{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-finite-commutative-ring-01",
  "mko_id": "mko-finite-commutative-ring",
  "title": {
    "zh-Hant": "有限交換環：逐元素對核對 ℤ₆ 的交換律，並以上三角矩陣環為非交換反例",
    "en": "Finite commutative ring: checking commutativity of ℤ₆ over all element pairs, with an upper-triangular matrix ring as a non-commutative counterexample"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-finite-commutative-ring-01",
    "source": {
      "mko_id": "mko-finite-commutative-ring",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "5a488ffcb874034a7b2144c7585ccef8981c9d6bc0d2959107aa85650113234a",
      "english_sha256": "99c36775e32ec1ae2f5f4db3a9f3366e90e9bbb558ea3d9b1c720c3bd62b70fb"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "mul_comm",
      "eq_refl_symm_trans",
      "matrix_product_def",
      "quantifier_negation",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "（一）在已通過有限含一環資格的 ℤ₆={0,1,2,3,4,5}（a⊕b=(a+b) mod 6、a⊗b=(ab) mod 6）上，核對全部 36 個有序元素對的 a⊗b=b⊗a，判定 ℤ₆ 是否為有限交換環。（二）說明 ℤ₆ 有非零零因子且 2 不可逆，所以不是整環或域。（三）在已通過資格的 ℱ₂ 上 2×2 上三角矩陣環 U 中，說明若干乘積可交換，但 U 不是交換環。",
      "en": "(1) On ℤ₆={0,1,2,3,4,5} (a⊕b=(a+b) mod 6, a⊗b=(ab) mod 6), already qualified as a finite unital ring, check a⊗b=b⊗a for all 36 ordered element pairs and decide whether ℤ₆ is a finite commutative ring. (2) Show that ℤ₆ has nonzero zero divisors and that 2 is not invertible, so it is neither an integral domain nor a field. (3) In the qualified ring U of 2-by-2 upper-triangular matrices over 𝔽₂, show that several products commute and yet U is not commutative."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "ℤ₆ 的載體為 {0,1,2,3,4,5}，a⊕b=(a+b) mod 6，a⊗b=(ab) mod 6。乘法表各行：第 0 行 (0,0,0,0,0,0)；第 1 行 (0,1,2,3,4,5)；第 2 行 (0,2,4,0,2,4)；第 3 行 (0,3,0,3,0,3)；第 4 行 (0,4,2,0,4,2)；第 5 行 (0,5,4,3,2,1)；第 a 行、列索引 b 的項是 a⊗b。",
          "en": "ℤ₆ has carrier {0,1,2,3,4,5}, a⊕b=(a+b) mod 6 and a⊗b=(ab) mod 6. The rows of the multiplication table are: row 0 (0,0,0,0,0,0); row 1 (0,1,2,3,4,5); row 2 (0,2,4,0,2,4); row 3 (0,3,0,3,0,3); row 4 (0,4,2,0,4,2); row 5 (0,5,4,3,2,1); the entry at column index b in row a is a⊗b."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "(ℤ₆,⊕,⊗) 已由父節點以這兩張完整表通過有限含一環資格，0 為加法單位元，1 為乘法單位元；本稿不重新核對。",
          "en": "(ℤ₆,⊕,⊗) has already qualified, with these two complete tables, as a finite unital ring in the parent node, with additive identity 0 and multiplicative identity 1; this guide does not recheck it."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "U 的載體是 8 個三元組 (a,b,c)∈{0,1}³，代表第一行 (a,b)、第二行 (0,c) 的矩陣；加法逐項 mod 2，乘法為矩陣乘積再 mod 2。U 已由父節點通過有限含一環資格，單位元為 I=(1,0,1)；本稿不重新核對。",
          "en": "U has as carrier the 8 triples (a,b,c)∈{0,1}³, standing for the matrix with first row (a,b) and second row (0,c); addition is entrywise mod 2 and multiplication is the matrix product mod 2. U has already qualified as a finite unital ring in the parent node, with identity I=(1,0,1); this guide does not recheck it."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_comm_ring_def",
          "mko_unital_ring_def"
        ],
        "before_tex": "\\mathbb{Z}_6\\ \\text{finite commutative ring}",
        "after_tex": "\\mathbb{Z}_6\\ \\text{qualified unital ring}\\ \\land\\ \\forall a,b\\in\\mathbb{Z}_6:\\ a\\otimes b=b\\otimes a",
        "relation": "definition",
        "rule": "mko_comm_ring_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，有限交換環＝父環資格＋乘法交換律，兩者用同一組表。父環資格由 G2 提供（保留同一個 0、1 與全部環公理），所以剩下的責任只有：對每個有序對 (a,b) 核對 a⊗b=b⊗a。",
          "en": "By the MKO, a finite commutative ring is the parent-ring qualification plus commutativity of multiplication, both on the same tables. G2 supplies the parent qualification (keeping the same 0, 1 and all ring axioms), so the only remaining obligation is to check a⊗b=b⊗a for every ordered pair (a,b)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "mko_comm_ring_def",
            "mko_unital_ring_def"
          ],
          "explanation": {
            "zh-Hant": "此步只展開定義；若父環資格不成立，即使乘法表對稱也不能稱為交換環。",
            "en": "This step only unfolds the definition; if the parent qualification failed, a symmetric table would still not make a commutative ring."
          }
        },
        "leaf_basis": "mko_comm_ring_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "G1",
          "finite_counting_basics"
        ],
        "before_tex": "\\mathbb{Z}_6\\times\\mathbb{Z}_6",
        "after_tex": "36=6\\ \\text{diagonal pairs}+30\\ \\text{off-diagonal pairs}",
        "relation": "equivalent",
        "rule": "finite_counting_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "有序對共 6·6=36 個，分成 a=b 的 6 個對角對與 a≠b 的 30 個非對角對；30 個非對角對又兩兩配成 (a,b) 與 (b,a)，共 15 組。這個分割不漏掉任何一對。",
          "en": "There are 6·6=36 ordered pairs, split into the 6 diagonal pairs with a=b and the 30 off-diagonal pairs with a≠b; the 30 off-diagonal pairs match up as (a,b) and (b,a), giving 15 groups. The split omits no pair."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "finite_counting_basics",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "對角與非對角是不交且窮盡的分割，6+30=36，兩邊是同一個集合（N1、N2）。",
            "en": "Diagonal and off-diagonal form a disjoint and exhaustive split, 6+30=36, so both sides describe the same set (N1, N2)."
          }
        },
        "leaf_basis": "finite_counting_basics + cartesian_product_def + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S02"
        ],
        "before_tex": "(a,a),\\ a\\in\\mathbb{Z}_6",
        "after_tex": "a\\otimes a=a\\otimes a",
        "relation": "implies",
        "rule": "eq_refl_symm_trans",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對角對的交換條件是 a⊗a=a⊗a，由等號的自反性對 6 個對角對全部成立，不需查表。",
          "en": "For a diagonal pair the commutativity condition reads a⊗a=a⊗a, which holds by reflexivity of equality for all 6 diagonal pairs, with no table lookup."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "eq_refl_symm_trans"
          ],
          "explanation": {
            "zh-Hant": "只用自反性；這 6 對不需要任何環的性質。",
            "en": "Only reflexivity is used; these 6 pairs need no ring property."
          }
        },
        "leaf_basis": "eq_refl_symm_trans"
      },
      {
        "id": "S04",
        "uses": [
          "S02"
        ],
        "before_tex": "a\\otimes b=b\\otimes a\\ \\ \\text{and}\\ \\ b\\otimes a=a\\otimes b",
        "after_tex": "\\text{one equation per group}\\ \\{a,b\\},\\ a<b:\\ 15\\ \\text{checks}",
        "relation": "equivalent",
        "rule": "eq_refl_symm_trans",
        "conditions": [],
        "explanation": {
          "zh-Hant": "有序對 (a,b) 的條件 a⊗b=b⊗a 與 (b,a) 的條件 b⊗a=a⊗b 依等號對稱性互相等價，所以每組只需核對一個等式；這不是省略，而是同一個等式同時涵蓋兩個有序對。",
          "en": "The condition a⊗b=b⊗a for (a,b) and the condition b⊗a=a⊗b for (b,a) are equivalent by symmetry of equality, so each group needs one equation; nothing is skipped, since one equation covers both ordered pairs."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_refl_symm_trans"
          ],
          "explanation": {
            "zh-Hant": "等號對稱性雙向成立，所以兩個條件同真同假。",
            "en": "Symmetry of equality works both ways, so the two conditions are true or false together."
          }
        },
        "leaf_basis": "eq_refl_symm_trans"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "mul_comm",
          "mod_reduce"
        ],
        "before_tex": "a\\otimes b=(ab)\\bmod 6",
        "after_tex": "a\\otimes b=(ab)\\bmod 6=(ba)\\bmod 6=b\\otimes a\\ \\ (\\forall a,b\\in\\mathbb{Z}_6)",
        "relation": "implies",
        "rule": "mul_comm",
        "conditions": [],
        "explanation": {
          "zh-Hant": "全稱證書：表由公式 (ab) mod 6 產生。整數乘法交換，ab=ba；相等的整數有相同的模 6 餘數，所以 (ab) mod 6=(ba) mod 6，即 a⊗b=b⊗a。這對每一對都成立，不依賴樣本。",
          "en": "Universal certificate: the table is generated by the formula (ab) mod 6. Integer multiplication commutes, ab=ba; equal integers have equal residues mod 6, so (ab) mod 6=(ba) mod 6, that is a⊗b=b⊗a. This holds for every pair and does not depend on samples."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mul_comm",
            "mod_reduce",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "此公式證書使用 G1 明示的運算定義 (ab) mod 6。S06 另核列出表格的 15 組非對角對稱等式，只證表格對稱，不是 36 個表格值逐項等於公式的核對；若要從任意獨立表格移交公式證書，須另驗證該一致性。",
            "en": "The formula certificate uses the operation explicitly defined by G1 as (ab) mod 6. S06 separately checks the 15 off-diagonal symmetry equations of the listed table. It establishes table symmetry, not equality of all 36 stored entries with the formula; transferring a formula certificate to an arbitrary independent table would require that additional consistency check."
          }
        },
        "leaf_basis": "mul_comm + mod_reduce + eq_replace"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "S04"
        ],
        "before_tex": "\\text{row}\\ a\\ \\text{vs. column}\\ a\\ \\text{of the listed table}",
        "after_tex": "\\text{the 15 entries above the diagonal equal the 15 below}",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在列出的表上逐組核對 15 個等式：含 0 的 5 組都是 0=0；{1,b} 的 4 組是 b=b（第 1 行是 (0,1,2,3,4,5)）；{2,3}:0=0，{2,4}:2=2，{2,5}:4=4；{3,4}:0=0，{3,5}:3=3；{4,5}:2=2。5+4+3+2+1=15 組全部相等。",
          "en": "Check the 15 equations on the listed table group by group: the 5 groups containing 0 all read 0=0; the 4 groups {1,b} read b=b (row 1 is (0,1,2,3,4,5)); {2,3}: 0=0, {2,4}: 2=2, {2,5}: 4=4; {3,4}: 0=0, {3,5}: 3=3; {4,5}: 2=2. All 5+4+3+2+1=15 groups are equal."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S04",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "15 組涵蓋全部非對角有序對（S02、S04）；N3 在全部 36 對上以公式重算，因載體恰為這六個元素，這個有限核對窮盡了所有對。",
            "en": "The 15 groups cover every off-diagonal ordered pair (S02, S04); N3 recomputes the formula at all 36 pairs, and since the carrier is exactly these six elements this finite check exhausts every pair."
          }
        },
        "leaf_basis": "case_analysis + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S01",
          "S03",
          "S05",
          "S06"
        ],
        "before_tex": "\\text{6 diagonal pairs (S03)},\\ \\text{30 off-diagonal pairs (S05, S06)}",
        "after_tex": "\\mathbb{Z}_6\\ \\text{is a finite commutative ring}",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "36 個有序對分成對角與非對角兩類（S02），兩類都已核對，依分情形證明，所有對都滿足交換律；加上 G2 的父環資格，ℤ₆ 是有限交換環。",
          "en": "The 36 ordered pairs fall into the diagonal and off-diagonal classes (S02), both of which are checked, so by proof by cases every pair commutes; together with the parent qualification of G2, ℤ₆ is a finite commutative ring."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S01",
            "S02",
            "S03",
            "S05",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "目標 G：ℤ₆ 交換。足夠條件 H：父環資格且所有 36 對可交換；H⇒G 由 S01 的定義。H 由 G2、S03、S05 正向證得，沒有用到 G。",
            "en": "Goal G: ℤ₆ is commutative. Sufficient condition H: parent qualification and all 36 pairs commute; H⇒G by the definition in S01. H is proved forwards from G2, S03 and S05, without using G."
          }
        },
        "leaf_basis": "case_analysis + mko_comm_ring_def"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "S07",
          "domain_field_def"
        ],
        "before_tex": "2\\otimes 3=(6)\\bmod 6",
        "after_tex": "2\\otimes 3=0,\\ 2\\neq 0,\\ 3\\neq 0\\ \\Rightarrow\\ \\mathbb{Z}_6\\ \\text{not an integral domain}",
        "relation": "implies",
        "rule": "domain_field_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第 2 行、列索引 3 的項是 0：2⊗3=6 mod 6=0，而 2、3 都非零，所以 2 與 3 是零因子。ℤ₆ 雖然交換，卻不是整環。MKO 提醒：交換環仍可能有非零零因子。",
          "en": "The entry at column index 3 in row 2 is 0: 2⊗3=6 mod 6=0, while 2 and 3 are nonzero, so 2 and 3 are zero divisors. ℤ₆ is commutative but not an integral domain. The MKO warns that a commutative ring may still have nonzero zero divisors."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "domain_field_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一組零因子足以否定整環資格（N4）；交換性本身沒有被否定。",
            "en": "One pair of zero divisors suffices to deny the integral-domain property (N4); commutativity itself is not denied."
          }
        },
        "leaf_basis": "domain_field_def + quantifier_negation + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "G1",
          "domain_field_def"
        ],
        "before_tex": "\\text{row 2}=(0,2,4,0,2,4)",
        "after_tex": "\\nexists x:\\ 2\\otimes x=1\\ \\Rightarrow\\ \\mathbb{Z}_6\\ \\text{not a field}",
        "relation": "implies",
        "rule": "domain_field_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "2 的逆元 x 必須使 2⊗x=1；第 2 行的六個值是 0、2、4、0、2、4，沒有 1。所以非零元 2 不可逆，ℤ₆ 不是域。這是在完整的一行上窮舉，不是抽樣。",
          "en": "An inverse x of 2 would need 2⊗x=1; the six values of row 2 are 0, 2, 4, 0, 2, 4, with no 1. So the nonzero element 2 is not invertible and ℤ₆ is not a field. This is an exhaustive check over a complete row, not a sample."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "domain_field_def"
          ],
          "explanation": {
            "zh-Hant": "否定存在需要核對所有 x；第 2 行列出了全部六個 x 的值（N5 在整個載體上重算）。",
            "en": "Denying existence needs every x to be checked; row 2 lists the values for all six x (N5 recomputes over the whole carrier)."
          }
        },
        "leaf_basis": "domain_field_def + quantifier_negation + case_analysis"
      },
      {
        "id": "S10",
        "uses": [
          "G3",
          "matrix_product_def"
        ],
        "before_tex": "\\begin{pmatrix}a&b\\\\0&c\\end{pmatrix}\\begin{pmatrix}a'&b'\\\\0&c'\\end{pmatrix}",
        "after_tex": "(a,b,c)(a',b',c')=(aa',\\ ab'+bc',\\ cc')\\bmod 2",
        "relation": "equivalent",
        "rule": "matrix_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依矩陣乘積：左上 aa'+b·0=aa'；右上 ab'+bc'；左下 0·a'+c·0=0；右下 0·b'+cc'=cc'。所以乘積仍是上三角，記為三元組 (aa', ab'+bc', cc')，各項再取 mod 2。這個公式產生 U 的完整 8×8 乘法表。",
          "en": "By the matrix product: top left aa'+b·0=aa'; top right ab'+bc'; bottom left 0·a'+c·0=0; bottom right 0·b'+cc'=cc'. So the product is again upper triangular, written as the triple (aa', ab'+bc', cc'), each entry taken mod 2. This formula generates the complete 8-by-8 multiplication table of U."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "matrix_product_def",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "三元組與上三角矩陣一一對應，公式只是同一乘積的另一種寫法，雙向成立。",
            "en": "Triples correspond one-to-one to upper-triangular matrices, and the formula is another way of writing the same product, so the step holds both ways."
          }
        },
        "leaf_basis": "matrix_product_def + mul_zero + add_zero"
      },
      {
        "id": "S11",
        "uses": [
          "S10"
        ],
        "before_tex": "E_{11}E_{12}=(1,0,0)(0,1,0)",
        "after_tex": "E_{11}E_{12}=(1\\cdot0,\\ 1\\cdot1+0\\cdot0,\\ 0\\cdot0)=(0,1,0)=E_{12}",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代入 S10 的公式，a=1、b=0、c=0，a'=0、b'=1、c'=0：第一項 0，第二項 1·1+0·0=1，第三項 0。",
          "en": "Substitute into the formula of S10 with a=1, b=0, c=0 and a'=0, b'=1, c'=0: the first entry is 0, the second 1·1+0·0=1 and the third 0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S10",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確計算，兩邊同一元素（N6）。",
            "en": "Exact computation; both sides are the same element (N6)."
          }
        },
        "leaf_basis": "numeric_evaluation + matrix_product_def"
      },
      {
        "id": "S12",
        "uses": [
          "S10"
        ],
        "before_tex": "E_{12}E_{11}=(0,1,0)(1,0,0)",
        "after_tex": "E_{12}E_{11}=(0\\cdot1,\\ 0\\cdot0+1\\cdot0,\\ 0\\cdot0)=(0,0,0)",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "交換順序：a=0、b=1、c=0，a'=1、b'=0、c'=0。第二項 ab'+bc'=0·0+1·0=0，所以乘積是零矩陣。",
          "en": "Reverse the order: a=0, b=1, c=0 and a'=1, b'=0, c'=0. The second entry ab'+bc'=0·0+1·0=0, so the product is the zero matrix."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S10",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確計算，兩邊同一元素（N7）。",
            "en": "Exact computation; both sides are the same element (N7)."
          }
        },
        "leaf_basis": "numeric_evaluation + matrix_product_def"
      },
      {
        "id": "S13",
        "uses": [
          "S11",
          "S12",
          "G3",
          "mko_comm_ring_def"
        ],
        "before_tex": "E_{11}E_{12}=(0,1,0)\\neq(0,0,0)=E_{12}E_{11}",
        "after_tex": "\\exists a,b\\in U:\\ ab\\neq ba\\ \\Rightarrow\\ U\\ \\text{not commutative}",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個三元組的第二項 1≠0，所以 E₁₁E₁₂≠E₁₂E₁₁。交換律是全稱命題，一組失敗的有序對就足以否定：U 是有限含一環，但不是有限交換環。",
          "en": "The second entries of the two triples are 1≠0, so E₁₁E₁₂≠E₁₂E₁₁. Commutativity is a universal statement, so one failing ordered pair refutes it: U is a finite unital ring but not a finite commutative ring."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "S12",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "否定全稱只需一個見證；這不影響 U 的含一環資格（G3）。",
            "en": "Refuting a universal statement needs one witness; this does not affect U's qualification as a unital ring (G3)."
          }
        },
        "leaf_basis": "quantifier_negation + mko_comm_ring_def"
      },
      {
        "id": "S14",
        "uses": [
          "S10",
          "S13",
          "G3"
        ],
        "before_tex": "IX=XI,\\ E_{11}E_{22}=(0,0,0)=E_{22}E_{11},\\ E_{11}E_{11}=E_{11}",
        "after_tex": "\\text{some pairs commute, yet}\\ U\\ \\text{is not commutative}",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "U 中有許多對可交換：單位元 I 與每個元素交換；E₁₁E₂₂=(0,1·0+0·1,0)=(0,0,0)，E₂₂E₁₁=(0,0·0+0·0,0)=(0,0,0)；每個元素與自己交換。若只抽這些樣本，會誤以為 U 交換，但 S13 已給出反例。這正是 MKO 所說的「看到若干乘積可交換就宣稱整個環交換」的誤解。",
          "en": "Many pairs in U commute: the identity I commutes with every element; E₁₁E₂₂=(0,1·0+0·1,0)=(0,0,0) and E₂₂E₁₁=(0,0·0+0·0,0)=(0,0,0); every element commutes with itself. Sampling only these would suggest U is commutative, but S13 already gives a counterexample. This is the misconception identified in the MKO: declaring a whole ring commutative after seeing several commuting products."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "S13",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "樣本可交換與 S13 的反例不矛盾：全稱命題被一個反例否定，不被尚未窮盡定義域的正例樣本證明（N8）；若定義域有限且全部有序對都通過，完整窮舉就能建立該有限全稱命題。",
            "en": "The commuting samples do not contradict S13: a universal statement is refuted by one counterexample and is not proved by positive samples that leave cases unexamined (N8). For a finite domain, a successful exhaustive check of all ordered pairs does establish that finite universal statement."
          }
        },
        "leaf_basis": "quantifier_negation + numeric_evaluation"
      },
      {
        "id": "S15",
        "uses": [
          "S07",
          "S08",
          "S09",
          "S13"
        ],
        "before_tex": "\\text{S07},\\ \\text{S08},\\ \\text{S09},\\ \\text{S13}",
        "after_tex": "\\mathbb{Z}_6\\ \\text{commutative, not a domain, not a field};\\ U\\ \\text{not commutative}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "合併：ℤ₆ 經 36 個有序對全部核對，是有限交換環；它有零因子 2、3，2 不可逆，所以不是整環或域。U 有可交換的樣本，但 E₁₁E₁₂≠E₁₂E₁₁，不是交換環。",
          "en": "Combine: after all 36 ordered pairs are checked, ℤ₆ is a finite commutative ring; it has the zero divisors 2 and 3 and 2 is not invertible, so it is neither an integral domain nor a field. U has commuting samples but E₁₁E₁₂≠E₁₂E₁₁, so it is not commutative."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S07",
            "S08",
            "S09",
            "S13",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G 是三個任務的合取；H 由 S07、S08、S09、S13 正向證得，H⇒G 由合取引入，沒有把 G 當作前提。",
            "en": "Goal G is the conjunction of the three tasks; H is proved forwards by S07, S08, S09 and S13, H⇒G by conjunction introduction, and G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "ℤ₆ 的乘法對全部 36 個有序對都交換（6 個對角對由自反性，30 個非對角對由 15 個等式與公式證書），所以 ℤ₆ 是有限交換環；但 2⊗3=0 且 2 沒有乘法逆元，所以它不是整環也不是域。𝔽₂ 上的上三角矩陣環 U 中 I、E₁₁、E₂₂ 等樣本可交換，但 E₁₁E₁₂=E₁₂≠0=E₁₂E₁₁，所以 U 不是交換環。",
      "en": "Multiplication in ℤ₆ commutes for all 36 ordered pairs (the 6 diagonal pairs by reflexivity, the 30 off-diagonal pairs by 15 equations and the formula certificate), so ℤ₆ is a finite commutative ring; but 2⊗3=0 and 2 has no multiplicative inverse, so it is neither an integral domain nor a field. In the upper-triangular matrix ring U over 𝔽₂, samples such as I, E₁₁, E₂₂ commute, but E₁₁E₁₂=E₁₂≠0=E₁₂E₁₁, so U is not commutative."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "U 中 I 與所有元素交換，E₁₁ 與 E₂₂ 也交換，所以 U 是交換環",
          "en": "In U, I commutes with everything and E₁₁ commutes with E₂₂, so U is commutative"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "E₁₁E₁₂=E₁₂，E₁₂E₁₁=0，兩者不同。",
          "en": "E₁₁E₁₂=E₁₂ while E₁₂E₁₁=0, and these differ."
        }
      },
      {
        "claim": {
          "zh-Hant": "ℤ₆ 是交換環，所以沒有零因子",
          "en": "ℤ₆ is a commutative ring, so it has no zero divisors"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "2⊗3=0，而 2、3 都非零。",
          "en": "2⊗3=0 with both 2 and 3 nonzero."
        }
      },
      {
        "claim": {
          "zh-Hant": "ℤ₆ 是有限交換環，所以每個非零元都可逆",
          "en": "ℤ₆ is a finite commutative ring, so every nonzero element is invertible"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "第 2 行 (0,2,4,0,2,4) 中沒有 1，2 不可逆。",
          "en": "Row 2, (0,2,4,0,2,4), contains no 1, so 2 is not invertible."
        }
      },
      {
        "claim": {
          "zh-Hant": "若乘法表的某一行與對應的列相同，就足以判定環交換",
          "en": "If one row of the multiplication table equals the corresponding column, the ring is commutative"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "U 中單位元 I 的行與列相同，但 U 不交換；必須所有行都與對應的列相同。",
          "en": "In U the row and column of the identity I agree, yet U is not commutative; every row must equal its column."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "matrix_product_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "矩陣乘法：若 A 為 m×n、B 為 n×p，則 (AB)_{ij}=Σ_{k=1}^{n} A_{ik}B_{kj}。矩陣乘法一般不可交換。",
          "en": "Matrix product: for A of size m×n and B of size n×p, (AB)_{ij}=Σ_{k=1}^{n} A_{ik}B_{kj}. Matrix multiplication is in general not commutative."
        }
      },
      {
        "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": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "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": "cartesian_product_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "笛卡兒積：A×B 是所有有序對 (a,b)（a∈A、b∈B）的集合；(a,b)=(c,d) ⇔ a=c 且 b=d。",
          "en": "Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d."
        }
      },
      {
        "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": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "mul_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "id": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "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_comm_ring_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-finite-commutative-ring",
          "sha256": "5a488ffcb874034a7b2144c7585ccef8981c9d6bc0d2959107aa85650113234a",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在已通過完整有限含一環資格的同一加法與乘法表上，若每個 a,b 皆有 a 乘 b=b 乘 a，則為有限交換環。逐一核完整元素對，不由幾個樣本或表面對稱猜測。",
          "en": "Using the same complete addition and multiplication tables of an already qualified finite unital ring, require a*b=b*a for every a,b. All element pairs are checked rather than guessed from a few samples or apparent table symmetry."
        }
      },
      {
        "id": "mko_unital_ring_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-finite-unital-ring",
          "sha256": "e930e72ae14b44b56feb441e48526b614a1bfb2f281a1518ec6c7b19b1de48bf",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在同一有限載體上給完整加法表與乘法表：加法為阿貝爾群；乘法結合且對加法左右分配；存在唯一乘法單位元 1 且 1≠0。不要求乘法交換。",
          "en": "Complete addition and multiplication tables on one finite carrier: addition forms an abelian group; multiplication is associative and distributes over addition on both sides; there is a unique multiplicative identity 1 with 1≠0. Commutativity of multiplication is not required."
        }
      },
      {
        "id": "mod_reduce",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "對整數 m 與正整數 n，m mod n 是唯一滿足 0≤r<n 且 n 整除 m−r 的整數 r；相等的整數有相同的餘數。",
          "en": "For an integer m and a positive integer n, m mod n is the unique integer r with 0≤r<n such that n divides m−r; equal integers have equal residues."
        }
      },
      {
        "id": "domain_field_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零元 a、b 若 ab=0，稱為零因子。整環是 1≠0 且無零因子的交換環；域是每個非零元都有乘法逆元的交換環。",
          "en": "Nonzero elements a, b with ab=0 are zero divisors. An integral domain is a commutative ring with 1≠0 and no zero divisors; a field is a commutative ring in which every nonzero element has a multiplicative inverse."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "6*6",
        "rhs": "6+30"
      },
      {
        "id": "N2",
        "step": "S02",
        "lhs": "30/2",
        "rhs": "5+4+3+2+1"
      },
      {
        "id": "N3",
        "step": "S06",
        "vars": [
          "a",
          "b"
        ],
        "range": [
          0,
          5
        ],
        "lhs": "(a*b)%6",
        "rhs": "(b*a)%6"
      },
      {
        "id": "N4",
        "step": "S08",
        "lhs": "(2*3)%6",
        "rhs": "0"
      },
      {
        "id": "N5",
        "step": "S09",
        "vars": [
          "x"
        ],
        "range": [
          0,
          5
        ],
        "lhs": "(2*x)%6",
        "relation": "!=",
        "rhs": "1"
      },
      {
        "id": "N6",
        "step": "S11",
        "lhs": "(1*1+0*0)%2",
        "rhs": "1"
      },
      {
        "id": "N7",
        "step": "S12",
        "lhs": "(0*0+1*0)%2",
        "rhs": "0"
      },
      {
        "id": "N8",
        "step": "S14",
        "lhs": "(1*0+0*1)%2",
        "rhs": "(0*0+0*0)%2"
      },
      {
        "id": "N9",
        "step": "S06",
        "lhs": "(4*5)%6",
        "rhs": "(5*4)%6"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：使用已通過有限含一環資格的同一組加法與乘法表（假設 a1），本稿不重新核對環公理；交換律量化所有有序元素對，不額外假設無零因子或非零元可逆（假設 a2）。整環與域只用來說明交換環不必具備這些性質。不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: it uses the same addition and multiplication tables of rings already qualified as finite unital rings (assumption a1), and does not recheck the ring axioms; commutativity quantifies over every ordered element pair, with no extra premise of no zero divisors or invertibility of nonzero elements (assumption a2). Integral domains and fields appear only to show that a commutative ring need not have these properties. The canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "\\mathbb{Z}_6=\\{0,1,2,3,4,5\\}",
        "模 6 剩餘的載體",
        "the carrier of residues mod 6"
      ],
      [
        "a\\otimes b=(ab)\\bmod 6",
        "ℤ₆ 的乘法表",
        "the multiplication table of ℤ₆"
      ],
      [
        "U",
        "𝔽₂ 上 2×2 上三角矩陣環，元素記為 (a,b,c)",
        "the ring of 2-by-2 upper-triangular matrices over 𝔽₂, elements written (a,b,c)"
      ],
      [
        "(a,b,c)",
        "代表矩陣，第一行 (a,b)、第二行 (0,c)",
        "stands for the matrix with first row (a,b) and second row (0,c)"
      ],
      [
        "E_{11},E_{12},E_{22},I",
        "U 中的 (1,0,0)、(0,1,0)、(0,0,1)、(1,0,1)",
        "the elements (1,0,0), (0,1,0), (0,0,1), (1,0,1) of U"
      ]
    ],
    "md_error_controls": [
      {
        "title": "可交換的樣本不證明交換環 / Commuting samples do not prove commutativity",
        "zh": "MKO 指出的誤解是看到若干乘積可交換就宣稱整個環交換。U 中單位元、E₁₁ 與 E₂₂、每個元素與自身都可交換，但 E₁₁E₁₂≠E₁₂E₁₁。交換律必須量化所有有序對：ℤ₆ 用公式證書加 15 個等式涵蓋全部 36 對。",
        "en": "The misconception identified in the MKO is to declare a ring commutative after seeing several commuting products. In U the identity, E₁₁ with E₂₂, and every element with itself commute, but E₁₁E₁₂≠E₁₂E₁₁. Commutativity must quantify over every ordered pair: for ℤ₆ the formula certificate plus 15 equations cover all 36 pairs."
      },
      {
        "title": "交換環不等於整環或域 / A commutative ring is not automatically a domain or field",
        "zh": "MKO 的另一個誤解是把交換環直接當整環或域。ℤ₆ 交換，但 2⊗3=0（零因子），且 2 不可逆。交換律只說 ab=ba，不說 ab=0 推出 a=0 或 b=0，也不給逆元。",
        "en": "Another misconception identified in the MKO is to treat a commutative ring as an integral domain or field. ℤ₆ is commutative, but 2⊗3=0 (zero divisors) and 2 is not invertible. Commutativity only says ab=ba; it does not say that ab=0 forces a=0 or b=0, and it supplies no inverses."
      },
      {
        "title": "表面對稱要逐項核對 / Apparent symmetry must be checked entry by entry",
        "zh": "憑眼睛看表「大致對稱」不夠，抄錯一個數字就會改變結論。本稿用兩條獨立路線：S05 的公式證書，以及 S06 對列出的表逐組核對 15 個等式；N3 再在全部 36 對上重算公式。",
        "en": "Seeing that a table looks roughly symmetric is not enough; one miscopied entry would change the conclusion. This guide uses two independent routes: the formula certificate of S05 and the group-by-group check of the 15 equations on the listed table in S06; N3 recomputes the formula at all 36 pairs."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "ℤ₆ 與 U 的有限含一環資格取自父節點（G2、G3），本稿未重新核對環公理。",
        "en": "The finite-unital-ring qualifications of ℤ₆ and U are taken from the parent node (G2, G3); this guide does not recheck the ring axioms."
      },
      {
        "zh-Hant": "U 的完整 8×8 乘法表由 S10 的公式產生，本稿未逐格印出，只計算了用到的乘積。",
        "en": "The complete 8-by-8 multiplication table of U is generated by the formula of S10; the guide does not print every cell and computes only the products it uses."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "71ebaf592c6af1889ee73c2b6cda7d5ccfdbf9107995322905339d85c5f7b990",
    "original_source_sha256": "2aae99393024c2da54cb50cf7b82689e6617f3ce5fb9975a73db4ae5405b5182",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M077",
      "BR-M078"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}