{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-triangle-01",
  "mko_id": "mko-triangle",
  "title": {
    "zh-Hant": "三角形：在座標模型中確認三點界定非退化三角形",
    "en": "Triangle: confirming in the coordinate model that three points determine a nondegenerate triangle"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-triangle-01",
    "source": {
      "mko_id": "mko-triangle",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "1556dabd42bf5d132f41296684eb2b037900a511e659c48627bee6d0c4dcd461",
      "english_sha256": "e4a28d8e7440ec3005d4c989132d0cfa5a0c60bf2440ebcd24ca5c2e886f4b5f"
    },
    "mode": "definition_walkthrough",
    "audience_prerequisites": [
      "cartesian_product_def",
      "image_preimage_def",
      "set_operations_def",
      "eq_congruence",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在座標平面模型 ℝ² 中，設 A=(0,0)、B=(4,0)、C=(1,3)。確認 A、B、C 兩兩相異且不共線，因而依 MKO 界定非退化三角形 ABC；寫出三條邊與各頂點的對邊，並說明三頂點的平均點 G=(5/3,1) 不在三角形 ABC 上。另設 D=(2,1)、E=(6,3)，說明相異三點 A、D、E 共線，不界定非退化三角形。",
      "en": "In the coordinate-plane model ℝ², let A=(0,0), B=(4,0), C=(1,3). Confirm that A, B, C are pairwise distinct and noncollinear, so that by the MKO they determine the nondegenerate triangle ABC; write out its three sides and the side opposite each vertex, and show that the average of the vertices, G=(5/3,1), is not on triangle ABC. Then, with D=(2,1) and E=(6,3), show that the distinct points A, D, E are collinear and so do not determine a nondegenerate triangle."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "模型：點是 ℝ² 中的有序實數對，加法、減法與實數倍數逐坐標進行。對相異點 P≠Q，直線 L(P,Q)={P+t(Q−P) ∣ t∈ℝ}；模型中的直線恰好是這些集合。對任意點 P、Q，線段 [P,Q]={P+t(Q−P) ∣ 0≤t≤1}。",
          "en": "Model: points are ordered pairs of reals in ℝ², with addition, subtraction and real multiples acting coordinatewise. For distinct points P≠Q, the line L(P,Q)={P+t(Q−P) ∣ t∈ℝ}; the lines of the model are exactly these sets. For any points P, Q, the segment [P,Q]={P+t(Q−P) ∣ 0≤t≤1}."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "共線：若模型中有一條直線同時包含若干點，這些點稱為共線。",
          "en": "Collinear: points are collinear if some line of the model contains all of them."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "A=(0,0)、B=(4,0)、C=(1,3)、G=(5/3,1)；D=(2,1)、E=(6,3)。",
          "en": "A=(0,0), B=(4,0), C=(1,3), G=(5/3,1); D=(2,1), E=(6,3)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_model_dependence"
        ],
        "before_tex": "\\text{three points }A,B,C\\text{ of the model}",
        "after_tex": "\\triangle ABC=[A,B]\\cup[B,C]\\cup[C,A]\\ \\text{when}\\ A,B,C\\ \\text{are distinct and noncollinear}",
        "relation": "definition",
        "rule": "mko_triangle_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 的定義有兩個前提：三點相異，且不共線；在這兩個前提下，三條線段的聯集就是非退化三角形 ABC。線、線段與共線的精確意義取決於所選模型，所以先固定 ℝ² 的座標模型（G1、G2）。之後的工作是逐一核對兩個前提，再寫出聯集。",
          "en": "The MKO's definition has two hypotheses: the three points are distinct, and they are not collinear; under both, the union of the three segments is the nondegenerate triangle ABC. The precise meaning of lines, segments and collinearity depends on the chosen model, so the coordinate model of ℝ² (G1, G2) is fixed first. The rest of the work checks the two hypotheses one at a time and then writes out the union."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_triangle_def",
            "mko_model_dependence"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出定義與模型，不推出新事實；兩個前提尚未核對，還不能寫「ABC 是三角形」。",
            "en": "This step only states the definition and the model and derives no new fact; the two hypotheses are not yet checked, so 'ABC is a triangle' may not be written yet."
          }
        },
        "leaf_basis": "mko_triangle_def + mko_model_dependence"
      },
      {
        "id": "S02",
        "uses": [
          "G3",
          "cartesian_product_def"
        ],
        "before_tex": "A=(0,0),\\ B=(4,0),\\ C=(1,3)",
        "after_tex": "A\\neq B,\\ B\\neq C,\\ C\\neq A",
        "relation": "implies",
        "rule": "cartesian_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "有序對相等必須兩個坐標都相等。A 與 B 的第一坐標 0≠4，B 與 C 的第一坐標 4≠1，C 與 A 的第一坐標 1≠0，所以三點兩兩相異。這是 MKO 的第一個前提，也提供了 G1 中 L(A,B) 有定義所需的 A≠B。",
          "en": "Ordered pairs are equal only if both coordinates agree. The first coordinates of A and B are 0≠4, of B and C are 4≠1, and of C and A are 1≠0, so the three points are pairwise distinct. This is the MKO's first hypothesis, and it also supplies A≠B, which G1 needs for L(A,B) to be defined."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "每一對只需一個不同的坐標；只檢查其中兩對不夠，三對都要檢查。反向（兩點不等推出某個指定坐標不等）並未宣稱。",
            "en": "Each pair needs just one differing coordinate; checking only two of the three pairs would not be enough, so all three are checked. The converse (distinct points differ in a particular coordinate) is not claimed."
          }
        },
        "leaf_basis": "cartesian_product_def + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "G3",
          "S02"
        ],
        "before_tex": "L(A,B)=\\{A+t(B-A)\\mid t\\in\\mathbb{R}\\}",
        "after_tex": "L(A,B)=\\{(4t,\\,0)\\mid t\\in\\mathbb{R}\\}",
        "relation": "equivalent",
        "rule": "coord_ops",
        "conditions": [
          "A\\neq B"
        ],
        "explanation": {
          "zh-Hant": "逐坐標計算：B−A=(4−0, 0−0)=(4,0)，所以 A+t(B−A)=(0+4t, 0+0·t)=(4t,0)。對每個 t 兩種寫法是同一點，所以兩個集合相同。條件 A≠B 已在 S02 確認。",
          "en": "Compute coordinatewise: B−A=(4−0, 0−0)=(4,0), so A+t(B−A)=(0+4t, 0+0·t)=(4t,0). For every t the two expressions are the same point, so the two sets are equal. The condition A≠B was checked in S02."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "coord_ops",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "反向：t=1 時 (4,0)=B，t=0 時 (0,0)=A；每個 t 在兩種寫法中給出同一點，雙向成立。",
            "en": "Backward: t=1 gives (4,0)=B and t=0 gives (0,0)=A; each t gives the same point in both forms, so the step holds both ways."
          }
        },
        "leaf_basis": "coord_ops + mul_zero + add_zero"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "G3"
        ],
        "before_tex": "C\\in L(A,B)",
        "after_tex": "\\exists t\\in\\mathbb{R}:\\ 1=4t\\ \\land\\ 3=0",
        "relation": "equivalent",
        "rule": "image_preimage_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "L(A,B) 是 t↦(4t,0) 的像，所以 C∈L(A,B) 恰好表示有某個實數 t 使 (1,3)=(4t,0)；有序對相等拆成兩個坐標方程 1=4t 與 3=0，兩個方程用同一個 t。",
          "en": "L(A,B) is the image of t↦(4t,0), so C∈L(A,B) means exactly that some real t gives (1,3)=(4t,0); equality of ordered pairs splits this into the coordinate equations 1=4t and 3=0, both with the same t."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "image_preimage_def",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "像集合的展開與有序對相等的條件都是雙向的。",
            "en": "Unfolding membership in an image and the equality condition for ordered pairs are both two-way."
          }
        },
        "leaf_basis": "image_preimage_def + cartesian_product_def"
      },
      {
        "id": "S05",
        "uses": [
          "S04"
        ],
        "before_tex": "\\exists t\\in\\mathbb{R}:\\ 1=4t\\ \\land\\ 3=0",
        "after_tex": "C\\notin L(A,B)",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第二個方程 3=0 不含 t，對每個 t 都不成立，所以合取對每個 t 都不成立，不存在這樣的 t；依 S04 的等價，C∉L(A,B)。注意第一個方程單獨有解 t=1/4，只看第一坐標會誤判。",
          "en": "The second equation 3=0 does not involve t and fails for every t, so the conjunction fails for every t and no such t exists; by the equivalence S04, C∉L(A,B). Note that the first equation alone has the solution t=1/4, so looking at the first coordinate alone would mislead."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "contradiction",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "結論只需「沒有任何 t 同時滿足兩式」；3≠0 與 t 無關，所以存在命題為假。",
            "en": "The conclusion needs only that no t satisfies both equations; 3≠0 holds whatever t is, so the existential statement is false."
          }
        },
        "leaf_basis": "contradiction + quantifier_negation + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "A,B,C\\ \\text{collinear}",
        "after_tex": "\\exists P,Q,\\ P\\neq Q:\\ A,B,C\\in L(P,Q)",
        "relation": "equivalent",
        "rule": "collinear_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 G2，三點共線表示有一條直線同時包含三點；依 G1，模型中的直線恰好是相異兩點 P、Q 給出的 L(P,Q)。兩個定義合起來就是右式。下一步暫時假設三點共線，看會推出什麼。",
          "en": "By G2, the three points are collinear when one line contains all three; by G1, the lines of the model are exactly the sets L(P,Q) for distinct P, Q. Together the two definitions give the right-hand side. The next step temporarily assumes the points are collinear and sees what follows."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "collinear_def",
            "line_def"
          ],
          "explanation": {
            "zh-Hant": "只是展開兩個定義，雙向成立。",
            "en": "This only unfolds two definitions, so it holds both ways."
          }
        },
        "leaf_basis": "collinear_def + line_def + def_unfold"
      },
      {
        "id": "S07",
        "uses": [
          "S02",
          "S06",
          "L_line_unique"
        ],
        "before_tex": "A,B,C\\in L(P,Q),\\ P\\neq Q,\\ A\\neq B",
        "after_tex": "L(P,Q)=L(A,B),\\ C\\in L(A,B)",
        "relation": "implies",
        "rule": "L_line_unique",
        "conditions": [
          "P\\neq Q",
          "A\\neq B"
        ],
        "explanation": {
          "zh-Hant": "假設 A、B、C 共線；依 S06 取 P≠Q 使 A、B、C∈L(P,Q)（∃消去，P、Q 是新名字）。A、B 是 L(P,Q) 上的兩點且 A≠B（S02），由引理 L_line_unique（X=A、Y=B）得 L(P,Q)=L(A,B)。於是 C∈L(P,Q) 推出 C∈L(A,B)。引理在 S07.1 到 S07.3 證明。",
          "en": "Assume A, B, C are collinear; by S06 take P≠Q with A, B, C∈L(P,Q) (∃-elimination; P and Q are fresh names). A and B are two points of L(P,Q) and A≠B (S02), so Lemma L_line_unique (X=A, Y=B) gives L(P,Q)=L(A,B). Then C∈L(P,Q) gives C∈L(A,B). The lemma is proved in S07.1 to S07.3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S06",
            "L_line_unique",
            "exists_elim"
          ],
          "explanation": {
            "zh-Hant": "引理需要 A≠B；若 A=B，通過 A 的直線有很多條，推不出 L(P,Q)=L(A,B)。此步在共線假設之下，結論是條件式，不宣稱 C∈L(A,B) 真的成立。",
            "en": "The lemma needs A≠B; if A=B, many lines pass through A and L(P,Q)=L(A,B) would not follow. This step is under the collinearity assumption, so its conclusion is conditional; it does not claim that C∈L(A,B) is actually true."
          }
        },
        "leaf_basis": "L_line_unique + exists_elim + eq_replace"
      },
      {
        "id": "S07.1",
        "uses": [
          "G1"
        ],
        "before_tex": "X,Y\\in L(P,Q),\\ X\\neq Y,\\ D=Q-P",
        "after_tex": "X=P+sD,\\ Y=P+uD,\\ s\\neq u",
        "relation": "implies",
        "rule": "exists_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "X∈L(P,Q) 給出某個實數 s 使 X=P+sD；Y∈L(P,Q) 給出某個實數 u 使 Y=P+uD（∃消去，s、u 是新名字）。若 s=u，則 X=P+sD=P+uD=Y，與 X≠Y 矛盾；所以 s≠u。",
          "en": "X∈L(P,Q) gives some real s with X=P+sD, and Y∈L(P,Q) gives some real u with Y=P+uD (∃-elimination; s and u are fresh names). If s=u, then X=P+sD=P+uD=Y, contradicting X≠Y; so s≠u."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "exists_elim",
            "contrapositive"
          ],
          "explanation": {
            "zh-Hant": "s≠u 由 X≠Y 的逆否得出；反向（s≠u 推出 X≠Y）需要 D≠0，這裡不需要也不宣稱。",
            "en": "s≠u comes from X≠Y by contraposition; the converse (s≠u implies X≠Y) would need D≠0, and is neither needed nor claimed here."
          }
        },
        "leaf_basis": "exists_elim + contrapositive + eq_replace"
      },
      {
        "id": "S07.2",
        "uses": [
          "S07.1"
        ],
        "before_tex": "X+t(Y-X)",
        "after_tex": "P+\\bigl(s+t(u-s)\\bigr)D",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先算 Y−X=(P+uD)−(P+sD)=(u−s)D；再代入 X+t(Y−X)=P+sD+t(u−s)D=P+(s+t(u−s))D。每個坐標都只用到加法與乘法的結合律、交換律與分配律，所以對每個實數 t 成立（N14 以多項式恆等式核對）。",
          "en": "First Y−X=(P+uD)−(P+sD)=(u−s)D; then X+t(Y−X)=P+sD+t(u−s)D=P+(s+t(u−s))D. In each coordinate only associativity, commutativity and distributivity are used, so it holds for every real t (N14 checks it as a polynomial identity)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "coord_ops"
          ],
          "explanation": {
            "zh-Hant": "兩式對每個 t 給出同一點，是恆等式，雙向成立。",
            "en": "Both expressions give the same point for every t; it is an identity and holds both ways."
          }
        },
        "leaf_basis": "distributive + add_assoc + add_comm + mul_comm + coord_ops"
      },
      {
        "id": "S07.3",
        "uses": [
          "S07.1",
          "S07.2"
        ],
        "before_tex": "L(X,Y)=\\{P+(s+t(u-s))D\\mid t\\in\\mathbb{R}\\}",
        "after_tex": "L(X,Y)=\\{P+rD\\mid r\\in\\mathbb{R}\\}=L(P,Q)",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [
          "u-s\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "⊆：每個 s+t(u−s) 都是實數 r，所以 L(X,Y) 的每一點都在 L(P,Q) 中。⊇：給定實數 r，取 t=(r−s)/(u−s)，因 u−s≠0 這個 t 有定義，且 s+t(u−s)=r，所以 P+rD∈L(X,Y)。兩個包含合起來得集合相等。S07.1 到 S07.3 合起來證得引理 L_line_unique。",
          "en": "⊆: each s+t(u−s) is a real number r, so every point of L(X,Y) is in L(P,Q). ⊇: given a real r, take t=(r−s)/(u−s), which is defined because u−s≠0, and then s+t(u−s)=r, so P+rD∈L(X,Y). The two inclusions give equality of the sets. S07.1 to S07.3 together prove Lemma L_line_unique."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07.1",
            "set_extensionality",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "⊇ 方向要除以 u−s，正需要 S07.1 的 s≠u。若 s=u，集合只剩一點 P+sD，不是直線。",
            "en": "The ⊇ direction divides by u−s, which needs s≠u from S07.1. If s=u the set would be the single point P+sD, not a line."
          }
        },
        "leaf_basis": "set_extensionality + subset_def + mul_inverse_nonzero + exists_intro"
      },
      {
        "id": "S08",
        "uses": [
          "S05",
          "S07"
        ],
        "before_tex": "C\\in L(A,B)\\ \\text{(under the assumption)},\\ C\\notin L(A,B)",
        "after_tex": "A,B,C\\ \\text{noncollinear}",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "共線的假設推出 C∈L(A,B)（S07），但 S05 已證 C∉L(A,B)，矛盾。依反證法，假設不成立：沒有任何直線同時包含 A、B、C，即三點不共線。這是 MKO 的第二個前提。",
          "en": "The collinearity assumption leads to C∈L(A,B) (S07), but S05 proved C∉L(A,B): a contradiction. By proof by contradiction the assumption fails: no line contains A, B and C, that is, the three points are noncollinear. This is the MKO's second hypothesis."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S07",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "反證需要一個命題與其否定同時成立：C∈L(A,B) 與 C∉L(A,B)。只有 C∉L(A,B) 還不夠；必須靠 S07 的唯一性才能排除其他直線。",
            "en": "Proof by contradiction needs a statement and its negation together: C∈L(A,B) and C∉L(A,B). C∉L(A,B) alone is not enough; the uniqueness used in S07 is what rules out every other line."
          }
        },
        "leaf_basis": "contradiction + imp_intro"
      },
      {
        "id": "S09",
        "uses": [
          "S02",
          "S08",
          "mko_triangle_def"
        ],
        "before_tex": "A,B,C\\ \\text{distinct and noncollinear}",
        "after_tex": "\\triangle ABC=[A,B]\\cup[B,C]\\cup[C,A]\\ \\text{nondegenerate}",
        "relation": "theorem_application",
        "rule": "mko_triangle_def",
        "conditions": [
          "A\\neq B,\\ B\\neq C,\\ C\\neq A",
          "A,B,C\\ \\text{noncollinear}"
        ],
        "explanation": {
          "zh-Hant": "兩個前提都已成立（S02、S08），依 MKO，三條線段的聯集界定非退化三角形 ABC。頂點是 A、B、C 三個點；邊是 [A,B]、[B,C]、[C,A] 三條線段。每個頂點的對邊是不以它為端點的那條邊：A 的對邊是 BC，B 的對邊是 CA，C 的對邊是 AB。",
          "en": "Both hypotheses hold (S02, S08), so by the MKO the union of the three segments determines the nondegenerate triangle ABC. Its vertices are the three points A, B, C; its sides are the three segments [A,B], [B,C], [C,A]. The side opposite a vertex is the side that does not have it as an endpoint: BC is opposite A, CA is opposite B, and AB is opposite C."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S08",
            "mko_triangle_def"
          ],
          "explanation": {
            "zh-Hant": "MKO 的兩個前提缺一不可；這裡不宣稱三角形的任何長度、角度或面積性質。",
            "en": "Neither of the MKO's two hypotheses can be dropped; no property of lengths, angles or area of the triangle is claimed here."
          }
        },
        "leaf_basis": "mko_triangle_def + and_intro_elim"
      },
      {
        "id": "S10",
        "uses": [
          "S09",
          "G1",
          "G3"
        ],
        "before_tex": "[A,B],\\ [B,C],\\ [C,A]",
        "after_tex": "\\{(4t,0)\\},\\ \\{(4-3t,\\,3t)\\},\\ \\{(1-t,\\,3-3t)\\}\\quad(0\\le t\\le 1)",
        "relation": "equivalent",
        "rule": "segment_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依線段定義與逐坐標運算：B−A=(4,0)，[A,B]={(4t,0)}；C−B=(1−4, 3−0)=(−3,3)，[B,C]={(4−3t, 3t)}；A−C=(0−1, 0−3)=(−1,−3)，[C,A]={(1−t, 3−3t)}；三者都取 0≤t≤1。",
          "en": "By the segment definition and coordinatewise operations: B−A=(4,0), so [A,B]={(4t,0)}; C−B=(1−4, 3−0)=(−3,3), so [B,C]={(4−3t, 3t)}; A−C=(0−1, 0−3)=(−1,−3), so [C,A]={(1−t, 3−3t)}; all with 0≤t≤1."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "segment_def",
            "coord_ops"
          ],
          "explanation": {
            "zh-Hant": "反向核對端點：t=0 與 t=1 分別給出 A 與 B；B 與 C=(4−3,3)=(1,3)；C 與 A=(1−1,3−3)=(0,0)。每個 t 兩種寫法是同一點。",
            "en": "Backward, check the endpoints: t=0 and t=1 give A and B; then B and C=(4−3,3)=(1,3); then C and A=(1−1,3−3)=(0,0). For each t both forms give the same point."
          }
        },
        "leaf_basis": "segment_def + coord_ops + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S09",
          "L_seg_reverse"
        ],
        "before_tex": "\\triangle ACB=[A,C]\\cup[C,B]\\cup[B,A]",
        "after_tex": "[C,A]\\cup[B,C]\\cup[A,B]=\\triangle ABC",
        "relation": "equivalent",
        "rule": "L_seg_reverse",
        "conditions": [],
        "explanation": {
          "zh-Hant": "改變頂點順序只改變名稱：依引理 L_seg_reverse，[A,C]=[C,A]、[C,B]=[B,C]、[B,A]=[A,B]；聯集與次序無關，所以 ACB 與 ABC 是同一個集合。引理在 S11.1、S11.2 證明。",
          "en": "Changing the order of the vertices only changes the name: by Lemma L_seg_reverse, [A,C]=[C,A], [C,B]=[B,C] and [B,A]=[A,B]; a union does not depend on order, so ACB and ABC are the same set. The lemma is proved in S11.1 and S11.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_seg_reverse",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "每個替換都是集合相等，重排聯集也是相等，雙向成立。",
            "en": "Each replacement is an equality of sets and reordering a union is an equality, so the step holds both ways."
          }
        },
        "leaf_basis": "L_seg_reverse + set_operations_def + set_extensionality"
      },
      {
        "id": "S11.1",
        "uses": [
          "G1"
        ],
        "before_tex": "P+t(Q-P)",
        "after_tex": "Q+(1-t)(P-Q)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "展開右式：Q+(1−t)(P−Q)=Q+P−Q−tP+tQ=P+t(Q−P)。每個坐標都是實數的多項式恆等式，對所有 t、P、Q 成立（N15）。",
          "en": "Expand the right side: Q+(1−t)(P−Q)=Q+P−Q−tP+tQ=P+t(Q−P). In each coordinate this is a polynomial identity of reals, valid for every t, P and Q (N15)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "恆等式，雙向成立。",
            "en": "It is an identity and holds both ways."
          }
        },
        "leaf_basis": "distributive + add_inverse + add_comm + add_assoc + coord_ops"
      },
      {
        "id": "S11.2",
        "uses": [
          "S11.1"
        ],
        "before_tex": "0\\le t\\le 1",
        "after_tex": "[P,Q]=[Q,P]",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "0≤t≤1 ⇔ 0≤1−t≤1（乘以 −1 使不等號反向，再加 1）。所以 t 走遍 [0,1] 時，1−t 也恰好走遍 [0,1]；配合 S11.1，[P,Q] 的每一點都是 [Q,P] 的點，反之亦然。S11.1 與 S11.2 合起來證得引理 L_seg_reverse。",
          "en": "0≤t≤1 ⇔ 0≤1−t≤1 (multiplying by −1 reverses the inequalities, then add 1). So as t runs over [0,1], 1−t also runs over exactly [0,1]; with S11.1, every point of [P,Q] is a point of [Q,P] and conversely. S11.1 and S11.2 together prove Lemma L_seg_reverse."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11.1",
            "order_basics",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "此證明需要所用參數範圍在 t↦1−t 下不變；[0,1] 確實如此。換成 [0,2] 時不再不變，但不能由「不是 [0,1]」一概判定失敗，例如 [1/4,3/4] 也不變。本稿的線段仍依既定 [0,1] 定義。",
            "en": "This proof needs the chosen parameter range to be invariant under t↦1−t, as [0,1] is. The range [0,2] is not invariant, but being different from [0,1] does not by itself cause failure: [1/4,3/4] is invariant too. The guide still defines segments using [0,1]."
          }
        },
        "leaf_basis": "order_basics + set_extensionality + subset_def"
      },
      {
        "id": "S12",
        "uses": [
          "S10",
          "G3"
        ],
        "before_tex": "A=(4-3t,\\,3t),\\ 0\\le t\\le 1",
        "after_tex": "A\\notin[B,C]",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若 A∈[B,C]，則某個 t 使 (0,0)=(4−3t, 3t)。第二坐標 0=3t 給出 t=0，代入第一坐標得 4−3·0=4≠0，矛盾。所以頂點 A 不在它的對邊 BC 上。一般理由：[B,C]⊆L(B,C)，而 A∈L(B,C) 會使三點共線，與 S08 矛盾。",
          "en": "If A∈[B,C], some t gives (0,0)=(4−3t, 3t). The second coordinate 0=3t gives t=0, and then the first coordinate is 4−3·0=4≠0, a contradiction. So the vertex A is not on its opposite side BC. The general reason: [B,C]⊆L(B,C), and A∈L(B,C) would make the three points collinear, contradicting S08."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "第二坐標唯一決定 t=0；只需檢查這個 t 是否也滿足第一坐標。",
            "en": "The second coordinate forces t=0; it remains only to check whether this t also satisfies the first coordinate."
          }
        },
        "leaf_basis": "contradiction + cartesian_product_def + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S09",
          "G3"
        ],
        "before_tex": "G\\in\\triangle ABC",
        "after_tex": "G\\in[A,B]\\ \\lor\\ G\\in[B,C]\\ \\lor\\ G\\in[C,A]",
        "relation": "equivalent",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "G=((0+4+1)/3, (0+0+3)/3)=(5/3,1) 是三個頂點的平均。依 MKO，三角形 ABC 是三條邊的聯集，所以 G 在三角形上恰好表示 G 至少在其中一條邊上。以下三個子步逐邊檢查。",
          "en": "G=((0+4+1)/3, (0+0+3)/3)=(5/3,1) is the average of the three vertices. By the MKO, triangle ABC is the union of its three sides, so G is on the triangle exactly when it is on at least one side. The three sub-steps check each side."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_operations_def",
            "S09"
          ],
          "explanation": {
            "zh-Hant": "聯集的成員條件是雙向的。",
            "en": "The membership condition for a union is two-way."
          }
        },
        "leaf_basis": "set_operations_def + def_unfold"
      },
      {
        "id": "S13.1",
        "uses": [
          "S10"
        ],
        "before_tex": "(5/3,\\,1)=(4t,\\,0)",
        "after_tex": "G\\notin[A,B]",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "[A,B] 上每一點的第二坐標都是 0，G 的第二坐標是 1，而 1≠0，所以 G∉[A,B]。",
          "en": "Every point of [A,B] has second coordinate 0, while G has second coordinate 1, and 1≠0, so G∉[A,B]."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "一個坐標不等就足以排除；不需解第一坐標。",
            "en": "One unequal coordinate is enough to exclude it; the first coordinate need not be solved."
          }
        },
        "leaf_basis": "cartesian_product_def + contradiction"
      },
      {
        "id": "S13.2",
        "uses": [
          "S10"
        ],
        "before_tex": "(5/3,\\,1)=(4-3t,\\,3t)",
        "after_tex": "G\\notin[B,C]",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第二坐標 1=3t ⇔ t=1/3（3≠0）；代入第一坐標得 4−3·(1/3)=3，而 3≠5/3，所以沒有 t 同時滿足兩式，G∉[B,C]。",
          "en": "The second coordinate gives 1=3t ⇔ t=1/3 (3≠0); in the first coordinate this gives 4−3·(1/3)=3, and 3≠5/3, so no t satisfies both, and G∉[B,C]."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "contradiction",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "第二坐標唯一決定 t；兩個坐標必須用同一個 t。",
            "en": "The second coordinate determines t uniquely; both coordinates must use the same t."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero + contradiction + numeric_evaluation"
      },
      {
        "id": "S13.3",
        "uses": [
          "S10"
        ],
        "before_tex": "(5/3,\\,1)=(1-t,\\,3-3t)",
        "after_tex": "G\\notin[C,A]",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第二坐標 1=3−3t ⇔ t=(3−1)/3=2/3；代入第一坐標得 1−2/3=1/3，而 1/3≠5/3，所以 G∉[C,A]。",
          "en": "The second coordinate gives 1=3−3t ⇔ t=(3−1)/3=2/3; in the first coordinate this gives 1−2/3=1/3, and 1/3≠5/3, so G∉[C,A]."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "contradiction",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "與 S13.2 相同：第二坐標決定 t，再看第一坐標是否吻合。",
            "en": "As in S13.2: the second coordinate determines t, then the first coordinate is checked against it."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero + contradiction + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S13",
          "S13.1",
          "S13.2",
          "S13.3"
        ],
        "before_tex": "G\\notin[A,B],\\ G\\notin[B,C],\\ G\\notin[C,A]",
        "after_tex": "G\\notin\\triangle ABC",
        "relation": "implies",
        "rule": "de_morgan_logic",
        "conditions": [],
        "explanation": {
          "zh-Hant": "三個析取項都不成立，依德摩根律析取為假；依 S13 的等價，G∉△ABC。示意圖上 G 位於三角形圍住的區域內，但 MKO 所說的三角形是三條邊的聯集，不包含內部點。相對地，M=(2,0)=(4·(1/2),0) 在 [A,B] 上，所以 M∈△ABC。",
          "en": "All three disjuncts fail, so by De Morgan's law the disjunction is false, and by the equivalence S13, G∉△ABC. In a sketch G lies in the region the triangle encloses, but the triangle in the MKO is the union of the three sides and contains no interior point. By contrast M=(2,0)=(4·(1/2),0) is on [A,B], so M∈△ABC."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S13",
            "S13.1",
            "S13.2",
            "S13.3",
            "de_morgan_logic"
          ],
          "explanation": {
            "zh-Hant": "要證不在聯集中，三條邊都必須排除；只排除一條不夠。",
            "en": "To show a point is not in the union, all three sides must be excluded; excluding one is not enough."
          }
        },
        "leaf_basis": "de_morgan_logic + set_operations_def"
      },
      {
        "id": "S15",
        "uses": [
          "G1",
          "G2",
          "G3"
        ],
        "before_tex": "A=(0,0),\\ D=(2,1),\\ E=(6,3)",
        "after_tex": "A,D,E\\in L(A,D)",
        "relation": "implies",
        "rule": "collinear_def",
        "conditions": [
          "A\\neq D"
        ],
        "explanation": {
          "zh-Hant": "A≠D（第一坐標 0≠2），所以 L(A,D) 有定義；D−A=(2,1)。t=0 給出 A，t=1 給出 D，t=3 給出 A+3(D−A)=(6,3)=E。三點都在同一條直線 L(A,D) 上，依 G2 共線。三點也兩兩相異（E 的第一坐標 6 與 0、2 都不同）。",
          "en": "A≠D (first coordinates 0≠2), so L(A,D) is defined, and D−A=(2,1). t=0 gives A, t=1 gives D, and t=3 gives A+3(D−A)=(6,3)=E. All three points lie on the one line L(A,D), so by G2 they are collinear. They are also pairwise distinct (the first coordinate 6 of E differs from 0 and from 2)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "collinear_def",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "共線只需找到一條包含三點的直線，各點各給一個參數值即可；不需要唯一性。",
            "en": "Collinearity needs only one line containing the three points, with one parameter value for each point; no uniqueness is needed."
          }
        },
        "leaf_basis": "collinear_def + exists_intro + coord_ops + numeric_evaluation"
      },
      {
        "id": "S16",
        "uses": [
          "S09",
          "S10",
          "S12",
          "S14",
          "S15"
        ],
        "before_tex": "\\text{S09, S10, S14, S15}",
        "after_tex": "\\triangle ABC\\ \\text{nondegenerate};\\ A,D,E\\ \\text{give no nondegenerate triangle}",
        "relation": "implies",
        "rule": "mko_triangle_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "A、B、C 相異且不共線，界定非退化三角形 ABC，三邊如 S10，對邊關係如 S09，且 G 不在三角形上。A、D、E 雖然相異，卻共線，違反 MKO 的第二個前提，所以不界定本條目所指的非退化三角形。",
          "en": "A, B, C are distinct and noncollinear and determine the nondegenerate triangle ABC, with sides as in S10 and opposite sides as in S09, and G is not on the triangle. A, D, E are distinct but collinear, violating the MKO's second hypothesis, so they do not determine a nondegenerate triangle in this entry's sense."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S02",
            "S08",
            "S09",
            "S14",
            "S15",
            "mko_triangle_def"
          ],
          "explanation": {
            "zh-Hant": "目標：ABC 是非退化三角形，而 A、D、E 不界定非退化三角形。足夠條件 H：A、B、C 相異（S02）且不共線（S08），而 A、D、E 共線（S15）；H 推出目標由 MKO 的定義。H 由 S02 到 S15 從已知正向證成，沒有把目標當作前提。",
            "en": "Goal: ABC is a nondegenerate triangle, while A, D, E determine none. Sufficient condition H: A, B, C are distinct (S02) and noncollinear (S08), and A, D, E are collinear (S15); H gives the goal by the MKO's definition. H is proved forwards from the givens in S02 to S15; the goal is never used as a premise."
          }
        },
        "leaf_basis": "mko_triangle_def + and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "A=(0,0)、B=(4,0)、C=(1,3) 兩兩相異且不共線，依 MKO 界定非退化三角形 ABC=[A,B]∪[B,C]∪[C,A]；三邊為 {(4t,0)}、{(4−3t,3t)}、{(1−t,3−3t)}（0≤t≤1），A、B、C 的對邊分別是 BC、CA、AB。三頂點的平均點 G=(5/3,1) 不在任何一條邊上，所以不在三角形上。A、D=(2,1)、E=(6,3) 相異但共線（E=A+3(D−A)），不界定非退化三角形。",
      "en": "A=(0,0), B=(4,0), C=(1,3) are pairwise distinct and noncollinear, so by the MKO they determine the nondegenerate triangle ABC=[A,B]∪[B,C]∪[C,A]; its sides are {(4t,0)}, {(4−3t,3t)}, {(1−t,3−3t)} (0≤t≤1), and the sides opposite A, B, C are BC, CA, AB. The average of the vertices, G=(5/3,1), lies on none of the sides and so is not on the triangle. A, D=(2,1), E=(6,3) are distinct but collinear (E=A+3(D−A)), so they determine no nondegenerate triangle."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "A、D、E 三點兩兩相異，所以界定一個非退化三角形",
          "en": "A, D, E are pairwise distinct, so they determine a nondegenerate triangle"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "E=A+3(D−A)，三點共線（S15），違反不共線的前提。",
          "en": "E=A+3(D−A), so the three points are collinear (S15), violating the noncollinearity hypothesis."
        }
      },
      {
        "claim": {
          "zh-Hant": "G=(5/3,1) 在三角形 ABC 上，因為它在三角形圍住的區域內",
          "en": "G=(5/3,1) is on triangle ABC because it lies in the region the triangle encloses"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "依 MKO，三角形是三條邊的聯集；G 不在任何一條邊上（S13.1 到 S13.3）。",
          "en": "By the MKO the triangle is the union of its three sides; G lies on none of them (S13.1 to S13.3)."
        }
      },
      {
        "claim": {
          "zh-Hant": "AB 是三角形 ABC 的一個頂點",
          "en": "AB is a vertex of triangle ABC"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "AB 是邊，即線段 [A,B]；頂點是 A、B、C 三個點。",
          "en": "AB is a side, the segment [A,B]; the vertices are the three points A, B, C."
        }
      },
      {
        "claim": {
          "zh-Hant": "1=4t 有解 t=1/4，所以 C 在直線 AB 上",
          "en": "1=4t has the solution t=1/4, so C is on line AB"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "同一個 t 還須滿足第二坐標方程 3=0，對任何 t 都不成立（S05）。",
          "en": "The same t must also satisfy the second coordinate equation 3=0, which fails for every t (S05)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "cartesian_product_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "笛卡兒積：A×B 是所有有序對 (a,b)（a∈A、b∈B）的集合；(a,b)=(c,d) ⇔ a=c 且 b=d。",
          "en": "Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d."
        }
      },
      {
        "id": "image_preimage_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "像與原像：f(S)={f(a) ∣ a∈S}；f⁻¹(T)={a∈A ∣ f(a)∈T}。原像記號不預設 f 有反函數。",
          "en": "Image and preimage: f(S)={f(a) ∣ a∈S}; f⁻¹(T)={a∈A ∣ f(a)∈T}. The preimage notation does not presuppose that f has an inverse."
        }
      },
      {
        "id": "set_operations_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合運算：x∈A∪B ⇔ x∈A 或 x∈B；x∈A∩B ⇔ x∈A 且 x∈B；x∈A∖B ⇔ x∈A 且 x∉B；在全集 U 中 Aᶜ=U∖A。",
          "en": "Set operations: x∈A∪B ⇔ x∈A or x∈B; x∈A∩B ⇔ x∈A and x∈B; x∈A∖B ⇔ x∈A and x∉B; in a universe U, Aᶜ=U∖A."
        }
      },
      {
        "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": "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": "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": "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": "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": "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": "exists_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃消去：由 ∃x∈D, P(x)，可取一個新名字 a 表示某個滿足 P(a) 的元素；只要最後結論不含 a，就可用 a 推理。",
          "en": "∃-elimination: from ∃x∈D, P(x) we may name some element a with P(a) by a fresh name; reasoning with a is allowed as long as the final conclusion does not mention a."
        }
      },
      {
        "id": "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": "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": "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": "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": "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": "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": "subset_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "子集：A⊆B 表示 A 的每個元素都是 B 的元素，即 ∀x (x∈A⇒x∈B)。",
          "en": "Subset: A⊆B means every element of A is an element of B, i.e. ∀x (x∈A⇒x∈B)."
        }
      },
      {
        "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": "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": "imp_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "→引入（釋放假設）：若在暫時假設 P 的作用域中推出 Q，則得到 P⇒Q；此後 P 不再可當作已知使用。",
          "en": "→-introduction (discharging a hypothesis): if Q is derived inside a scope that temporarily assumes P, then P⇒Q follows, and P may no longer be used as known."
        }
      },
      {
        "id": "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": "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": "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": "de_morgan_logic",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "邏輯的德摩根律：¬(P∧Q) ⇔ ¬P∨¬Q；¬(P∨Q) ⇔ ¬P∧¬Q。",
          "en": "De Morgan's laws for logic: ¬(P∧Q) ⇔ ¬P∨¬Q and ¬(P∨Q) ⇔ ¬P∧¬Q."
        }
      },
      {
        "id": "mko_triangle_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-triangle",
          "sha256": "1556dabd42bf5d132f41296684eb2b037900a511e659c48627bee6d0c4dcd461",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "給定平面中三個相異且不共線的點 A、B、C，線段 AB、BC、CA 的聯合界定非退化三角形 ABC。",
          "en": "Given three distinct noncollinear points A, B, and C in the plane, the union of segments AB, BC, and CA determines the nondegenerate triangle ABC."
        }
      },
      {
        "id": "mko_model_dependence",
        "kind": "cited",
        "source": {
          "mko_id": "mko-point-line-plane",
          "sha256": "abf9d0f665f79eaddd4998e3b6080630ebd88fba8faec86475aa135d364f1efd",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在指定的幾何模型中，點表示位置；線與面是承載點並具有相應維度的幾何對象，其精確性質由該模型的公理與結構決定。",
          "en": "In a specified geometric model, a point represents a position. Lines and planes are geometric objects carrying points with their respective dimensions, and their precise properties are determined by the axioms and structure of that model."
        }
      },
      {
        "id": "coord_ops",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逐坐標運算：(a,b)+(c,d)=(a+c,b+d)，(a,b)−(c,d)=(a−c,b−d)，t(a,b)=(ta,tb)。",
          "en": "Coordinatewise operations: (a,b)+(c,d)=(a+c,b+d), (a,b)−(c,d)=(a−c,b−d), t(a,b)=(ta,tb)."
        }
      },
      {
        "id": "line_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "本模型的直線：對相異點 P≠Q，L(P,Q)={P+t(Q−P) ∣ t∈ℝ}；模型中的每一條直線都是某個這樣的 L(P,Q)。",
          "en": "Lines of this model: for distinct points P≠Q, L(P,Q)={P+t(Q−P) ∣ t∈ℝ}; every line of the model is such an L(P,Q)."
        }
      },
      {
        "id": "segment_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "線段：對點 P、Q，[P,Q]={P+t(Q−P) ∣ 0≤t≤1}；P、Q 是它的端點。",
          "en": "Segment: for points P, Q, [P,Q]={P+t(Q−P) ∣ 0≤t≤1}; P and Q are its endpoints."
        }
      },
      {
        "id": "collinear_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "共線：若存在一條直線同時包含若干點，這些點稱為共線。",
          "en": "Collinear: points are collinear if some line contains all of them."
        }
      },
      {
        "id": "L_line_unique",
        "kind": "lemma",
        "proved_in": [
          "S07.1",
          "S07.2",
          "S07.3"
        ],
        "statement": {
          "zh-Hant": "若 P≠Q，且 X、Y 是 L(P,Q) 上的相異兩點，則 L(X,Y)=L(P,Q)。",
          "en": "If P≠Q and X, Y are distinct points of L(P,Q), then L(X,Y)=L(P,Q)."
        }
      },
      {
        "id": "L_seg_reverse",
        "kind": "lemma",
        "proved_in": [
          "S11.1",
          "S11.2"
        ],
        "statement": {
          "zh-Hant": "對任意點 P、Q：[P,Q]=[Q,P]。",
          "en": "For any points P, Q: [P,Q]=[Q,P]."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "4-0",
        "rhs": "4"
      },
      {
        "id": "N2",
        "step": "S05",
        "lhs": "4*(1/4)",
        "rhs": "1"
      },
      {
        "id": "N3",
        "step": "S10",
        "lhs": "1-4",
        "rhs": "-3"
      },
      {
        "id": "N4",
        "step": "S10",
        "lhs": "0-3",
        "rhs": "-3"
      },
      {
        "id": "N5",
        "step": "S10",
        "lhs": "4-3*1",
        "rhs": "1"
      },
      {
        "id": "N6",
        "step": "S12",
        "lhs": "4-3*0",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N7",
        "step": "S13",
        "lhs": "(0+4+1)/3",
        "rhs": "5/3"
      },
      {
        "id": "N8",
        "step": "S13",
        "lhs": "(0+0+3)/3",
        "rhs": "1"
      },
      {
        "id": "N9",
        "step": "S13.2",
        "lhs": "4-3*(1/3)",
        "rhs": "3"
      },
      {
        "id": "N10",
        "step": "S13.2",
        "lhs": "3",
        "relation": "!=",
        "rhs": "5/3"
      },
      {
        "id": "N11",
        "step": "S13.3",
        "lhs": "(3-1)/3",
        "rhs": "2/3"
      },
      {
        "id": "N12",
        "step": "S13.3",
        "lhs": "1-2/3",
        "rhs": "1/3"
      },
      {
        "id": "N13",
        "step": "S13.3",
        "lhs": "1/3",
        "relation": "!=",
        "rhs": "5/3"
      },
      {
        "id": "N14",
        "step": "S07.2",
        "vars": [
          "p",
          "d",
          "s",
          "u",
          "t"
        ],
        "degree": 2,
        "lhs": "p+s*d+t*((p+u*d)-(p+s*d))",
        "rhs": "p+(s+t*(u-s))*d"
      },
      {
        "id": "N15",
        "step": "S11.1",
        "vars": [
          "p",
          "q",
          "t"
        ],
        "degree": 2,
        "lhs": "p+t*(q-p)",
        "rhs": "q+(1-t)*(p-q)"
      },
      {
        "id": "N16",
        "step": "S15",
        "lhs": "0+3*(2-0)",
        "rhs": "6"
      },
      {
        "id": "N17",
        "step": "S15",
        "lhs": "0+3*(1-0)",
        "rhs": "3"
      },
      {
        "id": "N18",
        "step": "S08",
        "lhs": "(4-0)*(3-0)-(1-0)*(0-0)",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N19",
        "step": "S15",
        "lhs": "(2-0)*(3-0)-(6-0)*(1-0)",
        "rhs": "0"
      },
      {
        "id": "N20",
        "lhs": "(-1)**2+(-3)**2",
        "rhs": "10"
      },
      {
        "id": "N21",
        "lhs": "(-3)**2+3**2",
        "rhs": "18"
      },
      {
        "id": "N22",
        "step": "S14",
        "lhs": "4*(1/2)",
        "rhs": "2"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只在 ℝ² 的座標模型中工作（直線與線段以參數式定義），處理 MKO 所建立的頂點、邊、對邊與非退化條件。不處理按邊或角的分類、內角和、面積、全等與相似（MKO 指出這些需要額外資料與定理）；不由綜合幾何公理推導座標模型；不改動 canonical MKO 或英文 edition。",
      "en": "Works only in the coordinate model of ℝ² (lines and segments defined parametrically) and treats what the MKO sets up: vertices, sides, opposite sides and the nondegeneracy condition. It does not treat classification by sides or angles, angle sums, area, congruence or similarity (the MKO says these need further data and theorems), does not derive the coordinate model from synthetic axioms, and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "A,B,C",
        "三角形的三個頂點",
        "the three vertices of the triangle"
      ],
      [
        "L(P,Q)",
        "通過相異點 P、Q 的直線",
        "the line through distinct points P and Q"
      ],
      [
        "[P,Q]",
        "以 P、Q 為端點的線段",
        "the segment with endpoints P and Q"
      ],
      [
        "\\triangle ABC",
        "三條邊的聯集 [A,B]∪[B,C]∪[C,A]",
        "the union of the three sides [A,B]∪[B,C]∪[C,A]"
      ],
      [
        "t,s,u,r",
        "實數參數",
        "real parameters"
      ],
      [
        "G",
        "三個頂點的平均點 (5/3,1)",
        "the average of the three vertices, (5/3,1)"
      ]
    ],
    "md_error_controls": [
      {
        "title": "任三點不一定構成三角形 / Three points need not form a triangle",
        "zh": "MKO 指出常見誤解是以為任何三點都形成非退化三角形。A、D、E 兩兩相異，粗略的示意圖上也可能像一個很扁的三角形，但 E=A+3(D−A)，三點在同一直線上（S15）。若三點中有重合，例如 A、A、B，連「相異」都不成立。所以 S02 與 S08 兩個前提都必須明確核對，只說「有三個點」不夠。",
        "en": "The MKO notes the common mistake of thinking that any three points form a nondegenerate triangle. A, D, E are pairwise distinct and in a rough sketch may look like a very flat triangle, but E=A+3(D−A), so the three points lie on one line (S15). If two of the points coincide, as in A, A, B, even distinctness fails. So both hypotheses, S02 and S08, must be checked explicitly; saying 'there are three points' is not enough."
      },
      {
        "title": "頂點、邊與三角形本身 / Vertices, sides and the triangle itself",
        "zh": "頂點是點（例如 A），邊是線段（AB 指 [A,B]），兩者不能互換：說「頂點 AB」是把邊的名稱當成頂點。依 MKO，三角形 ABC 是三條邊的聯集：G=(5/3,1) 在圖上看起來「在三角形裡」，卻不在任何一條邊上（S13 到 S14），所以 G∉△ABC。",
        "en": "A vertex is a point (for example A), a side is a segment (AB means [A,B]), and the two may not be swapped: 'the vertex AB' uses a side's name as if it were a vertex. By the MKO, triangle ABC is the union of the three sides: G=(5/3,1) looks 'inside the triangle' in a picture, yet lies on none of the sides (S13 to S14), so G∉△ABC."
      },
      {
        "title": "不從圖形比例猜邊長或角度 / Do not read lengths or angles off the sketch",
        "zh": "MKO 提醒不要從畫圖比例猜性質。示意圖上 AC 與 BC 也許看起來差不多長，但另行計算長度平方，|AC|²=(−1)²+(−3)²=10、|BC|²=(−3)²+3²=18（N20、N21），兩者不同。這類長度與角度結論屬於其他 MKO，需要各自的定義與定理；本稿只建立頂點、邊與對邊。",
        "en": "The MKO warns against guessing properties from the proportions of a drawing. In a sketch AC and BC may look about equally long, but computing squared lengths separately gives |AC|²=(−1)²+(−3)²=10 and |BC|²=(−3)²+3²=18 (N20, N21), which differ. Such length and angle conclusions belong to other MKOs with their own definitions and theorems; this guide establishes only vertices, sides and opposite sides."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "N18、N19 是行列式形式的獨立交叉核對，只對這組數字有效；本稿論證不依賴它們。",
        "en": "N18 and N19 are independent cross-checks in determinant form, valid for these numbers only; the argument does not depend on them."
      },
      {
        "zh-Hant": "本稿只在 ℝ² 座標模型中工作；座標模型與綜合幾何公理系統之間的對應未在此處理。",
        "en": "This guide works only in the coordinate model of ℝ²; the correspondence between the coordinate model and a synthetic axiom system is not treated here."
      },
      {
        "zh-Hant": "N20、N21 的長度平方只用於錯誤控制，屬於歐幾里得長度的 MKO，不屬於本稿論證。",
        "en": "The squared lengths in N20 and N21 serve only an error control; they belong to the Euclidean length MKO and are not part of this argument."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "579b72812beece3e12baad63e611a795eb814972eceec9d3b1dd1abbe85ae559",
    "original_source_sha256": "c4b2a9675620225c60a7ccfcc9072afda7b8e81861a9cd1a613e7b2b997c9dcc",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M002"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}