{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-quantifier-01",
  "mko_id": "mko-quantifier",
  "title": {
    "zh-Hant": "量詞與明示論域：在有限整數論域上判定全稱、存在及其否定",
    "en": "Quantifiers and explicit domains: deciding universal and existential statements and their negations on a finite integer domain"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-quantifier-01",
    "source": {
      "mko_id": "mko-quantifier",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "3b2f73bd57aa43b5441edef2b27d44470f8cbf04a229fb31e98243708e818341",
      "english_sha256": "cf34ab77acaa732f1cc6926b00bee27e73188f1108aaad401ca0a12fc391cc57"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "forall_elim",
      "forall_intro",
      "exists_intro",
      "quantifier_negation",
      "numeric_evaluation",
      "restricted_quantifier"
    ],
    "goal": {
      "zh-Hant": "在明示有限論域 U={−2,−1,0,1,2,3} 上，判定 φ₁：∀x∈U, x²<10；φ₂：∀x∈U, x²>x；φ₃：∃x∈U, x²=2 的真假；正確寫出 φ₂ 與 φ₃ 的否定；並說明論域為空集合時全稱與存在陳述的真假。",
      "en": "On the explicit finite domain U={−2,−1,0,1,2,3}, decide the truth of φ₁: ∀x∈U, x²<10; φ₂: ∀x∈U, x²>x; φ₃: ∃x∈U, x²=2; write the negations of φ₂ and φ₃ correctly; and settle the truth of universal and existential statements when the domain is the empty set."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "論域 U={−2,−1,0,1,2,3} 以列舉方式明示，是有限整數集合；採古典真值語義（MKO 假設 a1、a2）。",
          "en": "The domain U={−2,−1,0,1,2,3} is given explicitly by listing and is a finite set of integers; classical truth semantics is used (MKO assumptions a1, a2)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "述詞：P(x) 為 x²<10，Q(x) 為 x²>x，E(x) 為 x²=2；每一個都能以精確整數計算判定。",
          "en": "Predicates: P(x) is x²<10, Q(x) is x²>x, E(x) is x²=2; each is decided by exact integer computation."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "待判定：φ₁=∀x∈U P(x)、φ₂=∀x∈U Q(x)、φ₃=∃x∈U E(x)；另判定以空集合 ∅ 為論域時的 ∀x∈∅ R(x) 與 ∃x∈∅ R(x)。",
          "en": "To decide: φ₁=∀x∈U P(x), φ₂=∀x∈U Q(x), φ₃=∃x∈U E(x); and, with the empty set ∅ as domain, ∀x∈∅ R(x) and ∃x∈∅ R(x)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "G3",
          "mko_quantifier_def"
        ],
        "before_tex": "\\forall x\\in U\\,P(x),\\quad \\exists x\\in U\\,E(x)",
        "after_tex": "\\forall x\\,(x\\in U\\Rightarrow P(x)),\\quad \\exists x\\,(x\\in U\\land E(x))",
        "relation": "definition",
        "rule": "restricted_quantifier",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 把論域與述詞都當作陳述本身的一部分，所以先固定 U={−2,−1,0,1,2,3} 與三個述詞 P(x)：x²<10、Q(x)：x²>x、E(x)：x²=2，採古典真值語義。受限量詞展開為：∀x∈U P(x) 指「對每個 x，若 x∈U 則 P(x)」；∃x∈U E(x) 指「有某個 x 同時滿足 x∈U 與 E(x)」。之後每一步都在這個明示論域上判定。",
          "en": "The MKO makes the domain and the predicate part of the statement itself, so first fix U={−2,−1,0,1,2,3} and the three predicates P(x): x²<10, Q(x): x²>x, E(x): x²=2, under classical truth semantics. Restricted quantifiers unfold as follows: ∀x∈U P(x) means 'for every x, if x∈U then P(x)'; ∃x∈U E(x) means 'some x satisfies both x∈U and E(x)'. Every later step is decided on this explicit domain."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_quantifier_def",
            "restricted_quantifier"
          ],
          "explanation": {
            "zh-Hant": "此步只固定論域、述詞與展開方式，不推出新事實。若論域沒有明示，後面的枚舉與否定都沒有確定的意義。",
            "en": "This step only fixes the domain, the predicates and the unfolding, and derives no new fact. Without an explicit domain, the later enumeration and negations would have no definite meaning."
          }
        },
        "leaf_basis": "mko_quantifier_def + restricted_quantifier"
      },
      {
        "id": "S02",
        "uses": [
          "G1"
        ],
        "before_tex": "(-2)^2,\\ (-1)^2,\\ 0^2,\\ 1^2,\\ 2^2,\\ 3^2",
        "after_tex": "4,\\ 1,\\ 0,\\ 1,\\ 4,\\ 9",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對 U 的六個元素逐一精確計算平方：(−2)²=4、(−1)²=1、0²=0、1²=1、2²=4、3²=9。負數的平方用 (−u)(−u)=u·u。這張表之後由 P、Q、E 三個述詞共用。",
          "en": "Square each of the six elements of U exactly: (−2)²=4, (−1)²=1, 0²=0, 1²=1, 2²=4, 3²=9. Squares of negatives use (−u)(−u)=u·u. This table is shared later by the three predicates P, Q and E."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "每個等式都可獨立重算（N1、N2）；等號兩邊是同一個整數，雙向成立。",
            "en": "Each equality can be recomputed independently (N1, N2); both sides are the same integer, so the step holds both ways."
          }
        },
        "leaf_basis": "numeric_evaluation + neg_mul"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "G2"
        ],
        "before_tex": "x^2\\in\\{4,1,0,1,4,9\\}\\ \\text{for the six}\\ x\\in U",
        "after_tex": "P(-2)\\land P(-1)\\land P(0)\\land P(1)\\land P(2)\\land P(3)",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "P(x) 是 x²<10。由 S02，六個平方值 4、1、0、1、4、9 都小於 10（最大的是 9<10），所以 P 在 U 的每一個列出元素上都成立；再用合取把六個事實合在一起。",
          "en": "P(x) is x²<10. By S02 the six squares 4, 1, 0, 1, 4, 9 are all less than 10 (the largest is 9<10), so P holds at every listed element of U; conjunction then combines the six facts."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "numeric_evaluation",
            "order_basics",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "需要的是 U 的每一個元素都被檢查到；六個比較都是精確的整數比較。這一步不宣稱反向。",
            "en": "What is needed is that every element of U is checked; the six comparisons are exact integer comparisons. No converse is claimed."
          }
        },
        "leaf_basis": "numeric_evaluation + order_basics + and_intro_elim"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "G1",
          "L_finite_forall"
        ],
        "before_tex": "P(-2)\\land P(-1)\\land P(0)\\land P(1)\\land P(2)\\land P(3)",
        "after_tex": "\\forall x\\in U\\,P(x)",
        "relation": "equivalent",
        "rule": "L_finite_forall",
        "conditions": [
          "U=\\{-2,-1,0,1,2,3\\}\\ \\text{listed completely}"
        ],
        "explanation": {
          "zh-Hant": "U 以列舉給出：a₁=−2、a₂=−1、a₃=0、a₄=1、a₅=2、a₆=3。依引理 L_finite_forall（n=6），∀x∈U P(x) 等價於六個 P(a_i) 的合取，而 S03 已證得這個合取，所以 φ₁ 為真。這是完整枚舉：結論成立正是因為 U 的每個元素都被檢查過。",
          "en": "U is given by listing: a₁=−2, a₂=−1, a₃=0, a₄=1, a₅=2, a₆=3. By Lemma L_finite_forall (n=6), ∀x∈U P(x) is equivalent to the conjunction of the six P(a_i), which S03 established, so φ₁ is true. This is complete enumeration: the conclusion holds precisely because every element of U was checked."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_finite_forall"
          ],
          "explanation": {
            "zh-Hant": "引理兩個方向都在 S04.1、S04.2 證明；反向即由 φ₁ 取回每個 P(a_i)。",
            "en": "Both directions of the lemma are proved in S04.1 and S04.2; backwards, φ₁ gives back each P(a_i)."
          }
        },
        "leaf_basis": "forall_elim + forall_intro + case_analysis + roster_membership"
      },
      {
        "id": "S04.1",
        "uses": [
          "roster_membership",
          "forall_elim"
        ],
        "before_tex": "\\forall x\\in U\\,R(x)",
        "after_tex": "\\bigwedge_{i=1}^{n}R(a_i)",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [
          "U=\\{a_i\\mid 1\\le i\\le n\\}"
        ],
        "explanation": {
          "zh-Hant": "設 U={a_i ∣ 1≤i≤n}，R 為任意述詞。對每個 i，由列舉成員關係得 a_i∈U；對全稱 ∀x∈U R(x) 用 ∀消去代入 x=a_i，得 R(a_i)。把 n 個結果用合取合在一起。這是引理的「⇒」方向。",
          "en": "Let U={a_i ∣ 1≤i≤n} and let R be any predicate. For each i, roster membership gives a_i∈U; ∀-elimination of ∀x∈U R(x) at x=a_i gives R(a_i). Combine the n results by conjunction. This is the '⇒' direction of the lemma."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "roster_membership",
            "forall_elim",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "每次代入都需要 a_i∈U，這由列舉保證；另一個方向在 S04.2 另行證明，此步不宣稱。",
            "en": "Each instantiation needs a_i∈U, which the listing guarantees; the other direction is proved separately in S04.2 and is not claimed here."
          }
        },
        "leaf_basis": "forall_elim + roster_membership + and_intro_elim"
      },
      {
        "id": "S04.2",
        "uses": [
          "roster_membership",
          "case_analysis",
          "eq_replace",
          "forall_intro"
        ],
        "before_tex": "\\bigwedge_{i=1}^{n}R(a_i)",
        "after_tex": "\\forall x\\in U\\,R(x)",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [
          "U=\\{a_i\\mid 1\\le i\\le n\\}"
        ],
        "explanation": {
          "zh-Hant": "假設合取成立。任取 x∈U，對 x 不加其他假設。由列舉成員關係，x=a_i 對某個 i 成立，共有 n 種情況。在情況 x=a_i 中，合取給出 R(a_i)，以相等的 x 代換 a_i 得 R(x)。n 種情況涵蓋 U 的全部元素，所以 R(x) 成立；x 是任意的，依 ∀引入得 ∀x∈U R(x)。S04.1 與 S04.2 合起來證得引理 L_finite_forall。",
          "en": "Assume the conjunction. Take an arbitrary x∈U with no further assumption on x. By roster membership, x=a_i for some i, giving n cases. In the case x=a_i the conjunction gives R(a_i), and replacing a_i by the equal x gives R(x). The n cases cover every element of U, so R(x) holds; since x was arbitrary, ∀-introduction gives ∀x∈U R(x). S04.1 and S04.2 together prove Lemma L_finite_forall."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "roster_membership",
            "case_analysis",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "關鍵是情況分析涵蓋全部可能，這只在 U 被完整列出時成立；若 U 無限或只列出一部分，有限個情況涵蓋不了，引理就不能用。",
            "en": "The key is that the case analysis covers every possibility, which holds only because U is listed completely; if U were infinite or only partly listed, finitely many cases could not cover it and the lemma would not apply."
          }
        },
        "leaf_basis": "forall_intro + case_analysis + roster_membership + eq_replace"
      },
      {
        "id": "S05",
        "uses": [
          "S04",
          "forall_elim",
          "contradiction"
        ],
        "before_tex": "\\forall x\\in U\\,P(x)\\ \\text{is true}",
        "after_tex": "\\neg\\forall x\\in\\mathbb{Z}\\,P(x)",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [
          "4\\in\\mathbb{Z}",
          "4\\notin U"
        ],
        "explanation": {
          "zh-Hant": "φ₁ 為真不能外推到 ℤ。若 ∀x∈ℤ P(x) 成立，代入 4∈ℤ 得 4²<10，即 16<10，與 16≥10 矛盾，所以 ∀x∈ℤ, x²<10 為假。MKO 強調完整枚舉只判定指定有限論域上的陳述：4 不在 U 中，所以它不影響 φ₁，但它推翻了把論域換成 ℤ 之後的另一個陳述。",
          "en": "The truth of φ₁ does not extend to ℤ. If ∀x∈ℤ P(x) held, instantiating at 4∈ℤ would give 4²<10, that is 16<10, contradicting 16≥10; so ∀x∈ℤ, x²<10 is false. The MKO stresses that complete enumeration settles the statement only on the specified finite domain: 4 is not in U, so it does not affect φ₁, but it refutes the different statement obtained by changing the domain to ℤ."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "forall_elim",
            "contradiction",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "此步需要 4∈ℤ 與精確比較 16≥10（N4）；它不改變 φ₁ 的真假，只說明換論域就是換陳述。",
            "en": "The step needs 4∈ℤ and the exact comparison 16≥10 (N4); it does not change the truth of φ₁, it only shows that changing the domain changes the statement."
          }
        },
        "leaf_basis": "forall_elim + contradiction + numeric_evaluation + order_basics"
      },
      {
        "id": "S06",
        "uses": [
          "S02",
          "G2"
        ],
        "before_tex": "Q(0):\\ 0^2>0",
        "after_tex": "0\\in U\\land\\neg Q(0)",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "Q(x) 是 x²>x。取 x=0：0²=0，而 0>0 不成立（序的三分律：0=0 時 0>0 與 0<0 都不成立）。0 是 U 的列出元素，所以 0∈U 且 ¬Q(0)。x=1 也有 1>1 不成立，但一個見證已足夠。",
          "en": "Q(x) is x²>x. Take x=0: 0²=0, and 0>0 fails (by trichotomy, 0=0 rules out both 0>0 and 0<0). 0 is a listed element of U, so 0∈U and ¬Q(0). x=1 also gives 1>1, which fails, but one witness is enough."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "numeric_evaluation",
            "order_basics",
            "roster_membership"
          ],
          "explanation": {
            "zh-Hant": "見證必須同時在 U 中並使 Q 失敗，兩項都已核對（N5）；不宣稱 0 是唯一見證。",
            "en": "A witness must both lie in U and make Q fail; both are checked (N5). It is not claimed that 0 is the only witness."
          }
        },
        "leaf_basis": "numeric_evaluation + order_basics + roster_membership"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "exists_intro"
        ],
        "before_tex": "0\\in U\\land\\neg Q(0)",
        "after_tex": "\\exists x\\in U\\,\\neg Q(x)",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 ∃引入，U 中的 0 使 ¬Q(0) 成立，所以 ∃x∈U ¬Q(x) 為真。存在陳述只需一個允許的見證。",
          "en": "By ∃-introduction, the element 0 of U satisfies ¬Q(0), so ∃x∈U ¬Q(x) is true. An existential statement needs only one admissible witness."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "需要見證在論域 U 中；反向（由存在陳述推回見證就是 0）並未宣稱。",
            "en": "The witness must be in the domain U; the converse (recovering from the existential that the witness is 0) is not claimed."
          }
        },
        "leaf_basis": "exists_intro"
      },
      {
        "id": "S08",
        "uses": [
          "S07",
          "L_restricted_negation"
        ],
        "before_tex": "\\exists x\\in U\\,\\neg Q(x)",
        "after_tex": "\\neg\\forall x\\in U\\,Q(x)",
        "relation": "equivalent",
        "rule": "L_restricted_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_restricted_negation 第 (i) 部分，∃x∈U ¬Q(x) 與 ¬∀x∈U Q(x) 等價。所以 φ₂ 的否定為真，φ₂ 為假。否定全稱時要同時把 ∀ 換成 ∃ 並否定述詞，論域 U 保持不變。",
          "en": "By part (i) of Lemma L_restricted_negation, ∃x∈U ¬Q(x) is equivalent to ¬∀x∈U Q(x). So the negation of φ₂ is true and φ₂ is false. Negating a universal switches ∀ to ∃ and negates the predicate at the same time, while the domain U stays the same."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_restricted_negation"
          ],
          "explanation": {
            "zh-Hant": "引理第 (i) 部分的兩個方向都在 S08.1、S08.2 證明。",
            "en": "Both directions of part (i) of the lemma are proved in S08.1 and S08.2."
          }
        },
        "leaf_basis": "quantifier_negation + classical_implication + restricted_quantifier"
      },
      {
        "id": "S08.1",
        "uses": [
          "restricted_quantifier",
          "quantifier_negation"
        ],
        "before_tex": "\\neg\\forall x\\in U\\,R(x)",
        "after_tex": "\\exists x\\,\\neg(x\\in U\\Rightarrow R(x))",
        "relation": "equivalent",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對任意述詞 R：先把受限全稱展開成 ∀x(x∈U⇒R(x))，再用無限制的量詞否定律 ¬∀x ψ(x) ⇔ ∃x ¬ψ(x)，其中 ψ(x) 是 x∈U⇒R(x)。",
          "en": "For any predicate R: first unfold the restricted universal to ∀x(x∈U⇒R(x)), then apply the unrestricted negation law ¬∀x ψ(x) ⇔ ∃x ¬ψ(x) with ψ(x) being x∈U⇒R(x)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "restricted_quantifier",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "展開定義與量詞否定律都是雙向等價。",
            "en": "Unfolding the definition and the quantifier negation law are both two-way equivalences."
          }
        },
        "leaf_basis": "restricted_quantifier + quantifier_negation"
      },
      {
        "id": "S08.2",
        "uses": [
          "S08.1",
          "classical_implication"
        ],
        "before_tex": "\\exists x\\,\\neg(x\\in U\\Rightarrow R(x))",
        "after_tex": "\\exists x\\,(x\\in U\\land\\neg R(x))=\\exists x\\in U\\,\\neg R(x)",
        "relation": "equivalent",
        "rule": "classical_implication",
        "conditions": [],
        "explanation": {
          "zh-Hant": "古典語義下 ¬(A⇒B) ⇔ A∧¬B（因 A⇒B 等價於 ¬A∨B，再用德摩根律與雙重否定）。對每個 x 取 A 為 x∈U、B 為 R(x)，得 ∃x(x∈U∧¬R(x))，也就是 ∃x∈U ¬R(x)。S08.1 與 S08.2 證得引理第 (i) 部分。",
          "en": "Under classical semantics ¬(A⇒B) ⇔ A∧¬B (since A⇒B is equivalent to ¬A∨B; then apply De Morgan and double negation). Taking A as x∈U and B as R(x) for each x gives ∃x(x∈U∧¬R(x)), which is ∃x∈U ¬R(x). S08.1 and S08.2 prove part (i) of the lemma."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "classical_implication",
            "de_morgan_logic"
          ],
          "explanation": {
            "zh-Hant": "每個 x 處的替換都是等價，所以在存在量詞之下也雙向成立。此步依賴古典語義，MKO 假設 a1 已明示採用。",
            "en": "The replacement at each x is an equivalence, so it holds both ways under the existential. The step relies on classical semantics, which MKO assumption a1 adopts explicitly."
          }
        },
        "leaf_basis": "classical_implication + de_morgan_logic + restricted_quantifier"
      },
      {
        "id": "S08.3",
        "uses": [
          "restricted_quantifier",
          "quantifier_negation",
          "de_morgan_logic",
          "classical_implication"
        ],
        "before_tex": "\\neg\\exists x\\in U\\,R(x)",
        "after_tex": "\\forall x\\in U\\,\\neg R(x)",
        "relation": "equivalent",
        "rule": "de_morgan_logic",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第 (ii) 部分：¬∃x(x∈U∧R(x)) ⇔ ∀x ¬(x∈U∧R(x))（量詞否定律）⇔ ∀x(x∉U ∨ ¬R(x))（德摩根律）⇔ ∀x(x∈U⇒¬R(x))（古典蘊涵 ¬A∨B ⇔ A⇒B），也就是 ∀x∈U ¬R(x)。S08.3 證得引理第 (ii) 部分。",
          "en": "Part (ii): ¬∃x(x∈U∧R(x)) ⇔ ∀x ¬(x∈U∧R(x)) (quantifier negation) ⇔ ∀x(x∉U ∨ ¬R(x)) (De Morgan) ⇔ ∀x(x∈U⇒¬R(x)) (classical implication ¬A∨B ⇔ A⇒B), which is ∀x∈U ¬R(x). S08.3 proves part (ii) of the lemma."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "quantifier_negation",
            "de_morgan_logic",
            "classical_implication",
            "logic_chain"
          ],
          "explanation": {
            "zh-Hant": "鏈中每一環都是等價，依推論的傳遞，整條鏈雙向成立。",
            "en": "Every link of the chain is an equivalence, so by chaining the whole chain holds both ways."
          }
        },
        "leaf_basis": "quantifier_negation + de_morgan_logic + classical_implication + restricted_quantifier + logic_chain"
      },
      {
        "id": "S09",
        "uses": [
          "S02",
          "G2",
          "forall_elim",
          "contradiction"
        ],
        "before_tex": "\\forall x\\in U\\,\\neg Q(x)",
        "after_tex": "\\neg\\forall x\\in U\\,\\neg Q(x)\\quad(\\text{since } 2\\in U,\\ 2^2>2)",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [
          "2\\in U"
        ],
        "explanation": {
          "zh-Hant": "常見錯誤是把 ¬∀x∈U Q(x) 寫成 ∀x∈U ¬Q(x)（「所有元素都不滿足」）。核對後者：若它成立，代入 2∈U 得 ¬Q(2)，即 4>2 不成立；但 2²=4>2 成立，矛盾。所以 ∀x∈U ¬Q(x) 為假，而 S07、S08 證得正確的否定 ∃x∈U ¬Q(x) 為真。兩者真假不同，因此不等價。",
          "en": "A common error writes ¬∀x∈U Q(x) as ∀x∈U ¬Q(x) ('every element fails'). Test the latter: if it held, instantiating at 2∈U would give ¬Q(2), that is, 4>2 would fail; but 2²=4>2 holds, a contradiction. So ∀x∈U ¬Q(x) is false, whereas S07 and S08 showed that the correct negation ∃x∈U ¬Q(x) is true. The two have different truth values, so they are not equivalent."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "forall_elim",
            "contradiction",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "需要 2∈U 與精確比較 4>2（N6）。此步只判定錯誤寫法為假，不影響 S08 的結論。",
            "en": "It needs 2∈U and the exact comparison 4>2 (N6). The step only shows the wrong form is false and does not affect the conclusion of S08."
          }
        },
        "leaf_basis": "forall_elim + contradiction + numeric_evaluation + order_basics"
      },
      {
        "id": "S10",
        "uses": [
          "S02",
          "G2"
        ],
        "before_tex": "x^2\\in\\{4,1,0,1,4,9\\}\\ \\text{for the six}\\ x\\in U",
        "after_tex": "\\neg E(-2)\\land\\neg E(-1)\\land\\neg E(0)\\land\\neg E(1)\\land\\neg E(2)\\land\\neg E(3)",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "E(x) 是 x²=2。由 S02，六個平方值 4、1、0、1、4、9 都不等於 2，這些都是精確的整數比較。所以 E 在每個列出元素上都失敗。",
          "en": "E(x) is x²=2. By S02 none of the six squares 4, 1, 0, 1, 4, 9 equals 2; these are exact integer comparisons. So E fails at every listed element."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "numeric_evaluation",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "U 的六個元素都必須檢查；漏掉任何一個，S11 就不能進行。",
            "en": "All six elements of U must be checked; if any were omitted, S11 could not be taken."
          }
        },
        "leaf_basis": "numeric_evaluation + and_intro_elim"
      },
      {
        "id": "S11",
        "uses": [
          "S10",
          "L_finite_forall"
        ],
        "before_tex": "\\neg E(-2)\\land\\neg E(-1)\\land\\neg E(0)\\land\\neg E(1)\\land\\neg E(2)\\land\\neg E(3)",
        "after_tex": "\\forall x\\in U\\,\\neg E(x)",
        "relation": "equivalent",
        "rule": "L_finite_forall",
        "conditions": [
          "U\\ \\text{listed completely}"
        ],
        "explanation": {
          "zh-Hant": "把引理 L_finite_forall 用在述詞 ¬E 上（引理對任何述詞 R 成立）：六個 ¬E(a_i) 的合取等價於 ∀x∈U ¬E(x)。",
          "en": "Apply Lemma L_finite_forall to the predicate ¬E (the lemma holds for any predicate R): the conjunction of the six ¬E(a_i) is equivalent to ∀x∈U ¬E(x)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_finite_forall"
          ],
          "explanation": {
            "zh-Hant": "同 S04，引理兩個方向都已證。",
            "en": "As in S04, both directions of the lemma are proved."
          }
        },
        "leaf_basis": "forall_elim + forall_intro + case_analysis + roster_membership"
      },
      {
        "id": "S12",
        "uses": [
          "S11",
          "L_restricted_negation"
        ],
        "before_tex": "\\forall x\\in U\\,\\neg E(x)",
        "after_tex": "\\neg\\exists x\\in U\\,E(x)",
        "relation": "equivalent",
        "rule": "L_restricted_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理第 (ii) 部分，∀x∈U ¬E(x) 等價於 ¬∃x∈U E(x)。所以 φ₃ 為假，它的否定 ∀x∈U, x²≠2 為真。要證存在陳述為假，必須檢查全部允許元素；只找到一個失敗值不夠。",
          "en": "By part (ii) of the lemma, ∀x∈U ¬E(x) is equivalent to ¬∃x∈U E(x). So φ₃ is false, and its negation ∀x∈U, x²≠2 is true. To show an existential false, every admissible element must be checked; finding one failure is not enough."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_restricted_negation"
          ],
          "explanation": {
            "zh-Hant": "第 (ii) 部分在 S08.3 證明，雙向成立。",
            "en": "Part (ii) is proved in S08.3 and holds both ways."
          }
        },
        "leaf_basis": "quantifier_negation + de_morgan_logic + classical_implication + restricted_quantifier"
      },
      {
        "id": "S13",
        "uses": [
          "G2",
          "sqrt_def",
          "exists_intro"
        ],
        "before_tex": "\\neg\\exists x\\in U\\,E(x)",
        "after_tex": "\\exists x\\in\\mathbb{R}\\,E(x)",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [
          "2\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "把同一個述詞 E 放到論域 ℝ：2≥0，所以 √2 存在，√2∈ℝ 且 (√2)²=2；依 ∃引入，∃x∈ℝ, x²=2 為真。φ₃ 在 U 上為假、在 ℝ 上為真，說明論域是陳述本身的一部分，不能省略或任意替換。",
          "en": "Put the same predicate E on the domain ℝ: since 2≥0, √2 exists, √2∈ℝ and (√2)²=2; by ∃-introduction, ∃x∈ℝ, x²=2 is true. φ₃ is false on U and true on ℝ, which shows that the domain is part of the statement and may not be dropped or swapped freely."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "sqrt_def",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "需要 √2 的存在（主平方根定義）；此步不改變 φ₃ 在 U 上的結論。",
            "en": "It needs the existence of √2 (principal square root); it does not change the verdict on φ₃ over U."
          }
        },
        "leaf_basis": "sqrt_def + exists_intro"
      },
      {
        "id": "S14",
        "uses": [
          "restricted_quantifier",
          "empty_set_def",
          "classical_implication",
          "forall_intro"
        ],
        "before_tex": "\\forall x\\in\\varnothing\\,R(x)",
        "after_tex": "\\forall x\\,(x\\in\\varnothing\\Rightarrow R(x))\\ \\text{holds}",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把論域換成空集合。展開為 ∀x(x∈∅⇒R(x))。任取 x：x∈∅ 為假，前件為假的蘊涵在古典語義下為真；x 是任意的，依 ∀引入，全稱為真。沒有任何元素能成為反例，所以不論 R 是什麼，∀x∈∅ R(x) 都為真。",
          "en": "Replace the domain by the empty set. Unfold to ∀x(x∈∅⇒R(x)). For an arbitrary x, x∈∅ is false, and an implication with a false antecedent is true under classical semantics; since x is arbitrary, ∀-introduction makes the universal true. No element can be a counterexample, so ∀x∈∅ R(x) is true whatever R is."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "empty_set_def",
            "classical_implication",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要「∅ 沒有元素」與古典蘊涵的真值；這不表示 R 對任何東西成立。",
            "en": "It needs that ∅ has no elements and the classical truth value of implication; it does not mean R holds of anything."
          }
        },
        "leaf_basis": "restricted_quantifier + empty_set_def + classical_implication + forall_intro"
      },
      {
        "id": "S15",
        "uses": [
          "restricted_quantifier",
          "empty_set_def",
          "and_intro_elim",
          "quantifier_negation"
        ],
        "before_tex": "\\exists x\\in\\varnothing\\,R(x)",
        "after_tex": "\\neg\\exists x\\,(x\\in\\varnothing\\land R(x))",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "∃x∈∅ R(x) 展開為 ∃x(x∈∅∧R(x))。對每個 x，合取的第一項 x∈∅ 為假，所以合取為假，即 ∀x ¬(x∈∅∧R(x))；依量詞否定律，這就是 ¬∃x(x∈∅∧R(x))。空論域沒有可用的見證，存在陳述為假。",
          "en": "∃x∈∅ R(x) unfolds to ∃x(x∈∅∧R(x)). For every x the first conjunct x∈∅ is false, so the conjunction is false, that is ∀x ¬(x∈∅∧R(x)); by quantifier negation this is ¬∃x(x∈∅∧R(x)). The empty domain offers no witness, so the existential is false."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "empty_set_def",
            "and_intro_elim",
            "forall_intro",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "需要對每個 x 證合取為假（任意 x 後用 ∀引入），再用量詞否定律。",
            "en": "It needs the conjunction to be false for every x (arbitrary x, then ∀-introduction), followed by quantifier negation."
          }
        },
        "leaf_basis": "restricted_quantifier + empty_set_def + and_intro_elim + forall_intro + quantifier_negation"
      },
      {
        "id": "S16",
        "uses": [
          "S04",
          "S08",
          "S09",
          "S12",
          "S14",
          "S15"
        ],
        "before_tex": "\\text{S04, S08, S12, S14, S15}",
        "after_tex": "\\varphi_1\\ \\text{true},\\ \\varphi_2\\ \\text{false},\\ \\varphi_3\\ \\text{false};\\ \\forall x\\in\\varnothing\\ \\text{true},\\ \\exists x\\in\\varnothing\\ \\text{false}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整：φ₁ 為真（S04，完整枚舉）；φ₂ 為假，其否定是 ∃x∈U, x²≤x，見證 x=0（S08），而不是 ∀x∈U, x²≤x（S09 證其為假）；φ₃ 為假，其否定是 ∀x∈U, x²≠2（S12）；空論域上全稱為真、存在為假（S14、S15）。",
          "en": "Summary: φ₁ is true (S04, complete enumeration); φ₂ is false, and its negation is ∃x∈U, x²≤x with witness x=0 (S08), not ∀x∈U, x²≤x (shown false in S09); φ₃ is false, and its negation is ∀x∈U, x²≠2 (S12); over the empty domain the universal is true and the existential is false (S14, S15)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S04",
            "S08",
            "S12",
            "S14",
            "S15",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：判定 φ₁、φ₂、φ₃，寫出正確否定，並處理空論域邊界。足夠條件 H：S04、S08、S12、S14、S15 五個結論。H⇒G 只是合取；H 的每一部分都由前面的步驟正向證成，沒有把 G 當作前提。",
            "en": "Goal G: decide φ₁, φ₂, φ₃, write the correct negations and settle the empty-domain boundary. Sufficient condition H: the five conclusions S04, S08, S12, S14, S15. H⇒G is just conjunction; each part of H is proved forwards by earlier steps, and G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在 U={−2,−1,0,1,2,3} 上：φ₁（∀x∈U, x²<10）為真，由完整枚舉證得；φ₂（∀x∈U, x²>x）為假，其否定為 ∃x∈U, x²≤x，見證 x=0；φ₃（∃x∈U, x²=2）為假，其否定為 ∀x∈U, x²≠2。論域為空集合時，∀x∈∅ R(x) 為真，∃x∈∅ R(x) 為假。這些結論只對所述論域成立：x²<10 在 ℤ 上不全成立（x=4），x²=2 在 ℝ 上有解（√2）。",
      "en": "On U={−2,−1,0,1,2,3}: φ₁ (∀x∈U, x²<10) is true, proved by complete enumeration; φ₂ (∀x∈U, x²>x) is false, and its negation is ∃x∈U, x²≤x with witness x=0; φ₃ (∃x∈U, x²=2) is false, and its negation is ∀x∈U, x²≠2. With the empty set as domain, ∀x∈∅ R(x) is true and ∃x∈∅ R(x) is false. These conclusions hold only for the stated domains: x²<10 does not hold on all of ℤ (x=4), and x²=2 has a solution in ℝ (√2)."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "−2、0、3 都滿足 x²<10，所以 ∀x∈ℤ, x²<10",
          "en": "−2, 0 and 3 all satisfy x²<10, so ∀x∈ℤ, x²<10"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x=4∈ℤ 時 4²=16≥10；幾個通過例子不證明無限論域上的全稱。",
          "en": "At x=4∈ℤ, 4²=16≥10; a few passing examples do not prove a universal over an infinite domain."
        }
      },
      {
        "claim": {
          "zh-Hant": "¬∀x∈U, x²>x 的意思是每個 x∈U 都有 x²≤x",
          "en": "¬∀x∈U, x²>x means every x∈U has x²≤x"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "正確否定是 ∃x∈U, x²≤x；而 x=2 時 4>2，所以「每個都 x²≤x」為假。",
          "en": "The correct negation is ∃x∈U, x²≤x; and at x=2, 4>2, so 'every x has x²≤x' is false."
        }
      },
      {
        "claim": {
          "zh-Hant": "∃x∈U, x²=2 為假，所以 ∃x∈ℝ, x²=2 也為假",
          "en": "∃x∈U, x²=2 is false, so ∃x∈ℝ, x²=2 is false too"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "論域不同就是陳述不同；√2∈ℝ 且 (√2)²=2。",
          "en": "A different domain is a different statement; √2∈ℝ and (√2)²=2."
        }
      },
      {
        "claim": {
          "zh-Hant": "∀x∈U ∃y∈U, x+y=0 為真，因為每個整數都有加法反元素",
          "en": "∀x∈U ∃y∈U, x+y=0 is true, because every integer has an additive inverse"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x=3 時唯一滿足 3+y=0 的 y 是 −3，但 −3∉U（−3<−2）；論域限制了可用的見證。",
          "en": "At x=3 the only y with 3+y=0 is −3, but −3∉U (−3<−2); the domain restricts which witnesses are available."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "exists_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。",
          "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)."
        }
      },
      {
        "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": "neg_mul",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。",
          "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "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": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "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": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "de_morgan_logic",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "邏輯的德摩根律：¬(P∧Q) ⇔ ¬P∨¬Q；¬(P∨Q) ⇔ ¬P∧¬Q。",
          "en": "De Morgan's laws for logic: ¬(P∧Q) ⇔ ¬P∨¬Q and ¬(P∨Q) ⇔ ¬P∧¬Q."
        }
      },
      {
        "id": "logic_chain",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "推論的傳遞：由 P⇒Q 與 Q⇒R 得 P⇒R；⇔ 同理。",
          "en": "Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔."
        }
      },
      {
        "id": "sqrt_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "主平方根：對 a≥0，√a 是唯一滿足 b≥0 且 b²=a 的實數 b。負數在實數中沒有平方根。",
          "en": "Principal square root: for a≥0, √a is the unique real b with b≥0 and b²=a. Negative numbers have no real square root."
        }
      },
      {
        "id": "mko_quantifier_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-quantifier",
          "sha256": "3b2f73bd57aa43b5441edef2b27d44470f8cbf04a229fb31e98243708e818341",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在明示論域 U 與述詞 P 下，∀x∈U P(x) 表示每個 U 中元素滿足 P，∃x∈U P(x) 表示至少一個 U 中元素滿足 P。採古典真值語義，¬∀x∈U P(x) 等價於 ∃x∈U ¬P(x)，¬∃x∈U P(x) 等價於 ∀x∈U ¬P(x)。",
          "en": "For an explicit domain U and predicate P, ∀x∈U P(x) states that every element of U satisfies P, while ∃x∈U P(x) states that at least one element of U satisfies P. Under classical truth semantics, ¬∀x∈U P(x) is equivalent to ∃x∈U ¬P(x), and ¬∃x∈U P(x) is equivalent to ∀x∈U ¬P(x)."
        }
      },
      {
        "id": "restricted_quantifier",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "受限量詞的意義：∀x∈U R(x) 表示 ∀x(x∈U⇒R(x))；∃x∈U R(x) 表示 ∃x(x∈U∧R(x))。",
          "en": "Meaning of restricted quantifiers: ∀x∈U R(x) means ∀x(x∈U⇒R(x)); ∃x∈U R(x) means ∃x(x∈U∧R(x))."
        }
      },
      {
        "id": "roster_membership",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "列舉集合的成員關係：對以列舉給出的有限集合 U={a_i ∣ 1≤i≤n}，x∈U 當且僅當存在 i（1≤i≤n）使 x=a_i。",
          "en": "Membership in a listed set: for a finite set given by listing, U={a_i ∣ 1≤i≤n}, x∈U holds exactly when x=a_i for some i with 1≤i≤n."
        }
      },
      {
        "id": "classical_implication",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "古典蘊涵：(A⇒B) ⇔ (¬A∨B)；因此 ¬(A⇒B) ⇔ (A∧¬B)，且前件 A 為假時 A⇒B 為真。",
          "en": "Classical implication: (A⇒B) ⇔ (¬A∨B); hence ¬(A⇒B) ⇔ (A∧¬B), and A⇒B is true whenever the antecedent A is false."
        }
      },
      {
        "id": "empty_set_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "空集合：對每個 x，x∉∅。",
          "en": "Empty set: for every x, x∉∅."
        }
      },
      {
        "id": "L_finite_forall",
        "kind": "lemma",
        "proved_in": [
          "S04.1",
          "S04.2"
        ],
        "statement": {
          "zh-Hant": "對以列舉給出的有限集合 U={a_i ∣ 1≤i≤n} 與任意述詞 R：∀x∈U R(x) ⇔ ⋀_{i=1}^{n} R(a_i)。",
          "en": "For a finite set given by listing, U={a_i ∣ 1≤i≤n}, and any predicate R: ∀x∈U R(x) ⇔ ⋀_{i=1}^{n} R(a_i)."
        }
      },
      {
        "id": "L_restricted_negation",
        "kind": "lemma",
        "proved_in": [
          "S08.1",
          "S08.2",
          "S08.3"
        ],
        "statement": {
          "zh-Hant": "古典語義下，對任意論域 U 與述詞 R：(i) ¬∀x∈U R(x) ⇔ ∃x∈U ¬R(x)；(ii) ¬∃x∈U R(x) ⇔ ∀x∈U ¬R(x)。",
          "en": "Under classical semantics, for any domain U and predicate R: (i) ¬∀x∈U R(x) ⇔ ∃x∈U ¬R(x); (ii) ¬∃x∈U R(x) ⇔ ∀x∈U ¬R(x)."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "(-2)**2",
        "rhs": "4"
      },
      {
        "id": "N2",
        "step": "S02",
        "lhs": "3**2",
        "rhs": "9"
      },
      {
        "id": "N3",
        "step": "S03",
        "vars": [
          "x"
        ],
        "range": [
          -2,
          3
        ],
        "lhs": "x*x",
        "relation": "<",
        "rhs": "10"
      },
      {
        "id": "N4",
        "step": "S05",
        "lhs": "4**2",
        "relation": ">=",
        "rhs": "10"
      },
      {
        "id": "N5",
        "step": "S06",
        "lhs": "0**2",
        "relation": "<=",
        "rhs": "0"
      },
      {
        "id": "N6",
        "step": "S09",
        "lhs": "2**2",
        "relation": ">",
        "rhs": "2"
      },
      {
        "id": "N7",
        "step": "S10",
        "vars": [
          "x"
        ],
        "range": [
          -2,
          3
        ],
        "lhs": "x*x",
        "relation": "!=",
        "rhs": "2"
      },
      {
        "id": "N8",
        "lhs": "3+(-3)",
        "rhs": "0"
      },
      {
        "id": "N9",
        "lhs": "-3",
        "relation": "<",
        "rhs": "-2"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：明示有限整數論域、可判定述詞（大小與相等比較）、古典真值語義。用完整枚舉判定有限論域上的陳述，並由無限制的量詞否定律證明受限量詞的否定律。枚舉的結論只針對指定的有限論域，不外推到 ℤ、ℝ 或無限論域；不宣稱一階邏輯的完備性已形式證明；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: explicit finite integer domains, decidable predicates (order and equality comparisons), classical truth semantics. Decides statements on a finite domain by complete enumeration, and proves the negation laws for restricted quantifiers from the unrestricted ones. The conclusions of the enumeration apply only to the specified finite domain and are not extended to ℤ, ℝ or infinite domains; no claim is made that completeness of first-order logic has been formally proved; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "U",
        "明示有限整數論域 {−2,−1,0,1,2,3}",
        "the explicit finite integer domain {−2,−1,0,1,2,3}"
      ],
      [
        "P(x),\\ Q(x),\\ E(x)",
        "述詞 x²<10、x²>x、x²=2",
        "the predicates x²<10, x²>x, x²=2"
      ],
      [
        "R(x)",
        "引理中的任意述詞",
        "an arbitrary predicate in the lemmas"
      ],
      [
        "a_i",
        "U 的第 i 個列出元素",
        "the i-th listed element of U"
      ],
      [
        "\\varphi_1,\\ \\varphi_2,\\ \\varphi_3",
        "三個待判定的量詞陳述",
        "the three quantified statements to decide"
      ]
    ],
    "md_error_controls": [
      {
        "title": "通過的例子不是無限全稱的證明 / Passing examples do not prove an infinite universal",
        "zh": "MKO 指出的常見誤解是把幾個通過例子當成無限全稱證明。φ₁ 在 U 上為真，是因為 U 只有六個元素且全部檢查過（S04 的引理需要完整列舉）。同樣六個檢查對 ℤ 沒有任何結論：x=4 時 16≥10，∀x∈ℤ, x²<10 為假（S05）。",
        "en": "The misconception identified in the MKO is treating a few passing examples as a proof of an infinite universal. φ₁ is true on U because U has only six elements and all were checked (the lemma in S04 needs a complete listing). The same six checks say nothing about ℤ: at x=4, 16≥10, so ∀x∈ℤ, x²<10 is false (S05)."
      },
      {
        "title": "否定全稱不是「全部不滿足」 / Negating a universal is not 'all fail'",
        "zh": "MKO 的第二個誤解是把否定全稱寫成所有元素皆不滿足。¬∀x∈U Q(x) 等價於 ∃x∈U ¬Q(x)（S08，為真，見證 0），而 ∀x∈U ¬Q(x) 為假（S09，x=2 時 4>2）。否定時要同時切換 ∀／∃ 並否定述詞，只改符號或只否定述詞都不對。",
        "en": "The second misconception identified in the MKO is writing a negated universal as 'every element fails'. ¬∀x∈U Q(x) is equivalent to ∃x∈U ¬Q(x) (S08, true, witness 0), while ∀x∈U ¬Q(x) is false (S09, since 4>2 at x=2). Negation switches ∀/∃ and negates the predicate together; changing only the quantifier or only the predicate is wrong."
      },
      {
        "title": "空論域的邊界 / The empty-domain boundary",
        "zh": "∀x∈∅ R(x) 為真不是因為 R 被某個元素滿足，而是因為沒有元素能成為反例（S14）；∃x∈∅ R(x) 為假，因為沒有可用的見證（S15）。所以由「全稱為真」推出「存在為真」需要論域非空。",
        "en": "∀x∈∅ R(x) is true not because some element satisfies R but because no element can be a counterexample (S14); ∃x∈∅ R(x) is false because there is no witness (S15). So passing from a true universal to a true existential needs a nonempty domain."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "N3 與 N7 是對 U 的六個元素的有限檢查，恰好覆蓋 U，用來重算 S03 與 S10；它們對其他論域沒有任何結論。",
        "en": "N3 and N7 are finite checks over exactly the six elements of U and recompute S03 and S10; they settle nothing about any other domain."
      },
      {
        "zh-Hant": "本稿只採古典真值語義（MKO 假設 a1）；S08.2 與 S08.3 用到的古典蘊涵在非古典語義下不一定成立，本稿不處理。也不宣稱一階邏輯完備性已形式證明。",
        "en": "This guide uses classical truth semantics only (MKO assumption a1); the classical implication laws used in S08.2 and S08.3 need not hold under non-classical semantics, which this guide does not treat. Nor does it claim that completeness of first-order logic has been formally proved."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "52748a3a639b1ac5e04a07bec894402431f75d818c68e49a83f15fee1b2c1952",
    "original_source_sha256": "52748a3a639b1ac5e04a07bec894402431f75d818c68e49a83f15fee1b2c1952",
    "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."
  }
}