{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-linear-function-01",
  "mko_id": "mko-linear-function",
  "title": {
    "zh-Hant": "仿射函數與一次函數：由兩點求 m、b，判定次數、零點與是否為線性映射",
    "en": "Affine and degree-one functions: finding m and b from two points, and deciding degree, zero and linearity"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-linear-function-01",
    "source": {
      "mko_id": "mko-linear-function",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "718236b5b7dfecd9f4d276590f1ddc012cfd5df96864540a3840a2b4e6650abe",
      "english_sha256": "15d0d8ac9fe6a30744f0bbe21d2d94a8273271f0e4c8c1d4ed982b99f72e098b"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "distributive",
      "eq_congruence",
      "mul_inverse_nonzero",
      "function_def",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "仿射函數 f:ℝ→ℝ，f(x)=mx+b 的圖形通過 (1,4) 與 (3,0)。求 m、b；判定 f 是否為次數一的函數、求它的零點，並判定 f 是否為線性代數意義下的 ℝ 線性映射。再證明兩個仿射函數在 ℝ 上相等當且僅當斜率與截距都相同，據此說明 f(2)=k(2)（k(x)=x）不能推出 f=k；最後處理 m=0 的兩個退化例子 g(x)=5 與 h(x)=0。",
      "en": "The graph of the affine function f:ℝ→ℝ, f(x)=mx+b, passes through (1,4) and (3,0). Find m and b; decide whether f has degree one, find its zero, and decide whether f is an ℝ-linear map in the sense of linear algebra. Then prove that two affine functions are equal on ℝ exactly when their slopes and intercepts agree, and use it to explain why f(2)=k(2) (with k(x)=x) does not give f=k; finally treat the two degenerate examples with m=0, g(x)=5 and h(x)=0."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "f:ℝ→ℝ，f(x)=mx+b，m、b 為實常數（MKO 假設 a1）。m≠0 時稱次數一，m=0 時為常數函數；線性代數中的 ℝ 線性映射另要求可加性與齊次性。",
          "en": "f:ℝ→ℝ with f(x)=mx+b and real constants m, b (MKO assumption a1). It has degree one when m≠0 and is constant when m=0; an ℝ-linear map in linear algebra additionally requires additivity and homogeneity."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "f 的圖形通過 (1,4) 與 (3,0)，即 f(1)=4、f(3)=0。",
          "en": "The graph of f passes through (1,4) and (3,0), that is, f(1)=4 and f(3)=0."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "比較用的函數：k(x)=x、g(x)=0·x+5、h(x)=0·x+0，皆為 ℝ→ℝ。",
          "en": "Functions for comparison: k(x)=x, g(x)=0·x+5 and h(x)=0·x+0, all ℝ→ℝ."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "f:\\mathbb{R}\\to\\mathbb{R}",
        "after_tex": "f(x)=mx+b;\\ m\\neq0\\Rightarrow\\deg f=1;\\ m=0\\Rightarrow f\\ \\text{constant}",
        "relation": "definition",
        "rule": "mko_affine",
        "conditions": [
          "m,b\\in\\mathbb{R}"
        ],
        "explanation": {
          "zh-Hant": "依 MKO，所有 mx+b 形式的函數都是仿射函數；其中只有 m≠0 的稱為次數一，m=0 的是常數函數。線性代數的「線性映射」是另一個概念，要求可加性與齊次性（linear_map_def），MKO 指出它另需 b=0。三個名稱不可混用。",
          "en": "By the MKO, every function of the form mx+b is affine; only those with m≠0 have degree one, and those with m=0 are constant. A linear map in linear algebra is a separate notion requiring additivity and homogeneity (linear_map_def), and the MKO notes that it additionally needs b=0. The three names must not be mixed up."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_affine",
            "linear_map_def"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出定義與分類，不推出新事實。",
            "en": "This step only states definitions and the classification; no new fact is derived."
          }
        },
        "leaf_basis": "mko_affine + def_intro"
      },
      {
        "id": "S02",
        "uses": [
          "S01"
        ],
        "before_tex": "f(x_2)-f(x_1)",
        "after_tex": "m(x_2-x_1)",
        "relation": "equivalent",
        "rule": "L_diff",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_diff，仿射函數的輸出差等於斜率乘輸入差；截距 b 在相減時消去。證明在 S02.1。",
          "en": "By Lemma L_diff, the difference of outputs of an affine function is the slope times the difference of inputs; the intercept b cancels in the subtraction. The proof is S02.1."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_diff"
          ],
          "explanation": {
            "zh-Hant": "兩邊是同一個實數的兩種寫法，對所有 x_1、x_2 都成立（N1）。",
            "en": "Both sides name the same real number for all x_1, x_2 (N1)."
          }
        },
        "leaf_basis": "distributive + add_inverse + add_comm"
      },
      {
        "id": "S02.1",
        "uses": [
          "distributive",
          "add_inverse"
        ],
        "before_tex": "(mx_2+b)-(mx_1+b)",
        "after_tex": "mx_2-mx_1=m(x_2-x_1)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "展開減法：(mx_2+b)−(mx_1+b)=mx_2+b−mx_1−b；b−b=0，剩下 mx_2−mx_1，依分配律提出 m 得 m(x_2−x_1)。這證得引理 L_diff。",
          "en": "Expand the subtraction: (mx_2+b)−(mx_1+b)=mx_2+b−mx_1−b; b−b=0 leaves mx_2−mx_1, and distributivity takes out m to give m(x_2−x_1). This proves Lemma L_diff."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "add_inverse",
            "add_comm"
          ],
          "explanation": {
            "zh-Hant": "每一步都是實數恆等式，雙向成立。",
            "en": "Every move is an identity of real numbers, valid both ways."
          }
        },
        "leaf_basis": "distributive + add_inverse + add_comm + add_assoc"
      },
      {
        "id": "S03",
        "uses": [
          "G2",
          "S02"
        ],
        "before_tex": "f(3)-f(1)=m(3-1)",
        "after_tex": "0-4=2m,\\ \\text{i.e.}\\ -4=2m",
        "relation": "theorem_application",
        "rule": "forall_elim",
        "conditions": [
          "x_1=1\\neq 3=x_2"
        ],
        "explanation": {
          "zh-Hant": "在 L_diff 中取 x_1=1、x_2=3，代入 f(1)=4、f(3)=0：0−4=m(3−1)，即 −4=2m。兩點的橫坐標不同（1≠3），所以 3−1=2≠0，下一步可以除。",
          "en": "Take x_1=1 and x_2=3 in L_diff and substitute f(1)=4 and f(3)=0: 0−4=m(3−1), that is −4=2m. The two inputs differ (1≠3), so 3−1=2≠0 and the next step may divide."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "S02",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "需要兩點都在圖形上（G2）；若 x_1=x_2，等式變成 0=0，不能決定 m（MKO 假設 a2）。",
            "en": "It needs both points on the graph (G2); if x_1=x_2 the equation would read 0=0 and could not determine m (MKO assumption a2)."
          }
        },
        "leaf_basis": "forall_elim + eq_replace + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "mul_inverse_nonzero"
        ],
        "before_tex": "-4=2m",
        "after_tex": "m=\\frac{-4}{2}=-2",
        "relation": "equivalent",
        "rule": "mul_inverse_nonzero",
        "conditions": [
          "2\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "因 2≠0，兩邊同乘 1/2 得 m=−4/2=−2。這就是兩點斜率公式 m=(y_2−y_1)/(x_2−x_1)=(0−4)/(3−1)。",
          "en": "Since 2≠0, multiplying both sides by 1/2 gives m=−4/2=−2. This is the two-point slope formula m=(y_2−y_1)/(x_2−x_1)=(0−4)/(3−1)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_inverse_nonzero",
            "eq_congruence",
            "two_nonzero"
          ],
          "explanation": {
            "zh-Hant": "乘 1/2 與乘 2 互為反運算，前提 2≠0；代回 2·(−2)=−4。",
            "en": "Multiplying by 1/2 and by 2 undo each other, given 2≠0; substituting back, 2·(−2)=−4."
          }
        },
        "leaf_basis": "mul_inverse_nonzero + eq_congruence + two_nonzero + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "G2",
          "S04"
        ],
        "before_tex": "f(1)=4:\\ -2\\cdot 1+b=4",
        "after_tex": "b=4+2=6",
        "relation": "equivalent",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 m=−2 與 f(1)=4 代入 f(x)=mx+b：−2·1+b=4。兩邊同加 2 得 b=6。",
          "en": "Substitute m=−2 and f(1)=4 into f(x)=mx+b: −2·1+b=4. Adding 2 to both sides gives b=6."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "同加 2 與同減 2 互為反運算；代回 −2+6=4。",
            "en": "Adding 2 and subtracting 2 undo each other; substituting back, −2+6=4."
          }
        },
        "leaf_basis": "eq_congruence + add_inverse + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S04",
          "S05",
          "G2"
        ],
        "before_tex": "m=-2,\\ b=6",
        "after_tex": "f(x)=-2x+6,\\quad f(3)=-6+6=0",
        "relation": "implies",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "所以 f(x)=−2x+6。核對第二個點：f(3)=−2·3+6=0，與 G2 一致；f(1)=−2+6=4 也一致。S03 到 S05 每一步都是等價或由已知直接推出，所以通過這兩點的仿射函數只有這一個。",
          "en": "So f(x)=−2x+6. Check the second point: f(3)=−2·3+6=0, consistent with G2, and f(1)=−2+6=4 as well. Every move in S03 to S05 was an equivalence or followed directly from the data, so this is the only affine function through the two points."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S05",
            "G2",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "核對兩點只是確認計算無誤（N5、N6）；唯一性來自 S03 至 S05 的推導，不是來自這兩次代入。",
            "en": "Checking the two points only confirms the arithmetic (N5, N6); uniqueness comes from the derivation in S03 to S05, not from these two substitutions."
          }
        },
        "leaf_basis": "eq_replace + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "S02"
        ],
        "before_tex": "f(x)=-2x+6",
        "after_tex": "b=f(0)=6,\\quad f(x+1)-f(x)=m=-2",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO para0 的解讀：b=f(0)=−2·0+6=6；在 L_diff 中取 x_2=x_1+1，得 f(x+1)−f(x)=m·1=−2，即輸入每增加一單位，輸出減少 2。",
          "en": "Following MKO para0: b=f(0)=−2·0+6=6; taking x_2=x_1+1 in L_diff gives f(x+1)−f(x)=m·1=−2, so each unit increase of the input lowers the output by 2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S06",
            "forall_elim"
          ],
          "explanation": {
            "zh-Hant": "兩個解讀都是已證式子的特例，不另加前提（N7）。",
            "en": "Both readings are special cases of results already proved, with no extra premise (N7)."
          }
        },
        "leaf_basis": "forall_elim + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S01",
          "S04"
        ],
        "before_tex": "m=-2",
        "after_tex": "m\\neq 0\\Rightarrow f\\ \\text{has degree one}",
        "relation": "implies",
        "rule": "mko_affine",
        "conditions": [
          "m\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "m=−2≠0，依 MKO，f 作為多項式的次數是一，是一次函數。若 m 為 0，即使寫成 mx+b 的形式也不是一次函數（見 S13）。",
          "en": "m=−2≠0, so by the MKO f has polynomial degree one and is a degree-one function. If m were 0 it would not be degree one, even though it is written as mx+b (see S13)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "mko_affine"
          ],
          "explanation": {
            "zh-Hant": "判準只看 m 是否為零；截距 b 不影響次數。",
            "en": "The test looks only at whether m is zero; the intercept b does not affect the degree."
          }
        },
        "leaf_basis": "mko_affine + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S06",
          "mul_inverse_nonzero"
        ],
        "before_tex": "f(x)=0:\\ -2x+6=0",
        "after_tex": "x=\\frac{-6}{-2}=3=-\\frac{b}{m}",
        "relation": "equivalent",
        "rule": "mul_inverse_nonzero",
        "conditions": [
          "m=-2\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "−2x+6=0 兩邊同減 6 得 −2x=−6；因 −2≠0，兩邊同乘 −1/2 得 x=3。每一步都是等價，所以零點唯一，就是 −b/m=−6/(−2)=3，與 G2 的點 (3,0) 一致。",
          "en": "Subtract 6 from both sides of −2x+6=0 to get −2x=−6; since −2≠0, multiply both sides by −1/2 to get x=3. Every move is an equivalence, so the zero is unique and equals −b/m=−6/(−2)=3, in agreement with the point (3,0) of G2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_inverse_nonzero",
            "eq_congruence"
          ],
          "explanation": {
            "zh-Hant": "同減 6 與同乘 −1/2 都可逆，前提是 m≠0；代回 −2·3+6=0（N8）。",
            "en": "Subtracting 6 and multiplying by −1/2 are both reversible, given m≠0; substituting back, −2·3+6=0 (N8)."
          }
        },
        "leaf_basis": "mul_inverse_nonzero + eq_congruence + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "S06",
          "linear_map_def"
        ],
        "before_tex": "f(0)=6",
        "after_tex": "f\\ \\text{is not}\\ \\mathbb{R}\\text{-linear}",
        "relation": "implies",
        "rule": "contrapositive",
        "conditions": [
          "b=6\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_lin_zero，每個 ℝ 線性映射都把 0 送到 0（證明在 S10.1）。f(0)=6≠0，依逆否命題，f 不是 ℝ 線性映射。直接也可看出可加性失敗：f(1+1)=f(2)=2，但 f(1)+f(1)=4+4=8。所以帶非零截距的仿射函數不是線性代數的線性映射。",
          "en": "By Lemma L_lin_zero, every ℝ-linear map sends 0 to 0 (proof in S10.1). Since f(0)=6≠0, the contrapositive shows that f is not ℝ-linear. Additivity also fails directly: f(1+1)=f(2)=2, but f(1)+f(1)=4+4=8. So an affine function with nonzero intercept is not a linear map in the sense of linear algebra."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_lin_zero",
            "contrapositive",
            "linear_map_def"
          ],
          "explanation": {
            "zh-Hant": "只用到「線性 ⇒ T(0)=0」的逆否；不宣稱反向「T(0)=0 ⇒ 線性」對一般函數成立。",
            "en": "Only the contrapositive of 'linear ⇒ T(0)=0' is used; the converse 'T(0)=0 ⇒ linear' is not claimed for general functions."
          }
        },
        "leaf_basis": "L_lin_zero + contrapositive + numeric_evaluation"
      },
      {
        "id": "S10.1",
        "uses": [
          "linear_map_def",
          "mul_zero"
        ],
        "before_tex": "T\\ \\mathbb{R}\\text{-linear}",
        "after_tex": "T(0)=T(0\\cdot 0)=0\\cdot T(0)=0",
        "relation": "implies",
        "rule": "linear_map_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "0=0·0，依齊次性（c=0、u=0）T(0)=T(0·0)=0·T(0)，再依 u·0=0 得 0。這證得引理 L_lin_zero。",
          "en": "0=0·0, so homogeneity with c=0 and u=0 gives T(0)=T(0·0)=0·T(0), which is 0 by u·0=0. This proves Lemma L_lin_zero."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "linear_map_def",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "只用齊次性與乘以零；不需要可加性。",
            "en": "Uses only homogeneity and multiplication by zero; additivity is not needed."
          }
        },
        "leaf_basis": "linear_map_def + mul_zero + mul_comm"
      },
      {
        "id": "S11",
        "uses": [
          "S01"
        ],
        "before_tex": "\\forall x\\in\\mathbb{R}:\\ mx+b=nx+c",
        "after_tex": "m=n\\ \\land\\ b=c",
        "relation": "equivalent",
        "rule": "L_affine_eq",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_affine_eq，兩個仿射函數在整個 ℝ 上相等，當且僅當斜率與截距都相同（MKO para2）。兩個方向分別在 S11.1、S11.2 證明。",
          "en": "By Lemma L_affine_eq, two affine functions are equal on all of ℝ exactly when their slopes and intercepts both agree (MKO para2). The two directions are proved in S11.1 and S11.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_affine_eq"
          ],
          "explanation": {
            "zh-Hant": "引理兩個方向都在子步證明，所以可作為等價使用。",
            "en": "Both directions of the lemma are proved in the sub-steps, so it may be used as an equivalence."
          }
        },
        "leaf_basis": "forall_elim + eq_congruence + iff_intro"
      },
      {
        "id": "S11.1",
        "uses": [
          "forall_elim",
          "eq_congruence"
        ],
        "before_tex": "\\forall x:\\ mx+b=nx+c",
        "after_tex": "b=c\\ (x=0),\\quad m+b=n+c\\ (x=1)\\ \\Rightarrow\\ m=n",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "等式對所有 x 成立，取 x=0 得 b=c；取 x=1 得 m+b=n+c，兩邊同減 b（=c）得 m=n。這是 ⇒ 方向。注意需要兩個不同的輸入；只知道一個點的等式不夠。",
          "en": "The equation holds for every x; x=0 gives b=c, and x=1 gives m+b=n+c, and subtracting b (=c) from both sides gives m=n. This is the ⇒ direction. Note that two different inputs are needed; an equation at one input is not enough."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "forall_elim",
            "eq_congruence"
          ],
          "explanation": {
            "zh-Hant": "前提是對所有 x 成立；若只在 x=2 成立，S12 顯示結論可以失敗。",
            "en": "The premise is that the equation holds for every x; if it held only at x=2, S12 shows the conclusion can fail."
          }
        },
        "leaf_basis": "forall_elim + eq_congruence + add_inverse"
      },
      {
        "id": "S11.2",
        "uses": [
          "eq_replace"
        ],
        "before_tex": "m=n\\ \\land\\ b=c",
        "after_tex": "\\forall x:\\ mx+b=nx+c",
        "relation": "implies",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若 m=n 且 b=c，對任意 x 把 m 換成 n、b 換成 c，得 mx+b=nx+c；x 是任意的，依 forall_intro 對所有 x 成立。這是 ⇐ 方向。S11.1 與 S11.2 合起來證得引理 L_affine_eq。",
          "en": "If m=n and b=c, replacing m by n and b by c for an arbitrary x gives mx+b=nx+c; since x was arbitrary, it holds for every x by forall_intro. This is the ⇐ direction. S11.1 and S11.2 together prove Lemma L_affine_eq."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "eq_replace",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "等值代換對每個 x 都可做，沒有附加條件。",
            "en": "Replacement of equals can be done for every x, with no side condition."
          }
        },
        "leaf_basis": "eq_replace + forall_intro"
      },
      {
        "id": "S12",
        "uses": [
          "S06",
          "S11",
          "G3"
        ],
        "before_tex": "f(2)=-4+6=2=k(2)",
        "after_tex": "f\\neq k\\quad((-2,6)\\neq(1,0))",
        "relation": "implies",
        "rule": "L_affine_eq",
        "conditions": [],
        "explanation": {
          "zh-Hant": "f(2)=−2·2+6=2，k(2)=2，在輸入 2 的輸出相同，這只是單點事實（MKO 假設 a2）。f 的 (m,b)=(−2,6)，k 的 (m,b)=(1,0)，斜率不同，依 S11，f 與 k 在 ℝ 上不相等；例如 f(0)=6≠0=k(0)。",
          "en": "f(2)=−2·2+6=2 and k(2)=2, so the outputs at input 2 agree; this is a fact about one input only (MKO assumption a2). f has (m,b)=(−2,6) and k has (m,b)=(1,0); the slopes differ, so by S11 f and k are not equal on ℝ; for example f(0)=6≠0=k(0)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "S06",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "用到 S11 中 ⇒ 方向的逆否：係數不同推出函數不同；單點的吻合（N9）不能推出全域相等。",
            "en": "It uses the contrapositive of the ⇒ direction of S11: different coefficients give different functions; agreement at one input (N9) does not give global equality."
          }
        },
        "leaf_basis": "L_affine_eq + contrapositive + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "G3",
          "S01",
          "linear_map_def"
        ],
        "before_tex": "g(x)=0x+5,\\ h(x)=0x+0",
        "after_tex": "g\\ \\text{has no zero};\\ \\text{every real is a zero of}\\ h;\\ \\text{neither has degree one}",
        "relation": "case_split",
        "rule": "case_analysis",
        "conditions": [
          "m=0"
        ],
        "explanation": {
          "zh-Hant": "m=0 時零點分兩種情形。g：對所有 x，g(x)=0·x+5=5≠0，沒有零點。h：對所有 x，h(x)=0，每個實數都是零點。兩者都是仿射（常數）函數，但 m=0，所以都不是次數一的函數。h 還滿足 h(u+v)=0=h(u)+h(v)、h(cu)=0=c·h(u)，是 ℝ 線性映射；g 因 g(0)=5≠0 不是。",
          "en": "With m=0 the zeros split into two cases. g: for every x, g(x)=0·x+5=5≠0, so there is no zero. h: h(x)=0 for every x, so every real is a zero. Both are affine (constant) functions, but m=0, so neither has degree one. Moreover h satisfies h(u+v)=0=h(u)+h(v) and h(cu)=0=c·h(u), so it is ℝ-linear; g is not, since g(0)=5≠0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "case_analysis",
            "linear_map_def",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "m=0 的情形依 b 是否為零分成兩支，涵蓋全部可能；結論與 MKO para1 一致。",
            "en": "The case m=0 splits by whether b is zero, which covers every possibility; the conclusions agree with MKO para1."
          }
        },
        "leaf_basis": "case_analysis + mul_zero + linear_map_def"
      },
      {
        "id": "S14",
        "uses": [
          "S06",
          "S08",
          "S09",
          "S10",
          "S12",
          "S13"
        ],
        "before_tex": "f,\\ k,\\ g,\\ h",
        "after_tex": "f(x)=-2x+6:\\ \\deg 1,\\ \\text{zero}\\ 3,\\ \\text{not linear};\\ f\\neq k;\\ g,h\\ \\text{constant}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整：f(x)=−2x+6 是次數一的仿射函數，唯一零點 3，但不是 ℝ 線性映射；f 與 k 只在輸入 2 吻合，並不相等；g、h 是常數函數，g 無零點，h 處處為零。",
          "en": "Summary: f(x)=−2x+6 is an affine function of degree one with unique zero 3, but it is not ℝ-linear; f and k agree only at input 2 and are not equal; g and h are constant functions, g with no zero and h zero everywhere."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S06",
            "S08",
            "S09",
            "S10",
            "S12",
            "S13",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：各項結論。足夠條件 H：S06 至 S13 的各別結論；H⇒G 只是合取。H 由給定資料正向證成，沒有把 G 當作前提。",
            "en": "Goal G: all the stated conclusions. Sufficient condition H: the separate conclusions of S06 to S13; H⇒G is just conjunction. H is proved forwards from the data; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "由 (1,4)、(3,0) 得 m=(0−4)/(3−1)=−2、b=6，f(x)=−2x+6。因 m≠0，f 次數為一，唯一零點是 −b/m=3；但 f(0)=6≠0，f 不是 ℝ 線性映射。兩個仿射函數在 ℝ 上相等當且僅當斜率與截距都相同，所以 f(2)=k(2)=2 並不推出 f=k。m=0 時，g(x)=5 無零點，h(x)=0 處處為零，兩者都不是次數一的函數。",
      "en": "From (1,4) and (3,0), m=(0−4)/(3−1)=−2 and b=6, so f(x)=−2x+6. Since m≠0, f has degree one with unique zero −b/m=3; but f(0)=6≠0, so f is not ℝ-linear. Two affine functions are equal on ℝ exactly when their slopes and intercepts agree, so f(2)=k(2)=2 does not give f=k. With m=0, g(x)=5 has no zero and h(x)=0 vanishes everywhere, and neither has degree one."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "g(x)=5 是一次函數，因為它可以寫成 0·x+5",
          "en": "g(x)=5 is a degree-one function because it can be written as 0·x+5"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "m=0，g 是常數函數，屬於仿射族但次數不是一。",
          "en": "m=0, so g is a constant function: affine, but not of degree one."
        }
      },
      {
        "claim": {
          "zh-Hant": "f(x)=−2x+6 是 ℝ 線性映射",
          "en": "f(x)=−2x+6 is an ℝ-linear map"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "f(0)=6≠0；而且 f(1+1)=2≠8=f(1)+f(1)。",
          "en": "f(0)=6≠0; moreover f(1+1)=2≠8=f(1)+f(1)."
        }
      },
      {
        "claim": {
          "zh-Hant": "通過 (2,1) 與 (2,5) 的仿射函數斜率是 (5−1)/(2−2)",
          "en": "The affine function through (2,1) and (2,5) has slope (5−1)/(2−2)"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x_1=x_2 時斜率公式除以零；也沒有函數 ℝ→ℝ 在 2 同時取 1 與 5。",
          "en": "With x_1=x_2 the slope formula divides by zero; and no function ℝ→ℝ takes both 1 and 5 at 2."
        }
      },
      {
        "claim": {
          "zh-Hant": "f(2)=k(2)=2，所以 f=k",
          "en": "f(2)=k(2)=2, so f=k"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "單點相等不推出全域相等：f(0)=6≠0=k(0)，斜率 −2≠1。",
          "en": "Equality at one input does not give equality everywhere: f(0)=6≠0=k(0), and the slopes −2≠1."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "eq_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": "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": "function_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "函數：f:A→B 對每個 a∈A 指定恰好一個 f(a)∈B。兩函數相等 ⇔ 定義域、對應域相同且對每個 a 都有相同值。",
          "en": "Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a."
        }
      },
      {
        "id": "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_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。",
          "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable."
        }
      },
      {
        "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": "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": "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": "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": "eq_replace",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。",
          "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true."
        }
      },
      {
        "id": "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": "contrapositive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逆否命題：P⇒Q 與 ¬Q⇒¬P 等價。逆命題 Q⇒P 一般不與 P⇒Q 等價。",
          "en": "Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q."
        }
      },
      {
        "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_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": "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_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": "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": "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_affine",
        "kind": "cited",
        "source": {
          "mko_id": "mko-linear-function",
          "sha256": "718236b5b7dfecd9f4d276590f1ddc012cfd5df96864540a3840a2b4e6650abe",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對 m、b∈R，f:R→R 由 f(x)=mx+b 定義為仿射函數。當 m≠0 時多項式次數為一；m=0 時為常數函數。線性代數中的 R 線性映射另要求 b=0，不能與初等教材的 linear function 稱呼混同。",
          "en": "For m,b∈R, the function f:R→R defined by f(x)=mx+b is affine. If m≠0 its polynomial degree is one; if m=0 it is constant. An R-linear map in linear algebra additionally requires b=0, and must be distinguished from the elementary textbook term linear function."
        }
      },
      {
        "id": "linear_map_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "T:ℝ→ℝ 是 ℝ 線性映射，指對所有實數 u、v、c 都有 T(u+v)=T(u)+T(v) 與 T(cu)=cT(u)。",
          "en": "T:ℝ→ℝ is an ℝ-linear map when T(u+v)=T(u)+T(v) and T(cu)=cT(u) for all reals u, v, c."
        }
      },
      {
        "id": "L_diff",
        "kind": "lemma",
        "proved_in": [
          "S02.1"
        ],
        "statement": {
          "zh-Hant": "對仿射 f(x)=mx+b 與所有實數 x_1、x_2：f(x_2)−f(x_1)=m(x_2−x_1)。",
          "en": "For affine f(x)=mx+b and all reals x_1, x_2: f(x_2)−f(x_1)=m(x_2−x_1)."
        }
      },
      {
        "id": "L_lin_zero",
        "kind": "lemma",
        "proved_in": [
          "S10.1"
        ],
        "statement": {
          "zh-Hant": "每個 ℝ 線性映射 T 都有 T(0)=0。",
          "en": "Every ℝ-linear map T has T(0)=0."
        }
      },
      {
        "id": "L_affine_eq",
        "kind": "lemma",
        "proved_in": [
          "S11.1",
          "S11.2"
        ],
        "statement": {
          "zh-Hant": "對實數 m、b、n、c：（對所有 x∈ℝ，mx+b=nx+c）當且僅當（m=n 且 b=c）。",
          "en": "For reals m, b, n, c: (mx+b=nx+c for every x∈ℝ) if and only if (m=n and b=c)."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "vars": [
          "m",
          "b",
          "x1",
          "x2"
        ],
        "degree": 1,
        "lhs": "(m*x2+b)-(m*x1+b)",
        "rhs": "m*(x2-x1)"
      },
      {
        "id": "N2",
        "step": "S03",
        "lhs": "0-4",
        "rhs": "-4"
      },
      {
        "id": "N3",
        "step": "S03",
        "lhs": "3-1",
        "rhs": "2"
      },
      {
        "id": "N4",
        "step": "S04",
        "lhs": "-4/2",
        "rhs": "-2"
      },
      {
        "id": "N5",
        "step": "S05",
        "lhs": "4-(-2)*1",
        "rhs": "6"
      },
      {
        "id": "N6",
        "step": "S06",
        "lhs": "-2*3+6",
        "rhs": "0"
      },
      {
        "id": "N7",
        "step": "S06",
        "lhs": "-2*1+6",
        "rhs": "4"
      },
      {
        "id": "N8",
        "step": "S07",
        "lhs": "-2*0+6",
        "rhs": "6"
      },
      {
        "id": "N9",
        "step": "S09",
        "lhs": "-6/(-2)",
        "rhs": "3"
      },
      {
        "id": "N10",
        "step": "S12",
        "lhs": "-2*2+6",
        "rhs": "2"
      },
      {
        "id": "N11",
        "step": "S10",
        "lhs": "(-2*1+6)+(-2*1+6)",
        "rhs": "8"
      },
      {
        "id": "N12",
        "step": "S10",
        "lhs": "-2*2+6",
        "relation": "!=",
        "rhs": "(-2*1+6)+(-2*1+6)"
      },
      {
        "id": "N13",
        "step": "S12",
        "lhs": "-2*0+6",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N14",
        "step": "S13",
        "lhs": "0*0+5",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N15",
        "step": "S04",
        "lhs": "2*(-2)",
        "rhs": "-4"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內：定義域與陪域都是 ℝ，m、b 是實常數。兩點斜率只在 x_1≠x_2 時使用；單點比較只主張該點的輸出相同。不討論多變數仿射映射或其他域；不改動 canonical MKO 或英文 edition。",
      "en": "Within the MKO's scope: domain and codomain are both ℝ, and m, b are real constants. The two-point slope is used only when x_1≠x_2, and a comparison at one input asserts only that the outputs there agree. Multivariable affine maps and other fields are not discussed; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "f(x)=mx+b",
        "仿射函數，m 為斜率、b 為截距",
        "an affine function with slope m and intercept b"
      ],
      [
        "(x_1,y_1),\\ (x_2,y_2)",
        "圖形上的兩點",
        "two points on the graph"
      ],
      [
        "k(x)=x",
        "比較用的函數",
        "a function for comparison"
      ],
      [
        "g(x)=5,\\ h(x)=0",
        "m=0 的兩個退化例子",
        "two degenerate examples with m=0"
      ]
    ],
    "md_error_controls": [
      {
        "title": "不是每個 mx+b 都是一次函數 / Not every mx+b is a degree-one function",
        "zh": "MKO 指出的第一個誤解是把所有 mx+b 都叫一次函數。只有 m≠0 時次數為一（S08）；g(x)=0·x+5 雖然寫成 mx+b 的形式，卻是常數函數，沒有零點（S13）。",
        "en": "The first error named in the MKO is to call every mx+b a degree-one function. Only m≠0 gives degree one (S08); g(x)=0·x+5, although written in the form mx+b, is a constant function with no zero (S13)."
      },
      {
        "title": "仿射不等於線性映射 / Affine is not linear in the linear-algebra sense",
        "zh": "MKO 指出的第二個誤解是把帶非零截距的仿射函數當成線性代數的線性映射。線性映射必有 T(0)=0（S10.1），而 f(0)=6；可加性也失敗：f(2)=2≠8=f(1)+f(1)（S10）。初等教材的 linear function 與線性代數的 linear map 是不同概念。",
        "en": "The second error named in the MKO is to treat an affine function with nonzero intercept as a linear map of linear algebra. A linear map must have T(0)=0 (S10.1), but f(0)=6; additivity fails too: f(2)=2≠8=f(1)+f(1) (S10). The elementary 'linear function' and the linear-algebra 'linear map' are different notions."
      },
      {
        "title": "兩點斜率要 x_1≠x_2；單點相等不是全域相等 / The two-point slope needs x_1≠x_2; equality at one input is not equality everywhere",
        "zh": "S03、S04 能除以 x_2−x_1 是因為 1≠3。若兩點橫坐標相同，例如 (2,1) 與 (2,5)，(5−1)/(2−2) 無定義，而且沒有函數 ℝ→ℝ 能在 2 同時取 1 與 5。另一方面，f(2)=k(2) 只說明輸入 2 的輸出相同；全域相等要求斜率與截距都相同（S11、S12）。",
        "en": "S03 and S04 may divide by x_2−x_1 because 1≠3. If two points share the first coordinate, say (2,1) and (2,5), then (5−1)/(2−2) is undefined, and no function ℝ→ℝ takes both values 1 and 5 at 2. On the other hand, f(2)=k(2) says only that the outputs at input 2 agree; equality everywhere needs equal slopes and equal intercepts (S11, S12)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿只處理 ℝ→ℝ 的仿射函數；多變數仿射映射與一般向量空間上的線性映射不在範圍內。",
        "en": "This guide handles only affine functions ℝ→ℝ; multivariable affine maps and linear maps on general vector spaces are out of scope."
      },
      {
        "zh-Hant": "N1 是 degree 型格點核對（各變數次數至多一），作為 L_diff 代數恆等式的輔助；引理的證明是 S02.1。",
        "en": "N1 is a degree-type grid check (degree at most one in each variable) supporting the algebraic identity L_diff; the lemma's proof is S02.1."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "fb722ebaf4dc7b60839292d104faf6dc837ca15a0a3ed5015f7c9c626d96a85a",
    "original_source_sha256": "fb722ebaf4dc7b60839292d104faf6dc837ca15a0a3ed5015f7c9c626d96a85a",
    "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."
  }
}