{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-euclid-pythagorean-theorem-01",
  "mko_id": "mko-euclid-pythagorean-theorem",
  "title": {
    "zh-Hant": "畢達哥拉斯定理：在座標模型中以內積證明",
    "en": "Pythagorean theorem: a proof by dot products in the coordinate model"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-euclid-pythagorean-theorem-01",
    "source": {
      "mko_id": "mko-euclid-pythagorean-theorem",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "f91873f82805e9a6d9ee8358fa64afa78e7594fc6568999814e34c1205ad51c6",
      "english_sha256": "1ad53a259b7aa8574f8579d208fda5c28c60bbd7d60b814200022c5917b7a849"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "square_nonneg",
      "sqrt_def",
      "distributive",
      "cartesian_product_def",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在以 ℝ² 表示的歐幾里得平面中（距離為歐幾里得長度），證明：對任何直角三角形 ABC，若直角在 C，兩股長 a=d(C,B)、b=d(C,A)，斜邊長 c=d(B,A)，則 a²+b²=c²。再以 C=(1,2)、B=(4,6)、A=(−7,8) 驗算（25+100=125），並以非直角三角形 E=(0,0)、F=(4,0)、H=(1,2) 說明沒有直角時關係一般不成立。",
      "en": "In the Euclidean plane represented by ℝ² (with Euclidean length as distance), prove: for every right triangle ABC with the right angle at C, legs of lengths a=d(C,B) and b=d(C,A), and hypotenuse of length c=d(B,A), a²+b²=c². Then check the example C=(1,2), B=(4,6), A=(−7,8) (25+100=125), and use the non-right triangle E=(0,0), F=(4,0), H=(1,2) to show that without a right angle the relation fails in general."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "模型：歐幾里得平面以 ℝ² 表示（MKO 假設 a1），點的坐標為實數，運算逐坐標；距離 d(P,Q)=√((x₂−x₁)²+(y₂−y₁)²)；內積 u·v=u₁v₁+u₂v₂；對相異點 P≠Q，直線 L(P,Q)={P+t(Q−P) ∣ t∈ℝ}。",
          "en": "Model: the Euclidean plane is represented by ℝ² (MKO assumption a1), with real coordinates and coordinatewise operations; distance d(P,Q)=√((x₂−x₁)²+(y₂−y₁)²); dot product u·v=u₁v₁+u₂v₂; for distinct points P≠Q, the line L(P,Q)={P+t(Q−P) ∣ t∈ℝ}."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "內角量值：非退化三角形的頂點 X 處，令 u=Y−X、v=Z−X（皆非零），內角是唯一滿足 cos θ=(u·v)/(|u||v|) 的 θ∈[0,π]，其中 |u|=√(u·u)。",
          "en": "Interior angle measure: at a vertex X of a nondegenerate triangle, with u=Y−X and v=Z−X (both nonzero), the interior angle is the unique θ∈[0,π] with cos θ=(u·v)/(|u||v|), where |u|=√(u·u)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "ABC 是任意一個非退化三角形，其 C 處內角 γ=π/2（MKO 假設 a2）；a=d(C,B)、b=d(C,A) 是兩股長，c=d(B,A) 是斜邊長（假設 a3），三者都是非負實數（假設 a4）。除此之外對 ABC 不作任何假設。",
          "en": "ABC is an arbitrary nondegenerate triangle whose interior angle at C is γ=π/2 (MKO assumption a2); a=d(C,B) and b=d(C,A) are the leg lengths and c=d(B,A) is the hypotenuse length (assumption a3), all nonnegative reals (assumption a4). Nothing else is assumed about ABC."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "例子：C=(1,2)、B=(4,6)、A=(−7,8)；對照：E=(0,0)、F=(4,0)、H=(1,2)。",
          "en": "Example: C=(1,2), B=(4,6), A=(−7,8); comparison: E=(0,0), F=(4,0), H=(1,2)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G3",
          "mko_length_def",
          "mko_right_triangle_def"
        ],
        "before_tex": "\\text{right triangle }ABC,\\ \\gamma=\\tfrac{\\pi}{2}",
        "after_tex": "a=d(C,B),\\ b=d(C,A),\\ c=d(B,A);\\ \\text{goal}\\ a^2+b^2=c^2",
        "relation": "definition",
        "rule": "def_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把定理的資料寫成模型中的量：依直角三角形的 MKO，ABC 有一個內角等於 π/2；把直角的頂點命名為 C（頂點名稱可以任意調換，所以不失一般性）。兩股是以 C 為端點的兩邊，斜邊是 C 的對邊，三者的長度依歐幾里得長度的 MKO 計算，都是非負實數（假設 a4）。目標是 a²+b²=c²。",
          "en": "Write the data of the theorem as quantities of the model: by the right-triangle MKO, ABC has an interior angle equal to π/2; call its vertex C (vertex names can be permuted freely, so no generality is lost). The legs are the two sides with endpoint C, the hypotenuse is the side opposite C, and their lengths are computed by the Euclidean length MKO, all nonnegative reals (assumption a4). The goal is a²+b²=c²."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "def_intro",
            "mko_right_triangle_def",
            "mko_length_def"
          ],
          "explanation": {
            "zh-Hant": "此步只命名；沒有對 ABC 加任何額外假設，這是 S09 能推廣到所有直角三角形的前提。",
            "en": "This step only names things; it puts no extra assumption on ABC, which is what lets S09 generalize to every right triangle."
          }
        },
        "leaf_basis": "def_intro + mko_right_triangle_def + mko_length_def"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "G3"
        ],
        "before_tex": "B-C,\\ A-C",
        "after_tex": "u=(u_1,u_2)=B-C,\\ v=(v_1,v_2)=A-C,\\ u\\neq 0,\\ v\\neq 0",
        "relation": "definition",
        "rule": "def_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "以 C 為起點引入兩個邊向量 u=B−C、v=A−C。三角形非退化，所以 B≠C、A≠C，於是 u、v 都不是零向量。",
          "en": "Introduce the two side vectors from C: u=B−C and v=A−C. The triangle is nondegenerate, so B≠C and A≠C, hence u and v are nonzero."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "def_intro",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "非零性來自非退化；若 B=C，u=0，C 處沒有角。",
            "en": "Nonzero-ness comes from nondegeneracy; if B=C then u=0 and there is no angle at C."
          }
        },
        "leaf_basis": "def_intro + coord_ops"
      },
      {
        "id": "S03",
        "uses": [
          "S02"
        ],
        "before_tex": "A-B",
        "after_tex": "v-u",
        "relation": "equivalent",
        "rule": "coord_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "v−u=(A−C)−(B−C)=A−C−B+C=A−B，逐坐標都是恆等式（N3）。所以斜邊方向的差向量可以用兩個股向量表示。",
          "en": "v−u=(A−C)−(B−C)=A−C−B+C=A−B, an identity in each coordinate (N3). So the difference vector along the hypotenuse is expressed through the two leg vectors."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "coord_ops",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "恆等式，雙向成立。",
            "en": "It is an identity and holds both ways."
          }
        },
        "leaf_basis": "coord_ops + add_assoc + add_comm + add_inverse"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "G2",
          "L_right_iff_dot"
        ],
        "before_tex": "\\gamma=\\tfrac{\\pi}{2}",
        "after_tex": "u\\cdot v=u_1v_1+u_2v_2=0",
        "relation": "equivalent",
        "rule": "L_right_iff_dot",
        "conditions": [
          "u\\neq 0",
          "v\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依 G2，C 處內角 γ∈[0,π] 滿足 cos γ=(u·v)/(|u||v|)。依引理 L_right_iff_dot，γ=π/2 ⇔ u·v=0。這把幾何的「直角」轉成代數的「內積為零」。引理在 S04.1、S04.2 證明。",
          "en": "By G2, the interior angle γ∈[0,π] at C satisfies cos γ=(u·v)/(|u||v|). By Lemma L_right_iff_dot, γ=π/2 ⇔ u·v=0. This turns the geometric 'right angle' into the algebraic 'dot product zero'. The lemma is proved in S04.1 and S04.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_right_iff_dot",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "引理是雙條件；u、v 非零（S02）不可省。",
            "en": "The lemma is a biconditional; u and v being nonzero (S02) cannot be dropped."
          }
        },
        "leaf_basis": "L_right_iff_dot + angle_cos_def"
      },
      {
        "id": "S04.1",
        "uses": [
          "dot_def",
          "square_nonneg"
        ],
        "before_tex": "\\cos\\theta=\\frac{u\\cdot v}{|u|\\,|v|},\\ u\\neq 0,\\ v\\neq 0",
        "after_tex": "\\cos\\theta=0\\iff u\\cdot v=0",
        "relation": "equivalent",
        "rule": "mul_inverse_nonzero",
        "conditions": [
          "|u|\\,|v|\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "u≠0 表示 u₁、u₂ 不全為 0；平方非負，且非零數的平方大於 0，所以 u₁²+u₂²>0，|u|>0。同理 |v|>0，故 |u||v|>0，特別是 ≠0。於是 cos θ=0 時兩邊乘 |u||v| 得 u·v=0；反之 u·v=0 時 cos θ=0/(|u||v|)=0。",
          "en": "u≠0 means u₁ and u₂ are not both 0; squares are nonnegative and the square of a nonzero number is positive, so u₁²+u₂²>0 and |u|>0. Likewise |v|>0, so |u||v|>0, in particular ≠0. If cos θ=0, multiplying by |u||v| gives u·v=0; conversely, if u·v=0 then cos θ=0/(|u||v|)=0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_inverse_nonzero",
            "square_nonneg",
            "sqrt_def"
          ],
          "explanation": {
            "zh-Hant": "乘以與除以 |u||v| 互為反運算，前提 |u||v|≠0；若 u=0，分母為 0，式子沒有定義。",
            "en": "Multiplying and dividing by |u||v| undo each other, given |u||v|≠0; if u=0 the denominator is 0 and the formula is undefined."
          }
        },
        "leaf_basis": "square_nonneg + sqrt_def + order_basics + mul_inverse_nonzero + mul_zero"
      },
      {
        "id": "S04.2",
        "uses": [
          "cos_facts",
          "S04.1"
        ],
        "before_tex": "\\theta\\in[0,\\pi]",
        "after_tex": "\\theta=\\tfrac{\\pi}{2}\\iff\\cos\\theta=0\\iff u\\cdot v=0",
        "relation": "equivalent",
        "rule": "cos_facts",
        "conditions": [
          "\\theta\\in[0,\\pi]"
        ],
        "explanation": {
          "zh-Hant": "⇒：cos(π/2)=0。⇐：cos 在 [0,π] 上嚴格遞減，所以在那裡單射；若 θ∈[0,π] 且 cos θ=0=cos(π/2)，則 θ=π/2。再與 S04.1 串接得 θ=π/2 ⇔ u·v=0。S04.1 與 S04.2 合起來證得引理 L_right_iff_dot。",
          "en": "⇒: cos(π/2)=0. ⇐: cos is strictly decreasing on [0,π], hence injective there; if θ∈[0,π] and cos θ=0=cos(π/2), then θ=π/2. Chaining with S04.1 gives θ=π/2 ⇔ u·v=0. S04.1 and S04.2 together prove Lemma L_right_iff_dot."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "cos_facts",
            "logic_chain",
            "S04.1"
          ],
          "explanation": {
            "zh-Hant": "⇐ 方向需要 θ 限制在 [0,π]；若允許 θ=3π/2，cos θ=0 也成立。",
            "en": "The ⇐ direction needs θ restricted to [0,π]; if θ=3π/2 were allowed, cos θ=0 would hold too."
          }
        },
        "leaf_basis": "cos_facts + logic_chain + injective_surjective_def"
      },
      {
        "id": "S05",
        "uses": [
          "S02",
          "S03",
          "G1",
          "sqrt_def"
        ],
        "before_tex": "a=d(C,B),\\ b=d(C,A),\\ c=d(B,A)",
        "after_tex": "a^2=u_1^2+u_2^2,\\ b^2=v_1^2+v_2^2,\\ c^2=(v_1-u_1)^2+(v_2-u_2)^2",
        "relation": "equivalent",
        "rule": "sqrt_def",
        "conditions": [
          "a,b,c\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "依距離定義，a=√((x_B−x_C)²+(y_B−y_C)²)=√(u₁²+u₂²)；被開方數 ≥0，主平方根滿足 (√s)²=s，所以 a²=u₁²+u₂²。同理 b²=v₁²+v₂²。對 c，A−B=v−u（S03），所以 c²=(v₁−u₁)²+(v₂−u₂)²。",
          "en": "By the distance definition, a=√((x_B−x_C)²+(y_B−y_C)²)=√(u₁²+u₂²); the radicand is ≥0 and the principal root satisfies (√s)²=s, so a²=u₁²+u₂². Likewise b²=v₁²+v₂². For c, A−B=v−u (S03), so c²=(v₁−u₁)²+(v₂−u₂)²."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "sqrt_def",
            "mko_length_def"
          ],
          "explanation": {
            "zh-Hant": "對非負的 x 與 s，x=√s ⇔ x²=s；三個長度都非負（假設 a4），所以平方前後資訊相同。",
            "en": "For nonnegative x and s, x=√s ⇔ x²=s; all three lengths are nonnegative (assumption a4), so squaring loses no information."
          }
        },
        "leaf_basis": "sqrt_def + mko_length_def + eq_replace"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "L_sq_diff"
        ],
        "before_tex": "(v_1-u_1)^2+(v_2-u_2)^2",
        "after_tex": "(v_1^2-2u_1v_1+u_1^2)+(v_2^2-2u_2v_2+u_2^2)",
        "relation": "equivalent",
        "rule": "L_sq_diff",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對每個坐標使用引理 L_sq_diff（p=v_i、q=u_i）：(v_i−u_i)²=v_i²−2v_iu_i+u_i²，再用乘法交換律把中間項寫成 −2u_iv_i。引理在 S06.1 證明。",
          "en": "Apply Lemma L_sq_diff in each coordinate (p=v_i, q=u_i): (v_i−u_i)²=v_i²−2v_iu_i+u_i², and commutativity writes the middle term as −2u_iv_i. The lemma is proved in S06.1."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_sq_diff",
            "mul_comm"
          ],
          "explanation": {
            "zh-Hant": "恆等式，雙向成立。",
            "en": "It is an identity and holds both ways."
          }
        },
        "leaf_basis": "L_sq_diff + mul_comm + eq_replace"
      },
      {
        "id": "S06.1",
        "uses": [
          "distributive",
          "neg_mul"
        ],
        "before_tex": "(p-q)^2=(p-q)(p-q)",
        "after_tex": "p^2-2pq+q^2",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "用分配律展開：(p−q)(p−q)=p·p−p·q−q·p+q·q；依交換律 qp=pq，兩個 −pq 合併成 −2pq（2pq=pq+pq）；(−q)(−q)=q²（負號與乘法）。得 p²−2pq+q²，是多項式恆等式（N2）。這證得引理 L_sq_diff。",
          "en": "Expand with distributivity: (p−q)(p−q)=p·p−p·q−q·p+q·q; by commutativity qp=pq, and the two −pq combine to −2pq (2pq=pq+pq); (−q)(−q)=q² (signs and multiplication). The result p²−2pq+q² is a polynomial identity (N2). This proves Lemma L_sq_diff."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "mul_comm",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "恆等式，雙向成立。",
            "en": "It is an identity and holds both ways."
          }
        },
        "leaf_basis": "distributive + mul_comm + neg_mul + add_assoc"
      },
      {
        "id": "S07",
        "uses": [
          "S05",
          "S06"
        ],
        "before_tex": "(v_1^2-2u_1v_1+u_1^2)+(v_2^2-2u_2v_2+u_2^2)",
        "after_tex": "c^2=a^2+b^2-2\\,u\\cdot v",
        "relation": "equivalent",
        "rule": "add_comm",
        "conditions": [],
        "explanation": {
          "zh-Hant": "重新分組（加法交換律與結合律、分配律）：=(u₁²+u₂²)+(v₁²+v₂²)−2(u₁v₁+u₂v₂)。依 S05 與內積定義，這是 a²+b²−2u·v。注意這個等式對任何三角形都成立（N1 是四個變數的多項式恆等式），尚未用到直角。",
          "en": "Regroup (commutativity and associativity of addition, distributivity): =(u₁²+u₂²)+(v₁²+v₂²)−2(u₁v₁+u₂v₂). By S05 and the definition of the dot product, this is a²+b²−2u·v. Note that this equation holds for every triangle (N1 is a polynomial identity in four variables); the right angle has not been used yet."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "add_comm",
            "add_assoc",
            "distributive",
            "dot_def"
          ],
          "explanation": {
            "zh-Hant": "恆等式，雙向成立。",
            "en": "It is an identity and holds both ways."
          }
        },
        "leaf_basis": "add_comm + add_assoc + distributive + dot_def"
      },
      {
        "id": "S08",
        "uses": [
          "S04",
          "S07"
        ],
        "before_tex": "c^2=a^2+b^2-2\\,u\\cdot v,\\ u\\cdot v=0",
        "after_tex": "a^2+b^2=c^2",
        "relation": "implies",
        "rule": "eq_replace",
        "conditions": [
          "u\\cdot v=0"
        ],
        "explanation": {
          "zh-Hant": "直角在這裡才用上：依 S04，u·v=0，代入 S07 得 c²=a²+b²−2·0=a²+b²；由等號的對稱即 a²+b²=c²。",
          "en": "The right angle is used only here: by S04, u·v=0, and substituting into S07 gives c²=a²+b²−2·0=a²+b²; by symmetry of equality, a²+b²=c²."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S07",
            "eq_replace",
            "mul_zero"
          ],
          "explanation": {
            "zh-Hant": "需要 u·v=0；沒有直角時多出的 −2u·v 一般不為 0（見 S13）。此步不宣稱逆命題。",
            "en": "It needs u·v=0; without a right angle the extra term −2u·v is in general not 0 (see S13). No converse is claimed in this step."
          }
        },
        "leaf_basis": "eq_replace + mul_zero + add_zero + eq_refl_symm_trans"
      },
      {
        "id": "S09",
        "uses": [
          "S01",
          "S08"
        ],
        "before_tex": "a^2+b^2=c^2\\ \\text{for the arbitrary right triangle of G3}",
        "after_tex": "\\text{every right triangle of the model}:\\ a^2+b^2=c^2",
        "relation": "implies",
        "rule": "forall_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "S01 到 S08 只用到 G3 所說的「非退化、C 處為直角」，沒有用到任何具體坐標，所以 ABC 是任意的；依全稱推廣，模型中每個直角三角形都有 a²+b²=c²。若直角在其他頂點，只要重新命名頂點即可。",
          "en": "S01 to S08 use only what G3 says, nondegenerate with a right angle at C, and no particular coordinates, so ABC is arbitrary; by ∀-introduction every right triangle of the model satisfies a²+b²=c². If the right angle is at another vertex, it suffices to rename the vertices."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S08",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "全稱推廣要求沒有對 ABC 加額外假設；有限個例子（如 S12）不能取代這一步。",
            "en": "∀-introduction requires that no extra assumption was put on ABC; finitely many examples (such as S12) cannot replace this step."
          }
        },
        "leaf_basis": "forall_intro"
      },
      {
        "id": "S10",
        "uses": [
          "G4",
          "line_unique"
        ],
        "before_tex": "C=(1,2),\\ B=(4,6),\\ A=(-7,8)",
        "after_tex": "\\triangle ABC\\ \\text{nondegenerate}",
        "relation": "implies",
        "rule": "line_unique",
        "conditions": [
          "C\\neq B"
        ],
        "explanation": {
          "zh-Hant": "三點兩兩相異（第一坐標 1、4、−7 互不相同）。B−C=(3,4)，A∈L(C,B) ⇔ 有 t 使 (−7,8)=(1+3t, 2+4t) ⇔ t=−8/3 且 t=3/2，不可能，所以 A∉L(C,B)；依直線唯一性三點不共線，三角形非退化。",
          "en": "The points are pairwise distinct (first coordinates 1, 4, −7 all differ). B−C=(3,4), and A∈L(C,B) ⇔ some t gives (−7,8)=(1+3t, 2+4t) ⇔ t=−8/3 and t=3/2, which is impossible, so A∉L(C,B); by uniqueness of the line the points are noncollinear and the triangle is nondegenerate."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "line_unique",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "例子也必須先確認非退化，否則 G3 不適用。",
            "en": "The example must also be checked to be nondegenerate first; otherwise G3 does not apply."
          }
        },
        "leaf_basis": "line_unique + image_preimage_def + contradiction + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S10",
          "S04",
          "G4"
        ],
        "before_tex": "u=B-C=(3,4),\\ v=A-C=(-8,6)",
        "after_tex": "u\\cdot v=-24+24=0,\\ \\gamma=\\tfrac{\\pi}{2}",
        "relation": "equivalent",
        "rule": "L_right_iff_dot",
        "conditions": [
          "u\\neq 0",
          "v\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "u=(4−1, 6−2)=(3,4)，v=(−7−1, 8−2)=(−8,6)；u·v=3·(−8)+4·6=−24+24=0。依 S04 的等價，C 處是直角，例子符合 G3。",
          "en": "u=(4−1, 6−2)=(3,4) and v=(−7−1, 8−2)=(−8,6); u·v=3·(−8)+4·6=−24+24=0. By the equivalence S04 the angle at C is right, so the example satisfies G3."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_right_iff_dot",
            "dot_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確計算可重算（N7）；引理雙向成立。",
            "en": "The exact computation can be redone (N7); the lemma holds both ways."
          }
        },
        "leaf_basis": "L_right_iff_dot + dot_def + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S05",
          "S11",
          "G4"
        ],
        "before_tex": "a^2,\\ b^2,\\ c^2\\ \\text{for the example}",
        "after_tex": "25+100=125",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "a²=3²+4²=25，b²=(−8)²+6²=64+36=100；A−B=(−7−4, 8−6)=(−11,2)，c²=121+4=125；25+100=125，與 S08 一致。a=5、b=10、c=√125=5√5（因 5√5≥0 且 (5√5)²=25·5=125）。這只是一個例子的核對，不是定理的證明。",
          "en": "a²=3²+4²=25 and b²=(−8)²+6²=64+36=100; A−B=(−7−4, 8−6)=(−11,2), so c²=121+4=125; 25+100=125, agreeing with S08. a=5, b=10, c=√125=5√5 (since 5√5≥0 and (5√5)²=25·5=125). This is a check of one example, not a proof of the theorem."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S11",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "可獨立重算（N8 到 N12）；它支持 S08，但不能取代 S01 到 S09 的一般證明。",
            "en": "Independently recomputable (N8 to N12); it supports S08 but cannot replace the general proof in S01 to S09."
          }
        },
        "leaf_basis": "numeric_evaluation + sqrt_def + power_laws"
      },
      {
        "id": "S13",
        "uses": [
          "S07",
          "G4"
        ],
        "before_tex": "E=(0,0),\\ F=(4,0),\\ H=(1,2)",
        "after_tex": "d(E,F)^2=16\\neq 13+5=d(H,F)^2+d(H,E)^2",
        "relation": "implies",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "非直角的對照：H 不在 x 軸 L(E,F)={(4t,0)} 上（第二坐標 2≠0），所以 EFH 非退化。在 H：u=F−H=(3,−2)、v=E−H=(−1,−2)，u·v=−3+4=1≠0，所以 H 處不是直角。d(H,F)²=9+4=13，d(H,E)²=1+4=5，d(E,F)²=16；13+5=18≠16。S07 的一般等式給出 16=13+5−2·1，正好說明少掉的是 −2u·v 這一項。E、F 處的內積分別是 4 與 12，也都不是直角；三種配對 13+5=16、13+16=5、5+16=13 都不成立。",
          "en": "A non-right comparison: H is not on the x-axis L(E,F)={(4t,0)} (second coordinate 2≠0), so EFH is nondegenerate. At H: u=F−H=(3,−2), v=E−H=(−1,−2), and u·v=−3+4=1≠0, so the angle at H is not right. d(H,F)²=9+4=13, d(H,E)²=1+4=5, d(E,F)²=16; 13+5=18≠16. The general equation of S07 gives 16=13+5−2·1, showing that what is missing is exactly the term −2u·v. The dot products at E and F are 4 and 12, so those angles are not right either; none of the three pairings 13+5=16, 13+16=5, 5+16=13 holds."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "dot_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "這說明定理的直角前提不可省；它不是定理的反例，因為 EFH 不是直角三角形。",
            "en": "This shows the right-angle hypothesis cannot be dropped; it is not a counterexample to the theorem, because EFH is not a right triangle."
          }
        },
        "leaf_basis": "eq_replace + dot_def + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S08",
          "S09",
          "S12",
          "S13"
        ],
        "before_tex": "\\text{S08, S09, S12, S13}",
        "after_tex": "\\forall\\ \\text{right triangles}:\\ a^2+b^2=c^2",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "定理在模型中成立：每個直角三角形的兩股平方和等於斜邊平方。例子 25+100=125 與之一致；EFH 顯示沒有直角時一般不成立。",
          "en": "The theorem holds in the model: for every right triangle the sum of the squares of the legs equals the square of the hypotenuse. The example 25+100=125 agrees; EFH shows that without a right angle the relation fails in general."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S04",
            "S07",
            "S08",
            "S09"
          ],
          "explanation": {
            "zh-Hant": "目標：模型中所有直角三角形 a²+b²=c²。足夠條件 H：對任意直角三角形，c²=a²+b²−2u·v（S07）且 u·v=0（S04）；H 推出目標由 S08 與 S09。H 由定義與恆等式正向證成，沒有把目標當作前提。",
            "en": "Goal: a²+b²=c² for all right triangles of the model. Sufficient condition H: for an arbitrary right triangle, c²=a²+b²−2u·v (S07) and u·v=0 (S04); H gives the goal through S08 and S09. H is proved forwards from definitions and identities; the goal is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim + forall_intro"
      }
    ],
    "result": {
      "zh-Hant": "在 ℝ² 座標模型中，對任意直角三角形（直角在 C），以 u=B−C、v=A−C 表示，恆有 c²=a²+b²−2u·v，而直角等價於 u·v=0，所以 a²+b²=c²。例子 C=(1,2)、B=(4,6)、A=(−7,8) 給出 25+100=125；非直角的 EFH 給出 13+5=18≠16，差額正是 2u·v=2。",
      "en": "In the coordinate model of ℝ², for any right triangle (right angle at C), writing u=B−C and v=A−C, always c²=a²+b²−2u·v, and the right angle is equivalent to u·v=0, so a²+b²=c². The example C=(1,2), B=(4,6), A=(−7,8) gives 25+100=125; the non-right triangle EFH gives 13+5=18≠16, and the difference is exactly 2u·v=2."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "對三角形 EFH，兩短邊平方和等於最長邊平方：13+5=16",
          "en": "For triangle EFH, the squares of the two shorter sides add to the square of the longest: 13+5=16"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "13+5=18≠16；EFH 沒有直角（S13）。",
          "en": "13+5=18≠16; EFH has no right angle (S13)."
        }
      },
      {
        "claim": {
          "zh-Hant": "例子中 25+100=125 成立，所以定理對所有直角三角形得證",
          "en": "25+100=125 holds in the example, so the theorem is proved for all right triangles"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "有限個例子只是核對；普遍性來自 S01 到 S09 的任意三角形論證。",
          "en": "Finitely many examples are only checks; generality comes from the argument for an arbitrary triangle in S01 to S09."
        }
      },
      {
        "claim": {
          "zh-Hant": "在例子中把 b=10 當作斜邊：25+125=100",
          "en": "In the example, treat b=10 as the hypotenuse: 25+125=100"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "150≠100（N23）；斜邊是直角 C 的對邊 AB。",
          "en": "150≠100 (N23); the hypotenuse is AB, the side opposite the right angle C."
        }
      },
      {
        "claim": {
          "zh-Hant": "直角三角形的兩股長相加等於斜邊長：5+10=√125",
          "en": "In a right triangle the leg lengths add to the hypotenuse length: 5+10=√125"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "(5+10)²=225≠125（N24）；定理說的是平方的關係。",
          "en": "(5+10)²=225≠125 (N24); the theorem is a relation between squares."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "square_nonneg",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數平方非負：對所有實數 u，u²≥0，且 u²=0 ⇔ u=0。",
          "en": "Squares are nonnegative: for every real u, u²≥0, and u²=0 ⇔ u=0."
        }
      },
      {
        "id": "sqrt_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "主平方根：對 a≥0，√a 是唯一滿足 b≥0 且 b²=a 的實數 b。負數在實數中沒有平方根。",
          "en": "Principal square root: for a≥0, √a is the unique real b with b≥0 and b²=a. Negative numbers have no real square root."
        }
      },
      {
        "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": "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": "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_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": "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": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "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": "logic_chain",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "推論的傳遞：由 P⇒Q 與 Q⇒R 得 P⇒R；⇔ 同理。",
          "en": "Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔."
        }
      },
      {
        "id": "injective_surjective_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "單射：f(a)=f(a′) ⇒ a=a′。滿射：每個 b∈B 都有 a∈A 使 f(a)=b。雙射：既單又滿，此時有唯一反函數 f⁻¹ 使 f⁻¹∘f=id_A、f∘f⁻¹=id_B。",
          "en": "Injective: f(a)=f(a′) ⇒ a=a′. Surjective: every b∈B has some a∈A with f(a)=b. Bijective: both; then there is a unique inverse f⁻¹ with f⁻¹∘f=id_A and f∘f⁻¹=id_B."
        }
      },
      {
        "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": "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": "neg_mul",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。",
          "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv."
        }
      },
      {
        "id": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "id": "eq_refl_symm_trans",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。",
          "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w."
        }
      },
      {
        "id": "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": "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": "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": "power_laws",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。",
          "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1."
        }
      },
      {
        "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_right_triangle_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-right-triangle",
          "sha256": "189cad28fbc4231852d6848d7f5209ff5d91a3b2dabd6380708fc678de0a758e",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "若三角形的一個內角等於九十度，或以弧度表示為 π/2，則此三角形稱為直角三角形。",
          "en": "A triangle is called a right triangle if one interior angle equals ninety degrees, equivalently π/2 radians."
        }
      },
      {
        "id": "mko_length_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-euclidean-length",
          "sha256": "485948172a6b39e7ec2cf3170ab03489fed98190d0c868161fbbd037f7151e91",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "獨立來源追加式勘誤（使用者已採用於本地修訂；未合併或發布）：在歐幾里得平面中，若 P=(x₁,y₁)、Q=(x₂,y₂)，則兩點距離取座標差平方和的非負主平方根；P=Q 時此值為 0。原來源的「正平方根」字樣保留於父版與修訂紀錄。",
          "en": "Independent append-only source erratum (owner-adopted in this local revision; not merged or published): in the Euclidean plane, for P=(x₁,y₁) and Q=(x₂,y₂), their distance is the nonnegative principal square root of the sum of the squared coordinate differences; it is 0 when P=Q. The original Chinese positive-root wording remains in the parent and revision history."
        }
      },
      {
        "id": "dot_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "內積與長度：u·v=u₁v₁+u₂v₂，|u|=√(u₁²+u₂²)。",
          "en": "Dot product and length: u·v=u₁v₁+u₂v₂ and |u|=√(u₁²+u₂²)."
        }
      },
      {
        "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": "angle_cos_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "座標模型中的內角量值：頂點 X、u=Y−X、v=Z−X 非零時，X 處內角是唯一滿足 cos θ=(u·v)/(|u||v|) 的 θ∈[0,π]。",
          "en": "Interior angle measure in the coordinate model: at a vertex X with nonzero u=Y−X and v=Z−X, the interior angle is the unique θ∈[0,π] with cos θ=(u·v)/(|u||v|)."
        }
      },
      {
        "id": "cos_facts",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "餘弦函數在 [0,π] 上嚴格遞減（因此在 [0,π] 上單射），且 cos(π/2)=0。",
          "en": "The cosine function is strictly decreasing on [0,π] (hence injective there), and cos(π/2)=0."
        }
      },
      {
        "id": "line_unique",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "在 ℝ² 中，通過相異兩點 X、Y 的直線恰有一條，就是 L(X,Y)。",
          "en": "In ℝ², exactly one line passes through two distinct points X and Y, namely L(X,Y)."
        }
      },
      {
        "id": "L_right_iff_dot",
        "kind": "lemma",
        "proved_in": [
          "S04.1",
          "S04.2"
        ],
        "statement": {
          "zh-Hant": "對非零向量 u、v 與 θ∈[0,π]，若 cos θ=(u·v)/(|u||v|)，則 θ=π/2 ⇔ u·v=0。",
          "en": "For nonzero vectors u, v and θ∈[0,π] with cos θ=(u·v)/(|u||v|): θ=π/2 ⇔ u·v=0."
        }
      },
      {
        "id": "L_sq_diff",
        "kind": "lemma",
        "proved_in": [
          "S06.1"
        ],
        "statement": {
          "zh-Hant": "對實數 p、q：(p−q)²=p²−2pq+q²。",
          "en": "For reals p, q: (p−q)²=p²−2pq+q²."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S07",
        "vars": [
          "u1",
          "u2",
          "v1",
          "v2"
        ],
        "degree": 2,
        "lhs": "(v1-u1)**2+(v2-u2)**2",
        "rhs": "(u1**2+u2**2)+(v1**2+v2**2)-2*(u1*v1+u2*v2)"
      },
      {
        "id": "N2",
        "step": "S06.1",
        "vars": [
          "p",
          "q"
        ],
        "degree": 2,
        "lhs": "(p-q)**2",
        "rhs": "p**2-2*p*q+q**2"
      },
      {
        "id": "N3",
        "step": "S03",
        "vars": [
          "a",
          "b",
          "c"
        ],
        "degree": 1,
        "lhs": "(a-c)-(b-c)",
        "rhs": "a-b"
      },
      {
        "id": "N4",
        "step": "S10",
        "lhs": "(-7-1)/3",
        "rhs": "-8/3"
      },
      {
        "id": "N5",
        "step": "S10",
        "lhs": "(8-2)/4",
        "rhs": "3/2"
      },
      {
        "id": "N6",
        "step": "S10",
        "lhs": "-8/3",
        "relation": "!=",
        "rhs": "3/2"
      },
      {
        "id": "N7",
        "step": "S11",
        "lhs": "3*(-8)+4*6",
        "rhs": "0"
      },
      {
        "id": "N8",
        "step": "S12",
        "lhs": "3**2+4**2",
        "rhs": "25"
      },
      {
        "id": "N9",
        "step": "S12",
        "lhs": "(-8)**2+6**2",
        "rhs": "100"
      },
      {
        "id": "N10",
        "step": "S12",
        "lhs": "(-7-4)**2+(8-6)**2",
        "rhs": "125"
      },
      {
        "id": "N11",
        "step": "S12",
        "lhs": "25+100",
        "rhs": "125"
      },
      {
        "id": "N12",
        "step": "S12",
        "lhs": "5**2*5",
        "rhs": "125"
      },
      {
        "id": "N13",
        "step": "S13",
        "lhs": "3*(-1)+(-2)*(-2)",
        "rhs": "1"
      },
      {
        "id": "N14",
        "step": "S13",
        "lhs": "3**2+(-2)**2",
        "rhs": "13"
      },
      {
        "id": "N15",
        "step": "S13",
        "lhs": "(-1)**2+(-2)**2",
        "rhs": "5"
      },
      {
        "id": "N16",
        "step": "S13",
        "lhs": "(4-0)**2+(0-0)**2",
        "rhs": "16"
      },
      {
        "id": "N17",
        "step": "S13",
        "lhs": "13+5",
        "relation": "!=",
        "rhs": "16"
      },
      {
        "id": "N18",
        "step": "S13",
        "lhs": "13+5-2*1",
        "rhs": "16"
      },
      {
        "id": "N19",
        "step": "S13",
        "lhs": "13+16",
        "relation": "!=",
        "rhs": "5"
      },
      {
        "id": "N20",
        "step": "S13",
        "lhs": "5+16",
        "relation": "!=",
        "rhs": "13"
      },
      {
        "id": "N21",
        "step": "S13",
        "lhs": "(4-0)*(1-0)+(0-0)*(2-0)",
        "rhs": "4"
      },
      {
        "id": "N22",
        "step": "S13",
        "lhs": "(0-4)*(1-4)+(0-0)*(2-0)",
        "rhs": "12"
      },
      {
        "id": "N23",
        "lhs": "25+125",
        "relation": "!=",
        "rhs": "100"
      },
      {
        "id": "N24",
        "lhs": "(5+10)**2",
        "relation": "!=",
        "rhs": "125"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 ℝ² 座標模型中證明定理：長度用歐幾里得長度的 MKO，直角用直角三角形的 MKO，內角量值用 G2 的餘弦式。證明是內積版本，相當於 MKO 所說的向量夾角版本；不證明 MKO 摘要中的面積拼補法，也不證明逆定理；不處理綜合幾何模型或其他頂點表示法與本模型的等價（MKO formal next）；不改動 canonical MKO 或英文 edition。",
      "en": "Proves the theorem in the coordinate model of ℝ²: lengths come from the Euclidean length MKO, right angles from the right-triangle MKO, and angle measures from the cosine formula in G2. The proof is the dot-product version, corresponding to the vector-angle version mentioned by the MKO; it does not prove the area-rearrangement argument summarized in the MKO, nor the converse; it does not treat the equivalence of this model with synthetic geometry or with other vertex representations (MKO formal next); and it does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "a,b,c",
        "兩股長與斜邊長",
        "the leg lengths and the hypotenuse length"
      ],
      [
        "u=B-C,\\ v=A-C",
        "從直角頂點 C 出發的兩個邊向量",
        "the two side vectors from the right-angle vertex C"
      ],
      [
        "u\\cdot v",
        "內積 u₁v₁+u₂v₂",
        "dot product u₁v₁+u₂v₂"
      ],
      [
        "d(P,Q)",
        "歐幾里得距離",
        "Euclidean distance"
      ],
      [
        "\\gamma",
        "C 處的內角量值",
        "the interior angle measure at C"
      ]
    ],
    "md_error_controls": [
      {
        "title": "只對直角三角形成立 / It holds only for right triangles",
        "zh": "MKO 指出的誤解是把 a²+b²=c² 套用到任意三角形。S07 顯示一般三角形滿足的是 c²=a²+b²−2u·v；只有直角（u·v=0）時才去掉最後一項。EFH 沒有直角，13+5=18≠16（S13），所以不能把定理用在它身上。",
        "en": "The misconception the MKO names is applying a²+b²=c² to any triangle. S07 shows that a general triangle satisfies c²=a²+b²−2u·v; only a right angle (u·v=0) removes the last term. EFH has no right angle and 13+5=18≠16 (S13), so the theorem may not be applied to it."
      },
      {
        "title": "模型的範圍 / The scope of the model",
        "zh": "MKO 的另一個誤解是把已形式化的指定向量邊長模型，誤認為所有仿射表示法都已自動完成等價證明。本稿同樣只在 ℝ² 座標模型中、以 C 出發的邊向量證明；其他頂點表示法、或綜合幾何的歐幾里得平面與本模型的等價，需要另外的定理（MKO formal next），本稿沒有證明。",
        "en": "Another misconception identified in the MKO is taking the formalized model with specified vector side lengths as if every affine representation were automatically proved equivalent. Likewise this guide proves the theorem only in the coordinate model of ℝ², with side vectors from C; the equivalence of this model with other vertex representations or with a synthetic Euclidean plane needs a separate theorem (MKO formal next) and is not proved here."
      },
      {
        "title": "斜邊必須是直角的對邊；例子不是證明 / The hypotenuse must be opposite the right angle; an example is not a proof",
        "zh": "公式中的 c 必須是直角所對的邊。在例子中若把股 b=10 誤當斜邊，會寫出 25+125=100，這不成立（N23）。另外，MKO 提醒有限檢查不能取代普遍證明：S12 的 25+100=125 只核對一個例子，定理的普遍性來自 S01 到 S09。",
        "en": "The c in the formula must be the side opposite the right angle. In the example, mistaking the leg b=10 for the hypotenuse gives 25+125=100, which is false (N23). Also, as the MKO warns, finite checks cannot replace a general proof: 25+100=125 in S12 checks one example, and the theorem's generality comes from S01 to S09."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "內角量值以 G2 的餘弦式定義；cos_facts 與 line_unique 是前提，未在本稿推導。",
        "en": "Angle measure is defined by the cosine formula of G2; cos_facts and line_unique are premises and are not derived here."
      },
      {
        "zh-Hant": "本稿不證明逆定理（a²+b²=c² 推出直角），也不證明 MKO 摘要中的面積拼補法。",
        "en": "This guide proves neither the converse (a²+b²=c² implies a right angle) nor the area-rearrangement argument summarized in the MKO."
      },
      {
        "zh-Hant": "本稿的模型是 ℝ² 與歐幾里得長度；與綜合歐幾里得幾何或其他仿射表示法的等價需要另外的定理（MKO formal next）。",
        "en": "The model here is ℝ² with Euclidean length; its equivalence with synthetic Euclidean geometry or other affine representations needs a separate theorem (MKO formal next)."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "8723c1e344103a9b4fb116bcfed73b10bf9b0fbc16c5935a181e826d20f1685b",
    "original_source_sha256": "54d2c3d929a38adb25ccbcded51b2cfc53d2a26489ddb50f38f6f0d5fd2b93e7",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M003"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}