{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-finite-group-01",
  "mko_id": "mko-finite-group",
  "title": {
    "zh-Hant": "有限群：以完整 Cayley 表核對群公理",
    "en": "Finite group: auditing the group axioms from a complete Cayley table"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-finite-group-01",
    "source": {
      "mko_id": "mko-finite-group",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "491d28b1773ac105187d1f17664e8e55b1ab4831613c354693901e40fae75f01",
      "english_sha256": "dc483579d967de45b346797aa7d8b882baaf395c1c61b932fcf9072babf13bc2"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "division_algorithm",
      "eq_refl_symm_trans",
      "case_analysis",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "給兩個完整運算表。（甲）G={0,1,2,3} 配上表 T，證明 (G,*) 是有限群：封閉、全部 4³=64 個三元組的結合律、雙側單位元、雙側逆元。（乙）K={0,1,2} 配上表 ∘：它對稱、有雙側單位元、每個元素都有雙側逆元，卻在三元組 (1,1,2) 違反結合律，因此不是群；同時說明含 0 的 19 個三元組全部通過，抽樣會誤判。",
      "en": "Two complete operation tables are given. (A) For G={0,1,2,3} with table T, prove that (G,*) is a finite group: closure, associativity for all 4³=64 triples, a two-sided identity and two-sided inverses. (B) K={0,1,2} with table ∘ is symmetric, has a two-sided identity and two-sided inverses for every element, yet violates associativity at the triple (1,1,2), so it is not a group; also show that all 19 triples containing 0 pass, so sampling would misjudge it."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "候選甲：G={0,1,2,3}，運算表 T 的第 a 行（row）依序列出 a*0、a*1、a*2、a*3：第 0 行 0,1,2,3；第 1 行 1,2,3,0；第 2 行 2,3,0,1；第 3 行 3,0,1,2。",
          "en": "Candidate A: G={0,1,2,3}; row a of the table T lists a*0, a*1, a*2, a*3 in order: row 0 is 0,1,2,3; row 1 is 1,2,3,0; row 2 is 2,3,0,1; row 3 is 3,0,1,2."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "候選乙：K={0,1,2}，運算表的第 a 行（row）依序列出 a∘0、a∘1、a∘2：第 0 行 0,1,2；第 1 行 1,0,1；第 2 行 2,1,0。",
          "en": "Candidate B: K={0,1,2}; row a of its table lists a∘0, a∘1, a∘2 in order: row 0 is 0,1,2; row 1 is 1,0,1; row 2 is 2,1,0."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "抽樣行（row）或少數三元組不證群；結合律必須對全部 n³ 個有序三元組成立（MKO 假設 a2）。",
          "en": "Sampled rows or a few triples do not prove a group; associativity must hold for every one of the n³ ordered triples (MKO assumption a2)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "mko_finite_group_def"
        ],
        "before_tex": "(G,*)\\ \\text{given by table } T",
        "after_tex": "\\text{closure},\\ \\forall a,b,c:\\ (a*b)*c=a*(b*c),\\ \\exists e\\ \\text{two-sided identity},\\ \\forall a\\ \\exists a'\\ \\text{two-sided inverse}",
        "relation": "definition",
        "rule": "mko_finite_group_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，(G,*) 是有限群需要四件事：封閉、對所有三元組結合、存在雙側單位元、每個元素有雙側逆元。交換律不在其中；表是否對稱與群無關。",
          "en": "By the MKO, (G,*) is a finite group when four things hold: closure, associativity for every triple, a two-sided identity, and a two-sided inverse for every element. Commutativity is not among them; whether the table is symmetric is irrelevant to being a group."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_finite_group_def"
          ],
          "explanation": {
            "zh-Hant": "此步只列出要核的條件，不推出新事實。四個條件缺一不可，之後逐一核對。",
            "en": "This step only lists the conditions to check and derives no new fact. None of the four may be skipped; they are checked one by one below."
          }
        },
        "leaf_basis": "mko_finite_group_def + group_axioms"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "mod_notation"
        ],
        "before_tex": "T(a,b)\\ \\text{for all } a,b\\in G",
        "after_tex": "a*b=(a+b)\\bmod 4",
        "relation": "implies",
        "rule": "finite_exhaustion",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐行（row）比對 16 格：第 a 行是 a+0、a+1、a+2、a+3 除以 4 的餘數。例如第 1 行 1,2,3,4 的餘數是 1,2,3,0；第 2 行 2,3,4,5 的餘數是 2,3,0,1；第 3 行 3,4,5,6 的餘數是 3,0,1,2；第 0 行不變。16 格全部相符，所以表 T 就是 a*b=(a+b) mod 4。",
          "en": "Compare all 16 cells row by row: row a consists of the remainders of a+0, a+1, a+2, a+3 on division by 4. For example row 1, from 1,2,3,4, gives 1,2,3,0; row 2, from 2,3,4,5, gives 2,3,0,1; row 3, from 3,4,5,6, gives 3,0,1,2; row 0 is unchanged. All 16 cells match, so the table T is a*b=(a+b) mod 4."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "finite_exhaustion",
            "mod_notation"
          ],
          "explanation": {
            "zh-Hant": "公式只是表的另一種寫法，靠的是 16 格逐一相符；這只對表 T 成立，不宣稱任何表都能寫成這種公式。",
            "en": "The formula is another description of the table, resting on all 16 cells matching; it holds for T only and does not claim that every table has such a formula."
          }
        },
        "leaf_basis": "finite_exhaustion + mod_notation + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S02"
        ],
        "before_tex": "a,b\\in G",
        "after_tex": "a*b\\in G",
        "relation": "implies",
        "rule": "finite_exhaustion",
        "conditions": [],
        "explanation": {
          "zh-Hant": "封閉：表 T 的 16 格輸出都在 {0,1,2,3} 中。也可由 S02 看出：餘數 r 滿足 0≤r<4，必為 0、1、2、3 之一。",
          "en": "Closure: all 16 entries of T lie in {0,1,2,3}. It also follows from S02: a remainder r satisfies 0≤r<4, so it is one of 0, 1, 2, 3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "finite_exhaustion",
            "division_algorithm"
          ],
          "explanation": {
            "zh-Hant": "封閉只要求每格輸出仍屬於 G；它本身不推出結合律或單位元。",
            "en": "Closure only requires every output to lie in G; on its own it gives neither associativity nor an identity."
          }
        },
        "leaf_basis": "finite_exhaustion + division_algorithm"
      },
      {
        "id": "S04",
        "uses": [
          "division_algorithm",
          "mod_notation"
        ],
        "before_tex": "x-y=mk,\\ k\\in\\mathbb{Z}",
        "after_tex": "x\\bmod m=y\\bmod m",
        "relation": "implies",
        "rule": "L_same_rem",
        "conditions": [
          "m>0"
        ],
        "explanation": {
          "zh-Hant": "結合律的論證需要一個關於餘數的引理 L_same_rem：相差 m 的整數倍的兩個整數，除以 m 的餘數相同。證明在子步 S04.1、S04.2，只用除法演算法的存在與唯一性。",
          "en": "The associativity argument needs a lemma on remainders, L_same_rem: two integers that differ by an integer multiple of m leave the same remainder on division by m. The proof is in sub-steps S04.1 and S04.2 and uses only existence and uniqueness in the division algorithm."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_same_rem"
          ],
          "explanation": {
            "zh-Hant": "引理需要 m>0 與 k 為整數；此處只宣稱這個方向，反向（餘數相同推出差為 m 的倍數）雖然也成立，但本稿不需要也不使用。",
            "en": "The lemma needs m>0 and an integer k; only this direction is claimed. The converse (equal remainders imply the difference is a multiple of m) is also true but is neither needed nor used here."
          }
        },
        "leaf_basis": "division_algorithm + mod_notation"
      },
      {
        "id": "S04.1",
        "uses": [
          "division_algorithm",
          "mod_notation"
        ],
        "before_tex": "x\\in\\mathbb{Z},\\ m>0",
        "after_tex": "x=mq+r,\\ 0\\le r<m,\\ r=x\\bmod m",
        "relation": "theorem_application",
        "rule": "division_algorithm",
        "conditions": [
          "m>0"
        ],
        "explanation": {
          "zh-Hant": "對 x 與正整數 m 使用除法演算法：存在整數 q 與 r，使 x=mq+r 且 0≤r<m；依記號，r=x mod m。",
          "en": "Apply the division algorithm to x and the positive integer m: there are integers q and r with x=mq+r and 0≤r<m; by the notation, r=x mod m."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "division_algorithm",
            "mod_notation"
          ],
          "explanation": {
            "zh-Hant": "除法演算法要求除數為正整數，條件 m>0 正是為此。",
            "en": "The division algorithm needs a positive divisor, which is what the condition m>0 provides."
          }
        },
        "leaf_basis": "division_algorithm + mod_notation"
      },
      {
        "id": "S04.2",
        "uses": [
          "S04.1",
          "division_algorithm"
        ],
        "before_tex": "y=x-mk=m(q-k)+r,\\ 0\\le r<m",
        "after_tex": "y\\bmod m=r=x\\bmod m",
        "relation": "implies",
        "rule": "division_algorithm",
        "conditions": [
          "q-k\\in\\mathbb{Z}"
        ],
        "explanation": {
          "zh-Hant": "由 x−y=mk 得 y=x−mk=mq+r−mk=m(q−k)+r。q−k 是整數且 0≤r<m，所以 (q−k, r) 正是 y 除以 m 的商與餘數；依除法演算法的唯一性，y mod m=r=x mod m。S04.1 與 S04.2 合起來證得引理 L_same_rem。",
          "en": "From x−y=mk we get y=x−mk=mq+r−mk=m(q−k)+r. Since q−k is an integer and 0≤r<m, the pair (q−k, r) is exactly the quotient and remainder of y on division by m; by uniqueness in the division algorithm, y mod m=r=x mod m. S04.1 and S04.2 together prove Lemma L_same_rem."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04.1",
            "division_algorithm",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "關鍵是唯一性：若沒有唯一性，y 可能有另一種寫法給出不同的餘數。0≤r<m 的界由 S04.1 帶入。",
            "en": "The key is uniqueness: without it, y might have another representation with a different remainder. The bound 0≤r<m is carried over from S04.1."
          }
        },
        "leaf_basis": "division_algorithm + distributive + eq_replace"
      },
      {
        "id": "S05",
        "uses": [
          "S02",
          "S04",
          "division_algorithm"
        ],
        "before_tex": "(a*b)*c",
        "after_tex": "(a+b+c)\\bmod 4",
        "relation": "equivalent",
        "rule": "L_same_rem",
        "conditions": [
          "a,b,c\\in G"
        ],
        "explanation": {
          "zh-Hant": "由 S02，(a*b)*c=((a+b) mod 4 + c) mod 4。設 a+b=4q+r，r=(a+b) mod 4；則 (r+c)−(a+b+c)=r−(a+b)=−4q 是 4 的整數倍，依引理 L_same_rem（m=4），(r+c) mod 4=(a+b+c) mod 4。a、b、c 是 G 中任意元素。",
          "en": "By S02, (a*b)*c=((a+b) mod 4 + c) mod 4. Write a+b=4q+r with r=(a+b) mod 4; then (r+c)−(a+b+c)=r−(a+b)=−4q is an integer multiple of 4, so by Lemma L_same_rem (m=4), (r+c) mod 4=(a+b+c) mod 4. Here a, b, c are arbitrary elements of G."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S02",
            "L_same_rem",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個整數的兩種寫法；把相等的式子互換，兩個方向都成立。",
            "en": "Both sides name the same integer; replacing equals by equals works in both directions."
          }
        },
        "leaf_basis": "L_same_rem + division_algorithm + eq_replace"
      },
      {
        "id": "S06",
        "uses": [
          "S02",
          "S04",
          "division_algorithm"
        ],
        "before_tex": "a*(b*c)",
        "after_tex": "(a+b+c)\\bmod 4",
        "relation": "equivalent",
        "rule": "L_same_rem",
        "conditions": [
          "a,b,c\\in G"
        ],
        "explanation": {
          "zh-Hant": "同理，a*(b*c)=(a+(b+c) mod 4) mod 4。設 b+c=4p+s，s=(b+c) mod 4；則 (a+s)−(a+b+c)=−4p，依引理得 (a+s) mod 4=(a+b+c) mod 4。",
          "en": "Likewise a*(b*c)=(a+(b+c) mod 4) mod 4. Write b+c=4p+s with s=(b+c) mod 4; then (a+s)−(a+b+c)=−4p, and the lemma gives (a+s) mod 4=(a+b+c) mod 4."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S02",
            "L_same_rem",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "與 S05 相同，是相等式子的互換，雙向成立。",
            "en": "As in S05, this replaces equals by equals and holds both ways."
          }
        },
        "leaf_basis": "L_same_rem + division_algorithm + eq_replace"
      },
      {
        "id": "S07",
        "uses": [
          "S05",
          "S06",
          "G3"
        ],
        "before_tex": "(a*b)*c=(a+b+c)\\bmod 4=a*(b*c)",
        "after_tex": "\\forall a,b,c\\in G:\\ (a*b)*c=a*(b*c)",
        "relation": "implies",
        "rule": "eq_refl_symm_trans",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊都等於 (a+b+c) mod 4，依等號的傳遞得 (a*b)*c=a*(b*c)。S05、S06 對 G 中任意 a、b、c 成立，所以這一次論證涵蓋全部 4³=64 個有序三元組，符合 G3 的要求。N4 另外把 64 個三元組逐一窮舉作交叉核對；兩者都是對這個有限模型的完整核對，不是抽樣。",
          "en": "Both sides equal (a+b+c) mod 4, so transitivity of equality gives (a*b)*c=a*(b*c). S05 and S06 hold for arbitrary a, b, c in G, so this one argument covers all 4³=64 ordered triples, as G3 requires. N4 separately enumerates the 64 triples one by one as a cross-check; both are complete checks of this finite model, not samples."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S06",
            "eq_refl_symm_trans",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "∀ 引入要求 a、b、c 沒有被特別取值；S05、S06 確實沒有。結合律與交換律不同，此步沒有用到 a*b=b*a。",
            "en": "∀-introduction needs a, b, c to carry no special values, and in S05 and S06 they carry none. Associativity differs from commutativity; this step does not use a*b=b*a."
          }
        },
        "leaf_basis": "eq_refl_symm_trans + forall_intro"
      },
      {
        "id": "S08",
        "uses": [
          "G1"
        ],
        "before_tex": "\\text{row } 0 \\text{ and column } 0 \\text{ of } T",
        "after_tex": "\\forall a\\in G:\\ 0*a=a\\ \\land\\ a*0=a",
        "relation": "implies",
        "rule": "finite_exhaustion",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第 0 行（row）是 0,1,2,3，表示 0*a=a 對四個 a 都成立（左單位）；第 0 列（column，每行的第一個數）依序是 0,1,2,3，表示 a*0=a（右單位）。左右都成立，0 是雙側單位元。",
          "en": "Row 0 is 0,1,2,3, so 0*a=a for all four a (left identity); column 0 (the first entry of each row) is 0,1,2,3 in order, so a*0=a (right identity). Both sides hold, so 0 is a two-sided identity."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "finite_exhaustion"
          ],
          "explanation": {
            "zh-Hant": "必須同時查對應的行（row）與列（column）；只查一邊只能得到單側單位元（見錯誤控制三）。",
            "en": "Both the row and the column must be inspected; inspecting one side only yields a one-sided identity (see error control 3)."
          }
        },
        "leaf_basis": "finite_exhaustion"
      },
      {
        "id": "S09",
        "uses": [
          "G1",
          "S08"
        ],
        "before_tex": "a\\in G",
        "after_tex": "0'=0,\\ 1'=3,\\ 2'=2,\\ 3'=1;\\ a*a'=a'*a=0",
        "relation": "implies",
        "rule": "finite_exhaustion",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在每一行（row）找出值為單位元 0 的位置：0*0=0；1*3=0 且 3*1=0；2*2=0；3*1=0 且 1*3=0。每一對都檢查左右兩個乘積，所以 0、3、2、1 分別是 0、1、2、3 的雙側逆元。",
          "en": "Find where each row takes the identity value 0: 0*0=0; 1*3=0 and 3*1=0; 2*2=0; 3*1=0 and 1*3=0. Both products are checked for each pair, so 0, 3, 2, 1 are two-sided inverses of 0, 1, 2, 3 respectively."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S08",
            "finite_exhaustion"
          ],
          "explanation": {
            "zh-Hant": "逆元的定義依賴 S08 的單位元，且必須左右乘都得到同一個單位元；四個元素逐一核完。",
            "en": "Inverses are defined relative to the identity of S08, and both products must give that same identity; all four elements are checked."
          }
        },
        "leaf_basis": "finite_exhaustion + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "S03",
          "S07",
          "S08",
          "S09"
        ],
        "before_tex": "\\text{closure, associativity, identity, inverses}",
        "after_tex": "(G,*)\\ \\text{is a finite group}",
        "relation": "implies",
        "rule": "mko_finite_group_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S03、S07、S08、S09 分別給出四個條件，依 MKO 的定義，(G,*) 是有限群。",
          "en": "S03, S07, S08 and S09 supply the four conditions, so by the MKO's definition (G,*) is a finite group."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S03",
            "S07",
            "S08",
            "S09",
            "mko_finite_group_def"
          ],
          "explanation": {
            "zh-Hant": "目標 G：(G,*) 是群。足夠條件 H：四個條件同時成立；H⇒G 由定義。H 的每一部分都由表 T 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: (G,*) is a group. Sufficient condition H: the four conditions hold together; H⇒G by the definition. Each part of H is proved forwards from the table T; G is never used as a premise."
          }
        },
        "leaf_basis": "mko_finite_group_def + and_intro_elim"
      },
      {
        "id": "S11",
        "uses": [
          "G2"
        ],
        "before_tex": "(K,\\circ)\\ \\text{given by its table}",
        "after_tex": "\\text{closed},\\ a\\circ b=b\\circ a,\\ 0\\ \\text{two-sided identity},\\ a\\circ a=0",
        "relation": "implies",
        "rule": "finite_exhaustion",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐格查候選乙的 9 格：輸出都在 {0,1,2}（封閉）；1∘2=1=2∘1，其餘對稱位置也相等（表對稱）；第 0 行（row）與第 0 列（column）都是 0,1,2（0 是雙側單位元）；0∘0=0、1∘1=0、2∘2=0（每個元素是自己的雙側逆元）。",
          "en": "Inspect the 9 cells of candidate B: every output lies in {0,1,2} (closed); 1∘2=1=2∘1 and the other mirror cells agree too (the table is symmetric); row 0 and column 0 are both 0,1,2 (0 is a two-sided identity); 0∘0=0, 1∘1=0 and 2∘2=0 (each element is its own two-sided inverse)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "finite_exhaustion"
          ],
          "explanation": {
            "zh-Hant": "這些性質都由逐格檢查得到；它們都不涉及三元組，所以對結合律沒有任何保證。",
            "en": "Each property comes from cell-by-cell inspection; none of them involves triples, so they guarantee nothing about associativity."
          }
        },
        "leaf_basis": "finite_exhaustion"
      },
      {
        "id": "S12",
        "uses": [
          "S11"
        ],
        "before_tex": "(a,b,c)\\in K^3,\\ 0\\in\\{a,b,c\\}",
        "after_tex": "(a\\circ b)\\circ c=a\\circ(b\\circ c)",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若三元組含 0，依位置分情形，每種情形都用 0 是雙側單位元：a=0 時兩邊都是 b∘c；b=0 時兩邊都是 a∘c；c=0 時兩邊都是 a∘b。這樣的三元組有 3³−2³=27−8=19 個，全部通過。所以若只抽查含單位元的三元組，會誤以為 ∘ 結合。",
          "en": "If a triple contains 0, split by its position, each case using that 0 is a two-sided identity: if a=0 both sides are b∘c; if b=0 both are a∘c; if c=0 both are a∘b. There are 3³−2³=27−8=19 such triples, and all pass. So a sample of triples containing the identity would wrongly suggest that ∘ is associative."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "三種情形可能重疊（例如兩個 0），但都涵蓋到；此步只宣稱含 0 的三元組，不涉及其餘 8 個。",
            "en": "The three cases may overlap (for example two zeros), but all such triples are covered; the step claims nothing about the remaining 8 triples."
          }
        },
        "leaf_basis": "case_analysis + finite_counting_basics + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "G2"
        ],
        "before_tex": "(1\\circ 1)\\circ 2\\ \\text{vs}\\ 1\\circ(1\\circ 2)",
        "after_tex": "0\\circ 2=2\\neq 0=1\\circ 1",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "查表：1∘1=0，所以 (1∘1)∘2=0∘2=2；1∘2=1，所以 1∘(1∘2)=1∘1=0。2≠0，三元組 (1,1,2) 違反結合律。",
          "en": "Read the table: 1∘1=0, so (1∘1)∘2=0∘2=2; 1∘2=1, so 1∘(1∘2)=1∘1=0. Since 2≠0, the triple (1,1,2) violates associativity."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "def_unfold",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "四次查表都可在 G2 直接核對；一個反例就足以否定全稱的結合律。",
            "en": "All four table lookups can be checked directly in G2; one counterexample suffices to refute the universal associativity law."
          }
        },
        "leaf_basis": "def_unfold + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S11",
          "S12",
          "S13",
          "G3"
        ],
        "before_tex": "\\exists (a,b,c):\\ (a\\circ b)\\circ c\\neq a\\circ(b\\circ c)",
        "after_tex": "(K,\\circ)\\ \\text{is not a group}",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "群需要對所有三元組結合；S13 給出反例，依量詞否定這個全稱命題為假，所以 (K,∘) 不是群。這發生在 S11 的對稱、單位元與逆元全部通過、S12 的 19 個三元組也全部通過之後：只看表的對稱或少數案例會誤判。",
          "en": "A group needs associativity for every triple; S13 gives a counterexample, so by quantifier negation the universal statement is false and (K,∘) is not a group. This happens even though symmetry, identity and inverses in S11 and the 19 triples of S12 all pass: judging by table symmetry or a few cases would be wrong."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S13",
            "quantifier_negation",
            "mko_finite_group_def"
          ],
          "explanation": {
            "zh-Hant": "目標 G：(K,∘) 不是群。足夠條件 H：某個三元組違反結合律（H⇒G 由定義）。H 由 S13 直接查表得到，沒有把 G 當作前提。",
            "en": "Goal G: (K,∘) is not a group. Sufficient condition H: some triple violates associativity (H⇒G by the definition). H is read off the table in S13; G is never used as a premise."
          }
        },
        "leaf_basis": "quantifier_negation + mko_finite_group_def"
      }
    ],
    "result": {
      "zh-Hant": "候選甲 (G,*) 的表就是 a*b=(a+b) mod 4；它封閉，全部 64 個三元組結合，0 是雙側單位元，0、1、2、3 的雙側逆元分別是 0、3、2、1，所以是有限群。候選乙 (K,∘) 封閉、對稱、以 0 為雙側單位元且每個元素自逆，含 0 的 19 個三元組也都結合，但 (1∘1)∘2=2≠0=1∘(1∘2)，所以不是群。",
      "en": "The table of candidate A (G,*) is a*b=(a+b) mod 4; it is closed, all 64 triples are associative, 0 is a two-sided identity, and the two-sided inverses of 0, 1, 2, 3 are 0, 3, 2, 1, so it is a finite group. Candidate B (K,∘) is closed, symmetric, has 0 as a two-sided identity and every element is its own inverse, and the 19 triples containing 0 are associative, but (1∘1)∘2=2≠0=1∘(1∘2), so it is not a group."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "候選乙的表對稱，且有單位元與逆元，所以 (K,∘) 是群",
          "en": "Candidate B's table is symmetric and has an identity and inverses, so (K,∘) is a group"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "結合律在 (1,1,2) 失敗：(1∘1)∘2=2，1∘(1∘2)=0。",
          "en": "Associativity fails at (1,1,2): (1∘1)∘2=2 while 1∘(1∘2)=0."
        }
      },
      {
        "claim": {
          "zh-Hant": "已核 (0,1,2)、(1,0,2)、(2,1,0)、(1,2,0) 四個三元組都結合，所以 ∘ 結合",
          "en": "The triples (0,1,2), (1,0,2), (2,1,0) and (1,2,0) are all associative, so ∘ is associative"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "這四個三元組都含 0，必然通過；不含 0 的 (1,1,2) 失敗。若要肯定結合律，論證必須涵蓋全部 27 個三元組。",
          "en": "All four contain 0 and pass automatically; (1,1,2), which avoids 0, fails. Establishing associativity requires an argument covering all 27 triples."
        }
      },
      {
        "claim": {
          "zh-Hant": "在 {0,1,2} 上 a•b=(a−b) mod 3，因為 a•0=a 對所有 a 成立，所以 0 是單位元",
          "en": "On {0,1,2} with a•b=(a−b) mod 3, 0 is the identity because a•0=a for every a"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "0 只是右單位元：0•1=2≠1。",
          "en": "0 is only a right identity: 0•1=2≠1."
        }
      },
      {
        "claim": {
          "zh-Hant": "只要每行（row）每列（column）都是集合的一個排列（拉丁方），運算表就定義一個群",
          "en": "Any table whose rows and columns are permutations of the set (a Latin square) defines a group"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(a−b) mod 3 的表是拉丁方，但 (0•1)•1=2•1=1，而 0•(1•1)=0•0=0。",
          "en": "The table of (a−b) mod 3 is a Latin square, but (0•1)•1=2•1=1 while 0•(1•1)=0•0=0."
        }
      }
    ],
    "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": "eq_refl_symm_trans",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。",
          "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w."
        }
      },
      {
        "id": "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": "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": "group_axioms",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "群公理：集合 G 配二元運算，滿足結合律、有單位元 e（eg=ge=g）、每個元素 g 有反元素 g⁻¹（gg⁻¹=g⁻¹g=e）。",
          "en": "Group axioms: a set G with a binary operation that is associative, has an identity e (eg=ge=g), and in which every g has an inverse g⁻¹ (gg⁻¹=g⁻¹g=e)."
        }
      },
      {
        "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": "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_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": "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": "finite_counting_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限計數：互斥有限集合 A、B 有 |A∪B|=|A|+|B|；|A×B|=|A|·|B|；有限集之間存在雙射 ⇔ 元素個數相同。",
          "en": "Finite counting: for disjoint finite sets A, B, |A∪B|=|A|+|B|; |A×B|=|A|·|B|; there is a bijection between finite sets ⇔ they have the same number of elements."
        }
      },
      {
        "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": "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": "mko_finite_group_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-finite-group",
          "sha256": "491d28b1773ac105187d1f17664e8e55b1ab4831613c354693901e40fae75f01",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "有限集合 G 配備二元運算 *。若運算對 G 封閉、對所有三元組結合、存在雙側單位元且每元素有雙側逆元，則 (G,*) 為有限群。完整 Cayley 表可窮舉核對指定有限模型。",
          "en": "Let a finite set G carry a binary operation *. If it is closed on G, associative for every triple, has a two-sided identity and every element has a two-sided inverse, then (G,*) is a finite group. A complete Cayley table allows exhaustive checking of the specified finite model."
        }
      },
      {
        "id": "mod_notation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "對整數 x 與正整數 m，x mod m 表示除法演算法中的餘數 r：即唯一滿足 0≤r<m 且 x=mq+r（q 為整數）的 r。",
          "en": "For an integer x and a positive integer m, x mod m denotes the remainder r of the division algorithm: the unique r with 0≤r<m and x=mq+r for some integer q."
        }
      },
      {
        "id": "finite_exhaustion",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "在有限集合上，對全部元素（或全部有序對、三元組）的命題成立，當且僅當它在列出的每一個情形都成立；逐一核完每個情形是該有限模型的完整證明，但不推及其他集合。",
          "en": "On a finite set, a statement about all elements (or all ordered pairs or triples) holds exactly when it holds in each of the finitely many listed cases; checking every case is a complete proof for that finite model, but says nothing about other sets."
        }
      },
      {
        "id": "L_same_rem",
        "kind": "lemma",
        "proved_in": [
          "S04.1",
          "S04.2"
        ],
        "statement": {
          "zh-Hant": "設 m 為正整數、x、y 為整數。若 x−y=mk（k 為整數），則 x mod m=y mod m。",
          "en": "Let m be a positive integer and x, y integers. If x−y=mk for an integer k, then x mod m=y mod m."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "(1+3)%4",
        "rhs": "0"
      },
      {
        "id": "N2",
        "step": "S02",
        "lhs": "(3+3)%4",
        "rhs": "2"
      },
      {
        "id": "N3",
        "step": "S02",
        "lhs": "(2+3)%4",
        "rhs": "1"
      },
      {
        "id": "N4",
        "step": "S07",
        "vars": [
          "a",
          "b",
          "c"
        ],
        "range": [
          0,
          3
        ],
        "lhs": "((a+b)%4+c)%4",
        "rhs": "(a+(b+c)%4)%4"
      },
      {
        "id": "N5",
        "step": "S04.2",
        "lhs": "13%4",
        "rhs": "(13-4*5)%4"
      },
      {
        "id": "N6",
        "step": "S07",
        "lhs": "4**3",
        "rhs": "64"
      },
      {
        "id": "N7",
        "step": "S08",
        "vars": [
          "a"
        ],
        "range": [
          0,
          3
        ],
        "lhs": "(0+a)%4",
        "rhs": "a"
      },
      {
        "id": "N8",
        "step": "S09",
        "lhs": "(3+1)%4",
        "rhs": "0"
      },
      {
        "id": "N9",
        "step": "S12",
        "lhs": "3**3-2**3",
        "rhs": "19"
      },
      {
        "id": "N10",
        "step": "S13",
        "lhs": "2",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N11",
        "lhs": "((0-1)%3-1)%3",
        "relation": "!=",
        "rhs": "(0-(1-1)%3)%3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理題目給定的有限模型：元素標為 0 到 n−1、運算由完整 n×n 表給出（MKO 假設 a1）。結合律以涵蓋全部三元組的論證完成，並以窮舉全部 n³ 個三元組作獨立交叉核對（MKO 假設 a2）。不處理子群、同態或無限群；不改動 canonical MKO 或英文 edition。",
      "en": "Treats only the specified finite models: elements labeled 0 to n−1 and the operation given by a complete n×n table (MKO assumption a1). Associativity is established by an argument covering every triple and cross-checked by enumerating all n³ triples (MKO assumption a2). Subgroups, homomorphisms and infinite groups are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "G=\\{0,1,2,3\\}",
        "候選甲的載體",
        "the carrier of candidate A"
      ],
      [
        "a*b",
        "表 T 中第 a 行（row）第 b 列（column）的值",
        "the entry of table T in row a, column b"
      ],
      [
        "K=\\{0,1,2\\}",
        "候選乙的載體",
        "the carrier of candidate B"
      ],
      [
        "a\\circ b",
        "表 ∘ 中第 a 行（row）第 b 列（column）的值",
        "the entry of table ∘ in row a, column b"
      ],
      [
        "x\\bmod m",
        "x 除以正整數 m 的餘數，0≤r<m",
        "the remainder of x on division by the positive integer m, 0≤r<m"
      ]
    ],
    "md_error_controls": [
      {
        "title": "對稱的表不等於群 / A symmetric table is not thereby a group",
        "zh": "MKO 指出的誤解是只看表的對稱性就宣稱是群。對稱只表示交換律，而交換律不是群公理，也不推出結合律。候選乙的表完全對稱、有單位元與逆元，仍在 (1,1,2) 違反結合律。",
        "en": "The MKO names the mistake of declaring a group from the symmetry of the table. Symmetry means only commutativity, which is not a group axiom and does not imply associativity. Candidate B's table is fully symmetric with an identity and inverses, yet violates associativity at (1,1,2)."
      },
      {
        "title": "抽樣三元組不夠 / Sampling triples is not enough",
        "zh": "候選乙的 27 個三元組中，含 0 的 19 個全部通過（S12）。抽查幾個三元組很可能全部落在這 19 個之中而誤判。結合律必須涵蓋全部 n³ 個三元組：本稿以對任意 a、b、c 成立的論證（S05 到 S07）涵蓋候選甲，並以 N4 窮舉 64 個三元組交叉核對。",
        "en": "Of the 27 triples of candidate B, the 19 containing 0 all pass (S12). A sample of a few triples may well fall entirely among those 19 and mislead. Associativity must cover all n³ triples: for candidate A this guide does so with an argument valid for arbitrary a, b, c (S05 to S07) and cross-checks by enumerating all 64 triples in N4."
      },
      {
        "title": "單側單位元或逆元 / One-sided identity or inverse",
        "zh": "MKO 也提醒不要把單側單位元或逆元當雙側。例：在 {0,1,2} 上令 a•b=(a−b) mod 3。對每個 a，a•0=a，所以 0 是右單位元；但 0•1=2≠1，0 不是左單位元，因此沒有雙側單位元。S08 必須同時查第 0 行（row）與第 0 列（column），S09 必須同時查 a*a' 與 a'*a。",
        "en": "The MKO also warns against treating a one-sided identity or inverse as two-sided. Example: on {0,1,2} let a•b=(a−b) mod 3. For every a, a•0=a, so 0 is a right identity; but 0•1=2≠1, so 0 is not a left identity and there is no two-sided identity. S08 must inspect both row 0 and column 0, and S09 both a*a' and a'*a."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "結論只屬於兩個指定的有限模型（MKO 假設 a2）；本稿不處理子群、同態、陪集等（MKO 的 formal next）。",
        "en": "The conclusions concern only the two specified finite models (MKO assumption a2); subgroups, homomorphisms, cosets and so on (the MKO's formal next) are not treated."
      },
      {
        "zh-Hant": "N4 是對候選甲全部 64 個三元組的有限窮舉，對這個模型是完整的，但它依賴 S02 的公式重現表 T；S05 到 S07 的論證不依賴 N4。",
        "en": "N4 is a finite enumeration of all 64 triples of candidate A, complete for this model, but it relies on the formula of S02 reproducing table T; the argument of S05 to S07 does not depend on N4."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "df045187fbff83ea6ba906772f9d4bccd3acd1bc60dcb87c4028857733f76d42",
    "original_source_sha256": "df045187fbff83ea6ba906772f9d4bccd3acd1bc60dcb87c4028857733f76d42",
    "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."
  }
}