{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-proposition-01",
  "mko_id": "mko-proposition",
  "title": {
    "zh-Hant": "命題：判定候選語句是否為命題，並區分「是命題」與「已證明為真」",
    "en": "Proposition: deciding which candidate sentences are propositions, and separating 'is a proposition' from 'is proved true'"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-proposition-01",
    "source": {
      "mko_id": "mko-proposition",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "77a5e0d8403fa47d1ee57431b11f6fc52bfdde80a575db9dac3589fb0b9e8510",
      "english_sha256": "3d649ce5ebed9f16e8aea38557e60c43c771430b81bdc1be7533001a956f0608"
    },
    "mode": "definition_walkthrough",
    "audience_prerequisites": [
      "divides_def",
      "division_algorithm",
      "order_basics",
      "quantifier_negation",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在入門古典語境中，逐一判定下列候選是否為命題；是命題者再判定真假並給出理由：s1「7 是質數」；s2「4 是質數」；s3「7 是質數嗎？」；s4「計算 3+4。」；s5「x>0」（x 是未指定的整數變數），以及 s5 的封閉版本：代入 x=2 與 x=−1、加上 ∀x∈ℤ、加上 ∃x∈ℤ。",
      "en": "In an introductory classical context, decide for each candidate whether it is a proposition, and for each proposition decide whether it is true, with reasons: s1 '7 is prime'; s2 '4 is prime'; s3 'Is 7 prime?'; s4 'Compute 3+4.'; s5 'x>0' (x an unassigned integer variable), together with closed versions of s5: substituting x=2 and x=−1, prefixing ∀x∈ℤ, and prefixing ∃x∈ℤ."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "語境：入門古典邏輯；陳述對象是整數，使用一般的算術與大小順序。候選語句是命題，當且僅當它是良構的直述句且沒有未指定的自由變數；這樣的命題在此語境中恰為真或假之一（MKO 假設 a1、a2）。",
          "en": "Context: introductory classical logic about integers with the usual arithmetic and order. A candidate sentence is a proposition exactly when it is a well-formed declarative sentence with no unassigned free variable; such a proposition is, in this context, exactly one of true or false (MKO assumptions a1 and a2)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "候選：s1「7 是質數」；s2「4 是質數」；s3「7 是質數嗎？」；s4「計算 3+4。」；s5「x>0」，其中 x 是未指定的整數變數。",
          "en": "Candidates: s1 '7 is prime'; s2 '4 is prime'; s3 'Is 7 prime?'; s4 'Compute 3+4.'; s5 'x>0', where x is an unassigned integer variable."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "質數的定義：整數 p 是質數，當且僅當 p≥2 且沒有整數 d 滿足 2≤d≤p−1 且 d∣p。",
          "en": "Definition of prime: an integer p is prime exactly when p≥2 and no integer d with 2≤d≤p−1 satisfies d∣p."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "\\text{a candidate sentence}",
        "after_tex": "\\text{proposition}\\iff\\text{well-formed declarative sentence with no unassigned free variable}",
        "relation": "definition",
        "rule": "mko_prop",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，本稿採入門古典語境：命題是可以被斷言或否定的陳述，在此語境中為真或為假；但「是命題」不表示「已被證明為真」。判定某個候選是不是命題，看的是它的形式（是否為封閉的直述句，prop_test），而不是它的真假。之後每個候選都先判定「是不是命題」，再另外判定真假。",
          "en": "By the MKO, this guide uses an introductory classical context: a proposition is a statement that can be asserted or denied, and in this context it is true or false; but being a proposition does not mean being proved true. Whether a candidate is a proposition is decided by its form (whether it is a closed declarative sentence, prop_test), not by its truth. For each candidate, first decide whether it is a proposition, then separately decide its truth."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_prop",
            "prop_test"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出判準，不推出新事實；真假二分只限本語境（MKO 假設 a2）。",
            "en": "This step only states the test and derives no new fact; the true-or-false dichotomy is limited to this context (MKO assumption a2)."
          }
        },
        "leaf_basis": "mko_prop + prop_test"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "G2"
        ],
        "before_tex": "s_1:\\ 7\\ \\text{is prime}",
        "after_tex": "s_1\\ \\text{is a proposition}",
        "relation": "implies",
        "rule": "prop_test",
        "conditions": [],
        "explanation": {
          "zh-Hant": "s1 是直述句，斷言確定整數 7 的一個性質，其中沒有未指定的變數，依 prop_test 它是命題。此時還沒有判定它的真假，更沒有證明它；「是命題」在這一步就已確定。",
          "en": "s1 is a declarative sentence asserting a property of the definite integer 7, with no unassigned variable, so by prop_test it is a proposition. At this point its truth has not been decided, let alone proved; its status as a proposition is settled already here."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "prop_test"
          ],
          "explanation": {
            "zh-Hant": "只檢查形式（直述、封閉）；不宣稱 s1 為真，真假在 S03 到 S05 另行證明。",
            "en": "Only the form (declarative, closed) is checked; s1 is not claimed to be true here, its truth is proved separately in S03 to S05."
          }
        },
        "leaf_basis": "prop_test"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "G3"
        ],
        "before_tex": "7\\ \\text{is prime}",
        "after_tex": "7\\ge 2\\ \\land\\ \\neg\\exists d\\in\\mathbb{Z}\\,(2\\le d\\le 6\\ \\land\\ d\\mid 7)",
        "relation": "equivalent",
        "rule": "prime_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 G3 的質數定義（p=7，p−1=6）展開 s1。第一個合取項 7≥2 成立；剩下要證的是：沒有整數 d 滿足 2≤d≤6 且 d∣7。",
          "en": "Unfold s1 by the definition of prime in G3 (p=7, p−1=6). The first conjunct, 7≥2, holds; what remains is that no integer d with 2≤d≤6 satisfies d∣7."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "prime_def"
          ],
          "explanation": {
            "zh-Hant": "展開與收回定義是雙向的。",
            "en": "Unfolding and folding a definition are two-way."
          }
        },
        "leaf_basis": "prime_def + def_unfold"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "int_no_gap",
          "division_algorithm"
        ],
        "before_tex": "d\\in\\{2,3,4,5,6\\}",
        "after_tex": "d\\nmid 7\\ \\text{for each such}\\ d",
        "relation": "implies",
        "rule": "L_rem",
        "conditions": [],
        "explanation": {
          "zh-Hant": "滿足 2≤d≤6 的整數恰是 2、3、4、5、6（相鄰整數之間沒有整數，int_no_gap），這是一個有限且完整列出的範圍。對每個 d 寫出帶餘除法：7=2·3+1、7=3·2+1、7=4·1+3、7=5·1+2、7=6·1+1，餘數依序為 1、1、3、2、1，都滿足 0<r<d。依引理 L_rem，這五個 d 都不整除 7。",
          "en": "The integers with 2≤d≤6 are exactly 2, 3, 4, 5, 6 (there is no integer between consecutive integers, int_no_gap), a finite range listed completely. For each d write the division with remainder: 7=2·3+1, 7=3·2+1, 7=4·1+3, 7=5·1+2, 7=6·1+1, with remainders 1, 1, 3, 2, 1, each satisfying 0<r<d. By Lemma L_rem none of these five values of d divides 7."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_rem",
            "int_no_gap",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "五個算式可獨立重算（N1 到 N5）。這裡逐一檢查構成證明，正是因為論域有限而且已完整列出；這不同於用有限個例子去支持一個無窮論域上的全稱命題。",
            "en": "The five identities can be recomputed (N1 to N5). Checking case by case is a proof here precisely because the range is finite and listed completely; this differs from supporting a universal statement over an infinite domain by finitely many examples."
          }
        },
        "leaf_basis": "division_algorithm + divides_def + int_no_gap + numeric_evaluation"
      },
      {
        "id": "S04.1",
        "uses": [
          "divides_def"
        ],
        "before_tex": "n=dq+r,\\ 0<r<d;\\ \\text{assume}\\ d\\mid n",
        "after_tex": "n=dk+0,\\ 0\\le 0<d",
        "relation": "implies",
        "rule": "divides_def",
        "conditions": [
          "d>0"
        ],
        "explanation": {
          "zh-Hant": "證明引理 L_rem：設 n=dq+r 且 0<r<d，並為了反證暫時假設 d∣n。依整除定義有整數 k 使 n=dk，也就是 n=dk+0，而且 0≤0<d。",
          "en": "Proof of Lemma L_rem: let n=dq+r with 0<r<d, and for contradiction temporarily assume d∣n. By the definition of divisibility there is an integer k with n=dk, that is n=dk+0, and 0≤0<d."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "divides_def",
            "add_zero"
          ],
          "explanation": {
            "zh-Hant": "只是把 d∣n 寫成帶餘除法的形式（餘數為 0）。",
            "en": "This only rewrites d∣n in the form of a division with remainder (remainder 0)."
          }
        },
        "leaf_basis": "divides_def + add_zero"
      },
      {
        "id": "S04.2",
        "uses": [
          "S04.1",
          "division_algorithm"
        ],
        "before_tex": "n=dq+r\\ (0<r<d),\\quad n=dk+0",
        "after_tex": "r=0,\\ \\text{contradicting}\\ r>0;\\ \\text{hence}\\ d\\nmid n",
        "relation": "implies",
        "rule": "division_algorithm",
        "conditions": [
          "d>0"
        ],
        "explanation": {
          "zh-Hant": "依帶餘除法的唯一性（d 是正整數），n 除以 d 的商與餘數唯一；n=dq+r 與 n=dk+0 的餘數都在 0 與 d 之間（含 0、不含 d），所以 r=0，與 r>0 矛盾。因此暫時假設不成立，d∤n。S04.1 與 S04.2 合起來證得引理 L_rem。",
          "en": "By uniqueness in the division algorithm (d a positive integer), the quotient and remainder of n by d are unique; both n=dq+r and n=dk+0 have remainders in the range from 0 inclusive to d exclusive, so r=0, contradicting r>0. Hence the temporary assumption fails and d∤n. S04.1 and S04.2 together prove Lemma L_rem."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "division_algorithm",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "關鍵是餘數唯一；只有存在性不足以導出矛盾。",
            "en": "The key is uniqueness of the remainder; existence alone would not give the contradiction."
          }
        },
        "leaf_basis": "division_algorithm + contradiction"
      },
      {
        "id": "S05",
        "uses": [
          "S03",
          "S04"
        ],
        "before_tex": "7\\ge 2,\\ \\text{and no}\\ d\\ \\text{with}\\ 2\\le d\\le 6\\ \\text{divides}\\ 7",
        "after_tex": "s_1\\ \\text{is true, and now proved}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個合取項都已成立：7≥2，且 S04 對範圍內每個 d 都證明 d∤7，所以不存在這樣的 d。依 S03 的等價，s1 為真，而且現在有了證明。",
          "en": "Both conjuncts now hold: 7≥2, and S04 shows d∤7 for every d in the range, so no such d exists. By the equivalence S03, s1 is true, and it now has a proof."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S03",
            "S04",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：s1 為真。足夠條件 H：S03 展開後的兩個合取項；H⇒G 由 S03 的等價。H 由 S04 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: s1 is true. Sufficient condition H: the two conjuncts from S03; H⇒G by the equivalence S03. H is proved forwards in S04; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim + quantifier_negation"
      },
      {
        "id": "S06",
        "uses": [
          "G2",
          "G3",
          "prop_test"
        ],
        "before_tex": "s_2:\\ 4\\ \\text{is prime}",
        "after_tex": "s_2\\ \\text{is a proposition, and}\\ \\neg s_2\\ \\text{holds}",
        "relation": "implies",
        "rule": "prime_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "s2 也是封閉的直述句，依 prop_test 是命題。依質數定義，4 是質數需要沒有整數 d 滿足 2≤d≤3 且 d∣4；但 d=2 滿足 2≤2≤3，且 4=2·2，所以 2∣4。因此 s2 為假，被證明的是它的否定 ¬s2。s2 是一個假的命題：「是命題」與「為真」是兩回事。",
          "en": "s2 is also a closed declarative sentence, so by prop_test it is a proposition. By the definition, 4 is prime only if no integer d with 2≤d≤3 divides 4; but d=2 satisfies 2≤2≤3 and 4=2·2, so 2∣4. Hence s2 is false, and what is proved is its negation ¬s2. s2 is a false proposition: being a proposition and being true are different things."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "prop_test",
            "prime_def",
            "divides_def",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "否定 s2 只需一個見證 d=2；s2 仍是命題，因為命題身分只看形式。",
            "en": "Refuting s2 needs only the witness d=2; s2 is still a proposition, because that status depends only on form."
          }
        },
        "leaf_basis": "prime_def + divides_def + exists_intro + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "G2",
          "prop_test"
        ],
        "before_tex": "s_3:\\ \\text{Is 7 prime?}",
        "after_tex": "s_3\\ \\text{is not a proposition}",
        "relation": "implies",
        "rule": "prop_test",
        "conditions": [],
        "explanation": {
          "zh-Hant": "s3 是疑問句，不斷言任何事情，本身無所謂真假，依 prop_test 不是命題。它詢問的內容對應到命題 s1；回答「是」所依據的是 S05 對 s1 的證明，而不是疑問句本身為真。",
          "en": "s3 is a question; it asserts nothing and is neither true nor false, so by prop_test it is not a proposition. What it asks about corresponds to the proposition s1; answering 'yes' rests on the proof of s1 in S05, not on the question itself being true."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "prop_test"
          ],
          "explanation": {
            "zh-Hant": "判定只依句型；不宣稱 s3 與 s1 是同一個對象。",
            "en": "The verdict depends only on the sentence type; s3 is not claimed to be the same object as s1."
          }
        },
        "leaf_basis": "prop_test"
      },
      {
        "id": "S08",
        "uses": [
          "G2",
          "prop_test"
        ],
        "before_tex": "s_4:\\ \\text{Compute}\\ 3+4.",
        "after_tex": "s_4\\ \\text{is not a proposition}",
        "relation": "implies",
        "rule": "prop_test",
        "conditions": [],
        "explanation": {
          "zh-Hant": "s4 是命令句，要求做一個計算，不斷言任何事情，依 prop_test 不是命題。相對地，直述句「3+4=7」是一個為真的命題（N14）。",
          "en": "s4 is a command asking for a computation; it asserts nothing, so by prop_test it is not a proposition. By contrast, the declarative sentence '3+4=7' is a true proposition (N14)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "prop_test",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "判定只依句型；對照的「3+4=7」可獨立重算。",
            "en": "The verdict depends only on the sentence type; the contrasting '3+4=7' can be recomputed."
          }
        },
        "leaf_basis": "prop_test + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "G2",
          "S01",
          "prop_test"
        ],
        "before_tex": "s_5:\\ x>0",
        "after_tex": "s_5\\ \\text{is an open sentence, not a proposition}",
        "relation": "implies",
        "rule": "prop_test",
        "conditions": [],
        "explanation": {
          "zh-Hant": "s5 是直述句，但 x 是未指定的自由變數，依 prop_test 它是開放句：它的真假要看 x 取什麼值（S10 會看到 x=2 時為真、x=−1 時為假），所以在封閉語境中它本身不是命題。要得到命題，必須代入或量化（MKO 第三段）。",
          "en": "s5 is declarative, but x is an unassigned free variable, so by prop_test it is an open sentence: its truth depends on the value of x (S10 shows it true at x=2 and false at x=−1), so in a closed context it is not itself a proposition. To get a proposition it must be closed by substitution or quantification (the MKO's third paragraph)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "prop_test",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "需要確認 x 沒有在語境中被指定；若語境先指定了 x，結論會不同。",
            "en": "It must be confirmed that the context assigns no value to x; if it did, the verdict would differ."
          }
        },
        "leaf_basis": "prop_test"
      },
      {
        "id": "S10",
        "uses": [
          "S09"
        ],
        "before_tex": "x>0\\ \\text{with}\\ x:=2\\ \\text{or}\\ x:=-1",
        "after_tex": "2>0\\ (\\mathrm{true}),\\quad -1>0\\ (\\mathrm{false})",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代入是一種封閉方式。x:=2 得命題「2>0」，為真（0<1<2）；x:=−1 得命題「−1>0」，為假：因 0<1，兩邊加 −1 得 −1<0，依三一律排除 −1>0。同一個開放句經不同代入得到兩個不同的命題，真假也不同。",
          "en": "Substitution is one way to close it. x:=2 gives the proposition '2>0', which is true (0<1<2); x:=−1 gives '−1>0', which is false: from 0<1, adding −1 gives −1<0, and trichotomy rules out −1>0. Different substitutions into one open sentence give two different propositions with different truth values."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "two_nonzero",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "每個代入結果都是封閉的直述句，因此是命題；真假由順序性質判定。",
            "en": "Each substitution result is a closed declarative sentence and hence a proposition; its truth is decided by the order properties."
          }
        },
        "leaf_basis": "order_basics + two_nonzero"
      },
      {
        "id": "S11",
        "uses": [
          "S09"
        ],
        "before_tex": "\\forall x\\in\\mathbb{Z},\\ x>0",
        "after_tex": "\\text{false: counterexample}\\ x=0",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "全稱量化是另一種封閉方式：「∀x∈ℤ, x>0」沒有自由變數，是命題。取 x=0：0=0，依三一律 0>0 不成立，所以 ∃x∈ℤ, ¬(x>0)，依量詞否定得 ¬∀x∈ℤ, x>0，因此原先的全稱命題為假。1>0、2>0、3>0 等正例再多也不能證明這個全稱命題，而一個反例就足以否定它。",
          "en": "Universal quantification is another way to close it: '∀x∈ℤ, x>0' has no free variable and is a proposition. Take x=0: 0=0, so by trichotomy 0>0 fails; hence ∃x∈ℤ, ¬(x>0), and by negation of quantifiers ¬∀x∈ℤ, x>0, so the original universally quantified proposition is false. However many positive cases such as 1>0, 2>0, 3>0 are listed, they cannot prove this universal statement, while one counterexample refutes it."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "quantifier_negation",
            "order_basics",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "一個反例足以否定全稱；不宣稱 x=0 是唯一反例（x=−1 也是）。",
            "en": "One counterexample suffices to refute a universal statement; x=0 is not claimed to be the only one (x=−1 is another)."
          }
        },
        "leaf_basis": "quantifier_negation + exists_intro + order_basics"
      },
      {
        "id": "S12",
        "uses": [
          "S09"
        ],
        "before_tex": "\\exists x\\in\\mathbb{Z},\\ x>0",
        "after_tex": "\\text{true: witness}\\ x=1",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "存在量化：「∃x∈ℤ, x>0」是命題。取 x=1，1>0 成立（0<1），依 ∃引入，這個命題為真。",
          "en": "Existential quantification: '∃x∈ℤ, x>0' is a proposition. Take x=1; 1>0 holds (0<1), so by ∃-introduction the proposition is true."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "exists_intro",
            "two_nonzero"
          ],
          "explanation": {
            "zh-Hant": "一個見證就足以證明存在命題。",
            "en": "One witness is enough to prove an existential statement."
          }
        },
        "leaf_basis": "exists_intro + two_nonzero"
      },
      {
        "id": "S13",
        "uses": [
          "S02",
          "S05",
          "S06",
          "S07",
          "S08",
          "S09",
          "S10",
          "S11",
          "S12"
        ],
        "before_tex": "\\text{verdicts of S02 to S12}",
        "after_tex": "\\text{propositions: }s_1,s_2,\\text{closures of }s_5;\\ \\text{not propositions: }s_3,s_4,s_5",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "命題：s1（真，S05 已證）、s2（假，S06 已證其否定）、s5 的四個封閉版本（2>0 真、−1>0 假、∀ 版本假、∃ 版本真）。不是命題：s3（疑問句）、s4（命令句）、s5 本身（開放句）。每個候選的命題身分都在判定真假之前確定。",
          "en": "Propositions: s1 (true, proved in S05), s2 (false, its negation proved in S06) and the four closed versions of s5 (2>0 true, −1>0 false, the ∀ version false, the ∃ version true). Not propositions: s3 (a question), s4 (a command) and s5 itself (an open sentence). For every candidate, proposition status was settled before truth."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S02",
            "S05",
            "S06",
            "S07",
            "S08",
            "S09",
            "S10",
            "S11",
            "S12",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：對每個候選給出判定。足夠條件 H：S02 到 S12 的各項結論；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: a verdict for each candidate. Sufficient condition H: the conclusions of S02 to S12; 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": "s1「7 是質數」是命題且為真（已證明）；s2「4 是質數」是命題但為假（已證明其否定）；s3 疑問句與 s4 命令句不是命題；s5「x>0」是開放句，本身不是命題。s5 的封閉版本都是命題：2>0 真、−1>0 假、∀x∈ℤ, x>0 假（反例 x=0）、∃x∈ℤ, x>0 真（見證 x=1）。是否為命題由形式決定，與是否已證明為真無關。",
      "en": "s1 '7 is prime' is a proposition and true (proved); s2 '4 is prime' is a proposition but false (its negation proved); the question s3 and the command s4 are not propositions; s5 'x>0' is an open sentence and not itself a proposition. Its closed versions are propositions: 2>0 true, −1>0 false, ∀x∈ℤ, x>0 false (counterexample x=0), ∃x∈ℤ, x>0 true (witness x=1). Being a proposition is decided by form and is independent of having been proved true."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "「7 是質數嗎？」是命題，因為它的答案是肯定的",
          "en": "'Is 7 prime?' is a proposition because its answer is yes"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "疑問句不斷言任何事情，本身無所謂真假；為真的是命題 s1（S07）。",
          "en": "A question asserts nothing and is neither true nor false; what is true is the proposition s1 (S07)."
        }
      },
      {
        "claim": {
          "zh-Hant": "「4 是質數」是假的，所以它不是命題",
          "en": "'4 is prime' is false, so it is not a proposition"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "假的命題仍是命題；命題身分只看形式（S06）。",
          "en": "A false proposition is still a proposition; the status depends only on form (S06)."
        }
      },
      {
        "claim": {
          "zh-Hant": "x>0 為真",
          "en": "x>0 is true"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x 未指定；x=−1 時為假，x=2 時為真（S10）。",
          "en": "x is unassigned; at x=−1 it is false and at x=2 it is true (S10)."
        }
      },
      {
        "claim": {
          "zh-Hant": "因為 1>0、2>0、3>0，所以 ∀x∈ℤ, x>0",
          "en": "Since 1>0, 2>0 and 3>0, ∀x∈ℤ, x>0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x=0 是反例（S11）；有限個正例不能建立無窮論域上的全稱命題。",
          "en": "x=0 is a counterexample (S11); finitely many positive cases cannot establish a universal statement over an infinite domain."
        }
      }
    ],
    "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": "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": "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": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=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": "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": "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": "two_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數中 0<1，故 0<1+1=2，所以 2≠0。",
          "en": "In the reals 0<1, hence 0<1+1=2, so 2≠0."
        }
      },
      {
        "id": "mko_prop",
        "kind": "cited",
        "source": {
          "mko_id": "mko-proposition",
          "sha256": "77a5e0d8403fa47d1ee57431b11f6fc52bfdde80a575db9dac3589fb0b9e8510",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "本條目採入門古典邏輯語境：命題 P 可以為真或為假；P 是命題不表示 P 已被證明為真。",
          "en": "This entry uses an introductory classical context: a proposition P may be true or false, but being a proposition does not mean that P has been proved true."
        }
      },
      {
        "id": "prop_test",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "命題判準（本語境）：候選是命題，當且僅當它是良構的直述句且沒有未指定的自由變數。疑問句與命令句不是直述句；含未指定自由變數的直述句是開放句，經代入或量化封閉後才成為命題。",
          "en": "Test for propositions (this context): a candidate is a proposition exactly when it is a well-formed declarative sentence with no unassigned free variable. Questions and commands are not declarative; a declarative sentence with an unassigned free variable is an open sentence, which becomes a proposition only after substitution or quantification closes it."
        }
      },
      {
        "id": "prime_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "質數：整數 p 是質數 ⇔ p≥2 且沒有整數 d 滿足 2≤d≤p−1 且 d∣p。",
          "en": "Prime: an integer p is prime ⇔ p≥2 and no integer d with 2≤d≤p−1 satisfies d∣p."
        }
      },
      {
        "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."
        }
      },
      {
        "id": "L_rem",
        "kind": "lemma",
        "proved_in": [
          "S04.1",
          "S04.2"
        ],
        "statement": {
          "zh-Hant": "對整數 n 與正整數 d：若 n=dq+r，其中 q 是整數且 0<r<d，則 d∤n。",
          "en": "For an integer n and a positive integer d: if n=dq+r with q an integer and 0<r<d, then d∤n."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S04",
        "lhs": "2*3+1",
        "rhs": "7"
      },
      {
        "id": "N2",
        "step": "S04",
        "lhs": "3*2+1",
        "rhs": "7"
      },
      {
        "id": "N3",
        "step": "S04",
        "lhs": "4*1+3",
        "rhs": "7"
      },
      {
        "id": "N4",
        "step": "S04",
        "lhs": "5*1+2",
        "rhs": "7"
      },
      {
        "id": "N5",
        "step": "S04",
        "lhs": "6*1+1",
        "rhs": "7"
      },
      {
        "id": "N6",
        "step": "S04",
        "vars": [
          "d"
        ],
        "range": [
          2,
          6
        ],
        "lhs": "7 % d",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N7",
        "step": "S06",
        "lhs": "2*2",
        "rhs": "4"
      },
      {
        "id": "N8",
        "step": "S10",
        "lhs": "2",
        "relation": ">",
        "rhs": "0"
      },
      {
        "id": "N9",
        "step": "S10",
        "lhs": "-1",
        "relation": "<",
        "rhs": "0"
      },
      {
        "id": "N10",
        "step": "S11",
        "lhs": "0",
        "relation": "<=",
        "rhs": "0"
      },
      {
        "id": "N11",
        "step": "S12",
        "lhs": "1",
        "relation": ">",
        "rhs": "0"
      },
      {
        "id": "N12",
        "step": "S05",
        "lhs": "isprime(7)",
        "rhs": "1"
      },
      {
        "id": "N13",
        "step": "S06",
        "lhs": "isprime(4)",
        "rhs": "0"
      },
      {
        "id": "N14",
        "step": "S08",
        "lhs": "3+4",
        "rhs": "7"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "採 MKO 的入門古典語境：良構且封閉的直述句是命題，在此語境中非真即假（MKO 假設 a2）。只處理關於整數的少數具體候選；不討論直覺主義邏輯、自指悖論或形式語言的完整語法；不改動 canonical MKO 或英文 edition。",
      "en": "Uses the MKO's introductory classical context: a well-formed closed declarative sentence is a proposition and in this context is either true or false (MKO assumption a2). Treats only a few concrete candidates about integers; it does not discuss intuitionistic logic, self-referential paradoxes or the full syntax of a formal language, and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "P",
        "一個命題",
        "a proposition"
      ],
      [
        "s_1,\\ s_2,\\ s_3,\\ s_4,\\ s_5",
        "五個候選語句",
        "the five candidate sentences"
      ],
      [
        "x>0",
        "含自由變數 x 的開放句",
        "an open sentence with the free variable x"
      ],
      [
        "\\forall x\\in\\mathbb{Z},\\ \\exists x\\in\\mathbb{Z}",
        "全稱與存在量詞，封閉開放句的兩種方式",
        "universal and existential quantifiers, two ways of closing an open sentence"
      ],
      [
        "d\\mid n",
        "d 整除 n：存在整數 k 使 n=dk",
        "d divides n: n=dk for some integer k"
      ]
    ],
    "md_error_controls": [
      {
        "title": "是命題不等於已證明為真 / Being a proposition is not being proved true",
        "zh": "MKO 指出常見誤解是把「是命題」理解成「已證明為真」。s1 在 S02 就已是命題，直到 S05 才有證明；s2 是命題，卻是假的（S06），被證明的是它的否定 ¬s2。所以「P 是命題」只是說 P 可以被斷言或否定，真假與證明要另外處理。",
        "en": "The MKO names the common mistake of reading 'is a proposition' as 'has been proved true'. s1 was a proposition already at S02 and got its proof only at S05; s2 is a proposition but false (S06), and what is proved is its negation ¬s2. So 'P is a proposition' only says that P can be asserted or denied; truth and proof are separate matters."
      },
      {
        "title": "疑問句、命令句與開放句 / Questions, commands and open sentences",
        "zh": "MKO 的另一個誤解是把疑問句、命令句或未封閉的開放表達式直接當作命題。s3 與 s4 不斷言任何事情（S07、S08）；s5 雖是直述句，但 x 未指定，真假隨 x 改變（S09、S10）。判定時先看句型，再看有沒有未指定的自由變數。",
        "en": "Another mistake identified in the MKO is treating questions, commands or unclosed open expressions as propositions. s3 and s4 assert nothing (S07, S08); s5 is declarative but x is unassigned, so its truth changes with x (S09, S10). Check the sentence type first, then whether any free variable is unassigned."
      },
      {
        "title": "封閉方式要說清楚 / Say how an open sentence is closed",
        "zh": "同一個開放句 x>0 可以封閉成不同的命題：代入 x=2 為真、代入 x=−1 為假、加 ∀ 為假、加 ∃ 為真（S10 到 S12）。只說「x>0 是真的」沒有意義，必須說明是哪一種封閉。",
        "en": "The single open sentence x>0 closes into different propositions: substituting x=2 gives a true one, x=−1 a false one, prefixing ∀ a false one and prefixing ∃ a true one (S10 to S12). Saying 'x>0 is true' is meaningless until the way of closing it is stated."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿的命題判準（prop_test）是入門古典語境中的工作定義；自指語句與形式語言中良構性的完整定義不在此處理。",
        "en": "The test prop_test is a working definition for the introductory classical context; self-referential sentences and a full definition of well-formedness in a formal language are not treated here."
      },
      {
        "zh-Hant": "N6 是在 d=2 到 6 上的有限核對，只重算 S04 中已完整列出的五個情形；證明本身在 S04 與其子步。",
        "en": "N6 is a finite check over d=2 to 6 and only recomputes the five cases already listed completely in S04; the proof itself is S04 with its sub-steps."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "f92ed859a98fb4b70ab6353e1cfab4a7c1931edf858182744aa06e993740ad07",
    "original_source_sha256": "f92ed859a98fb4b70ab6353e1cfab4a7c1931edf858182744aa06e993740ad07",
    "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."
  }
}