{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-proof-01",
  "mko_id": "mko-proof",
  "title": {
    "zh-Hant": "數學證明：證明 n²+n 恆為偶數，並以反例否定「n²+n+41 恆為質數」",
    "en": "Mathematical proof: proving n²+n is always even, and refuting 'n²+n+41 is always prime' by a counterexample"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-proof-01",
    "source": {
      "mko_id": "mko-proof",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "72f6009b912caa3c05a2948a97ca422367574a19ad5fc665964a540875417b91",
      "english_sha256": "1c8ec335b48f7d913d3057b2d632c371b092ca869f799b056b72464d948504fe"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "division_algorithm",
      "divides_def",
      "distributive",
      "power_laws",
      "case_analysis",
      "forall_intro",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "(1) 在整數的指定語境中，由明示的前提與規則證明定理 T：對每個整數 n，2∣(n²+n)；再由 T 以反證法推出沒有整數 n 使 n²+n 為奇數。(2) 檢驗主張 C：「對每個自然數 n，n²+n+41 是質數」；說明 n=0 到 39 的 40 個正例為何不構成證明，並以 n=40 的反例否定 C。",
      "en": "(1) In a stated context for the integers, prove from explicit premises and rules the theorem T: for every integer n, 2∣(n²+n); then derive from T by contradiction that no integer n makes n²+n odd. (2) Test the claim C: 'for every natural number n, n²+n+41 is prime'; explain why the 40 positive cases n=0 to 39 do not constitute a proof, and refute C by the counterexample n=40."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "語境與可用前提：整數 ℤ 與 ℕ={n∈ℤ ∣ n≥0}，使用規則表中的環運算、順序、整除定義與帶餘除法（含唯一性）；推理規則為古典邏輯的分情況、全稱推廣、反證法與量詞否定。",
          "en": "Context and usable premises: the integers ℤ and ℕ={n∈ℤ ∣ n≥0}, with the ring operations, order, definition of divisibility and division algorithm (including uniqueness) from the glossary; inference rules are the classical ones for cases, generalization, contradiction and negating quantifiers."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "待證定理 T：對每個整數 n，2∣(n²+n)。",
          "en": "Theorem T to prove: for every integer n, 2∣(n²+n)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "被檢驗的主張 C：對每個 n∈ℕ，n²+n+41 是質數；質數指整數 p≥2，且 p 的正因數只有 1 與 p。",
          "en": "Claim C under test: for every n∈ℕ, n²+n+41 is prime, where a prime is an integer p≥2 whose only positive divisors are 1 and p."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "\\text{prove}\\ T",
        "after_tex": "\\text{premises of G1}\\ +\\ \\text{valid rules}\\ \\Rightarrow\\ T",
        "relation": "definition",
        "rule": "mko_proof",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，證明要在指定語境中，由已接受的前提、定義或先前結果，經合法推理得到待證命題。所以先寫清楚：待證命題是 T（G2）；可用前提是 G1 所列的整數運算、順序、整除定義與帶餘除法；可用的推理是古典邏輯規則。之後每一步都要能指出所用的前提與規則。",
          "en": "By the MKO, a proof derives the proposition within a stated context from accepted premises, definitions or earlier results by valid inference. So state first: the proposition is T (G2); the usable premises are the integer operations, order, definition of divisibility and division algorithm of G1; the usable inferences are the rules of classical logic. Every later step must name the premises and rules it uses."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_proof"
          ],
          "explanation": {
            "zh-Hant": "此步只確定證明的語境，不推出新事實。",
            "en": "This step only fixes the context of the proof and derives no new fact."
          }
        },
        "leaf_basis": "mko_proof + def_intro"
      },
      {
        "id": "S02",
        "uses": [
          "G1"
        ],
        "before_tex": "n\\in\\mathbb{Z}\\ \\text{arbitrary}",
        "after_tex": "n=2q+r,\\ q,r\\in\\mathbb{Z},\\ 0\\le r<2",
        "relation": "theorem_application",
        "rule": "division_algorithm",
        "conditions": [
          "2>0"
        ],
        "explanation": {
          "zh-Hant": "任取一個整數 n，不對它加任何額外假設（為了最後的全稱推廣）。依帶餘除法（a=n、b=2，2 是正整數），存在整數 q、r 使 n=2q+r 且 0≤r<2。這對負整數也成立，例如 −3=2·(−2)+1。",
          "en": "Take an arbitrary integer n with no extra assumption (so that generalization is allowed at the end). By the division algorithm (a=n, b=2, and 2 is a positive integer) there are integers q, r with n=2q+r and 0≤r<2. This holds for negative integers too, for example −3=2·(−2)+1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "division_algorithm",
            "two_nonzero"
          ],
          "explanation": {
            "zh-Hant": "帶餘除法只要求除數是正整數；n 可以是任何整數。",
            "en": "The division algorithm only requires a positive divisor; n may be any integer."
          }
        },
        "leaf_basis": "division_algorithm"
      },
      {
        "id": "S03",
        "uses": [
          "S02"
        ],
        "before_tex": "0\\le r<2,\\ r\\in\\mathbb{Z}",
        "after_tex": "r=0\\ \\lor\\ r=1",
        "relation": "case_split",
        "rule": "int_no_gap",
        "conditions": [],
        "explanation": {
          "zh-Hant": "r 是整數且 0≤r<2。依 int_no_gap（m=0），r≤0 或 r≥1。若 r≤0，配合 0≤r 得 r=0；若 r≥1，再依 int_no_gap（m=1），r≤1 或 r≥2，而 r<2 排除後者，所以 r=1。因此 r=0 或 r=1，兩個情形涵蓋所有可能。",
          "en": "r is an integer with 0≤r<2. By int_no_gap (m=0), r≤0 or r≥1. If r≤0, together with 0≤r this gives r=0; if r≥1, int_no_gap (m=1) gives r≤1 or r≥2, and r<2 excludes the latter, so r=1. Hence r=0 or r=1, and the two cases cover every possibility."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "int_no_gap",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "兩支涵蓋全部可能，由 r 的範圍與整數沒有空隙保證，不是靠看例子。",
            "en": "That the two branches cover everything follows from the range of r and the absence of gaps in the integers, not from looking at examples."
          }
        },
        "leaf_basis": "int_no_gap + order_basics + case_analysis"
      },
      {
        "id": "S04",
        "uses": [
          "S03"
        ],
        "before_tex": "r=0:\\ n=2q",
        "after_tex": "n^2+n=(2q)^2+2q=4q^2+2q",
        "relation": "equivalent",
        "rule": "power_laws",
        "conditions": [],
        "explanation": {
          "zh-Hant": "情形 r=0：n=2q。代入得 n²+n=(2q)²+2q；依指數律 (2q)²=2²q²=4q²，所以 n²+n=4q²+2q。",
          "en": "Case r=0: n=2q. Substituting gives n²+n=(2q)²+2q; by the exponent laws (2q)²=2²q²=4q², so n²+n=4q²+2q."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "power_laws",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "只是等值代換，兩邊是同一個整數。",
            "en": "Only a replacement of equals; both sides are the same integer."
          }
        },
        "leaf_basis": "power_laws + eq_replace + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "S04"
        ],
        "before_tex": "4q^2+2q",
        "after_tex": "2(2q^2+q),\\ \\text{so}\\ 2\\mid n^2+n",
        "relation": "implies",
        "rule": "divides_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依分配律 4q²+2q=2(2q²+q)；2q²+q 是整數（整數對加法與乘法封閉），所以依整除定義 2∣(n²+n)。",
          "en": "By distributivity 4q²+2q=2(2q²+q); 2q²+q is an integer (the integers are closed under addition and multiplication), so 2∣(n²+n) by the definition of divisibility."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "distributive",
            "divides_def",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "需要一個整數見證 k=2q²+q；N1 是對所用恆等式的獨立多項式核對。",
            "en": "An integer witness k=2q²+q is needed; N1 is an independent polynomial check of the identity used."
          }
        },
        "leaf_basis": "distributive + divides_def + exists_intro"
      },
      {
        "id": "S06",
        "uses": [
          "S03"
        ],
        "before_tex": "r=1:\\ n=2q+1",
        "after_tex": "n^2+n=(2q+1)^2+(2q+1)=4q^2+6q+2",
        "relation": "equivalent",
        "rule": "L_odd_case",
        "conditions": [],
        "explanation": {
          "zh-Hant": "情形 r=1：n=2q+1。代入並依引理 L_odd_case 展開，得 n²+n=4q²+6q+2。",
          "en": "Case r=1: n=2q+1. Substituting and expanding by Lemma L_odd_case gives n²+n=4q²+6q+2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_odd_case"
          ],
          "explanation": {
            "zh-Hant": "等值代換，雙向成立；引理在子步中證明。",
            "en": "A replacement of equals, two-way; the lemma is proved in the sub-steps."
          }
        },
        "leaf_basis": "distributive + power_laws + add_assoc + add_comm + numeric_evaluation"
      },
      {
        "id": "S06.1",
        "uses": [
          "distributive",
          "power_laws"
        ],
        "before_tex": "(2q+1)^2",
        "after_tex": "4q^2+4q+1",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(2q+1)²=(2q+1)(2q+1)=2q·2q+2q·1+1·2q+1·1=4q²+2q+2q+1=4q²+4q+1（分配律兩次、乘法交換律、合併同類項）。",
          "en": "(2q+1)²=(2q+1)(2q+1)=2q·2q+2q·1+1·2q+1·1=4q²+2q+2q+1=4q²+4q+1 (distributivity twice, commutativity of multiplication, collecting like terms)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "mul_comm",
            "add_assoc"
          ],
          "explanation": {
            "zh-Hant": "恆等式，倒著讀同樣成立；N3 是獨立核對。",
            "en": "An identity, valid read backwards; N3 is an independent check."
          }
        },
        "leaf_basis": "distributive + power_laws + mul_comm + add_assoc + numeric_evaluation"
      },
      {
        "id": "S06.2",
        "uses": [
          "S06.1"
        ],
        "before_tex": "(4q^2+4q+1)+(2q+1)",
        "after_tex": "4q^2+6q+2",
        "relation": "equivalent",
        "rule": "add_assoc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依加法結合律與交換律重排：4q²+(4q+2q)+(1+1)；依分配律 4q+2q=(4+2)q=6q，且 1+1=2，得 4q²+6q+2。S06.1 與 S06.2 合起來證得引理 L_odd_case。",
          "en": "Rearrange by associativity and commutativity of addition: 4q²+(4q+2q)+(1+1); by distributivity 4q+2q=(4+2)q=6q, and 1+1=2, giving 4q²+6q+2. S06.1 and S06.2 together prove Lemma L_odd_case."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "add_assoc",
            "add_comm",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "只用恆等式重排，雙向成立。",
            "en": "Only identities are used to rearrange, so the step is two-way."
          }
        },
        "leaf_basis": "add_assoc + add_comm + distributive + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S06"
        ],
        "before_tex": "4q^2+6q+2",
        "after_tex": "2(2q^2+3q+1),\\ \\text{so}\\ 2\\mid n^2+n",
        "relation": "implies",
        "rule": "divides_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依分配律 4q²+6q+2=2(2q²+3q+1)；2q²+3q+1 是整數，所以依整除定義 2∣(n²+n)。",
          "en": "By distributivity 4q²+6q+2=2(2q²+3q+1); 2q²+3q+1 is an integer, so 2∣(n²+n) by the definition of divisibility."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "distributive",
            "divides_def",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "需要整數見證 2q²+3q+1；N2 獨立核對整個恆等式。",
            "en": "The integer witness 2q²+3q+1 is needed; N2 checks the whole identity independently."
          }
        },
        "leaf_basis": "distributive + divides_def + exists_intro"
      },
      {
        "id": "S08",
        "uses": [
          "S03",
          "S05",
          "S07"
        ],
        "before_tex": "r=0\\Rightarrow 2\\mid n^2+n,\\quad r=1\\Rightarrow 2\\mid n^2+n",
        "after_tex": "2\\mid n^2+n",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S03 的兩個情形涵蓋全部可能，兩個情形都推出 2∣(n²+n)，依分情況論證，對這個任取的 n 有 2∣(n²+n)。",
          "en": "The two cases of S03 cover every possibility and both yield 2∣(n²+n), so by proof by cases 2∣(n²+n) for this arbitrary n."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S05",
            "S07",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "若只處理 r=0，就只證明了偶數 n 的情形；兩支都必須做。",
            "en": "Treating only r=0 would prove the claim only for even n; both branches are required."
          }
        },
        "leaf_basis": "case_analysis"
      },
      {
        "id": "S09",
        "uses": [
          "S02",
          "S08"
        ],
        "before_tex": "2\\mid n^2+n\\ \\text{for the arbitrary}\\ n",
        "after_tex": "\\forall n\\in\\mathbb{Z}:\\ 2\\mid n^2+n\\quad(T)",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S02 中的 n 是任取的，S02 到 S08 沒有對 n 加任何額外假設，所以依全稱推廣，對每個整數 n 都有 2∣(n²+n)，定理 T 得證。",
          "en": "The n of S02 was arbitrary and S02 to S08 placed no extra assumption on it, so by universal generalization 2∣(n²+n) for every integer n: theorem T is proved."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S02",
            "S08",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "目標 G：T。足夠條件 H：對一個沒有額外假設的任意整數 n 證得 2∣(n²+n)；H⇒G 由全稱推廣。H 由 S02 到 S08 從 G1 的前提正向證成，沒有把 T 當作前提。N4 在 n=−20 到 20 的有限核對只是交叉核對，不是證明的一部分。",
            "en": "Goal G: T. Sufficient condition H: 2∣(n²+n) proved for an arbitrary integer n with no extra assumption; H⇒G by generalization. H is proved forwards from the premises of G1 in S02 to S08, never using T as a premise. The finite check N4 for n=−20 to 20 is only a cross-check, not part of the proof."
          }
        },
        "leaf_basis": "forall_intro"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "division_algorithm"
        ],
        "before_tex": "\\text{suppose}\\ n^2+n=2m+1\\ \\text{for some}\\ n,m\\in\\mathbb{Z}",
        "after_tex": "\\neg\\exists n,m\\in\\mathbb{Z}:\\ n^2+n=2m+1",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "反證法：假設有整數 n、m 使 n²+n=2m+1。由 T（S09）又有整數 k 使 n²+n=2k=2k+0。於是同一個整數 n²+n 除以 2 有兩種帶餘表示，餘數分別是 1 與 0，都滿足 0≤r<2，與帶餘除法的唯一性矛盾。所以沒有整數 n 使 n²+n 為奇數。這是 MKO 所說的反證法：加入結論的否定並導出矛盾。",
          "en": "By contradiction: suppose integers n, m satisfy n²+n=2m+1. By T (S09) there is also an integer k with n²+n=2k=2k+0. Then the single integer n²+n has two divisions by 2 with remainders 1 and 0, both satisfying 0≤r<2, contradicting uniqueness in the division algorithm. So no integer n makes n²+n odd. This is the method of contradiction the MKO describes: add the negation of the conclusion and derive a contradiction."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "division_algorithm",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "矛盾來自餘數唯一；暫時假設中的 n、m 是新名字，結論中沒有它們的自由出現。",
            "en": "The contradiction comes from uniqueness of the remainder; the n, m of the temporary assumption are fresh names with no free occurrence in the final conclusion."
          }
        },
        "leaf_basis": "contradiction + division_algorithm + divides_def"
      },
      {
        "id": "S11",
        "uses": [
          "G3"
        ],
        "before_tex": "C:\\ \\forall n\\in\\mathbb{N},\\ n^2+n+41\\ \\text{is prime}",
        "after_tex": "n^2+n+41\\ \\text{is prime for}\\ 0\\le n\\le 39\\ \\text{(finite check only)}",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對 n=0 到 39 逐一計算 n²+n+41 並檢查是否為質數（N5，在 40 個整數點上的有限核對），全部是質數。但 C 是對所有自然數的全稱命題，ℕ 是無窮的；這 40 個正例既沒有證明論域有限，也沒有窮盡論域，因此不能建立 C。MKO 第一段正是這樣說的。",
          "en": "For n=0 to 39 compute n²+n+41 and test for primality (N5, a finite check at 40 integer points): all are prime. But C is a universal statement about all natural numbers and ℕ is infinite; these 40 positive cases neither show the domain is finite nor exhaust it, so they cannot establish C. This is exactly what the MKO's first paragraph says."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "numeric_evaluation",
            "prime_def"
          ],
          "explanation": {
            "zh-Hant": "此步只記錄 40 個有限的計算結果；它不推出 C，後面也沒有把它當作 C 的證明。",
            "en": "This step only records 40 finite computations; it does not imply C, and no later step uses it as a proof of C."
          }
        },
        "leaf_basis": "numeric_evaluation + prime_def"
      },
      {
        "id": "S12",
        "uses": [
          "G3"
        ],
        "before_tex": "n=40:\\ 40^2+40+41",
        "after_tex": "1681=41\\cdot 41",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取 n=40∈ℕ：40²+40+41=1600+40+41=1681；另一方面 41·41=1681。也看得出結構：40²+40+41=40·(40+1)+41=40·41+41=41·41。",
          "en": "Take n=40∈ℕ: 40²+40+41=1600+40+41=1681; and 41·41=1681. The structure is visible too: 40²+40+41=40·(40+1)+41=40·41+41=41·41."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個整數，可重算（N6、N7、N10）。",
            "en": "Both sides are the same integer and can be recomputed (N6, N7, N10)."
          }
        },
        "leaf_basis": "numeric_evaluation + distributive"
      },
      {
        "id": "S13",
        "uses": [
          "S12"
        ],
        "before_tex": "1681=41\\cdot 41",
        "after_tex": "1681\\ \\text{is not prime}",
        "relation": "implies",
        "rule": "prime_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "41 是 1681 的正因數（1681=41·41），而 41≠1 且 41≠1681。依質數定義，1681 的正因數不只 1 與自身，所以 1681 不是質數。",
          "en": "41 is a positive divisor of 1681 (1681=41·41), and 41≠1 and 41≠1681. By the definition, 1681 has positive divisors other than 1 and itself, so 1681 is not prime."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "prime_def",
            "divides_def"
          ],
          "explanation": {
            "zh-Hant": "只需一個非平凡的正因數。",
            "en": "One nontrivial positive divisor is enough."
          }
        },
        "leaf_basis": "prime_def + divides_def + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S12",
          "S13",
          "G3"
        ],
        "before_tex": "40\\in\\mathbb{N},\\ 40^2+40+41\\ \\text{not prime}",
        "after_tex": "\\neg C",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "n=40∈ℕ（40≥0）使 n²+n+41 不是質數，所以 ∃n∈ℕ 使 n²+n+41 不是質數，依量詞否定 C 為假。一個有效反例就足以否定全稱命題，不論之前有多少正例。",
          "en": "n=40∈ℕ (40≥0) makes n²+n+41 not prime, so ∃n∈ℕ with n²+n+41 not prime, and by negation of quantifiers C is false. One valid counterexample refutes a universal statement, however many positive cases came before."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13",
            "exists_intro",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "需要反例確實在論域 ℕ 中，且確實不滿足性質（S13）。",
            "en": "The counterexample must lie in the domain ℕ and must really fail the property (S13)."
          }
        },
        "leaf_basis": "quantifier_negation + exists_intro"
      },
      {
        "id": "S15",
        "uses": [
          "S09",
          "S10",
          "S11",
          "S14"
        ],
        "before_tex": "\\text{conclusions of S09, S10, S11, S14}",
        "after_tex": "T\\ \\text{proved};\\ C\\ \\text{refuted}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "T 由明示前提經合法推理證得，對所有整數成立（S09），並推出 n²+n 永不為奇數（S10）。C 雖有 40 個正例（S11），卻被 n=40 的單一反例否定（S14）。前者是證明；後者說明抽樣不是證明。",
          "en": "T is derived from explicit premises by valid inference and holds for every integer (S09), and it follows that n²+n is never odd (S10). C has 40 positive cases (S11) yet is refuted by the single counterexample n=40 (S14). The first is a proof; the second shows that sampling is not proof."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S09",
            "S10",
            "S14",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：證明 T 並判定 C。足夠條件 H：S09、S10 與 S14；H⇒G 只是並列。H 由 G1 到 G3 正向證成，沒有把 G 當作前提；S11 只是對照，不在 H 中。",
            "en": "Goal G: prove T and decide C. Sufficient condition H: S09, S10 and S14; H⇒G just lists them. H is proved forwards from G1 to G3 without using G as a premise; S11 is only a contrast and is not part of H."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "定理 T 成立：對每個整數 n，2∣(n²+n)；證明以帶餘除法把 n 分成 n=2q 與 n=2q+1 兩個情形，分別得 n²+n=2(2q²+q) 與 n²+n=2(2q²+3q+1)，再全稱推廣。由 T 以反證法得：沒有整數 n 使 n²+n 為奇數。主張 C 為假：雖然 n=0 到 39 時 n²+n+41 都是質數，但 n=40 時它等於 1681=41·41。",
      "en": "Theorem T holds: for every integer n, 2∣(n²+n); the proof splits n by the division algorithm into n=2q and n=2q+1, obtaining n²+n=2(2q²+q) and n²+n=2(2q²+3q+1), and then generalizes. From T, by contradiction, no integer n makes n²+n odd. Claim C is false: although n²+n+41 is prime for n=0 to 39, at n=40 it equals 1681=41·41."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "已計算 n=1 到 100 時 n²+n 都是偶數，所以 T 得證",
          "en": "n²+n was computed to be even for n=1 to 100, so T is proved"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "有限個情形不涵蓋 ℤ（無窮，且含負整數）；T 的證明是 S02 到 S09。",
          "en": "Finitely many cases do not cover ℤ (infinite, with negatives); the proof of T is S02 to S09."
        }
      },
      {
        "claim": {
          "zh-Hant": "n=0 到 39 時 n²+n+41 都是質數，所以對所有自然數都是質數",
          "en": "n²+n+41 is prime for n=0 to 39, so it is prime for every natural number"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "n=40 時得 1681=41·41，不是質數（S12、S13）。",
          "en": "At n=40 it is 1681=41·41, not prime (S12, S13)."
        }
      },
      {
        "claim": {
          "zh-Hant": "只證 r=0 的情形就足以證明 T",
          "en": "Proving the case r=0 alone proves T"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "n=1=2·0+1 屬於 r=1，沒有被涵蓋（S08）。",
          "en": "n=1=2·0+1 has r=1 and is not covered (S08)."
        }
      },
      {
        "claim": {
          "zh-Hant": "權威教科書斷言 C，所以 C 為真",
          "en": "An authoritative textbook asserts C, so C is true"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "權威不是推理規則；n=40 是有效反例（S14）。",
          "en": "Authority is not an inference rule; n=40 is a valid counterexample (S14)."
        }
      }
    ],
    "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": "divides_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整除：對整數 a、b，a∣b 表示存在整數 k 使 b=ak。",
          "en": "Divisibility: for integers a, b, a∣b means there is an integer k with b=ak."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "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": "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": "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": "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": "two_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數中 0<1，故 0<1+1=2，所以 2≠0。",
          "en": "In the reals 0<1, hence 0<1+1=2, so 2≠0."
        }
      },
      {
        "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": "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": "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": "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": "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": "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": "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_proof",
        "kind": "cited",
        "source": {
          "mko_id": "mko-proof",
          "sha256": "72f6009b912caa3c05a2948a97ca422367574a19ad5fc665964a540875417b91",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "給定待證命題 P，證明必須在指定語境中由已接受的前提、定義、公理或先前結果，經合法推理得到 P。",
          "en": "Given a proposition P to be proved, a proof must derive P within a specified context from accepted premises, definitions, axioms, or earlier results by valid inference."
        }
      },
      {
        "id": "prime_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "質數：整數 p 是質數 ⇔ p≥2 且 p 的正因數只有 1 與 p。",
          "en": "Prime: an integer p is prime ⇔ p≥2 and the only positive divisors of p are 1 and p."
        }
      },
      {
        "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。",
          "en": "No gaps in the integers: for integers k and m, k≤m or k≥m+1."
        }
      },
      {
        "id": "L_odd_case",
        "kind": "lemma",
        "proved_in": [
          "S06.1",
          "S06.2"
        ],
        "statement": {
          "zh-Hant": "對每個整數 q：(2q+1)²+(2q+1)=4q²+6q+2。",
          "en": "For every integer q: (2q+1)²+(2q+1)=4q²+6q+2."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S05",
        "vars": [
          "q"
        ],
        "degree": 2,
        "lhs": "(2*q)**2+2*q",
        "rhs": "2*(2*q**2+q)"
      },
      {
        "id": "N2",
        "step": "S07",
        "vars": [
          "q"
        ],
        "degree": 2,
        "lhs": "(2*q+1)**2+(2*q+1)",
        "rhs": "2*(2*q**2+3*q+1)"
      },
      {
        "id": "N3",
        "step": "S06.1",
        "vars": [
          "q"
        ],
        "degree": 2,
        "lhs": "(2*q+1)**2",
        "rhs": "4*q**2+4*q+1"
      },
      {
        "id": "N4",
        "step": "S09",
        "vars": [
          "n"
        ],
        "range": [
          -20,
          20
        ],
        "lhs": "(n*n+n) % 2",
        "rhs": "0"
      },
      {
        "id": "N5",
        "step": "S11",
        "vars": [
          "n"
        ],
        "range": [
          0,
          39
        ],
        "lhs": "isprime(n*n+n+41)",
        "rhs": "1"
      },
      {
        "id": "N6",
        "step": "S12",
        "lhs": "40*40+40+41",
        "rhs": "1681"
      },
      {
        "id": "N7",
        "step": "S12",
        "lhs": "41*41",
        "rhs": "1681"
      },
      {
        "id": "N8",
        "step": "S13",
        "lhs": "isprime(1681)",
        "rhs": "0"
      },
      {
        "id": "N9",
        "step": "S02",
        "lhs": "2*(-2)+1",
        "rhs": "-3"
      },
      {
        "id": "N10",
        "step": "S12",
        "lhs": "40*41+41",
        "rhs": "41*41"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "依 MKO：證明須在指定語境中，由已接受的前提、定義與先前結果，經合法推理得到待證命題。語境是 ℤ 的環運算與順序、整除與帶餘除法，以及古典邏輯（G1）。示範直接證明、分情況、全稱推廣、反證法，以及單一反例否定全稱命題。不形式化，不執行 Lean；不改動 canonical MKO 或英文 edition。",
      "en": "Follows the MKO: a proof must derive the proposition within a stated context from accepted premises, definitions and earlier results by valid inference. The context is the ring operations and order of ℤ, divisibility and the division algorithm, and classical logic (G1). It illustrates direct proof, proof by cases, universal generalization, proof by contradiction, and refuting a universal statement by one counterexample. It is not formalized, Lean is not run, and the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "T",
        "待證定理：∀n∈ℤ, 2∣(n²+n)",
        "the theorem to prove: ∀n∈ℤ, 2∣(n²+n)"
      ],
      [
        "C",
        "被檢驗的主張：∀n∈ℕ, n²+n+41 是質數",
        "the claim under test: ∀n∈ℕ, n²+n+41 is prime"
      ],
      [
        "q,\\ r",
        "n 除以 2 的商與餘數",
        "the quotient and remainder of n by 2"
      ],
      [
        "d\\mid m",
        "存在整數 k 使 m=dk",
        "there is an integer k with m=dk"
      ],
      [
        "\\mathbb{N}",
        "非負整數 {n∈ℤ ∣ n≥0}",
        "the nonnegative integers {n∈ℤ ∣ n≥0}"
      ]
    ],
    "md_error_controls": [
      {
        "title": "數值抽樣不是證明 / Numerical sampling is not proof",
        "zh": "MKO 指出常見誤解是把數值抽樣當成一般證明。C 在 n=0 到 39 都成立，卻在 n=40 失敗（S11 到 S14）。同理，N4 在 n=−20 到 20 核對 n²+n 為偶數也只支持那 41 個點；T 對所有整數成立的理由是 S02 到 S09 的推理。",
        "en": "The MKO names the mistake of taking numerical sampling as a general proof. C holds for n=0 to 39 yet fails at n=40 (S11 to S14). Likewise N4, which checks that n²+n is even for n=−20 to 20, supports only those 41 points; the reason T holds for every integer is the reasoning of S02 to S09."
      },
      {
        "title": "分情況必須涵蓋全部可能 / Cases must cover every possibility",
        "zh": "S03 的兩個情形由帶餘除法與 0≤r<2 保證涵蓋所有整數，包括負整數（例如 −3=2·(−2)+1 落在 r=1）。若只證明 r=0，例如 n=1=2·0+1 就沒有被涵蓋。",
        "en": "The two cases of S03 are guaranteed by the division algorithm and 0≤r<2 to cover every integer, negatives included (for example −3=2·(−2)+1 falls under r=1). Proving only r=0 would leave, for example, n=1=2·0+1 uncovered."
      },
      {
        "title": "權威與直覺不是推理規則 / Authority and intuition are not inference rules",
        "zh": "MKO 也提到把權威引用或未交代規則的直覺當作證明。「某本書說 C 為真」或「看起來 n²+n+41 總是質數」都不是 G1 中的推理規則；n=40 的計算直接否定 C。證明中每一步都要能指出所用的前提與規則（S01）。",
        "en": "The MKO also mentions treating appeals to authority or unexplained intuition as proof. 'A book says C is true' or 'n²+n+41 looks always prime' are not inference rules of G1; the computation at n=40 refutes C outright. Every step of a proof must name the premises and rules it uses (S01)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "N4 是 n 從 −20 到 20 的有限核對，N5 是 n 從 0 到 39 的有限核對；兩者都不證明任何全稱命題，T 的證明在 S02 到 S09。",
        "en": "N4 is a finite check for n from −20 to 20 and N5 a finite check for n from 0 to 39; neither proves any universal statement, and the proof of T is S02 to S09."
      },
      {
        "zh-Hant": "帶餘除法（含唯一性）與整數的環、序性質作為前提使用，不在此由更基本的公理推出。",
        "en": "The division algorithm (with uniqueness) and the ring and order properties of the integers are used as premises and are not derived here from more basic axioms."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "2c088a296a9a9f16755f489459feaca7850399ca52c46b4d09bc6cc6e9d310f7",
    "original_source_sha256": "2c088a296a9a9f16755f489459feaca7850399ca52c46b4d09bc6cc6e9d310f7",
    "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."
  }
}