{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-finite-unital-ring-01", "mko_id": "mko-finite-unital-ring", "title": { "zh-Hant": "有限含一環：證明模 4 的加法表與乘法表構成含一環", "en": "Finite unital ring: proving that the mod-4 addition and multiplication tables form a unital ring" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-finite-unital-ring-01", "source": { "mko_id": "mko-finite-unital-ring", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "e930e72ae14b44b56feb441e48526b614a1bfb2f281a1518ec6c7b19b1de48bf", "english_sha256": "0e72ca06145d9acc6b850e3f01c4fc86eec243a747e18cc1dcc09f6abfc706f6" }, "mode": "theorem_proof", "audience_prerequisites": [ "division_algorithm", "group_axioms", "distributive", "finite_forall" ], "goal": { "zh-Hant": "在載體 X={0,1,2,3} 上給定完整的加法表 ⊕ 與乘法表 ⊙（即模 4 的加法與乘法）。對所有元素對與三元組證明：(X,⊕) 是阿貝爾群；⊙ 結合；⊙ 對 ⊕ 左右分配；存在唯一的雙側乘法單位元 1，且 1≠0。並指出 2⊙2=0 不妨礙它是含一環。", "en": "On the carrier X={0,1,2,3}, complete addition and multiplication tables ⊕ and ⊙ (addition and multiplication mod 4) are given. Prove for all pairs and triples: (X,⊕) is an abelian group; ⊙ is associative; ⊙ distributes over ⊕ on both sides; there is a unique two-sided multiplicative identity 1, and 1≠0. Note also that 2⊙2=0 does not stop it from being a unital ring." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "加法表 ⊕（行 a、列 b）：第 0 行 0,1,2,3；第 1 行 1,2,3,0；第 2 行 2,3,0,1；第 3 行 3,0,1,2。", "en": "Addition table ⊕ (row a, column b): row 0 is 0,1,2,3; row 1 is 1,2,3,0; row 2 is 2,3,0,1; row 3 is 3,0,1,2." } }, { "id": "G2", "statement": { "zh-Hant": "乘法表 ⊙（行 a、列 b）：第 0 行 0,0,0,0；第 1 行 0,1,2,3；第 2 行 0,2,0,2；第 3 行 0,3,2,1。", "en": "Multiplication table ⊙ (row a, column b): row 0 is 0,0,0,0; row 1 is 0,1,2,3; row 2 is 0,2,0,2; row 3 is 0,3,2,1." } }, { "id": "G3", "statement": { "zh-Hant": "r(m) 表示整數 m 除以 4 的餘數：依除法演算法 m=4q+r(m)、0≤r(m)<4，q、r(m) 唯一。", "en": "r(m) is the remainder of the integer m on division by 4: by the division algorithm m=4q+r(m) with 0≤r(m)<4, and q, r(m) are unique." } } ], "steps": [ { "id": "S01", "uses": [ "G1", "G2" ], "before_tex": "(X,\\oplus,\\odot)\\ \\text{finite unital ring}", "after_tex": "(X,\\oplus)\\ \\text{abelian group};\\ \\odot\\ \\text{associative};\\ \\text{left and right distributive};\\ \\exists!\\,1,\\ 1\\neq 0", "relation": "definition", "rule": "mko_ring_def", "conditions": [], "explanation": { "zh-Hant": "依 MKO 列出全部責任：加法是阿貝爾群（封閉、結合、單位元 0、逆元、交換）；乘法結合；乘法對加法左分配與右分配分開核；存在唯一的雙側乘法單位元 1，且 1≠0。乘法交換不是責任之一。", "en": "List every obligation from the MKO: addition is an abelian group (closure, associativity, identity 0, inverses, commutativity); multiplication is associative; left and right distributivity are checked separately; there is a unique two-sided multiplicative identity 1, and 1≠0. Commutativity of multiplication is not an obligation." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_ring_def" ], "explanation": { "zh-Hant": "此步只展開定義，不推出新事實。", "en": "This step only unfolds the definition and derives no new fact." } }, "leaf_basis": "mko_ring_def + def_unfold" }, { "id": "S02", "uses": [ "G1", "G2", "G3", "finite_forall" ], "before_tex": "\\text{the 32 entries of the two tables}", "after_tex": "a\\oplus b=r(a+b),\\ a\\odot b=r(ab)\\ \\text{for all}\\ a,b\\in X", "relation": "implies", "rule": "finite_forall", "conditions": [], "explanation": { "zh-Hant": "逐格比對 16+16 個表格項與公式。例如乘法第 3 行：3·0=0、3·1=3、3·2=6 餘 2、3·3=9 餘 1，與表中 0,3,2,1 相同；第 2 行：0、2、4 餘 0、6 餘 2；加法第 3 行：3、4 餘 0、5 餘 1、6 餘 2。全部 32 格都相符，所以兩個表格運算恰是模 4 的加法與乘法。", "en": "Compare the 16+16 table entries with the formulas cell by cell. For example row 3 of ⊙: 3·0=0, 3·1=3, 3·2=6 with remainder 2, 3·3=9 with remainder 1, matching 0,3,2,1 in the table; row 2: 0, 2, 4 with remainder 0, 6 with remainder 2; row 3 of ⊕: 3, 4 with remainder 0, 5 with remainder 1, 6 with remainder 2. All 32 cells agree, so the two table operations are exactly addition and multiplication mod 4." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "G2", "G3", "finite_forall" ], "explanation": { "zh-Hant": "32 格全部比對，所以這是對全部元素對的完整核對，不是抽樣；之後的公理證明都建立在這個識別上。", "en": "All 32 cells are compared, so this is a complete check over every pair, not a sample; the axiom proofs below rest on this identification." } }, "leaf_basis": "finite_forall + division_algorithm + numeric_evaluation" }, { "id": "S03", "uses": [ "G3" ], "before_tex": "m,\\ k\\in\\mathbb{Z}", "after_tex": "r(m+k)=r(r(m)+r(k)),\\ r(mk)=r(r(m)r(k))", "relation": "theorem_application", "rule": "L_mod_compat", "conditions": [], "explanation": { "zh-Hant": "引理 L_mod_compat（S03.1 到 S03.3 證明）說明：先取餘再運算再取餘，與直接運算後取餘相同。之後每條公理都化成整數的對應恆等式。", "en": "Lemma L_mod_compat (proved in S03.1 to S03.3) says that reducing, operating and reducing again gives the same as operating and then reducing. Every axiom below is reduced to the corresponding identity of integers." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G3", "L_mod_compat" ], "explanation": { "zh-Hant": "只依賴除法演算法的存在與唯一性；對所有整數成立，不是有限樣本。", "en": "It relies only on existence and uniqueness in the division algorithm; it holds for all integers, not a finite sample." } }, "leaf_basis": "L_mod_compat" }, { "id": "S03.1", "uses": [ "G3", "division_algorithm" ], "before_tex": "m-m'=4t,\\ m'=4s+r(m')", "after_tex": "r(m)=r(m')", "relation": "implies", "rule": "division_algorithm", "conditions": [ "0\\le r(m')<4" ], "explanation": { "zh-Hant": "先證輔助事實：若 m−m′ 是 4 的倍數，則兩者餘數相同。由 m′=4s+r(m′) 得 m=m′+4t=4(s+t)+r(m′)，且 0≤r(m′)<4。除法演算法的唯一性說明 m 的餘數只能是 r(m′)。", "en": "First an auxiliary fact: if m−m′ is a multiple of 4, the two have the same remainder. From m′=4s+r(m′) we get m=m′+4t=4(s+t)+r(m′) with 0≤r(m′)<4. Uniqueness in the division algorithm says the remainder of m can only be r(m′)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "division_algorithm" ], "explanation": { "zh-Hant": "關鍵是唯一性：找到一個符合 0≤r<4 的表示式就確定了餘數。此子步不使用 L_mod_compat。", "en": "The key is uniqueness: one representation with 0≤r<4 fixes the remainder. This sub-step does not use L_mod_compat." } }, "leaf_basis": "division_algorithm + distributive + add_assoc" }, { "id": "S03.2", "uses": [ "S03.1", "G3" ], "before_tex": "m=4q+r(m),\\ k=4p+r(k)", "after_tex": "(m+k)-(r(m)+r(k))=4(q+p)\\Rightarrow r(m+k)=r(r(m)+r(k))", "relation": "implies", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "相加得 m+k=4(q+p)+(r(m)+r(k))，所以 m+k 與 r(m)+r(k) 相差 4 的倍數；依 S03.1 兩者餘數相同。N3 核對所用的代數恆等式。", "en": "Adding gives m+k=4(q+p)+(r(m)+r(k)), so m+k and r(m)+r(k) differ by a multiple of 4; by S03.1 they have the same remainder. N3 checks the algebraic identity used." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03.1", "distributive" ], "explanation": { "zh-Hant": "只用整數的分配律與 S03.1；此子步不使用 L_mod_compat。", "en": "It uses only distributivity of integers and S03.1; this sub-step does not use L_mod_compat." } }, "leaf_basis": "distributive + add_assoc + add_comm" }, { "id": "S03.3", "uses": [ "S03.1", "G3" ], "before_tex": "m=4q+r(m),\\ k=4p+r(k)", "after_tex": "mk-r(m)r(k)=4(4qp+q\\,r(k)+p\\,r(m))\\Rightarrow r(mk)=r(r(m)r(k))", "relation": "implies", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "展開 mk=(4q+r(m))(4p+r(k))=16qp+4q·r(k)+4p·r(m)+r(m)r(k)，所以 mk 與 r(m)r(k) 相差 4(4qp+q·r(k)+p·r(m))，是 4 的倍數；依 S03.1 餘數相同。S03.1 到 S03.3 合起來證得引理 L_mod_compat。N2 核對展開式。", "en": "Expand mk=(4q+r(m))(4p+r(k))=16qp+4q·r(k)+4p·r(m)+r(m)r(k), so mk and r(m)r(k) differ by 4(4qp+q·r(k)+p·r(m)), a multiple of 4; by S03.1 the remainders agree. S03.1 to S03.3 together prove Lemma L_mod_compat. N2 checks the expansion." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03.1", "distributive" ], "explanation": { "zh-Hant": "只用整數的分配律、交換律與 S03.1；此子步不使用 L_mod_compat。", "en": "It uses only distributivity and commutativity of integers and S03.1; this sub-step does not use L_mod_compat." } }, "leaf_basis": "distributive + mul_comm + mul_assoc + add_assoc" }, { "id": "S04", "uses": [ "S02", "G3" ], "before_tex": "a,b\\in X", "after_tex": "a\\oplus b\\in X,\\ a\\odot b\\in X,\\ r(a)=a", "relation": "implies", "rule": "division_algorithm", "conditions": [], "explanation": { "zh-Hant": "餘數滿足 0≤r<4，所以每個運算結果都在 X 中：兩個運算是 X 上的封閉二元運算。另外，a∈X 時 a=4·0+a 且 0≤a<4，依唯一性 r(a)=a；之後用它把 c 換成 r(c)。", "en": "A remainder satisfies 0≤r<4, so every result lies in X: both operations are closed binary operations on X. Also, for a∈X, a=4·0+a with 0≤a<4, so r(a)=a by uniqueness; this is used below to replace c by r(c)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "division_algorithm" ], "explanation": { "zh-Hant": "封閉性只需值域在 X；r(a)=a 只對 X 中的 a 宣稱。", "en": "Closure needs only that values lie in X; r(a)=a is claimed only for a in X." } }, "leaf_basis": "division_algorithm" }, { "id": "S05", "uses": [ "S02", "S03", "S04" ], "before_tex": "(a\\oplus b)\\oplus c=r(r(a+b)+c)", "after_tex": "r(a+b+c)=a\\oplus(b\\oplus c),\\quad a\\oplus b=b\\oplus a", "relation": "equivalent", "rule": "L_mod_compat", "conditions": [ "a,b,c\\in X" ], "explanation": { "zh-Hant": "因 c=r(c)，r(r(a+b)+c)=r(r(a+b)+r(c))=r((a+b)+c)（引理）。同理 a⊕(b⊕c)=r(a+r(b+c))=r(a+(b+c))。整數加法結合，所以兩者相等：⊕ 結合。又 r(a+b)=r(b+a)，⊕ 交換。a、b、c 是 X 的任意元素，所以涵蓋全部 64 個三元組。", "en": "Since c=r(c), r(r(a+b)+c)=r(r(a+b)+r(c))=r((a+b)+c) by the lemma. Likewise a⊕(b⊕c)=r(a+r(b+c))=r(a+(b+c)). Integer addition is associative, so the two agree: ⊕ is associative. Also r(a+b)=r(b+a), so ⊕ is commutative. a, b, c are arbitrary elements of X, so all 64 triples are covered." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_mod_compat", "S04", "add_assoc" ], "explanation": { "zh-Hant": "每個等號都是等值改寫（引理是等式），可以倒著讀，雙向成立。", "en": "Every equality is a rewrite of equal values (the lemma is an equation), so the chain reads backwards as well." } }, "leaf_basis": "L_mod_compat + add_assoc + add_comm + forall_intro" }, { "id": "S06", "uses": [ "S02", "S04", "G1" ], "before_tex": "a\\oplus 0,\\ a\\oplus(\\ominus a)", "after_tex": "a\\oplus 0=a;\\ \\ominus0=0,\\ \\ominus1=3,\\ \\ominus2=2,\\ \\ominus3=1", "relation": "implies", "rule": "group_axioms", "conditions": [], "explanation": { "zh-Hant": "a⊕0=r(a)=a，0⊕a=a，所以 0 是加法單位元（加法表第 0 行與第 0 列都是 0,1,2,3）。逆元：1⊕3=r(4)=0，2⊕2=r(4)=0，3⊕1=0，0⊕0=0；加法表每一行都恰含一個 0。所以每個元素都有加法逆元。", "en": "a⊕0=r(a)=a and 0⊕a=a, so 0 is the additive identity (row 0 and column 0 of the addition table are 0,1,2,3). Inverses: 1⊕3=r(4)=0, 2⊕2=r(4)=0, 3⊕1=0, 0⊕0=0; each row of the addition table contains exactly one 0. So every element has an additive inverse." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S04", "G1", "group_axioms" ], "explanation": { "zh-Hant": "四個元素的逆元逐一列出；因 ⊕ 交換，單側即雙側。", "en": "The inverses of all four elements are listed; since ⊕ is commutative, one-sided means two-sided." } }, "leaf_basis": "group_axioms + finite_forall + numeric_evaluation" }, { "id": "S07", "uses": [ "S04", "S05", "S06" ], "before_tex": "\\text{closure, associativity, identity 0, inverses, commutativity}", "after_tex": "(X,\\oplus)\\ \\text{abelian group}", "relation": "implies", "rule": "group_axioms", "conditions": [], "explanation": { "zh-Hant": "合併 S04（封閉）、S05（結合、交換）、S06（單位元、逆元），(X,⊕) 是阿貝爾群，0 是加法單位元。", "en": "Combining S04 (closure), S05 (associativity, commutativity) and S06 (identity, inverses), (X,⊕) is an abelian group with additive identity 0." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S04", "S05", "S06", "group_axioms" ], "explanation": { "zh-Hant": "目標：加法是阿貝爾群。足夠條件：群公理加交換律，全部由前面的步驟正向證成。", "en": "Goal: addition is an abelian group. Sufficient condition: the group axioms plus commutativity, all proved forwards in the preceding steps." } }, "leaf_basis": "group_axioms + and_intro_elim" }, { "id": "S08", "uses": [ "S02", "S03", "S04" ], "before_tex": "(a\\odot b)\\odot c=r(r(ab)\\,c)", "after_tex": "r(abc)=a\\odot(b\\odot c)", "relation": "equivalent", "rule": "L_mod_compat", "conditions": [ "a,b,c\\in X" ], "explanation": { "zh-Hant": "因 c=r(c)，r(r(ab)c)=r(r(ab)r(c))=r((ab)c)（引理）；a⊙(b⊙c)=r(a·r(bc))=r(r(a)r(bc))=r(a(bc))。整數乘法結合，所以 ⊙ 結合，對全部三元組成立。", "en": "Since c=r(c), r(r(ab)c)=r(r(ab)r(c))=r((ab)c) by the lemma; a⊙(b⊙c)=r(a·r(bc))=r(r(a)r(bc))=r(a(bc)). Integer multiplication is associative, so ⊙ is associative, for every triple." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_mod_compat", "S04", "mul_assoc" ], "explanation": { "zh-Hant": "等式鏈的每一環都是等值改寫，可倒推。", "en": "Every link of the chain is a rewrite of equal values and can be reversed." } }, "leaf_basis": "L_mod_compat + mul_assoc + forall_intro" }, { "id": "S09", "uses": [ "S02", "S03", "S04" ], "before_tex": "a\\odot(b\\oplus c)=r(a\\,r(b+c))", "after_tex": "r(ab+ac)=(a\\odot b)\\oplus(a\\odot c)", "relation": "equivalent", "rule": "L_mod_compat", "conditions": [ "a,b,c\\in X" ], "explanation": { "zh-Hant": "左分配：r(a·r(b+c))=r(r(a)r(b+c))=r(a(b+c))=r(ab+ac)=r(r(ab)+r(ac))=(a⊙b)⊕(a⊙c)。第二個等號與最後一步用引理的乘法與加法部分，中間用整數分配律。", "en": "Left distributivity: r(a·r(b+c))=r(r(a)r(b+c))=r(a(b+c))=r(ab+ac)=r(r(ab)+r(ac))=(a⊙b)⊕(a⊙c). The second equality and the last use the multiplicative and additive parts of the lemma; the middle uses integer distributivity." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_mod_compat", "distributive" ], "explanation": { "zh-Hant": "等式鏈可倒推；只證左分配，右分配另在 S10 證明。", "en": "The chain can be reversed; only the left law is proved here, the right law separately in S10." } }, "leaf_basis": "L_mod_compat + distributive + forall_intro" }, { "id": "S10", "uses": [ "S02", "S03", "S04" ], "before_tex": "(a\\oplus b)\\odot c=r(r(a+b)\\,c)", "after_tex": "r(ac+bc)=(a\\odot c)\\oplus(b\\odot c)", "relation": "equivalent", "rule": "L_mod_compat", "conditions": [ "a,b,c\\in X" ], "explanation": { "zh-Hant": "右分配要分開證：r(r(a+b)c)=r(r(a+b)r(c))=r((a+b)c)=r(ac+bc)=r(r(ac)+r(bc))=(a⊙c)⊕(b⊙c)。這裡用的是整數右分配 (a+b)c=ac+bc，沒有借用 ⊙ 的交換性。", "en": "The right law is proved separately: r(r(a+b)c)=r(r(a+b)r(c))=r((a+b)c)=r(ac+bc)=r(r(ac)+r(bc))=(a⊙c)⊕(b⊙c). It uses integer right distributivity (a+b)c=ac+bc and does not borrow commutativity of ⊙." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_mod_compat", "distributive" ], "explanation": { "zh-Hant": "等式鏈可倒推；本稿分別直接證明左右分配，不借用乘法交換性；不由非交換矩陣例子推論兩條公理的獨立性。", "en": "The chain can be reversed; this guide directly proves each distributive law without using multiplicative commutativity; the noncommutative matrix example is not used to infer logical independence." } }, "leaf_basis": "L_mod_compat + distributive + forall_intro" }, { "id": "S11", "uses": [ "S02", "S04", "G2" ], "before_tex": "1\\odot a,\\ a\\odot 1", "after_tex": "1\\odot a=a=a\\odot 1\\ \\text{for all}\\ a\\in X", "relation": "implies", "rule": "mul_one", "conditions": [], "explanation": { "zh-Hant": "1⊙a=r(a)=a，a⊙1=r(a)=a；乘法表第 1 行與第 1 列都是 0,1,2,3。所以 1 是雙側乘法單位元（MKO 要求雙側作用）。", "en": "1⊙a=r(a)=a and a⊙1=r(a)=a; row 1 and column 1 of the multiplication table are both 0,1,2,3. So 1 is a two-sided multiplicative identity (the MKO requires it to act on both sides)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S04", "G2", "mul_one" ], "explanation": { "zh-Hant": "行與列都核過，所以是雙側；只有一側不夠。", "en": "Both the row and the column are checked, so it is two-sided; one side alone would not suffice." } }, "leaf_basis": "mul_one + finite_forall" }, { "id": "S12", "uses": [ "S11" ], "before_tex": "e\\ \\text{a two-sided identity for}\\ \\odot", "after_tex": "e=1", "relation": "theorem_application", "rule": "L_unique_identity", "conditions": [], "explanation": { "zh-Hant": "依引理 L_unique_identity（S12.1 證明），任何雙側單位元 e 都等於 1，所以乘法單位元唯一。也可對照表：只有第 1 行等於 0,1,2,3。", "en": "By Lemma L_unique_identity (proved in S12.1), any two-sided identity e equals 1, so the multiplicative identity is unique. The table agrees: only row 1 equals 0,1,2,3." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S11", "L_unique_identity" ], "explanation": { "zh-Hant": "需要 1 本身是雙側單位元（S11）；唯一性不依賴表的大小。", "en": "It needs 1 itself to be a two-sided identity (S11); uniqueness does not depend on the size of the table." } }, "leaf_basis": "L_unique_identity" }, { "id": "S12.1", "uses": [ "eq_refl_symm_trans" ], "before_tex": "e\\odot x=x=x\\odot e,\\ e'\\odot x=x=x\\odot e'\\ \\forall x", "after_tex": "e=e\\odot e'=e'", "relation": "implies", "rule": "eq_refl_symm_trans", "conditions": [], "explanation": { "zh-Hant": "e′ 是右單位元，所以 e⊙e′=e；e 是左單位元，所以 e⊙e′=e′。依等號的對稱與傳遞得 e=e′。這證得引理 L_unique_identity。", "en": "e′ is a right identity, so e⊙e′=e; e is a left identity, so e⊙e′=e′. Symmetry and transitivity of equality give e=e′. This proves Lemma L_unique_identity." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "eq_refl_symm_trans" ], "explanation": { "zh-Hant": "用到一個是左單位元、另一個是右單位元；此子步不使用 L_unique_identity。", "en": "It uses that one is a left identity and the other a right identity; this sub-step does not use L_unique_identity." } }, "leaf_basis": "eq_refl_symm_trans + eq_replace" }, { "id": "S13", "uses": [ "S06", "S11" ], "before_tex": "0\\ \\text{additive identity},\\ 1\\ \\text{multiplicative identity}", "after_tex": "1\\neq 0", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "1 與 0 是 X 中不同的元素；而且它們作用不同：1⊙1=1，而 0⊙1=0。MKO 要求 1≠0，排除只有一個元素的零環。", "en": "1 and 0 are different elements of X, and they act differently: 1⊙1=1 while 0⊙1=0. The MKO requires 1≠0, which rules out the one-element zero ring." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S06", "S11" ], "explanation": { "zh-Hant": "只需比較兩個元素；不涉及其他公理。", "en": "It needs only a comparison of two elements and no other axiom." } }, "leaf_basis": "numeric_evaluation + mul_one" }, { "id": "S14", "uses": [ "S01", "S07", "S08", "S09", "S10", "S11", "S12", "S13" ], "before_tex": "\\text{all obligations of S01}", "after_tex": "(X,\\oplus,\\odot)\\ \\text{is a finite unital ring};\\ 2\\odot2=0", "relation": "implies", "rule": "mko_ring_def", "conditions": [], "explanation": { "zh-Hant": "S01 的全部責任都已證成：S07（加法阿貝爾群）、S08（乘法結合）、S09 與 S10（左右分配）、S11 到 S13（唯一雙側單位元且 1≠0）。所以它是有限含一環。另外 2⊙2=r(4)=0 而 2≠0，2 是零因子：它不是整環或域，但 MKO 不要求這些（它們另立節點）。", "en": "Every obligation of S01 is proved: S07 (additive abelian group), S08 (associative multiplication), S09 and S10 (both distributive laws), S11 to S13 (unique two-sided identity with 1≠0). So it is a finite unital ring. Also 2⊙2=r(4)=0 with 2≠0, so 2 is a zero divisor: the ring is not an integral domain or a field, but the MKO does not require those (they are separate nodes)." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S01", "S07", "S08", "S09", "S10", "S11", "S12", "S13" ], "explanation": { "zh-Hant": "目標 G：含一環。足夠條件 H：S01 的全部責任；H⇒G 由定義。H 由 S02 到 S13 正向證成，沒有把 G 當作前提。", "en": "Goal G: a unital ring. Sufficient condition H: all obligations of S01; H⇒G by the definition. H is proved forwards in S02 to S13; G is never used as a premise." } }, "leaf_basis": "mko_ring_def + and_intro_elim" } ], "result": { "zh-Hant": "X={0,1,2,3} 上的兩個表恰是模 4 的加法與乘法；由模運算相容引理，(X,⊕) 是阿貝爾群，⊙ 結合且左右分配，1 是唯一的雙側乘法單位元且 1≠0。所以它是有限含一環；2⊙2=0 顯示它有零因子，但這不違反含一環的定義。", "en": "The two tables on X={0,1,2,3} are exactly addition and multiplication mod 4; by the remainder-compatibility lemma, (X,⊕) is an abelian group, ⊙ is associative and distributes on both sides, and 1 is the unique two-sided multiplicative identity with 1≠0. So it is a finite unital ring; 2⊙2=0 shows it has zero divisors, which does not violate the definition." }, "negative_controls": [ { "claim": { "zh-Hant": "單點載體 {0}（0⊕0=0、0⊙0=0）是本 MKO 意義下的含一環", "en": "The one-point carrier {0} with 0⊕0=0 and 0⊙0=0 is a unital ring in the sense of this MKO" }, "expected": "reject", "reason": { "zh-Hant": "唯一可能的乘法單位元是 0，但 MKO 要求 1≠0。", "en": "The only possible multiplicative identity is 0, but the MKO requires 1≠0." } }, { "claim": { "zh-Hant": "模 4 的表不是含一環，因為 2⊙2=0", "en": "The mod-4 tables are not a unital ring because 2⊙2=0" }, "expected": "reject", "reason": { "zh-Hant": "含一環不要求沒有零因子；S14 已證全部責任成立。", "en": "A unital ring need not be free of zero divisors; S14 shows every obligation holds." } }, { "claim": { "zh-Hant": "含一環的乘法必須交換", "en": "The multiplication of a unital ring must be commutative" }, "expected": "reject", "reason": { "zh-Hant": "ℤ/2 上的 2×2 矩陣構成含一環，但 E11·E12=E12≠0=E12·E11。", "en": "2×2 matrices over ℤ/2 form a unital ring, but E11·E12=E12≠0=E12·E11." } }, { "claim": { "zh-Hant": "(X,⊕,⊙′)（⊙′ 為次序 0<2<3<1 下的最小值）有單位元 1，且 a=0、1 的分配樣本都通過，所以是含一環", "en": "(X,⊕,⊙′), with ⊙′ the minimum for the order 0<2<3<1, has identity 1 and passes the distributivity samples with a=0, 1, so it is a unital ring" }, "expected": "reject", "reason": { "zh-Hant": "2⊙′(1⊕1)=2，而 (2⊙′1)⊕(2⊙′1)=0。", "en": "2⊙′(1⊕1)=2 while (2⊙′1)⊕(2⊙′1)=0." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "division_algorithm", "kind": "prerequisite", "statement": { "zh-Hant": "帶餘除法：對整數 a 與正整數 b，存在唯一的整數 q、r 使 a=bq+r 且 0≤r