{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-finite-abelian-group-01",
  "mko_id": "mko-finite-abelian-group",
  "title": {
    "zh-Hant": "有限阿貝爾群：由完整 Cayley 表先核群資格、再核交換性",
    "en": "Finite abelian group: from a complete Cayley table, group qualification first, then commutativity"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-finite-abelian-group-01",
    "source": {
      "mko_id": "mko-finite-abelian-group",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "1486476569952aee082941d6f91d2921526de39dd732a11bacce98c252b95415",
      "english_sha256": "7b784df7deb082eb20d9dc02a3cc71c900becdfafb7f8a2828225c8c53f99407"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "group_axioms",
      "add_comm",
      "add_assoc",
      "forall_intro",
      "quantifier_negation",
      "abs_rules"
    ],
    "goal": {
      "zh-Hant": "載體 G={0,1,2,3} 上給出完整的 4×4 Cayley 表（行 a、列 b 的格子是 a*b）：第 0 行 (1,2,3,0)，第 1 行 (2,3,0,1)，第 2 行 (3,0,1,2)，第 3 行 (0,1,2,3)。依序核對封閉、結合、唯一單位元與逆元，再對所有 a、b 核對 a*b=b*a，判定 (G,*) 是否為有限阿貝爾群；求元素 0 與 1 的階與它們生成的循環子群；並以 {0,1,2} 上的對稱表 a∘b=|a−b| 說明「表對稱」不能代替群資格。",
      "en": "On the carrier G={0,1,2,3} a complete 4×4 Cayley table is given (the cell in row a, column b is a*b): row 0 is (1,2,3,0), row 1 is (2,3,0,1), row 2 is (3,0,1,2), row 3 is (0,1,2,3). Check closure, associativity, a unique identity and inverses in that order, then check a*b=b*a for all a, b, to decide whether (G,*) is a finite abelian group; find the orders of the elements 0 and 1 and the cyclic subgroups they generate; and use the symmetric table a∘b=|a−b| on {0,1,2} to show that a symmetric table cannot replace group qualification."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "載體 G={0,1,2,3}，標籤為 0 到 n−1，n=4。",
          "en": "The carrier G={0,1,2,3}, with labels 0 to n−1 and n=4."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "完整的 Cayley 表（行 a、列 b 的格子是 a*b）：第 0 行 (1,2,3,0)；第 1 行 (2,3,0,1)；第 2 行 (3,0,1,2)；第 3 行 (0,1,2,3)。",
          "en": "The complete Cayley table (the cell in row a, column b is a*b): row 0 (1,2,3,0); row 1 (2,3,0,1); row 2 (3,0,1,2); row 3 (0,1,2,3)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "對照表：載體 K={0,1,2}，a∘b=|a−b|，即第 0 行 (0,1,2)、第 1 行 (1,0,1)、第 2 行 (2,1,0)。",
          "en": "Contrast table: carrier K={0,1,2}, a∘b=|a−b|, that is row 0 (0,1,2), row 1 (1,0,1), row 2 (2,1,0)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "mko_fag",
          "group_axioms"
        ],
        "before_tex": "(G,*)\\ \\text{finite abelian group}",
        "after_tex": "\\text{closure}\\land\\text{assoc}\\land\\text{unique } e\\land\\text{inverses},\\ \\text{then}\\ \\forall a,b:\\ a*b=b*a",
        "relation": "definition",
        "rule": "mko_fag",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，分類分兩階段：先用完整表核群資格（封閉、結合、唯一單位元、逆元），全部通過後才核交換性 a*b=b*a 對所有 a、b。順序不能顛倒：交換性只對已是群的結構才構成「阿貝爾群」。",
          "en": "By the MKO, classification has two stages: first use the complete table to check group qualification (closure, associativity, a unique identity, inverses), and only after all of these pass check commutativity a*b=b*a for all a, b. The order cannot be reversed: commutativity makes an 'abelian group' only of a structure that is already a group."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_fag",
            "group_axioms"
          ],
          "explanation": {
            "zh-Hant": "此步只列出義務，不推出新事實。",
            "en": "This step only lists the obligations and derives no new fact."
          }
        },
        "leaf_basis": "mko_fag + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "G2",
          "table_read",
          "mod_arith"
        ],
        "before_tex": "a*b\\ \\text{read from the table}",
        "after_tex": "a*b=(a+b+1)\\bmod 4\\ \\text{for all 16 pairs}",
        "relation": "equivalent",
        "rule": "table_read",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐格比較：第 a 行依 b=0,1,2,3 列出 a+1、a+2、a+3、a+4 除以 4 的餘數。例如第 0 行 1,2,3,4 mod 4=0，得 (1,2,3,0)；第 2 行 3,4,5,6 的餘數為 (3,0,1,2)。四行共 16 格全部與 (a+b+1) mod 4 相符。這是對完整表的窮舉比較，不是抽樣；之後可用模算術對所有元素同時論證。",
          "en": "Compare cell by cell: row a lists, for b=0,1,2,3, the remainders of a+1, a+2, a+3, a+4 on division by 4. For example row 0 gives 1, 2, 3 and 4 mod 4=0, that is (1,2,3,0); row 2 gives the remainders of 3, 4, 5, 6, that is (3,0,1,2). All 16 cells of the four rows agree with (a+b+1) mod 4. This is an exhaustive comparison over the complete table, not a sample; afterwards modular arithmetic can argue for all elements at once."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "table_read",
            "mod_arith",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "公式在 16 對上產生的值恰好就是表；由公式重算表或由表讀回公式值，兩者一致。",
            "en": "The values of the formula on the 16 pairs are exactly the table; recomputing the table from the formula or reading the formula values off the table gives the same thing."
          }
        },
        "leaf_basis": "table_read + mod_arith + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "mod_arith"
        ],
        "before_tex": "a,b\\in G",
        "after_tex": "a*b=(a+b+1)\\bmod 4\\in\\{0,1,2,3\\}=G",
        "relation": "implies",
        "rule": "mod_arith",
        "conditions": [],
        "explanation": {
          "zh-Hant": "封閉：除以 4 的餘數一定在 0 到 3 之間，所以每個 a*b 都在 G 中。直接看表也一樣：16 個格子都是 0、1、2、3 之一。",
          "en": "Closure: a remainder on division by 4 always lies between 0 and 3, so every a*b lies in G. Reading the table gives the same: all 16 cells are among 0, 1, 2, 3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "mod_arith"
          ],
          "explanation": {
            "zh-Hant": "需要 S02 的公式與餘數的範圍；不宣稱反向。",
            "en": "It needs the formula of S02 and the range of remainders; no converse is claimed."
          }
        },
        "leaf_basis": "mod_arith + forall_intro"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "S03",
          "mod_arith"
        ],
        "before_tex": "(a*b)*c,\\ a*(b*c)",
        "after_tex": "(a*b)*c=a*(b*c)\\ \\text{for all } a,b,c\\in G",
        "relation": "implies",
        "rule": "mod_arith",
        "conditions": [],
        "explanation": {
          "zh-Hant": "結合律對 64 組 (a,b,c) 都要成立。不逐一列舉，而用模算術一次證明：兩邊都與 a+b+c+2 模 4 同餘（S04.1、S04.2），而且都在 0 到 3 之間，所以相等（S04.3）。N3 再對全部 64 組做窮舉交叉核對。",
          "en": "Associativity must hold for all 64 triples (a,b,c). Instead of listing them, modular arithmetic proves it at once: both sides are congruent to a+b+c+2 modulo 4 (S04.1, S04.2), and both lie between 0 and 3, so they are equal (S04.3). N3 additionally cross-checks all 64 triples exhaustively."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S03",
            "mod_arith",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要 S02 的公式；a、b、c 任意，所以是全稱結論。",
            "en": "It needs the formula of S02; a, b, c are arbitrary, so the conclusion is universal."
          }
        },
        "leaf_basis": "mod_arith + add_assoc + add_comm + forall_intro"
      },
      {
        "id": "S04.1",
        "uses": [
          "S02",
          "mod_arith",
          "add_assoc"
        ],
        "before_tex": "(a*b)*c=\\big((a*b)+c+1\\big)\\bmod 4",
        "after_tex": "(a*b)*c\\equiv(a+b+1)+c+1=a+b+c+2\\pmod 4",
        "relation": "implies",
        "rule": "mod_arith",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依公式 (a*b)*c=((a*b)+c+1) mod 4，它與 (a*b)+c+1 同餘；又 a*b≡a+b+1，同餘與加法相容，所以 (a*b)*c≡a+b+1+c+1=a+b+c+2 (mod 4)。",
          "en": "By the formula (a*b)*c=((a*b)+c+1) mod 4, it is congruent to (a*b)+c+1; also a*b≡a+b+1, and congruence is compatible with addition, so (a*b)*c≡a+b+1+c+1=a+b+c+2 (mod 4)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mod_arith",
            "add_assoc"
          ],
          "explanation": {
            "zh-Hant": "只用同餘的性質與加法結合律。",
            "en": "Only properties of congruence and associativity of addition are used."
          }
        },
        "leaf_basis": "mod_arith + add_assoc + add_comm"
      },
      {
        "id": "S04.2",
        "uses": [
          "S02",
          "mod_arith",
          "add_assoc"
        ],
        "before_tex": "a*(b*c)=\\big(a+(b*c)+1\\big)\\bmod 4",
        "after_tex": "a*(b*c)\\equiv a+(b+c+1)+1=a+b+c+2\\pmod 4",
        "relation": "implies",
        "rule": "mod_arith",
        "conditions": [],
        "explanation": {
          "zh-Hant": "同理，a*(b*c)≡a+(b*c)+1≡a+(b+c+1)+1=a+b+c+2 (mod 4)。",
          "en": "Likewise, a*(b*c)≡a+(b*c)+1≡a+(b+c+1)+1=a+b+c+2 (mod 4)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mod_arith",
            "add_assoc"
          ],
          "explanation": {
            "zh-Hant": "與 S04.1 對稱。",
            "en": "Symmetric to S04.1."
          }
        },
        "leaf_basis": "mod_arith + add_assoc"
      },
      {
        "id": "S04.3",
        "uses": [
          "S04.1",
          "S04.2",
          "S03",
          "mod_arith"
        ],
        "before_tex": "(a*b)*c\\equiv a*(b*c)\\pmod 4,\\ \\text{both in}\\ \\{0,1,2,3\\}",
        "after_tex": "(a*b)*c=a*(b*c)",
        "relation": "implies",
        "rule": "mod_arith",
        "conditions": [],
        "explanation": {
          "zh-Hant": "同餘的傳遞性給出 (a*b)*c≡a*(b*c)；依 S03 兩者都在 0 到 3 之間，兩個在此範圍內且同餘的整數相等。",
          "en": "Transitivity of congruence gives (a*b)*c≡a*(b*c); by S03 both lie between 0 and 3, and two integers in this range that are congruent are equal."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04.1",
            "S04.2",
            "S03",
            "mod_arith"
          ],
          "explanation": {
            "zh-Hant": "關鍵是兩邊都已化成餘數；只有同餘而不在 0 到 3 之間的數不一定相等。",
            "en": "The key is that both sides are already remainders; numbers that are merely congruent, not both between 0 and 3, need not be equal."
          }
        },
        "leaf_basis": "mod_arith"
      },
      {
        "id": "S05",
        "uses": [
          "S02",
          "mod_arith"
        ],
        "before_tex": "3*b,\\ a*3",
        "after_tex": "3*b=(b+4)\\bmod 4=b,\\ a*3=(a+4)\\bmod 4=a",
        "relation": "implies",
        "rule": "mod_arith",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取 e=3：3*b=(3+b+1) mod 4=(b+4) mod 4=b，因為 b 在 0 到 3 之間且 b+4≡b；同樣 a*3=a。表中第 3 行 (0,1,2,3) 與第 3 列都重現標籤，與此一致。注意單位元是 3，不是標籤 0：0*0=1≠0。",
          "en": "Take e=3: 3*b=(3+b+1) mod 4=(b+4) mod 4=b, since b lies between 0 and 3 and b+4≡b; likewise a*3=a. Row 3 of the table, (0,1,2,3), and column 3 both reproduce the labels, consistently. Note that the identity is 3, not the label 0: 0*0=1≠0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "mod_arith",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "雙邊都要核（左乘與右乘）；本步兩邊都核了。",
            "en": "Both sides must be checked (left and right multiplication); this step checks both."
          }
        },
        "leaf_basis": "mod_arith + forall_intro"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "L_id_unique"
        ],
        "before_tex": "e=3\\ \\text{is an identity}",
        "after_tex": "e=3\\ \\text{is the unique identity}",
        "relation": "implies",
        "rule": "L_id_unique",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 要求單位元唯一。依引理 L_id_unique（S06.1 證明），任何另一個雙邊單位元 e′ 都等於 3。",
          "en": "The MKO requires the identity to be unique. By Lemma L_id_unique (proved in S06.1), any other two-sided identity e′ equals 3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "L_id_unique"
          ],
          "explanation": {
            "zh-Hant": "需要 S05 的存在性與引理的唯一性。",
            "en": "It needs existence from S05 and uniqueness from the lemma."
          }
        },
        "leaf_basis": "L_id_unique"
      },
      {
        "id": "S06.1",
        "uses": [
          "eq_refl_symm_trans"
        ],
        "before_tex": "e,\\ e'\\ \\text{two-sided identities}",
        "after_tex": "e=e*e'=e'",
        "relation": "implies",
        "rule": "eq_refl_symm_trans",
        "conditions": [],
        "explanation": {
          "zh-Hant": "因 e′ 是單位元，e*e′=e；因 e 是單位元，e*e′=e′。依等號的對稱與傳遞，e=e′。這證明了引理 L_id_unique，對任何二元運算都成立，不需要結合律。",
          "en": "Since e′ is an identity, e*e′=e; since e is an identity, e*e′=e′. By symmetry and transitivity of equality, e=e′. This proves Lemma L_id_unique, valid for any binary operation and without associativity."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "eq_refl_symm_trans"
          ],
          "explanation": {
            "zh-Hant": "只用兩個單位元的定義與等號性質；不使用 L_id_unique 本身。",
            "en": "It uses only the definition of the two identities and properties of equality; L_id_unique itself is not used."
          }
        },
        "leaf_basis": "eq_refl_symm_trans + def_unfold"
      },
      {
        "id": "S07",
        "uses": [
          "S02",
          "S05",
          "mod_arith"
        ],
        "before_tex": "a\\in G",
        "after_tex": "a^{-1}=(2-a)\\bmod 4:\\ 0\\leftrightarrow 2,\\ 1\\leftrightarrow 1,\\ 3\\leftrightarrow 3",
        "relation": "implies",
        "rule": "mod_arith",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對 b=(2−a) mod 4：a*b≡a+(2−a)+1=3 (mod 4)，而 a*b 在 0 到 3 之間，所以 a*b=3=e；同樣 b*a=3。具體：0*2=3、1*1=3、2*0=3、3*3=3，與表相符。每個元素都有逆元。",
          "en": "For b=(2−a) mod 4: a*b≡a+(2−a)+1=3 (mod 4), and a*b lies between 0 and 3, so a*b=3=e; likewise b*a=3. Concretely: 0*2=3, 1*1=3, 2*0=3, 3*3=3, matching the table. Every element has an inverse."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S05",
            "mod_arith"
          ],
          "explanation": {
            "zh-Hant": "逆元以 S05 的 e=3 為準；若誤用標籤 0 當單位元，會找錯逆元。",
            "en": "Inverses are taken relative to e=3 from S05; wrongly using the label 0 as the identity would give wrong inverses."
          }
        },
        "leaf_basis": "mod_arith + forall_intro"
      },
      {
        "id": "S08",
        "uses": [
          "S03",
          "S04",
          "S06",
          "S07",
          "group_axioms"
        ],
        "before_tex": "\\text{closure, associativity, unique } e,\\ \\text{inverses}",
        "after_tex": "(G,*)\\ \\text{is a group}",
        "relation": "implies",
        "rule": "group_axioms",
        "conditions": [],
        "explanation": {
          "zh-Hant": "群資格的四項都已證明（S03、S04、S06、S07），依群公理 (G,*) 是群。到這裡才進入第二階段。",
          "en": "All four items of group qualification are proved (S03, S04, S06, S07), so by the group axioms (G,*) is a group. Only now does the second stage begin."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S04",
            "S06",
            "S07",
            "group_axioms",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "四項缺一不可；交換性尚未使用。",
            "en": "None of the four items may be missing; commutativity has not been used yet."
          }
        },
        "leaf_basis": "group_axioms + and_intro_elim"
      },
      {
        "id": "S09",
        "uses": [
          "S02",
          "S08",
          "add_comm"
        ],
        "before_tex": "a*b=(a+b+1)\\bmod 4",
        "after_tex": "a*b=(b+a+1)\\bmod 4=b*a\\ \\text{for all } a,b\\in G",
        "relation": "implies",
        "rule": "add_comm",
        "conditions": [],
        "explanation": {
          "zh-Hant": "整數加法可交換，a+b+1=b+a+1，所以 a*b=b*a；a、b 任意，涵蓋全部 16 對（包括 6 個非對角的無序對），而非只看幾個樣本。等價地說，在行與列用同一順序排行時，表對主對角線對稱；但這個對稱只有在 S08 的群資格之後才代表阿貝爾群。",
          "en": "Integer addition is commutative, a+b+1=b+a+1, so a*b=b*a; a and b are arbitrary, covering all 16 pairs (including the 6 unordered off-diagonal pairs), not just a few samples. Equivalently, with rows and columns in the same order, the table is symmetric about its main diagonal; but this symmetry signals an abelian group only after the group qualification of S08."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S08",
            "add_comm",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要對所有 a、b 成立的公式；只檢查例如 0*1=1*0 與 0*2=2*0 兩個樣本不夠。",
            "en": "It needs the formula valid for all a, b; checking only samples such as 0*1=1*0 and 0*2=2*0 would not be enough."
          }
        },
        "leaf_basis": "add_comm + forall_intro"
      },
      {
        "id": "S10",
        "uses": [
          "S08",
          "S09",
          "mko_fag"
        ],
        "before_tex": "(G,*)\\ \\text{group},\\ \\forall a,b:\\ a*b=b*a",
        "after_tex": "(G,*)\\ \\text{is a finite abelian group of order } 4",
        "relation": "implies",
        "rule": "mko_fag",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩階段都通過：(G,*) 是群（S08），且所有元素兩兩交換（S09）。依 MKO，(G,*) 是有限阿貝爾群，階數 |G|=4。",
          "en": "Both stages pass: (G,*) is a group (S08) and all elements commute pairwise (S09). By the MKO, (G,*) is a finite abelian group of order |G|=4."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "S09",
            "mko_fag"
          ],
          "explanation": {
            "zh-Hant": "只合取兩階段的結論；分類只關於這一張明示表。",
            "en": "It only conjoins the two stages; the classification concerns this one supplied table."
          }
        },
        "leaf_basis": "mko_fag + and_intro_elim"
      },
      {
        "id": "S11",
        "uses": [
          "S02",
          "S05",
          "power_def"
        ],
        "before_tex": "0^1=0",
        "after_tex": "0^2=0*0=1,\\ 0^3=1*0=2,\\ 0^4=2*0=3=e;\\ \\operatorname{ord}(0)=4",
        "relation": "implies",
        "rule": "power_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依冪的定義用同一個群運算計算：0²=0*0=1，0³=0²*0=1*0=2，0⁴=0³*0=2*0=3=e。0¹、0²、0³ 都不等於 e=3，所以 0 的階是 4。階必須相對於單位元 3 計算，而不是標籤 0（MKO para1）。",
          "en": "Compute with the same group operation, by the definition of powers: 0²=0*0=1, 0³=0²*0=1*0=2, 0⁴=0³*0=2*0=3=e. None of 0¹, 0², 0³ equals e=3, so the order of 0 is 4. The order must be computed relative to the identity 3, not the label 0 (MKO para1)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S05",
            "power_def"
          ],
          "explanation": {
            "zh-Hant": "需要表的值與 e=3；最小性由逐一核對 k=1、2、3 得到。",
            "en": "It needs the table values and e=3; minimality comes from checking k=1, 2, 3 one by one."
          }
        },
        "leaf_basis": "power_def + table_read"
      },
      {
        "id": "S12",
        "uses": [
          "S11",
          "S02",
          "S05",
          "power_def"
        ],
        "before_tex": "\\langle 0\\rangle,\\ 1^2=1*1",
        "after_tex": "\\langle 0\\rangle=\\{0,1,2,3\\}=G;\\ 1^2=3=e,\\ \\operatorname{ord}(1)=2,\\ \\langle 1\\rangle=\\{1,3\\}",
        "relation": "implies",
        "rule": "power_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "0 的冪是 0、1、2、3 後又循環回來，所以 ⟨0⟩=G，這張表定義的群是循環群。另一方面 1¹=1≠e，1²=1*1=3=e，所以 1 的階是 2，⟨1⟩={1,3}。標籤 1 並不是生成元；這些子群只屬於這個指定的有限群（MKO para2）。",
          "en": "The powers of 0 are 0, 1, 2, 3 and then repeat, so ⟨0⟩=G and the group defined by this table is cyclic. On the other hand 1¹=1≠e and 1²=1*1=3=e, so the order of 1 is 2 and ⟨1⟩={1,3}. The label 1 is not a generator; these subgroups belong only to this specified finite group (MKO para2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "power_def",
            "table_read"
          ],
          "explanation": {
            "zh-Hant": "需要表的值與 e=3；不宣稱任何關於其他群的結論。",
            "en": "It needs the table values and e=3; no claim is made about any other group."
          }
        },
        "leaf_basis": "power_def + table_read"
      },
      {
        "id": "S13",
        "uses": [
          "G3",
          "abs_rules",
          "quantifier_negation"
        ],
        "before_tex": "a\\circ b=|a-b|\\ \\text{on}\\ K=\\{0,1,2\\}",
        "after_tex": "(1\\circ 1)\\circ 2=2\\neq 0=1\\circ(1\\circ 2);\\ (K,\\circ)\\ \\text{is not a group}",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對照表：|a−b|=|b−a|，所以表對角線對稱；0 是單位元（|0−b|=b）；每個元素是自己的逆（a∘a=0）。但 (1∘1)∘2=0∘2=2，而 1∘(1∘2)=1∘1=0，結合律失敗。一組反例就否定「對所有三元組結合」，所以 (K,∘) 不是群，自然也不是阿貝爾群。表對稱不能代替群資格。",
          "en": "Contrast table: |a−b|=|b−a|, so the table is symmetric about its diagonal; 0 is an identity (|0−b|=b); every element is its own inverse (a∘a=0). But (1∘1)∘2=0∘2=2 while 1∘(1∘2)=1∘1=0, so associativity fails. One counterexample refutes 'associative for all triples', so (K,∘) is not a group, and hence not an abelian group. A symmetric table cannot replace group qualification."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "abs_rules",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "否證只需一組三元組；不宣稱其他三元組也失敗。",
            "en": "A refutation needs one triple; it is not claimed that other triples fail too."
          }
        },
        "leaf_basis": "quantifier_negation + abs_rules + numeric_evaluation"
      }
    ],
    "result": {
      "zh-Hant": "給定的 4×4 表就是 a*b=(a+b+1) mod 4。它封閉、結合（兩邊都與 a+b+c+2 模 4 同餘）、有唯一單位元 e=3、每個 a 有逆元 (2−a) mod 4，且對所有 a、b 交換，所以 (G,*) 是 4 階有限阿貝爾群。元素 0 的階為 4，⟨0⟩=G；元素 1 的階為 2，⟨1⟩={1,3}。對照的對稱表 |a−b| 在 {0,1,2} 上不結合，不是群。",
      "en": "The given 4×4 table is a*b=(a+b+1) mod 4. It is closed, associative (both sides are congruent to a+b+c+2 modulo 4), has the unique identity e=3, every a has the inverse (2−a) mod 4, and it is commutative for all a, b, so (G,*) is a finite abelian group of order 4. The element 0 has order 4 and ⟨0⟩=G; the element 1 has order 2 and ⟨1⟩={1,3}. The symmetric contrast table |a−b| on {0,1,2} is not associative and is not a group."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "{0,1,2} 上 |a−b| 的表對主對角線對稱，所以它是阿貝爾群",
          "en": "The table of |a−b| on {0,1,2} is symmetric about the main diagonal, so it is an abelian group"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "結合律失敗：(1∘1)∘2=2，1∘(1∘2)=0。",
          "en": "Associativity fails: (1∘1)∘2=2 and 1∘(1∘2)=0."
        }
      },
      {
        "claim": {
          "zh-Hant": "由 0*1=1*0 與 0*2=2*0 兩個樣本即可推出 (G,*) 交換",
          "en": "From the two samples 0*1=1*0 and 0*2=2*0 one may infer that (G,*) is commutative"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "兩個樣本不涵蓋其餘四個非對角無序對；結論要靠 S09 對所有 a、b 的論證。",
          "en": "Two samples do not cover the other four unordered off-diagonal pairs; the conclusion needs the argument of S09 for all a, b."
        }
      },
      {
        "claim": {
          "zh-Hant": "標籤 0 是這個群的單位元",
          "en": "The label 0 is the identity of this group"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "0*0=1≠0；單位元是 3。",
          "en": "0*0=1≠0; the identity is 3."
        }
      },
      {
        "claim": {
          "zh-Hant": "元素 1 生成整個群 G",
          "en": "The element 1 generates the whole group G"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "1*1=3=e，所以 ⟨1⟩={1,3}≠G。",
          "en": "1*1=3=e, so ⟨1⟩={1,3}≠G."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "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": "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": "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": "abs_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "絕對值（實數）：|u|=u（u≥0）、|u|=−u（u<0）；|u|≥0；|uv|=|u||v|；三角不等式 |u+v|≤|u|+|v|；|u|<r ⇔ −r<u<r。",
          "en": "Absolute value (reals): |u|=u if u≥0 and |u|=−u if u<0; |u|≥0; |uv|=|u||v|; triangle inequality |u+v|≤|u|+|v|; |u|<r ⇔ −r<u<r."
        }
      },
      {
        "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": "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": "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": "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_fag",
        "kind": "cited",
        "source": {
          "mko_id": "mko-finite-abelian-group",
          "sha256": "1486476569952aee082941d6f91d2921526de39dd732a11bacce98c252b95415",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "有限阿貝爾群先須以完整 Cayley 表通過封閉、結合、唯一單位元與逆元資格，再對所有 a、b 核 a*b=b*a。本輪只分類明示有限表；不由幾個樣本或生成元看似交換外推整群交換。",
          "en": "A finite abelian group must first qualify as a group using its complete Cayley table: closure, associativity, a unique identity and inverses. Then check a*b=b*a for all a, b. This slice classifies only explicitly supplied finite tables; a few commuting samples or apparently commuting generators are not used to infer commutativity of the whole group."
        }
      },
      {
        "id": "table_read",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "讀表：a*b 是明示 Cayley 表第 a 行第 b 列的格子；表完整表示每一對 (a,b) 都有恰好一個格子。",
          "en": "Reading the table: a*b is the cell in row a, column b of the supplied Cayley table; a complete table has exactly one cell for every pair (a,b)."
        }
      },
      {
        "id": "mod_arith",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "模 n 算術（n≥1）：x mod n 是唯一滿足 0≤r≤n−1 且 n∣(x−r) 的整數 r。x≡y (mod n) 表示 n∣(x−y)；它是等價關係且與加法相容；x mod n≡x；兩個都在 0 到 n−1 之間且模 n 同餘的整數相等。",
          "en": "Arithmetic modulo n (n≥1): x mod n is the unique integer r with 0≤r≤n−1 and n∣(x−r). x≡y (mod n) means n∣(x−y); it is an equivalence relation compatible with addition; x mod n≡x; two integers between 0 and n−1 that are congruent modulo n are equal."
        }
      },
      {
        "id": "power_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "群中的冪與階：a¹=a，a^(k+1)=a^k*a；a 的階是使 a^k=e 的最小正整數 k；⟨a⟩ 是 a 的所有冪組成的集合。",
          "en": "Powers and order in a group: a¹=a and a^(k+1)=a^k*a; the order of a is the least positive integer k with a^k=e; ⟨a⟩ is the set of all powers of a."
        }
      },
      {
        "id": "L_id_unique",
        "kind": "lemma",
        "proved_in": [
          "S06.1"
        ],
        "statement": {
          "zh-Hant": "若 e 與 e′ 都是運算 * 的雙邊單位元，則 e=e′。",
          "en": "If e and e′ are both two-sided identities for the operation *, then e=e′."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "vars": [
          "a",
          "b"
        ],
        "range": [
          0,
          3
        ],
        "lhs": "(a+b+1)%4",
        "relation": "<=",
        "rhs": "3"
      },
      {
        "id": "N2",
        "step": "S03",
        "vars": [
          "a",
          "b"
        ],
        "range": [
          0,
          3
        ],
        "lhs": "(a+b+1)%4",
        "relation": ">=",
        "rhs": "0"
      },
      {
        "id": "N3",
        "step": "S04",
        "vars": [
          "a",
          "b",
          "c"
        ],
        "range": [
          0,
          3
        ],
        "lhs": "((a+b+1)%4+c+1)%4",
        "rhs": "(a+(b+c+1)%4+1)%4"
      },
      {
        "id": "N4",
        "step": "S04.1",
        "vars": [
          "a",
          "b",
          "c"
        ],
        "range": [
          0,
          3
        ],
        "lhs": "(((a+b+1)%4+c+1)-(a+b+c+2))%4",
        "rhs": "0"
      },
      {
        "id": "N5",
        "step": "S05",
        "vars": [
          "a"
        ],
        "range": [
          0,
          3
        ],
        "lhs": "(3+a+1)%4",
        "rhs": "a"
      },
      {
        "id": "N6",
        "step": "S05",
        "vars": [
          "a"
        ],
        "range": [
          0,
          3
        ],
        "lhs": "(a+3+1)%4",
        "rhs": "a"
      },
      {
        "id": "N7",
        "step": "S07",
        "vars": [
          "a"
        ],
        "range": [
          0,
          3
        ],
        "lhs": "(a+(2-a)%4+1)%4",
        "rhs": "3"
      },
      {
        "id": "N8",
        "step": "S07",
        "vars": [
          "a"
        ],
        "range": [
          0,
          3
        ],
        "lhs": "((2-a)%4+a+1)%4",
        "rhs": "3"
      },
      {
        "id": "N9",
        "step": "S09",
        "vars": [
          "a",
          "b"
        ],
        "range": [
          0,
          3
        ],
        "lhs": "(a+b+1)%4",
        "rhs": "(b+a+1)%4"
      },
      {
        "id": "N10",
        "step": "S11",
        "lhs": "(0+0+1)%4",
        "rhs": "1"
      },
      {
        "id": "N11",
        "step": "S11",
        "lhs": "(1+0+1)%4",
        "rhs": "2"
      },
      {
        "id": "N12",
        "step": "S11",
        "lhs": "(2+0+1)%4",
        "rhs": "3"
      },
      {
        "id": "N13",
        "step": "S12",
        "lhs": "(1+1+1)%4",
        "rhs": "3"
      },
      {
        "id": "N14",
        "step": "S13",
        "lhs": "abs(abs(1-1)-2)",
        "relation": "!=",
        "rhs": "abs(1-abs(1-2))"
      },
      {
        "id": "N15",
        "step": "S02",
        "lhs": "(0+3+1)%4",
        "rhs": "0"
      },
      {
        "id": "N16",
        "step": "S02",
        "lhs": "(2+1+1)%4",
        "rhs": "0"
      },
      {
        "id": "N17",
        "step": "S05",
        "lhs": "(0+0+1)%4",
        "relation": "!=",
        "rhs": "0"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "停在 MKO 的切片內：載體標籤 0 到 n−1（此處 n=4）與完整的 n×n 表，只分類明示給出的有限表。用模 4 算術給出全稱證明，並把對整個載體的窮舉核對作為交叉核對。不涉及有限阿貝爾群的結構定理或其他群；不改動 canonical MKO 或英文 edition。",
      "en": "Stays inside the MKO's slice: carrier labels 0 to n−1 (here n=4) and a complete n×n table, classifying only an explicitly supplied finite table. Modular arithmetic modulo 4 gives the universal proofs, and exhaustive checks over the whole carrier serve as cross-checks. The structure theorem for finite abelian groups and other groups are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "G=\\{0,1,2,3\\}",
        "載體（標籤）",
        "the carrier (labels)"
      ],
      [
        "a*b",
        "表中第 a 行第 b 列的格子",
        "the entry in row a, column b of the table"
      ],
      [
        "x\\bmod 4",
        "x 除以 4 的餘數，取值 0 到 3",
        "the remainder of x on division by 4, a value from 0 to 3"
      ],
      [
        "e",
        "單位元（此表中 e=3）",
        "the identity element (here e=3)"
      ],
      [
        "a^k,\\ \\operatorname{ord}(a)",
        "a 的 k 次方與 a 的階",
        "the k-th power of a and the order of a"
      ],
      [
        "\\langle a\\rangle",
        "a 生成的循環子群",
        "the cyclic subgroup generated by a"
      ]
    ],
    "md_error_controls": [
      {
        "title": "表對稱不足以判定阿貝爾群 / A symmetric table alone does not establish an abelian group",
        "zh": "MKO 指出的誤解是因表對角線對稱就宣稱阿貝爾群。|a−b| 的表對稱、有單位元 0、每個元素自逆，但 (1∘1)∘2=2≠0=1∘(1∘2)，不是群（S13）。必須先完成群資格（S03 至 S08），交換性才有分類意義。",
        "en": "The mistake named by the MKO is to declare an abelian group because the table is symmetric about its diagonal. The table of |a−b| is symmetric, has identity 0 and every element is self-inverse, yet (1∘1)∘2=2≠0=1∘(1∘2), so it is not a group (S13). Group qualification (S03 to S08) must be completed first; only then does commutativity classify anything."
      },
      {
        "title": "幾個交換樣本不夠 / A few commuting samples are not enough",
        "zh": "MKO 的另一個誤解是由幾個元素交換外推整群交換。S09 用對所有 a、b 成立的公式。反例模式：在 {1,2,3} 的置換群中，r=(1 2 3) 與 r² 交換、單位元與一切交換，但 s=(1 2) 與 r 不交換：(r∘s)(1)=r(2)=3，而 (s∘r)(1)=s(2)=1。",
        "en": "Another misconception identified in the MKO is to infer that the whole group commutes from a few commuting elements. S09 uses a formula valid for all a, b. A counterexample to the pattern: in the permutation group of {1,2,3}, r=(1 2 3) commutes with r² and the identity commutes with everything, but s=(1 2) does not commute with r: (r∘s)(1)=r(2)=3 while (s∘r)(1)=s(2)=1."
      },
      {
        "title": "單位元不一定是標籤 0 / The identity need not be the label 0",
        "zh": "標籤 0 到 n−1 只是名字。此表的單位元是 3（0*0=1≠0）。若誤把 0 當單位元，會找錯逆元，也會算錯階：例如以 0 為單位元時 0 的階會被誤判為 1。",
        "en": "The labels 0 to n−1 are only names. The identity of this table is 3 (0*0=1≠0). Wrongly taking 0 as the identity gives wrong inverses and wrong orders: for example, the order of 0 would wrongly be taken as 1."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "N1 至 N9 是對載體 {0,1,2,3} 的每個元素（或每對、每組三元組）做的有限窮舉核對。因為載體恰有這四個元素，它們涵蓋這一張表的所有情形；它們是 S03 至 S09 模算術證明的交叉核對，對其他表不說明任何事。",
        "en": "N1 to N9 are finite exhaustive checks over every element (or pair, or triple) of the carrier {0,1,2,3}. Since the carrier has exactly these four elements they cover every case of this one table; they cross-check the modular proofs of S03 to S09 and say nothing about other tables."
      },
      {
        "zh-Hant": "錯誤控制中的置換群反例沒有寫成完整 6×6 表，只用置換合成計算一組不交換的元素。",
        "en": "The permutation-group counterexample in the error controls is not written as a complete 6×6 table; only one non-commuting pair is computed by composing permutations."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "8364c2e3fd58e2b9251219c8186c64a27ff0c6e88d59156b8ccb03f562a4beda",
    "original_source_sha256": "8364c2e3fd58e2b9251219c8186c64a27ff0c6e88d59156b8ccb03f562a4beda",
    "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."
  }
}