{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-variable-expression-01",
  "mko_id": "mko-variable-expression",
  "title": {
    "zh-Hant": "變數與代數式：式子的值由取值、作用域與數系共同決定",
    "en": "Variables and expressions: a value depends on the assignment, the scope and the number system"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-variable-expression-01",
    "source": {
      "mko_id": "mko-variable-expression",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "da4b6e4a4edfd379a38b6535fd3030cc5c552e2b549572c0d4733d68974a292c",
      "english_sha256": "4a8cfbcc1fa5d73c360d18a95f35f69578250d0104baf76036f4efc8de9d7e81"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "add_assoc",
      "add_comm",
      "distributive",
      "mul_inverse_nonzero",
      "order_basics",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "設 E=x+1、F=(x+1)÷2、H=Σ_{x=1}^{3}(x+1)。(1) 在 ℤ 中求 E 在 x=2 與 x=5 的值，並證明 E 沒有與 x 無關的固定值；(2) 求 F 在 x=4 時於 ℚ 與 ℤ 中的值；(3) 在外部指定 x=5 時求 H；(4) 判定 (x+1)+x 與 2x+1、x·x 與 2x 在 ℤ 上是否等價；(5) 區分表達式 x+1 與方程 x+1=3，並求方程在 ℤ 中的解集。",
      "en": "Let E=x+1, F=(x+1)÷2 and H=Σ_{x=1}^{3}(x+1). (1) In ℤ, evaluate E at x=2 and at x=5 and prove that E has no fixed value independent of x; (2) evaluate F at x=4 in ℚ and in ℤ; (3) evaluate H when x=5 is assigned from outside; (4) decide whether (x+1)+x and 2x+1, and whether x·x and 2x, are equivalent on ℤ; (5) distinguish the expression x+1 from the equation x+1=3 and find the solution set of the equation in ℤ."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "數系與運算：ℤ⊂ℚ，使用一般的加法與乘法；精確除法 u÷v 在數系 S 中定義為使 v·w=u 的唯一 w∈S，若沒有這樣的 w 則在 S 中沒有定義（在 ℚ 中對 v≠0 有 u÷v=u·(1/v)）。",
          "en": "Systems and operations: ℤ⊂ℚ with the usual addition and multiplication; exact division u÷v in a system S is the unique w∈S with v·w=u, and is undefined in S when there is no such w (in ℚ, for v≠0, u÷v=u·(1/v))."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "式子：E=x+1，F=(x+1)÷2，H=Σ_{x=1}^{3}(x+1)；其中 Σ_{x=1}^{3} g(x) 表示 g(1)+g(2)+g(3)，H 中的 x 由求和符號約束。",
          "en": "Expressions: E=x+1, F=(x+1)÷2 and H=Σ_{x=1}^{3}(x+1), where Σ_{x=1}^{3} g(x) means g(1)+g(2)+g(3) and the x in H is bound by the summation sign."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "比較的對象：(x+1)+x 與 2x+1；x·x 與 2x；以及 ℤ 上的方程 x+1=3。",
          "en": "Objects to compare: (x+1)+x and 2x+1; x·x and 2x; and the equation x+1=3 over ℤ."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "x+1,\\ (x+1)\\div 2,\\ \\sum_{x=1}^{3}(x+1)",
        "after_tex": "\\text{value}=U[x:=c]\\ \\text{computed in a stated}\\ S",
        "relation": "definition",
        "rule": "mko_var_expr",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 說代數式的值由變數取值、作用域、數系與運算規則共同決定。所以求值前要說清四件事：x 取哪個值（指派）、哪些 x 是自由出現（作用域）、在哪個數系計算，以及用哪些運算規則（G1 的加法、乘法與精確除法）。缺任何一項，「式子的值」都沒有確定意義。",
          "en": "The MKO says the value of an algebraic expression is determined jointly by the assignment, the scope, the number system and the operation rules. So before evaluating, four things must be stated: which value x takes (the assignment), which occurrences of x are free (the scope), which system the computation is in, and which operation rules apply (addition, multiplication and exact division of G1). Without any one of them, 'the value of the expression' has no definite meaning."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_var_expr",
            "eval_assign"
          ],
          "explanation": {
            "zh-Hant": "此步只列出求值所需的資料，不推出新事實。",
            "en": "This step only lists the data that evaluation needs and derives no new fact."
          }
        },
        "leaf_basis": "mko_var_expr + eval_assign"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "G2"
        ],
        "before_tex": "E[x:=2]\\ \\text{in}\\ \\mathbb{Z}",
        "after_tex": "2+1=3",
        "relation": "equivalent",
        "rule": "eval_assign",
        "conditions": [],
        "explanation": {
          "zh-Hant": "E=x+1 中 x 自由出現一次；依 eval_assign 把它換成 2 得 2+1，在 ℤ 中計算得 3。",
          "en": "In E=x+1 the variable x occurs free once; by eval_assign it is replaced by 2, giving 2+1, which computes to 3 in ℤ."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eval_assign",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "E[x:=2]、2+1 與 3 是同一個整數的不同寫法，可獨立重算（N1）；這只是 x=2 時的值。",
            "en": "E[x:=2], 2+1 and 3 are different names for the same integer and can be recomputed independently (N1); this is only the value at x=2."
          }
        },
        "leaf_basis": "eval_assign + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "G2"
        ],
        "before_tex": "E[x:=5]\\ \\text{in}\\ \\mathbb{Z}",
        "after_tex": "5+1=6",
        "relation": "equivalent",
        "rule": "eval_assign",
        "conditions": [],
        "explanation": {
          "zh-Hant": "同一個式子，換一個指派：把 x 換成 5 得 5+1，在 ℤ 中等於 6。",
          "en": "The same expression under a different assignment: replacing x by 5 gives 5+1, which equals 6 in ℤ."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eval_assign",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個整數，可重算（N2）。",
            "en": "Both sides are the same integer and can be recomputed (N2)."
          }
        },
        "leaf_basis": "eval_assign + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "S03"
        ],
        "before_tex": "E[x:=2]=3,\\ E[x:=5]=6",
        "after_tex": "\\neg\\exists v\\in\\mathbb{Z}\\ \\forall c\\in\\mathbb{Z}:\\ E[x:=c]=v",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "假設有一個固定的整數 v，使每個指派都有 E[x:=c]=v。取 c=2 得 v=3，取 c=5 得 v=6，依等號的對稱與傳遞得 3=6，與 3≠6 矛盾。所以 E 沒有與 x 無關的固定值；說「x+1 的值」時必須先指定 x。",
          "en": "Suppose some fixed integer v had E[x:=c]=v for every assignment. Taking c=2 gives v=3 and taking c=5 gives v=6, so symmetry and transitivity of equality give 3=6, contradicting 3≠6. So E has no fixed value independent of x; 'the value of x+1' only makes sense once x is assigned."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S03",
            "contradiction",
            "eq_refl_symm_trans"
          ],
          "explanation": {
            "zh-Hant": "兩個不同的值就足以推翻「有固定值」；這一步不需要也不宣稱 E 的值域是什麼。",
            "en": "Two different values are enough to refute 'has a fixed value'; the step neither needs nor claims anything about the full range of E."
          }
        },
        "leaf_basis": "contradiction + eq_refl_symm_trans + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "S01",
          "G1",
          "G2"
        ],
        "before_tex": "F[x:=4]\\ \\text{in}\\ \\mathbb{Q}",
        "after_tex": "(4+1)\\div 2=5\\cdot\\tfrac{1}{2}=\\tfrac{5}{2}",
        "relation": "equivalent",
        "rule": "eval_assign",
        "conditions": [
          "2\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "把 x 換成 4 得 (4+1)÷2=5÷2。在 ℚ 中 2≠0，所以依 G1，5÷2=5·(1/2)=5/2；它是 2w=5 在 ℚ 中唯一的解（代回 2·(5/2)=5）。",
          "en": "Replacing x by 4 gives (4+1)÷2=5÷2. In ℚ, 2≠0, so by G1, 5÷2=5·(1/2)=5/2; it is the unique solution of 2w=5 in ℚ (substituting back, 2·(5/2)=5)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eval_assign",
            "mul_inverse_nonzero",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個有理數；條件 2≠0 使 1/2 存在，唯一性來自兩邊同乘 1/2 與同乘 2 互為反運算。",
            "en": "Both sides are the same rational number; the condition 2≠0 makes 1/2 exist, and uniqueness comes from multiplying by 1/2 and by 2 undoing each other."
          }
        },
        "leaf_basis": "eval_assign + mul_inverse_nonzero + two_nonzero + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "F[x:=4]\\ \\text{in}\\ \\mathbb{Z}:\\ w\\in\\mathbb{Z}\\ \\text{with}\\ 2w=5\\ ?",
        "after_tex": "w\\le 2\\ \\lor\\ w\\ge 3",
        "relation": "case_split",
        "rule": "int_no_gap",
        "conditions": [
          "w\\in\\mathbb{Z}"
        ],
        "explanation": {
          "zh-Hant": "在 ℤ 中，(4+1)÷2=5÷2 需要一個整數 w 使 2w=5。依 int_no_gap（m=2），每個整數 w 滿足 w≤2 或 w≥3，兩支涵蓋全部整數，在 S06.1、S06.2 處理。",
          "en": "In ℤ, (4+1)÷2=5÷2 needs an integer w with 2w=5. By int_no_gap (m=2), every integer w satisfies w≤2 or w≥3; the two branches cover all integers and are handled in S06.1 and S06.2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "int_no_gap"
          ],
          "explanation": {
            "zh-Hant": "分情況依賴 w 是整數；在 ℚ 中 w=5/2 就落在兩支之間，這正是 S05 與 S07 結論不同的原因。",
            "en": "The case split relies on w being an integer; in ℚ, w=5/2 lies between the two branches, which is exactly why S05 and S07 reach different conclusions."
          }
        },
        "leaf_basis": "int_no_gap + case_analysis"
      },
      {
        "id": "S06.1",
        "uses": [
          "S06"
        ],
        "before_tex": "w\\le 2",
        "after_tex": "2w\\le 4<5,\\ \\text{so}\\ 2w\\neq 5",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "2>0"
        ],
        "explanation": {
          "zh-Hant": "若 w<2，乘以正數 2 保持順序，得 2w<4；若 w=2，則 2w=4。所以 2w≤4<5，於是 2w≠5。",
          "en": "If w<2, multiplying by the positive number 2 preserves order, so 2w<4; if w=2 then 2w=4. So 2w≤4<5, and therefore 2w≠5."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics",
            "two_nonzero"
          ],
          "explanation": {
            "zh-Hant": "用到 2>0 保持不等號方向；只推出 2w≠5，不宣稱反向。",
            "en": "2>0 keeps the direction of the inequality; only 2w≠5 is concluded and no converse is claimed."
          }
        },
        "leaf_basis": "order_basics + numeric_evaluation"
      },
      {
        "id": "S06.2",
        "uses": [
          "S06"
        ],
        "before_tex": "w\\ge 3",
        "after_tex": "2w\\ge 6>5,\\ \\text{so}\\ 2w\\neq 5",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "2>0"
        ],
        "explanation": {
          "zh-Hant": "若 w>3，乘以 2>0 得 2w>6；若 w=3，則 2w=6。所以 2w≥6>5，於是 2w≠5。",
          "en": "If w>3, multiplying by 2>0 gives 2w>6; if w=3 then 2w=6. So 2w≥6>5, and therefore 2w≠5."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "同 S06.1；不宣稱反向。",
            "en": "As in S06.1; no converse is claimed."
          }
        },
        "leaf_basis": "order_basics + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "S06.1",
          "S06.2",
          "S05",
          "G1"
        ],
        "before_tex": "\\forall w\\in\\mathbb{Z}:\\ 2w\\neq 5",
        "after_tex": "F[x:=4]\\ \\text{has no value in}\\ \\mathbb{Z},\\ \\text{while}\\ F[x:=4]=\\tfrac{5}{2}\\ \\text{in}\\ \\mathbb{Q}",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩支都得 2w≠5，所以 ℤ 中沒有 w 使 2w=5；依 G1，5÷2 在 ℤ 中沒有定義，再依 eval_assign，F 在 x=4 時於 ℤ 中沒有值。對照 S05：同一個式子、同一個指派，換了數系，從「值為 5/2」變成「沒有值」。數系是求值資料的一部分。",
          "en": "Both branches give 2w≠5, so no w in ℤ has 2w=5; by G1, 5÷2 is undefined in ℤ, and by eval_assign, F has no value at x=4 in ℤ. Compare S05: the same expression with the same assignment goes from 'value 5/2' to 'no value' when the system changes. The number system is part of the data of evaluation."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S06.1",
            "S06.2",
            "case_analysis",
            "eval_assign"
          ],
          "explanation": {
            "zh-Hant": "需要兩支涵蓋所有整數；結論只關於 x=4，不宣稱 F 在 ℤ 中對任何 x 都沒有值（例如 x=3 時 (3+1)÷2=2）。",
            "en": "It needs the two branches to cover all integers; the conclusion is only about x=4 and does not claim that F has no value in ℤ for every x (for example at x=3, (3+1)÷2=2)."
          }
        },
        "leaf_basis": "case_analysis + eval_assign + quantifier_negation"
      },
      {
        "id": "S08",
        "uses": [
          "G2",
          "bound_var"
        ],
        "before_tex": "H=\\sum_{x=1}^{3}(x+1)",
        "after_tex": "H=(1+1)+(2+1)+(3+1)",
        "relation": "equivalent",
        "rule": "finite_sum_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依求和記號的意義（G2，亦即 finite_sum_rules 的遞迴定義展開三次），H 是把約束變數 x 依序換成 1、2、3 後相加。這裡的 x 只取 1、2、3，與外部是否指定 x 無關。",
          "en": "By the meaning of the summation sign (G2, that is, the recursive definition in finite_sum_rules unfolded three times), H is the sum obtained by setting the bound variable x to 1, 2 and 3 in turn. This x takes only the values 1, 2, 3, whatever is assigned to x outside."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "finite_sum_rules",
            "bound_var"
          ],
          "explanation": {
            "zh-Hant": "展開與收回求和記號是同一定義的兩個方向。",
            "en": "Unfolding and folding the summation sign are the two directions of one definition."
          }
        },
        "leaf_basis": "finite_sum_rules + bound_var"
      },
      {
        "id": "S09",
        "uses": [
          "S08"
        ],
        "before_tex": "(1+1)+(2+1)+(3+1)",
        "after_tex": "2+3+4=9",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐項計算 1+1=2、2+1=3、3+1=4，再相加得 9。",
          "en": "Compute term by term, 1+1=2, 2+1=3 and 3+1=4, then add to get 9."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個整數，可重算（N8）。",
            "en": "Both sides are the same integer and can be recomputed (N8)."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "bound_var",
          "eval_assign"
        ],
        "before_tex": "H[x:=5]",
        "after_tex": "H[x:=5]=H=9\\neq 18=(5+1)+(5+1)+(5+1)",
        "relation": "implies",
        "rule": "bound_var",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 eval_assign，指派 x:=5 只替換自由出現的 x；H 中的 x 全是約束出現（bound_var），沒有自由出現，所以 H[x:=5] 就是 H 本身，值仍為 9。若錯把約束的 x 也換成 5，會得到 (5+1)+(5+1)+(5+1)=18，與 9 不同；這正是忽略作用域的錯誤。",
          "en": "By eval_assign, the assignment x:=5 replaces only free occurrences of x; every x in H is bound (bound_var) and none is free, so H[x:=5] is H itself and its value is still 9. Wrongly replacing the bound x by 5 as well would give (5+1)+(5+1)+(5+1)=18, which differs from 9; this is exactly the error of ignoring scope."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "bound_var",
            "eval_assign"
          ],
          "explanation": {
            "zh-Hant": "需要辨認 H 中每個 x 都被 Σ 約束；結論只關於 H，不宣稱外部的 x 在其他式子中沒有作用。",
            "en": "It needs every x in H to be recognised as bound by the Σ; the conclusion is about H only and does not claim that the outside x has no effect in other expressions."
          }
        },
        "leaf_basis": "bound_var + eval_assign + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "G3"
        ],
        "before_tex": "(x+1)+x",
        "after_tex": "2x+1",
        "relation": "equivalent",
        "rule": "L_double",
        "conditions": [
          "x\\in\\mathbb{Z}"
        ],
        "explanation": {
          "zh-Hant": "任取 x∈ℤ，不加任何額外假設。依引理 L_double，(x+1)+x=2x+1。引理在 S11.1、S11.2 中只用 ℤ 的運算規則證明，而不是看兩式外觀相似。",
          "en": "Take any x∈ℤ, with no extra assumption. By Lemma L_double, (x+1)+x=2x+1. The lemma is proved in S11.1 and S11.2 from the operation rules of ℤ only, not from the two expressions looking alike."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_double"
          ],
          "explanation": {
            "zh-Hant": "兩式是同一個整數的兩種寫法；引理的每一步都是等值代換，可逆。",
            "en": "The two expressions name the same integer; every step of the lemma is a replacement of equals and can be reversed."
          }
        },
        "leaf_basis": "add_assoc + add_comm + distributive + mul_one + numeric_evaluation"
      },
      {
        "id": "S11.1",
        "uses": [
          "add_assoc",
          "add_comm"
        ],
        "before_tex": "(x+1)+x",
        "after_tex": "(x+x)+1",
        "relation": "equivalent",
        "rule": "add_assoc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(x+1)+x=x+(1+x)（結合律）=x+(x+1)（交換律）=(x+x)+1（結合律）。",
          "en": "(x+1)+x=x+(1+x) (associativity) =x+(x+1) (commutativity) =(x+x)+1 (associativity)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "add_assoc",
            "add_comm"
          ],
          "explanation": {
            "zh-Hant": "結合律與交換律都是對所有整數成立的等式，倒著讀同樣成立。",
            "en": "Associativity and commutativity are identities for all integers and hold read backwards too."
          }
        },
        "leaf_basis": "add_assoc + add_comm"
      },
      {
        "id": "S11.2",
        "uses": [
          "S11.1"
        ],
        "before_tex": "(x+x)+1",
        "after_tex": "2x+1",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x+x=1·x+1·x（乘法單位元）=(1+1)x（分配律與交換律）=2x，所以 (x+x)+1=2x+1。S11.1 與 S11.2 合起來證得引理 L_double。",
          "en": "x+x=1·x+1·x (multiplicative identity) =(1+1)x (distributivity with commutativity) =2x, so (x+x)+1=2x+1. S11.1 and S11.2 together prove Lemma L_double."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "mul_one",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "所用的都是恆等式，等值代換雙向成立。",
            "en": "Only identities are used, so the replacement of equals holds both ways."
          }
        },
        "leaf_basis": "distributive + mul_one + mul_comm + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S11",
          "expr_equiv"
        ],
        "before_tex": "\\text{for arbitrary}\\ x\\in\\mathbb{Z}:\\ (x+1)+x=2x+1",
        "after_tex": "(x+1)+x\\equiv_{\\mathbb{Z}}2x+1",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S11 對任意 x∈ℤ 成立，沒有對 x 加任何條件；依全稱推廣，對所有 x∈ℤ 兩式的值相等，而且兩式在 ℤ 中總有值（只用加法與乘法），依 expr_equiv 兩式在 ℤ 上等價。語法不同的式子可以等價，但理由是運算規則。",
          "en": "S11 holds for an arbitrary x∈ℤ with no condition on x; by universal generalisation the two expressions have equal values for every x∈ℤ, and both always have values in ℤ (only addition and multiplication occur), so by expr_equiv they are equivalent on ℤ. Syntactically different expressions can be equivalent, but the reason is the operation rules."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "forall_intro",
            "expr_equiv"
          ],
          "explanation": {
            "zh-Hant": "需要 S11 中的 x 是任意的；N11 只是在少數格點上的獨立交叉核對，一般性來自 S11 的規則推導。",
            "en": "It needs the x of S11 to be arbitrary; N11 is only an independent cross-check on a few grid points, and the generality comes from the rule-based derivation in S11."
          }
        },
        "leaf_basis": "forall_intro + expr_equiv"
      },
      {
        "id": "S13",
        "uses": [
          "G3",
          "expr_equiv"
        ],
        "before_tex": "x\\cdot x\\ \\text{vs}\\ 2x",
        "after_tex": "1\\cdot 1=1\\neq 2=2\\cdot 1,\\ \\text{so}\\ x\\cdot x\\not\\equiv_{\\mathbb{Z}}2x",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "等價要求每個 x 都相等。取 x=1：x·x=1，2x=2，1≠2，所以存在使兩式不等的 x，依量詞否定兩式在 ℤ 上不等價。注意兩式在 x=0（都為 0）與 x=2（都為 4）相等：在部分取值上相等不足以推出等價。",
          "en": "Equivalence requires equality at every x. Take x=1: x·x=1 and 2x=2, and 1≠2, so some x makes the two differ, and by negation of quantifiers they are not equivalent on ℤ. Note that they agree at x=0 (both 0) and at x=2 (both 4): agreeing at some values is not enough for equivalence."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "expr_equiv",
            "quantifier_negation",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一個反例足以否定「對所有 x 相等」；不宣稱兩式只在 x=1 不等。",
            "en": "One counterexample is enough to refute 'equal for all x'; it is not claimed that x=1 is the only place where they differ."
          }
        },
        "leaf_basis": "quantifier_negation + expr_equiv + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "G3",
          "S02",
          "mko_equation"
        ],
        "before_tex": "x+1\\quad\\text{vs}\\quad x+1=3",
        "after_tex": "\\{x\\in\\mathbb{Z}\\mid x+1=3\\}=\\{2\\}",
        "relation": "implies",
        "rule": "mko_equation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x+1 沒有等號，是表達式：它在每個指派下有一個值（S02、S03），但本身沒有真假，也沒有解集。x+1=3 有等號，是方程：依 mko_equation，它的解集是使兩側都有定義且相等的 x∈ℤ。兩側在 ℤ 中總有定義；兩邊同加 −1 得 x+1=3 ⇔ x=2（反向是同加 1），所以解集是 {2}。S02 的 E[x:=2]=3 正是這個解的代回核對。",
          "en": "x+1 has no equality sign, so it is an expression: it has a value under each assignment (S02, S03), but by itself it is neither true nor false and has no solution set. x+1=3 has an equality sign, so it is an equation: by mko_equation its solution set consists of the x∈ℤ at which both sides are defined and equal. Both sides are always defined in ℤ; adding −1 to both sides gives x+1=3 ⇔ x=2 (backwards, add 1), so the solution set is {2}. E[x:=2]=3 from S02 is exactly the substitution check of this solution."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_equation",
            "eq_congruence",
            "S02",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "解集的計算用了等價變形（同加 −1 與同加 1 互逆），而且兩側在 ℤ 中都有定義；這一步沒有把表達式 x+1 本身當作方程。",
            "en": "The solution set is computed by an equivalence (adding −1 and adding 1 undo each other), and both sides are defined in ℤ; the step never treats the expression x+1 itself as an equation."
          }
        },
        "leaf_basis": "mko_equation + eq_congruence + add_inverse + set_extensionality"
      },
      {
        "id": "S15",
        "uses": [
          "S04",
          "S07",
          "S10",
          "S12",
          "S13",
          "S14"
        ],
        "before_tex": "\\text{conclusions of S04, S07, S10, S12, S13, S14}",
        "after_tex": "\\text{value of }U=\\text{function of assignment, scope, system and rules}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把結論合在一起：取值改變值（S04）；數系改變有無值（S07）；作用域決定哪些 x 被指派取代（S10）；等價由運算規則對所有取值證明（S12），部分取值相等不夠（S13）；表達式與方程角色不同（S14）。",
          "en": "Together: changing the assignment changes the value (S04); changing the system changes whether there is a value (S07); scope decides which occurrences of x an assignment replaces (S10); equivalence is proved for all values from the operation rules (S12), and agreement at some values is not enough (S13); an expression and an equation play different roles (S14)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S04",
            "S07",
            "S10",
            "S12",
            "S13",
            "S14",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：回答題目的五個部分。足夠條件 H：S04、S07、S10、S12、S13、S14 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 of the problem. Sufficient condition H: the conclusions of S04, S07, S10, S12, S13 and S14; 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": "在 ℤ 中 E 在 x=2 時為 3、在 x=5 時為 6，因此沒有固定值。F 在 x=4 時於 ℚ 中為 5/2，於 ℤ 中沒有值。外部指定 x=5 時 H 仍為 9，因為 H 中的 x 是約束變數。(x+1)+x 與 2x+1 在 ℤ 上等價；x·x 與 2x 不等價（x=1 時 1≠2）。x+1 是表達式，沒有解集；x+1=3 是方程，在 ℤ 中的解集為 {2}。",
      "en": "In ℤ, E equals 3 at x=2 and 6 at x=5, so it has no fixed value. F at x=4 equals 5/2 in ℚ and has no value in ℤ. With x=5 assigned outside, H is still 9, because the x in H is bound. (x+1)+x and 2x+1 are equivalent on ℤ; x·x and 2x are not (at x=1, 1≠2). x+1 is an expression with no solution set; x+1=3 is an equation whose solution set in ℤ is {2}."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "x+1 的值是 3",
          "en": "The value of x+1 is 3"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "只在指派 x=2 時成立；x=5 時值為 6（S03、S04）。",
          "en": "Only under the assignment x=2; at x=5 the value is 6 (S03, S04)."
        }
      },
      {
        "claim": {
          "zh-Hant": "x+1 是方程，它的解是 x=−1",
          "en": "x+1 is an equation and its solution is x=−1"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x+1 沒有等號；x=−1 是另一個方程 x+1=0 的解，而那個方程不是題目給的。",
          "en": "x+1 has no equality sign; x=−1 solves a different equation, x+1=0, which the problem never stated."
        }
      },
      {
        "claim": {
          "zh-Hant": "x·x 與 2x 在 x=0 與 x=2 時相等，所以它們在 ℤ 上等價",
          "en": "x·x and 2x agree at x=0 and x=2, so they are equivalent on ℤ"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x=1 時 1≠2（S13）；部分取值相等不足以推出等價。",
          "en": "At x=1, 1≠2 (S13); agreement at some values does not give equivalence."
        }
      },
      {
        "claim": {
          "zh-Hant": "(x+1)÷2 在 x=4 時的值是 5/2，在任何數系都一樣",
          "en": "(x+1)÷2 at x=4 is 5/2 in every number system"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "在 ℤ 中沒有整數 w 使 2w=5，所以沒有值（S07）。",
          "en": "In ℤ no integer w has 2w=5, so there is no value (S07)."
        }
      },
      {
        "claim": {
          "zh-Hant": "外部指定 x=5 時，Σ_{x=1}^{3}(x+1)=18",
          "en": "With x=5 assigned outside, Σ_{x=1}^{3}(x+1)=18"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "H 中的 x 是約束變數，不被外部指派取代，值仍為 9（S10）。",
          "en": "The x in H is bound and is not replaced by the outside assignment; the value is still 9 (S10)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "add_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。",
          "en": "Associativity of addition: (u+v)+w=u+(v+w)."
        }
      },
      {
        "id": "add_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法交換律：u+v=v+u。",
          "en": "Commutativity of addition: u+v=v+u."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "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": "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": "eq_refl_symm_trans",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。",
          "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w."
        }
      },
      {
        "id": "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": "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": "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": "finite_sum_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限和：Σ_{i=1}^{0} a_i=0、Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}；Σ(a_i+b_i)=Σa_i+Σb_i；Σ c·a_i=c·Σa_i。",
          "en": "Finite sums: Σ_{i=1}^{0} a_i=0 and Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}; Σ(a_i+b_i)=Σa_i+Σb_i; Σ c·a_i=c·Σa_i."
        }
      },
      {
        "id": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "id": "mul_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。",
          "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)."
        }
      },
      {
        "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": "eq_congruence",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。",
          "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule."
        }
      },
      {
        "id": "add_inverse",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。",
          "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)."
        }
      },
      {
        "id": "set_extensionality",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合相等：A=B 當且僅當 A⊆B 且 B⊆A，即兩者有相同元素。",
          "en": "Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements."
        }
      },
      {
        "id": "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_var_expr",
        "kind": "cited",
        "source": {
          "mko_id": "mko-variable-expression",
          "sha256": "da4b6e4a4edfd379a38b6535fd3030cc5c552e2b549572c0d4733d68974a292c",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "代數式的值由其中變數的取值、作用域、型別或數系以及運算規則共同決定；沒有等號的代數式本身不是方程式。",
          "en": "The value of an algebraic expression is jointly determined by its variable assignments, scopes, types or number systems, and operation rules. An algebraic expression without an equality sign is not itself an equation."
        }
      },
      {
        "id": "mko_equation",
        "kind": "cited",
        "source": {
          "mko_id": "mko-equation",
          "sha256": "6f8d6ca379da35727c14f320d99d0bc8c53c47ad478d0f462d30a4127aea0a78",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "給定兩個具有共同語境的表達式 L(x)、R(x)，方程式 L(x)=R(x) 的解集由所有使兩側皆有定義且值相等的允許 x 組成。",
          "en": "Given two expressions L(x) and R(x) in a common context, the solution set of L(x)=R(x) consists of all allowed x for which both sides are defined and equal."
        }
      },
      {
        "id": "eval_assign",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "求值規則：式子 U 在指派 x:=c（c∈S）下的值，是把 U 中每個自由出現的 x 換成 c，再用 S 的運算計算；若過程中某個運算在 S 中沒有定義，U 在此指派下沒有值。",
          "en": "Evaluation: the value of an expression U under the assignment x:=c (c∈S) is obtained by replacing every free occurrence of x in U by c and computing with the operations of S; if some operation along the way is undefined in S, U has no value under that assignment."
        }
      },
      {
        "id": "bound_var",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "約束變數：Σ_{x=1}^{3} g(x) 中的 x 由求和符號約束，只取 1、2、3；外部對 x 的指派不替換約束出現；把約束變數一致改名（例如改成 k）不改變值。",
          "en": "Bound variables: the x in Σ_{x=1}^{3} g(x) is bound by the summation sign and takes only the values 1, 2, 3; an outside assignment to x does not replace bound occurrences; renaming the bound variable consistently (for example to k) does not change the value."
        }
      },
      {
        "id": "expr_equiv",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "式子等價：U 與 V 在 S 上等價，表示對每個 c∈S，U[x:=c] 與 V[x:=c] 都有值且相等。",
          "en": "Equivalence of expressions: U and V are equivalent on S when, for every c∈S, U[x:=c] and V[x:=c] both have values and these are equal."
        }
      },
      {
        "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_double",
        "kind": "lemma",
        "proved_in": [
          "S11.1",
          "S11.2"
        ],
        "statement": {
          "zh-Hant": "對每個 x∈ℤ：(x+1)+x=2x+1。",
          "en": "For every x∈ℤ: (x+1)+x=2x+1."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "2+1",
        "rhs": "3"
      },
      {
        "id": "N2",
        "step": "S03",
        "lhs": "5+1",
        "rhs": "6"
      },
      {
        "id": "N3",
        "step": "S04",
        "lhs": "3",
        "relation": "!=",
        "rhs": "6"
      },
      {
        "id": "N4",
        "step": "S05",
        "lhs": "(4+1)/2",
        "rhs": "5/2"
      },
      {
        "id": "N5",
        "step": "S05",
        "lhs": "2*(5/2)",
        "rhs": "5"
      },
      {
        "id": "N6",
        "step": "S06.1",
        "lhs": "2*2",
        "relation": "<",
        "rhs": "5"
      },
      {
        "id": "N7",
        "step": "S06.2",
        "lhs": "2*3",
        "relation": ">",
        "rhs": "5"
      },
      {
        "id": "N8",
        "step": "S09",
        "lhs": "(1+1)+(2+1)+(3+1)",
        "rhs": "9"
      },
      {
        "id": "N9",
        "step": "S10",
        "lhs": "(5+1)+(5+1)+(5+1)",
        "rhs": "18"
      },
      {
        "id": "N10",
        "step": "S10",
        "lhs": "18",
        "relation": "!=",
        "rhs": "9"
      },
      {
        "id": "N11",
        "step": "S11",
        "vars": [
          "x"
        ],
        "degree": 1,
        "lhs": "(x+1)+x",
        "rhs": "2*x+1"
      },
      {
        "id": "N12",
        "step": "S13",
        "lhs": "1*1",
        "relation": "!=",
        "rhs": "2*1"
      },
      {
        "id": "N13",
        "step": "S13",
        "lhs": "0*0",
        "rhs": "2*0"
      },
      {
        "id": "N14",
        "step": "S13",
        "lhs": "2*2",
        "rhs": "4"
      },
      {
        "id": "N15",
        "step": "S14",
        "lhs": "3-1",
        "rhs": "2"
      },
      {
        "id": "N16",
        "step": "S07",
        "lhs": "(3+1)/2",
        "rhs": "2"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只用 ℤ、ℚ 的一般運算，除法取精確除法（G1），一個自由變數 x 與一個有界求和中的約束變數。示範 MKO：代數式的值由取值、作用域、數系與運算規則共同決定；沒有等號的代數式不是方程。不建立形式語法或求值函數，不處理多變數或函數型別；不改動 canonical MKO 或英文 edition。",
      "en": "Uses only the usual operations of ℤ and ℚ with exact division (G1), one free variable x and one bound variable of a finite sum. It illustrates the MKO: the value of an algebraic expression is determined jointly by the assignment, the scope, the number system and the operation rules, and an expression without an equality sign is not an equation. It does not build a formal syntax or an evaluation function, does not treat several variables or function types, and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "x",
        "變數；作用域與數系在各部分指定",
        "the variable; its scope and number system are stated in each part"
      ],
      [
        "U[x:=c]",
        "把 U 中自由出現的 x 換成 c 後求得的值",
        "the value of U after replacing the free occurrences of x by c"
      ],
      [
        "u\\div v",
        "精確除法：使 v·w=u 的唯一 w",
        "exact division: the unique w with v·w=u"
      ],
      [
        "\\sum_{x=1}^{3} g(x)",
        "g(1)+g(2)+g(3)，其中 x 是約束變數",
        "g(1)+g(2)+g(3), where x is a bound variable"
      ],
      [
        "U\\equiv_{\\mathbb{Z}} V",
        "U 與 V 在 ℤ 上等價",
        "U and V are equivalent on ℤ"
      ]
    ],
    "md_error_controls": [
      {
        "title": "本例 E=x+1 的值依賴 x 的取值 / The value of E=x+1 depends on the assignment to x",
        "zh": "MKO 指出的常見誤解是把含變數的表達式當成已有固定值。x+1 在 x=2 時是 3，在 x=5 時是 6（S02、S03），S04 證明它沒有與 x 無關的值。所以「x+1=3」只能讀成對 x 的要求（方程），不能讀成「x+1 這個式子就是 3」。",
        "en": "The MKO names the common mistake of treating an expression with a variable as if it already had a fixed value. x+1 is 3 at x=2 and 6 at x=5 (S02, S03), and S04 proves it has no value independent of x. So 'x+1=3' can only be read as a requirement on x (an equation), not as 'the expression x+1 is 3'."
      },
      {
        "title": "沒有等號就不是方程 / Without an equality sign there is no equation",
        "zh": "MKO 的另一個誤解是把任何含 x 的文字都當作方程。x+1 沒有等號，不能「解」；若有人說「解 x+1 得 x=−1」，其實是自行補上了方程 x+1=0。方程需要兩個表達式與等號，解集還要求兩側在指定數系中有定義（S14）。",
        "en": "Another mistake identified in the MKO is calling any text containing x an equation. x+1 has no equality sign and cannot be 'solved'; whoever says 'solving x+1 gives x=−1' has silently supplied the equation x+1=0. An equation needs two expressions and an equality sign, and its solution set also requires both sides to be defined in the stated system (S14)."
      },
      {
        "title": "約束變數不接受外部指派 / Bound variables do not take outside values",
        "zh": "H=Σ_{x=1}^{3}(x+1) 中的 x 被求和符號約束。外部指派 x=5 時 H 仍是 9；把約束出現也換成 5 會錯得 18（S10）。把約束變數改名為 k，寫成 Σ_{k=1}^{3}(k+1)，就看得出它與外部的 x 無關。",
        "en": "The x in H=Σ_{x=1}^{3}(x+1) is bound by the summation sign. With x=5 assigned outside, H is still 9; replacing the bound occurrences by 5 as well wrongly gives 18 (S10). Renaming the bound variable to k, as Σ_{k=1}^{3}(k+1), shows that it has nothing to do with the outside x."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿以 eval_assign、bound_var、expr_equiv 的文字定義作為前提，未建立形式語法、自由變數的遞迴定義或求值函數（MKO 的 formal next）。",
        "en": "This guide takes the verbal definitions eval_assign, bound_var and expr_equiv as premises; it does not build a formal syntax, a recursive definition of free variables or an evaluation function (the MKO's formal next step)."
      },
      {
        "zh-Hant": "N11 是在格點 {0,1} 上的一次多項式恆等式核對，只作為 S11 的獨立交叉核對；本稿的一般性來自 S11 的規則推導。",
        "en": "N11 is a degree-one polynomial identity check on the grid {0,1}, used only as an independent cross-check of S11; the generality in this guide comes from the rule-based derivation of S11."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "7bae56c281b5a8252bb4564f48957ca0b1b4d33f1093e8dfcaed3df6d8e85309",
    "original_source_sha256": "4e7848b5ae3d6ec21f3cf5b0f6297ac58148c0066c80cbcda81b003b28355a4a",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M005"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}