{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-finite-integral-domain-01",
  "mko_id": "mko-finite-integral-domain",
  "title": {
    "zh-Hant": "有限整環：以完整乘法表判定 ℤ/5ℤ 與 ℤ/6ℤ，再由有限性推出可逆",
    "en": "Finite integral domain: deciding ℤ/5ℤ and ℤ/6ℤ from complete tables, then deriving invertibility from finiteness"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-finite-integral-domain-01",
    "source": {
      "mko_id": "mko-finite-integral-domain",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "5aefd54faece27112b7f0cc473cb7465bf2c77336f1d8ebfa180261027ba56fc",
      "english_sha256": "c032ef1004cdaaf34161bef5e966394605373aef8940ce31a696453c59a97a93"
    },
    "mode": "definition_walkthrough",
    "audience_prerequisites": [
      "divides_def",
      "division_algorithm",
      "distributive",
      "finite_counting_basics",
      "quantifier_negation",
      "injective_surjective_def",
      "case_analysis"
    ],
    "goal": {
      "zh-Hant": "以完整乘法表判定 ℤ/5ℤ 是有限整環、ℤ/6ℤ 不是；再把「每個非零元可逆」作為定義之後的定理，對任何有限整環用有限性證明，並在 ℤ/5ℤ 中求出全部逆元，同時說明 ℤ/6ℤ 中 2 為何沒有逆元。",
      "en": "Using complete multiplication tables, decide that ℤ/5ℤ is a finite integral domain and ℤ/6ℤ is not; then prove, as a theorem that comes after the definition, that every nonzero element of any finite integral domain is invertible, using finiteness; find all inverses in ℤ/5ℤ and explain why 2 has no inverse in ℤ/6ℤ."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "R5=ℤ/5ℤ={0,1,2,3,4}，加法與乘法是整數運算後取除以 5 的餘數；完整的加法表與乘法表由此確定，R5 已是有限交換含一環（MKO 假設 a1）。",
          "en": "R5=ℤ/5ℤ={0,1,2,3,4}, with addition and multiplication given by the integer operation followed by the remainder on division by 5; this fixes the complete addition and multiplication tables, and R5 already is a finite commutative unital ring (MKO assumption a1)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "R6=ℤ/6ℤ={0,1,2,3,4,5}，運算同上但取模 6；R6 也已是有限交換含一環。",
          "en": "R6=ℤ/6ℤ={0,1,2,3,4,5}, with the same operations taken modulo 6; R6 also is a finite commutative unital ring."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "定義 D(R)：有限交換含一環 R 是有限整環，若 1≠0，且對所有 a,b∈R，a≠0、b≠0 ⇒ ab≠0（量化全部非零元素對，MKO 假設 a2）。定義中沒有「非零元可逆」。",
          "en": "Definition D(R): a finite commutative unital ring R is a finite integral domain if 1≠0 and, for all a,b∈R, a≠0 and b≠0 ⇒ ab≠0 (quantified over every pair of nonzero elements, MKO assumption a2). The definition contains no clause about nonzero elements being invertible."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G3"
        ],
        "before_tex": "D(\\mathbb{Z}/5\\mathbb{Z})",
        "after_tex": "\\text{(i) ring (G1)}\\ \\land\\ \\text{(ii) }1\\neq 0\\ \\land\\ \\text{(iii) }\\forall a,b\\in R_5\\setminus\\{0\\}:\\ ab\\neq 0",
        "relation": "definition",
        "rule": "mko_fid_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "展開定義：要核三件事。(i) 父結構已是完整有限交換含一環（G1，MKO 假設 a1）；(ii) 1≠0；(iii) 每一對非零元素的積非零。定義中沒有「每個非零元可逆」，那是 S08 到 S10 才證明的定理。",
          "en": "Unfold the definition: three things must be checked. (i) The parent structure already is a complete finite commutative unital ring (G1, MKO assumption a1); (ii) 1≠0; (iii) the product of every pair of nonzero elements is nonzero. The definition has no clause 'every nonzero element is invertible'; that is a theorem proved only in S08 to S10."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G3",
            "mko_fid_def"
          ],
          "explanation": {
            "zh-Hant": "此步只展開定義，不推出新事實；三項之一不成立，D(R5) 就不成立。",
            "en": "This step only unfolds the definition and derives no new fact; if any one of the three fails, D(R5) fails."
          }
        },
        "leaf_basis": "def_unfold + mko_fid_def"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "mod_reduce"
        ],
        "before_tex": "1,0\\in\\mathbb{Z}/5\\mathbb{Z}",
        "after_tex": "1\\neq 0",
        "relation": "implies",
        "rule": "mod_reduce",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個整數在 ℤ/5ℤ 中代表同一元素當且僅當 5 整除其差。1−0=1，而 5 不整除 1（0<1<5，1 不是 5 的倍數）。所以 1≠0，條件 (ii) 成立；這是父含一環保留下來的條件（MKO 段落 0）。",
          "en": "Two integers stand for the same element of ℤ/5ℤ exactly when 5 divides their difference. 1−0=1, and 5 does not divide 1 (0<1<5, so 1 is not a multiple of 5). Hence 1≠0 and condition (ii) holds; it is inherited from the parent unital ring (MKO paragraph 0)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mod_reduce",
            "divides_def"
          ],
          "explanation": {
            "zh-Hant": "只需要 5∤1。在 ℤ/1ℤ 中 1−0=1 是 1 的倍數，1=0，條件 (ii) 會失敗。",
            "en": "Only 5∤1 is needed. In ℤ/1ℤ, 1−0=1 is a multiple of 1, so 1=0 and condition (ii) would fail."
          }
        },
        "leaf_basis": "mod_reduce + divides_def + division_algorithm"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S01"
        ],
        "before_tex": "\\forall (a,b)\\in\\{1,2,3,4\\}^2:\\ ab\\neq 0",
        "after_tex": "\\forall a\\le b\\ \\text{in}\\ \\{1,2,3,4\\}:\\ ab\\neq 0",
        "relation": "equivalent",
        "rule": "mul_comm",
        "conditions": [],
        "explanation": {
          "zh-Hant": "非零元素是 1、2、3、4，有序對共 16 個。⇒：右邊只是左邊的特例。⇐：對任一有序對 (a,b)，若 a≤b 已在右邊清單；若 a>b，則 ab=ba 且 (b,a) 滿足 b<a，也在清單中。所以只需算 4 個平方加 C(4,2)=6 個相異元素的乘積，共 10 個（N4）。",
          "en": "The nonzero elements are 1, 2, 3, 4, giving 16 ordered pairs. ⇒: the right side is a special case of the left. ⇐: for any ordered pair (a,b), if a≤b it is on the right-hand list; if a>b, then ab=ba and (b,a) has b<a, so it is on the list too. Hence only 4 squares plus C(4,2)=6 products of distinct elements, 10 products in all, need computing (N4)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_comm",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "兩個方向都有理由：特例化給 ⇒，交換律給 ⇐。化約只減少計算量，不減少被覆蓋的情況。",
            "en": "Both directions have reasons: specialization gives ⇒ and commutativity gives ⇐. The reduction lowers the amount of computation, not the set of cases covered."
          }
        },
        "leaf_basis": "mul_comm + forall_elim + finite_counting_basics"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "G1"
        ],
        "before_tex": "ab\\ (1\\le a\\le b\\le 4)\\ \\text{in}\\ \\mathbb{Z}/5\\mathbb{Z}",
        "after_tex": "1\\cdot b=b;\\ 2\\cdot2=4,\\ 2\\cdot3=1,\\ 2\\cdot4=3,\\ 3\\cdot3=4,\\ 3\\cdot4=2,\\ 4\\cdot4=1",
        "relation": "equivalent",
        "rule": "mod_reduce",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐一算出 S03 的 10 個乘積：含 1 的 4 個在 S04.1，其餘 6 個在 S04.2。這是乘法表中所需部分的完整列出。",
          "en": "Compute the 10 products of S03 one by one: the 4 involving 1 in S04.1 and the other 6 in S04.2. This lists the required part of the multiplication table completely."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mod_reduce",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "每個結果都可反向核對：整數積減去所寫的餘數是 5 的倍數；兩邊表示同一元素。",
            "en": "Each result can be checked backwards: the integer product minus the stated remainder is a multiple of 5, so both sides name the same element."
          }
        },
        "leaf_basis": "mod_reduce + numeric_evaluation"
      },
      {
        "id": "S04.1",
        "uses": [
          "G1",
          "mul_one"
        ],
        "before_tex": "1\\cdot b,\\ b\\in\\{1,2,3,4\\}",
        "after_tex": "1\\cdot b=b\\neq 0",
        "relation": "equivalent",
        "rule": "mul_one",
        "conditions": [],
        "explanation": {
          "zh-Hant": "1 是乘法單位元，1·b=b；b 非零，所以 1·1、1·2、1·3、1·4 都非零。",
          "en": "1 is the multiplicative identity, so 1·b=b; b is nonzero, so 1·1, 1·2, 1·3 and 1·4 are all nonzero."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_one"
          ],
          "explanation": {
            "zh-Hant": "1·b=b 是恆等式，兩邊互換即反向。",
            "en": "1·b=b is an identity; reading it right to left is the backward direction."
          }
        },
        "leaf_basis": "mul_one"
      },
      {
        "id": "S04.2",
        "uses": [
          "G1",
          "mod_reduce"
        ],
        "before_tex": "2\\cdot2,\\ 2\\cdot3,\\ 2\\cdot4,\\ 3\\cdot3,\\ 3\\cdot4,\\ 4\\cdot4",
        "after_tex": "4,\\ 1,\\ 3,\\ 4,\\ 2,\\ 1",
        "relation": "equivalent",
        "rule": "mod_reduce",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先在 ℤ 中相乘得 4、6、8、9、12、16，再取除以 5 的餘數：4=0·5+4、6=1·5+1、8=1·5+3、9=1·5+4、12=2·5+2、16=3·5+1。所以六個積依序為 4、1、3、4、2、1（N1、N2）。",
          "en": "Multiply in ℤ to get 4, 6, 8, 9, 12, 16, then take remainders on division by 5: 4=0·5+4, 6=1·5+1, 8=1·5+3, 9=1·5+4, 12=2·5+2, 16=3·5+1. So the six products are 4, 1, 3, 4, 2, 1 in order (N1, N2)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mod_reduce",
            "division_algorithm"
          ],
          "explanation": {
            "zh-Hant": "反向核對：4−4、6−1、8−3、9−4、12−2、16−1 依序為 0、5、5、5、10、15，都是 5 的倍數；餘數由帶餘除法唯一決定。",
            "en": "Backward check: 4−4, 6−1, 8−3, 9−4, 12−2, 16−1 are 0, 5, 5, 5, 10, 15, all multiples of 5; the remainder is unique by the division algorithm."
          }
        },
        "leaf_basis": "mod_reduce + division_algorithm + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "S03",
          "S04",
          "S04.1",
          "S04.2"
        ],
        "before_tex": "\\text{the 10 products lie in }\\{1,2,3,4\\}",
        "after_tex": "\\forall a,b\\in R_5\\setminus\\{0\\}:\\ ab\\neq 0",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "10 個乘積都在 {1,2,3,4} 中，沒有一個是 0。由 S03，這 10 種情況涵蓋全部 16 個非零有序對，所以全稱的條件 (iii) 成立。這是對整個有限集合的窮舉，每一對都被核到，不是抽樣；N3 另把 16 個有序對逐一重算，它恰好涵蓋此有限集合的全部情況。",
          "en": "All 10 products lie in {1,2,3,4}; none is 0. By S03 these 10 cases cover all 16 nonzero ordered pairs, so the universal condition (iii) holds. This is an exhaustion of the whole finite set in which every pair is checked, not a sample; N3 recomputes the 16 ordered pairs one by one, and it covers exactly all cases of this finite set."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S04",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "需要情況完整（S03 的化約與 S04 的完整列表）；只核其中幾個乘積不能推出全稱命題。",
            "en": "It needs the cases to be complete (the reduction of S03 and the full list of S04); checking only some of the products cannot establish the universal statement."
          }
        },
        "leaf_basis": "case_analysis + forall_intro"
      },
      {
        "id": "S06",
        "uses": [
          "S01",
          "S02",
          "S05",
          "G1"
        ],
        "before_tex": "\\text{(i)},\\ \\text{(ii)},\\ \\text{(iii) for }R_5",
        "after_tex": "D(\\mathbb{Z}/5\\mathbb{Z})",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(i) 由 G1，(ii) 由 S02，(iii) 由 S05。三者合取，依 S01 的展開，ℤ/5ℤ 是有限整環。",
          "en": "(i) comes from G1, (ii) from S02 and (iii) from S05. Their conjunction gives, by the unfolding in S01, that ℤ/5ℤ is a finite integral domain."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S02",
            "S05",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "三項都需要；此步沒有用到可逆性。",
            "en": "All three items are needed; invertibility is not used in this step."
          }
        },
        "leaf_basis": "and_intro_elim + def_unfold"
      },
      {
        "id": "S07",
        "uses": [
          "G2",
          "G3",
          "mod_reduce"
        ],
        "before_tex": "2,3\\in R_6\\setminus\\{0\\}",
        "after_tex": "2\\cdot 3=0\\ \\text{in}\\ \\mathbb{Z}/6\\mathbb{Z},\\ \\text{hence}\\ \\neg D(\\mathbb{Z}/6\\mathbb{Z})",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 ℤ/6ℤ 中 2≠0、3≠0（6 不整除 2 或 3），但 2·3=6=1·6+0，所以 2·3=0（N5）。這一對使 (iii) 的否定「存在非零 a、b 使 ab=0」成立；依量詞否定，全稱的 (iii) 不成立，所以 ℤ/6ℤ 不是有限整環，雖然它的 1≠0 仍成立。另有 3·4=12≡0（N6）。否定一個全稱命題，一個反例就夠。",
          "en": "In ℤ/6ℤ we have 2≠0 and 3≠0 (6 divides neither 2 nor 3), but 2·3=6=1·6+0, so 2·3=0 (N5). This pair makes the negation of (iii), 'some nonzero a, b have ab=0', true; by quantifier negation the universal (iii) fails, so ℤ/6ℤ is not a finite integral domain, even though 1≠0 still holds there. Also 3·4=12≡0 (N6). One counterexample suffices to refute a universal statement."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "mod_reduce",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "只需一個零因子對；本步不宣稱找出了全部零因子。",
            "en": "One zero-divisor pair is enough; this step does not claim to list all zero divisors."
          }
        },
        "leaf_basis": "quantifier_negation + mod_reduce + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "G3"
        ],
        "before_tex": "D(R),\\ a\\in R\\setminus\\{0\\},\\ \\lambda_a(x)=ax",
        "after_tex": "\\lambda_a:R\\to R\\ \\text{injective}",
        "relation": "implies",
        "rule": "L_lambda_injective",
        "conditions": [
          "a\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "以下對任何有限整環 R 與任何 a≠0 工作，不再只看 ℤ/5ℤ。定義 λ_a:R→R，x↦ax。依引理 L_lambda_injective（在 S08.1、S08.2 證明），λ_a 是單射。這一步只用到無零因子，尚未用到有限性。",
          "en": "From here on work with any finite integral domain R and any a≠0, not just ℤ/5ℤ. Define λ_a:R→R by x↦ax. By Lemma L_lambda_injective (proved in S08.1 and S08.2), λ_a is injective. This step uses only the absence of zero divisors, not yet finiteness."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "L_lambda_injective"
          ],
          "explanation": {
            "zh-Hant": "需要 a≠0 與定義的 (iii)；在 ℤ/6ℤ 中 a=2 時結論失敗（S12）。",
            "en": "It needs a≠0 and clause (iii) of the definition; in ℤ/6ℤ with a=2 the conclusion fails (S12)."
          }
        },
        "leaf_basis": "distributive + contrapositive + injective_surjective_def"
      },
      {
        "id": "S08.1",
        "uses": [
          "ring_axioms"
        ],
        "before_tex": "ax=ay",
        "after_tex": "a(x-y)=0",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊同減 ay 得 ax−ay=0；由分配律 ax−ay=ax+a(−y)=a(x−y)。反向：由 a(x−y)=0 得 ax−ay=0，兩邊同加 ay 得 ax=ay。",
          "en": "Subtract ay from both sides: ax−ay=0; by distributivity ax−ay=ax+a(−y)=a(x−y). Backward: a(x−y)=0 gives ax−ay=0, and adding ay to both sides gives ax=ay."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "eq_congruence",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "加減 ay 互為反運算，分配律是恆等式，所以雙向成立。",
            "en": "Adding and subtracting ay undo each other and distributivity is an identity, so the step holds both ways."
          }
        },
        "leaf_basis": "distributive + eq_congruence + add_inverse + neg_mul"
      },
      {
        "id": "S08.2",
        "uses": [
          "S08.1",
          "G3"
        ],
        "before_tex": "a(x-y)=0,\\ a\\neq 0",
        "after_tex": "x=y",
        "relation": "implies",
        "rule": "contrapositive",
        "conditions": [
          "a\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "若 x−y≠0，則 a 與 x−y 都非零，依定義 (iii) a(x−y)≠0，與 S08.1 矛盾。所以 x−y=0，兩邊同加 y 得 x=y。S08.1 與 S08.2 合起來證得引理 L_lambda_injective。",
          "en": "If x−y≠0, then a and x−y are both nonzero, and clause (iii) gives a(x−y)≠0, contradicting S08.1. So x−y=0, and adding y to both sides gives x=y. S08.1 and S08.2 together prove Lemma L_lambda_injective."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08.1",
            "G3",
            "contrapositive"
          ],
          "explanation": {
            "zh-Hant": "這裡用的是 (iii) 的逆否形式；沒有 (iii) 時，ℤ/6ℤ 的 2(3−0)=0 而 3−0≠0 正是反例。",
            "en": "This uses the contrapositive form of (iii); without (iii), 2(3−0)=0 with 3−0≠0 in ℤ/6ℤ is exactly a counterexample."
          }
        },
        "leaf_basis": "contrapositive + add_inverse + eq_congruence"
      },
      {
        "id": "S09",
        "uses": [
          "S08"
        ],
        "before_tex": "\\lambda_a:R\\to R\\ \\text{injective},\\ R\\ \\text{finite}",
        "after_tex": "\\lambda_a(R)=R",
        "relation": "implies",
        "rule": "L_finite_inj_surj",
        "conditions": [
          "R\\ \\text{finite}"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_finite_inj_surj（在 S09.1、S09.2 證明），有限集上的單射自映射是滿射，所以 λ_a 的像是整個 R。這是整個論證中第一次、也是唯一一次用到 R 有限。",
          "en": "By Lemma L_finite_inj_surj (proved in S09.1 and S09.2), an injective self-map of a finite set is surjective, so the image of λ_a is all of R. This is the first and only place where the finiteness of R is used."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "L_finite_inj_surj"
          ],
          "explanation": {
            "zh-Hant": "需要單射與有限兩個前提；對無限集不成立：ℤ 中 x↦2x 是單射但不是滿射。",
            "en": "It needs both injectivity and finiteness; for infinite sets it fails: x↦2x on ℤ is injective but not surjective."
          }
        },
        "leaf_basis": "finite_counting_basics + injective_surjective_def"
      },
      {
        "id": "S09.1",
        "uses": [
          "S08",
          "finite_counting_basics",
          "injective_surjective_def"
        ],
        "before_tex": "\\lambda_a\\ \\text{injective}",
        "after_tex": "|\\lambda_a(R)|=|R|",
        "relation": "implies",
        "rule": "finite_counting_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 λ_a 看成 R→λ_a(R) 的函數：依像的定義它是滿射，又是單射，所以是雙射。有限集之間有雙射當且僅當元素個數相同，故 |λ_a(R)|=|R|。論證對任何有限集上的單射自映射都相同。",
          "en": "Regard λ_a as a map R→λ_a(R): by the definition of image it is surjective, and it is injective, so it is a bijection. Finite sets admit a bijection exactly when they have the same number of elements, so |λ_a(R)|=|R|. The argument is the same for any injective self-map of a finite set."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "finite_counting_basics",
            "image_preimage_def"
          ],
          "explanation": {
            "zh-Hant": "需要單射；非單射的映射（如 ℤ/6ℤ 的 λ_2）其像較小。",
            "en": "It needs injectivity; a non-injective map (such as λ_2 on ℤ/6ℤ) has a smaller image."
          }
        },
        "leaf_basis": "finite_counting_basics + injective_surjective_def + image_preimage_def"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "subset_def"
        ],
        "before_tex": "\\lambda_a(R)\\subseteq R,\\ |\\lambda_a(R)|=|R|",
        "after_tex": "\\lambda_a(R)=R",
        "relation": "implies",
        "rule": "finite_counting_basics",
        "conditions": [
          "R\\ \\text{finite}"
        ],
        "explanation": {
          "zh-Hant": "若 λ_a(R)≠R，則 R∖λ_a(R) 非空，R 是不交有限集 λ_a(R) 與 R∖λ_a(R) 的聯集，|R|=|λ_a(R)|+|R∖λ_a(R)|≥|λ_a(R)|+1，與 S09.1 矛盾。所以 λ_a(R)=R。S09.1 與 S09.2 合起來證得引理 L_finite_inj_surj。",
          "en": "If λ_a(R)≠R, then R∖λ_a(R) is nonempty and R is the union of the disjoint finite sets λ_a(R) and R∖λ_a(R), so |R|=|λ_a(R)|+|R∖λ_a(R)|≥|λ_a(R)|+1, contradicting S09.1. Hence λ_a(R)=R. S09.1 and S09.2 together prove Lemma L_finite_inj_surj."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09.1",
            "finite_counting_basics",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "加法公式 |A∪B|=|A|+|B| 只對有限集給出矛盾；無限集的真子集可以與全集等大。",
            "en": "The count |A∪B|=|A|+|B| gives a contradiction only for finite sets; an infinite set can be as large as a proper subset."
          }
        },
        "leaf_basis": "finite_counting_basics + set_operations_def + contradiction"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "ring_axioms"
        ],
        "before_tex": "1\\in R=\\lambda_a(R)",
        "after_tex": "\\exists b\\in R:\\ ab=ba=1",
        "relation": "implies",
        "rule": "image_preimage_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "1∈R=λ_a(R)，依像的定義存在 b∈R 使 λ_a(b)=ab=1；由交換律 ba=1。所以 a 可逆。逆元唯一：若 ab=ac=1，則 b=b(ac)=(ba)c=c。這是定義之後、用到有限性的定理（MKO 段落 2）。",
          "en": "1∈R=λ_a(R), so by the definition of image there is b∈R with λ_a(b)=ab=1, and commutativity gives ba=1. Hence a is invertible. The inverse is unique: if ab=ac=1 then b=b(ac)=(ba)c=c. This is a theorem after the definition, and it uses finiteness (MKO paragraph 2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "image_preimage_def",
            "ring_axioms"
          ],
          "explanation": {
            "zh-Hant": "需要 S09 的滿射；不宣稱反向，即可逆性不能反過來當作 D(R) 的定義條款。",
            "en": "It needs the surjectivity of S09; no converse is claimed, and invertibility is not turned back into a clause of D(R)."
          }
        },
        "leaf_basis": "image_preimage_def + exists_elim + mul_comm + mul_assoc"
      },
      {
        "id": "S11",
        "uses": [
          "S06",
          "S10",
          "S04.2"
        ],
        "before_tex": "\\lambda_2\\ \\text{on}\\ \\mathbb{Z}/5\\mathbb{Z}",
        "after_tex": "(0,1,2,3,4)\\mapsto(0,2,4,1,3);\\ 2^{-1}=3,\\ 3^{-1}=2,\\ 4^{-1}=4,\\ 1^{-1}=1",
        "relation": "implies",
        "rule": "mod_reduce",
        "conditions": [],
        "explanation": {
          "zh-Hant": "λ_2 的值：2·0=0、2·1=2、2·2=4、2·3=6≡1、2·4=8≡3（N7）。五個值互異，是 R5 的一個排列，正如 S08、S09 所預測。1=λ_2(3)，所以 2^{-1}=3。由 S04.2：3·2=1、4·4=1；又 1·1=1。故 R5 的每個非零元都可逆，這裡逐一算出，與 S10 一致。",
          "en": "Values of λ_2: 2·0=0, 2·1=2, 2·2=4, 2·3=6≡1, 2·4=8≡3 (N7). The five values are distinct, a permutation of R5, as S08 and S09 predict. 1=λ_2(3), so 2^{-1}=3. From S04.2: 3·2=1 and 4·4=1; also 1·1=1. So every nonzero element of R5 is invertible, computed here one by one, in agreement with S10."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S10",
            "S04.2",
            "mod_reduce"
          ],
          "explanation": {
            "zh-Hant": "每個逆元都可直接乘回核對；這些計算是 S10 在一個例子上的實例，不取代 S10 的一般證明。",
            "en": "Each inverse can be checked by multiplying back; these computations are an instance of S10 in one example and do not replace its general proof."
          }
        },
        "leaf_basis": "mod_reduce + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S07",
          "G2"
        ],
        "before_tex": "\\lambda_2\\ \\text{on}\\ \\mathbb{Z}/6\\mathbb{Z}",
        "after_tex": "(0,1,2,3,4,5)\\mapsto(0,2,4,0,2,4);\\ 1\\notin\\lambda_2(R_6)",
        "relation": "implies",
        "rule": "mod_reduce",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 ℤ/6ℤ 中 2·3=6≡0=2·0、2·4=8≡2=2·1、2·5=10≡4=2·2（N8）。λ_2 不是單射，像只有 {0,2,4}，不含 1，所以 2 在 ℤ/6ℤ 中沒有逆元。S08 的論證在 S08.2 斷掉：2(3−0)=0 而 3−0≠0，正是 S07 的零因子。",
          "en": "In ℤ/6ℤ, 2·3=6≡0=2·0, 2·4=8≡2=2·1 and 2·5=10≡4=2·2 (N8). λ_2 is not injective, and its image is only {0,2,4}, which does not contain 1, so 2 has no inverse in ℤ/6ℤ. The argument of S08 breaks at S08.2: 2(3−0)=0 while 3−0≠0, which is exactly the zero divisor of S07."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "mod_reduce"
          ],
          "explanation": {
            "zh-Hant": "六個值逐一可重算；1 不在像中這一點對 6 個輸入是完整核對。",
            "en": "The six values can be recomputed one by one; that 1 is not in the image is a complete check over all 6 inputs."
          }
        },
        "leaf_basis": "mod_reduce + numeric_evaluation + injective_surjective_def"
      },
      {
        "id": "S13",
        "uses": [
          "divides_def"
        ],
        "before_tex": "2b=1,\\ b\\in\\mathbb{Z}",
        "after_tex": "\\text{no solution: }2\\ \\text{is not invertible in}\\ \\mathbb{Z}",
        "relation": "implies",
        "rule": "divides_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "ℤ 是無零因子的交換含一環（零乘積性質），但不是有限的。若 2b=1 有整數解，則 2∣1；但 1 不是 2 的倍數（0<1<2）。所以 2 在 ℤ 中不可逆。這說明 S09 的有限性不可省略：無零因子本身推不出可逆，所以可逆性不能偷放入定義。",
          "en": "ℤ is a commutative unital ring without zero divisors (zero-product property), but it is not finite. If 2b=1 had an integer solution, then 2∣1; but 1 is not a multiple of 2 (0<1<2). So 2 is not invertible in ℤ. This shows that the finiteness used in S09 cannot be dropped: the absence of zero divisors alone does not give invertibility, so invertibility must not be slipped into the definition."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "divides_def",
            "zero_product"
          ],
          "explanation": {
            "zh-Hant": "ℤ 不屬於 D 的論域（D 只談有限環），這裡只作為對照，說明有限性的作用。",
            "en": "ℤ is outside the domain of D (which concerns finite rings); it is used here only as a contrast showing what finiteness does."
          }
        },
        "leaf_basis": "divides_def + division_algorithm + zero_product"
      },
      {
        "id": "S14",
        "uses": [
          "S06",
          "S07",
          "S10",
          "S11",
          "S12"
        ],
        "before_tex": "D(\\mathbb{Z}/5\\mathbb{Z}),\\ \\neg D(\\mathbb{Z}/6\\mathbb{Z}),\\ \\text{S08 to S10}",
        "after_tex": "\\mathbb{Z}/5\\mathbb{Z}\\ \\text{finite integral domain};\\ \\mathbb{Z}/6\\mathbb{Z}\\ \\text{not};\\ D(R)\\Rightarrow\\text{every }a\\neq0\\ \\text{invertible}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "完整表的窮舉核對給出 D(ℤ/5ℤ)；一個零因子對給出 ¬D(ℤ/6ℤ)；對任何有限整環，S08 到 S10 證明每個非零元可逆，S11 在 ℤ/5ℤ 中逐一算出逆元。",
          "en": "Exhaustive checking of the complete table gives D(ℤ/5ℤ); one zero-divisor pair gives ¬D(ℤ/6ℤ); for any finite integral domain S08 to S10 prove that every nonzero element is invertible, and S11 computes the inverses in ℤ/5ℤ one by one."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S06",
            "S07",
            "S10",
            "S11"
          ],
          "explanation": {
            "zh-Hant": "目標 G：判定兩個環並證明可逆性定理。足夠條件 H：S06、S07、S10、S11；H⇒G 由合取。H 由 G1 到 G3 正向證成，沒有把 G 當作前提，也沒有把可逆性當作定義的一部分。",
            "en": "Goal G: decide the two rings and prove the invertibility theorem. Sufficient condition H: S06, S07, S10 and S11; H⇒G by conjunction. H is proved forwards from G1 to G3; G is never used as a premise, and invertibility is never made part of the definition."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "ℤ/5ℤ 是有限整環：1≠0，且完整窮舉 16 個非零有序對（化為 10 個乘積）沒有一個積為 0。ℤ/6ℤ 不是：2·3=0。對任何有限整環，乘以非零 a 的映射是單射，有限性使它成為滿射，所以 a 可逆；在 ℤ/5ℤ 中 2^{-1}=3、3^{-1}=2、4^{-1}=4、1^{-1}=1。可逆性是定義之後的定理，ℤ 顯示沒有有限性時它可以失敗。",
      "en": "ℤ/5ℤ is a finite integral domain: 1≠0, and exhausting all 16 nonzero ordered pairs (reduced to 10 products) gives no zero product. ℤ/6ℤ is not: 2·3=0. In any finite integral domain multiplication by a nonzero a is injective, finiteness makes it surjective, and so a is invertible; in ℤ/5ℤ, 2^{-1}=3, 3^{-1}=2, 4^{-1}=4 and 1^{-1}=1. Invertibility is a theorem after the definition, and ℤ shows it can fail without finiteness."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "在 ℤ/6ℤ 中抽查 1·5、5·5、2·2、3·3 都非零，所以 ℤ/6ℤ 是整環",
          "en": "In ℤ/6ℤ the spot checks 1·5, 5·5, 2·2 and 3·3 are nonzero, so ℤ/6ℤ is an integral domain"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "2·3=6≡0，2、3 都非零；抽樣不能證明全稱命題。",
          "en": "2·3=6≡0 with 2 and 3 nonzero; sampling cannot prove a universal statement."
        }
      },
      {
        "claim": {
          "zh-Hant": "依定義，整環的每個非零元都可逆",
          "en": "By definition, every nonzero element of an integral domain is invertible"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "定義只要求無零因子；ℤ 中 2 不可逆。有限整環的可逆性是用有限性證出的定理。",
          "en": "The definition only requires no zero divisors; 2 is not invertible in ℤ. Invertibility in a finite integral domain is a theorem proved using finiteness."
        }
      },
      {
        "claim": {
          "zh-Hant": "乘法表中 0 的那一行全是 0，所以每個環都有零因子",
          "en": "The row of 0 in the multiplication table is all zeros, so every ring has zero divisors"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "零因子要求兩個因子都非零；a·0=0 不算。ℤ/5ℤ 沒有零因子（S05）。",
          "en": "Zero divisors need both factors nonzero; a·0=0 does not count. ℤ/5ℤ has none (S05)."
        }
      },
      {
        "claim": {
          "zh-Hant": "零環 {0} 沒有非零元素對，所以是有限整環",
          "en": "The zero ring {0} has no pairs of nonzero elements, so it is a finite integral domain"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "定義要求 1≠0；零環中 1=0。",
          "en": "The definition requires 1≠0; in the zero ring 1=0."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "divides_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整除：對整數 a、b，a∣b 表示存在整數 k 使 b=ak。",
          "en": "Divisibility: for integers a, b, a∣b means there is an integer k with b=ak."
        }
      },
      {
        "id": "division_algorithm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "帶餘除法：對整數 a 與正整數 b，存在唯一的整數 q、r 使 a=bq+r 且 0≤r<b。",
          "en": "Division algorithm: for an integer a and a positive integer b there are unique integers q, r with a=bq+r and 0≤r<b."
        }
      },
      {
        "id": "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": "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": "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": "injective_surjective_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "單射：f(a)=f(a′) ⇒ a=a′。滿射：每個 b∈B 都有 a∈A 使 f(a)=b。雙射：既單又滿，此時有唯一反函數 f⁻¹ 使 f⁻¹∘f=id_A、f∘f⁻¹=id_B。",
          "en": "Injective: f(a)=f(a′) ⇒ a=a′. Surjective: every b∈B has some a∈A with f(a)=b. Bijective: both; then there is a unique inverse f⁻¹ with f⁻¹∘f=id_A and f∘f⁻¹=id_B."
        }
      },
      {
        "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": "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": "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": "forall_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。",
          "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds."
        }
      },
      {
        "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": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "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": "contrapositive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逆否命題：P⇒Q 與 ¬Q⇒¬P 等價。逆命題 Q⇒P 一般不與 P⇒Q 等價。",
          "en": "Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q."
        }
      },
      {
        "id": "eq_congruence",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。",
          "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule."
        }
      },
      {
        "id": "add_inverse",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。",
          "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)."
        }
      },
      {
        "id": "neg_mul",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。",
          "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv."
        }
      },
      {
        "id": "image_preimage_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "像與原像：f(S)={f(a) ∣ a∈S}；f⁻¹(T)={a∈A ∣ f(a)∈T}。原像記號不預設 f 有反函數。",
          "en": "Image and preimage: f(S)={f(a) ∣ a∈S}; f⁻¹(T)={a∈A ∣ f(a)∈T}. The preimage notation does not presuppose that f has an inverse."
        }
      },
      {
        "id": "subset_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "子集：A⊆B 表示 A 的每個元素都是 B 的元素，即 ∀x (x∈A⇒x∈B)。",
          "en": "Subset: A⊆B means every element of A is an element of B, i.e. ∀x (x∈A⇒x∈B)."
        }
      },
      {
        "id": "set_operations_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合運算：x∈A∪B ⇔ x∈A 或 x∈B；x∈A∩B ⇔ x∈A 且 x∈B；x∈A∖B ⇔ x∈A 且 x∉B；在全集 U 中 Aᶜ=U∖A。",
          "en": "Set operations: x∈A∪B ⇔ x∈A or x∈B; x∈A∩B ⇔ x∈A and x∈B; x∈A∖B ⇔ x∈A and x∉B; in a universe U, Aᶜ=U∖A."
        }
      },
      {
        "id": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "exists_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃消去：由 ∃x∈D, P(x)，可取一個新名字 a 表示某個滿足 P(a) 的元素；只要最後結論不含 a，就可用 a 推理。",
          "en": "∃-elimination: from ∃x∈D, P(x) we may name some element a with P(a) by a fresh name; reasoning with a is allowed as long as the final conclusion does not mention a."
        }
      },
      {
        "id": "mul_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法結合律：(uv)w=u(vw)。",
          "en": "Associativity of multiplication: (uv)w=u(vw)."
        }
      },
      {
        "id": "zero_product",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "零乘積性質：在 ℤ、ℚ、ℝ、ℂ（或任何整域）中，uv=0 ⇔ u=0 或 v=0。",
          "en": "Zero-product property: in ℤ, ℚ, ℝ, ℂ (or any integral domain), uv=0 ⇔ u=0 or v=0."
        }
      },
      {
        "id": "mko_fid_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-finite-integral-domain",
          "sha256": "5aefd54faece27112b7f0cc473cb7465bf2c77336f1d8ebfa180261027ba56fc",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在同一完整有限交換含一環表上，若對所有非零 a,b 皆有 a 乘 b 不等於 0，則為有限整環。完整表可逐對核零因子；每個非零元可逆是利用有限性的後續定理步驟，不偷放入定義。",
          "en": "Using the same complete finite commutative unital-ring tables, require a*b to differ from zero for every pair of nonzero elements. The complete table supports an exhaustive zero-divisor audit. Invertibility of every nonzero element is a subsequent theorem using finiteness, not silently inserted into this definition."
        }
      },
      {
        "id": "mod_reduce",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "在 ℤ/nℤ 中，整數 m 代表它除以 n 的餘數 r（0≤r<n）；兩個整數代表同一元素當且僅當 n 整除兩者之差；兩元素的積是整數積的餘數。",
          "en": "In ℤ/nℤ an integer m stands for its remainder r on division by n (0≤r<n); two integers stand for the same element exactly when n divides their difference; the product of two elements is the remainder of the integer product."
        }
      },
      {
        "id": "ring_axioms",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "交換含一環：加法成交換群，乘法結合、交換、有單位元 1，且分配律成立；由此 a·0=0 且 a(x−y)=ax−ay。",
          "en": "Commutative unital ring: addition forms a commutative group, multiplication is associative and commutative with identity 1, and distributivity holds; hence a·0=0 and a(x−y)=ax−ay."
        }
      },
      {
        "id": "L_lambda_injective",
        "kind": "lemma",
        "proved_in": [
          "S08.1",
          "S08.2"
        ],
        "statement": {
          "zh-Hant": "在整環 R 中，若 a≠0，則 λ_a(x)=ax 是單射：ax=ay ⇒ x=y。",
          "en": "In an integral domain R, if a≠0 then λ_a(x)=ax is injective: ax=ay ⇒ x=y."
        }
      },
      {
        "id": "L_finite_inj_surj",
        "kind": "lemma",
        "proved_in": [
          "S09.1",
          "S09.2"
        ],
        "statement": {
          "zh-Hant": "有限集 S 上的單射 f:S→S 必為滿射。",
          "en": "An injective map f:S→S on a finite set S is surjective."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S04.2",
        "lhs": "(2*3)%5",
        "rhs": "1"
      },
      {
        "id": "N2",
        "step": "S04.2",
        "lhs": "(4*4)%5",
        "rhs": "1"
      },
      {
        "id": "N3",
        "step": "S05",
        "vars": [
          "a",
          "b"
        ],
        "range": [
          1,
          4
        ],
        "lhs": "(a*b)%5",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N4",
        "step": "S03",
        "lhs": "4+comb(4,2)",
        "rhs": "10"
      },
      {
        "id": "N5",
        "step": "S07",
        "lhs": "(2*3)%6",
        "rhs": "0"
      },
      {
        "id": "N6",
        "step": "S07",
        "lhs": "(3*4)%6",
        "rhs": "0"
      },
      {
        "id": "N7",
        "step": "S11",
        "lhs": "(2*4)%5",
        "rhs": "3"
      },
      {
        "id": "N8",
        "step": "S12",
        "lhs": "(2*3)%6",
        "rhs": "(2*0)%6"
      },
      {
        "id": "N9",
        "lhs": "5+comb(5,2)",
        "rhs": "15"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只在 MKO 範圍內工作：父結構已是完整有限交換含一環（ℤ/5ℤ 與 ℤ/6ℤ，加法與乘法表取模給出，假設 a1）；無零因子的條件對全部非零元素對量化（假設 a2）。可逆性定理對一般有限整環證明。不證明「ℤ/nℤ 是整環當且僅當 n 為質數」的一般判準，不建立有限域節點；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: the parent structure already is a complete finite commutative unital ring (ℤ/5ℤ and ℤ/6ℤ, with tables given by reduction modulo n, assumption a1), and the no-zero-divisor condition is quantified over every pair of nonzero elements (assumption a2). The invertibility theorem is proved for a general finite integral domain. It does not prove the general criterion that ℤ/nℤ is a domain exactly when n is prime, and does not set up the finite-field node; it does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "\\mathbb{Z}/n\\mathbb{Z}",
        "模 n 的剩餘類環，元素記為 0 到 n−1",
        "integers modulo n, elements written 0 to n−1"
      ],
      [
        "D(R)",
        "R 是有限整環",
        "R is a finite integral domain"
      ],
      [
        "\\lambda_a",
        "乘以 a 的映射 x↦ax",
        "the multiplication map x↦ax"
      ],
      [
        "a^{-1}",
        "a 的乘法逆元",
        "the multiplicative inverse of a"
      ],
      [
        "R\\setminus\\{0\\}",
        "R 的非零元素",
        "the nonzero elements of R"
      ]
    ],
    "md_error_controls": [
      {
        "title": "抽查乘積不是證書 / Sampled products are not a certificate",
        "zh": "MKO 指出的誤解：抽查後宣稱無零因子。在 ℤ/6ℤ 中 1·5=5、5·5=25≡1、2·2=4、3·3=9≡3、4·4=16≡4、2·5=10≡4 都非零，但 2·3≡0、3·4≡0。本稿選用對全部非零元素對完整窮舉的證書（ℤ/6ℤ 有 5+C(5,2)=15 個無序對，N9），ℤ/5ℤ 的 S03 到 S05 就採此法；但它不是唯一可能的證法。例如可先直接核出每個非零元的逆元，再把 ab=0 乘以 a 的逆元推出 b=0。這個替代證書的逆元必須獨立核得，不能循環引用本稿稍後先假設整環才證得的可逆性定理。",
        "en": "The misconception identified in the MKO: declaring no zero divisors after spot checks. In ℤ/6ℤ, 1·5=5, 5·5=25≡1, 2·2=4, 3·3=9≡3, 4·4=16≡4 and 2·5=10≡4 are all nonzero, yet 2·3≡0 and 3·4≡0. This guide selects exhaustive checking of all nonzero pairs as its certificate (ℤ/6ℤ has 5+C(5,2)=15 unordered pairs, N9), as in S03 to S05 for ℤ/5ℤ. It is not the only possible proof: independently verify an inverse for every nonzero a, then multiply ab=0 by that inverse to get b=0. Those inverse witnesses must be checked independently, not obtained circularly from the later theorem that first assumes an integral domain."
      },
      {
        "title": "可逆不放入定義 / Invertibility is not part of the definition",
        "zh": "MKO 的第二個誤解是把「非零元可逆」直接塞進定義。本稿把它放在 S08 到 S10 作為定理，並指出唯一用到有限性的是 S09。ℤ 無零因子但 2 不可逆（S13），所以沒有有限性這個定理就不成立。",
        "en": "The second misconception identified in the MKO is to put 'nonzero elements are invertible' straight into the definition. Here it is a theorem in S08 to S10, and the only use of finiteness is S09. ℤ has no zero divisors but 2 is not invertible (S13), so without finiteness the theorem fails."
      },
      {
        "title": "在 ℤ/nℤ 中計算，而不是在 ℤ 中 / Compute in ℤ/nℤ, not in ℤ",
        "zh": "2·3=6 在 ℤ 中非零，但在 ℤ/6ℤ 中 6 的餘數是 0。每個乘積都要取模後再判斷是否為 0（S04.2、S07）。",
        "en": "2·3=6 is nonzero in ℤ, but in ℤ/6ℤ the remainder of 6 is 0. Every product must be reduced modulo n before deciding whether it is 0 (S04.2, S07)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿不證明一般判準「ℤ/nℤ 是整環當且僅當 n 為質數」，只對 n=5、6 用完整表判定。有限域節點（MKO 的下一步）另立；本稿只指出 S10 使 ℤ/5ℤ 的非零元可逆。",
        "en": "This guide does not prove the general criterion 'ℤ/nℤ is a domain exactly when n is prime'; it decides only n=5 and n=6 from complete tables. The finite-field node (the MKO's next step) is separate; this guide only notes that S10 makes the nonzero elements of ℤ/5ℤ invertible."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "6842601da20f360629d3dee64daf3c9fc262cafcaa31e16bfde9af109e74d478",
    "original_source_sha256": "c6d77ede514c66536e880f8031f33ef853dd3bf2f798b17a549fcac1447b8f09",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M088"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}