{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-subgroup-homomorphism-01",
  "mko_id": "mko-subgroup-homomorphism",
  "title": {
    "zh-Hant": "子群與群同態：在 ℤ/4 與 ℤ/2 的完整表上核對",
    "en": "Subgroups and homomorphisms: auditing on the complete tables of ℤ/4 and ℤ/2"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-subgroup-homomorphism-01",
    "source": {
      "mko_id": "mko-subgroup-homomorphism",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "2121cf9c71fab6aa9bfa10d881cdffbab5367fdfebbee9381b35fefec60837c0",
      "english_sha256": "d8b3efcc34a6efe50514c12c02d1650c151d671a36a64e0511bf4a65cf744d59"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "division_algorithm",
      "function_def",
      "injective_surjective_def",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "G={0,1,2,3} 以 (a+b) mod 4 為運算，K={0,1} 以 (a+b) mod 2 為運算，兩者都給出完整 Cayley 表。（一）證明 H={0,2} 是 G 的子群，H'={0,1} 不是；（二）證明映射表 f(0)=0、f(1)=1、f(2)=0、f(3)=1 是同態 G→K（涵蓋全部 16 個輸入對）；（三）求 kernel 與 image，說明 f 滿射但不單射，因此不是同構；（四）說明常數映射 h≡1 不是同態。",
      "en": "G={0,1,2,3} with operation (a+b) mod 4 and K={0,1} with operation (a+b) mod 2 are given by complete Cayley tables. (1) Prove that H={0,2} is a subgroup of G and H'={0,1} is not; (2) prove that the mapping table f(0)=0, f(1)=1, f(2)=0, f(3)=1 is a homomorphism G→K, covering all 16 input pairs; (3) find the kernel and the image, and show that f is surjective but not injective, hence not an isomorphism; (4) show that the constant map h≡1 is not a homomorphism."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "來源群 G={0,1,2,3}，Cayley 表的第 a 行（row）依序為 a*0、a*1、a*2、a*3：0,1,2,3；1,2,3,0；2,3,0,1；3,0,1,2，即 a*b=(a+b) mod 4；單位元為 0。(G,*) 是有限群。",
          "en": "Source group G={0,1,2,3}; row a of its Cayley table lists a*0, a*1, a*2, a*3: 0,1,2,3; 1,2,3,0; 2,3,0,1; 3,0,1,2, that is a*b=(a+b) mod 4; the identity is 0. (G,*) is a finite group."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "目標群 K={0,1}，Cayley 表兩行（row）為 0,1；1,0，即 a⊕b=(a+b) mod 2；單位元為 0。(K,⊕) 是有限群。",
          "en": "Target group K={0,1}; its Cayley table has rows 0,1 and 1,0, that is a⊕b=(a+b) mod 2; the identity is 0. (K,⊕) is a finite group."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "映射表：f(0)=0，f(1)=1，f(2)=0，f(3)=1；h(0)=h(1)=h(2)=h(3)=1。",
          "en": "Mapping tables: f(0)=0, f(1)=1, f(2)=0, f(3)=1; h(0)=h(1)=h(2)=h(3)=1."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "候選子集 H={0,2} 與 H'={0,1}。",
          "en": "Candidate subsets H={0,2} and H'={0,1}."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "mko_subhom_def",
          "G1",
          "G2"
        ],
        "before_tex": "H\\subseteq G,\\ f:G\\to K",
        "after_tex": "H\\neq\\varnothing,\\ HH\\subseteq H,\\ e\\in H,\\ H^{-1}\\subseteq H;\\ \\forall a,b:\\ f(a*b)=f(a)\\oplus f(b)",
        "relation": "definition",
        "rule": "mko_subhom_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，子群要核四件事：非空、封閉、含單位元、對逆元封閉；同態要對全部輸入對滿足 f(a*b)=f(a)⊕f(b)。MKO 也提醒同態不自動為單射。",
          "en": "By the MKO a subgroup needs four checks: nonempty, closed, contains the identity, closed under inverses; a homomorphism must satisfy f(a*b)=f(a)⊕f(b) for every input pair. The MKO also warns that a homomorphism is not automatically injective."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_subhom_def",
            "G1",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "此步只列出要核的條件，不推出新事實。",
            "en": "This step only lists the conditions to check and derives no new fact."
          }
        },
        "leaf_basis": "mko_subhom_def + group_axioms"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "G4"
        ],
        "before_tex": "H=\\{0,2\\}",
        "after_tex": "H\\neq\\varnothing,\\ H\\subseteq G",
        "relation": "implies",
        "rule": "subset_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "H 含 0，所以非空；它的兩個元素 0、2 都在 G 中，所以 H⊆G。",
          "en": "H contains 0, so it is nonempty; both of its elements 0 and 2 lie in G, so H⊆G."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "subset_def",
            "G4"
          ],
          "explanation": {
            "zh-Hant": "兩個元素逐一核對；這一步不涉及運算。",
            "en": "Both elements are checked; the operation plays no role here."
          }
        },
        "leaf_basis": "subset_def"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S02"
        ],
        "before_tex": "a,b\\in H",
        "after_tex": "0*0=0,\\ 0*2=2,\\ 2*0=2,\\ 2*2=0\\ \\in H",
        "relation": "implies",
        "rule": "finite_exhaustion",
        "conditions": [],
        "explanation": {
          "zh-Hant": "H 有 2²=4 個有序對，查 G 的表：0*0=0，0*2=2，2*0=2，2*2=(2+2) mod 4=0。四個輸出都在 H，所以 H 封閉。",
          "en": "H has 2²=4 ordered pairs; reading G's table: 0*0=0, 0*2=2, 2*0=2 and 2*2=(2+2) mod 4=0. All four outputs lie in H, so H is closed."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "finite_exhaustion"
          ],
          "explanation": {
            "zh-Hant": "4 個有序對都查過；封閉只是四個條件之一，S04、S05 另外核。",
            "en": "All 4 ordered pairs are checked; closure is only one of the four conditions, and S04 and S05 check the rest."
          }
        },
        "leaf_basis": "finite_exhaustion + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S02"
        ],
        "before_tex": "e_G=0",
        "after_tex": "e_G\\in H",
        "relation": "implies",
        "rule": "set_builder",
        "conditions": [],
        "explanation": {
          "zh-Hant": "G 的單位元是 0（表的第 0 行（row）與第 0 列（column）都是 0,1,2,3），而 0∈H。",
          "en": "The identity of G is 0 (row 0 and column 0 of the table are both 0,1,2,3), and 0∈H."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G4"
          ],
          "explanation": {
            "zh-Hant": "必須是 G 的單位元在 H 中，而不是 H 自己另選一個元素。",
            "en": "It must be the identity of G that lies in H, not some element H picks for itself."
          }
        },
        "leaf_basis": "group_axioms + set_builder"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S04"
        ],
        "before_tex": "a\\in H",
        "after_tex": "0^{-1}=0\\in H,\\ 2^{-1}=2\\in H",
        "relation": "implies",
        "rule": "finite_exhaustion",
        "conditions": [],
        "explanation": {
          "zh-Hant": "0*0=0，所以 0 的逆元是 0；2*2=0 而且運算對稱位置相同，所以 2 的雙側逆元是 2。兩個逆元都在 H 中。",
          "en": "0*0=0, so the inverse of 0 is 0; 2*2=0 on both sides, so the two-sided inverse of 2 is 2. Both inverses lie in H."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "finite_exhaustion",
            "group_axioms"
          ],
          "explanation": {
            "zh-Hant": "逆元是在 G 中取的，然後檢查它是否落在 H；H 的兩個元素都查過。",
            "en": "Inverses are taken in G and then tested for membership in H; both elements of H are checked."
          }
        },
        "leaf_basis": "finite_exhaustion + group_axioms"
      },
      {
        "id": "S06",
        "uses": [
          "S02",
          "S03",
          "S04",
          "S05"
        ],
        "before_tex": "\\text{nonempty, closed, identity, inverses}",
        "after_tex": "H\\ \\text{is a subgroup of } G",
        "relation": "implies",
        "rule": "mko_subhom_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "四個條件分別由 S02 到 S05 給出，依 MKO 的定義 H 是 G 的子群。",
          "en": "The four conditions come from S02 to S05, so by the MKO's definition H is a subgroup of G."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S02",
            "S03",
            "S04",
            "S05",
            "mko_subhom_def"
          ],
          "explanation": {
            "zh-Hant": "目標 G：H 是子群。足夠條件：四個條件同時成立；由定義推出目標。四者都由表正向證成，沒有把目標當作前提。",
            "en": "Goal: H is a subgroup. Sufficient condition: the four conditions together, which give the goal by the definition. All four are proved forwards from the table; the goal is never used as a premise."
          }
        },
        "leaf_basis": "mko_subhom_def + and_intro_elim"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "G4"
        ],
        "before_tex": "H'=\\{0,1\\},\\ 1*1",
        "after_tex": "1*1=2\\notin H'\\ \\Rightarrow\\ H'\\ \\text{not a subgroup}",
        "relation": "implies",
        "rule": "mko_subhom_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "H' 非空且含單位元 0，但 1*1=(1+1) mod 4=2 不在 H' 中，封閉失敗，所以 H' 不是子群。另外 1 的逆元是 3，也不在 H' 中。",
          "en": "H' is nonempty and contains the identity 0, but 1*1=(1+1) mod 4=2 is not in H', so closure fails and H' is not a subgroup. Moreover the inverse of 1 is 3, which is not in H' either."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "G1",
            "mko_subhom_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "目標：H' 不是子群。足夠條件：某個條件失敗；一個不在 H' 的乘積就夠了。含單位元並不能補救。",
            "en": "Goal: H' is not a subgroup. Sufficient condition: one condition fails; a single product outside H' suffices. Containing the identity does not rescue it."
          }
        },
        "leaf_basis": "mko_subhom_def + quantifier_negation + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "G3",
          "G2",
          "mod_notation"
        ],
        "before_tex": "f(0)=0,\\ f(1)=1,\\ f(2)=0,\\ f(3)=1",
        "after_tex": "f:G\\to K,\\ f(a)=a\\bmod 2",
        "relation": "implies",
        "rule": "function_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "映射表對 G 的每個元素恰給一個 K 中的值，所以 f 是函數 G→K。四格逐一比對：0、1、2、3 除以 2 的餘數是 0、1、0、1，所以 f(a)=a mod 2。",
          "en": "The mapping table assigns exactly one value in K to each element of G, so f is a function G→K. Comparing the four cells: the remainders of 0, 1, 2, 3 on division by 2 are 0, 1, 0, 1, so f(a)=a mod 2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "function_def",
            "mod_notation",
            "finite_exhaustion"
          ],
          "explanation": {
            "zh-Hant": "公式只是映射表的另一種寫法，靠四格逐一相符；這只對這張表成立。",
            "en": "The formula is another description of the mapping table, resting on all four cells matching; it holds for this table only."
          }
        },
        "leaf_basis": "function_def + finite_exhaustion + mod_notation"
      },
      {
        "id": "S09",
        "uses": [
          "division_algorithm",
          "mod_notation"
        ],
        "before_tex": "x-y=mk,\\ k\\in\\mathbb{Z}",
        "after_tex": "x\\bmod m=y\\bmod m",
        "relation": "implies",
        "rule": "L_same_rem",
        "conditions": [
          "m>0"
        ],
        "explanation": {
          "zh-Hant": "同態的論證需要引理 L_same_rem：相差 m 的整數倍的兩個整數，除以 m 的餘數相同。證明在 S09.1、S09.2，只用除法演算法的存在與唯一性。",
          "en": "The homomorphism argument needs Lemma L_same_rem: two integers that differ by an integer multiple of m leave the same remainder on division by m. The proof is in S09.1 and S09.2 and uses only existence and uniqueness in the division algorithm."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_same_rem"
          ],
          "explanation": {
            "zh-Hant": "需要 m>0 與 k 為整數；只宣稱這個方向。",
            "en": "It needs m>0 and an integer k; only this direction is claimed."
          }
        },
        "leaf_basis": "division_algorithm + mod_notation"
      },
      {
        "id": "S09.1",
        "uses": [
          "division_algorithm",
          "mod_notation"
        ],
        "before_tex": "x\\in\\mathbb{Z},\\ m>0",
        "after_tex": "x=mq+r,\\ 0\\le r<m,\\ r=x\\bmod m",
        "relation": "theorem_application",
        "rule": "division_algorithm",
        "conditions": [
          "m>0"
        ],
        "explanation": {
          "zh-Hant": "對 x 與正整數 m 使用除法演算法：有整數 q、r 使 x=mq+r 且 0≤r<m；依記號 r=x mod m。",
          "en": "Apply the division algorithm to x and the positive integer m: there are integers q, r with x=mq+r and 0≤r<m; by the notation r=x mod m."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "division_algorithm",
            "mod_notation"
          ],
          "explanation": {
            "zh-Hant": "除法演算法要求除數為正整數。",
            "en": "The division algorithm needs a positive divisor."
          }
        },
        "leaf_basis": "division_algorithm + mod_notation"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "division_algorithm"
        ],
        "before_tex": "y=x-mk=m(q-k)+r,\\ 0\\le r<m",
        "after_tex": "y\\bmod m=r=x\\bmod m",
        "relation": "implies",
        "rule": "division_algorithm",
        "conditions": [
          "q-k\\in\\mathbb{Z}"
        ],
        "explanation": {
          "zh-Hant": "y=x−mk=m(q−k)+r，q−k 是整數且 0≤r<m，所以 (q−k, r) 正是 y 除以 m 的商與餘數；依唯一性，y mod m=r=x mod m。S09.1 與 S09.2 合起來證得引理 L_same_rem。",
          "en": "y=x−mk=m(q−k)+r with q−k an integer and 0≤r<m, so (q−k, r) is exactly the quotient and remainder of y on division by m; by uniqueness, y mod m=r=x mod m. S09.1 and S09.2 together prove Lemma L_same_rem."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09.1",
            "division_algorithm",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "關鍵是除法演算法的唯一性。",
            "en": "The key is uniqueness in the division algorithm."
          }
        },
        "leaf_basis": "division_algorithm + distributive + eq_replace"
      },
      {
        "id": "S10",
        "uses": [
          "S08",
          "S09",
          "G1",
          "division_algorithm"
        ],
        "before_tex": "f(a*b)=((a+b)\\bmod 4)\\bmod 2",
        "after_tex": "(a+b)\\bmod 2",
        "relation": "equivalent",
        "rule": "L_same_rem",
        "conditions": [
          "a,b\\in G"
        ],
        "explanation": {
          "zh-Hant": "設 a+b=4q+s，s=(a+b) mod 4。則 s−(a+b)=−4q=2·(−2q) 是 2 的整數倍，依引理（m=2），s mod 2=(a+b) mod 2。a、b 是 G 中任意元素。",
          "en": "Write a+b=4q+s with s=(a+b) mod 4. Then s−(a+b)=−4q=2·(−2q) is an integer multiple of 2, so by the lemma (m=2), s mod 2=(a+b) mod 2. Here a and b are arbitrary elements of G."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_same_rem",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個整數，相等式子互換，雙向成立。關鍵是 2 整除 4。",
            "en": "Both sides are the same integer, so replacing equals works both ways. The key fact is that 2 divides 4."
          }
        },
        "leaf_basis": "L_same_rem + division_algorithm + divides_def + eq_replace"
      },
      {
        "id": "S11",
        "uses": [
          "S08",
          "S09",
          "G2",
          "division_algorithm"
        ],
        "before_tex": "f(a)\\oplus f(b)=(a\\bmod 2+b\\bmod 2)\\bmod 2",
        "after_tex": "(a+b)\\bmod 2",
        "relation": "equivalent",
        "rule": "L_same_rem",
        "conditions": [
          "a,b\\in G"
        ],
        "explanation": {
          "zh-Hant": "設 a=2p+r₁、b=2p'+r₂（r₁=a mod 2、r₂=b mod 2）。則 (r₁+r₂)−(a+b)=−2(p+p') 是 2 的倍數，依引理 (r₁+r₂) mod 2=(a+b) mod 2。",
          "en": "Write a=2p+r₁ and b=2p'+r₂ (r₁=a mod 2, r₂=b mod 2). Then (r₁+r₂)−(a+b)=−2(p+p') is a multiple of 2, so by the lemma (r₁+r₂) mod 2=(a+b) mod 2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_same_rem",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "與 S10 相同，是相等式子的互換，雙向成立。",
            "en": "As in S10, this replaces equals by equals and holds both ways."
          }
        },
        "leaf_basis": "L_same_rem + division_algorithm + eq_replace"
      },
      {
        "id": "S12",
        "uses": [
          "S10",
          "S11"
        ],
        "before_tex": "f(a*b)=(a+b)\\bmod 2=f(a)\\oplus f(b)",
        "after_tex": "\\forall a,b\\in G:\\ f(a*b)=f(a)\\oplus f(b)",
        "relation": "implies",
        "rule": "eq_refl_symm_trans",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊都等於 (a+b) mod 2，依等號的傳遞相等。S10、S11 對任意 a、b 成立，所以涵蓋全部 4²=16 個輸入對（MKO 第二段）。N5 另外逐一窮舉 16 對作交叉核對。f 是同態。",
          "en": "Both sides equal (a+b) mod 2, so they are equal by transitivity. S10 and S11 hold for arbitrary a, b, so all 4²=16 input pairs are covered (MKO paragraph 2). N5 separately enumerates the 16 pairs as a cross-check. f is a homomorphism."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "S11",
            "eq_refl_symm_trans",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "∀ 引入要求 a、b 沒有被特別取值，S10、S11 確實沒有；這不是抽樣。",
            "en": "∀-introduction needs a, b to carry no special values, and in S10 and S11 they carry none; this is not sampling."
          }
        },
        "leaf_basis": "eq_refl_symm_trans + forall_intro"
      },
      {
        "id": "S13",
        "uses": [
          "S08",
          "G2",
          "kernel_image_def"
        ],
        "before_tex": "\\ker f=\\{a\\in G\\mid f(a)=0\\}",
        "after_tex": "\\{0,2\\}=H",
        "relation": "equivalent",
        "rule": "kernel_image_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "目標單位元是 0。逐一查：f(0)=0、f(1)=1、f(2)=0、f(3)=1，所以 ker f={0,2}，正是 S06 的子群 H。",
          "en": "The target identity is 0. Checking each element: f(0)=0, f(1)=1, f(2)=0, f(3)=1, so ker f={0,2}, which is exactly the subgroup H of S06."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "kernel_image_def",
            "set_builder",
            "finite_exhaustion"
          ],
          "explanation": {
            "zh-Hant": "四個元素逐一判定，集合建構記號的成員條件是雙向的。",
            "en": "All four elements are tested, and set-builder membership is two-way."
          }
        },
        "leaf_basis": "kernel_image_def + set_builder + finite_exhaustion"
      },
      {
        "id": "S14",
        "uses": [
          "S08",
          "S13",
          "kernel_image_def"
        ],
        "before_tex": "\\operatorname{im} f,\\ f(0)=f(2)",
        "after_tex": "\\operatorname{im} f=\\{0,1\\}=K;\\ f\\ \\text{surjective, not injective}",
        "relation": "implies",
        "rule": "injective_surjective_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "像集是所有映射值 {0,1}=K，所以 f 滿射。但 f(0)=f(2)=0 而 0≠2，所以 f 不是單射；kernel 不只含單位元也反映了這點。|im f|=2<4=|G|，MKO 第三段說 image 大小等於來源大小才可能單射。f 不是雙射，因此不是同構。",
          "en": "The image, all mapped values, is {0,1}=K, so f is surjective. But f(0)=f(2)=0 with 0≠2, so f is not injective; the kernel containing more than the identity reflects this. |im f|=2<4=|G|, and the MKO's third paragraph says injectivity is possible only when the image is as large as the source. f is not a bijection, hence not an isomorphism."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "injective_surjective_def",
            "kernel_image_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "否定單射只需一對不同元素有相同像；滿射需要 K 的每個元素都被取到，兩者都已逐一核對。",
            "en": "Refuting injectivity needs one pair of distinct elements with the same image; surjectivity needs every element of K to be hit; both are checked explicitly."
          }
        },
        "leaf_basis": "injective_surjective_def + kernel_image_def + finite_counting_basics"
      },
      {
        "id": "S15",
        "uses": [
          "G3",
          "G1",
          "G2"
        ],
        "before_tex": "h(0*0)\\ \\text{vs}\\ h(0)\\oplus h(0)",
        "after_tex": "1\\neq 0=1\\oplus 1\\ \\Rightarrow\\ h\\ \\text{not a homomorphism}",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "h(0*0)=h(0)=1，但 h(0)⊕h(0)=1⊕1=(1+1) mod 2=0。1≠0，同態條件在 (0,0) 失敗；依量詞否定，h 不是同態。一個失敗對就足以否定，但要肯定同態，論證必須涵蓋全部 16 對。",
          "en": "h(0*0)=h(0)=1, but h(0)⊕h(0)=1⊕1=(1+1) mod 2=0. Since 1≠0, the homomorphism condition fails at (0,0); by quantifier negation h is not a homomorphism. One failing pair suffices to reject, whereas establishing a homomorphism requires an argument covering all 16 pairs."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "G3",
            "quantifier_negation",
            "mko_subhom_def"
          ],
          "explanation": {
            "zh-Hant": "目標：h 不是同態。足夠條件：一個輸入對違反條件；由定義推出目標。此對由查表直接得到，沒有把目標當作前提。",
            "en": "Goal: h is not a homomorphism. Sufficient condition: one input pair violates the condition, which gives the goal by the definition. That pair is read directly from the tables; the goal is never used as a premise."
          }
        },
        "leaf_basis": "quantifier_negation + numeric_evaluation"
      },
      {
        "id": "S16",
        "uses": [
          "S06",
          "S07",
          "S12",
          "S13",
          "S14",
          "S15"
        ],
        "before_tex": "H,\\ H',\\ f,\\ h",
        "after_tex": "H\\le G,\\ H'\\not\\le G,\\ f\\ \\text{hom.},\\ \\ker f=H,\\ \\operatorname{im} f=K,\\ f\\ \\text{not injective},\\ h\\ \\text{not hom.}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "合併全部結論；它們只屬於 G 與 K 這兩個指定的有限群。",
          "en": "Combine all conclusions; they concern only the two specified finite groups G and K."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S06",
            "S07",
            "S12",
            "S13",
            "S14",
            "S15",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標：全部判定。足夠條件：各步已證結論；由合取引入推出目標，沒有把目標當作前提。",
            "en": "Goal: all decisions. Sufficient condition: the conclusions proved in the listed steps, which give the goal by conjunction introduction; the goal is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "H={0,2} 非空、封閉、含單位元 0 且對逆元封閉，是 G 的子群；H'={0,1} 因 1*1=2∉H' 不是子群。f(a)=a mod 2 對全部 16 個輸入對滿足 f(a*b)=f(a)⊕f(b)，是同態；ker f={0,2}=H，im f={0,1}=K，f 滿射但因 f(0)=f(2) 不是單射，也不是同構。常數映射 h≡1 在 (0,0) 違反同態條件。",
      "en": "H={0,2} is nonempty, closed, contains the identity 0 and is closed under inverses, so it is a subgroup of G; H'={0,1} is not, since 1*1=2∉H'. f(a)=a mod 2 satisfies f(a*b)=f(a)⊕f(b) for all 16 input pairs, so it is a homomorphism; ker f={0,2}=H and im f={0,1}=K; f is surjective but, since f(0)=f(2), not injective and not an isomorphism. The constant map h≡1 violates the homomorphism condition at (0,0)."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "H'={0,1} 含單位元 0，所以是子群",
          "en": "H'={0,1} contains the identity 0, so it is a subgroup"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "1*1=2∉H'，封閉失敗；1 的逆元 3 也不在 H'。",
          "en": "1*1=2∉H', so closure fails; the inverse 3 of 1 is not in H' either."
        }
      },
      {
        "claim": {
          "zh-Hant": "f 是同態，所以 f 是單射",
          "en": "f is a homomorphism, so f is injective"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "f(0)=f(2)=0 而 0≠2。",
          "en": "f(0)=f(2)=0 although 0≠2."
        }
      },
      {
        "claim": {
          "zh-Hant": "f 是同構 G≅K",
          "en": "f is an isomorphism G≅K"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "|G|=4≠2=|K|，沒有雙射。",
          "en": "|G|=4≠2=|K|, so there is no bijection."
        }
      },
      {
        "claim": {
          "zh-Hant": "h'（0↦0、1↦1、2↦1、3↦0）是同態，因為對每個 a 都有 h'(0*a)=h'(0)⊕h'(a)",
          "en": "h' (0↦0, 1↦1, 2↦1, 3↦0) is a homomorphism because h'(0*a)=h'(0)⊕h'(a) for every a"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "h'(1*1)=h'(2)=1，而 h'(1)⊕h'(1)=0。",
          "en": "h'(1*1)=h'(2)=1 while h'(1)⊕h'(1)=0."
        }
      },
      {
        "claim": {
          "zh-Hant": "常數映射 h≡1 是同態",
          "en": "The constant map h≡1 is a homomorphism"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "h(0*0)=1≠0=h(0)⊕h(0)。",
          "en": "h(0*0)=1≠0=h(0)⊕h(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<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": "function_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "函數：f:A→B 對每個 a∈A 指定恰好一個 f(a)∈B。兩函數相等 ⇔ 定義域、對應域相同且對每個 a 都有相同值。",
          "en": "Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a."
        }
      },
      {
        "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": "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": "group_axioms",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "群公理：集合 G 配二元運算，滿足結合律、有單位元 e（eg=ge=g）、每個元素 g 有反元素 g⁻¹（gg⁻¹=g⁻¹g=e）。",
          "en": "Group axioms: a set G with a binary operation that is associative, has an identity e (eg=ge=g), and in which every g has an inverse g⁻¹ (gg⁻¹=g⁻¹g=e)."
        }
      },
      {
        "id": "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_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": "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": "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": "eq_replace",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。",
          "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true."
        }
      },
      {
        "id": "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": "eq_refl_symm_trans",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。",
          "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w."
        }
      },
      {
        "id": "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": "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": "mko_subhom_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-subgroup-homomorphism",
          "sha256": "2121cf9c71fab6aa9bfa10d881cdffbab5367fdfebbee9381b35fefec60837c0",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "有限群 G 的非空子集 H 若對運算封閉、含單位元且對逆元封閉，則為子群。群同態 f:G→K 滿足 f(a*b)=f(a)*f(b)；同態不自動為單射。",
          "en": "A nonempty subset H of a finite group G is a subgroup if it is closed under the operation, contains the identity and is closed under inverses. A homomorphism f:G→K satisfies f(a*b)=f(a)*f(b). A homomorphism is not automatically injective."
        }
      },
      {
        "id": "mod_notation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "對整數 x 與正整數 m，x mod m 表示除法演算法中的餘數 r：唯一滿足 0≤r<m 且 x=mq+r（q 為整數）的 r。",
          "en": "For an integer x and a positive integer m, x mod m denotes the remainder r of the division algorithm: the unique r with 0≤r<m and x=mq+r for some integer q."
        }
      },
      {
        "id": "finite_exhaustion",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "在有限集合上，對全部元素（或全部有序對）的命題成立，當且僅當它在列出的每一個情形都成立；逐一核完是該有限模型的完整證明，但不推及其他集合。",
          "en": "On a finite set, a statement about all elements (or all ordered pairs) holds exactly when it holds in each listed case; checking every case is a complete proof for that finite model, but says nothing about other sets."
        }
      },
      {
        "id": "kernel_image_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "對 f:G→K，ker f={a∈G ∣ f(a)=e_K}（映到目標單位元的來源元素），im f={f(a) ∣ a∈G}（所有映射值）。",
          "en": "For f:G→K, ker f={a∈G ∣ f(a)=e_K} (the source elements sent to the target identity) and im f={f(a) ∣ a∈G} (all mapped values)."
        }
      },
      {
        "id": "L_same_rem",
        "kind": "lemma",
        "proved_in": [
          "S09.1",
          "S09.2"
        ],
        "statement": {
          "zh-Hant": "設 m 為正整數、x、y 為整數。若 x−y=mk（k 為整數），則 x mod m=y mod m。",
          "en": "Let m be a positive integer and x, y integers. If x−y=mk for an integer k, then x mod m=y mod m."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "(2+2)%4",
        "rhs": "0"
      },
      {
        "id": "N2",
        "step": "S07",
        "lhs": "(1+1)%4",
        "rhs": "2"
      },
      {
        "id": "N3",
        "step": "S08",
        "lhs": "2%2",
        "rhs": "0"
      },
      {
        "id": "N4",
        "step": "S08",
        "lhs": "3%2",
        "rhs": "1"
      },
      {
        "id": "N5",
        "step": "S12",
        "vars": [
          "a",
          "b"
        ],
        "range": [
          0,
          3
        ],
        "lhs": "((a+b)%4)%2",
        "rhs": "(a%2+b%2)%2"
      },
      {
        "id": "N6",
        "step": "S09.2",
        "lhs": "17%2",
        "rhs": "(17-2*3)%2"
      },
      {
        "id": "N7",
        "step": "S15",
        "lhs": "(1+1)%2",
        "rhs": "0"
      },
      {
        "id": "N8",
        "step": "S15",
        "lhs": "1",
        "relation": "!=",
        "rhs": "(1+1)%2"
      },
      {
        "id": "N9",
        "step": "S14",
        "lhs": "4",
        "relation": "!=",
        "rhs": "2"
      },
      {
        "id": "N10",
        "step": "S12",
        "lhs": "4**2",
        "rhs": "16"
      },
      {
        "id": "N11",
        "lhs": "(1+1)%2",
        "relation": "!=",
        "rhs": "1"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理題目給定的兩個有限群與映射表（MKO 假設 a1）；結論只屬於這兩個指定的有限群，不外推到任意群，也不把同態當同構（MKO 假設 a2）。G 與 K 是群作為已知前提（可依 mko-finite-group 的方法窮舉核對）。不處理陪集、正規子群或同構定理；不改動 canonical MKO 或英文 edition。",
      "en": "Treats only the two given finite groups and mapping tables (MKO assumption a1); conclusions concern these two specified finite groups only, are not extended to arbitrary groups, and do not treat homomorphisms as isomorphisms (MKO assumption a2). That G and K are groups is a given premise (checkable exhaustively as in mko-finite-group). Cosets, normal subgroups and the isomorphism theorems are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "G=\\{0,1,2,3\\},\\ a*b=(a+b)\\bmod 4",
        "來源群",
        "the source group"
      ],
      [
        "K=\\{0,1\\},\\ a\\oplus b=(a+b)\\bmod 2",
        "目標群",
        "the target group"
      ],
      [
        "H,\\ H'",
        "兩個候選子集",
        "the two candidate subsets"
      ],
      [
        "f,\\ h",
        "兩個候選映射",
        "the two candidate maps"
      ],
      [
        "\\ker f,\\ \\operatorname{im} f",
        "核與像",
        "the kernel and the image"
      ]
    ],
    "md_error_controls": [
      {
        "title": "只核閉合不等於子群 / Checking closure alone does not define a subgroup",
        "zh": "MKO 指出的誤解之一是只核閉合就叫子群。空集合對運算（空泛地）封閉，卻不含單位元；在無限群 (ℤ,+) 中，自然數 ℕ 封閉且含 0，但 1 在 ℕ 中沒有逆元。對有限群，非空且封閉的子集其實必為子群（標準定理，本稿未證也未使用）；本稿依 MKO 逐一核四個條件。",
        "en": "One mistake the MKO names is calling a set a subgroup after checking closure only. The empty set is (vacuously) closed but contains no identity; in the infinite group (ℤ,+) the natural numbers ℕ are closed and contain 0, but 1 has no inverse in ℕ. For finite groups a nonempty closed subset is in fact a subgroup (a standard theorem, neither proved nor used here); this guide checks all four conditions as the MKO prescribes."
      },
      {
        "title": "同態不一定單射或同構 / A homomorphism need not be injective or an isomorphism",
        "zh": "MKO 的另一個誤解是把每個同態都當單射或同構。f 是同態，但 f(0)=f(2)，kernel={0,2} 不只含單位元；|G|=4≠2=|K|，兩者之間根本沒有雙射。",
        "en": "Another mistake identified in the MKO is treating every homomorphism as injective or an isomorphism. f is a homomorphism, but f(0)=f(2) and the kernel {0,2} contains more than the identity; since |G|=4≠2=|K| there is no bijection between them at all."
      },
      {
        "title": "同態要核全部輸入對 / A homomorphism proof must cover every input pair",
        "zh": "考慮 h'(0)=0、h'(1)=1、h'(2)=1、h'(3)=0。含 0 的輸入對都會通過（h'(0)=0 是目標單位元），但 h'(1*1)=h'(2)=1，而 h'(1)⊕h'(1)=0。只抽查含單位元的輸入對會誤判。本稿以對任意 a、b 成立的 S10 到 S12 涵蓋 f 的 16 對，並以 N5 窮舉交叉核對。",
        "en": "Consider h'(0)=0, h'(1)=1, h'(2)=1, h'(3)=0. Every input pair containing 0 passes (h'(0)=0 is the target identity), but h'(1*1)=h'(2)=1 while h'(1)⊕h'(1)=0. Sampling only pairs containing the identity would misjudge it. This guide covers the 16 pairs of f with S10 to S12, valid for arbitrary a, b, and cross-checks by enumeration in N5."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "G 與 K 是群作為前提（G1、G2），本稿沒有重新核它們的群公理；結論只屬於這兩個指定的有限群（MKO 假設 a2）。不處理陪集、正規子群、商群與同構定理（MKO 的 formal next）。",
        "en": "That G and K are groups is a premise (G1, G2); their group axioms are not re-checked here. The conclusions concern only these two specified finite groups (MKO assumption a2). Cosets, normal subgroups, quotient groups and the isomorphism theorems (the MKO's formal next) are not treated."
      },
      {
        "zh-Hant": "N5 是對全部 16 個輸入對的有限窮舉，對這個模型是完整的交叉核對；它依賴 S08 的公式 f(a)=a mod 2 重現映射表，S10 到 S12 的論證不依賴它。",
        "en": "N5 is a finite enumeration of all 16 input pairs, a complete cross-check for this model; it relies on the formula f(a)=a mod 2 of S08 reproducing the mapping table, and the argument of S10 to S12 does not depend on it."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "b770b6bc3f247912cf3bc8b6e85e939ce34d84eb71296cbfb81183aa78d3a291",
    "original_source_sha256": "b770b6bc3f247912cf3bc8b6e85e939ce34d84eb71296cbfb81183aa78d3a291",
    "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."
  }
}