{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-natural-divisibility-01",
  "mko_id": "mko-natural-divisibility",
  "title": {
    "zh-Hant": "自然數整除：以乘數見證判定、處理 0 的情形並證明遞移性",
    "en": "Natural-number divisibility: deciding by a multiplier witness, handling zero, and proving transitivity"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-natural-divisibility-01",
    "source": {
      "mko_id": "mko-natural-divisibility",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "2e5d00faa3ae3031b2cb2a650b12db0ab8e8aa1b59a7ea6b2350f49a792758d7",
      "english_sha256": "acae542d73cc84e0fd3fd3443b35f99bca7758dce17f8b8ab65ee9957193e395"
    },
    "mode": "definition_walkthrough",
    "audience_prerequisites": [
      "division_algorithm",
      "order_basics",
      "mul_zero",
      "mul_assoc",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "依 MKO 定義（a∣b ⇔ 存在自然數 k 使 b=a·k）：(1) 證明 6∣84；(2) 證明 6∤87；(3) 證明 4∤10，雖然 10/4=5/2 是一個數；(4) 證明每個 a 都整除 0；(5) 證明 0∣b ⇔ b=0，並說明 0∣0 的見證不唯一、0∤5；(6) 證明 a>0 時見證至多一個（若 a∣b，則既有見證唯一）；(7) 證明整除的自反性與遞移性，並由 3∣6、6∣84 求出 3∣84 的見證。",
      "en": "From the MKO's definition (a∣b ⇔ there is a natural k with b=a·k): (1) prove 6∣84; (2) prove 6∤87; (3) prove 4∤10, although 10/4=5/2 is a number; (4) prove that every a divides 0; (5) prove 0∣b ⇔ b=0, and explain why the witness for 0∣0 is not unique and why 0∤5; (6) prove that for a>0 there is at most one witness (and when a∣b that existing witness is unique); (7) prove reflexivity and transitivity of divisibility, and from 3∣6 and 6∣84 find a witness for 3∣84."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "ℕ 包含 0；乘法是自然數乘法，自然數的和與積仍是自然數（MKO 假設 a1）。",
          "en": "ℕ includes 0; multiplication is natural-number multiplication, and sums and products of naturals are natural (MKO assumption a1)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "整除的見證 k 必須是自然數；非整數的商不是見證（MKO 假設 a2）。",
          "en": "The witness k must be a natural number; a noninteger quotient is not a witness (MKO assumption a2)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "要判定的數對：(6,84)、(6,87)、(4,10)、(3,6)、(0,0)、(0,5)。",
          "en": "Pairs to decide: (6,84), (6,87), (4,10), (3,6), (0,0), (0,5)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_dvd_def"
        ],
        "before_tex": "a\\mid b",
        "after_tex": "\\exists k\\in\\mathbb{N}:\\ b=a\\cdot k",
        "relation": "definition",
        "rule": "mko_dvd_def",
        "conditions": [
          "a,b\\in\\mathbb{N}"
        ],
        "explanation": {
          "zh-Hant": "依 MKO 的定義，a∣b 的意思是存在一個自然數 k（乘數見證）使 b=a·k。要證 a∣b，就要給出這樣的 k；要證 a∤b，就要說明沒有任何自然數 k 能做到。",
          "en": "By the MKO's definition, a∣b means there is a natural number k (a multiplier witness) with b=a·k. To prove a∣b one must exhibit such a k; to prove a∤b one must show that no natural k works."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_dvd_def"
          ],
          "explanation": {
            "zh-Hant": "此步只展開定義；見證限定為自然數（G2），不是有理數。",
            "en": "This step only unfolds the definition; the witness is restricted to naturals (G2), not rationals."
          }
        },
        "leaf_basis": "mko_dvd_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "G3"
        ],
        "before_tex": "84=6\\cdot 14,\\ 14\\in\\mathbb{N}",
        "after_tex": "6\\mid 84",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "6·14=84（N1），且 14 是自然數，所以 k=14 是見證，6∣84。",
          "en": "6·14=84 (N1) and 14 is natural, so k=14 is a witness and 6∣84."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "exists_intro",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "存在性只需一個見證；這一步不宣稱見證唯一（唯一性見 S09）。",
            "en": "Existence needs only one witness; uniqueness is not claimed here (see S09)."
          }
        },
        "leaf_basis": "exists_intro + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "L_rem_test"
        ],
        "before_tex": "a\\mid b\\quad(a>0)",
        "after_tex": "b\\bmod a=0",
        "relation": "equivalent",
        "rule": "L_rem_test",
        "conditions": [
          "a>0"
        ],
        "explanation": {
          "zh-Hant": "a>0 時，整除可以用餘數判定：依引理 L_rem_test（S03.1 到 S03.3 證明），a∣b 恰好當 b 除以 a 的餘數為 0。這把「找見證」換成一次可重算的帶餘除法。",
          "en": "For a>0, divisibility can be decided by the remainder: by Lemma L_rem_test (proved in S03.1 to S03.3), a∣b exactly when the remainder of b on division by a is 0. This replaces 'search for a witness' by one recomputable division with remainder."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_rem_test"
          ],
          "explanation": {
            "zh-Hant": "引理的兩個方向（S03.2、S03.3）都已證；前提 a>0 是帶餘除法所需。",
            "en": "Both directions of the lemma (S03.2, S03.3) are proved; the premise a>0 is what the division algorithm needs."
          }
        },
        "leaf_basis": "division_algorithm + exists_intro + order_basics"
      },
      {
        "id": "S03.1",
        "uses": [
          "division_algorithm",
          "order_basics"
        ],
        "before_tex": "b=aq+r,\\ 0\\le r<a,\\ q\\in\\mathbb{Z},\\ b\\ge 0",
        "after_tex": "q\\ge 0,\\ \\text{so}\\ q\\in\\mathbb{N}",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "a>0"
        ],
        "explanation": {
          "zh-Hant": "帶餘除法給出整數 q 與 r。若 q≤−1，則 aq≤−a，於是 b=aq+r≤−a+r<−a+a=0，與 b≥0 矛盾。所以 q≥0，q 是自然數。",
          "en": "The division algorithm gives integers q and r. If q≤−1, then aq≤−a, so b=aq+r≤−a+r<−a+a=0, contradicting b≥0. Hence q≥0 and q is natural."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "division_algorithm",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要 a>0 與 b≥0；這保證商可以當自然數見證。",
            "en": "It needs a>0 and b≥0; this ensures the quotient can serve as a natural witness."
          }
        },
        "leaf_basis": "division_algorithm + order_basics + contradiction"
      },
      {
        "id": "S03.2",
        "uses": [
          "S03.1",
          "S01"
        ],
        "before_tex": "b=aq+0,\\ q\\in\\mathbb{N}",
        "after_tex": "a\\mid b",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [
          "r=0"
        ],
        "explanation": {
          "zh-Hant": "（⇐ 方向）若餘數 r=0，則 b=aq，而 q 是自然數（S03.1），所以 q 是見證，a∣b。",
          "en": "(The ⇐ direction) If the remainder r=0, then b=aq with q natural (S03.1), so q is a witness and a∣b."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03.1",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "只用到 q 是自然數；沒有使用引理本身。",
            "en": "Only the fact that q is a natural number is used; the lemma itself is not used."
          }
        },
        "leaf_basis": "exists_intro + add_zero"
      },
      {
        "id": "S03.3",
        "uses": [
          "S01",
          "division_algorithm"
        ],
        "before_tex": "b=a\\cdot k,\\ k\\in\\mathbb{N}",
        "after_tex": "b\\bmod a=0",
        "relation": "implies",
        "rule": "division_algorithm",
        "conditions": [
          "a>0"
        ],
        "explanation": {
          "zh-Hant": "（⇒ 方向）若 b=ak，則 b=a·k+0，且 0≤0<a。帶餘除法的商與餘數唯一，所以 b 除以 a 的餘數就是 0（商就是 k）。S03.1 到 S03.3 合起來證得引理 L_rem_test。",
          "en": "(The ⇒ direction) If b=ak, then b=a·k+0 with 0≤0<a. The quotient and remainder of the division algorithm are unique, so the remainder of b on division by a is 0 (and the quotient is k). S03.1 to S03.3 together prove Lemma L_rem_test."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "division_algorithm"
          ],
          "explanation": {
            "zh-Hant": "關鍵是帶餘除法的唯一性；若只有存在性，不能由一種寫法推出餘數為 0。沒有使用引理本身。",
            "en": "The key is uniqueness in the division algorithm; existence alone would not let one representation determine the remainder. The lemma itself is not used."
          }
        },
        "leaf_basis": "division_algorithm + add_zero"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "G3"
        ],
        "before_tex": "87=6\\cdot 14+3,\\ 0\\le 3<6",
        "after_tex": "6\\nmid 87",
        "relation": "implies",
        "rule": "L_rem_test",
        "conditions": [
          "6>0"
        ],
        "explanation": {
          "zh-Hant": "87 除以 6 的餘數是 3≠0（N2、N3），依 S03 的等價，6∤87。直接核對也一致：6·14=84<87<90=6·15，而每個自然數 k 都有 k≤14（則 6k≤84）或 k≥15（則 6k≥90），所以沒有 k 使 6k=87。在有理數中 87/6=29/2=14.5，不是自然數。",
          "en": "The remainder of 87 on division by 6 is 3≠0 (N2, N3), so by the equivalence of S03, 6∤87. A direct check agrees: 6·14=84<87<90=6·15, and every natural k has k≤14 (then 6k≤84) or k≥15 (then 6k≥90), so no k gives 6k=87. In the rationals 87/6=29/2=14.5, which is not natural."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "nat_discrete",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "否定需要排除所有 k，S03 的等價或 nat_discrete 的兩段夾擠都做到了這一點；只檢查幾個 k 不夠。",
            "en": "A negative answer must exclude every k; the equivalence of S03, or the two-sided squeeze with nat_discrete, does so. Checking a few values of k would not be enough."
          }
        },
        "leaf_basis": "division_algorithm + nat_discrete + order_basics + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "S01",
          "S03",
          "G2"
        ],
        "before_tex": "10=4k,\\ k\\in\\mathbb{N}\\ ?",
        "after_tex": "4\\nmid 10",
        "relation": "implies",
        "rule": "mul_inverse_nonzero",
        "conditions": [
          "4\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "在有理數中 4k=10 唯一的解是 k=10/4=5/2（4≠0，可除）。5/2 不是自然數，依 G2 不能當見證，所以沒有自然數見證，4∤10。用 S03 核對：10=4·2+2，餘數 2≠0，結論相同。",
          "en": "In the rationals the only solution of 4k=10 is k=10/4=5/2 (division is allowed since 4≠0). 5/2 is not natural, so by G2 it is not a witness; there is no natural witness, and 4∤10. Check with S03: 10=4·2+2 with remainder 2≠0, the same conclusion."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S03",
            "G2",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "論證依賴「有理解唯一」：若唯一的解不是自然數，就沒有自然數解。有一個數值商不代表整除。",
            "en": "The argument relies on the rational solution being unique: if the only solution is not natural, there is no natural solution. Having a numerical quotient does not mean divisibility."
          }
        },
        "leaf_basis": "mul_inverse_nonzero + frac_rules + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S01",
          "mul_zero"
        ],
        "before_tex": "a\\in\\mathbb{N}",
        "after_tex": "0=a\\cdot 0,\\ \\text{so}\\ a\\mid 0",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對任意自然數 a，a·0=0，而 0 是自然數，所以 k=0 是見證，a∣0。a 是任意的，包括 a=0。",
          "en": "For any natural a, a·0=0 and 0 is natural, so k=0 is a witness and a∣0. a was arbitrary, including a=0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "mul_zero",
            "exists_intro",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "只用乘以零的性質；對每個 a 都成立，所以可全稱推廣。",
            "en": "Only multiplication by zero is used; it holds for every a, so the statement generalizes."
          }
        },
        "leaf_basis": "mul_zero + exists_intro + forall_intro"
      },
      {
        "id": "S07",
        "uses": [
          "S01",
          "mul_zero"
        ],
        "before_tex": "0\\mid b",
        "after_tex": "b=0",
        "relation": "equivalent",
        "rule": "mul_zero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "0∣b 表示存在自然數 k 使 b=0·k。但對每個 k 都有 0·k=0，所以這句話等於說 b=0。",
          "en": "0∣b means there is a natural k with b=0·k. But 0·k=0 for every k, so this says exactly that b=0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_zero",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "反向：若 b=0，取 k=0 得 b=0·0，所以 0∣b。兩向都成立。",
            "en": "Backward: if b=0, take k=0 to get b=0·0, so 0∣b. Both directions hold."
          }
        },
        "leaf_basis": "mul_zero + mul_comm + exists_intro"
      },
      {
        "id": "S08",
        "uses": [
          "S07",
          "G3"
        ],
        "before_tex": "0=0\\cdot 0=0\\cdot 5,\\quad 5\\neq 0",
        "after_tex": "0\\mid 0\\ \\text{with non-unique witness};\\ 0\\nmid 5",
        "relation": "implies",
        "rule": "mul_zero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "0∣0 成立，而且 k=0 與 k=5 都是見證（0·5=0，N9），見證不唯一。5≠0，所以依 S07，0∤5。可見「除數是 0 就一律不整除」是錯的：0∣0 為真。",
          "en": "0∣0 holds, and both k=0 and k=5 are witnesses (0·5=0, N9), so the witness is not unique. Since 5≠0, S07 gives 0∤5. So 'zero never divides any number' is wrong: 0∣0 is true."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "只是 S07 的兩個實例與一個見證不唯一的例子。",
            "en": "Two instances of S07 and one example of a non-unique witness."
          }
        },
        "leaf_basis": "mul_zero + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S01",
          "order_basics"
        ],
        "before_tex": "b=a k=a k',\\ a>0,\\ k,k'\\in\\mathbb{N}",
        "after_tex": "k=k'",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "a>0"
        ],
        "explanation": {
          "zh-Hant": "依三分律，k<k′、k=k′、k>k′ 恰有一個成立。若 k<k′，因 a>0 得 ak<ak′，與 ak=ak′ 矛盾；k>k′ 同理。所以 k=k′：在見證存在時它唯一，亦即 a>0 時見證至多一個。這與 S08 中 a=0 的情形形成對比。",
          "en": "By trichotomy exactly one of k<k′, k=k′, k>k′ holds. If k<k′, then a>0 gives ak<ak′, contradicting ak=ak′; likewise for k>k′. So k=k′: an existing witness is unique, or equivalently there is at most one witness when a>0. This contrasts with the case a=0 in S08."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "需要 a>0 才能保持嚴格不等；a=0 時結論不成立（S08）。",
            "en": "a>0 is needed to preserve strict inequality; for a=0 the conclusion fails (S08)."
          }
        },
        "leaf_basis": "order_basics + case_analysis + contradiction"
      },
      {
        "id": "S10",
        "uses": [
          "S01",
          "mul_one"
        ],
        "before_tex": "a\\in\\mathbb{N}",
        "after_tex": "a=a\\cdot 1,\\ \\text{so}\\ a\\mid a",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "自反性：a·1=a，1 是自然數，所以 a∣a；包括 0∣0（0=0·1）。",
          "en": "Reflexivity: a·1=a and 1 is natural, so a∣a; this includes 0∣0 (0=0·1)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mul_one",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "只用乘法單位元。",
            "en": "Only the multiplicative identity is used."
          }
        },
        "leaf_basis": "mul_one + mul_comm + exists_intro"
      },
      {
        "id": "S11",
        "uses": [
          "S01"
        ],
        "before_tex": "a\\mid b,\\ b\\mid c",
        "after_tex": "a\\mid c",
        "relation": "implies",
        "rule": "exists_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "遞移性：由 a∣b 取見證 k 使 b=ak；由 b∣c 取見證 l 使 c=bl（S11.1）。代入並用結合律得 c=a(kl)，且 kl 是自然數（S11.2），所以 a∣c，見證是 kl。",
          "en": "Transitivity: from a∣b take a witness k with b=ak; from b∣c take a witness l with c=bl (S11.1). Substituting and using associativity gives c=a(kl), with kl natural (S11.2), so a∣c with witness kl."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "exists_elim",
            "mul_assoc"
          ],
          "explanation": {
            "zh-Hant": "結論不含暫取的名字 k、l，所以 ∃消去合法。不宣稱反向：a∣c 不推出 a∣b。",
            "en": "The conclusion does not mention the temporary names k and l, so ∃-elimination is legitimate. No converse is claimed: a∣c does not give a∣b."
          }
        },
        "leaf_basis": "exists_elim + mul_assoc + exists_intro"
      },
      {
        "id": "S11.1",
        "uses": [
          "S01",
          "exists_elim"
        ],
        "before_tex": "b=ak,\\ c=bl",
        "after_tex": "c=(ak)l",
        "relation": "implies",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 c=bl 中的 b 換成相等的 ak。",
          "en": "Replace b in c=bl by the equal expression ak."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "等值代換。",
            "en": "Replacement of equals."
          }
        },
        "leaf_basis": "eq_replace"
      },
      {
        "id": "S11.2",
        "uses": [
          "S11.1",
          "G1"
        ],
        "before_tex": "c=(ak)l",
        "after_tex": "c=a(kl),\\ kl\\in\\mathbb{N}",
        "relation": "implies",
        "rule": "mul_assoc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "乘法結合律給 (ak)l=a(kl)；自然數的積仍是自然數（G1），所以 kl 是合法見證。",
          "en": "Associativity gives (ak)l=a(kl); a product of naturals is natural (G1), so kl is a legitimate witness."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mul_assoc",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "需要 ℕ 對乘法封閉；在有理數中這一步會允許非自然數見證。",
            "en": "It needs ℕ to be closed under multiplication; over the rationals this step would allow non-natural witnesses."
          }
        },
        "leaf_basis": "mul_assoc"
      },
      {
        "id": "S12",
        "uses": [
          "S11",
          "G3"
        ],
        "before_tex": "6=3\\cdot 2,\\ 84=6\\cdot 14",
        "after_tex": "84=3\\cdot 28,\\ 3\\mid 84",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "3∣6 的見證 k=2，6∣84 的見證 l=14；依 S11，3∣84 的見證是 kl=28。核對：3·28=84，84 除以 3 的餘數為 0（N10 到 N12）。",
          "en": "3∣6 has witness k=2 and 6∣84 has witness l=14; by S11, 3∣84 has witness kl=28. Check: 3·28=84 and 84 leaves remainder 0 on division by 3 (N10 to N12)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "forall_elim",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "這是 S11 的一個實例並附數值核對；它不能取代一般證明。",
            "en": "An instance of S11 with a numeric check; it cannot replace the general proof."
          }
        },
        "leaf_basis": "forall_elim + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S03",
          "S07",
          "S02",
          "S04",
          "S05",
          "S08"
        ],
        "before_tex": "(a,b)\\in\\mathbb{N}^2",
        "after_tex": "a=0:\\ a\\mid b\\iff b=0;\\quad a>0:\\ a\\mid b\\iff b\\bmod a=0",
        "relation": "case_split",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "判定程序：先看 a=0，此時 a∣b ⇔ b=0（S07）；否則 a>0，a∣b ⇔ 餘數為 0（S03）。兩種情況涵蓋所有自然數 a。套用：(6,84) 餘數 0，整除；(6,87) 餘數 3，不整除；(4,10) 餘數 2，不整除；(0,0) 整除；(0,5) 不整除。",
          "en": "Decision procedure: first check a=0, where a∣b ⇔ b=0 (S07); otherwise a>0 and a∣b ⇔ the remainder is 0 (S03). The two cases cover every natural a. Applied: (6,84) remainder 0, divides; (6,87) remainder 3, does not; (4,10) remainder 2, does not; (0,0) divides; (0,5) does not."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S03",
            "S07",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "目標 G：對每個數對正確判定。足夠條件 H：a=0 與 a>0 兩種情況各有已證的判準；H⇒G 由分情況論證。H 由 S03、S07 正向證成。",
            "en": "Goal G: decide every pair correctly. Sufficient condition H: each of the cases a=0 and a>0 has a proved criterion; H⇒G by proof by cases. H is proved forwards in S03 and S07."
          }
        },
        "leaf_basis": "case_analysis + order_basics"
      }
    ],
    "result": {
      "zh-Hant": "6∣84（見證 14）；6∤87（餘數 3）；4∤10（唯一的有理解 5/2 不是自然數，餘數 2）；每個 a 都整除 0（見證 0）；0∣b ⇔ b=0，所以 0∣0 而 0∤5，且 0∣0 的見證不唯一；a>0 時見證至多一個，存在時唯一；整除具自反性與遞移性，3∣84 的見證為 2·14=28。",
      "en": "6∣84 (witness 14); 6∤87 (remainder 3); 4∤10 (the only rational solution 5/2 is not natural; remainder 2); every a divides 0 (witness 0); 0∣b ⇔ b=0, so 0∣0 but 0∤5, and the witness for 0∣0 is not unique; for a>0 there is at most one witness, unique when it exists; divisibility is reflexive and transitive, and 3∣84 has witness 2·14=28."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "4∣10，因為 10/4=2.5 是一個數",
          "en": "4∣10 because 10/4=2.5 is a number"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "見證必須是自然數；10=4·2+2，餘數 2≠0。",
          "en": "The witness must be natural; 10=4·2+2 with remainder 2≠0."
        }
      },
      {
        "claim": {
          "zh-Hant": "0 不整除任何數",
          "en": "0 divides no number"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "0=0·0，所以 0∣0。",
          "en": "0=0·0, so 0∣0."
        }
      },
      {
        "claim": {
          "zh-Hant": "0∣5",
          "en": "0∣5"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "對每個 k，0·k=0≠5。",
          "en": "For every k, 0·k=0≠5."
        }
      },
      {
        "claim": {
          "zh-Hant": "6∣3，因為 3=6·(1/2)",
          "en": "6∣3 because 3=6·(1/2)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "1/2 不是自然數；3=6·0+3，餘數 3≠0。",
          "en": "1/2 is not natural; 3=6·0+3 with remainder 3≠0."
        }
      },
      {
        "claim": {
          "zh-Hant": "0∣0 的見證唯一",
          "en": "The witness for 0∣0 is unique"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "0=0·0=0·5，k=0 與 k=5 都是見證。",
          "en": "0=0·0=0·5, so k=0 and k=5 are both witnesses."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "division_algorithm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "帶餘除法：對整數 a 與正整數 b，存在唯一的整數 q、r 使 a=bq+r 且 0≤r<b。",
          "en": "Division algorithm: for an integer a and a positive integer b there are unique integers q, r with a=bq+r and 0≤r<b."
        }
      },
      {
        "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": "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": "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": "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": "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": "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": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "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": "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": "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": "exists_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃消去：由 ∃x∈D, P(x)，可取一個新名字 a 表示某個滿足 P(a) 的元素；只要最後結論不含 a，就可用 a 推理。",
          "en": "∃-elimination: from ∃x∈D, P(x) we may name some element a with P(a) by a fresh name; reasoning with a is allowed as long as the final conclusion does not mention a."
        }
      },
      {
        "id": "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": "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": "mko_dvd_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-natural-divisibility",
          "sha256": "2e5d00faa3ae3031b2cb2a650b12db0ab8e8aa1b59a7ea6b2350f49a792758d7",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對自然數 a、b，a∣b 當且僅當存在自然數 k 使 b=a·k。",
          "en": "For natural numbers a and b, a∣b exactly when there is a natural number k such that b=a·k."
        }
      },
      {
        "id": "nat_discrete",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "自然數沒有中間值：對自然數 k 與 n，必有 k≤n 或 k≥n+1。",
          "en": "Naturals have no values in between: for naturals k and n, either k≤n or k≥n+1."
        }
      },
      {
        "id": "L_rem_test",
        "kind": "lemma",
        "proved_in": [
          "S03.1",
          "S03.2",
          "S03.3"
        ],
        "statement": {
          "zh-Hant": "對自然數 b 與正自然數 a：a∣b ⇔ b 除以 a 的餘數 r 為 0。",
          "en": "For a natural b and a positive natural a: a∣b ⇔ the remainder r of b on division by a is 0."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "6*14",
        "rhs": "84"
      },
      {
        "id": "N2",
        "step": "S04",
        "lhs": "6*14+3",
        "rhs": "87"
      },
      {
        "id": "N3",
        "step": "S04",
        "lhs": "87 % 6",
        "rhs": "3"
      },
      {
        "id": "N4",
        "step": "S04",
        "lhs": "6*15",
        "rhs": "90"
      },
      {
        "id": "N5",
        "step": "S04",
        "lhs": "87/6",
        "rhs": "29/2"
      },
      {
        "id": "N6",
        "step": "S05",
        "lhs": "10/4",
        "rhs": "5/2"
      },
      {
        "id": "N7",
        "step": "S05",
        "lhs": "4*2+2",
        "rhs": "10"
      },
      {
        "id": "N8",
        "step": "S05",
        "lhs": "10 % 4",
        "rhs": "2"
      },
      {
        "id": "N9",
        "step": "S08",
        "lhs": "0*5",
        "rhs": "0"
      },
      {
        "id": "N10",
        "step": "S12",
        "lhs": "2*14",
        "rhs": "28"
      },
      {
        "id": "N11",
        "step": "S12",
        "lhs": "3*28",
        "rhs": "84"
      },
      {
        "id": "N12",
        "step": "S12",
        "lhs": "84 % 3",
        "rhs": "0"
      },
      {
        "id": "N13",
        "step": "S13",
        "lhs": "84 % 6",
        "rhs": "0"
      },
      {
        "id": "N14",
        "lhs": "3 % 6",
        "rhs": "3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只在包含 0 的自然數中工作（MKO 假設 a1），見證必須是自然數（a2）。為判定 a>0 的情形，本稿由帶餘除法（引用的前置規則）證明「a∣b ⇔ 餘數為 0」。不處理負整數的整除、最大公因數或質數；不改動 canonical MKO 或英文 edition。",
      "en": "Works in the natural numbers including 0 (MKO assumption a1), with natural witnesses only (a2). To decide the case a>0, the guide proves 'a∣b ⇔ the remainder is 0' from the division algorithm (a cited prerequisite). Divisibility of negative integers, greatest common divisors and primes are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "a\\mid b",
        "a 整除 b",
        "a divides b"
      ],
      [
        "a\\nmid b",
        "a 不整除 b",
        "a does not divide b"
      ],
      [
        "k",
        "乘數見證，必須是自然數",
        "the multiplier witness, which must be natural"
      ],
      [
        "b\\bmod a",
        "b 除以 a（a>0）的餘數",
        "the remainder of b on division by a (a>0)"
      ],
      [
        "\\mathbb{N}",
        "自然數，包含 0",
        "the natural numbers, including 0"
      ]
    ],
    "md_error_controls": [
      {
        "title": "有數值商不等於整除 / Having a numerical quotient is not divisibility",
        "zh": "MKO 指出的誤解：看到 b/a 是一個數就宣稱 a∣b。10/4=5/2 是有理數，但見證必須是自然數，所以 4∤10（S05）。判定要看自然數見證或餘數是否為 0。",
        "en": "The misconception the MKO names: claiming a∣b because b/a is a number. 10/4=5/2 is a rational number, but the witness must be natural, so 4∤10 (S05). Decide by a natural witness or by whether the remainder is 0."
      },
      {
        "title": "除數為 0 不是一律不整除 / A divisor equal to zero does not make every divisibility claim false",
        "zh": "MKO 指出的另一個誤解：把除數為零一律判成沒有整除關係。依定義 0∣b ⇔ b=0（S07），所以 0∣0 成立，0∤5。注意這裡沒有做「除以 0」，只是找乘數見證。",
        "en": "The other misconception the MKO names: ruling that zero never divides any number. By the definition 0∣b ⇔ b=0 (S07), so 0∣0 holds and 0∤5. Note that no 'division by 0' is performed; one only looks for a multiplier witness."
      },
      {
        "title": "否定要排除所有見證 / A negative answer must exclude every witness",
        "zh": "證明 6∤87 時，試 k=14、15 兩個值不夠；必須用餘數引理或 nat_discrete 的夾擠排除所有 k（S04）。同理，6∣3 不成立：3=6·(1/2) 的 1/2 不是自然數，3 除以 6 的餘數是 3。",
        "en": "To prove 6∤87, trying k=14 and 15 is not enough; the remainder lemma or the squeeze with nat_discrete must exclude every k (S04). Likewise 6∣3 fails: in 3=6·(1/2) the factor 1/2 is not natural, and 3 leaves remainder 3 on division by 6."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "帶餘除法（存在與唯一）作為前置規則引用，本稿只由它推出餘數判準 L_rem_test。",
        "en": "The division algorithm (existence and uniqueness) is cited as a prerequisite; this guide only derives the remainder criterion L_rem_test from it."
      },
      {
        "zh-Hant": "本稿只處理 ℕ；含負數的整數整除不在範圍內。",
        "en": "Only ℕ is treated; divisibility among integers with negative values is out of scope."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "c7700b425af628e5a943830098f9039a8ba739a667d2577ec46189e0e2124324",
    "original_source_sha256": "1b68c27c071f5fe0cfe48fffd7605f00b4dfae82833e7c47a8a473157438e68f",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M031"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}