{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-counterexample-01",
  "mko_id": "mko-counterexample",
  "title": {
    "zh-Hant": "反例與論域資格：n²+n+11 的質數性、失敗集合與最小反例",
    "en": "Counterexamples and domain eligibility: primality of n²+n+11, the failure set and the least counterexample"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-counterexample-01",
    "source": {
      "mko_id": "mko-counterexample",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "c3528c0ad339526d31db8324f3a6cbcdc28e756c638ba57e2792a24d8cc5beef",
      "english_sha256": "8fb486a8d29e7dd741f3fc050973e1897157bfa5f03ee86c37e9b8e1b728a8e4"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "divides_def",
      "division_algorithm",
      "order_basics",
      "numeric_evaluation",
      "prime_def"
    ],
    "goal": {
      "zh-Hant": "設 U={n∈ℤ ∣ 0≤n≤12}、V={n∈ℤ ∣ 0≤n≤9}、m(n)=n²+n+11、P(n)：「m(n) 是質數」。判定 φ_U：∀n∈U P(n)；求失敗集合 F_U 與最小反例；判定候選值 −11 的資格；並判定 φ_V：∀n∈V P(n)。",
      "en": "Let U={n∈ℤ ∣ 0≤n≤12}, V={n∈ℤ ∣ 0≤n≤9}, m(n)=n²+n+11 and P(n): 'm(n) is prime'. Decide φ_U: ∀n∈U P(n); find the failure set F_U and the least counterexample; decide whether the candidate −11 is eligible; and decide φ_V: ∀n∈V P(n)."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "U={n∈ℤ ∣ 0≤n≤12}，V={n∈ℤ ∣ 0≤n≤9}；兩者都是明示的有限整數論域，且 V⊆U。",
          "en": "U={n∈ℤ ∣ 0≤n≤12} and V={n∈ℤ ∣ 0≤n≤9}; both are explicit finite integer domains, and V⊆U."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "m(n)=n²+n+11；P(n)：「m(n) 是質數」，對每個整數都可精確判定。φ_U：∀n∈U P(n)；φ_V：∀n∈V P(n)。",
          "en": "m(n)=n²+n+11; P(n): 'm(n) is prime', decidable exactly for every integer. φ_U: ∀n∈U P(n); φ_V: ∀n∈V P(n)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "待判定資格的候選值：c=−11（對 φ_U），以及 c=10（對 φ_V）。",
          "en": "Candidates whose eligibility is to be decided: c=−11 (for φ_U) and c=10 (for φ_V)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_counterexample_def"
        ],
        "before_tex": "c\\ \\text{is a valid counterexample to}\\ \\forall n\\in U\\,P(n)",
        "after_tex": "c\\in U\\ \\land\\ \\neg P(c)",
        "relation": "definition",
        "rule": "mko_counterexample_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，c 是 φ_U 的有效反例，當且僅當兩件事都成立：c 在論域 U 中，而且 P(c) 為假。檢查順序是先查論域資格，再查述詞是否失敗。",
          "en": "By the MKO, c is a valid counterexample to φ_U exactly when both hold: c lies in the domain U, and P(c) is false. The order of checking is domain eligibility first, then failure of the predicate."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_counterexample_def",
            "G1",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出定義，不推出新事實；兩個條件缺一不可。",
            "en": "This step only states the definition and derives no new fact; neither condition may be dropped."
          }
        },
        "leaf_basis": "mko_counterexample_def"
      },
      {
        "id": "S02",
        "uses": [
          "G2"
        ],
        "before_tex": "m(10)=10^2+10+11",
        "after_tex": "m(10)=121=11\\cdot 11",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "精確計算：10²+10+11=100+10+11=121，而 11·11=121。",
          "en": "Exact computation: 10²+10+11=100+10+11=121, and 11·11=121."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個整數，可獨立重算（N1、N2）。",
            "en": "Both sides are the same integer and can be recomputed (N1, N2)."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "prime_def"
        ],
        "before_tex": "m(10)=121=11\\cdot 11",
        "after_tex": "\\neg P(10)",
        "relation": "implies",
        "rule": "prime_def",
        "conditions": [
          "1<11<121",
          "11\\mid 121"
        ],
        "explanation": {
          "zh-Hant": "121≥2，且 a=11 滿足 1<11<121 與 11∣121（121=11·11）。依質數定義，121 是合數，所以 P(10) 為假。完全平方數 121 也是合數，沒有例外。",
          "en": "121≥2, and a=11 satisfies 1<11<121 and 11∣121 (121=11·11). By the definition of prime, 121 is composite, so P(10) is false. The perfect square 121 is composite too; there is no exception."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "prime_def",
            "divides_def"
          ],
          "explanation": {
            "zh-Hant": "只需要一個真因數 11；不宣稱反向。",
            "en": "One proper divisor, 11, is enough; no converse is claimed."
          }
        },
        "leaf_basis": "prime_def + divides_def + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S01",
          "S03",
          "G1"
        ],
        "before_tex": "0\\le 10\\le 12,\\ \\neg P(10)",
        "after_tex": "10\\in U\\land\\neg P(10)",
        "relation": "implies",
        "rule": "mko_counterexample_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先查資格：10 是整數，且 0≤10≤12，依集合建構式 10∈U。再查失敗：S03 給出 ¬P(10)。兩個條件都成立，所以 10 是 φ_U 的有效反例。",
          "en": "Eligibility first: 10 is an integer with 0≤10≤12, so 10∈U by set-builder notation. Then failure: S03 gives ¬P(10). Both conditions hold, so 10 is a valid counterexample to φ_U."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "set_builder",
            "S03",
            "mko_counterexample_def"
          ],
          "explanation": {
            "zh-Hant": "兩個條件必須同時成立；只有 ¬P(10) 而沒有 10∈U 不夠。",
            "en": "Both conditions are needed together; ¬P(10) without 10∈U would not be enough."
          }
        },
        "leaf_basis": "set_builder + mko_counterexample_def + and_intro_elim"
      },
      {
        "id": "S05",
        "uses": [
          "S04",
          "exists_intro",
          "mko_quantifier_neg"
        ],
        "before_tex": "10\\in U\\land\\neg P(10)",
        "after_tex": "\\neg\\forall n\\in U\\,P(n)",
        "relation": "implies",
        "rule": "mko_quantifier_neg",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由 ∃引入得 ∃n∈U ¬P(n)；依受限量詞的否定律（mko-quantifier），這等價於 ¬∀n∈U P(n)。一個有效反例就足以否定全稱陳述 φ_U，不需要先知道全部失敗值。",
          "en": "∃-introduction gives ∃n∈U ¬P(n); by the negation law for restricted quantifiers (mko-quantifier) this is equivalent to ¬∀n∈U P(n). One valid counterexample suffices to refute φ_U; the full list of failures is not needed for that."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "exists_intro",
            "mko_quantifier_neg"
          ],
          "explanation": {
            "zh-Hant": "需要見證在 U 中；否定全稱只需一個見證。不宣稱反向。",
            "en": "The witness must be in U; refuting a universal needs one witness. No converse is claimed."
          }
        },
        "leaf_basis": "exists_intro + mko_quantifier_neg"
      },
      {
        "id": "S06",
        "uses": [
          "S01",
          "G1",
          "G2",
          "G3"
        ],
        "before_tex": "c=-11:\\ m(-11)=121",
        "after_tex": "-11\\notin U,\\ \\text{so}\\ -11\\ \\text{is not a counterexample to}\\ \\varphi_U",
        "relation": "implies",
        "rule": "mko_counterexample_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "公式在域外也能計算：(−11)²+(−11)+11=121−11+11=121，是合數。但資格先於失敗：−11<0，不滿足 0≤n≤12，所以 −11∉U。依 S01 的定義，−11 不是 φ_U 的反例；它的失敗值與 φ_U 無關。",
          "en": "The formula can be evaluated outside the domain: (−11)²+(−11)+11=121−11+11=121, which is composite. But eligibility comes before failure: −11<0 fails 0≤n≤12, so −11∉U. By the definition in S01, −11 is not a counterexample to φ_U; its failing value is irrelevant to φ_U."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "set_builder",
            "mko_counterexample_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "判定只需第一個條件失敗（N3 只重算公式值）；不宣稱 φ_U 因此為真，φ_U 已由 S05 判為假。",
            "en": "The verdict needs only the first condition to fail (N3 just recomputes the formula); it is not claimed that φ_U is therefore true, since S05 already showed it false."
          }
        },
        "leaf_basis": "set_builder + mko_counterexample_def + order_basics"
      },
      {
        "id": "S07",
        "uses": [
          "G2",
          "L_no_small_factor"
        ],
        "before_tex": "n\\in V",
        "after_tex": "2\\nmid m(n),\\ 3\\nmid m(n),\\ 5\\nmid m(n),\\ 7\\nmid m(n)",
        "relation": "theorem_application",
        "rule": "L_no_small_factor",
        "conditions": [
          "n\\in\\mathbb{Z}"
        ],
        "explanation": {
          "zh-Hant": "要判定 φ_V，不能靠「搜尋沒找到」，必須對 V 的每個元素給出理由。第一個工具是引理 L_no_small_factor：對每個整數 n，2、3、5、7 都不整除 m(n)。V 的元素都是整數，所以引理適用。引理在子步 S07.1 到 S07.5 證明。",
          "en": "To decide φ_V we cannot rely on 'the search found nothing'; every element of V needs a reason. The first tool is Lemma L_no_small_factor: for every integer n, none of 2, 3, 5, 7 divides m(n). Elements of V are integers, so the lemma applies. It is proved in sub-steps S07.1 to S07.5."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_no_small_factor"
          ],
          "explanation": {
            "zh-Hant": "只需 n 是整數；引理不說 m(n) 是質數，只排除四個小質因數。",
            "en": "It needs only that n is an integer; the lemma does not say m(n) is prime, it only rules out four small prime factors."
          }
        },
        "leaf_basis": "division_algorithm + rem_divides + divides_combo + case_analysis"
      },
      {
        "id": "S07.1",
        "uses": [
          "division_algorithm",
          "distributive",
          "divides_combo"
        ],
        "before_tex": "n=pq+r,\\ 0\\le r<p",
        "after_tex": "m(n)=p(pq^2+2qr+q)+m(r),\\quad p\\mid m(n)\\Leftrightarrow p\\mid m(r)",
        "relation": "implies",
        "rule": "distributive",
        "conditions": [
          "p\\ge 1"
        ],
        "explanation": {
          "zh-Hant": "依帶餘除法寫 n=pq+r（0≤r<p）。展開：(pq+r)²+(pq+r)+11=p²q²+2pqr+r²+pq+r+11=p(pq²+2qr+q)+(r²+r+11)=p(pq²+2qr+q)+m(r)（N4 對這個多項式恆等式做格點核對）。因為 p 整除 p(pq²+2qr+q)，依整除的和差性質：若 p∣m(n)，相減得 p∣m(r)；若 p∣m(r)，相加得 p∣m(n)。這證得輔助引理 L_residue。",
          "en": "By the division algorithm write n=pq+r with 0≤r<p. Expanding: (pq+r)²+(pq+r)+11=p²q²+2pqr+r²+pq+r+11=p(pq²+2qr+q)+(r²+r+11)=p(pq²+2qr+q)+m(r) (N4 checks this polynomial identity on a grid). Since p divides p(pq²+2qr+q), the sum and difference property gives: if p∣m(n), subtracting yields p∣m(r); if p∣m(r), adding yields p∣m(n). This proves the auxiliary Lemma L_residue."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "division_algorithm",
            "distributive",
            "divides_combo"
          ],
          "explanation": {
            "zh-Hant": "需要帶餘除法的分解與展開恆等式；得到的 ⇔ 兩個方向分別由相減與相加給出理由。",
            "en": "It needs the division-algorithm decomposition and the expansion identity; the two directions of the resulting ⇔ are justified by subtracting and by adding."
          }
        },
        "leaf_basis": "division_algorithm + distributive + add_assoc + mul_comm + divides_combo"
      },
      {
        "id": "S07.2",
        "uses": [
          "L_residue",
          "rem_divides",
          "case_analysis"
        ],
        "before_tex": "p=2,\\ r\\in\\{0,1\\}",
        "after_tex": "m(0)=11=2\\cdot 5+1,\\ m(1)=13=2\\cdot 6+1;\\ 2\\nmid m(n)",
        "relation": "implies",
        "rule": "L_residue",
        "conditions": [],
        "explanation": {
          "zh-Hant": "p=2：餘數 r 只能是 0 或 1。m(0)=11=2·5+1，m(1)=13=2·6+1，餘數都是 1≠0，所以 2∤m(0)、2∤m(1)。依 L_residue，對每個整數 n 都有 2∤m(n)。",
          "en": "p=2: the remainder r is 0 or 1. m(0)=11=2·5+1 and m(1)=13=2·6+1 both leave remainder 1≠0, so 2∤m(0) and 2∤m(1). By L_residue, 2∤m(n) for every integer n."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_residue",
            "rem_divides",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "兩種餘數涵蓋所有整數（帶餘除法）；每種情況的餘數都已精確計算（N5）。",
            "en": "The two remainders cover every integer (division algorithm); the remainder in each case is computed exactly (N5)."
          }
        },
        "leaf_basis": "rem_divides + case_analysis + numeric_evaluation + division_algorithm"
      },
      {
        "id": "S07.3",
        "uses": [
          "L_residue",
          "rem_divides",
          "case_analysis"
        ],
        "before_tex": "p=3,\\ r\\in\\{0,1,2\\}",
        "after_tex": "m(r)\\in\\{11,13,17\\},\\ \\text{remainders}\\ 2,1,2;\\ 3\\nmid m(n)",
        "relation": "implies",
        "rule": "L_residue",
        "conditions": [],
        "explanation": {
          "zh-Hant": "p=3：r=0、1、2 時 m(r)=11、13、17；11=3·3+2、13=3·4+1、17=3·5+2，餘數 2、1、2 都不是 0，所以 3 不整除這三個值。依 L_residue，對每個整數 n 都有 3∤m(n)。",
          "en": "p=3: for r=0, 1, 2, m(r)=11, 13, 17; 11=3·3+2, 13=3·4+1, 17=3·5+2, with remainders 2, 1, 2, none of them 0, so 3 divides none of the three values. By L_residue, 3∤m(n) for every integer n."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_residue",
            "rem_divides",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "三種餘數涵蓋所有整數；三個餘數都已計算（N6）。",
            "en": "The three remainders cover every integer; all three remainders are computed (N6)."
          }
        },
        "leaf_basis": "rem_divides + case_analysis + numeric_evaluation + division_algorithm"
      },
      {
        "id": "S07.4",
        "uses": [
          "L_residue",
          "rem_divides",
          "case_analysis"
        ],
        "before_tex": "p=5,\\ r\\in\\{0,1,2,3,4\\}",
        "after_tex": "m(r)\\in\\{11,13,17,23,31\\},\\ \\text{remainders}\\ 1,3,2,3,1;\\ 5\\nmid m(n)",
        "relation": "implies",
        "rule": "L_residue",
        "conditions": [],
        "explanation": {
          "zh-Hant": "p=5：r=0 到 4 時 m(r)=11、13、17、23、31，除以 5 的餘數依序是 1、3、2、3、1，都不是 0。依 L_residue，對每個整數 n 都有 5∤m(n)。",
          "en": "p=5: for r=0 to 4, m(r)=11, 13, 17, 23, 31, whose remainders on division by 5 are 1, 3, 2, 3, 1, none of them 0. By L_residue, 5∤m(n) for every integer n."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_residue",
            "rem_divides",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "五種餘數涵蓋所有整數；五個餘數都已計算（N7）。",
            "en": "The five remainders cover every integer; all five remainders are computed (N7)."
          }
        },
        "leaf_basis": "rem_divides + case_analysis + numeric_evaluation + division_algorithm"
      },
      {
        "id": "S07.5",
        "uses": [
          "L_residue",
          "rem_divides",
          "case_analysis"
        ],
        "before_tex": "p=7,\\ r\\in\\{0,1,2,3,4,5,6\\}",
        "after_tex": "m(r)\\in\\{11,13,17,23,31,41,53\\},\\ \\text{remainders}\\ 4,6,3,2,3,6,4;\\ 7\\nmid m(n)",
        "relation": "implies",
        "rule": "L_residue",
        "conditions": [],
        "explanation": {
          "zh-Hant": "p=7：r=0 到 6 時 m(r)=11、13、17、23、31、41、53，除以 7 的餘數依序是 4、6、3、2、3、6、4，都不是 0。依 L_residue，對每個整數 n 都有 7∤m(n)。S07.2 到 S07.5 合起來證得引理 L_no_small_factor。",
          "en": "p=7: for r=0 to 6, m(r)=11, 13, 17, 23, 31, 41, 53, whose remainders on division by 7 are 4, 6, 3, 2, 3, 6, 4, none of them 0. By L_residue, 7∤m(n) for every integer n. S07.2 to S07.5 together prove Lemma L_no_small_factor."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_residue",
            "rem_divides",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "七種餘數涵蓋所有整數；七個餘數都已計算（N8）。普遍結論來自「有限個餘數情況涵蓋全部整數」這個論證，不是來自對 n 的抽樣。",
            "en": "The seven remainders cover every integer; all seven are computed (N8). The universal conclusion comes from the argument that finitely many remainder cases cover all integers, not from sampling values of n."
          }
        },
        "leaf_basis": "rem_divides + case_analysis + numeric_evaluation + division_algorithm"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "G2",
          "order_basics"
        ],
        "before_tex": "0\\le n\\le 9",
        "after_tex": "11\\le m(n)\\le 101",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "n\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "對 0≤n≤9：n≥0 給出 n²≥0 與 n²+n≥0，所以 m(n)≥11；由 n≤9 且 n≥0 得 n²=n·n≤9·n≤81，所以 m(n)≤81+9+11=101。",
          "en": "For 0≤n≤9: n≥0 gives n²≥0 and n²+n≥0, so m(n)≥11; from n≤9 and n≥0 we get n²=n·n≤9·n≤81, so m(n)≤81+9+11=101."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "square_nonneg"
          ],
          "explanation": {
            "zh-Hant": "乘以非負數保持不等式方向，這需要 n≥0；上界 101 可重算（N9）。不宣稱反向。",
            "en": "Multiplying by a nonnegative number keeps the inequality direction, which needs n≥0; the bound 101 can be recomputed (N9). No converse is claimed."
          }
        },
        "leaf_basis": "order_basics + square_nonneg + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S07",
          "S08",
          "L_trial"
        ],
        "before_tex": "11\\le m(n)\\le 101,\\quad 2,3,5,7\\nmid m(n)",
        "after_tex": "P(n)\\ \\text{for an arbitrary}\\ n\\in V",
        "relation": "theorem_application",
        "rule": "L_trial",
        "conditions": [
          "m(n)\\ge 2"
        ],
        "explanation": {
          "zh-Hant": "取任意 n∈V。若質數 p 滿足 p²≤m(n)≤101，則 p≤10（因為 11²=121>101），所以 p∈{2,3,5,7}；S07 說這四個都不整除 m(n)。又 m(n)≥11≥2（S08）。依引理 L_trial，m(n) 是質數，即 P(n)。",
          "en": "Take an arbitrary n∈V. If a prime p satisfies p²≤m(n)≤101, then p≤10 (since 11²=121>101), so p∈{2,3,5,7}; S07 says none of these divides m(n). Also m(n)≥11≥2 (S08). By Lemma L_trial, m(n) is prime, that is P(n)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "S08",
            "L_trial"
          ],
          "explanation": {
            "zh-Hant": "需要列全 p²≤101 的質數：10 以下的質數 2、3、5、7（N10 核對 11²>101）；漏掉任何一個都會使論證不完整。",
            "en": "It needs every prime with p²≤101, namely the primes up to 10: 2, 3, 5, 7 (N10 checks 11²>101); missing any would leave the argument incomplete."
          }
        },
        "leaf_basis": "prime_def + order_basics + numeric_evaluation"
      },
      {
        "id": "S09.1",
        "uses": [
          "prime_def",
          "order_basics"
        ],
        "before_tex": "m\\ge 2,\\ m\\ \\text{not prime}",
        "after_tex": "m=ab,\\ 2\\le a\\le b,\\ a^2\\le m",
        "relation": "implies",
        "rule": "prime_def",
        "conditions": [
          "a>0"
        ],
        "explanation": {
          "zh-Hant": "設 m≥2 不是質數。依定義有整數 a，1<a<m 且 a∣m；令 b=m/a，則 b 是整數，且由 a<m 得 b>1、由 a>1 得 b<m。若 a>b 就交換兩者，所以可設 2≤a≤b。兩邊乘正數 a 得 a²≤ab=m。",
          "en": "Suppose m≥2 is not prime. By definition there is an integer a with 1<a<m and a∣m; let b=m/a, an integer, with b>1 because a<m and b<m because a>1. Swap the two if a>b, so we may assume 2≤a≤b. Multiplying by the positive a gives a²≤ab=m."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "prime_def",
            "divides_def",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要 a>0 才能保持不等式方向；b 的範圍由 m=ab 與 1<a<m 推出。",
            "en": "a>0 is needed to keep the inequality direction; the range of b follows from m=ab and 1<a<m."
          }
        },
        "leaf_basis": "prime_def + divides_def + order_basics"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "prime_divisor_exists",
          "divides_combo"
        ],
        "before_tex": "m=ab,\\ 2\\le a,\\ a^2\\le m",
        "after_tex": "\\exists\\ \\text{prime}\\ p:\\ p\\mid m\\land p^2\\le m",
        "relation": "implies",
        "rule": "prime_divisor_exists",
        "conditions": [],
        "explanation": {
          "zh-Hant": "a≥2，所以 a 有質因數 p，且 p≤a。由 p∣a 與 a∣m 得 p∣m；由 0<p≤a 得 p²≤a²≤m。所以：若 m≥2 不是質數，就有質數 p 滿足 p²≤m 且 p∣m。",
          "en": "a≥2, so a has a prime divisor p, with p≤a. From p∣a and a∣m we get p∣m; from 0<p≤a we get p²≤a²≤m. Hence: if m≥2 is not prime, some prime p with p²≤m divides m."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "prime_divisor_exists",
            "divides_combo",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "質因數的存在是本稿的前提（可由良序原理證明），不在此重證。",
            "en": "The existence of a prime divisor is a premise of this guide (provable from well-ordering) and is not reproved here."
          }
        },
        "leaf_basis": "prime_divisor_exists + divides_combo + order_basics"
      },
      {
        "id": "S09.3",
        "uses": [
          "S09.2",
          "contrapositive"
        ],
        "before_tex": "m\\ \\text{not prime}\\Rightarrow\\exists\\ \\text{prime}\\ p:\\ p\\mid m\\land p^2\\le m",
        "after_tex": "(\\forall\\ \\text{prime}\\ p:\\ p^2\\le m\\Rightarrow p\\nmid m)\\Rightarrow m\\ \\text{prime}",
        "relation": "equivalent",
        "rule": "contrapositive",
        "conditions": [
          "m\\ge 2"
        ],
        "explanation": {
          "zh-Hant": "取 S09.2 的逆否命題：若沒有滿足 p²≤m 的質數整除 m（且 m≥2），則 m 是質數。否定「存在這樣的 p」用量詞否定律得到「每個 p²≤m 的質數都不整除 m」。S09.1 到 S09.3 證得引理 L_trial。",
          "en": "Take the contrapositive of S09.2: if no prime p with p²≤m divides m (and m≥2), then m is prime. Negating 'there is such a p' by quantifier negation gives 'every prime with p²≤m fails to divide m'. S09.1 to S09.3 prove Lemma L_trial."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "contrapositive",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "命題與其逆否命題等價，兩向成立；否定存在量詞的量詞否定律也是等價。",
            "en": "A statement and its contrapositive are equivalent, both ways; the quantifier negation used for the existential is an equivalence too."
          }
        },
        "leaf_basis": "contrapositive + quantifier_negation"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "forall_intro"
        ],
        "before_tex": "P(n)\\ \\text{for an arbitrary}\\ n\\in V",
        "after_tex": "\\forall n\\in V\\,P(n)",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S07 到 S09 對任意 n∈V 證明了 P(n)，除 0≤n≤9 之外沒有對 n 加任何假設。依 ∀引入，φ_V 為真。這是對有限論域 V 每個元素都有效的論證，不是「搜尋沒有找到反例」；N11 以 isprime 對 n=0 到 9 逐一重算，作為獨立的完整核對。",
          "en": "S07 to S09 proved P(n) for an arbitrary n∈V, with no assumption on n beyond 0≤n≤9. By ∀-introduction, φ_V is true. This is an argument valid for every element of the finite domain V, not 'a search found no counterexample'; N11 recomputes isprime for n=0 to 9 one by one as an independent complete check."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要 n 是 V 的任意元素；若證明偷用了特定 n 的性質，就不能推廣。",
            "en": "n must be an arbitrary element of V; had the proof used a property of a specific n, it could not be generalized."
          }
        },
        "leaf_basis": "forall_intro"
      },
      {
        "id": "S11",
        "uses": [
          "G1",
          "G2",
          "prime_def"
        ],
        "before_tex": "m(11)=11^2+11+11",
        "after_tex": "m(11)=143=11\\cdot 13,\\ 11\\in U,\\ \\neg P(11)",
        "relation": "implies",
        "rule": "prime_def",
        "conditions": [
          "1<11<143"
        ],
        "explanation": {
          "zh-Hant": "11²+11+11=121+22=143=11·13，而 1<11<143，所以 143 是合數；0≤11≤12，所以 11∈U。11 是 φ_U 的第二個有效反例。",
          "en": "11²+11+11=121+22=143=11·13 with 1<11<143, so 143 is composite; 0≤11≤12, so 11∈U. 11 is a second valid counterexample to φ_U."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "prime_def",
            "set_builder",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "兩個條件都已核對（N12、N13）。",
            "en": "Both conditions are checked (N12, N13)."
          }
        },
        "leaf_basis": "prime_def + set_builder + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "L_no_small_factor",
          "L_trial",
          "G2"
        ],
        "before_tex": "m(12)=12^2+12+11=167",
        "after_tex": "P(12)",
        "relation": "theorem_application",
        "rule": "L_trial",
        "conditions": [
          "167\\ge 2"
        ],
        "explanation": {
          "zh-Hant": "m(12)=144+12+11=167。滿足 p²≤167 的質數是 p≤12 的 2、3、5、7、11（13²=169>167）。依 L_no_small_factor，2、3、5、7 不整除 167；又 167=11·15+2，餘數 2≠0，所以 11∤167。依引理 L_trial，167 是質數，P(12) 成立。",
          "en": "m(12)=144+12+11=167. The primes with p²≤167 are those with p≤12: 2, 3, 5, 7, 11 (since 13²=169>167). By L_no_small_factor, 2, 3, 5, 7 do not divide 167; and 167=11·15+2 leaves remainder 2≠0, so 11∤167. By Lemma L_trial, 167 is prime and P(12) holds."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_no_small_factor",
            "L_trial",
            "rem_divides"
          ],
          "explanation": {
            "zh-Hant": "需要列全 p≤12 的質數；11 不在 L_no_small_factor 的範圍內，所以另以餘數判定（N15、N16）。",
            "en": "All primes up to 12 must be listed; 11 lies outside L_no_small_factor and is decided separately by its remainder (N15, N16)."
          }
        },
        "leaf_basis": "rem_divides + numeric_evaluation + order_basics"
      },
      {
        "id": "S13",
        "uses": [
          "S03",
          "S10",
          "S11",
          "S12",
          "G1"
        ],
        "before_tex": "U=V\\cup\\{10,11,12\\}",
        "after_tex": "F_U=\\{n\\in U\\mid\\neg P(n)\\}=\\{10,11\\}",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "U 的元素是 0 到 12 的整數，分成 V（0 到 9）與 10、11、12 三個元素，這涵蓋全部情況。P 在 V 上都成立（S10），在 12 成立（S12），在 10 與 11 失敗（S03、S11）。所以失敗集合恰為 {10,11}：兩個方向的包含都由逐一情況核對得到。這是有限論域的完整驗證。",
          "en": "The elements of U are the integers 0 to 12, split into V (0 to 9) and the three elements 10, 11, 12, which covers every case. P holds throughout V (S10) and at 12 (S12), and fails at 10 and 11 (S03, S11). So the failure set is exactly {10,11}: both inclusions follow from the case-by-case check. This is a complete verification of a finite domain."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S10",
            "S11",
            "S12",
            "set_extensionality",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "情況劃分必須涵蓋 U 的全部 13 個元素；只要漏掉一個，F_U 就沒有完整確定。",
            "en": "The case split must cover all 13 elements of U; missing one would leave F_U undetermined."
          }
        },
        "leaf_basis": "set_extensionality + case_analysis + set_builder"
      },
      {
        "id": "S14",
        "uses": [
          "S13"
        ],
        "before_tex": "F_U=\\{10,11\\}",
        "after_tex": "\\min F_U=10",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "F_U\\neq\\varnothing"
        ],
        "explanation": {
          "zh-Hant": "依 MKO 假設 a2，最小反例是有限失敗集合的數值最小值。F_U={10,11} 非空，且 10<11，所以最小反例是 10。這個值由完整的 F_U 算出，不是由抽樣中第一個遇到的失敗值猜出。",
          "en": "By MKO assumption a2, the least counterexample is the numerically smallest element of the finite failure set. F_U={10,11} is nonempty and 10<11, so the least counterexample is 10. The value comes from the complete F_U, not from the first failure met in a sample."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要 F_U 已完整確定（S13）；否則無法排除更小的失敗值。",
            "en": "It needs F_U to be fully determined (S13); otherwise a smaller failure could not be ruled out."
          }
        },
        "leaf_basis": "order_basics + numeric_evaluation"
      },
      {
        "id": "S15",
        "uses": [
          "S10",
          "S13",
          "G1",
          "G3"
        ],
        "before_tex": "10\\in F_U,\\ V\\subseteq U",
        "after_tex": "10\\notin V,\\ F_V=F_U\\cap V=\\varnothing",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "限制到 V：10>9，所以 10∉V，它對 φ_V 失去反例資格；11 也一樣。V 的失敗集合 F_V={n∈V ∣ ¬P(n)}=F_U∩V=∅，所以 φ_V 沒有反例，最小反例明示為「不存在」。注意 φ_V 為真是由 S10 的論證得到，不是因為 10 失去資格；失去見證本身不證明任何全稱陳述。",
          "en": "Restrict to V: 10>9, so 10∉V and it loses eligibility as a counterexample to φ_V; so does 11. The failure set of V is F_V={n∈V ∣ ¬P(n)}=F_U∩V=∅, so φ_V has no counterexample, and its least counterexample is explicitly reported as nonexistent. Note that φ_V is true because of the argument in S10, not because 10 lost eligibility; losing a witness proves no universal statement by itself."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "S13",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "F_V=∅ 需要 S10（V 中沒有失敗值）；只靠 10∉V 推不出 F_V=∅。",
            "en": "F_V=∅ needs S10 (no failure in V); 10∉V alone would not give F_V=∅."
          }
        },
        "leaf_basis": "set_operations_def + set_builder"
      },
      {
        "id": "S16",
        "uses": [
          "S05",
          "S06",
          "S10",
          "S13",
          "S14",
          "S15"
        ],
        "before_tex": "\\text{S05, S06, S10, S13, S14, S15}",
        "after_tex": "\\neg\\varphi_U,\\ F_U=\\{10,11\\},\\ \\min F_U=10,\\ -11\\ \\text{ineligible},\\ \\varphi_V\\ \\text{true}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整：φ_U 為假（S05，有效反例 10）；F_U={10,11}，最小反例 10（S13、S14）；−11 雖使公式失敗卻不在 U 中，不是反例（S06）；φ_V 為真（S10），F_V=∅，沒有最小反例（S15）。",
          "en": "Summary: φ_U is false (S05, valid counterexample 10); F_U={10,11} with least counterexample 10 (S13, S14); −11 makes the formula fail but is not in U, so it is not a counterexample (S06); φ_V is true (S10), F_V=∅, and there is no least counterexample (S15)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S05",
            "S06",
            "S10",
            "S13",
            "S14",
            "S15",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：四項判定。足夠條件 H：S05、S06、S10、S13、S14、S15 的結論；H⇒G 只是合取。H 的每一部分都由前面正向證成，沒有把 G 當作前提。",
            "en": "Goal G: the four verdicts. Sufficient condition H: the conclusions of S05, S06, S10, S13, S14 and S15; H⇒G is just conjunction. Each part of H is proved forwards earlier, and G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "φ_U（∀n∈U, n²+n+11 是質數，U 為 0 到 12）為假：失敗集合 F_U={10,11}（121=11·11、143=11·13），最小反例是 10。候選值 −11 雖然給出 121，但 −11∉U，不是反例。φ_V（V 為 0 到 9）為真：對每個 n∈V，11≤m(n)≤101，且 2、3、5、7 都不整除 m(n)，所以 m(n) 是質數；F_V=∅，沒有最小反例。",
      "en": "φ_U (∀n∈U, n²+n+11 is prime, with U the integers 0 to 12) is false: the failure set is F_U={10,11} (121=11·11, 143=11·13), and the least counterexample is 10. The candidate −11 gives 121 but −11∉U, so it is not a counterexample. φ_V (V the integers 0 to 9) is true: for every n∈V, 11≤m(n)≤101 and none of 2, 3, 5, 7 divides m(n), so m(n) is prime; F_V=∅ and there is no least counterexample."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "m(−11)=121 是合數，所以 −11 是 ∀n∈U P(n) 的反例",
          "en": "m(−11)=121 is composite, so −11 is a counterexample to ∀n∈U P(n)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "−11<0，所以 −11∉U；域外的失敗值不是反例。",
          "en": "−11<0, so −11∉U; a failing value outside the domain is not a counterexample."
        }
      },
      {
        "claim": {
          "zh-Hant": "n=0 到 9 都使 n²+n+11 為質數，所以對所有自然數 n 都成立",
          "en": "n=0 to 9 all make n²+n+11 prime, so it holds for every natural number n"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "n=10 時 n²+n+11=121=11·11，不是質數。",
          "en": "At n=10, n²+n+11=121=11·11, which is not prime."
        }
      },
      {
        "claim": {
          "zh-Hant": "10∉V，所以 ∀n∈V P(n) 已得證",
          "en": "10∉V, so ∀n∈V P(n) is proved"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "失去一個見證不證明全稱；W={11,12} 時 10 也不在 W 中，但 ∀n∈W P(n) 仍因 11 而為假。φ_V 需要 S07 到 S10 的論證。",
          "en": "Losing one witness proves no universal; with W={11,12}, 10 is also outside W, yet ∀n∈W P(n) is still false because of 11. φ_V needs the argument of S07 to S10."
        }
      },
      {
        "claim": {
          "zh-Hant": "φ_V 的最小反例是 10",
          "en": "The least counterexample to φ_V is 10"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "10∉V；F_V=∅，所以 φ_V 沒有反例，最小反例不存在。",
          "en": "10∉V; F_V=∅, so φ_V has no counterexample and no least counterexample exists."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "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": "set_builder",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。",
          "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and 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": "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": "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": "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": "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": "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": "square_nonneg",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數平方非負：對所有實數 u，u²≥0，且 u²=0 ⇔ u=0。",
          "en": "Squares are nonnegative: for every real u, u²≥0, and u²=0 ⇔ u=0."
        }
      },
      {
        "id": "contrapositive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逆否命題：P⇒Q 與 ¬Q⇒¬P 等價。逆命題 Q⇒P 一般不與 P⇒Q 等價。",
          "en": "Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q."
        }
      },
      {
        "id": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "id": "set_extensionality",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合相等：A=B 當且僅當 A⊆B 且 B⊆A，即兩者有相同元素。",
          "en": "Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements."
        }
      },
      {
        "id": "set_operations_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合運算：x∈A∪B ⇔ x∈A 或 x∈B；x∈A∩B ⇔ x∈A 且 x∈B；x∈A∖B ⇔ x∈A 且 x∉B；在全集 U 中 Aᶜ=U∖A。",
          "en": "Set operations: x∈A∪B ⇔ x∈A or x∈B; x∈A∩B ⇔ x∈A and x∈B; x∈A∖B ⇔ x∈A and x∉B; in a universe U, Aᶜ=U∖A."
        }
      },
      {
        "id": "mko_counterexample_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-counterexample",
          "sha256": "c3528c0ad339526d31db8324f3a6cbcdc28e756c638ba57e2792a24d8cc5beef",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對全稱陳述 ∀x∈U P(x)，c 是有效反例當且僅當 c∈U 且 P(c) 為假。一個有效反例即可否定該全稱陳述；沒有找到反例本身不是證明，除非論域有限且已完整驗證所有元素。",
          "en": "For ∀x∈U P(x), c is a valid counterexample exactly when c∈U and P(c) is false. One valid counterexample refutes the universal statement. Failure to find one is not itself a proof unless the domain is finite and every element has been checked completely."
        }
      },
      {
        "id": "mko_quantifier_neg",
        "kind": "cited",
        "source": {
          "mko_id": "mko-quantifier",
          "sha256": "3b2f73bd57aa43b5441edef2b27d44470f8cbf04a229fb31e98243708e818341",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在明示論域與古典真值語義下，¬∀x∈U P(x) 等價於 ∃x∈U ¬P(x)，¬∃x∈U P(x) 等價於 ∀x∈U ¬P(x)。",
          "en": "For an explicit domain under classical truth semantics, ¬∀x∈U P(x) is equivalent to ∃x∈U ¬P(x), and ¬∃x∈U P(x) is equivalent to ∀x∈U ¬P(x)."
        }
      },
      {
        "id": "prime_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數 m≥2 是質數，指它的正因數只有 1 與 m；m≥2 不是質數（合數）當且僅當存在整數 a 使 1<a<m 且 a∣m。",
          "en": "An integer m≥2 is prime when its only positive divisors are 1 and m; m≥2 is not prime (composite) exactly when some integer a has 1<a<m and a∣m."
        }
      },
      {
        "id": "prime_divisor_exists",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "每個整數 a≥2 都有質因數 p（p∣a），且正整數 a 的正因數不超過 a。（可由良序原理得到：a 的最小大於 1 的因數是質數；本稿當作前提。）",
          "en": "Every integer a≥2 has a prime divisor p (p∣a), and a positive divisor of a positive integer a is at most a. (This follows from well-ordering: the least divisor of a above 1 is prime; it is a premise here.)"
        }
      },
      {
        "id": "rem_divides",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "對整數 a 與正整數 b，b∣a 當且僅當帶餘除法 a=bq+r（0≤r<b）的餘數 r=0（由帶餘除法的唯一性）。",
          "en": "For an integer a and a positive integer b, b∣a exactly when the remainder r in the division a=bq+r (0≤r<b) is 0 (by uniqueness in the division algorithm)."
        }
      },
      {
        "id": "divides_combo",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "若 p∣u 且 p∣v，則 p∣u+v 且 p∣u−v；若 p∣a 且 a∣m，則 p∣m。",
          "en": "If p∣u and p∣v then p∣u+v and p∣u−v; if p∣a and a∣m then p∣m."
        }
      },
      {
        "id": "L_residue",
        "kind": "lemma",
        "proved_in": [
          "S07.1"
        ],
        "statement": {
          "zh-Hant": "對整數 p≥1 與整數 n，寫 n=pq+r（0≤r<p），則 m(n)=p(pq²+2qr+q)+m(r)；因此 p∣m(n) ⇔ p∣m(r)。",
          "en": "For an integer p≥1 and an integer n, writing n=pq+r with 0≤r<p gives m(n)=p(pq²+2qr+q)+m(r); hence p∣m(n) ⇔ p∣m(r)."
        }
      },
      {
        "id": "L_no_small_factor",
        "kind": "lemma",
        "proved_in": [
          "S07.2",
          "S07.3",
          "S07.4",
          "S07.5"
        ],
        "statement": {
          "zh-Hant": "對每個整數 n，2、3、5、7 都不整除 m(n)=n²+n+11。",
          "en": "For every integer n, none of 2, 3, 5, 7 divides m(n)=n²+n+11."
        }
      },
      {
        "id": "L_trial",
        "kind": "lemma",
        "proved_in": [
          "S09.1",
          "S09.2",
          "S09.3"
        ],
        "statement": {
          "zh-Hant": "若整數 m≥2，且沒有滿足 p²≤m 的質數 p 整除 m，則 m 是質數。",
          "en": "If an integer m≥2 is divisible by no prime p with p²≤m, then m is prime."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "10**2+10+11",
        "rhs": "121"
      },
      {
        "id": "N2",
        "step": "S02",
        "lhs": "11*11",
        "rhs": "121"
      },
      {
        "id": "N3",
        "step": "S06",
        "lhs": "(-11)**2+(-11)+11",
        "rhs": "121"
      },
      {
        "id": "N4",
        "step": "S07.1",
        "vars": [
          "p",
          "q",
          "r"
        ],
        "degree": 2,
        "lhs": "(p*q+r)**2+(p*q+r)+11",
        "rhs": "p*(p*q*q+2*q*r+q)+r*r+r+11"
      },
      {
        "id": "N5",
        "step": "S07.2",
        "vars": [
          "r"
        ],
        "range": [
          0,
          1
        ],
        "lhs": "(r*r+r+11) % 2",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N6",
        "step": "S07.3",
        "vars": [
          "r"
        ],
        "range": [
          0,
          2
        ],
        "lhs": "(r*r+r+11) % 3",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N7",
        "step": "S07.4",
        "vars": [
          "r"
        ],
        "range": [
          0,
          4
        ],
        "lhs": "(r*r+r+11) % 5",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N8",
        "step": "S07.5",
        "vars": [
          "r"
        ],
        "range": [
          0,
          6
        ],
        "lhs": "(r*r+r+11) % 7",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N9",
        "step": "S08",
        "lhs": "9**2+9+11",
        "rhs": "101"
      },
      {
        "id": "N10",
        "step": "S09",
        "lhs": "11*11",
        "relation": ">",
        "rhs": "101"
      },
      {
        "id": "N11",
        "step": "S10",
        "vars": [
          "n"
        ],
        "range": [
          0,
          9
        ],
        "lhs": "isprime(n*n+n+11)",
        "rhs": "1"
      },
      {
        "id": "N12",
        "step": "S11",
        "lhs": "11**2+11+11",
        "rhs": "143"
      },
      {
        "id": "N13",
        "step": "S11",
        "lhs": "11*13",
        "rhs": "143"
      },
      {
        "id": "N14",
        "step": "S12",
        "lhs": "12**2+12+11",
        "rhs": "167"
      },
      {
        "id": "N15",
        "step": "S12",
        "lhs": "167 % 11",
        "rhs": "2"
      },
      {
        "id": "N16",
        "step": "S12",
        "lhs": "13*13",
        "relation": ">",
        "rhs": "167"
      },
      {
        "id": "N17",
        "step": "S12",
        "lhs": "isprime(167)",
        "rhs": "1"
      },
      {
        "id": "N18",
        "step": "S14",
        "lhs": "min(10,11)",
        "rhs": "10"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：明示有限整數論域與可判定述詞。示範反例的兩個條件（在論域中、述詞失敗）、受限後的資格變化、完整失敗集合與最小反例，以及以對每個元素都有效的論證證明有限全稱。不對 ℕ 或 ℤ 上的 n²+n+11 作任何全稱宣稱；「每個整數 a≥2 有質因數」作為前提；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: explicit finite integer domains and decidable predicates. Shows the two conditions of a counterexample (in the domain, predicate fails), how eligibility changes under restriction, the complete failure set and the least counterexample, and a proof of a finite universal by an argument valid for every element. Makes no universal claim about n²+n+11 over ℕ or ℤ; takes 'every integer a≥2 has a prime divisor' as a premise; does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "U,\\ V",
        "論域：0≤n≤12 與 0≤n≤9 的整數",
        "domains: the integers with 0≤n≤12 and with 0≤n≤9"
      ],
      [
        "m(n)",
        "n²+n+11",
        "n²+n+11"
      ],
      [
        "P(n)",
        "述詞「m(n) 是質數」",
        "the predicate 'm(n) is prime'"
      ],
      [
        "F_U,\\ F_V",
        "失敗集合 {n∈U ∣ ¬P(n)} 與 {n∈V ∣ ¬P(n)}",
        "the failure sets {n∈U ∣ ¬P(n)} and {n∈V ∣ ¬P(n)}"
      ],
      [
        "p\\mid m,\\ p\\nmid m",
        "p 整除 m；p 不整除 m",
        "p divides m; p does not divide m"
      ]
    ],
    "md_error_controls": [
      {
        "title": "先查論域資格 / Check domain eligibility first",
        "zh": "MKO 指出的誤解之一是以域外值反駁受限全稱。−11 代入公式得 121，是合數，但 −11∉U，所以不是 φ_U 的反例（S06）。判定順序永遠是先查 c∈U，再查 P(c) 是否失敗。",
        "en": "One misconception named by the MKO is refuting a restricted universal with a value outside the domain. Substituting −11 gives 121, which is composite, but −11∉U, so it is not a counterexample to φ_U (S06). The order is always: check c∈U first, then whether P(c) fails."
      },
      {
        "title": "沒找到反例不是證明 / Not finding a counterexample is not a proof",
        "zh": "MKO 的另一個誤解是把一段搜尋沒找到反例當成普遍證明。逐一試 n=0 到 9 都得到質數，但 n=10 就失敗；若只搜尋到 9 就宣稱「對所有 n 成立」是錯的。φ_V 的證明是 S07 到 S10 對 V 的每個元素都有效的論證；對無限論域，任何有限搜尋都不構成證明。",
        "en": "Another misconception named by the MKO is treating an unsuccessful search as a general proof. Trying n=0 to 9 gives primes every time, yet n=10 fails; claiming 'true for all n' after searching up to 9 would be wrong. The proof of φ_V is the argument in S07 to S10, valid for every element of V; over an infinite domain no finite search is a proof."
      },
      {
        "title": "失去見證不等於新全稱成立 / Losing a witness does not prove the restricted universal",
        "zh": "MKO 第二段：限制到子集後，原反例可能離開子集而失去資格，但這不表示新全稱為真。例如 W={11,12}⊆U：10∉W 失去資格，可是 11∈W 仍使 P 失敗，所以 ∀n∈W P(n) 仍為假。φ_V 為真是因為 S10 的證明，不是因為 10∉V。",
        "en": "MKO paragraph two: after restricting to a subset, an old counterexample may leave the subset and lose eligibility, but that does not make the new universal true. For example W={11,12}⊆U: 10∉W loses eligibility, yet 11∈W still makes P fail, so ∀n∈W P(n) is still false. φ_V is true because of the proof in S10, not because 10∉V."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "N5 到 N8 是對每個 p 的全部餘數 r 的有限檢查；普遍結論 L_no_small_factor 依賴 S07.1 的 L_residue 與「餘數情況涵蓋全部整數」，不是只靠這些檢查。",
        "en": "N5 to N8 are finite checks over all remainders r for each p; the universal conclusion L_no_small_factor rests on L_residue (S07.1) and on the remainder cases covering every integer, not on these checks alone."
      },
      {
        "zh-Hant": "「每個整數 a≥2 有質因數」在本稿是前提，未在此證明。N11、N17 以 isprime 做獨立重算，論證不依賴它們。",
        "en": "'Every integer a≥2 has a prime divisor' is a premise here and is not proved. N11 and N17 are independent recomputations with isprime; the argument does not depend on them."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "50a0b8ce1009ddb04ac5c306440773799a1f5d16fecb299d252315f534a48568",
    "original_source_sha256": "50a0b8ce1009ddb04ac5c306440773799a1f5d16fecb299d252315f534a48568",
    "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."
  }
}