{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-absolute-value-01",
  "mko_id": "mko-absolute-value",
  "title": {
    "zh-Hant": "絕對值：由分段定義證明非負性、零點與 |a−b|=|b−a|",
    "en": "Absolute value: nonnegativity, the zero case and |a−b|=|b−a| from the piecewise definition"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-absolute-value-01",
    "source": {
      "mko_id": "mko-absolute-value",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "7df0aed9ca41a0e67b84a464f450422348753d1171802220b41a103488fb5f95",
      "english_sha256": "7f191b7dbbd24ac7ec2a252a08d3178f961ef31414f76098c51f5f01d63473f6"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "order_basics",
      "add_inverse",
      "add_assoc",
      "add_comm",
      "add_zero",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "只用分段定義（x≥0 時 |x|=x，x<0 時 |x|=−x）與實數的序和加法，證明對所有實數 x、a、b：(1) |x|≥0；(2) |x|=0 ⇔ x=0；(3) |−x|=|x|；(4) |a−b|=|b−a|。並計算 |−3|、|2−7|、|7−2|，說明「|x|=x 對所有 x 成立」是錯的。",
      "en": "Using only the piecewise definition (|x|=x when x≥0, |x|=−x when x<0) together with the order and addition of the reals, prove for all reals x, a, b: (1) |x|≥0; (2) |x|=0 ⇔ x=0; (3) |−x|=|x|; (4) |a−b|=|b−a|. Then compute |−3|, |2−7| and |7−2|, and show that '|x|=x for every x' is false."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "x、a、b 是實數；ℝ 具有標準線性序與加法反元素（MKO 假設 a1）。",
          "en": "x, a and b are real numbers; ℝ has its standard linear order and additive inverses (MKO assumption a1)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "定義：若 x≥0，則 |x|=x；若 x<0，則 |x|=−x。",
          "en": "Definition: if x≥0 then |x|=x; if x<0 then |x|=−x."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_abs_def"
        ],
        "before_tex": "|x|",
        "after_tex": "|x|=x\\ \\ (x\\geq 0),\\qquad |x|=-x\\ \\ (x<0)",
        "relation": "definition",
        "rule": "mko_abs_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO 寫出分段定義。依實數序的三一律，每個 x 恰好滿足 x<0、x=0、x>0 之一，而 x≥0 表示 x>0 或 x=0；所以每個 x 恰好落在一個分支，|x| 有唯一的值，A 確實是 ℝ→ℝ 的函數。判斷用哪個分支，看的是 x 這個數的正負，不是式子裡有沒有負號。",
          "en": "State the piecewise definition from the MKO. By trichotomy each real x satisfies exactly one of x<0, x=0, x>0, and x≥0 means x>0 or x=0; so each x falls in exactly one branch, |x| has a single value, and A really is a function ℝ→ℝ. Which branch applies depends on the sign of the number x, not on whether a minus sign appears in an expression."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "mko_abs_def",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出定義；分支互斥且涵蓋全部實數，靠的是三一律。",
            "en": "This step only states the definition; the branches are exclusive and cover every real because of trichotomy."
          }
        },
        "leaf_basis": "mko_abs_def + order_basics"
      },
      {
        "id": "S02",
        "uses": [
          "S01"
        ],
        "before_tex": "x\\geq 0",
        "after_tex": "|x|=x\\geq 0",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [
          "x\\geq 0"
        ],
        "explanation": {
          "zh-Hant": "情況一：x≥0。依第一分支 |x|=x，而 x≥0 是本情況的假設，所以 |x|≥0。",
          "en": "Case one: x≥0. The first branch gives |x|=x, and x≥0 is the assumption of this case, so |x|≥0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "只在 x≥0 的情況下使用第一分支；不宣稱 x<0 時 |x|=x。",
            "en": "The first branch is used only when x≥0; nothing is claimed about |x|=x when x<0."
          }
        },
        "leaf_basis": "def_unfold"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "L_sign"
        ],
        "before_tex": "x<0",
        "after_tex": "|x|=-x>0",
        "relation": "implies",
        "rule": "L_sign",
        "conditions": [
          "x<0"
        ],
        "explanation": {
          "zh-Hant": "情況二：x<0。依第二分支 |x|=−x；依引理 L_sign，x<0 推出 0<−x。所以此時 |x|>0。式子 −x 前面有負號，但它表示的數是正的。",
          "en": "Case two: x<0. The second branch gives |x|=−x, and by Lemma L_sign, x<0 implies 0<−x. So in this case |x|>0. The expression −x carries a minus sign, but the number it names is positive."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "L_sign"
          ],
          "explanation": {
            "zh-Hant": "需要本情況的假設 x<0 與引理；引理在 S03.1、S03.2 證明。",
            "en": "It needs the case assumption x<0 and the lemma, which is proved in S03.1 and S03.2."
          }
        },
        "leaf_basis": "def_unfold + order_basics + add_inverse"
      },
      {
        "id": "S03.1",
        "uses": [
          "order_basics"
        ],
        "before_tex": "x<0",
        "after_tex": "x+(-x)<0+(-x)",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 order_basics（u<v 推出 u+w<v+w），兩邊同加 w=−x。同理，對 0<x 兩邊同加 −x 得 0+(−x)<x+(−x)，這是引理的第二部分。",
          "en": "By order_basics (u<v gives u+w<v+w), add w=−x to both sides. Likewise, adding −x to both sides of 0<x gives 0+(−x)<x+(−x), the second part of the lemma."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "兩邊加同一個數保持嚴格不等號的方向，對任何實數 w 都成立，沒有附加條件。",
            "en": "Adding the same number to both sides keeps the direction of a strict inequality for every real w, with no side condition."
          }
        },
        "leaf_basis": "order_basics"
      },
      {
        "id": "S03.2",
        "uses": [
          "S03.1",
          "add_inverse",
          "add_zero",
          "add_comm"
        ],
        "before_tex": "x+(-x)<0+(-x)",
        "after_tex": "0<-x",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "化簡兩邊：x+(−x)=0（加法反元素），0+(−x)=(−x)+0=−x（交換律與單位元）。第二部分同理化簡為 −x<0。S03.1 與 S03.2 合起來證得引理 L_sign。",
          "en": "Simplify both sides: x+(−x)=0 (additive inverse) and 0+(−x)=(−x)+0=−x (commutativity and identity). The second part simplifies in the same way to −x<0. S03.1 and S03.2 together prove Lemma L_sign."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "add_inverse",
            "add_zero",
            "add_comm",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "只把式子換成相等的式子，不等式的真假不變，雙向成立。",
            "en": "Only equal expressions are exchanged, so the truth of the inequality is unchanged and the step is two-way."
          }
        },
        "leaf_basis": "eq_replace + add_inverse + add_zero + add_comm"
      },
      {
        "id": "S04",
        "uses": [
          "S02",
          "S03",
          "order_basics"
        ],
        "before_tex": "x\\geq 0\\ \\lor\\ x<0",
        "after_tex": "\\forall x\\in\\mathbb{R}:\\ |x|\\geq 0",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依三一律每個實數都滿足 x≥0 或 x<0；兩種情況下分別由 S02 得 |x|≥0、由 S03 得 |x|>0（因此也 ≥0）。x 是任意的，所以對所有實數 |x|≥0。這就是 MKO 中「|x|≥0」的證明。",
          "en": "By trichotomy every real satisfies x≥0 or x<0; S02 gives |x|≥0 in the first case and S03 gives |x|>0, hence ≥0, in the second. Since x was arbitrary, |x|≥0 for every real x. This proves the clause '|x|≥0' of the MKO."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S03",
            "case_analysis",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "分情況論證要求情況涵蓋全部可能；x≥0 與 x<0 由三一律涵蓋所有實數。",
            "en": "Proof by cases needs the cases to cover every possibility; x≥0 and x<0 cover all reals by trichotomy."
          }
        },
        "leaf_basis": "case_analysis + order_basics + forall_intro"
      },
      {
        "id": "S05",
        "uses": [
          "S02",
          "S03",
          "order_basics"
        ],
        "before_tex": "x\\neq 0",
        "after_tex": "|x|>0,\\ \\text{hence}\\ |x|\\neq 0;\\ \\text{so}\\ |x|=0\\Rightarrow x=0",
        "relation": "implies",
        "rule": "contrapositive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "設 x≠0，則 x>0 或 x<0。x>0 時 |x|=x>0（S02）；x<0 時 |x|=−x>0（S03）。兩種情況都有 |x|>0，所以 |x|≠0。這證明了 x≠0 ⇒ |x|≠0，其逆否命題就是 |x|=0 ⇒ x=0。",
          "en": "Suppose x≠0; then x>0 or x<0. If x>0 then |x|=x>0 (S02); if x<0 then |x|=−x>0 (S03). In both cases |x|>0, so |x|≠0. This proves x≠0 ⇒ |x|≠0, whose contrapositive is |x|=0 ⇒ x=0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "S03",
            "contrapositive",
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "逆否命題與原命題等價；此步只得到一個方向，另一方向在 S06。",
            "en": "A contrapositive is equivalent to the original implication; this step gives one direction only, and the other is in S06."
          }
        },
        "leaf_basis": "contrapositive + case_analysis + order_basics"
      },
      {
        "id": "S06",
        "uses": [
          "S01",
          "S05"
        ],
        "before_tex": "x=0",
        "after_tex": "|0|=0;\\ \\ \\text{hence}\\ |x|=0\\Leftrightarrow x=0",
        "relation": "implies",
        "rule": "iff_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "0≥0，所以 0 落在第一分支，|0|=0，即 x=0 ⇒ |x|=0。與 S05 合起來得 |x|=0 ⇔ x=0：絕對值只在 0 處為 0。",
          "en": "Since 0≥0, the number 0 falls in the first branch and |0|=0, so x=0 ⇒ |x|=0. Together with S05 this gives |x|=0 ⇔ x=0: the absolute value vanishes only at 0."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "iff_intro",
            "def_unfold"
          ],
          "explanation": {
            "zh-Hant": "雙條件需要兩個方向都已證：⇐ 在此步，⇒ 在 S05。",
            "en": "A biconditional needs both directions proved: ⇐ here and ⇒ in S05."
          }
        },
        "leaf_basis": "iff_intro + def_unfold"
      },
      {
        "id": "S07",
        "uses": [
          "S01",
          "L_sign",
          "L_inv_unique"
        ],
        "before_tex": "x>0",
        "after_tex": "|-x|=-(-x)=x=|x|",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [
          "x>0"
        ],
        "explanation": {
          "zh-Hant": "證 |−x|=|x| 的情況 x>0：依 L_sign，−x<0，所以對數 −x 用第二分支：|−x|=−(−x)。依引理 L_inv_unique，−(−x)=x；而 x>0 時 |x|=x。所以 |−x|=|x|。",
          "en": "Case x>0 of |−x|=|x|: by L_sign, −x<0, so the second branch applies to the number −x: |−x|=−(−x). By Lemma L_inv_unique, −(−x)=x, and for x>0, |x|=x. Hence |−x|=|x|."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_sign",
            "L_inv_unique",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "分支是依 −x 這個數的正負選的，不是依 x 的正負；這正是容易弄錯的地方。",
            "en": "The branch is chosen by the sign of the number −x, not of x; this is exactly where mistakes are easy."
          }
        },
        "leaf_basis": "def_unfold + L_sign + L_inv_unique"
      },
      {
        "id": "S07.1",
        "uses": [
          "add_zero",
          "add_inverse",
          "add_assoc",
          "add_comm"
        ],
        "before_tex": "u+v=0",
        "after_tex": "v=-u",
        "relation": "implies",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "v=v+0=v+(u+(−u))=(v+u)+(−u)=(u+v)+(−u)=0+(−u)=−u，依序用到單位元、反元素、結合律、交換律、假設 u+v=0 與單位元。取 u=−x、v=x：(−x)+x=0，所以 x=−(−x)。取 u=v=0：0+0=0，所以 0=−0。此子步證得引理 L_inv_unique。",
          "en": "v=v+0=v+(u+(−u))=(v+u)+(−u)=(u+v)+(−u)=0+(−u)=−u, using in turn the identity, inverses, associativity, commutativity, the hypothesis u+v=0, and the identity again. With u=−x and v=x: (−x)+x=0, so x=−(−x). With u=v=0: 0+0=0, so 0=−0. This sub-step proves Lemma L_inv_unique."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "add_zero",
            "add_inverse",
            "add_assoc",
            "add_comm",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "每一步都是對所有實數成立的恆等式或假設 u+v=0 的代換；結論只說反元素唯一。",
            "en": "Each link is an identity for all reals or a replacement using the hypothesis u+v=0; the conclusion only says that the additive inverse is unique."
          }
        },
        "leaf_basis": "eq_replace + add_zero + add_inverse + add_assoc + add_comm"
      },
      {
        "id": "S08",
        "uses": [
          "S01",
          "L_sign"
        ],
        "before_tex": "x<0",
        "after_tex": "|-x|=-x=|x|",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [
          "x<0"
        ],
        "explanation": {
          "zh-Hant": "情況 x<0：依 L_sign，−x>0，所以對 −x 用第一分支：|−x|=−x。而 x<0 時 |x|=−x（第二分支）。兩者相同，|−x|=|x|。",
          "en": "Case x<0: by L_sign, −x>0, so the first branch applies to −x: |−x|=−x. And for x<0, |x|=−x by the second branch. The two agree, so |−x|=|x|."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "L_sign"
          ],
          "explanation": {
            "zh-Hant": "兩次展開定義所用的分支不同（−x 用第一分支、x 用第二分支），各自由相應數值的正負判定。",
            "en": "The two unfoldings use different branches (the first for −x, the second for x), each chosen by the sign."
          }
        },
        "leaf_basis": "def_unfold + L_sign"
      },
      {
        "id": "S09",
        "uses": [
          "S07",
          "S08",
          "L_inv_unique"
        ],
        "before_tex": "x>0\\ \\lor\\ x<0\\ \\lor\\ x=0",
        "after_tex": "\\forall x\\in\\mathbb{R}:\\ |-x|=|x|",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "剩下的情況 x=0：依 L_inv_unique，−0=0，所以 |−0|=|0|。三種情況由三一律涵蓋全部實數，依分情況論證與全稱推廣，對所有實數 |−x|=|x|。",
          "en": "The remaining case x=0: by L_inv_unique, −0=0, so |−0|=|0|. The three cases cover every real by trichotomy, so by proof by cases and generalization, |−x|=|x| for every real x."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "S08",
            "case_analysis",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "三種情況必須都處理；只看 x>0 不能推出一般結論。",
            "en": "All three cases must be handled; the case x>0 alone would not give the general statement."
          }
        },
        "leaf_basis": "case_analysis + order_basics + forall_intro"
      },
      {
        "id": "S10",
        "uses": [
          "L_inv_unique",
          "add_assoc",
          "add_comm",
          "add_inverse",
          "add_zero"
        ],
        "before_tex": "(a-b)+(b-a)",
        "after_tex": "(a-b)+(b-a)=0,\\ \\text{hence}\\ b-a=-(a-b)",
        "relation": "implies",
        "rule": "L_inv_unique",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(a−b)+(b−a)=a+(−b)+b+(−a)，依結合律與交換律重排為 (a+(−a))+(b+(−b))=0+0=0。依引理 L_inv_unique（u=a−b、v=b−a），b−a=−(a−b)。",
          "en": "(a−b)+(b−a)=a+(−b)+b+(−a), which associativity and commutativity rearrange as (a+(−a))+(b+(−b))=0+0=0. By Lemma L_inv_unique with u=a−b and v=b−a, b−a=−(a−b)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_inv_unique",
            "add_assoc",
            "add_comm",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "重排只用加法的結合律與交換律；N5 以多項式恆等式核對 (a−b)+(b−a)=0。",
            "en": "The rearrangement uses only associativity and commutativity of addition; N5 checks (a−b)+(b−a)=0 as a polynomial identity."
          }
        },
        "leaf_basis": "L_inv_unique + add_assoc + add_comm + add_inverse + add_zero"
      },
      {
        "id": "S11",
        "uses": [
          "S09",
          "S10"
        ],
        "before_tex": "|b-a|",
        "after_tex": "|b-a|=|-(a-b)|=|a-b|",
        "relation": "implies",
        "rule": "forall_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 S10 把 b−a 換成相等的 −(a−b)，再把 S09 用在 x=a−b：|−(a−b)|=|a−b|。所以對所有實數 a、b，|a−b|=|b−a|：數線上兩點的距離不因交換端點而改變。",
          "en": "By S10 replace b−a by the equal number −(a−b), then apply S09 with x=a−b: |−(a−b)|=|a−b|. So |a−b|=|b−a| for all reals a, b: the distance between two points of the number line does not change when the endpoints are swapped."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "S10",
            "forall_elim",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "S09 對所有實數成立，所以可代入 x=a−b；N6 只是 a、b 在 −6 到 6 的有限核對，不是證明。",
            "en": "S09 holds for every real, so x=a−b may be substituted; N6 is only a finite check for a, b from −6 to 6, not a proof."
          }
        },
        "leaf_basis": "forall_elim + eq_replace"
      },
      {
        "id": "S12",
        "uses": [
          "S01",
          "L_inv_unique"
        ],
        "before_tex": "|-3|",
        "after_tex": "|-3|=-(-3)=3",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [
          "-3<0"
        ],
        "explanation": {
          "zh-Hant": "−3<0，所以用第二分支：|−3|=−(−3)=3（L_inv_unique）。結果 3≥0，符合 S04。",
          "en": "−3<0, so the second branch applies: |−3|=−(−3)=3 (L_inv_unique). The result 3 is ≥0, as S04 requires."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "L_inv_unique",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "先判定 −3 的正負才選分支；N1 重算 −(−3)=3。",
            "en": "The sign of −3 is decided before choosing the branch; N1 recomputes −(−3)=3."
          }
        },
        "leaf_basis": "def_unfold + L_inv_unique + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S01",
          "S11"
        ],
        "before_tex": "|2-7|,\\ |7-2|",
        "after_tex": "|2-7|=|-5|=5,\\ \\ |7-2|=|5|=5",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先算括號內的數：2−7=−5<0，用第二分支得 |−5|=−(−5)=5；7−2=5≥0，用第一分支得 |5|=5。兩者相等，符合 S11（a=2、b=7）：2 與 7 在數線上相距 5。",
          "en": "First compute the number inside: 2−7=−5<0, so the second branch gives |−5|=−(−5)=5; 7−2=5≥0, so the first branch gives |5|=5. They agree, as S11 says for a=2 and b=7: 2 and 7 are 5 apart on the number line."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S11",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "必須先求出 2−7 的值再判定分支；不能看到式子中的減號就直接刪掉。",
            "en": "The value of 2−7 must be found before choosing a branch; a minus sign in the expression may not simply be deleted."
          }
        },
        "leaf_basis": "def_unfold + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S12"
        ],
        "before_tex": "\\forall x\\in\\mathbb{R}:\\ |x|=x",
        "after_tex": "\\text{false, since}\\ |-3|=3\\neq -3",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "反例：命題「對所有實數 x，|x|=x」的否定是「存在 x 使 |x|≠x」。x=−3 就是這樣的 x：|−3|=3，而 3>0>−3，所以 3≠−3。這個反例已否定對所有 x 的等式，所以不能不分情況把 |x|=x 當作一般規則；本步不由單一反例推導全部成立情形。",
          "en": "Counterexample: the negation of 'for every real x, |x|=x' is 'some x has |x|≠x'. The number x=−3 is such an x: |−3|=3, and 3>0>−3, so 3≠−3. This counterexample refutes the equality for all x, so |x|=x cannot be used unconditionally as a general rule; this step does not derive every equality case from one example."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12",
            "quantifier_negation",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "推翻全稱命題只需要一個反例；這不宣稱 |x|=x 對其他 x 是否成立。",
            "en": "One counterexample suffices to refute a universal statement; nothing is claimed here about other values of x."
          }
        },
        "leaf_basis": "quantifier_negation + exists_intro + order_basics"
      },
      {
        "id": "S15",
        "uses": [
          "S04",
          "S06",
          "S09",
          "S11"
        ],
        "before_tex": "S04,\\ S06,\\ S09,\\ S11",
        "after_tex": "|x|\\geq 0,\\ \\ |x|=0\\Leftrightarrow x=0,\\ \\ |-x|=|x|,\\ \\ |a-b|=|b-a|",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "四個目標分別由 S04、S06、S09、S11 證成，全部只依賴分段定義、實數的序與加法。",
          "en": "The four goals are established by S04, S06, S09 and S11, all resting only on the piecewise definition and on the order and addition of the reals."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S04",
            "S06",
            "S09",
            "S11",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：(1) 到 (4) 同時成立。足夠條件 H：S04、S06、S09、S11 各自成立；H⇒G 由合取引入。H 由定義正向證成，沒有把 G 當作前提；S12 到 S14 是例子與反例。",
            "en": "Goal G: (1) to (4) together. Sufficient condition H: S04, S06, S09 and S11 each hold; H⇒G by conjunction introduction. H is proved forwards from the definition and G is never used as a premise; S12 to S14 are examples and a counterexample."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "由分段定義可證：對所有實數 |x|≥0；|x|=0 ⇔ x=0；|−x|=|x|；|a−b|=|b−a|。例子：|−3|=3，|2−7|=|7−2|=5；|x|=x 在 x=−3 時不成立，所以不是一般規則。",
      "en": "From the piecewise definition: for all reals, |x|≥0; |x|=0 ⇔ x=0; |−x|=|x|; |a−b|=|b−a|. Examples: |−3|=3 and |2−7|=|7−2|=5; |x|=x fails at x=−3, so it is not a general rule."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "對所有實數 x，|x|=x",
          "en": "For every real x, |x|=x"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x=−3 時 |x|=3≠−3。",
          "en": "At x=−3, |x|=3≠−3."
        }
      },
      {
        "claim": {
          "zh-Hant": "|2−7|=2+7=9",
          "en": "|2−7|=2+7=9"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "2−7=−5，|−5|=5。",
          "en": "2−7=−5 and |−5|=5."
        }
      },
      {
        "claim": {
          "zh-Hant": "對所有實數 x，|−x|=x",
          "en": "For every real x, |−x|=x"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "x=−2 時 |−x|=|2|=2≠−2；正確的是 |−x|=|x|。",
          "en": "At x=−2, |−x|=|2|=2≠−2; the correct statement is |−x|=|x|."
        }
      },
      {
        "claim": {
          "zh-Hant": "存在實數 x 使 |x|<0",
          "en": "Some real x has |x|<0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "S04 證明對所有實數 |x|≥0。",
          "en": "S04 proves |x|≥0 for every real x."
        }
      },
      {
        "claim": {
          "zh-Hant": "因為 a−b≠b−a（a≠b 時），所以 |a−b| 與 |b−a| 一般不相等",
          "en": "Since a−b≠b−a when a≠b, |a−b| and |b−a| are in general different"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "b−a=−(a−b)，而 |−x|=|x|（S11）；例如 |2−7|=|7−2|=5。",
          "en": "b−a=−(a−b) and |−x|=|x| (S11); for example |2−7|=|7−2|=5."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "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": "add_inverse",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。",
          "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)."
        }
      },
      {
        "id": "add_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_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "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_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "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": "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": "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": "iff_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "雙條件：P⇒Q 與 Q⇒P 都成立時，P⇔Q 成立；反之亦然。",
          "en": "Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold."
        }
      },
      {
        "id": "forall_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。",
          "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds."
        }
      },
      {
        "id": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "exists_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。",
          "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)."
        }
      },
      {
        "id": "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_abs_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-absolute-value",
          "sha256": "7df0aed9ca41a0e67b84a464f450422348753d1171802220b41a103488fb5f95",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "函數 A:ℝ→ℝ 定義為：若 x≥0，則 A(x)=x；若 x<0，則 A(x)=−x。慣用記號為 |x|，且 |x|≥0。",
          "en": "Define A:ℝ→ℝ by A(x)=x when x≥0 and A(x)=−x when x<0. Its customary notation is |x|, and |x|≥0."
        }
      },
      {
        "id": "L_sign",
        "kind": "lemma",
        "proved_in": [
          "S03.1",
          "S03.2"
        ],
        "statement": {
          "zh-Hant": "對實數 x：若 x<0 則 0<−x；若 0<x 則 −x<0。",
          "en": "For a real x: if x<0 then 0<−x, and if 0<x then −x<0."
        }
      },
      {
        "id": "L_inv_unique",
        "kind": "lemma",
        "proved_in": [
          "S07.1"
        ],
        "statement": {
          "zh-Hant": "對實數 u、v：若 u+v=0，則 v=−u。特別地 −(−x)=x，且 −0=0。",
          "en": "For reals u, v: if u+v=0 then v=−u. In particular −(−x)=x and −0=0."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S12",
        "lhs": "-(-3)",
        "rhs": "3"
      },
      {
        "id": "N2",
        "step": "S13",
        "lhs": "2-7",
        "rhs": "-5"
      },
      {
        "id": "N3",
        "step": "S13",
        "lhs": "-(-5)",
        "rhs": "5"
      },
      {
        "id": "N4",
        "step": "S13",
        "lhs": "7-2",
        "rhs": "5"
      },
      {
        "id": "N5",
        "step": "S10",
        "vars": [
          "a",
          "b"
        ],
        "degree": 1,
        "lhs": "(a-b)+(b-a)",
        "rhs": "0"
      },
      {
        "id": "N6",
        "step": "S11",
        "vars": [
          "a",
          "b"
        ],
        "range": [
          -6,
          6
        ],
        "lhs": "abs(a-b)",
        "rhs": "abs(b-a)"
      },
      {
        "id": "N7",
        "step": "S14",
        "lhs": "abs(-3)",
        "relation": "!=",
        "rhs": "-3"
      },
      {
        "id": "N8",
        "lhs": "2+7",
        "relation": "!=",
        "rhs": "abs(2-7)"
      },
      {
        "id": "N9",
        "lhs": "abs(-(-2))",
        "relation": "!=",
        "rhs": "-2"
      },
      {
        "id": "N10",
        "step": "S04",
        "vars": [
          "x"
        ],
        "range": [
          -10,
          10
        ],
        "lhs": "abs(x)",
        "relation": ">=",
        "rhs": "0"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內：實數及其標準線性序與加法反元素（假設 a1），以分段定義的函數 A(x)=|x|。所有性質都由分段定義推出，不使用現成的絕對值規則；不處理三角不等式、不等式解法、複數模或範數（MKO 指出這些是後續內容）；不改動 canonical MKO 或英文 edition。",
      "en": "Inside the MKO's scope: the real numbers with their standard linear order and additive inverses (assumption a1), and the piecewise function A(x)=|x|. Every property is derived from the piecewise definition, without using ready-made absolute-value rules. It does not treat the triangle inequality, solving inequalities, the complex modulus or norms (the MKO says these come later), and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "|x|",
        "x 的絕對值 A(x)",
        "the absolute value A(x) of x"
      ],
      [
        "-x",
        "x 的加法反元素",
        "the additive inverse of x"
      ],
      [
        "a,\\ b",
        "數線上的兩個實數",
        "two reals on the number line"
      ],
      [
        "|a-b|",
        "a 與 b 的距離",
        "the distance between a and b"
      ]
    ],
    "md_error_controls": [
      {
        "title": "|x| 不一律等於 x / |x| is not always x",
        "zh": "MKO 指出的第一個誤解是把 |x| 一律寫成 x。|x|=x 只是 x≥0 時的分支；x<0 時 |x|=−x。反例：x=−3 時 |x|=3，而 x=−3。使用絕對值前，先判定裡面那個數的正負。",
        "en": "The first mistake named by the MKO is always writing |x| as x. |x|=x is only the branch for x≥0; for x<0, |x|=−x. Counterexample: at x=−3, |x|=3 while x=−3. Before using an absolute value, decide the sign of the number inside."
      },
      {
        "title": "絕對值不是刪除負號的文字規則 / Absolute value is not a rule for deleting minus signs",
        "zh": "MKO 指出的第二個誤解是把絕對值當成刪除負號。把 |2−7| 的減號刪掉得 2+7=9，但 2−7=−5，|−5|=5。又如 x=−2 時 |−x|=|2|=2，而刪除負號會得到 x=−2。定義看的是數值的正負，不是符號的外觀。",
        "en": "The second mistake named by the MKO is treating the absolute value as deleting minus signs. Deleting the minus sign in |2−7| gives 2+7=9, but 2−7=−5 and |−5|=5. Likewise, for x=−2, |−x|=|2|=2, while deleting the minus sign would give x=−2. The definition looks at the sign of the value, not at how the symbols look."
      },
      {
        "title": "|x| 不會是負數，而 −x 不一定是負數 / |x| is never negative, and −x need not be negative",
        "zh": "MKO 指出的第三個誤解是認為 |x| 可以是負數。第二分支寫成 |x|=−x，看起來帶負號，但它只在 x<0 時使用，而依 L_sign，此時 −x>0；例如 x=−3 時 −x=3。所以 S04 證明 |x|≥0 對所有實數成立。",
        "en": "The third mistake named by the MKO is believing |x| can be negative. The second branch reads |x|=−x and looks negative, but it is used only when x<0, and by L_sign then −x>0; for example x=−3 gives −x=3. So S04 proves |x|≥0 for every real."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "N6 與 N10 是整數點上的有限核對（分別為 a、b 從 −6 到 6，x 從 −10 到 10），只支持這些點；一般結論來自 S04 與 S11 的證明。",
        "en": "N6 and N10 are finite checks at integer points (a, b from −6 to 6 and x from −10 to 10 respectively) and support only those points; the general statements come from the proofs in S04 and S11."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "ca5b2649a099d99d3fc2f80cd2245c776c98a9a136217dfab0d702504caa791f",
    "original_source_sha256": "34445117b994db885a5ef32034463f123a04ed40293eda3d29414fa6e7d493f6",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M007"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}