{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-congruence-01",
  "mko_id": "mko-congruence",
  "title": {
    "zh-Hant": "全等：以 SSS 切片判定有明示對應的三角形全等",
    "en": "Congruence: deciding congruence of labeled triangles by the SSS slice"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-congruence-01",
    "source": {
      "mko_id": "mko-congruence",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "493629f3decdf708c53ca64d9ba2a4332bee59a979caa6d4767e0577fac7fbec",
      "english_sha256": "e83086a05f66a3d37e6976814fd8902f9c3dcf2d8ea8312c74f38348425f3457"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "eq_congruence",
      "distributive",
      "zero_product",
      "order_basics",
      "sqrt_def",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在同一個等單位正交平面中，A1=(0,0)、A2=(4,0)、A3=(0,3)，B1=(1,1)、B2=(1,5)、B3=(4,1)，對應明示為 Ai↔Bi。先以行列式驗證兩組點都是非退化三角形，再依 12、13、23 的順序精確比較平方邊長，判定 G(A,B)；說明 B 的方向與 A 相反卻仍全等；最後找出另一個對應 A1↔B1、A2↔B3、A3↔B2 的首個不等見證，並說明退化點組不合格。",
      "en": "In one equal-unit orthonormal plane, let A1=(0,0), A2=(4,0), A3=(0,3) and B1=(1,1), B2=(1,5), B3=(4,1), with explicit correspondence Ai↔Bi. First verify by determinants that both triples are nondegenerate triangles, then compare squared side lengths exactly in the order 12, 13, 23 to decide G(A,B); explain why B has the opposite orientation and is still congruent; finally find the first unequal witness for the other correspondence A1↔B1, A2↔B3, A3↔B2, and show that a degenerate triple is not eligible."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "模型：同一個等單位正交歐氏平面 ℝ²，點以整數座標給出。",
          "en": "Model: one equal-unit orthonormal Euclidean plane ℝ², with points given by integer coordinates."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "A1=(0,0)、A2=(4,0)、A3=(0,3)；B1=(1,1)、B2=(1,5)、B3=(4,1)；明示對應 A1↔B1、A2↔B2、A3↔B3。",
          "en": "A1=(0,0), A2=(4,0), A3=(0,3); B1=(1,1), B2=(1,5), B3=(4,1); explicit correspondence A1↔B1, A2↔B2, A3↔B3."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "比較用資料：另一個對應 A1↔B1、A2↔B3、A3↔B2；退化候選點組 D=((0,0),(1,1),(2,2)) 與 D'=((5,5),(6,6),(7,7))。",
          "en": "Comparison data: the other correspondence A1↔B1, A2↔B3, A3↔B2; degenerate candidate triples D=((0,0),(1,1),(2,2)) and D'=((5,5),(6,6),(7,7))."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "G(A,B)\\ \\text{for labeled triangles}",
        "after_tex": "G(A,B)\\iff A,B\\ \\text{nondegenerate}\\ \\land\\ |A_iA_j|=|B_iB_j|\\ (ij=12,13,23)",
        "relation": "definition",
        "rule": "mko_sss_slice",
        "conditions": [
          "A_i\\leftrightarrow B_i\\ \\text{explicit}"
        ],
        "explanation": {
          "zh-Hant": "依 MKO 的 SSS 切片固定判準：在明示對應 Ai↔Bi 下，A 與 B 全等，當且僅當兩者都是非退化三角形，且 12、13、23 三組對應邊長都相等。判準有兩道關卡：先驗非退化，再比邊長；比較的是指定的對應邊，不是任意重新配對。",
          "en": "Fix the criterion from the MKO's SSS slice: under the explicit correspondence Ai↔Bi, A and B are congruent exactly when both are nondegenerate triangles and the three corresponding side lengths 12, 13, 23 agree. There are two gates: nondegeneracy first, then side lengths; and the sides compared are the specified corresponding ones, not an arbitrary re-pairing."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_sss_slice"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出本稿使用的判準，不推出新事實；判準只在 MKO 的切片（同一正交平面、明示對應）內使用。",
            "en": "This step only states the criterion used here and derives no new fact; it is used only inside the MKO's slice (one orthonormal plane, explicit correspondence)."
          }
        },
        "leaf_basis": "mko_sss_slice"
      },
      {
        "id": "S02",
        "uses": [
          "G2",
          "det2_def"
        ],
        "before_tex": "\\det(A_2-A_1,\\,A_3-A_1)",
        "after_tex": "\\det((4,0),(0,3))=4\\cdot 3-0\\cdot 0=12",
        "relation": "equivalent",
        "rule": "det2_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐坐標相減：A2−A1=(4−0, 0−0)=(4,0)，A3−A1=(0−0, 3−0)=(0,3)。依行列式公式 ad−bc：4·3−0·0=12。兩邊是同一個數，這是等值代換。",
          "en": "Subtract coordinatewise: A2−A1=(4−0, 0−0)=(4,0) and A3−A1=(0−0, 3−0)=(0,3). By the determinant formula ad−bc: 4·3−0·0=12. Both sides name the same number, so this is a replacement of equals."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "det2_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "反向重算（N1）：12=4·3−0·0 正是兩行（row）(4,0)、(0,3) 的行列式；兩式表示同一個數。",
            "en": "Recompute backwards (N1): 12=4·3−0·0 is exactly the determinant with rows (4,0) and (0,3); both expressions name the same number."
          }
        },
        "leaf_basis": "det2_def + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "det_collinear"
        ],
        "before_tex": "\\det(A_2-A_1,\\,A_3-A_1)=12\\neq 0",
        "after_tex": "A_1,A_2,A_3\\ \\text{distinct, noncollinear: }A\\ \\text{nondegenerate}",
        "relation": "implies",
        "rule": "det_collinear",
        "conditions": [
          "12\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依 det_collinear，三點共線當且僅當行列式為 0；這裡行列式為 12≠0，所以 A1、A2、A3 不共線，也就兩兩相異（有重點的三點必共線）。依 MKO 假設 a1，這是把 A 當作三角形使用之前必須先驗的條件。",
          "en": "By det_collinear, three points are collinear exactly when the determinant is 0; here it is 12≠0, so A1, A2, A3 are noncollinear and hence pairwise distinct (a triple with a repeated point is collinear). By MKO assumption a1, this must be established before A is treated as a triangle."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "det_collinear"
          ],
          "explanation": {
            "zh-Hant": "只用到「det≠0 ⇒ 不共線」這一方向；本步不宣稱其他三角形性質。",
            "en": "Only the direction det≠0 ⇒ noncollinear is used; no other property of triangles is claimed here."
          }
        },
        "leaf_basis": "det_collinear + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "G2",
          "det2_def",
          "det_collinear"
        ],
        "before_tex": "\\det(B_2-B_1,\\,B_3-B_1)",
        "after_tex": "\\det((0,4),(3,0))=0\\cdot 0-4\\cdot 3=-12\\neq 0:\\ B\\ \\text{nondegenerate}",
        "relation": "implies",
        "rule": "det_collinear",
        "conditions": [
          "-12\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "B2−B1=(1−1, 5−1)=(0,4)，B3−B1=(4−1, 1−1)=(3,0)，行列式 0·0−4·3=−12。它不是 0，所以 B1、B2、B3 相異且不共線，B 也是非退化三角形。行列式為負這件事留到 S11 解讀為方向；非退化只看它是否為 0。",
          "en": "B2−B1=(1−1, 5−1)=(0,4) and B3−B1=(4−1, 1−1)=(3,0), with determinant 0·0−4·3=−12. It is not 0, so B1, B2, B3 are distinct and noncollinear, and B is a nondegenerate triangle too. The negative sign is read as orientation in S11; nondegeneracy depends only on whether the determinant is 0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "det2_def",
            "det_collinear"
          ],
          "explanation": {
            "zh-Hant": "重算見 N2；判定只用 det≠0 ⇒ 不共線，負號不影響資格。",
            "en": "The recomputation is N2; the decision uses only det≠0 ⇒ noncollinear, and the minus sign does not affect eligibility."
          }
        },
        "leaf_basis": "det2_def + det_collinear + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "dist_formula"
        ],
        "before_tex": "|A_iA_j|=|B_iB_j|",
        "after_tex": "|A_iA_j|^2=|B_iB_j|^2",
        "relation": "equivalent",
        "rule": "L_sq_eq",
        "conditions": [
          "|A_iA_j|\\ge 0",
          "|B_iB_j|\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "距離都是非負實數（dist_formula），所以依引理 L_sq_eq（u=|AiAj|、v=|BiBj|），比較邊長等價於比較平方邊長。平方邊長是整數座標差的平方和，可精確計算，不需開根號或取近似值。引理在 S05.1 至 S05.3 證明。",
          "en": "Distances are nonnegative reals (dist_formula), so by Lemma L_sq_eq (u=|AiAj|, v=|BiBj|) comparing side lengths is equivalent to comparing squared side lengths. Squared lengths are sums of squares of integer coordinate differences and can be computed exactly, with no square roots or approximations. The lemma is proved in S05.1 to S05.3."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_sq_eq",
            "dist_formula"
          ],
          "explanation": {
            "zh-Hant": "引理兩個方向都在子步證明，前提 u、v≥0 由距離非負提供；若允許負數，平方相等不能推回原數相等（例如 2 與 −2）。",
            "en": "Both directions of the lemma are proved in the sub-steps, and its hypothesis u, v≥0 comes from distances being nonnegative; if negatives were allowed, equal squares would not give equal numbers (for example 2 and −2)."
          }
        },
        "leaf_basis": "eq_congruence + distributive + zero_product + order_basics"
      },
      {
        "id": "S05.1",
        "uses": [
          "eq_congruence"
        ],
        "before_tex": "u=v",
        "after_tex": "u^2=v^2",
        "relation": "implies",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對相等的兩數作同一個運算（平方），結果相等。這是引理的「⇒」方向，對任何實數都成立。",
          "en": "Applying the same operation (squaring) to equal numbers gives equal results. This is the '⇒' direction of the lemma and holds for all reals."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "eq_congruence"
          ],
          "explanation": {
            "zh-Hant": "此子步只宣稱 ⇒；反向需要非負條件，在 S05.3 處理。",
            "en": "This sub-step claims only ⇒; the converse needs nonnegativity and is handled in S05.3."
          }
        },
        "leaf_basis": "eq_congruence"
      },
      {
        "id": "S05.2",
        "uses": [
          "distributive"
        ],
        "before_tex": "u^2=v^2",
        "after_tex": "(u-v)(u+v)=0",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊同減 v² 得 u²−v²=0；再以分配律展開 (u−v)(u+v)=u²+uv−vu−v²=u²−v²（乘法交換使 uv−vu=0）。N9 以多項式恆等式核對這個展開。",
          "en": "Subtract v² from both sides to get u²−v²=0; distributivity expands (u−v)(u+v)=u²+uv−vu−v²=u²−v² (commutativity makes uv−vu=0). N9 checks this expansion as a polynomial identity."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "eq_congruence",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "減 v² 與加 v² 互為反運算，展開是恆等式，所以雙向成立。",
            "en": "Subtracting v² and adding v² undo each other, and the expansion is an identity, so the step holds both ways."
          }
        },
        "leaf_basis": "distributive + mul_comm + add_inverse + eq_congruence"
      },
      {
        "id": "S05.3",
        "uses": [
          "S05.2",
          "zero_product",
          "order_basics"
        ],
        "before_tex": "(u-v)(u+v)=0,\\ u\\ge 0,\\ v\\ge 0",
        "after_tex": "u=v",
        "relation": "case_split",
        "rule": "zero_product",
        "conditions": [
          "u\\ge 0",
          "v\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "依零積性質，u−v=0 或 u+v=0。情況一：u−v=0，即 u=v。情況二：u+v=0，則 u=−v≤0，又 u≥0，所以 u=0，進而 v=0，仍得 u=v。兩種情況涵蓋全部可能，所以 u²=v² 推出 u=v。S05.1 至 S05.3 合起來證得引理 L_sq_eq。",
          "en": "By the zero-product property, u−v=0 or u+v=0. Case 1: u−v=0, that is u=v. Case 2: u+v=0, so u=−v≤0, and with u≥0 we get u=0 and then v=0, so again u=v. The two cases cover every possibility, so u²=v² gives u=v. S05.1 to S05.3 together prove Lemma L_sq_eq."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05.2",
            "zero_product",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "情況二用到 u、v≥0；沒有這個條件，u=2、v=−2 會使 u²=v² 而 u≠v。",
            "en": "Case 2 uses u, v≥0; without it, u=2 and v=−2 give u²=v² with u≠v."
          }
        },
        "leaf_basis": "zero_product + case_analysis + order_basics"
      },
      {
        "id": "S06",
        "uses": [
          "G2",
          "dist_formula"
        ],
        "before_tex": "|A_1A_2|^2,\\ |A_1A_3|^2,\\ |A_2A_3|^2",
        "after_tex": "16,\\ 9,\\ 25",
        "relation": "equivalent",
        "rule": "dist_formula",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依距離平方公式，按 12、13、23 順序：|A1A2|²=(4−0)²+(0−0)²=16；|A1A3|²=(0−0)²+(3−0)²=9；|A2A3|²=(0−4)²+(3−0)²=16+9=25。每一項都是精確整數。",
          "en": "By the squared-distance formula, in the order 12, 13, 23: |A1A2|²=(4−0)²+(0−0)²=16; |A1A3|²=(0−0)²+(3−0)²=9; |A2A3|²=(0−4)²+(3−0)²=16+9=25. Each is an exact integer."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "dist_formula",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "每個值都可由公式重算（N3、N4、N5），左右兩邊是同一個數。",
            "en": "Each value can be recomputed from the formula (N3, N4, N5); both sides name the same number."
          }
        },
        "leaf_basis": "dist_formula + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "G2",
          "dist_formula"
        ],
        "before_tex": "|B_1B_2|^2,\\ |B_1B_3|^2,\\ |B_2B_3|^2",
        "after_tex": "16,\\ 9,\\ 25",
        "relation": "equivalent",
        "rule": "dist_formula",
        "conditions": [],
        "explanation": {
          "zh-Hant": "同一公式、同一順序：|B1B2|²=(1−1)²+(5−1)²=16；|B1B3|²=(4−1)²+(1−1)²=9；|B2B3|²=(4−1)²+(1−5)²=9+16=25。B 的邊指向不同方向（B1B2 鉛直，A1A2 水平），但平方長度只看座標差的平方和。",
          "en": "Same formula, same order: |B1B2|²=(1−1)²+(5−1)²=16; |B1B3|²=(4−1)²+(1−1)²=9; |B2B3|²=(4−1)²+(1−5)²=9+16=25. The sides of B point in other directions (B1B2 is vertical while A1A2 is horizontal), but a squared length depends only on the sum of squared coordinate differences."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "dist_formula",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "重算見 N6、N7、N8；兩邊是同一個數。",
            "en": "The recomputations are N6, N7, N8; both sides name the same number."
          }
        },
        "leaf_basis": "dist_formula + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S06",
          "S07"
        ],
        "before_tex": "(16,9,25)\\ \\text{vs}\\ (16,9,25)",
        "after_tex": "|A_iA_j|^2=|B_iB_j|^2\\ (ij=12,13,23)",
        "relation": "implies",
        "rule": "eq_refl_symm_trans",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 12、13、23 的順序逐一比較：16=16、9=9、25=25。三組都相等，所以沒有首個不等見證，答案是「無不等邊」。比較的是指定對應下的邊：A1A2 只和 B1B2 比，不和 B 的其他邊比。",
          "en": "Compare one by one in the order 12, 13, 23: 16=16, 9=9, 25=25. All three agree, so there is no first unequal witness and the answer is 'no unequal side'. The sides compared are those of the specified correspondence: A1A2 is compared only with B1B2, not with other sides of B."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S07",
            "eq_refl_symm_trans"
          ],
          "explanation": {
            "zh-Hant": "只需 S06 與 S07 的六個值。若有不等，依 12、13、23 的既定順序遇到的第一組不等才是首個見證；其後也不等的組不會因此成為首組。S13 的另一個對應正是這種情形。",
            "en": "Only the six values of S06 and S07 are needed. If any differ, the first unequal pair encountered in the declared order 12,13,23 is the first witness; later unequal pairs are not first. The alternative correspondence in S13 illustrates this distinction."
          }
        },
        "leaf_basis": "eq_refl_symm_trans + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S05",
          "S08"
        ],
        "before_tex": "|A_iA_j|^2=|B_iB_j|^2\\ (ij=12,13,23)",
        "after_tex": "|A_1A_2|=|B_1B_2|=4,\\ |A_1A_3|=|B_1B_3|=3,\\ |A_2A_3|=|B_2B_3|=5",
        "relation": "implies",
        "rule": "L_sq_eq",
        "conditions": [
          "\\text{lengths}\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "由 S05（引理 L_sq_eq 的 ⇐ 方向），平方相等的非負長度本身相等。具體數值：16、9、25 的非負平方根是 4、3、5（sqrt_def），所以三組對應邊長分別是 4、3、5。",
          "en": "By S05 (the ⇐ direction of Lemma L_sq_eq), nonnegative lengths with equal squares are equal. Concretely, the nonnegative square roots of 16, 9, 25 are 4, 3, 5 (sqrt_def), so the three corresponding side lengths are 4, 3 and 5."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S08",
            "sqrt_def"
          ],
          "explanation": {
            "zh-Hant": "需要長度非負，這由距離公式保證；具體長度值只是說明，判定本身只用平方相等。",
            "en": "It needs lengths to be nonnegative, which the distance formula guarantees; the explicit lengths are illustration, and the decision itself uses only equal squares."
          }
        },
        "leaf_basis": "L_sq_eq + sqrt_def"
      },
      {
        "id": "S10",
        "uses": [
          "S01",
          "S03",
          "S04",
          "S09"
        ],
        "before_tex": "A,B\\ \\text{nondegenerate},\\ |A_iA_j|=|B_iB_j|\\ (ij=12,13,23)",
        "after_tex": "G(A,B)\\ \\text{under}\\ A_i\\leftrightarrow B_i",
        "relation": "theorem_application",
        "rule": "mko_sss_slice",
        "conditions": [
          "A,B\\ \\text{nondegenerate}",
          "A_i\\leftrightarrow B_i\\ \\text{explicit}"
        ],
        "explanation": {
          "zh-Hant": "S01 判準的兩道關卡都已通過：非退化（S03、S04）與三組對應邊相等（S09）。所以在明示對應 Ai↔Bi 下 A 與 B 全等。判定沒有用到圖形外觀，也沒有猜測其他標籤排列。",
          "en": "Both gates of the criterion in S01 are passed: nondegeneracy (S03, S04) and three equal corresponding sides (S09). So A and B are congruent under the explicit correspondence Ai↔Bi. The decision used no visual appearance and guessed no other labeling."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S01",
            "S03",
            "S04",
            "S09",
            "mko_sss_slice"
          ],
          "explanation": {
            "zh-Hant": "目標 G：G(A,B)。足夠條件 H：兩者非退化且三組對應邊等長；H⇒G 由 SSS 切片。H 由 S03、S04、S09 從 G2 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: G(A,B). Sufficient condition H: both are nondegenerate and the three corresponding sides are equal; H⇒G by the SSS slice. H is proved forwards from G2 in S03, S04 and S09; G is never used as a premise."
          }
        },
        "leaf_basis": "mko_sss_slice + and_intro_elim"
      },
      {
        "id": "S11",
        "uses": [
          "S02",
          "S04",
          "S10"
        ],
        "before_tex": "\\det(A_2-A_1,A_3-A_1)=12>0,\\ \\det(B_2-B_1,B_3-B_1)=-12<0",
        "after_tex": "\\text{opposite orientations; }G(A,B)\\ \\text{still holds}",
        "relation": "implies",
        "rule": "mko_sss_slice",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個行列式異號：A1→A2→A3 是逆時針，B1→B2→B3 是順時針，B 的方向與 A 相反。MKO 指出鏡射改變方向但不破壞全等；S10 的判定只用到非退化與邊長，兩者都與方向無關，所以 G(A,B) 不受影響。",
          "en": "The two determinants have opposite signs: A1→A2→A3 runs counterclockwise and B1→B2→B3 clockwise, so B has the opposite orientation. The MKO says reflection reverses orientation without defeating congruence; the decision in S10 used only nondegeneracy and side lengths, neither of which depends on orientation, so G(A,B) is unaffected."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S04",
            "S10",
            "mko_sss_slice"
          ],
          "explanation": {
            "zh-Hant": "方向只由行列式的正負讀出；本步不宣稱「方向相同」是全等的必要條件，反而說明它不是。",
            "en": "Orientation is read only from the signs of the determinants; the step does not claim that equal orientation is necessary for congruence, and in fact shows that it is not."
          }
        },
        "leaf_basis": "mko_sss_slice + order_basics"
      },
      {
        "id": "S12",
        "uses": [
          "G2",
          "S10"
        ],
        "before_tex": "T(x,y)=(y+1,\\,x+1)",
        "after_tex": "T(A_1)=B_1,\\ T(A_2)=B_2,\\ T(A_3)=B_3",
        "relation": "implies",
        "rule": "L_T_iso",
        "conditions": [],
        "explanation": {
          "zh-Hant": "交叉核對：T 是先關於直線 y=x 鏡射、再平移 (1,1)。T(A1)=(0+1, 0+1)=(1,1)=B1；T(A2)=(0+1, 4+1)=(1,5)=B2；T(A3)=(3+1, 0+1)=(4,1)=B3。依引理 L_T_iso（S12.1），T 保持所有平方距離，所以 B 是 A 的鏡像搬移，與 S10、S11 一致。本步不是 S10 的前提。",
          "en": "Cross-check: T is reflection across the line y=x followed by translation by (1,1). T(A1)=(0+1, 0+1)=(1,1)=B1; T(A2)=(0+1, 4+1)=(1,5)=B2; T(A3)=(3+1, 0+1)=(4,1)=B3. By Lemma L_T_iso (S12.1), T preserves all squared distances, so B is a moved mirror image of A, consistent with S10 and S11. This step is not a premise of S10."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "L_T_iso"
          ],
          "explanation": {
            "zh-Hant": "三個像點可逐坐標重算；引理在 S12.1 證明。本步只作一致性核對，不宣稱 SSS 判定需要找到這樣的映射。",
            "en": "The three images can be recomputed coordinatewise; the lemma is proved in S12.1. The step is a consistency check only and does not claim that the SSS decision requires finding such a map."
          }
        },
        "leaf_basis": "L_T_iso + eq_replace + numeric_evaluation"
      },
      {
        "id": "S12.1",
        "uses": [
          "dist_formula",
          "add_comm"
        ],
        "before_tex": "|T(P)T(Q)|^2,\\ P=(x_1,y_1),\\ Q=(x_2,y_2)",
        "after_tex": "((y_2+1)-(y_1+1))^2+((x_2+1)-(x_1+1))^2=(x_2-x_1)^2+(y_2-y_1)^2=|PQ|^2",
        "relation": "equivalent",
        "rule": "dist_formula",
        "conditions": [],
        "explanation": {
          "zh-Hant": "T(P)=(y1+1, x1+1)、T(Q)=(y2+1, x2+1)。依距離平方公式，|T(P)T(Q)|²=((y2+1)−(y1+1))²+((x2+1)−(x1+1))²=(y2−y1)²+(x2−x1)²；依加法交換，這等於 (x2−x1)²+(y2−y1)²=|PQ|²。P、Q 是任意點，所以證得引理 L_T_iso（N10 以多項式恆等式核對）。",
          "en": "T(P)=(y1+1, x1+1) and T(Q)=(y2+1, x2+1). By the squared-distance formula, |T(P)T(Q)|²=((y2+1)−(y1+1))²+((x2+1)−(x1+1))²=(y2−y1)²+(x2−x1)²; by commutativity of addition this equals (x2−x1)²+(y2−y1)²=|PQ|². P and Q are arbitrary, so Lemma L_T_iso is proved (N10 checks it as a polynomial identity)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "dist_formula",
            "add_comm"
          ],
          "explanation": {
            "zh-Hant": "每個等號都是對所有實數成立的恆等式，所以整條鏈雙向成立；沒有對 P、Q 加任何額外條件。",
            "en": "Each equality is an identity for all reals, so the whole chain holds both ways; no extra condition is placed on P or Q."
          }
        },
        "leaf_basis": "dist_formula + add_comm + add_inverse + forall_intro"
      },
      {
        "id": "S13",
        "uses": [
          "G3",
          "S05",
          "S06",
          "S07"
        ],
        "before_tex": "A_1\\leftrightarrow B_1,\\ A_2\\leftrightarrow B_3,\\ A_3\\leftrightarrow B_2",
        "after_tex": "|A_1A_2|^2=16\\neq 9=|B_1B_3|^2:\\ \\text{first unequal witness}\\ 12",
        "relation": "implies",
        "rule": "L_sq_eq",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在另一個對應下，第 12 組對應邊是 A1A2 與 B1B3。由 S06、S07，平方長度是 16 與 9，不等；依 S05（長度相等必平方相等）的逆否，長度也不等。依 12、13、23 順序，首個不等見證是 12，不必再看 13、23。這只否定這個對應；在 Ai↔Bi 下的全等（S10）仍然成立。",
          "en": "Under the other correspondence, the pair of sides 12 is A1A2 with B1B3. By S06 and S07 their squared lengths are 16 and 9, which differ; by the contrapositive of S05 (equal lengths have equal squares), the lengths differ too. In the order 12, 13, 23 the first unequal witness is 12, and 13 and 23 need not be examined. This rejects only this correspondence; the congruence under Ai↔Bi (S10) still holds."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S06",
            "S07",
            "contrapositive"
          ],
          "explanation": {
            "zh-Hant": "用的是「長度相等 ⇒ 平方相等」的逆否；本步不宣稱 A、B 在任何對應下都不全等。",
            "en": "It uses the contrapositive of 'equal lengths ⇒ equal squares'; the step does not claim that A and B are congruent under no correspondence."
          }
        },
        "leaf_basis": "L_sq_eq + contrapositive + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "G3",
          "det2_def",
          "det_collinear"
        ],
        "before_tex": "D=((0,0),(1,1),(2,2))",
        "after_tex": "\\det((1,1),(2,2))=1\\cdot 2-1\\cdot 2=0:\\ D\\ \\text{collinear, not eligible}",
        "relation": "implies",
        "rule": "det_collinear",
        "conditions": [],
        "explanation": {
          "zh-Hant": "D 的兩個差向量是 (1,1) 與 (2,2)，行列式 0，所以 D 三點共線，不是非退化三角形；D' 是 D 平移 (5,5)，差向量相同，也共線。D 與 D' 的對應平方邊長都是 2、8、2（N13），三組都相等，但 SSS 切片的第一道關卡（非退化）不通過，所以不能說它們是全等三角形。這說明 S03、S04 必須在比較邊長之前完成。",
          "en": "The two difference vectors of D are (1,1) and (2,2), with determinant 0, so D is collinear and not a nondegenerate triangle; D' is D translated by (5,5), has the same difference vectors, and is collinear too. The corresponding squared sides of D and D' are both 2, 8, 2 (N13), all equal, yet the first gate of the SSS slice (nondegeneracy) fails, so they may not be called congruent triangles. This is why S03 and S04 must come before comparing sides."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "det2_def",
            "det_collinear"
          ],
          "explanation": {
            "zh-Hant": "用的是「det=0 ⇒ 共線」；邊長相等的事實在這裡不足以補救資格。",
            "en": "It uses det=0 ⇒ collinear; equal side lengths cannot repair eligibility here."
          }
        },
        "leaf_basis": "det2_def + det_collinear + numeric_evaluation"
      }
    ],
    "result": {
      "zh-Hant": "兩組點都是非退化三角形（行列式 12 與 −12）。在明示對應 Ai↔Bi 下，三組對應平方邊長依 12、13、23 順序都是 16、9、25，所以 A 與 B 全等，邊長為 4、3、5。B 的方向與 A 相反（B 是 A 關於 y=x 鏡射再平移 (1,1) 的像），但這不破壞全等。另一個對應 A1↔B1、A2↔B3、A3↔B2 的首個不等見證是第 12 組（16≠9），只否定該對應；共線點組 D、D' 雖然對應邊相等，但不合格。",
      "en": "Both triples are nondegenerate triangles (determinants 12 and −12). Under the explicit correspondence Ai↔Bi the corresponding squared sides in the order 12, 13, 23 are 16, 9, 25 for both, so A and B are congruent, with side lengths 4, 3, 5. B has the opposite orientation (it is the image of A under reflection across y=x followed by translation by (1,1)), which does not defeat congruence. For the other correspondence A1↔B1, A2↔B3, A3↔B2 the first unequal witness is pair 12 (16≠9), which rejects only that correspondence; the collinear triples D and D' have equal corresponding sides but are not eligible."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "det(A) 與 det(B) 異號，所以 A 與 B 不全等",
          "en": "det(A) and det(B) have opposite signs, so A and B are not congruent"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "異號只表示方向相反；SSS 切片的三組對應邊的平方長度 16、9、25 全部相等，A、B 全等（S10、S11）。",
          "en": "Opposite signs only mean opposite orientations; the three corresponding squared sides 16, 9, 25 agree, so A and B are congruent (S10, S11)."
        }
      },
      {
        "claim": {
          "zh-Hant": "對應 A1↔B1、A2↔B3、A3↔B2 在第 12 組不等，所以 A 與 B 不全等",
          "en": "The correspondence A1↔B1, A2↔B3, A3↔B2 fails at pair 12, so A and B are not congruent"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "這只否定該對應；在明示對應 Ai↔Bi 下 A、B 全等。",
          "en": "This rejects only that correspondence; under the explicit correspondence Ai↔Bi, A and B are congruent."
        }
      },
      {
        "claim": {
          "zh-Hant": "D 與 D' 的對應平方邊長都是 2、8、2，所以它們是全等三角形",
          "en": "D and D' both have corresponding squared sides 2, 8, 2, so they are congruent triangles"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "det((1,1),(2,2))=0，兩者共線，不是非退化三角形，不在本條目的適用範圍。",
          "en": "det((1,1),(2,2))=0, so both are collinear and not nondegenerate triangles; they are outside this entry."
        }
      },
      {
        "claim": {
          "zh-Hant": "兩組對應邊相等就足以判定全等",
          "en": "Two pairs of equal corresponding sides suffice for congruence"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "P=((0,0),(5,0),(3,4)) 與 Q=((0,0),(5,0),(4,3)) 都非退化，第 12、13 組平方邊長都是 25、25，但第 23 組是 20 與 10。",
          "en": "P=((0,0),(5,0),(3,4)) and Q=((0,0),(5,0),(4,3)) are both nondegenerate with squared sides 25, 25 in pairs 12 and 13, but pair 23 is 20 versus 10."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "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": "zero_product",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "零乘積性質：在 ℤ、ℚ、ℝ、ℂ（或任何整域）中，uv=0 ⇔ u=0 或 v=0。",
          "en": "Zero-product property: in ℤ, ℚ, ℝ, ℂ (or any integral domain), uv=0 ⇔ u=0 or v=0."
        }
      },
      {
        "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": "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": "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_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": "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": "case_analysis",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。",
          "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility."
        }
      },
      {
        "id": "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": "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": "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": "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": "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": "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": "mko_sss_slice",
        "kind": "cited",
        "source": {
          "mko_id": "mko-congruence",
          "sha256": "493629f3decdf708c53ca64d9ba2a4332bee59a979caa6d4767e0577fac7fbec",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在同一等單位正交歐氏平面，A、B 都為非退化三角形且對應明示為 Ai↔Bi 時，本輪以三組對應邊長全部相等的 SSS 切片判全等；可用平方長度精確比較，鏡射改變方向但不破壞全等。",
          "en": "In one equal-unit orthonormal Euclidean plane, for nondegenerate triangles A and B with explicit correspondence Ai↔Bi, this pulse uses the SSS congruence slice: all three corresponding side lengths agree. Compare their squares exactly; reflection reverses orientation without defeating congruence."
        }
      },
      {
        "id": "dist_formula",
        "kind": "cited",
        "source": {
          "mko_id": "mko-distance-formula",
          "sha256": "153d0d0feabadaf33e08081677be0f81709df28ffa5bffaf0176a0f7875e5904",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在同一等單位正交座標系，P=(x1,y1)、Q=(x2,y2) 的距離為 d(P,Q)=√((x2−x1)²+(y2−y1)²)≥0；距離平方是兩個座標差平方之和。",
          "en": "In one equal-unit orthonormal frame, P=(x1,y1) and Q=(x2,y2) have distance d(P,Q)=√((x2−x1)²+(y2−y1)²)≥0; its square is the sum of the squared coordinate differences."
        }
      },
      {
        "id": "det2_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-determinant",
          "sha256": "d49050d70b58e0d9d2cab2f5ddc64133813f5a99418865b5941af4f1ec240093",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "實二階方陣 M=[[a,b],[c,d]] 的行列式為 det(M)=ad−bc。本稿把向量 (a,b)、(c,d) 作為兩行（row），寫成 det((a,b),(c,d))。",
          "en": "A real 2×2 matrix M=[[a,b],[c,d]] has determinant det(M)=ad−bc. Here the vectors (a,b) and (c,d) are its two rows, written det((a,b),(c,d))."
        }
      },
      {
        "id": "det_collinear",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "平面上三點 P、Q、R 共線（有一條直線含三者）當且僅當 det(Q−P, R−P)=0。有重點的三點必共線，所以 det≠0 推出三點相異且不共線。",
          "en": "Three points P, Q, R of the plane are collinear (some line contains all three) exactly when det(Q−P, R−P)=0. A triple with a repeated point is collinear, so det≠0 gives three distinct noncollinear points."
        }
      },
      {
        "id": "L_sq_eq",
        "kind": "lemma",
        "proved_in": [
          "S05.1",
          "S05.2",
          "S05.3"
        ],
        "statement": {
          "zh-Hant": "對實數 u≥0、v≥0：u=v ⇔ u²=v²。",
          "en": "For reals u≥0 and v≥0: u=v ⇔ u²=v²."
        }
      },
      {
        "id": "L_T_iso",
        "kind": "lemma",
        "proved_in": [
          "S12.1"
        ],
        "statement": {
          "zh-Hant": "映射 T(x,y)=(y+1, x+1)（關於 y=x 鏡射後平移 (1,1)）保持平方距離：對所有點 P、Q，|T(P)T(Q)|²=|PQ|²。",
          "en": "The map T(x,y)=(y+1, x+1) (reflection across y=x followed by translation by (1,1)) preserves squared distance: |T(P)T(Q)|²=|PQ|² for all points P, Q."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "4*3-0*0",
        "rhs": "12"
      },
      {
        "id": "N2",
        "step": "S04",
        "lhs": "0*0-4*3",
        "rhs": "-12"
      },
      {
        "id": "N3",
        "step": "S06",
        "lhs": "(4-0)**2+(0-0)**2",
        "rhs": "16"
      },
      {
        "id": "N4",
        "step": "S06",
        "lhs": "(0-0)**2+(3-0)**2",
        "rhs": "9"
      },
      {
        "id": "N5",
        "step": "S06",
        "lhs": "(0-4)**2+(3-0)**2",
        "rhs": "25"
      },
      {
        "id": "N6",
        "step": "S07",
        "lhs": "(1-1)**2+(5-1)**2",
        "rhs": "16"
      },
      {
        "id": "N7",
        "step": "S07",
        "lhs": "(4-1)**2+(1-1)**2",
        "rhs": "9"
      },
      {
        "id": "N8",
        "step": "S07",
        "lhs": "(4-1)**2+(1-5)**2",
        "rhs": "25"
      },
      {
        "id": "N9",
        "step": "S05.2",
        "vars": [
          "u",
          "v"
        ],
        "degree": 2,
        "lhs": "(u-v)*(u+v)",
        "rhs": "u**2-v**2"
      },
      {
        "id": "N10",
        "step": "S12.1",
        "vars": [
          "x1",
          "y1",
          "x2",
          "y2"
        ],
        "degree": 2,
        "lhs": "((y2+1)-(y1+1))**2+((x2+1)-(x1+1))**2",
        "rhs": "(x2-x1)**2+(y2-y1)**2"
      },
      {
        "id": "N11",
        "step": "S13",
        "lhs": "(4-0)**2+(0-0)**2",
        "relation": "!=",
        "rhs": "(4-1)**2+(1-1)**2"
      },
      {
        "id": "N12",
        "step": "S14",
        "lhs": "(1-0)*(2-0)-(1-0)*(2-0)",
        "rhs": "0"
      },
      {
        "id": "N13",
        "step": "S14",
        "lhs": "(2-0)**2+(2-0)**2",
        "rhs": "(7-5)**2+(7-5)**2"
      },
      {
        "id": "N14",
        "step": "S11",
        "lhs": "(4*3-0*0)*(0*0-4*3)",
        "relation": "<",
        "rhs": "0"
      },
      {
        "id": "N15",
        "lhs": "(3-5)**2+(4-0)**2",
        "relation": "!=",
        "rhs": "(4-5)**2+(3-0)**2"
      },
      {
        "id": "N16",
        "step": "S12",
        "lhs": "4+1",
        "rhs": "5"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只在 MKO 的 SSS 切片內工作：同一等單位正交座標平面、整數座標、明示對應、以平方長度精確比較。不證明 SSS 判定本身，也不處理 SAS、ASA 等其他判定或無標籤的對應搜尋；鏡射映射只作交叉核對。不改動 canonical MKO 或英文 edition。",
      "en": "Works only inside the MKO's SSS slice: one equal-unit orthonormal coordinate plane, integer coordinates, an explicit correspondence, exact comparison of squared lengths. It does not prove the SSS criterion itself, does not treat other criteria such as SAS or ASA, and does not search over unlabeled correspondences; the reflection map serves only as a cross-check. It does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "A_i,\\ B_i",
        "兩個三角形的頂點，i=1,2,3",
        "the vertices of the two triangles, i=1,2,3"
      ],
      [
        "A_i\\leftrightarrow B_i",
        "明示的頂點對應",
        "the explicit vertex correspondence"
      ],
      [
        "|PQ|",
        "P、Q 間的歐氏距離",
        "the Euclidean distance between P and Q"
      ],
      [
        "\\det(u,v)",
        "以 u=(a,b)、v=(c,d) 為兩行（row）的二階行列式 ad−bc",
        "the 2×2 determinant ad−bc with rows u=(a,b) and v=(c,d)"
      ],
      [
        "G(A,B)",
        "在對應 Ai↔Bi 下 A 與 B 全等",
        "A and B are congruent under Ai↔Bi"
      ]
    ],
    "md_error_controls": [
      {
        "title": "方向相反不是不全等 / Opposite orientation is not non-congruence",
        "zh": "MKO 列出的誤解之一是把鏡射造成的方向反轉當成不全等。本例 det 為 12 與 −12，方向相反，但 SSS 切片只看非退化與三組對應邊長，與方向無關（S11）；S12 更找到把 A 送到 B 的鏡射加平移。方向只說明是否需要鏡射，不決定全等與否。",
        "en": "One misconception the MKO lists is treating the orientation reversal caused by a reflection as non-congruence. Here the determinants are 12 and −12, so the orientations are opposite, but the SSS slice looks only at nondegeneracy and the three corresponding side lengths, which do not depend on orientation (S11); S12 even exhibits a reflection plus translation sending A to B. Orientation only tells whether a reflection is needed; it does not decide congruence."
      },
      {
        "title": "先驗資格與對應，再比邊 / Eligibility and correspondence before comparison",
        "zh": "MKO 也警告忽略非退化與標記對應、或猜圖形全等。D、D' 的對應邊相等，但它們共線，不是三角形（S14）；所以非退化必須先驗。另一方面，對應是給定的，不能自行重排：在 A1↔B1、A2↔B3、A3↔B2 下第 12 組就不等（S13），而這只否定那個對應，不否定 Ai↔Bi 下的全等。",
        "en": "The MKO also warns against ignoring nondegeneracy and the labeled correspondence, or guessing congruence from a figure. D and D' have equal corresponding sides but are collinear, not triangles (S14), so nondegeneracy must be checked first. On the other hand, the correspondence is given and may not be rearranged at will: under A1↔B1, A2↔B3, A3↔B2 the pair 12 already differs (S13), and this rejects only that correspondence, not the congruence under Ai↔Bi."
      },
      {
        "title": "比較平方需要非負 / Comparing squares needs nonnegativity",
        "zh": "以平方長度代替長度的依據是引理 L_sq_eq，它要求兩數都非負。距離非負，所以可用；但同樣的推理不能用在可正可負的量上（2 與 −2 平方相等）。平方比較也避免了 √25 之類的近似誤差：所有比較都是整數相等。",
        "en": "Replacing lengths by squared lengths rests on Lemma L_sq_eq, which needs both numbers to be nonnegative. Distances are nonnegative, so it applies; the same reasoning may not be used for quantities that can be negative (2 and −2 have equal squares). Comparing squares also avoids approximation errors from taking roots such as √25: every comparison is an equality of integers."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "SSS 切片取自 MKO，本稿不從「全等即存在等距映射」的定義去證明它；S12 的映射只是與結論一致的交叉核對。",
        "en": "The SSS slice is taken from the MKO; this guide does not prove it from a definition of congruence as the existence of an isometry. The map of S12 is only a cross-check consistent with the conclusion."
      },
      {
        "zh-Hant": "det_collinear（三點共線 ⇔ 差向量行列式為 0）是本稿的前提，未在本稿中證明。",
        "en": "det_collinear (three points are collinear ⇔ the determinant of the difference vectors is 0) is a premise of this guide and is not proved here."
      },
      {
        "zh-Hant": "本稿只檢查題目給定的兩個對應，不搜尋其餘四個頂點排列。",
        "en": "This guide checks only the two correspondences given; it does not search the remaining four vertex permutations."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "83caa0f7f586d45c5a409ca376921926f4b747c37d11d7d8431e892c2d0e380c",
    "original_source_sha256": "727649daf1fc5c011c5f0341aae0d1188360a7ed505f2b639965409a85d8281c",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M020"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}