{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-coordinate-plane-01",
  "mko_id": "mko-coordinate-plane",
  "title": {
    "zh-Hant": "座標平面：判定點在哪條軸或哪個象限，並比較有序對",
    "en": "Coordinate plane: deciding axis or quadrant membership and comparing ordered pairs"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-coordinate-plane-01",
    "source": {
      "mko_id": "mko-coordinate-plane",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "eb010001cb85a8fe8eee5a876ee1ec40caddc67fe621f671741caacf48ab3c78",
      "english_sha256": "5da6135768256b641dcda707cae4d1c9b7a425ea489f9b1e850eaa6bf67fa0b3"
    },
    "mode": "definition_walkthrough",
    "audience_prerequisites": [
      "cartesian_product_def",
      "order_basics",
      "set_extensionality",
      "set_operations_def",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在固定的平面直角座標系中，給定 A=(−3,2)、B=(2,−3)、B′=(6/3,−3)、C=(0,−4)、O=(0,0)、E=(−1/2,−5)。先證明兩軸只交於原點、每個點要嘛在軸上、要嘛恰在一個象限中；再逐點判定它所在的軸或象限，並判定 A 與 B、B 與 B′ 是否為同一點。",
      "en": "In a fixed Cartesian coordinate system, let A=(−3,2), B=(2,−3), B′=(6/3,−3), C=(0,−4), O=(0,0) and E=(−1/2,−5). First prove that the axes meet only at the origin and that every point either lies on an axis or lies in exactly one quadrant; then decide for each point its axis or quadrant, and decide whether A and B, and B and B′, are the same point."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "平面上已固定原點、互相垂直的有向 x 軸與 y 軸及各軸單位；在這個座標系中，每個點對應一個實數有序對 (x,y)∈ℝ²，所有點都在同一座標系中比較（MKO 假設 a1、a2）。",
          "en": "An origin, perpendicular oriented x- and y-axes and a unit on each axis are fixed in the plane; in this coordinate system each point corresponds to an ordered pair of reals (x,y)∈ℝ², and all points are compared in the same system (MKO assumptions a1, a2)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "x 軸 𝒳={(x,y)∈ℝ² ∣ y=0}，y 軸 𝒴={(x,y)∈ℝ² ∣ x=0}；Q_I={(x,y) ∣ x>0 且 y>0}，Q_II={(x,y) ∣ x<0 且 y>0}，Q_III={(x,y) ∣ x<0 且 y<0}，Q_IV={(x,y) ∣ x>0 且 y<0}。",
          "en": "The x-axis is 𝒳={(x,y)∈ℝ² ∣ y=0} and the y-axis is 𝒴={(x,y)∈ℝ² ∣ x=0}; Q_I={(x,y) ∣ x>0 and y>0}, Q_II={(x,y) ∣ x<0 and y>0}, Q_III={(x,y) ∣ x<0 and y<0}, Q_IV={(x,y) ∣ x>0 and y<0}."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "A=(−3,2)、B=(2,−3)、B′=(6/3,−3)、C=(0,−4)、O=(0,0)、E=(−1/2,−5)，皆在 G1 的座標系中給出。",
          "en": "A=(−3,2), B=(2,−3), B′=(6/3,−3), C=(0,−4), O=(0,0) and E=(−1/2,−5), all given in the coordinate system of G1."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "\\text{a point in the fixed coordinate system}",
        "after_tex": "\\text{point}\\leftrightarrow(x,y)\\in\\mathbb{R}^2,\\quad (x,y)=(x',y')\\iff x=x'\\land y=y'",
        "relation": "definition",
        "rule": "mko_coord_plane",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先固定座標系（G1），之後「點」就以實數有序對表示：第一坐標沿 x 軸讀取，第二坐標沿 y 軸讀取。依笛卡兒積的相等條件，兩點相同當且僅當兩個對應坐標分別相等；這是 MKO para2 所說的比較方式，而且只在同一座標系中有意義。",
          "en": "First fix the coordinate system (G1); from then on a point is represented by an ordered pair of reals, the first coordinate read along the x-axis and the second along the y-axis. By the equality condition for ordered pairs, two points are the same exactly when both corresponding coordinates are equal; this is the comparison described in MKO para2, and it is meaningful only within one coordinate system."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_coord_plane",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "此步只固定表示方式，不推出新事實；若換了座標系，同一個點的坐標可能不同，所以後面的判定都限於 G1 的座標系。",
            "en": "This step only fixes the representation and derives no new fact; in a different coordinate system the same point could have other coordinates, so every later decision is confined to the system of G1."
          }
        },
        "leaf_basis": "mko_coord_plane + cartesian_product_def"
      },
      {
        "id": "S02",
        "uses": [
          "G2",
          "S01"
        ],
        "before_tex": "\\text{axes and quadrants}",
        "after_tex": "\\mathcal{X}=\\{y=0\\},\\ \\mathcal{Y}=\\{x=0\\},\\ Q_{\\mathrm{I}}=\\{x>0,y>0\\},\\ Q_{\\mathrm{II}}=\\{x<0,y>0\\},\\ Q_{\\mathrm{III}}=\\{x<0,y<0\\},\\ Q_{\\mathrm{IV}}=\\{x>0,y<0\\}",
        "relation": "definition",
        "rule": "def_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 寫出兩軸與四個象限的集合定義。注意象限用的是嚴格不等號：Q_I 要求 x>0 且 y>0，而不是 x≥0 且 y≥0；這正是軸上的點被排除在象限之外的原因。",
          "en": "Following the MKO, write the two axes and four quadrants as sets. Note that the quadrants use strict inequalities: Q_I needs x>0 and y>0, not x≥0 and y≥0; this is exactly why points on an axis are excluded from the quadrants."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "def_intro",
            "mko_quadrant_signs",
            "set_builder"
          ],
          "explanation": {
            "zh-Hant": "此步只命名集合；嚴格不等號是定義的一部分，不能改成 ≥ 或 ≤。",
            "en": "This step only names the sets; the strict inequalities are part of the definition and may not be relaxed to ≥ or ≤."
          }
        },
        "leaf_basis": "def_intro + mko_quadrant_signs + set_builder"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "order_basics"
        ],
        "before_tex": "(x,y)\\in\\mathbb{R}^2",
        "after_tex": "(x=0\\lor y=0\\Rightarrow\\text{no quadrant})\\ \\land\\ (x\\neq 0\\land y\\neq 0\\Rightarrow\\text{exactly one quadrant})",
        "relation": "implies",
        "rule": "L_classify",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依引理 L_classify，平面上的每個點都被「兩軸的聯集」與四個象限恰好分類一次：軸上的點不在象限中，軸外的點恰在一個象限中。證明在 S03.1（軸上）、S03.2（軸外至少一個）、S03.3（至多一個）。",
          "en": "By Lemma L_classify, every point of the plane is classified exactly once by the union of the two axes and the four quadrants: points on an axis are in no quadrant, and points off the axes are in exactly one quadrant. The proof is S03.1 (on an axis), S03.2 (off the axes, at least one) and S03.3 (at most one)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_classify",
            "order_basics",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "只依 S02 的定義與實數三一律；不涉及任何具體點。",
            "en": "Relies only on the definitions of S02 and trichotomy of the reals; no particular point is involved."
          }
        },
        "leaf_basis": "L_classify + order_basics + case_analysis"
      },
      {
        "id": "S03.1",
        "uses": [
          "S02",
          "order_basics"
        ],
        "before_tex": "x=0\\ \\lor\\ y=0",
        "after_tex": "(x,y)\\notin Q_{\\mathrm{I}}\\cup Q_{\\mathrm{II}}\\cup Q_{\\mathrm{III}}\\cup Q_{\\mathrm{IV}}",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "每個象限都要求 x>0 或 x<0，也都要求 y>0 或 y<0。依三一律，x=0 時 x>0 與 x<0 都不成立；y=0 時 y>0 與 y<0 都不成立。所以只要 x=0 或 y=0，四個象限的條件都不滿足，該點不在任何象限。",
          "en": "Every quadrant requires x>0 or x<0, and also y>0 or y<0. By trichotomy, x=0 makes both x>0 and x<0 false, and y=0 makes both y>0 and y<0 false. So whenever x=0 or y=0, none of the four quadrant conditions holds and the point lies in no quadrant."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "order_basics",
            "set_operations_def"
          ],
          "explanation": {
            "zh-Hant": "依賴象限定義中的嚴格不等號；若誤用 ≥，原點會同時落在四個「象限」中。",
            "en": "It relies on the strict inequalities in the quadrant definitions; with ≥ instead, the origin would fall into all four 'quadrants'."
          }
        },
        "leaf_basis": "order_basics + set_operations_def"
      },
      {
        "id": "S03.2",
        "uses": [
          "S02",
          "order_basics"
        ],
        "before_tex": "x\\neq 0\\ \\land\\ y\\neq 0",
        "after_tex": "(x,y)\\in Q_{\\mathrm{I}}\\cup Q_{\\mathrm{II}}\\cup Q_{\\mathrm{III}}\\cup Q_{\\mathrm{IV}}",
        "relation": "case_split",
        "rule": "case_analysis",
        "conditions": [
          "x\\neq 0",
          "y\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依三一律，x≠0 時 x>0 或 x<0；y≠0 時 y>0 或 y<0。四種組合 (+,+)、(−,+)、(−,−)、(+,−) 涵蓋全部可能，分別恰是 Q_I、Q_II、Q_III、Q_IV 的定義條件，所以該點至少在一個象限中。",
          "en": "By trichotomy, x≠0 gives x>0 or x<0, and y≠0 gives y>0 or y<0. The four combinations (+,+), (−,+), (−,−) and (+,−) cover every possibility and are exactly the defining conditions of Q_I, Q_II, Q_III and Q_IV, so the point lies in at least one quadrant."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "order_basics",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "分情況必須涵蓋全部可能：x、y 各兩種符號共四種，缺一不可。",
            "en": "The cases must cover every possibility: two signs for x and two for y give four cases, and none may be left out."
          }
        },
        "leaf_basis": "case_analysis + order_basics"
      },
      {
        "id": "S03.3",
        "uses": [
          "S02",
          "order_basics"
        ],
        "before_tex": "(x,y)\\in Q_i\\cap Q_j,\\ i\\neq j",
        "after_tex": "\\bot",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "任兩個不同象限對 x 或對 y 要求相反的符號：Q_I 與 Q_II、Q_III 與 Q_IV 在 x 上相反；Q_I 與 Q_IV、Q_II 與 Q_III 在 y 上相反；Q_I 與 Q_III、Q_II 與 Q_IV 兩者都相反。若一點同在兩個不同象限，就要同時有例如 x>0 與 x<0，違反三一律。所以至多在一個象限；配合 S03.2 得恰在一個。S03.1 至 S03.3 合起來證得引理 L_classify。",
          "en": "Any two different quadrants demand opposite signs of x or of y: Q_I and Q_II, and Q_III and Q_IV, differ in x; Q_I and Q_IV, and Q_II and Q_III, differ in y; Q_I and Q_III, and Q_II and Q_IV, differ in both. A point in two different quadrants would need, for example, both x>0 and x<0, contradicting trichotomy. So a point lies in at most one quadrant, and with S03.2 in exactly one. S03.1 to S03.3 together prove Lemma L_classify."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "order_basics",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "六對象限逐一檢查；每對都有一個坐標的符號相反，矛盾來自三一律。",
            "en": "All six pairs of quadrants are checked; each pair has one coordinate with opposite signs, and the contradiction comes from trichotomy."
          }
        },
        "leaf_basis": "contradiction + order_basics"
      },
      {
        "id": "S04",
        "uses": [
          "S02"
        ],
        "before_tex": "\\mathcal{X}\\cap\\mathcal{Y}",
        "after_tex": "\\{(0,0)\\}",
        "relation": "implies",
        "rule": "set_extensionality",
        "conditions": [],
        "explanation": {
          "zh-Hant": "證明 MKO 的「兩軸交於 (0,0)」，而且只交於這一點。兩個方向的包含分別在 S04.1 與 S04.2。",
          "en": "Prove the MKO's statement that the axes meet at (0,0), and only there. The two inclusions are S04.1 and S04.2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "集合相等需要兩個方向，在子步中分別證明。",
            "en": "Set equality needs both directions, proved separately in the sub-steps."
          }
        },
        "leaf_basis": "set_extensionality + set_operations_def"
      },
      {
        "id": "S04.1",
        "uses": [
          "S02",
          "set_operations_def"
        ],
        "before_tex": "(x,y)\\in\\mathcal{X}\\cap\\mathcal{Y}",
        "after_tex": "(x,y)=(0,0)",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(x,y)∈𝒳∩𝒴 表示 (x,y)∈𝒳 且 (x,y)∈𝒴，也就是 y=0 且 x=0；依有序對相等條件，(x,y)=(0,0)。",
          "en": "(x,y)∈𝒳∩𝒴 means (x,y)∈𝒳 and (x,y)∈𝒴, that is y=0 and x=0; by the equality condition for ordered pairs, (x,y)=(0,0)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "set_operations_def",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "只用交集與有序對的定義；反向包含在 S04.2。",
            "en": "Uses only the definitions of intersection and of ordered pairs; the reverse inclusion is S04.2."
          }
        },
        "leaf_basis": "set_operations_def + cartesian_product_def"
      },
      {
        "id": "S04.2",
        "uses": [
          "S02",
          "S04.1"
        ],
        "before_tex": "(0,0)",
        "after_tex": "(0,0)\\in\\mathcal{X}\\cap\\mathcal{Y}",
        "relation": "implies",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(0,0) 的第二坐標是 0，所以在 𝒳；第一坐標是 0，所以在 𝒴；因此在交集中。S04.1 與 S04.2 合起來，依 set_extensionality 得 𝒳∩𝒴={(0,0)}。",
          "en": "The second coordinate of (0,0) is 0, so it lies in 𝒳; the first coordinate is 0, so it lies in 𝒴; hence it lies in the intersection. S04.1 and S04.2 together give 𝒳∩𝒴={(0,0)} by set_extensionality."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04.1",
            "set_extensionality",
            "set_builder"
          ],
          "explanation": {
            "zh-Hant": "兩個方向都已證，集合相等成立。",
            "en": "Both directions are now proved, so the sets are equal."
          }
        },
        "leaf_basis": "set_builder + set_operations_def + set_extensionality"
      },
      {
        "id": "S05",
        "uses": [
          "G3",
          "S02",
          "S03"
        ],
        "before_tex": "A=(-3,2)",
        "after_tex": "A\\in Q_{\\mathrm{II}},\\ A\\notin\\mathcal{X}\\cup\\mathcal{Y}",
        "relation": "implies",
        "rule": "mko_quadrant_signs",
        "conditions": [
          "-3<0",
          "2>0"
        ],
        "explanation": {
          "zh-Hant": "A 的第一坐標 −3<0、第二坐標 2>0，符號為 (−,+)，符合 Q_II 的定義。因 −3≠0，A 不在 y 軸；因 2≠0，A 不在 x 軸。依 L_classify，A 不在其他象限。",
          "en": "The first coordinate of A is −3<0 and the second is 2>0, sign pattern (−,+), which is the definition of Q_II. Since −3≠0, A is not on the y-axis; since 2≠0, A is not on the x-axis. By L_classify, A is in no other quadrant."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "S02",
            "S03",
            "mko_quadrant_signs"
          ],
          "explanation": {
            "zh-Hant": "只需兩個坐標的符號（N1、N2）；判定不依賴任何圖。",
            "en": "Only the signs of the two coordinates are needed (N1, N2); the decision does not depend on any drawing."
          }
        },
        "leaf_basis": "mko_quadrant_signs + order_basics + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "G3",
          "S02",
          "S03"
        ],
        "before_tex": "B=(2,-3)",
        "after_tex": "B\\in Q_{\\mathrm{IV}},\\ B\\notin\\mathcal{X}\\cup\\mathcal{Y}",
        "relation": "implies",
        "rule": "mko_quadrant_signs",
        "conditions": [
          "2>0",
          "-3<0"
        ],
        "explanation": {
          "zh-Hant": "B 的第一坐標 2>0、第二坐標 −3<0，符號為 (+,−)，所以 B∈Q_IV。兩坐標都非零，所以 B 不在任何一軸上。",
          "en": "The first coordinate of B is 2>0 and the second is −3<0, sign pattern (+,−), so B∈Q_IV. Both coordinates are nonzero, so B is on neither axis."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "S02",
            "S03",
            "mko_quadrant_signs"
          ],
          "explanation": {
            "zh-Hant": "與 S05 相同，只用坐標符號（N3、N4）。",
            "en": "As in S05, only the signs of the coordinates are used (N3, N4)."
          }
        },
        "leaf_basis": "mko_quadrant_signs + order_basics + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S01",
          "S05",
          "S06"
        ],
        "before_tex": "(-3,2)\\ \\text{and}\\ (2,-3)",
        "after_tex": "A\\neq B,\\quad\\text{while}\\quad\\{-3,2\\}=\\{2,-3\\}",
        "relation": "implies",
        "rule": "cartesian_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "有序對相等需要第一坐標相等，而 −3≠2，所以 A≠B。相反地，無序集合 {−3,2} 與 {2,−3} 有相同元素，依 set_extensionality 相等。交換坐標改變了點，甚至改變了象限（Q_II 變 Q_IV），所以點不能當成無序集合。",
          "en": "Equal ordered pairs need equal first coordinates, and −3≠2, so A≠B. By contrast the unordered sets {−3,2} and {2,−3} have the same elements and are equal by set_extensionality. Swapping the coordinates changed the point and even its quadrant (Q_II became Q_IV), so a point cannot be treated as an unordered set."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "cartesian_product_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "一個坐標不同就足以推出兩點不同（N5）；反向（兩點不同推出某個指定坐標不同）不宣稱。",
            "en": "One differing coordinate is enough for the points to differ (N5); the converse, that distinct points differ in a particular coordinate, is not claimed."
          }
        },
        "leaf_basis": "cartesian_product_def + set_extensionality + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "G3",
          "S01",
          "S06"
        ],
        "before_tex": "B'=\\left(\\tfrac{6}{3},-3\\right)",
        "after_tex": "B'=(2,-3)=B",
        "relation": "equivalent",
        "rule": "cartesian_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "6/3=2 且 −3=−3，兩個對應坐標都相等，依有序對相等條件 B′=B，所以 B′ 也在 Q_IV。點是否相同由坐標值決定，不由寫法決定。",
          "en": "6/3=2 and −3=−3, so both corresponding coordinates are equal and B′=B by the equality condition for ordered pairs; hence B′ is also in Q_IV. Whether points coincide is decided by the coordinate values, not by how they are written."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "cartesian_product_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "6/3 與 2 是同一個有理數（N6），兩種寫法可雙向代換。",
            "en": "6/3 and 2 are the same rational number (N6), so the two forms may replace each other in either direction."
          }
        },
        "leaf_basis": "cartesian_product_def + numeric_evaluation + eq_replace"
      },
      {
        "id": "S09",
        "uses": [
          "G3",
          "S02",
          "S03"
        ],
        "before_tex": "C=(0,-4)",
        "after_tex": "C\\in\\mathcal{Y},\\ C\\notin\\mathcal{X},\\ C\\ \\text{in no quadrant}",
        "relation": "implies",
        "rule": "L_classify",
        "conditions": [
          "0=0",
          "-4\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "C 的第一坐標是 0，所以 C∈𝒴；第二坐標 −4≠0，所以 C∉𝒳。依 L_classify（x=0 的情形），C 不在任何象限，雖然它的第二坐標是負的。",
          "en": "The first coordinate of C is 0, so C∈𝒴; the second coordinate is −4≠0, so C∉𝒳. By L_classify (the case x=0), C lies in no quadrant, even though its second coordinate is negative."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S03",
            "L_classify"
          ],
          "explanation": {
            "zh-Hant": "需要象限定義中的嚴格不等號；C 不可被歸入第三或第四象限（N7）。",
            "en": "It needs the strict inequalities in the quadrant definitions; C may not be placed in the third or fourth quadrant (N7)."
          }
        },
        "leaf_basis": "L_classify + set_builder + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "G3",
          "S03",
          "S04"
        ],
        "before_tex": "O=(0,0)",
        "after_tex": "O\\in\\mathcal{X}\\cap\\mathcal{Y},\\ O\\ \\text{in no quadrant}",
        "relation": "implies",
        "rule": "L_classify",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由 S04，O 同時在兩軸上；由 L_classify（x=0），O 不在任何象限。原點是兩軸的唯一公共點，不是第一象限的「角」。",
          "en": "By S04, O lies on both axes; by L_classify (x=0), O lies in no quadrant. The origin is the only common point of the axes, not a corner of the first quadrant."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S04",
            "L_classify"
          ],
          "explanation": {
            "zh-Hant": "直接引用 S04 與引理，不需另外計算。",
            "en": "Directly cites S04 and the lemma; no further computation is needed."
          }
        },
        "leaf_basis": "L_classify + set_operations_def"
      },
      {
        "id": "S11",
        "uses": [
          "G3",
          "S02",
          "S03"
        ],
        "before_tex": "E=\\left(-\\tfrac{1}{2},-5\\right)",
        "after_tex": "E\\in Q_{\\mathrm{III}},\\ E\\notin\\mathcal{X}\\cup\\mathcal{Y}",
        "relation": "implies",
        "rule": "mko_quadrant_signs",
        "conditions": [
          "-\\tfrac12<0",
          "-5<0"
        ],
        "explanation": {
          "zh-Hant": "E 的兩坐標 −1/2<0、−5<0，符號為 (−,−)，所以 E∈Q_III。坐標不必是整數；只看符號。兩坐標都非零，E 不在軸上。",
          "en": "The coordinates of E are −1/2<0 and −5<0, sign pattern (−,−), so E∈Q_III. The coordinates need not be integers; only their signs matter. Both are nonzero, so E is on no axis."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "S02",
            "S03",
            "mko_quadrant_signs"
          ],
          "explanation": {
            "zh-Hant": "只用坐標符號（N8、N9）。",
            "en": "Uses only the signs of the coordinates (N8, N9)."
          }
        },
        "leaf_basis": "mko_quadrant_signs + order_basics + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S05",
          "S06",
          "S07",
          "S08",
          "S09",
          "S10",
          "S11"
        ],
        "before_tex": "A,B,B',C,O,E",
        "after_tex": "A\\in Q_{\\mathrm{II}},\\ B=B'\\in Q_{\\mathrm{IV}},\\ E\\in Q_{\\mathrm{III}},\\ C\\in\\mathcal{Y}\\setminus\\mathcal{X},\\ O\\in\\mathcal{X}\\cap\\mathcal{Y},\\ A\\neq B",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把各點的判定合在一起：A 在第二象限，B 與 B′ 是同一點、在第四象限，E 在第三象限，C 在 y 軸上而不在 x 軸上，O 在兩軸上；C 與 O 都不在任何象限；A≠B。所有結論都只由坐標值與 S02 的定義推出。",
          "en": "Collect the decisions: A is in the second quadrant, B and B′ are the same point in the fourth quadrant, E is in the third quadrant, C is on the y-axis but not the x-axis, and O is on both axes; C and O are in no quadrant; A≠B. Every conclusion follows from the coordinate values and the definitions of S02 alone."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S05",
            "S06",
            "S07",
            "S08",
            "S09",
            "S10",
            "S11",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：各點的分類與兩個比較。足夠條件 H：S05 至 S11 的各別結論；H⇒G 只是合取。H 由坐標與定義正向證成，沒有把 G 當作前提。",
            "en": "Goal G: the classification of every point and the two comparisons. Sufficient condition H: the separate conclusions of S05 to S11; H⇒G is just conjunction. H is proved forwards from the coordinates and the definitions; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "兩軸只交於原點；每個點要嘛在軸上（不在任何象限），要嘛恰在一個象限。A=(−3,2)∈Q_II，B=(2,−3)=B′∈Q_IV，E=(−1/2,−5)∈Q_III，C=(0,−4) 在 y 軸上、不在任何象限，O 在兩軸上、不在任何象限；A≠B，雖然 {−3,2}={2,−3}。",
      "en": "The axes meet only at the origin, and every point either lies on an axis (and then in no quadrant) or lies in exactly one quadrant. A=(−3,2)∈Q_II, B=(2,−3)=B′∈Q_IV, E=(−1/2,−5)∈Q_III, C=(0,−4) is on the y-axis and in no quadrant, and O is on both axes and in no quadrant; A≠B although {−3,2}={2,−3}."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "原點 (0,0) 在第一象限，因為它的兩個坐標都 ≥0",
          "en": "The origin (0,0) is in the first quadrant because both its coordinates are ≥0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "Q_I 要求 x>0 且 y>0；0>0 不成立，原點不在任何象限。",
          "en": "Q_I requires x>0 and y>0; 0>0 is false, so the origin is in no quadrant."
        }
      },
      {
        "claim": {
          "zh-Hant": "(−3,2)=(2,−3)，因為兩者用的是同樣兩個數",
          "en": "(−3,2)=(2,−3) because both use the same two numbers"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "有序對相等需要第一坐標相等，−3≠2；兩點甚至在不同象限。",
          "en": "Equal ordered pairs need equal first coordinates, and −3≠2; the two points are even in different quadrants."
        }
      },
      {
        "claim": {
          "zh-Hant": "C=(0,−4) 的第二坐標為負，所以 C 在第三或第四象限",
          "en": "C=(0,−4) has a negative second coordinate, so C is in the third or fourth quadrant"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "第三、第四象限都要求 x≠0（x<0 或 x>0），而 C 的 x=0，C 在 y 軸上。",
          "en": "The third and fourth quadrants both require x≠0 (x<0 or x>0), while C has x=0, so C is on the y-axis."
        }
      },
      {
        "claim": {
          "zh-Hant": "B′=(6/3,−3) 與 B=(2,−3) 寫法不同，所以是不同的點",
          "en": "B′=(6/3,−3) and B=(2,−3) are written differently, so they are different points"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "6/3=2，兩個對應坐標都相等，所以 B′=B。",
          "en": "6/3=2, so both corresponding coordinates agree and B′=B."
        }
      }
    ],
    "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": "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": "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": "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": "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": "set_builder",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。",
          "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x)."
        }
      },
      {
        "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": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "eq_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": "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_coord_plane",
        "kind": "cited",
        "source": {
          "mko_id": "mko-coordinate-plane",
          "sha256": "eb010001cb85a8fe8eee5a876ee1ec40caddc67fe621f671741caacf48ab3c78",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在指定平面直角座標系中，點的座標是有序對 (x,y)。x 軸滿足 y=0，y 軸滿足 x=0；兩者交於 (0,0)。軸外四象限由 x、y 的符號決定。",
          "en": "In a specified Cartesian coordinate system, a point has the ordered coordinates (x,y). The x-axis satisfies y=0 and the y-axis satisfies x=0, and they meet at (0,0). Away from the axes, the four quadrants are determined by the signs of x and y."
        }
      },
      {
        "id": "mko_quadrant_signs",
        "kind": "cited",
        "source": {
          "mko_id": "mko-coordinate-plane",
          "sha256": "eb010001cb85a8fe8eee5a876ee1ec40caddc67fe621f671741caacf48ab3c78",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "x、y 都非零時，(+,+)、(−,+)、(−,−)、(+,−) 分別為第一、二、三、四象限；軸上的點不屬於任何象限。原點同時在兩軸。",
          "en": "When x and y are both nonzero, the sign patterns (+,+), (−,+), (−,−) and (+,−) give the first, second, third and fourth quadrants; points on an axis belong to no quadrant. The origin lies on both axes."
        }
      },
      {
        "id": "L_classify",
        "kind": "lemma",
        "proved_in": [
          "S03.1",
          "S03.2",
          "S03.3"
        ],
        "statement": {
          "zh-Hant": "對每個 (x,y)∈ℝ²：若 x=0 或 y=0，它不在任何象限；若 x≠0 且 y≠0，它恰在一個象限中。",
          "en": "For every (x,y)∈ℝ²: if x=0 or y=0, it lies in no quadrant; if x≠0 and y≠0, it lies in exactly one quadrant."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S05",
        "lhs": "-3",
        "relation": "<",
        "rhs": "0"
      },
      {
        "id": "N2",
        "step": "S05",
        "lhs": "2",
        "relation": ">",
        "rhs": "0"
      },
      {
        "id": "N3",
        "step": "S06",
        "lhs": "2",
        "relation": ">",
        "rhs": "0"
      },
      {
        "id": "N4",
        "step": "S06",
        "lhs": "-3",
        "relation": "<",
        "rhs": "0"
      },
      {
        "id": "N5",
        "step": "S07",
        "lhs": "-3",
        "relation": "!=",
        "rhs": "2"
      },
      {
        "id": "N6",
        "step": "S08",
        "lhs": "6/3",
        "rhs": "2"
      },
      {
        "id": "N7",
        "step": "S09",
        "lhs": "-4",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N8",
        "step": "S11",
        "lhs": "-1/2",
        "relation": "<",
        "rhs": "0"
      },
      {
        "id": "N9",
        "step": "S11",
        "lhs": "-5",
        "relation": "<",
        "rhs": "0"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只用 MKO 的定義：固定原點、互相垂直的有向軸與各軸單位，座標是實數有序對，x 軸由 y=0、y 軸由 x=0 定義，軸外的象限由兩坐標的正負號決定。不推論距離、角度或圖形尺寸（MKO 假設 a2），也不從示意圖讀坐標；不改動 canonical MKO 或英文 edition。",
      "en": "Uses only the MKO's definitions: a fixed origin, perpendicular oriented axes and a unit on each, coordinates as ordered pairs of reals, the x-axis given by y=0 and the y-axis by x=0, and the quadrants off the axes given by the signs of the two coordinates. It infers no distance, angle or figure size (MKO assumption a2) and reads no coordinates from a sketch; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "(x,y)",
        "點的有序座標，第一坐標沿 x 軸、第二坐標沿 y 軸讀取",
        "the ordered coordinates of a point, the first read along the x-axis and the second along the y-axis"
      ],
      [
        "\\mathcal{X},\\ \\mathcal{Y}",
        "x 軸 {y=0} 與 y 軸 {x=0}",
        "the x-axis {y=0} and the y-axis {x=0}"
      ],
      [
        "Q_{\\mathrm{I}},\\ Q_{\\mathrm{II}},\\ Q_{\\mathrm{III}},\\ Q_{\\mathrm{IV}}",
        "第一、二、三、四象限",
        "the first, second, third and fourth quadrants"
      ],
      [
        "O",
        "原點 (0,0)",
        "the origin (0,0)"
      ]
    ],
    "md_error_controls": [
      {
        "title": "軸上的點不在任何象限 / Points on an axis are in no quadrant",
        "zh": "MKO 指出的第一個誤解是把軸上的點或原點列入象限。象限定義用嚴格不等號（S02），所以 C=(0,−4) 雖然第二坐標為負，仍不在第三或第四象限；原點也不是第一象限的點（S03.1、S09、S10）。",
        "en": "The first error named in the MKO is to put points on an axis, or the origin, into a quadrant. The quadrants are defined with strict inequalities (S02), so C=(0,−4), although its second coordinate is negative, is in neither the third nor the fourth quadrant, and the origin is not a point of the first quadrant (S03.1, S09, S10)."
      },
      {
        "title": "有序對不可交換 / Ordered pairs do not commute",
        "zh": "MKO 指出的第二個誤解是把有序對當成無序集合。{−3,2}={2,−3}，但 (−3,2)≠(2,−3)：前者在第二象限，後者在第四象限（S07）。判定兩點是否相同要逐一比較對應坐標。",
        "en": "The second error named in the MKO is to treat an ordered pair as an unordered set. {−3,2}={2,−3}, but (−3,2)≠(2,−3): the first is in the second quadrant and the second in the fourth (S07). Whether two points coincide is decided by comparing corresponding coordinates one by one."
      },
      {
        "title": "坐標是資料，不從圖上讀 / Coordinates are data, not read from a sketch",
        "zh": "B′=(6/3,−3) 與 B 寫法不同卻是同一點（S08），而 E 的坐標 −1/2 在粗略的圖上可能看起來接近 y 軸。判定只依明示的坐標值與定義；本稿也不推論兩點距離或圖形大小（MKO 假設 a2）。",
        "en": "B′=(6/3,−3) is written differently from B but is the same point (S08), and E with first coordinate −1/2 might look close to the y-axis in a rough sketch. Decisions rest only on the stated coordinate values and the definitions; this guide also infers no distance between points and no figure size (MKO assumption a2)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "點與實數有序對的對應（G1）是座標系的前提，本稿不由幾何公理推導它。",
        "en": "The correspondence between points and ordered pairs of reals (G1) is a premise of the coordinate system; this guide does not derive it from geometric axioms."
      },
      {
        "zh-Hant": "本稿不涉及距離、角度或圖形大小；這些需要另外的語義橋（MKO formal next）。",
        "en": "This guide does not deal with distance, angle or figure size; those need a separate semantic bridge (MKO formal next)."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "fd4a7cac8ca8a31eb2c39927c5e170926c77882e66ef91e074b8612e9fe3626b",
    "original_source_sha256": "fd4a7cac8ca8a31eb2c39927c5e170926c77882e66ef91e074b8612e9fe3626b",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}