{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-finite-two-sided-ideal-01",
  "mko_id": "mko-finite-two-sided-ideal",
  "title": {
    "zh-Hant": "有限雙邊理想：在上三角矩陣環中分開核對左吸收與右吸收",
    "en": "Finite two-sided ideal: checking left and right absorption separately in an upper-triangular matrix ring"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-finite-two-sided-ideal-01",
    "source": {
      "mko_id": "mko-finite-two-sided-ideal",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "56473b0d9ce61008676f55869388fbd0ce017b0555d5a29cc9a088626d1ef36c",
      "english_sha256": "0500394e305af84ba3defda4666f536eb3db5c4d769350cc73df22321811454e"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "matrix_product_def",
      "mul_zero",
      "and_intro_elim",
      "quantifier_negation",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "父環 U 是 𝔽₂ 上 2×2 上三角矩陣環（元素記為 (a,b,c)，代表第一行 (a,b)、第二行 (0,c)），已通過有限含一環資格。（一）證明 I={(0,0,0),(0,1,0)} 是 U 的雙邊理想。（二）證明 L={(0,0,0),(1,0,0)} 滿足左吸收，但不是雙邊理想。（三）證明對角子環 D={(a,0,c)} 含 1 卻不是理想。（四）說明 {0} 與 U 本身是雙邊理想。",
      "en": "The parent ring U is the ring of 2-by-2 upper-triangular matrices over 𝔽₂ (elements written (a,b,c) for the matrix with first row (a,b) and second row (0,c)), already qualified as a finite unital ring. (1) Prove that I={(0,0,0),(0,1,0)} is a two-sided ideal of U. (2) Prove that L={(0,0,0),(1,0,0)} satisfies left absorption but is not a two-sided ideal. (3) Prove that the diagonal subring D={(a,0,c)} contains 1 but is not an ideal. (4) Show that {0} and U itself are two-sided ideals."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "U 的載體是 8 個三元組 (a,b,c)∈{0,1}³；加法逐項 mod 2；乘法 (a,b,c)(a',b',c')=(aa', ab'+bc', cc') mod 2（由矩陣乘積得到）。零元 (0,0,0)，單位元 1_U=(1,0,1)。",
          "en": "U has as carrier the 8 triples (a,b,c)∈{0,1}³; addition is entrywise mod 2; multiplication is (a,b,c)(a',b',c')=(aa', ab'+bc', cc') mod 2 (from the matrix product). The zero is (0,0,0) and the identity is 1_U=(1,0,1)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "U 已由父節點以完整加法表與乘法表通過有限含一環資格；本稿不重新核對環公理。",
          "en": "U has already qualified, with its complete addition and multiplication tables, as a finite unital ring in the parent node; this guide does not recheck the ring axioms."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "候選子集：I={(0,0,0),(0,1,0)}；L={(0,0,0),(1,0,0)}；D={(0,0,0),(1,0,0),(0,0,1),(1,0,1)}，即所有 (a,0,c)。",
          "en": "Candidate subsets: I={(0,0,0),(0,1,0)}; L={(0,0,0),(1,0,0)}; D={(0,0,0),(1,0,0),(0,0,1),(1,0,1)}, that is all (a,0,c)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "mko_ideal_def",
          "G2"
        ],
        "before_tex": "I\\ \\text{two-sided ideal of}\\ U",
        "after_tex": "I\\subseteq U,\\ I\\neq\\varnothing,\\ I+I\\subseteq I,\\ -I\\subseteq I,\\ \\forall r\\,\\forall x:\\ rx\\in I,\\ \\forall r\\,\\forall x:\\ xr\\in I",
        "relation": "definition",
        "rule": "mko_ideal_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 展開雙邊理想的六項責任：是父環 U 的子集、非空、對加法封閉、對加法逆元封閉、左吸收、右吸收。兩個吸收條件各自量化全部 r∈U 與 x∈I，必須分開核對。父環固定為 G2 的 U。",
          "en": "Unfold the MKO's six obligations for a two-sided ideal: a subset of the parent ring U, nonempty, closed under addition, closed under additive inverses, left absorption and right absorption. Each absorption condition quantifies over all r∈U and x∈I and must be checked separately. The parent ring is fixed as the U of G2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_ideal_def",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "只展開定義；若換了父環（例如把 I 看成另一個環的子集），結論不能沿用。",
            "en": "It only unfolds the definition; if the parent ring were changed (for example viewing I inside another ring), the conclusion would not carry over."
          }
        },
        "leaf_basis": "mko_ideal_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "G3",
          "S01"
        ],
        "before_tex": "I=\\{(0,0,0),(0,1,0)\\}",
        "after_tex": "I=\\{(0,\\beta,0):\\beta\\in\\{0,1\\}\\}\\subseteq U,\\ \\ (0,0,0)\\in I",
        "relation": "equivalent",
        "rule": "set_builder",
        "conditions": [],
        "explanation": {
          "zh-Hant": "I 的兩個元素是互異的載體元素，都是 {0,1}³ 中的三元組，所以 I⊆U；用參數 β 把 I 寫成 {(0,β,0)}，β=0 給零元、β=1 給 E₁₂。零元在 I 中，所以 I 非空。",
          "en": "The two elements of I are distinct carrier elements, both triples in {0,1}³, so I⊆U; with the parameter β, I is written as {(0,β,0)}, where β=0 gives the zero and β=1 gives E₁₂. The zero is in I, so I is nonempty."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_builder",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "β 取遍 {0,1} 恰好列出 I 的兩個元素，兩種寫法是同一集合。",
            "en": "As β runs over {0,1} it lists exactly the two elements of I, so both descriptions are the same set."
          }
        },
        "leaf_basis": "set_builder + set_extensionality"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "G1",
          "f2_arith"
        ],
        "before_tex": "(0,\\beta,0)+(0,\\beta',0)",
        "after_tex": "(0,\\ (\\beta+\\beta')\\bmod 2,\\ 0)\\in I",
        "relation": "implies",
        "rule": "f2_arith",
        "conditions": [],
        "explanation": {
          "zh-Hant": "加法逐項 mod 2：第一、三項 0+0=0，第二項 β+β' mod 2 屬於 {0,1}，所以和仍是 (0,β'',0) 的形式，屬於 I。這涵蓋全部 4 個有序和（0+0=0、0+E₁₂=E₁₂、E₁₂+0=E₁₂、E₁₂+E₁₂=0）。",
          "en": "Addition is entrywise mod 2: the first and third entries are 0+0=0, and the second entry β+β' mod 2 lies in {0,1}, so the sum again has the form (0,β'',0) and lies in I. This covers all 4 ordered sums (0+0=0, 0+E₁₂=E₁₂, E₁₂+0=E₁₂, E₁₂+E₁₂=0)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "f2_arith",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "用到的是 𝔽₂ 加法封閉於 {0,1}（N1 核 1+1=0）；不宣稱反向。",
            "en": "It uses that addition in 𝔽₂ stays in {0,1} (N1 checks 1+1=0); no converse is claimed."
          }
        },
        "leaf_basis": "f2_arith + set_builder"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "f2_arith"
        ],
        "before_tex": "-(0,\\beta,0)",
        "after_tex": "(0,\\ (-\\beta)\\bmod 2,\\ 0)=(0,\\beta,0)\\in I",
        "relation": "implies",
        "rule": "f2_arith",
        "conditions": [],
        "explanation": {
          "zh-Hant": "加法逆元逐項取：−0=0，−β mod 2=β（因 −1≡1）。所以每個元素是自己的加法逆元，逆元仍在 I 中。核對：(0,β,0)+(0,β,0)=(0,2β mod 2,0)=(0,0,0)。",
          "en": "Additive inverses are taken entrywise: −0=0 and −β mod 2=β (since −1≡1). So every element is its own additive inverse, and the inverse stays in I. Check: (0,β,0)+(0,β,0)=(0,2β mod 2,0)=(0,0,0)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "f2_arith"
          ],
          "explanation": {
            "zh-Hant": "依賴特徵 2 的算術（N2）；在其他環中加法逆元不一定是自己，必須另外核對。",
            "en": "It relies on characteristic-2 arithmetic (N2); in other rings an additive inverse need not be the element itself and must be checked separately."
          }
        },
        "leaf_basis": "f2_arith + add_inverse"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S02"
        ],
        "before_tex": "r\\cdot x=(a,b,c)(0,\\beta,0),\\ \\ r\\in U,\\ x\\in I",
        "after_tex": "r\\cdot x=(0,\\ a\\beta\\bmod 2,\\ 0)\\in I",
        "relation": "implies",
        "rule": "matrix_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "左吸收：用 G1 的乘法公式，r=(a,b,c) 任意、x=(0,β,0) 任意，乘積是 (a·0, aβ+b·0, c·0)。子步 S05.1、S05.2 逐項計算：第一、三項為 0，第二項 aβ mod 2∈{0,1}。所以 rx 屬於 I。公式對全部 8 個 r 與 2 個 x（16 個乘積）同時成立。",
          "en": "Left absorption: by the multiplication formula of G1, with r=(a,b,c) and x=(0,β,0) arbitrary, the product is (a·0, aβ+b·0, c·0). Sub-steps S05.1 and S05.2 compute it entry by entry: the first and third entries are 0 and the second is aβ mod 2∈{0,1}. So rx lies in I. The formula covers all 8 choices of r and 2 of x (16 products) at once."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S02",
            "matrix_product_def"
          ],
          "explanation": {
            "zh-Hant": "a、b、c、β 沒有被指定數值，所以這是全稱證書，不是抽樣；N3 在全部 4 個 (a,c) 組合上重算第一、三項。",
            "en": "No values are assigned to a, b, c, β, so this is a universal certificate, not a sample; N3 recomputes the first and third entries at all 4 combinations of (a,c)."
          }
        },
        "leaf_basis": "matrix_product_def + mul_zero + f2_arith"
      },
      {
        "id": "S05.1",
        "uses": [
          "G1",
          "mul_zero"
        ],
        "before_tex": "(a\\cdot 0,\\ \\ast,\\ c\\cdot 0)",
        "after_tex": "(0,\\ \\ast,\\ 0)",
        "relation": "equivalent",
        "rule": "mul_zero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第一項 aa' 中 a'=0，第三項 cc' 中 c'=0；任何數乘 0 得 0。",
          "en": "In the first entry aa' we have a'=0, and in the third entry cc' we have c'=0; anything times 0 is 0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "a·0=0 是恆等式，兩邊同值。",
            "en": "a·0=0 is an identity, so both sides have the same value."
          }
        },
        "leaf_basis": "mul_zero"
      },
      {
        "id": "S05.2",
        "uses": [
          "S05.1",
          "f2_arith"
        ],
        "before_tex": "a\\beta+b\\cdot 0",
        "after_tex": "a\\beta\\bmod 2\\in\\{0,1\\}",
        "relation": "implies",
        "rule": "f2_arith",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第二項 ab'+bc' 中 b'=β、c'=0，所以是 aβ+0=aβ，取 mod 2 後屬於 {0,1}。因此乘積具有 (0,β'',0) 的形式，屬於 I。",
          "en": "In the second entry ab'+bc' we have b'=β and c'=0, so it is aβ+0=aβ, which mod 2 lies in {0,1}. Hence the product has the form (0,β'',0) and lies in I."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "f2_arith",
            "S05.1"
          ],
          "explanation": {
            "zh-Hant": "只需第二項落在 {0,1}；I 對第二項沒有其他限制。",
            "en": "Only that the second entry lies in {0,1} is needed; I places no other restriction on the second entry."
          }
        },
        "leaf_basis": "f2_arith + mul_zero + add_zero"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "S02"
        ],
        "before_tex": "x\\cdot r=(0,\\beta,0)(a,b,c),\\ \\ r\\in U,\\ x\\in I",
        "after_tex": "x\\cdot r=(0\\cdot a,\\ 0\\cdot b+\\beta c,\\ 0\\cdot c)=(0,\\ \\beta c\\bmod 2,\\ 0)\\in I",
        "relation": "implies",
        "rule": "matrix_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "右吸收必須另外核對，因為 U 不交換。乘法公式給 xr=(0·a, 0·b+βc, 0·c)：第一、三項為 0，第二項 βc mod 2∈{0,1}。所以對全部 r∈U、x∈I，xr∈I。",
          "en": "Right absorption must be checked separately because U is not commutative. The multiplication formula gives xr=(0·a, 0·b+βc, 0·c): the first and third entries are 0 and the second is βc mod 2∈{0,1}. So xr∈I for all r∈U and x∈I."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "matrix_product_def",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "與 S05 是不同的乘積（左右因子對調），不能由 S05 推出；這裡獨立計算（N4）。",
            "en": "This is a different product from S05 (the factors are swapped) and cannot be inferred from S05; it is computed independently here (N4)."
          }
        },
        "leaf_basis": "matrix_product_def + mul_zero + f2_arith"
      },
      {
        "id": "S07",
        "uses": [
          "S01",
          "S02",
          "S03",
          "S04",
          "S05",
          "S06"
        ],
        "before_tex": "\\text{S02 to S06}",
        "after_tex": "I=\\{0,E_{12}\\}\\ \\text{is a two-sided ideal of}\\ U",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "六項責任都已證明：子集與非空（S02）、加法封閉（S03）、逆元封閉（S04）、左吸收（S05）、右吸收（S06）。依合取引入與 S01 的定義，I 是 U 的雙邊理想。",
          "en": "All six obligations are proved: subset and nonempty (S02), closure under addition (S03), under inverses (S04), left absorption (S05) and right absorption (S06). By conjunction introduction and the definition in S01, I is a two-sided ideal of U."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S01",
            "S02",
            "S03",
            "S04",
            "S05",
            "S06",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：I 是雙邊理想。足夠條件 H：六項責任；H⇒G 由 S01。H 的每一項都由 G1 到 G3 正向證得，沒有把 G 當作前提。",
            "en": "Goal G: I is a two-sided ideal. Sufficient condition H: the six obligations; H⇒G by S01. Each part of H is proved forwards from G1 to G3; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim + mko_ideal_def"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "G3",
          "one_sided_subring_def",
          "f2_arith"
        ],
        "before_tex": "L=\\{(\\beta,0,0):\\beta\\in\\{0,1\\}\\},\\ \\ r=(a,b,c)",
        "after_tex": "r\\cdot(\\beta,0,0)=(a\\beta,\\ a\\cdot 0+b\\cdot 0,\\ c\\cdot 0)=(a\\beta\\bmod 2,0,0)\\in L",
        "relation": "implies",
        "rule": "matrix_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "L 非空，E₁₁+E₁₁=(0,0,0)、−E₁₁=E₁₁，所以加法部分與 I 相同地成立。左吸收：對任意 r=(a,b,c)，r(β,0,0)=(aβ,0,0)∈L。所以 L 是左理想。",
          "en": "L is nonempty, E₁₁+E₁₁=(0,0,0) and −E₁₁=E₁₁, so the additive conditions hold as for I. Left absorption: for any r=(a,b,c), r(β,0,0)=(aβ,0,0)∈L. So L is a left ideal."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "one_sided_subring_def",
            "matrix_product_def",
            "f2_arith"
          ],
          "explanation": {
            "zh-Hant": "這只證明了左吸收；MKO 要求右吸收另外量化，見 S09。",
            "en": "This proves only left absorption; the MKO requires right absorption to be quantified separately, see S09."
          }
        },
        "leaf_basis": "matrix_product_def + mul_zero + f2_arith + one_sided_subring_def"
      },
      {
        "id": "S09",
        "uses": [
          "S08",
          "G1"
        ],
        "before_tex": "E_{11}\\cdot E_{12}=(1,0,0)(0,1,0)",
        "after_tex": "E_{11}E_{12}=(0,\\ 1\\cdot1+0\\cdot0,\\ 0)=(0,1,0)\\notin L\\ \\Rightarrow\\ L\\ \\text{not two-sided}",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取 x=E₁₁∈L、r=E₁₂∈U，右乘積 xr=(1·0, 1·1+0·0, 0·0)=(0,1,0)。L 的元素第二項都是 0，所以 (0,1,0)∉L。右吸收的全稱命題有一個反例，L 不是雙邊理想。這正是 MKO 指出的「只核左吸收」的誤解。",
          "en": "Take x=E₁₁∈L and r=E₁₂∈U: the right product xr=(1·0, 1·1+0·0, 0·0)=(0,1,0). Every element of L has second entry 0, so (0,1,0)∉L. The universal statement of right absorption has a counterexample, so L is not a two-sided ideal. This illustrates the misconception identified in the MKO: checking only left absorption."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "一個反例足以否定（N5）；S08 的左吸收仍然成立，兩者不矛盾。",
            "en": "One counterexample suffices (N5); left absorption from S08 still holds, and the two do not conflict."
          }
        },
        "leaf_basis": "quantifier_negation + matrix_product_def + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "G1",
          "G3",
          "one_sided_subring_def"
        ],
        "before_tex": "D=\\{(a,0,c)\\},\\ \\ (a,0,c)(a',0,c')",
        "after_tex": "(aa',\\ a\\cdot0+0\\cdot c',\\ cc')=(aa',0,cc')\\in D,\\ \\ 1_U=(1,0,1)\\in D",
        "relation": "implies",
        "rule": "one_sided_subring_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "D 是子環：兩個對角元的乘積第二項為 a·0+0·c'=0，仍是對角；加法逐項保持第二項為 0；每個元素是自己的加法逆元；單位元 (1,0,1) 在 D 中。MKO 提醒：子環不能與理想混稱，S11 說明 D 不是理想。",
          "en": "D is a subring: the product of two diagonal elements has second entry a·0+0·c'=0 and stays diagonal; entrywise addition keeps the second entry 0; every element is its own additive inverse; the identity (1,0,1) is in D. The MKO warns that a subring must not be called an ideal; S11 shows D is not one."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "one_sided_subring_def",
            "matrix_product_def"
          ],
          "explanation": {
            "zh-Hant": "子環要求的是對 D 內元素的乘法封閉，不是對全部 r∈U 的吸收；不宣稱 D 是理想。",
            "en": "A subring requires closure under multiplication of elements of D, not absorption of every r∈U; D is not claimed to be an ideal."
          }
        },
        "leaf_basis": "one_sided_subring_def + matrix_product_def + mul_zero"
      },
      {
        "id": "S11",
        "uses": [
          "S10",
          "G1"
        ],
        "before_tex": "1_U\\in D,\\ \\ E_{12}\\notin D",
        "after_tex": "D\\ \\text{is not an ideal};\\ \\text{witness}\\ E_{12}\\cdot E_{22}=(0,1,0)\\notin D",
        "relation": "implies",
        "rule": "L_one_in_ideal",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_one_in_ideal，若 D 滿足右吸收且含 1，則 D=U；但 E₁₂=(0,1,0)∉D，所以 D 不滿足右吸收。具體見證：E₁₂E₂₂=(0·0, 0·0+1·1, 0·1)=(0,1,0)∉D（左吸收亦失敗，r=E₁₂、x=E₂₂）。引理在子步 S11.1、S11.2 證明。",
          "en": "By Lemma L_one_in_ideal, if D satisfied right absorption and contained 1, then D=U; but E₁₂=(0,1,0)∉D, so D fails right absorption. A concrete witness: E₁₂E₂₂=(0·0, 0·0+1·1, 0·1)=(0,1,0)∉D (left absorption fails too, with r=E₁₂ and x=E₂₂). The lemma is proved in sub-steps S11.1 and S11.2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "L_one_in_ideal",
            "contrapositive"
          ],
          "explanation": {
            "zh-Hant": "使用引理的逆否：J≠R 且 1∈J ⇒ J 不滿足右吸收；具體乘積由 N6 核對。",
            "en": "It uses the contrapositive of the lemma: J≠R and 1∈J ⇒ J fails right absorption; the concrete product is checked by N6."
          }
        },
        "leaf_basis": "contrapositive + matrix_product_def + numeric_evaluation"
      },
      {
        "id": "S11.1",
        "uses": [
          "one_sided_subring_def"
        ],
        "before_tex": "1\\in J,\\ \\ \\forall x\\in J\\ \\forall r\\in R:\\ xr\\in J",
        "after_tex": "\\forall r\\in R:\\ r=1\\cdot r\\in J",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在右吸收中取 x=1：對每個 r∈R，1·r∈J；而 1·r=r（單位元），所以 r∈J。",
          "en": "Specialize right absorption to x=1: for every r∈R, 1·r∈J; and 1·r=r (identity), so r∈J."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "forall_elim",
            "mul_one"
          ],
          "explanation": {
            "zh-Hant": "需要 1∈J 與單位元性質 1·r=r；只用右吸收的一個特例。",
            "en": "It needs 1∈J and the identity property 1·r=r; only one instance of right absorption is used."
          }
        },
        "leaf_basis": "forall_elim + mul_one"
      },
      {
        "id": "S11.2",
        "uses": [
          "S11.1",
          "set_extensionality"
        ],
        "before_tex": "R\\subseteq J,\\ \\ J\\subseteq R",
        "after_tex": "J=R",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S11.1 給 R⊆J，J 本來就是 R 的子集，依集合外延性 J=R。S11.1 與 S11.2 合起來證得引理 L_one_in_ideal。",
          "en": "S11.1 gives R⊆J, and J is a subset of R by assumption, so J=R by extensionality. S11.1 and S11.2 together prove Lemma L_one_in_ideal."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11.1",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "兩個方向的包含都已具備，集合相等。",
            "en": "Both inclusions are available, so the sets are equal."
          }
        },
        "leaf_basis": "set_extensionality + subset_def"
      },
      {
        "id": "S12",
        "uses": [
          "G1",
          "mko_ideal_def"
        ],
        "before_tex": "\\{0\\},\\ U",
        "after_tex": "r0=0r=0\\in\\{0\\};\\ \\ rx,xr\\in U\\ \\Rightarrow\\ \\{0\\},\\ U\\ \\text{two-sided ideals}",
        "relation": "implies",
        "rule": "mko_ideal_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "零理想：0+0=0、−0=0，且對所有 r，r·0=0·r=0（mul_zero 對矩陣逐項成立）。整個環：乘法對 U 封閉，所以吸收自動成立。MKO 指出這兩個都是雙邊理想。",
          "en": "Zero ideal: 0+0=0, −0=0, and r·0=0·r=0 for every r (mul_zero entrywise for matrices). The whole ring: multiplication is closed on U, so absorption is automatic. The MKO notes that both are two-sided ideals."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_ideal_def",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "零乘任何元素為零是環公理的推論；U 的封閉性來自父環資格（G2）。",
            "en": "That zero times anything is zero follows from the ring axioms; closure of U comes from the parent qualification (G2)."
          }
        },
        "leaf_basis": "mko_ideal_def + mul_zero + add_zero"
      },
      {
        "id": "S13",
        "uses": [
          "S07",
          "S08",
          "S09",
          "S10",
          "S11",
          "S12"
        ],
        "before_tex": "\\text{S07},\\ \\text{S09},\\ \\text{S11},\\ \\text{S12}",
        "after_tex": "I\\ \\text{two-sided};\\ L\\ \\text{left only};\\ D\\ \\text{subring, not ideal};\\ \\{0\\},U\\ \\text{two-sided}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "合併四個任務：I={0,E₁₂} 是雙邊理想；L={0,E₁₁} 是左理想但不是雙邊理想；對角子環 D 不是理想；{0} 與 U 是雙邊理想。所有判定都綁定同一父環 U。",
          "en": "Combine the four tasks: I={0,E₁₂} is a two-sided ideal; L={0,E₁₁} is a left ideal but not two-sided; the diagonal subring D is not an ideal; {0} and U are two-sided ideals. Every decision refers to the same parent ring U."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S07",
            "S09",
            "S11",
            "S12",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G 是四項判定的合取；H 由 S07、S09、S11、S12 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 four decisions; H is proved forwards by S07, S09, S11 and S12, H⇒G by conjunction introduction, and G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在 𝔽₂ 上 2×2 上三角矩陣環 U 中：I={(0,0,0),(0,1,0)} 非空、對加法與逆元封閉，且對所有 r∈U 有 rx=(0,aβ,0)∈I、xr=(0,βc,0)∈I，所以是雙邊理想。L={(0,0,0),(1,0,0)} 有左吸收，但 E₁₁E₁₂=E₁₂∉L，不是雙邊理想。對角子環 D 含 1 卻 D≠U，E₁₂E₂₂=E₁₂∉D，不是理想。{0} 與 U 是雙邊理想。",
      "en": "In the ring U of 2-by-2 upper-triangular matrices over 𝔽₂: I={(0,0,0),(0,1,0)} is nonempty, closed under addition and inverses, and for every r∈U has rx=(0,aβ,0)∈I and xr=(0,βc,0)∈I, so it is a two-sided ideal. L={(0,0,0),(1,0,0)} has left absorption but E₁₁E₁₂=E₁₂∉L, so it is not two-sided. The diagonal subring D contains 1 but D≠U, and E₁₂E₂₂=E₁₂∉D, so it is not an ideal. {0} and U are two-sided ideals."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "L={0,E₁₁} 對所有 r∈U 有 rx∈L，所以 L 是 U 的雙邊理想",
          "en": "L={0,E₁₁} has rx∈L for every r∈U, so L is a two-sided ideal of U"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "右吸收失敗：E₁₁E₁₂=(0,1,0)∉L。",
          "en": "Right absorption fails: E₁₁E₁₂=(0,1,0)∉L."
        }
      },
      {
        "claim": {
          "zh-Hant": "對角集合 D 對乘法封閉且含 1，所以是 U 的理想",
          "en": "The diagonal set D is closed under multiplication and contains 1, so it is an ideal of U"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "E₁₂E₂₂=(0,1,0)∉D；含 1 的理想必為整個 U。",
          "en": "E₁₂E₂₂=(0,1,0)∉D; an ideal containing 1 must be all of U."
        }
      },
      {
        "claim": {
          "zh-Hant": "抽查 r=1_U 與 r=E₁₁ 時 rx、xr 都在 I 中，所以 I 是雙邊理想",
          "en": "Spot checks with r=1_U and r=E₁₁ put rx and xr in I, so I is a two-sided ideal"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "結論雖然正確，但抽查不是證明；需要 S05、S06 對全部 8 個 r 的公式證書。",
          "en": "The conclusion happens to be true, but spot checks are not a proof; the formula certificates of S05 and S06 for all 8 choices of r are needed."
        }
      },
      {
        "claim": {
          "zh-Hant": "L 在 D 中是雙邊理想，所以 L 在 U 中也是雙邊理想",
          "en": "L is a two-sided ideal of D, so it is also a two-sided ideal of U"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "資格綁定父環；在 U 中 E₁₁E₁₂=E₁₂∉L。",
          "en": "Qualification is bound to the parent ring; in U, E₁₁E₁₂=E₁₂∉L."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "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": "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": "set_builder",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。",
          "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x)."
        }
      },
      {
        "id": "set_extensionality",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合相等：A=B 當且僅當 A⊆B 且 B⊆A，即兩者有相同元素。",
          "en": "Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements."
        }
      },
      {
        "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": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "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": "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": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "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": "mko_ideal_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-finite-two-sided-ideal",
          "sha256": "56473b0d9ce61008676f55869388fbd0ce017b0555d5a29cc9a088626d1ef36c",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "給有限含一環 R 與完整子集 I。I 須非空、對加法及加法逆元封閉，並對所有 r∈R、x∈I 同時有 rx∈I 與 xr∈I。資格永遠綁定同一父環；左吸收與右吸收分開量化。",
          "en": "Given a finite unital ring R and a fully specified subset I, require I to be nonempty and closed under addition and additive inverses, and for every r in R and x in I require both rx and xr to belong to I. Qualification always refers to the same parent ring; left and right absorption are quantified separately."
        }
      },
      {
        "id": "f2_arith",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "𝔽₂={0,1} 以 mod 2 運算：0+0=0、0+1=1、1+1=0；−0=0、−1=1；0·t=0、1·1=1。",
          "en": "𝔽₂={0,1} with arithmetic mod 2: 0+0=0, 0+1=1, 1+1=0; −0=0 and −1=1; 0·t=0 and 1·1=1."
        }
      },
      {
        "id": "one_sided_subring_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "左理想：非空、對加法與加法逆元封閉，且對所有 r∈R、x∈J 有 rx∈J（右理想對稱地要求 xr∈J）。子環：含 1、對加法、加法逆元與乘法封閉的子集。",
          "en": "Left ideal: nonempty, closed under addition and additive inverses, with rx∈J for all r∈R and x∈J (a right ideal symmetrically requires xr∈J). Subring: a subset containing 1 and closed under addition, additive inverses and multiplication."
        }
      },
      {
        "id": "L_one_in_ideal",
        "kind": "lemma",
        "proved_in": [
          "S11.1",
          "S11.2"
        ],
        "statement": {
          "zh-Hant": "在含一環 R 中，若滿足右吸收（xr∈J）的子集 J 含 1，則 J=R。",
          "en": "In a unital ring R, if a subset J satisfying right absorption (xr∈J) contains 1, then J=R."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "(1+1)%2",
        "rhs": "0"
      },
      {
        "id": "N2",
        "step": "S04",
        "lhs": "(-1)%2",
        "rhs": "1"
      },
      {
        "id": "N3",
        "step": "S05",
        "vars": [
          "a",
          "c"
        ],
        "range": [
          0,
          1
        ],
        "lhs": "(a*0)%2+(c*0)%2",
        "rhs": "0"
      },
      {
        "id": "N4",
        "step": "S06",
        "vars": [
          "a",
          "c"
        ],
        "range": [
          0,
          1
        ],
        "lhs": "(0*a)%2+(0*c)%2",
        "rhs": "0"
      },
      {
        "id": "N5",
        "step": "S09",
        "lhs": "(1*1+0*0)%2",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N6",
        "step": "S11",
        "lhs": "(0*0+1*1)%2",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N7",
        "step": "S05",
        "lhs": "2**3*2",
        "rhs": "16"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：父環以完整雙表通過有限含一環資格，候選子集由互異載體元素完整列出（假設 a1）；左吸收與右吸收分開量化，並以對全部父環元素成立的公式證書核對，不以抽樣代替（假設 a2）。不處理商環（下一節點），不重新核對父環公理，不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: the parent ring qualifies as a finite unital ring with both complete tables, and each candidate subset is fully listed by distinct carrier elements (assumption a1); left and right absorption are quantified separately and checked by formula certificates valid for every element of the parent ring, not by sampling (assumption a2). Quotient rings (the next node) are not treated, the parent ring's axioms are not rechecked, and the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "U",
        "𝔽₂ 上 2×2 上三角矩陣環，8 個元素",
        "the ring of 2-by-2 upper-triangular matrices over 𝔽₂, 8 elements"
      ],
      [
        "(a,b,c)",
        "第一行 (a,b)、第二行 (0,c) 的矩陣，a,b,c∈{0,1}",
        "the matrix with first row (a,b) and second row (0,c), a,b,c∈{0,1}"
      ],
      [
        "E_{11},E_{12},E_{22}",
        "(1,0,0)、(0,1,0)、(0,0,1)",
        "(1,0,0), (0,1,0), (0,0,1)"
      ],
      [
        "1_U=(1,0,1)",
        "U 的乘法單位元",
        "the multiplicative identity of U"
      ],
      [
        "I,\\ L,\\ D",
        "三個候選子集",
        "the three candidate subsets"
      ],
      [
        "\\beta\\in\\{0,1\\}",
        "I 或 L 中元素的參數",
        "the parameter of an element of I or L"
      ]
    ],
    "md_error_controls": [
      {
        "title": "只核左吸收不夠 / Left absorption alone is not enough",
        "zh": "MKO 指出的誤解之一是只核左吸收。L={0,E₁₁} 對所有 r 滿足 rx∈L，但 E₁₁E₁₂=E₁₂∉L。在非交換環中 rx 與 xr 是不同的乘積，必須分開對全部父環元素核對（S05、S06 分別處理）。",
        "en": "One misconception named by the MKO is to check only left absorption. L={0,E₁₁} satisfies rx∈L for every r, but E₁₁E₁₂=E₁₂∉L. In a non-commutative ring rx and xr are different products and must be checked separately against every element of the parent ring (S05 and S06 treat them separately)."
      },
      {
        "title": "乘法封閉不足以判定理想 / Multiplicative closure does not establish an ideal",
        "zh": "另一個誤解是只核乘法封閉。D 對 D 內乘法封閉且含 1，是子環，但理想要求吸收所有父環元素；E₁₂E₂₂=E₁₂∉D。更一般地，含 1 且吸收的子集必是整個環（引理 L_one_in_ideal），所以真子環若含 1 就不可能是理想。",
        "en": "Another misconception is to check only closure under multiplication. D is closed under products of its own elements and contains 1, so it is a subring, but an ideal must absorb every element of the parent ring; E₁₂E₂₂=E₁₂∉D. More generally, an absorbing subset containing 1 is the whole ring (Lemma L_one_in_ideal), so a proper subring containing 1 can never be an ideal."
      },
      {
        "title": "資格綁定同一父環 / Qualification is bound to one parent ring",
        "zh": "同一個子集在不同父環中結論可能不同：例如 L 若被看成對角子環 D 的子集，D 交換，L 在 D 中是雙邊理想（(a,0,c)(β,0,0)=(aβ,0,0)=(β,0,0)(a,0,c)）；但在 U 中不是。所以每個判定都必須寫明父環。",
        "en": "The same subset can behave differently in different parent rings: for example, viewed inside the diagonal subring D, which is commutative, L is a two-sided ideal ((a,0,c)(β,0,0)=(aβ,0,0)=(β,0,0)(a,0,c)); in U it is not. So every decision must name its parent ring."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "U 的有限含一環資格取自父節點（G2），本稿未重新核對環公理；乘法公式由矩陣乘積導出，8×8 完整表未逐格印出。",
        "en": "U's finite-unital-ring qualification is taken from the parent node (G2) and the ring axioms are not rechecked; the multiplication formula is derived from the matrix product, and the complete 8-by-8 table is not printed cell by cell."
      },
      {
        "zh-Hant": "第三個負向控制的主張結論本身為真，被拒絕的是「抽查即證明」的推理。",
        "en": "The conclusion in the third negative control is itself true; what is rejected is the reasoning that spot checks constitute a proof."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "2b3540b5dbb78a14560841645c8bd368269c59a972c859b94548f2dadb8878eb",
    "original_source_sha256": "2b3540b5dbb78a14560841645c8bd368269c59a972c859b94548f2dadb8878eb",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}