5q>5·1=5；若 q=1，則 5q=5。所以 5q≥5>3，於是 5q≠3。",
          "en": "If q>1, multiplying by 5>0 gives 5q>5·1=5; if q=1 then 5q=5. So 5q≥5>3, and therefore 5q≠3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "mul_one"
          ],
          "explanation": {
            "zh-Hant": "同 S08.1，用到 5>0；不宣稱反向。",
            "en": "As in S08.1, 5>0 is used; no converse is claimed."
          }
        },
        "leaf_basis": "order_basics + mul_one"
      },
      {
        "id": "S09",
        "uses": [
          "S08",
          "S08.1",
          "S08.2",
          "G1",
          "G2"
        ],
        "before_tex": "q\\le 0\\Rightarrow 5q\\neq 3,\\quad q\\ge 1\\Rightarrow 5q\\neq 3",
        "after_tex": "\\neg\\exists q\\in\\mathbb{Z}:\\ 5q=3;\\ 3\\div 5\\ \\text{undefined in}\\ \\mathbb{Z}\\ \\text{and in}\\ \\mathbb{N}",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩種情形都得 5q≠3，依分情況論證，每個整數 q 都有 5q≠3，即 ℤ 中沒有解；依 G2，3÷5 在 ℤ 中沒有定義。因 ℕ⊂ℤ 且運算相同，ℕ 中的解也會是 ℤ 中的解，所以 3÷5 在 ℕ 中也沒有定義。ℤ 對除法不封閉。",
          "en": "Both cases give 5q≠3, so by proof by cases every integer q has 5q≠3: there is no solution in ℤ, and by G2, 3÷5 is undefined in ℤ. Since ℕ⊂ℤ with the same operations, a solution in ℕ would also be a solution in ℤ, so 3÷5 is undefined in ℕ as well. ℤ is not closed under division."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "S08.1",
            "S08.2",
            "case_analysis",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "需要兩個情形涵蓋全部整數（S08）；若只檢查了幾個 q，不能推出沒有解。",
            "en": "It needs the two cases to cover all integers (S08); checking a few values of q would not show that there is no solution."
          }
        },
        "leaf_basis": "case_analysis + quantifier_negation + def_unfold"
      },
      {
        "id": "S10",
        "uses": [
          "G1",
          "G3"
        ],
        "before_tex": "5q=3,\\ q\\in\\mathbb{Q}",
        "after_tex": "q=3\\cdot\\tfrac{1}{5}=\\tfrac{3}{5}",
        "relation": "equivalent",
        "rule": "L_div",
        "conditions": [
          "5\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "在 ℚ 中依引理 L_div（a=3、b=5，條件 5≠0）：5q=3 ⇔ q=3·(1/5)=3/5。所以在 ℚ 中方程 5q=3 恰有一個解。",
          "en": "In ℚ, by Lemma L_div (a=3, b=5, with 5≠0): 5q=3 ⇔ q=3·(1/5)=3/5. So in ℚ the equation 5q=3 has exactly one solution."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_div",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "反向：代回 5·(3/5)=3。引理的條件 5≠0 成立，兩個方向都已證，所以解存在且唯一。",
            "en": "Backward: substitute, 5·(3/5)=3. The lemma's hypothesis 5≠0 holds and both directions are proved, so the solution exists and is unique."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero + mul_assoc + mul_comm + mul_one + numeric_evaluation"
      },
      {
        "id": "S10.1",
        "uses": [
          "eq_congruence",
          "mul_inverse_nonzero"
        ],
        "before_tex": "bq=a",
        "after_tex": "\\tfrac{1}{b}(bq)=\\tfrac{1}{b}a",
        "relation": "equivalent",
        "rule": "mul_inverse_nonzero",
        "conditions": [
          "b\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "因 b≠0，倒數 1/b 存在；兩邊同乘 1/b。反向是兩邊同乘 b，再用 S10.2 的化簡。若 b=0，1/b 不存在，這一步不能做。",
          "en": "Since b≠0 the reciprocal 1/b exists; multiply both sides by 1/b. The backward move is to multiply both sides by b and simplify as in S10.2. If b=0 there is no 1/b and the step cannot be taken."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "乘 1/b 與乘 b 互為反運算，前提正是 b≠0。",
            "en": "Multiplying by 1/b and multiplying by b undo each other, given exactly b≠0."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero"
      },
      {
        "id": "S10.2",
        "uses": [
          "S10.1"
        ],
        "before_tex": "\\tfrac{1}{b}(bq)=\\tfrac{1}{b}a",
        "after_tex": "q=a\\cdot\\tfrac{1}{b}",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [
          "b\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "化簡左邊：(1/b)(bq)=((1/b)b)q=1·q=q（乘法結合律、倒數、單位元）；右邊依交換律 (1/b)a=a·(1/b)。S10.1 與 S10.2 合起來證得引理 L_div。",
          "en": "Simplify the left side: (1/b)(bq)=((1/b)b)q=1·q=q (associativity, reciprocal, identity); on the right, commutativity gives (1/b)a=a·(1/b). S10.1 and S10.2 together prove Lemma L_div."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_assoc",
            "mul_comm",
            "mul_one",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "只把式子換成相等的式子，雙向成立；(1/b)b=1 需要 b≠0。",
            "en": "Expressions are only replaced by equal ones, so the step is two-way; (1/b)b=1 needs b≠0."
          }
        },
        "leaf_basis": "mul_assoc + mul_comm + mul_one + mul_inverse_nonzero + eq_replace"
      },
      {
        "id": "S11",
        "uses": [
          "S10",
          "G2",
          "int_no_gap"
        ],
        "before_tex": "q=\\tfrac{3}{5}\\ \\text{is the only solution in}\\ \\mathbb{Q}",
        "after_tex": "3\\div 5=\\tfrac{3}{5}\\ \\text{in}\\ \\mathbb{Q},\\quad 0<\\tfrac{3}{5}<1,\\ \\tfrac{3}{5}\\notin\\mathbb{Z}",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 G2 與 S10，3÷5 在 ℚ 中有定義且等於 3/5。又 1/5>0（因 5>0），把 0<3<5 乘以 1/5 得 0<3/5<1；依 int_no_gap 沒有整數嚴格介於 0 與 1 之間，所以 3/5∉ℤ。這與 S09 一致：ℚ 中唯一的解不是整數。",
          "en": "By G2 and S10, 3÷5 is defined in ℚ and equals 3/5. Also 1/5>0 (because 5>0), and multiplying 0<3<5 by 1/5 gives 0<3/5<1; by int_no_gap no integer lies strictly between 0 and 1, so 3/5∉ℤ. This agrees with S09: the only solution in ℚ is not an integer."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "G2",
            "int_no_gap",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要 S10 的存在與唯一，以及 0<3/5<1；結論只說這個商在 ℚ 中有定義，不說 ℚ 中任何除法都有定義（見 S13、S14）。",
            "en": "It needs existence and uniqueness from S10 and 0<3/5<1; it says only that this quotient is defined in ℚ, not that every division in ℚ is defined (see S13 and S14)."
          }
        },
        "leaf_basis": "def_unfold + int_no_gap + order_basics + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "G1"
        ],
        "before_tex": "q\\in\\mathbb{Q}",
        "after_tex": "0\\cdot q=0",
        "relation": "implies",
        "rule": "mul_zero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "轉向除數為零。依 mul_zero，q·0=0；再依乘法交換律 0·q=q·0=0。這對每個 q∈ℚ 成立，因而對 ℤ、ℕ 中的 q 也成立。",
          "en": "Turn to division with a divisor equal to zero. By mul_zero, q·0=0, and by commutativity 0·q=q·0=0. This holds for every q∈ℚ, hence also for q in ℤ or ℕ."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mul_zero",
            "mul_comm"
          ],
          "explanation": {
            "zh-Hant": "mul_zero 對所有 q 成立（理由見規則表）；這一步沒有用到任何關於 q 的額外假設。",
            "en": "mul_zero holds for every q (its reason is in the glossary); this step uses no extra assumption about q."
          }
        },
        "leaf_basis": "mul_zero + mul_comm"
      },
      {
        "id": "S13",
        "uses": [
          "S12",
          "G1",
          "G2"
        ],
        "before_tex": "\\forall q\\in\\mathbb{Q}:\\ 0\\cdot q=0\\neq 3",
        "after_tex": "\\neg\\exists q\\in\\mathbb{Q}:\\ 0\\cdot q=3;\\ 3\\div 0\\ \\text{undefined in}\\ \\mathbb{Q},\\mathbb{Z},\\mathbb{N}",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若有 q∈ℚ 使 0·q=3，則由 S12 得 3=0·q=0，與 3≠0 矛盾。所以沒有任何 q∈ℚ 滿足 0·q=3，依 G2，3÷0 在 ℚ 中沒有定義；ℤ、ℕ 中的解也會是 ℚ 中的解，所以 3÷0 在三個數系中都沒有定義。這是「沒有候選值」型的除零。",
          "en": "If some q∈ℚ had 0·q=3, then S12 would give 3=0·q=0, contradicting 3≠0. So no q∈ℚ satisfies 0·q=3, and by G2, 3÷0 is undefined in ℚ; a solution in ℤ or ℕ would also be one in ℚ, so 3÷0 is undefined in all three systems. This is division by zero of the 'no candidate' kind."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12",
            "contradiction",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "矛盾只依賴 S12 對所有 q 成立與 3≠0；結論是「不存在」，不是「結果為 0」，也不是「結果為無窮大」。",
            "en": "The contradiction depends only on S12 holding for all q and on 3≠0; the conclusion is non-existence, not 'the result is 0' and not 'the result is infinity'."
          }
        },
        "leaf_basis": "contradiction + mul_zero + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S12",
          "G2"
        ],
        "before_tex": "\\forall q:\\ 0\\cdot q=0",
        "after_tex": "0\\cdot 0=0,\\ 0\\cdot 1=0,\\ 0\\neq 1;\\ 0\\div 0\\ \\text{undefined in}\\ \\mathbb{N},\\mathbb{Z},\\mathbb{Q}",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由 S12，每個 q 都滿足 0·q=0，例如 q=0 與 q=1；兩者都在 ℕ（因而也在 ℤ、ℚ）中，而且 0≠1。所以在每個數系中，方程 0·q=0 的解都不只一個；G2 要求恰有一個，因此 0÷0 在三個數系中都沒有定義。這是「候選值太多」型的除零。",
          "en": "By S12 every q satisfies 0·q=0, for example q=0 and q=1; both lie in ℕ (hence in ℤ and ℚ), and 0≠1. So in each system the equation 0·q=0 has more than one solution; G2 requires exactly one, so 0÷0 is undefined in all three systems. This is division by zero of the 'too many candidates' kind."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12",
            "G2",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "兩個相異的解就足以推翻唯一性；不宣稱 0÷0 等於某個特定值。",
            "en": "Two distinct solutions are enough to defeat uniqueness; no particular value of 0÷0 is claimed."
          }
        },
        "leaf_basis": "def_unfold + mul_zero"
      },
      {
        "id": "S15",
        "uses": [
          "S02",
          "S05",
          "S07",
          "S09",
          "S11",
          "S13",
          "S14"
        ],
        "before_tex": "\\text{verdicts of S02, S05, S07, S09, S11, S13, S14}",
        "after_tex": "3+5=8,\\ 3\\cdot 5=15\\ (\\mathbb{N});\\ 3-5=-2\\ (\\mathbb{Z});\\ 3\\div 5=\\tfrac{3}{5}\\ (\\mathbb{Q});\\ 3\\div 0,\\ 0\\div 0\\ \\text{undefined}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把各步結論合在一起：在 ℕ 中 3+5=8、3·5=15 有定義，3−5、3÷5 沒有定義；在 ℤ 中 3−5=−2 有定義，3÷5 仍沒有定義；在 ℚ 中 3÷5=3/5 有定義；3÷0 與 0÷0 在三個數系中都沒有定義。",
          "en": "Put the conclusions together: in ℕ, 3+5=8 and 3·5=15 are defined while 3−5 and 3÷5 are not; in ℤ, 3−5=−2 is defined while 3÷5 still is not; in ℚ, 3÷5=3/5 is defined; 3÷0 and 0÷0 are undefined in all three systems."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S02",
            "S05",
            "S07",
            "S09",
            "S11",
            "S13",
            "S14",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：對每個數系與每個運算給出判定。足夠條件 H：S02、S05、S07、S09、S11、S13、S14 的結論；H⇒G 只是把它們並列。H 由 G1 到 G3 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: a verdict for each system and each operation. Sufficient condition H: the conclusions of S02, S05, S07, S09, S11, S13 and S14; H⇒G just lists them together. H is proved forwards from G1 to G3; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在 ℕ 中，3+5=8 與 3·5=15 有定義（在 ℤ、ℚ 中是同樣的值）；3−5 與 3÷5 在 ℕ 中沒有定義。在 ℤ（以及 ℚ）中 3−5=−2，但 −2∉ℕ；3÷5 在 ℤ 中沒有定義，在 ℚ 中等於 3/5，而 3/5∉ℤ。3÷0 沒有任何候選值，0÷0 的候選值不唯一，兩者在 ℕ、ℤ、ℚ 中都沒有定義。",
      "en": "In ℕ, 3+5=8 and 3·5=15 are defined (with the same values in ℤ and ℚ); 3−5 and 3÷5 are undefined in ℕ. In ℤ (and in ℚ), 3−5=−2, but −2∉ℕ; 3÷5 is undefined in ℤ and equals 3/5 in ℚ, while 3/5∉ℤ. 3÷0 has no candidate value and 0÷0 has no unique one, so both are undefined in ℕ, ℤ and ℚ."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "在 ℕ 中 3−5=0",
          "en": "In ℕ, 3−5=0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "代回得 5+0=5≠3；把負的差改寫成 0 的截斷減法是另一個運算，不是 G2 的減法。",
          "en": "Substituting back gives 5+0=5≠3; truncated subtraction, which replaces a negative difference by 0, is a different operation, not the subtraction of G2."
        }
      },
      {
        "claim": {
          "zh-Hant": "3÷0=0",
          "en": "3÷0=0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "代回得 0·0=0≠3；依 S13，任何 q 都有 0·q=0≠3，沒有候選值。",
          "en": "Substituting back gives 0·0=0≠3; by S13 every q has 0·q=0≠3, so there is no candidate."
        }
      },
      {
        "claim": {
          "zh-Hant": "0÷0=1，因為任何數除以自己都等於 1",
          "en": "0÷0=1, because any number divided by itself is 1"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "u÷u=1 需要 u≠0；0·q=0 對每個 q 成立，q=0 與 q=1 都是解，解不唯一（S14）。",
          "en": "u÷u=1 needs u≠0; 0·q=0 holds for every q, so q=0 and q=1 are both solutions and the solution is not unique (S14)."
        }
      },
      {
        "claim": {
          "zh-Hant": "6÷3=2 是整數，所以 ℤ 對除法封閉",
          "en": "6÷3=2 is an integer, so ℤ is closed under division"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "一個例子不能證明封閉性；3÷5 在 ℤ 中沒有定義（S09），一個反例就足以否定封閉。",
          "en": "One example cannot prove closure; 3÷5 is undefined in ℤ (S09), and one counterexample is enough to refute closure."
        }
      }
    ],
    "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": "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": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "mul_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "id": "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_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_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "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": "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": "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": "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": "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": "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": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "id": "mul_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法結合律：(uv)w=u(vw)。",
          "en": "Associativity of multiplication: (uv)w=u(vw)."
        }
      },
      {
        "id": "mul_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。",
          "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)."
        }
      },
      {
        "id": "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": "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_arith",
        "kind": "cited",
        "source": {
          "mko_id": "mko-arithmetic-operations",
          "sha256": "452d77028e2659b860c94200c1de159c4184dde86d6d73950e1998a0080e265a",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "四則運算以兩個輸入形成結果；運算是否有定義、結果是否仍在原數系中，必須連同數系與除零等邊界一起指定。",
          "en": "Each arithmetic operation forms a result from two inputs. Whether the operation is defined and whether the result remains in the original number system must be stated together with boundaries such as division by zero."
        }
      },
      {
        "id": "nat_closed",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "ℕ 對加法與乘法封閉：若 m,n∈ℕ，則 m+n∈ℕ 且 m·n∈ℕ（非負整數的和與積仍是非負整數）。",
          "en": "ℕ is closed under addition and multiplication: if m,n∈ℕ then m+n∈ℕ and m·n∈ℕ (sums and products of nonnegative integers are nonnegative integers)."
        }
      },
      {
        "id": "int_no_gap",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數沒有空隙：對整數 k 與 m，必有 k≤m 或 k≥m+1；特別是沒有整數嚴格介於 m 與 m+1 之間。",
          "en": "No gaps in the integers: for integers k and m, k≤m or k≥m+1; in particular no integer lies strictly between m and m+1."
        }
      },
      {
        "id": "L_sub",
        "kind": "lemma",
        "proved_in": [
          "S06.1",
          "S06.2"
        ],
        "statement": {
          "zh-Hant": "在 ℤ 或 ℚ 中，對該數系的 a、b、c：b+c=a ⇔ c=a+(−b)。",
          "en": "In ℤ or in ℚ, for a, b, c in that system: b+c=a ⇔ c=a+(−b)."
        }
      },
      {
        "id": "L_div",
        "kind": "lemma",
        "proved_in": [
          "S10.1",
          "S10.2"
        ],
        "statement": {
          "zh-Hant": "在 ℚ 中，對 a,q∈ℚ 與 b≠0：b·q=a ⇔ q=a·(1/b)。",
          "en": "In ℚ, for a,q∈ℚ and b≠0: b·q=a ⇔ q=a·(1/b)."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "3+5",
        "rhs": "8"
      },
      {
        "id": "N2",
        "step": "S02",
        "lhs": "3*5",
        "rhs": "15"
      },
      {
        "id": "N3",
        "step": "S03",
        "lhs": "5+0",
        "rhs": "5"
      },
      {
        "id": "N4",
        "step": "S04",
        "lhs": "5",
        "relation": ">",
        "rhs": "3"
      },
      {
        "id": "N5",
        "step": "S06",
        "lhs": "3+(-5)",
        "rhs": "-2"
      },
      {
        "id": "N6",
        "step": "S06",
        "lhs": "5+(-2)",
        "rhs": "3"
      },
      {
        "id": "N7",
        "step": "S08.1",
        "lhs": "5*0",
        "relation": "<",
        "rhs": "3"
      },
      {
        "id": "N8",
        "step": "S08.2",
        "lhs": "5*1",
        "relation": ">",
        "rhs": "3"
      },
      {
        "id": "N9",
        "step": "S10",
        "lhs": "3*(1/5)",
        "rhs": "3/5"
      },
      {
        "id": "N10",
        "step": "S10",
        "lhs": "5*(3/5)",
        "rhs": "3"
      },
      {
        "id": "N11",
        "step": "S11",
        "lhs": "3/5",
        "relation": ">",
        "rhs": "0"
      },
      {
        "id": "N12",
        "step": "S11",
        "lhs": "3/5",
        "relation": "<",
        "rhs": "1"
      },
      {
        "id": "N13",
        "step": "S13",
        "lhs": "3",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N14",
        "step": "S14",
        "lhs": "0*0",
        "rhs": "0"
      },
      {
        "id": "N15",
        "step": "S14",
        "lhs": "0*1",
        "rhs": "0"
      },
      {
        "id": "N16",
        "lhs": "5*0+3",
        "rhs": "3"
      },
      {
        "id": "N17",
        "lhs": "0*0",
        "relation": "!=",
        "rhs": "3"
      },
      {
        "id": "N18",
        "lhs": "5+0",
        "relation": "!=",
        "rhs": "3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理三個明示數系 ℕ⊂ℤ⊂ℚ 與一組明示輸入；差與商以「方程 b+c=a、b·q=a 在該數系中恰有一解」定義（G2）。示範 MKO 的要點：運算是否有定義、結果是否留在原數系，必須連同數系與除零邊界一起指定。不建構這些數系，不處理截斷減法或帶餘除法等其他運算，不改動 canonical MKO 或英文 edition。",
      "en": "Treats only the three stated systems ℕ⊂ℤ⊂ℚ and one stated pair of inputs; difference and quotient are defined by the equations b+c=a and b·q=a having exactly one solution in the system (G2). It illustrates the MKO's point that whether an operation is defined, and whether its result stays in the original system, must be stated together with the system and the division-by-zero boundary. It does not construct these systems, does not treat other operations such as truncated subtraction or division with remainder, and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "\\mathbb{N}",
        "非負整數 0、1、2 等組成的集合",
        "the set of nonnegative integers 0, 1, 2 and so on"
      ],
      [
        "\\mathbb{Z},\\ \\mathbb{Q}",
        "整數集與有理數集",
        "the integers and the rationals"
      ],
      [
        "a-b",
        "在指定數系中使 b+c=a 的唯一 c",
        "the unique c in the stated system with b+c=a"
      ],
      [
        "a\\div b",
        "在指定數系中使 b·q=a 的唯一 q",
        "the unique q in the stated system with b·q=a"
      ],
      [
        "\\tfrac{1}{u}",
        "非零有理數 u 的倒數",
        "the reciprocal of a nonzero rational u"
      ]
    ],
    "md_error_controls": [
      {
        "title": "封閉性取決於數系 / Closure depends on the number system",
        "zh": "MKO 指出常見誤解是以為四則運算在所有數系都無條件封閉。本稿顯示：ℕ 對加法與乘法封閉（S02），但 3−5 在 ℕ 中沒有定義（S05）；ℤ 補上了減法（S07），但 3÷5 在 ℤ 中仍沒有定義（S09）；ℚ 中 3÷5=3/5（S11），但除數為零仍然不行（S13、S14）。說「某運算有定義」時，必須同時說出數系。",
        "en": "The MKO names the common mistake of believing the four operations are closed in every number system without conditions. Here ℕ is closed under addition and multiplication (S02), but 3−5 is undefined in ℕ (S05); ℤ supplies subtraction (S07), but 3÷5 is still undefined in ℤ (S09); in ℚ, 3÷5=3/5 (S11), but division by zero is still excluded (S13, S14). Saying that an operation is defined always requires naming the system."
      },
      {
        "title": "除以零不是普通的除法 / Division by zero is not an ordinary division",
        "zh": "MKO 的另一個誤解是把除以零當作普通除法結果。依 G2，a÷b 是 b·q=a 的唯一解。b=0 時 0·q 永遠是 0：若 a≠0（如 3÷0），沒有任何候選值（S13）；若 a=0（0÷0），每個數都是候選值（S14）。兩種情形都不給出結果。「3÷0=0」代回得 0·0=0≠3；「0÷0=1」雖然代回成立，但 q=0 也成立，唯一性不成立。",
        "en": "Another mistake identified in the MKO is treating division by zero as an ordinary quotient. By G2, a÷b is the unique solution of b·q=a. When b=0, 0·q is always 0: if a≠0 (as in 3÷0) there is no candidate (S13); if a=0 (0÷0) every number is a candidate (S14). Neither case gives a result. '3÷0=0' fails on substitution, 0·0=0≠3; '0÷0=1' does satisfy 0·1=0, but so does q=0, so uniqueness fails."
      },
      {
        "title": "帶餘除法不是精確除法 / Division with remainder is not exact division",
        "zh": "MKO 提到 3÷2 不是自然數的整除結果。同理，帶餘除法給出 3=5·0+3（商 0、餘 3），但這是另一個運算：G2 的 3÷5 要求 5q=3 恰好成立，而 5·0=0≠3。把帶餘除法的商 0 當作 3÷5 在 ℕ 中的值，等於換了運算而沒有說明。",
        "en": "The MKO notes that 3÷2 is not an exact division in the natural numbers. Likewise, division with remainder gives 3=5·0+3 (quotient 0, remainder 3), but that is a different operation: 3÷5 in the sense of G2 requires 5q=3 exactly, and 5·0=0≠3. Taking the quotient 0 from division with remainder as the value of 3÷5 in ℕ silently changes the operation."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿把 ℕ、ℤ、ℚ 的運算與順序性質（nat_closed、int_no_gap、order_basics 等）當作前提，不在此建構這些數系。",
        "en": "This guide takes the operations and order properties of ℕ, ℤ and ℚ (nat_closed, int_no_gap, order_basics and the like) as premises and does not construct these systems."
      },
      {
        "zh-Hant": "G2 是以唯一解定義差與商的一種常見約定；截斷減法、帶餘除法，或程式語言中對 a÷0 的特別約定，是不同的運算，需要各自的定義。",
        "en": "G2 is one common convention that defines difference and quotient by unique solutions; truncated subtraction, division with remainder, or a programming language's special convention for a÷0 are different operations and need their own definitions."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "ef166229ecdb3b0f44715ad7924ab80f5ce26775fb834d7df17af4dbc883ea69",
    "original_source_sha256": "ef166229ecdb3b0f44715ad7924ab80f5ce26775fb834d7df17af4dbc883ea69",
    "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."
  }
}