{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-logical-connectives-01",
  "mko_id": "mko-logical-connectives",
  "title": {
    "zh-Hant": "邏輯連接詞：以整除命題檢驗 ∨、→、↔ 與括號",
    "en": "Logical connectives: testing ∨, →, ↔ and brackets on divisibility statements"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-logical-connectives-01",
    "source": {
      "mko_id": "mko-logical-connectives",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "d102be1d207477db8f00af868ed046578922ba9eeef220b0f6049bfca3964270",
      "english_sha256": "e154036107025c087a386a9c0cd8abed85ff1311e7f1e0ba3d6eea5c6357b87a"
    },
    "mode": "definition_walkthrough",
    "audience_prerequisites": [
      "divides_def",
      "imp_intro",
      "forall_intro",
      "quantifier_negation",
      "order_basics",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "對整數 n，令 P(n) 為「4∣n」，Q(n) 為「2∣n」。(1) 判定「4∣12 ∨ 6∣12」的真假，說明析取是含括或；(2) 證明 ∀n∈ℤ (P(n)→Q(n))；(3) 證明 ∀n∈ℤ (Q(n)→P(n)) 為假，因而 ∀n∈ℤ (P(n)↔Q(n)) 為假；(4) 判定 P(3)→Q(3) 與「2+2=4 → 7 是奇數」的真假，說明蘊涵不是因果；(5) 比較 ¬Q(2)∧P(2) 與 ¬(Q(2)∧P(2)) 的真假。",
      "en": "For an integer n let P(n) be '4∣n' and Q(n) be '2∣n'. (1) Decide the truth of '4∣12 ∨ 6∣12' and explain that disjunction is inclusive; (2) prove ∀n∈ℤ (P(n)→Q(n)); (3) prove that ∀n∈ℤ (Q(n)→P(n)) is false, and hence that ∀n∈ℤ (P(n)↔Q(n)) is false; (4) decide the truth of P(3)→Q(3) and of '2+2=4 → 7 is odd', explaining that implication is not causation; (5) compare the truth of ¬Q(2)∧P(2) and ¬(Q(2)∧P(2))."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "語境：入門古典二值邏輯；每個命題恰為真（T）或假（F）之一（mko-proposition 的語境）。連接詞依 MKO 定義。",
          "en": "Context: introductory classical two-valued logic; every proposition is exactly one of true (T) or false (F) (the context of mko-proposition). The connectives are as defined in the MKO."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "整除：對整數 d、n，d∣n 表示存在整數 k 使 n=dk。對每個整數 n，P(n) 為「4∣n」，Q(n) 為「2∣n」。",
          "en": "Divisibility: for integers d and n, d∣n means that n=dk for some integer k. For every integer n, P(n) is '4∣n' and Q(n) is '2∣n'."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "具體命題：「6∣12」；「2+2=4」；「7 是奇數」，其中奇數指存在整數 m 使該數等於 2m+1。",
          "en": "Concrete statements: '6∣12'; '2+2=4'; '7 is odd', where odd means equal to 2m+1 for some integer m."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "\\neg P,\\ P\\land Q,\\ P\\lor Q,\\ P\\to Q,\\ P\\leftrightarrow Q",
        "after_tex": "\\text{truth value of each compound determined by those of}\\ P,\\ Q",
        "relation": "definition",
        "rule": "mko_conn",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，¬P 是否定，P∧Q 要求兩者皆成立，P∨Q 要求至少一者成立，P→Q 表示若 P 則 Q，P↔Q 要求兩個方向。在古典二值語境（G1）中，這些定義寫成真值條件（truth_table）：複合命題的真假只由成分的真假決定，與成分之間有沒有因果或時間關係無關。",
          "en": "By the MKO, ¬P is negation, P∧Q requires both, P∨Q requires at least one, P→Q says if P then Q, and P↔Q requires both directions. In the classical two-valued context (G1) these definitions are truth conditions (truth_table): the truth of a compound depends only on the truth of its parts, not on any causal or temporal link between them."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_conn",
            "truth_table"
          ],
          "explanation": {
            "zh-Hant": "此步只固定定義，不推出新事實。",
            "en": "This step only fixes the definitions and derives no new fact."
          }
        },
        "leaf_basis": "mko_conn + truth_table"
      },
      {
        "id": "S02",
        "uses": [
          "G2",
          "G3"
        ],
        "before_tex": "4\\mid 12,\\ 6\\mid 12",
        "after_tex": "\\mathrm{T},\\ \\mathrm{T}",
        "relation": "implies",
        "rule": "divides_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "12=4·3，3 是整數，所以 4∣12 為真；12=6·2，所以 6∣12 為真。兩個成分都為真。",
          "en": "12=4·3 with 3 an integer, so 4∣12 is true; 12=6·2, so 6∣12 is true. Both components are true."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "divides_def",
            "exists_intro",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "每個整除各需一個整數見證（3 與 2），可重算（N1、N2）。",
            "en": "Each divisibility needs one integer witness (3 and 2), and both can be recomputed (N1, N2)."
          }
        },
        "leaf_basis": "divides_def + exists_intro + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "S02"
        ],
        "before_tex": "4\\mid 12\\ \\lor\\ 6\\mid 12",
        "after_tex": "\\mathrm{T}",
        "relation": "implies",
        "rule": "truth_table",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 ∨ 的真值條件，至少一者為真即為真；這裡兩者都為真，所以「4∣12 ∨ 6∣12」為真。若把 ∨ 誤讀成排他或（恰好一者為真），這個式子會被判為假；MKO 的 ∨ 是含括或，允許兩者同時成立。",
          "en": "By the truth condition for ∨, the disjunction is true when at least one part is true; here both are, so '4∣12 ∨ 6∣12' is true. Misreading ∨ as exclusive or (exactly one true) would wrongly make it false; the MKO's ∨ is inclusive and allows both to hold."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "truth_table"
          ],
          "explanation": {
            "zh-Hant": "只需一個成分為真；兩個都真不影響結論。",
            "en": "One true component is enough; both being true does not change the verdict."
          }
        },
        "leaf_basis": "truth_table"
      },
      {
        "id": "S04",
        "uses": [
          "G2"
        ],
        "before_tex": "\\text{assume}\\ 4\\mid n\\ (n\\in\\mathbb{Z}\\ \\text{arbitrary})",
        "after_tex": "n=4k\\ \\text{for some}\\ k\\in\\mathbb{Z}",
        "relation": "implies",
        "rule": "divides_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "要證 ∀n (P(n)→Q(n))：任取整數 n，不加額外假設，並在暫時假設 4∣n 的範圍內推理。依整除定義，有整數 k 使 n=4k（以新名字 k 表示這個見證）。",
          "en": "To prove ∀n (P(n)→Q(n)): take an arbitrary integer n with no extra assumption, and reason inside the temporary assumption 4∣n. By the definition of divisibility there is an integer k with n=4k (k is a fresh name for this witness)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "divides_def",
            "exists_elim",
            "imp_intro"
          ],
          "explanation": {
            "zh-Hant": "暫時假設只在 S04 到 S06 的範圍內使用，S06 會把它釋放。",
            "en": "The temporary assumption is used only within S04 to S06 and is discharged in S06."
          }
        },
        "leaf_basis": "divides_def + exists_elim"
      },
      {
        "id": "S05",
        "uses": [
          "S04"
        ],
        "before_tex": "n=4k",
        "after_tex": "n=2(2k),\\ 2k\\in\\mathbb{Z},\\ \\text{so}\\ 2\\mid n",
        "relation": "implies",
        "rule": "mul_assoc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "4=2·2，依乘法結合律 4k=(2·2)k=2(2k)；2k 是整數（整數對乘法封閉），所以依整除定義 2∣n。",
          "en": "4=2·2, so by associativity 4k=(2·2)k=2(2k); 2k is an integer (the integers are closed under multiplication), so 2∣n by the definition of divisibility."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mul_assoc",
            "divides_def",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "需要 2k 是整數作為見證；N3 是對恆等式 4k=2(2k) 的獨立核對。",
            "en": "2k must be an integer to serve as the witness; N3 is an independent check of the identity 4k=2(2k)."
          }
        },
        "leaf_basis": "mul_assoc + divides_def + exists_intro + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S04",
          "S05"
        ],
        "before_tex": "\\text{under the assumption}\\ 4\\mid n:\\ 2\\mid n",
        "after_tex": "\\forall n\\in\\mathbb{Z}:\\ P(n)\\to Q(n)",
        "relation": "implies",
        "rule": "imp_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在暫時假設 4∣n 下推出了 2∣n，依 →引入釋放假設得 P(n)→Q(n)；n 是任取的，依全稱推廣得 ∀n∈ℤ (P(n)→Q(n))。這只證了一個方向。",
          "en": "Under the temporary assumption 4∣n we derived 2∣n, so →-introduction discharges the assumption and gives P(n)→Q(n); n was arbitrary, so generalization gives ∀n∈ℤ (P(n)→Q(n)). Only one direction has been proved."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S05",
            "imp_intro",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "S04 的 n 沒有額外條件，所以可以推廣；結論只是 P→Q，不宣稱 Q→P。",
            "en": "The n of S04 carried no extra condition, so generalization is allowed; the conclusion is P→Q only, and Q→P is not claimed."
          }
        },
        "leaf_basis": "imp_intro + forall_intro"
      },
      {
        "id": "S07",
        "uses": [
          "G2"
        ],
        "before_tex": "n=2",
        "after_tex": "Q(2):\\ 2=2\\cdot 1,\\ \\mathrm{T}",
        "relation": "implies",
        "rule": "divides_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "檢查反方向 Q(n)→P(n)。取 n=2：2=2·1，1 是整數，所以 2∣2，Q(2) 為真。",
          "en": "Now examine the reverse direction Q(n)→P(n). Take n=2: 2=2·1 with 1 an integer, so 2∣2 and Q(2) is true."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "divides_def",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "一個整數見證 1 就足夠。",
            "en": "The single integer witness 1 suffices."
          }
        },
        "leaf_basis": "divides_def + exists_intro + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "G2"
        ],
        "before_tex": "k\\in\\mathbb{Z}\\ \\text{with}\\ 4k=2\\ ?",
        "after_tex": "k\\le 0\\ \\lor\\ k\\ge 1",
        "relation": "case_split",
        "rule": "int_no_gap",
        "conditions": [
          "k\\in\\mathbb{Z}"
        ],
        "explanation": {
          "zh-Hant": "P(2) 為真需要整數 k 使 4k=2。依 int_no_gap（m=0），每個整數 k 滿足 k≤0 或 k≥1，兩支在 S08.1、S08.2 處理。",
          "en": "P(2) would need an integer k with 4k=2. By int_no_gap (m=0) every integer k satisfies k≤0 or k≥1; the two branches are handled in S08.1 and S08.2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "int_no_gap"
          ],
          "explanation": {
            "zh-Hant": "兩支涵蓋全部整數；這依賴 k 是整數（k=1/2 不在任何一支中）。",
            "en": "The two branches cover all integers; this relies on k being an integer (k=1/2 lies in neither)."
          }
        },
        "leaf_basis": "int_no_gap + case_analysis"
      },
      {
        "id": "S08.1",
        "uses": [
          "S08"
        ],
        "before_tex": "k\\le 0",
        "after_tex": "4k\\le 0<2,\\ \\text{so}\\ 4k\\neq 2",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "4>0"
        ],
        "explanation": {
          "zh-Hant": "若 k<0，乘以正數 4 得 4k<0；若 k=0，4k=0。所以 4k≤0<2，4k≠2。",
          "en": "If k<0, multiplying by the positive number 4 gives 4k<0; if k=0 then 4k=0. So 4k≤0<2 and 4k≠2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "用到 4>0 保持不等號方向；不宣稱反向。",
            "en": "4>0 keeps the direction of the inequality; no converse is claimed."
          }
        },
        "leaf_basis": "order_basics + mul_zero"
      },
      {
        "id": "S08.2",
        "uses": [
          "S08"
        ],
        "before_tex": "k\\ge 1",
        "after_tex": "4k\\ge 4>2,\\ \\text{so}\\ 4k\\neq 2",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "4>0"
        ],
        "explanation": {
          "zh-Hant": "若 k>1，乘以 4 得 4k>4；若 k=1，4k=4。所以 4k≥4>2，4k≠2。",
          "en": "If k>1, multiplying by 4 gives 4k>4; if k=1 then 4k=4. So 4k≥4>2 and 4k≠2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "mul_one"
          ],
          "explanation": {
            "zh-Hant": "同 S08.1。",
            "en": "As in S08.1."
          }
        },
        "leaf_basis": "order_basics + mul_one"
      },
      {
        "id": "S09",
        "uses": [
          "S07",
          "S08",
          "S08.1",
          "S08.2"
        ],
        "before_tex": "Q(2)=\\mathrm{T},\\ \\neg\\exists k\\in\\mathbb{Z}:\\ 4k=2",
        "after_tex": "P(2)=\\mathrm{F};\\ Q(2)\\to P(2)=\\mathrm{F};\\ \\neg\\forall n\\,(Q(n)\\to P(n))",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩支都得 4k≠2，所以沒有整數 k 使 4k=2，P(2) 為假。依 → 的真值條件，前件真、後件假時蘊涵為假，所以 Q(2)→P(2) 為假。n=2 是反例，依量詞否定 ∀n∈ℤ (Q(n)→P(n)) 為假。",
          "en": "Both branches give 4k≠2, so no integer k has 4k=2 and P(2) is false. By the truth condition for →, a true antecedent with a false consequent makes the implication false, so Q(2)→P(2) is false. n=2 is a counterexample, so by negation of quantifiers ∀n∈ℤ (Q(n)→P(n)) is false."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "S08",
            "case_analysis",
            "truth_table",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "一個反例足以否定全稱蘊涵；case_analysis 涵蓋了所有整數 k。",
            "en": "One counterexample suffices to refute the universal implication; case_analysis covered every integer k."
          }
        },
        "leaf_basis": "case_analysis + truth_table + quantifier_negation"
      },
      {
        "id": "S10",
        "uses": [
          "S06",
          "S09"
        ],
        "before_tex": "\\forall n\\,(P(n)\\to Q(n))\\ \\text{true},\\ \\forall n\\,(Q(n)\\to P(n))\\ \\text{false}",
        "after_tex": "\\forall n\\in\\mathbb{Z}:\\ P(n)\\leftrightarrow Q(n)\\ \\text{is false}",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "假設 ∀n (P(n)↔Q(n)) 為真。取 n=2 得 P(2)↔Q(2)，依雙條件的定義推出 Q(2)→P(2)，與 S09 矛盾。所以這個全稱雙條件為假。S06 證明的方向對所有 n 都成立，但只有一個方向不足以得到雙條件。",
          "en": "Suppose ∀n (P(n)↔Q(n)) were true. Taking n=2 gives P(2)↔Q(2), and by the definition of the biconditional Q(2)→P(2), contradicting S09. So the universal biconditional is false. The direction proved in S06 holds for every n, but one direction is not enough for the biconditional."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S09",
            "iff_intro",
            "forall_elim",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "用到雙條件包含兩個方向（iff_intro）；不宣稱 P(n)↔Q(n) 對每個 n 都假：例如 n=4 時 4=4·1、4=2·2，兩者皆真，雙條件為真。",
            "en": "It uses that a biconditional contains both directions (iff_intro); it does not claim P(n)↔Q(n) is false for every n: at n=4, 4=4·1 and 4=2·2, both parts are true and the biconditional is true."
          }
        },
        "leaf_basis": "contradiction + iff_intro + forall_elim"
      },
      {
        "id": "S11",
        "uses": [
          "G2"
        ],
        "before_tex": "n=3",
        "after_tex": "P(3)=\\mathrm{F},\\ Q(3)=\\mathrm{F}",
        "relation": "implies",
        "rule": "divides_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "n=3 的兩個成分都為假：S11.1 證明沒有整數 k 使 4k=3，S11.2 證明沒有整數 k 使 2k=3。",
          "en": "Both components at n=3 are false: S11.1 shows no integer k has 4k=3, and S11.2 shows no integer k has 2k=3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "divides_def",
            "int_no_gap",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "兩個「沒有見證」都在子步中以涵蓋全部整數的分情況證明。",
            "en": "Both non-existence claims are proved in the sub-steps by case splits that cover all integers."
          }
        },
        "leaf_basis": "divides_def + int_no_gap + order_basics"
      },
      {
        "id": "S11.1",
        "uses": [
          "S11"
        ],
        "before_tex": "4k=3,\\ k\\in\\mathbb{Z}",
        "after_tex": "\\text{impossible: } k\\le 0\\Rightarrow 4k\\le 0<3;\\ k\\ge 1\\Rightarrow 4k\\ge 4>3",
        "relation": "implies",
        "rule": "int_no_gap",
        "conditions": [
          "4>0"
        ],
        "explanation": {
          "zh-Hant": "依 int_no_gap，k≤0 或 k≥1。k≤0 時乘以 4>0 得 4k≤0<3；k≥1 時得 4k≥4>3。兩支都得 4k≠3，所以 4∤3，P(3) 為假。",
          "en": "By int_no_gap, k≤0 or k≥1. If k≤0, multiplying by 4>0 gives 4k≤0<3; if k≥1, 4k≥4>3. Both branches give 4k≠3, so 4∤3 and P(3) is false."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "int_no_gap",
            "order_basics",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "兩支涵蓋全部整數。",
            "en": "The two branches cover every integer."
          }
        },
        "leaf_basis": "int_no_gap + order_basics + case_analysis"
      },
      {
        "id": "S11.2",
        "uses": [
          "S11"
        ],
        "before_tex": "2k=3,\\ k\\in\\mathbb{Z}",
        "after_tex": "\\text{impossible: } k\\le 1\\Rightarrow 2k\\le 2<3;\\ k\\ge 2\\Rightarrow 2k\\ge 4>3",
        "relation": "implies",
        "rule": "int_no_gap",
        "conditions": [
          "2>0"
        ],
        "explanation": {
          "zh-Hant": "依 int_no_gap（m=1），k≤1 或 k≥2。k≤1 時 2k≤2<3；k≥2 時 2k≥4>3。兩支都得 2k≠3，所以 2∤3，Q(3) 為假。",
          "en": "By int_no_gap (m=1), k≤1 or k≥2. If k≤1 then 2k≤2<3; if k≥2 then 2k≥4>3. Both branches give 2k≠3, so 2∤3 and Q(3) is false."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "int_no_gap",
            "order_basics",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "兩支涵蓋全部整數。",
            "en": "The two branches cover every integer."
          }
        },
        "leaf_basis": "int_no_gap + order_basics + case_analysis"
      },
      {
        "id": "S12",
        "uses": [
          "S11",
          "S11.1",
          "S11.2",
          "S06"
        ],
        "before_tex": "P(3)=\\mathrm{F},\\ Q(3)=\\mathrm{F}",
        "after_tex": "P(3)\\to Q(3)=\\mathrm{T}",
        "relation": "implies",
        "rule": "truth_table",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 → 的真值條件，前件為假時蘊涵為真，所以 P(3)→Q(3) 為真；這也與 S06 在 n=3 的特例一致（∀消去）。雖然 2∤3，這個蘊涵仍為真：它的真假只看真值，不表示 4∣3 會「導致」什麼。",
          "en": "By the truth condition for →, a false antecedent makes the implication true, so P(3)→Q(3) is true; this agrees with S06 instantiated at n=3 (∀-elimination). Although 2∤3, the implication is true: its truth depends only on truth values and does not mean that 4∣3 'causes' anything."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "truth_table",
            "S06",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "只需前件為假；Q(3) 為假用來說明「後件為假，蘊涵仍為真」。",
            "en": "Only the false antecedent is needed; Q(3) being false illustrates that the implication is true even with a false consequent."
          }
        },
        "leaf_basis": "truth_table + forall_elim"
      },
      {
        "id": "S13",
        "uses": [
          "G3",
          "S01"
        ],
        "before_tex": "2+2=4,\\ 7=2\\cdot 3+1",
        "after_tex": "(2+2=4)\\to(7\\ \\text{is odd})=\\mathrm{T}",
        "relation": "implies",
        "rule": "truth_table",
        "conditions": [],
        "explanation": {
          "zh-Hant": "2+2=4 為真；7=2·3+1，3 是整數，所以 7 是奇數，為真。依 → 的真值條件，真→真為真。兩個命題之間沒有任何因果或推導關係，蘊涵仍為真：材料蘊涵只是一個真值條件，不表示因果，也不表示時間先後。",
          "en": "2+2=4 is true; 7=2·3+1 with 3 an integer, so 7 is odd, which is true. By the truth condition for →, true→true is true. There is no causal or inferential link between the two statements, yet the implication is true: material implication is a truth condition, not causation and not temporal order."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "truth_table",
            "numeric_evaluation",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "只用到兩個成分的真值。",
            "en": "Only the truth values of the two components are used."
          }
        },
        "leaf_basis": "truth_table + exists_intro + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S07",
          "S09"
        ],
        "before_tex": "\\neg Q(2)\\land P(2)\\quad\\text{vs}\\quad\\neg(Q(2)\\land P(2))",
        "after_tex": "\\mathrm{F}\\quad\\text{vs}\\quad\\mathrm{T}",
        "relation": "implies",
        "rule": "truth_table",
        "conditions": [],
        "explanation": {
          "zh-Hant": "Q(2) 為真（S07），P(2) 為假（S09）。在 ¬Q(2)∧P(2) 中，¬ 只作用於 Q(2)，最外層連接詞是 ∧：¬T∧F=F∧F=F。在 ¬(Q(2)∧P(2)) 中，最外層連接詞是 ¬：Q(2)∧P(2)=T∧F=F，取否定得 T。括號改變了最外層連接詞，也改變了真值。",
          "en": "Q(2) is true (S07) and P(2) is false (S09). In ¬Q(2)∧P(2), ¬ applies only to Q(2) and the main connective is ∧: ¬T∧F=F∧F=F. In ¬(Q(2)∧P(2)) the main connective is ¬: Q(2)∧P(2)=T∧F=F, whose negation is T. The brackets change the main connective and with it the truth value."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "S09",
            "truth_table"
          ],
          "explanation": {
            "zh-Hant": "兩式用的成分真值相同，差別只在結構。",
            "en": "Both use the same component truth values; they differ only in structure."
          }
        },
        "leaf_basis": "truth_table"
      },
      {
        "id": "S15",
        "uses": [
          "S03",
          "S06",
          "S09",
          "S10",
          "S12",
          "S13",
          "S14"
        ],
        "before_tex": "\\text{conclusions of S03, S06, S09, S10, S12, S13, S14}",
        "after_tex": "\\text{answers to parts (1) to (5)}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(1) 「4∣12 ∨ 6∣12」為真（S03）；(2) ∀n (P(n)→Q(n)) 已證（S06）；(3) ∀n (Q(n)→P(n)) 為假（S09），所以 ∀n (P(n)↔Q(n)) 為假（S10）；(4) P(3)→Q(3) 與「2+2=4 → 7 是奇數」都為真（S12、S13）；(5) ¬Q(2)∧P(2) 為假、¬(Q(2)∧P(2)) 為真（S14）。",
          "en": "(1) '4∣12 ∨ 6∣12' is true (S03); (2) ∀n (P(n)→Q(n)) is proved (S06); (3) ∀n (Q(n)→P(n)) is false (S09), so ∀n (P(n)↔Q(n)) is false (S10); (4) P(3)→Q(3) and '2+2=4 → 7 is odd' are both true (S12, S13); (5) ¬Q(2)∧P(2) is false and ¬(Q(2)∧P(2)) is true (S14)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S03",
            "S06",
            "S09",
            "S10",
            "S12",
            "S13",
            "S14",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：五個部分的答案。足夠條件 H：所列各步的結論；H⇒G 只是並列。H 由 G1 到 G3 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: answers to the five parts. Sufficient condition H: the conclusions of the listed steps; H⇒G just lists them. H is proved forwards from G1 to G3; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "「4∣12 ∨ 6∣12」為真，因為 ∨ 是含括或。∀n∈ℤ (4∣n → 2∣n) 成立，但 ∀n∈ℤ (2∣n → 4∣n) 為假（反例 n=2），所以 ∀n∈ℤ (4∣n ↔ 2∣n) 為假。4∣3 → 2∣3 與「2+2=4 → 7 是奇數」都為真，蘊涵只看真值而不表示因果。¬(2∣2)∧(4∣2) 為假，¬((2∣2)∧(4∣2)) 為真，括號決定最外層連接詞。",
      "en": "'4∣12 ∨ 6∣12' is true, because ∨ is inclusive. ∀n∈ℤ (4∣n → 2∣n) holds, but ∀n∈ℤ (2∣n → 4∣n) is false (counterexample n=2), so ∀n∈ℤ (4∣n ↔ 2∣n) is false. 4∣3 → 2∣3 and '2+2=4 → 7 is odd' are both true; implication depends only on truth values and does not express causation. ¬(2∣2)∧(4∣2) is false and ¬((2∣2)∧(4∣2)) is true; the brackets fix the main connective."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "「4∣12 或 6∣12」為假，因為兩者都成立",
          "en": "'4∣12 or 6∣12' is false because both hold"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "∨ 是含括或，兩者皆真時析取為真（S03）。",
          "en": "∨ is inclusive; a disjunction with both parts true is true (S03)."
        }
      },
      {
        "claim": {
          "zh-Hant": "已證 ∀n(4∣n→2∣n)，所以 ∀n(4∣n↔2∣n)",
          "en": "∀n(4∣n→2∣n) is proved, so ∀n(4∣n↔2∣n)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "n=2 時 2∣2 真而 4∣2 假，反方向失敗（S09、S10）。",
          "en": "At n=2, 2∣2 is true and 4∣2 is false, so the converse fails (S09, S10)."
        }
      },
      {
        "claim": {
          "zh-Hant": "「4∣3 → 2∣3」為假，因為 2∤3",
          "en": "'4∣3 → 2∣3' is false because 2∤3"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "前件 4∣3 為假，蘊涵為真（S12）。",
          "en": "The antecedent 4∣3 is false, so the implication is true (S12)."
        }
      },
      {
        "claim": {
          "zh-Hant": "¬A∧B 與 ¬(A∧B) 的意思相同",
          "en": "¬A∧B and ¬(A∧B) mean the same"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "取 A 為 2∣2（真）、B 為 4∣2（假）：前者為假，後者為真（S14）。",
          "en": "Take A to be 2∣2 (true) and B to be 4∣2 (false): the first is false and the second true (S14)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "divides_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整除：對整數 a、b，a∣b 表示存在整數 k 使 b=ak。",
          "en": "Divisibility: for integers a, b, a∣b means there is an integer k with b=ak."
        }
      },
      {
        "id": "imp_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "→引入（釋放假設）：若在暫時假設 P 的作用域中推出 Q，則得到 P⇒Q；此後 P 不再可當作已知使用。",
          "en": "→-introduction (discharging a hypothesis): if Q is derived inside a scope that temporarily assumes P, then P⇒Q follows, and P may no longer be used as known."
        }
      },
      {
        "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": "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": "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": "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": "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": "exists_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃消去：由 ∃x∈D, P(x)，可取一個新名字 a 表示某個滿足 P(a) 的元素；只要最後結論不含 a，就可用 a 推理。",
          "en": "∃-elimination: from ∃x∈D, P(x) we may name some element a with P(a) by a fresh name; reasoning with a is allowed as long as the final conclusion does not mention a."
        }
      },
      {
        "id": "mul_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法結合律：(uv)w=u(vw)。",
          "en": "Associativity of multiplication: (uv)w=u(vw)."
        }
      },
      {
        "id": "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": "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": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "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": "iff_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "雙條件：P⇒Q 與 Q⇒P 都成立時，P⇔Q 成立；反之亦然。",
          "en": "Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold."
        }
      },
      {
        "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": "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": "mko_conn",
        "kind": "cited",
        "source": {
          "mko_id": "mko-logical-connectives",
          "sha256": "d102be1d207477db8f00af868ed046578922ba9eeef220b0f6049bfca3964270",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對命題 P、Q，¬P 表示否定，P∧Q 表示兩者皆成立，P∨Q 表示至少一者成立，P→Q 表示若 P 則 Q，P↔Q 表示兩方向皆成立。",
          "en": "For propositions P and Q, ¬P denotes negation, P∧Q says that both hold, P∨Q says that at least one holds, P→Q says that if P then Q, and P↔Q denotes the biconditional requiring both directions."
        }
      },
      {
        "id": "truth_table",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "古典真值條件：¬P 為真 ⇔ P 為假；P∧Q 為真 ⇔ 兩者皆真；P∨Q 為真 ⇔ 至少一者為真（含兩者皆真）；P→Q 為假 ⇔ P 真且 Q 假，其餘情形為真；P↔Q 為真 ⇔ P→Q 與 Q→P 皆真，即 P、Q 真值相同。",
          "en": "Classical truth conditions: ¬P is true ⇔ P is false; P∧Q is true ⇔ both are true; P∨Q is true ⇔ at least one is true (including both); P→Q is false ⇔ P is true and Q is false, and true otherwise; P↔Q is true ⇔ P→Q and Q→P are both true, that is, P and Q have the same truth value."
        }
      },
      {
        "id": "int_no_gap",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數沒有空隙：對整數 k 與 m，必有 k≤m 或 k≥m+1。",
          "en": "No gaps in the integers: for integers k and m, k≤m or k≥m+1."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "4*3",
        "rhs": "12"
      },
      {
        "id": "N2",
        "step": "S02",
        "lhs": "6*2",
        "rhs": "12"
      },
      {
        "id": "N3",
        "step": "S05",
        "vars": [
          "k"
        ],
        "degree": 1,
        "lhs": "4*k",
        "rhs": "2*(2*k)"
      },
      {
        "id": "N4",
        "step": "S07",
        "lhs": "2*1",
        "rhs": "2"
      },
      {
        "id": "N5",
        "step": "S08.1",
        "lhs": "4*0",
        "relation": "<",
        "rhs": "2"
      },
      {
        "id": "N6",
        "step": "S08.2",
        "lhs": "4*1",
        "relation": ">",
        "rhs": "2"
      },
      {
        "id": "N7",
        "step": "S11.1",
        "lhs": "4*0",
        "relation": "<",
        "rhs": "3"
      },
      {
        "id": "N8",
        "step": "S11.1",
        "lhs": "4*1",
        "relation": ">",
        "rhs": "3"
      },
      {
        "id": "N9",
        "step": "S11.2",
        "lhs": "2*1",
        "relation": "<",
        "rhs": "3"
      },
      {
        "id": "N10",
        "step": "S11.2",
        "lhs": "2*2",
        "relation": ">",
        "rhs": "3"
      },
      {
        "id": "N11",
        "step": "S13",
        "lhs": "2+2",
        "rhs": "4"
      },
      {
        "id": "N12",
        "step": "S13",
        "lhs": "2*3+1",
        "rhs": "7"
      },
      {
        "id": "N13",
        "step": "S10",
        "lhs": "4*1",
        "rhs": "2*2"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "依 MKO 的連接詞定義，在入門古典二值語境中以真值條件判定複合命題的真假。只用整除與奇偶的具體命題；不處理直覺主義、模態或相關邏輯；不改動 canonical MKO 或英文 edition。",
      "en": "Uses the MKO's definitions of the connectives, read as truth conditions in an introductory classical two-valued context, to decide the truth of compound propositions. Only concrete statements about divisibility and parity are used; intuitionistic, modal or relevance logics are not treated, and the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "\\neg,\\ \\land,\\ \\lor",
        "否定、合取、（含括）析取",
        "negation, conjunction, (inclusive) disjunction"
      ],
      [
        "\\to,\\ \\leftrightarrow",
        "（材料）蘊涵、雙條件",
        "(material) implication and the biconditional"
      ],
      [
        "P(n),\\ Q(n)",
        "4∣n 與 2∣n",
        "4∣n and 2∣n"
      ],
      [
        "d\\mid n",
        "存在整數 k 使 n=dk",
        "there is an integer k with n=dk"
      ],
      [
        "\\mathrm{T},\\ \\mathrm{F}",
        "真、假",
        "true, false"
      ]
    ],
    "md_error_controls": [
      {
        "title": "析取是含括或 / Disjunction is inclusive",
        "zh": "MKO 指出常見誤解是把析取當成排他或。4∣12 與 6∣12 都為真，所以「4∣12 ∨ 6∣12」為真（S03）；若當成排他或就會錯判為假。日常語言中「或」有時帶排他意味，數學的 ∨ 不帶。",
        "en": "The MKO names the mistake of reading disjunction as exclusive or. 4∣12 and 6∣12 are both true, so '4∣12 ∨ 6∣12' is true (S03); reading ∨ as exclusive would wrongly make it false. Everyday 'or' is sometimes exclusive; the mathematical ∨ is not."
      },
      {
        "title": "蘊涵不是因果 / Implication is not causation",
        "zh": "MKO 的另一個誤解是把蘊涵當成因果。P(3)→Q(3) 為真只因前件為假（S12）；「2+2=4 → 7 是奇數」為真只因兩者皆真（S13），兩者之間沒有因果關係。要證明「若 P 則 Q」對所有情形成立，仍需像 S04 到 S06 那樣的推理；但單一蘊涵的真假只看真值。",
        "en": "The next mistake identified in the MKO is reading implication as causation. P(3)→Q(3) is true only because its antecedent is false (S12); '2+2=4 → 7 is odd' is true only because both parts are true (S13), with no causal link. Proving 'if P then Q' in every case still requires reasoning like S04 to S06, but the truth of a single implication depends only on truth values."
      },
      {
        "title": "雙條件要兩個方向 / A biconditional needs both directions",
        "zh": "MKO 的第三個誤解是只證一個方向便宣稱雙條件。S06 對所有整數證明了 4∣n → 2∣n，但 n=2 使反方向失敗（S09），所以 ∀n (4∣n ↔ 2∣n) 為假（S10）。證明 P↔Q 必須分別證明 P→Q 與 Q→P。",
        "en": "The third mistake identified in the MKO is proving one direction and claiming the biconditional. S06 proves 4∣n → 2∣n for every integer, but n=2 breaks the converse (S09), so ∀n (4∣n ↔ 2∣n) is false (S10). Proving P↔Q requires proving P→Q and Q→P separately."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "真值條件是古典二值語境中的定義；在直覺主義邏輯中 → 與 ∨ 的意義不同，本稿不處理。",
        "en": "The truth conditions are definitions for the classical two-valued context; in intuitionistic logic → and ∨ have different meanings, which this guide does not treat."
      },
      {
        "zh-Hant": "N3 是在格點上的一次多項式恆等式核對，只交叉核對 S05 的 4k=2(2k)；一般性來自 S04 到 S06 的推理。",
        "en": "N3 is a degree-one polynomial identity check on a grid and only cross-checks 4k=2(2k) in S05; the generality comes from the reasoning of S04 to S06."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "8c4378aa84c06b4b7dd718820bdb2b5a2f72f9a464ab78ec19745a2813a16d7a",
    "original_source_sha256": "8c4378aa84c06b4b7dd718820bdb2b5a2f72f9a464ab78ec19745a2813a16d7a",
    "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."
  }
}