{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-orbit-fixed-periodic-01",
  "mko_id": "mko-orbit-fixed-periodic",
  "title": {
    "zh-Hant": "軌道、固定點與週期點：完整分類 T(x)=(2x+1) mod 6 的六個狀態",
    "en": "Orbits, fixed points and periodic points: classifying all six states of T(x)=(2x+1) mod 6"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-orbit-fixed-periodic-01",
    "source": {
      "mko_id": "mko-orbit-fixed-periodic",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "702889d6acab1a9c1c6c8d26447c0b2a0201751460d118874f55c0486060083f",
      "english_sha256": "1b1c09c21ac73cce7e94f37cf573cdf19a2f3da65188ed5919f72ca01cbc909a"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "division_algorithm",
      "divides_def",
      "nat_induction",
      "composition_def",
      "image_preimage_def",
      "case_analysis",
      "contradiction"
    ],
    "goal": {
      "zh-Hant": "在狀態空間 X={0,1,2,3,4,5} 上取自映射 T(x)=(2x+1) mod 6。求出全部固定點與最小週期 2 的週期點；證明對最小週期 p 的點，T^n(x)=x 當且僅當 p 整除 n，並用它說明 T^4(1)=1 不表示 1 的最小週期是 4；最後證明 0、2、4 不是週期點，雖然它們的軌道一步之後就進入循環。",
      "en": "On the state space X={0,1,2,3,4,5} take the self-map T(x)=(2x+1) mod 6. Find all fixed points and all periodic points of least period 2; prove that for a point of least period p, T^n(x)=x exactly when p divides n, and use this to explain why T^4(1)=1 does not make 4 the least period of 1; finally prove that 0, 2 and 4 are not periodic, although their orbits enter a cycle after one step."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "X={0,1,2,3,4,5}，以最小非負餘數代表模 6 的六個剩餘類（完整有限剩餘類狀態，MKO 假設 a1）。",
          "en": "X={0,1,2,3,4,5}, the least nonnegative representatives of the six residue classes modulo 6 (complete finite residue-class states, MKO assumption a1)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "T:X→X，T(x)=(2x+1) mod 6，其中 m mod 6 是整數 m 除以 6 的餘數。",
          "en": "T:X→X, T(x)=(2x+1) mod 6, where m mod 6 is the remainder of the integer m on division by 6."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "迭代：T^0 是 X 上的恆等映射，T^(n+1)=T∘T^n。x 的前向軌道是 x、T(x)、T²(x)，依此類推。",
          "en": "Iterates: T^0 is the identity map of X and T^(n+1)=T∘T^n. The forward orbit of x is x, T(x), T²(x), and so on."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "定義（MKO）：T(x)=x 時 x 是固定點；若存在正整數 n 使 T^n(x)=x，x 是週期點，其中最小的這種 n 稱為最小週期 p(x)。本稿稱 x 為前週期點，指 x 不是週期點，但某個 T^k(x)（k≥1）是週期點。",
          "en": "Definitions (MKO): x is a fixed point if T(x)=x; x is periodic if T^n(x)=x for some positive integer n, and the least such n is the least period p(x). Here x is called preperiodic if x is not periodic but some T^k(x) with k≥1 is periodic."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "G3",
          "G4"
        ],
        "before_tex": "\\text{self-map and its iterates}",
        "after_tex": "T:X\\to X,\\ T^0=\\mathrm{id}_X,\\ T^{n+1}=T\\circ T^n",
        "relation": "definition",
        "rule": "mko_orbit_defs",
        "conditions": [],
        "explanation": {
          "zh-Hant": "固定模型與定義：狀態空間 X、自映射 T、迭代 T^n、軌道、固定點、最小週期與前週期點。MKO 強調兩件事：T^p(x)=x 只說 p 是一個返回時間，最小週期還要排除更小的正返回時間；最終進入循環不使起點成為週期點。之後的每一步都對照這兩點。",
          "en": "Fix the model and the definitions: state space X, self-map T, iterates T^n, orbit, fixed point, least period and preperiodic point. The MKO stresses two points: T^p(x)=x only makes p a return time, and the least period requires excluding smaller positive return times; and eventually entering a cycle does not make the starting point periodic. Every later step is checked against these two points."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "G3",
            "G4",
            "mko_orbit_defs"
          ],
          "explanation": {
            "zh-Hant": "此步只陳述模型與定義，不推出新事實；所有結論只在這個有限模型中成立。",
            "en": "This step only states the model and definitions and derives no new fact; every conclusion holds in this finite model only."
          }
        },
        "leaf_basis": "mko_orbit_defs + def_intro"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "G2",
          "division_algorithm"
        ],
        "before_tex": "x\\in X",
        "after_tex": "T(x)=(2x+1)\\bmod 6\\in X",
        "relation": "implies",
        "rule": "division_algorithm",
        "conditions": [
          "6>0"
        ],
        "explanation": {
          "zh-Hant": "帶餘除法：整數 2x+1 與正整數 6 有唯一的 q、r 使 2x+1=6q+r 且 0≤r<6。T(x)=r 落在 {0,1,2,3,4,5}=X，所以 T 確實是 X 到 X 的自映射，迭代 T^n 都有定義。",
          "en": "Division algorithm: for the integer 2x+1 and the positive integer 6 there are unique q, r with 2x+1=6q+r and 0≤r<6. T(x)=r lies in {0,1,2,3,4,5}=X, so T really is a self-map of X and every iterate T^n is defined."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "division_algorithm"
          ],
          "explanation": {
            "zh-Hant": "只需要除數 6 為正整數；不宣稱 T 是雙射（S04 會顯示它不是）。",
            "en": "It needs only that the divisor 6 is a positive integer; it is not claimed that T is a bijection (S04 shows it is not)."
          }
        },
        "leaf_basis": "division_algorithm"
      },
      {
        "id": "S03",
        "uses": [
          "S02"
        ],
        "before_tex": "T(0),T(1),T(2),T(3),T(4),T(5)",
        "after_tex": "1,\\ 3,\\ 5,\\ 1,\\ 3,\\ 5",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐點計算 T 的完整值表，分兩個子步：先算 2x+1，再取模 6 的餘數。",
          "en": "Compute the complete value table of T point by point, in two sub-steps: first 2x+1, then the remainder modulo 6."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation",
            "division_algorithm"
          ],
          "explanation": {
            "zh-Hant": "每個值都是同一數的兩種寫法，可逐一重算（N1 至 N4）。",
            "en": "Each value is the same number written two ways and can be recomputed one by one (N1 to N4)."
          }
        },
        "leaf_basis": "numeric_evaluation + division_algorithm"
      },
      {
        "id": "S03.1",
        "uses": [
          "G2"
        ],
        "before_tex": "2x+1\\ \\text{for}\\ x=0,1,2,3,4,5",
        "after_tex": "1,\\ 3,\\ 5,\\ 7,\\ 9,\\ 11",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "直接計算：2·0+1=1，2·1+1=3，2·2+1=5，2·3+1=7，2·4+1=9，2·5+1=11。",
          "en": "Direct computation: 2·0+1=1, 2·1+1=3, 2·2+1=5, 2·3+1=7, 2·4+1=9, 2·5+1=11."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "整數算術的值唯一，兩邊同值。",
            "en": "Integer arithmetic has unique values, so both sides agree."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S03.2",
        "uses": [
          "S03.1",
          "division_algorithm"
        ],
        "before_tex": "1,3,5,7,9,11\\ \\bmod 6",
        "after_tex": "1,\\ 3,\\ 5,\\ 1,\\ 3,\\ 5",
        "relation": "equivalent",
        "rule": "division_algorithm",
        "conditions": [
          "6>0"
        ],
        "explanation": {
          "zh-Hant": "1、3、5 已在 0 到 5 之間，餘數是自己；7=6·1+1、9=6·1+3、11=6·1+5，餘數分別是 1、3、5。所以 T(0)=1、T(1)=3、T(2)=5、T(3)=1、T(4)=3、T(5)=5。",
          "en": "1, 3 and 5 already lie between 0 and 5 and are their own remainders; 7=6·1+1, 9=6·1+3 and 11=6·1+5 give remainders 1, 3 and 5. So T(0)=1, T(1)=3, T(2)=5, T(3)=1, T(4)=3 and T(5)=5."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "division_algorithm"
          ],
          "explanation": {
            "zh-Hant": "反向：6·1+1=7 等式還原原數；帶餘除法的唯一性保證餘數唯一。",
            "en": "Backward: 6·1+1=7 and so on restore the original numbers; uniqueness in the division algorithm makes each remainder unique."
          }
        },
        "leaf_basis": "division_algorithm + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "image_preimage_def"
        ],
        "before_tex": "T(X)=\\{T(x)\\mid x\\in X\\}",
        "after_tex": "T(X)=\\{1,3,5\\}",
        "relation": "equivalent",
        "rule": "image_preimage_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "像集合是所有 T(x) 的集合。值表 S03 中出現的值只有 1、3、5，而且每個都出現（T(0)=1、T(1)=3、T(2)=5）。所以 T(X)={1,3,5}；0、2、4 都不是任何狀態的像。",
          "en": "The image is the set of all values T(x). The only values in the table S03 are 1, 3 and 5, and each occurs (T(0)=1, T(1)=3, T(2)=5). So T(X)={1,3,5}; none of 0, 2, 4 is the image of any state."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S03",
            "image_preimage_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "兩個包含都成立：每個 T(x) 屬於 {1,3,5}，每個 1、3、5 都有原像；集合依外延相等，兩種描述指同一集合。",
            "en": "Both inclusions hold: every T(x) belongs to {1,3,5}, and each of 1, 3, 5 has a preimage; by extensionality the two descriptions name the same set."
          }
        },
        "leaf_basis": "image_preimage_def + set_extensionality"
      },
      {
        "id": "S05",
        "uses": [
          "S02",
          "mod6_arith"
        ],
        "before_tex": "T(x)=x,\\ x\\in X",
        "after_tex": "x=5",
        "relation": "equivalent",
        "rule": "mod6_arith",
        "conditions": [
          "x\\in X"
        ],
        "explanation": {
          "zh-Hant": "用同餘代數一次判定所有固定點，分三個子步。結論：X 中唯一的固定點是 5。",
          "en": "Decide all fixed points at once by congruence algebra, in three sub-steps. Conclusion: the only fixed point in X is 5."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mod6_arith",
            "divides_def"
          ],
          "explanation": {
            "zh-Hant": "三個子步都是雙向的等價，所以 T(x)=x 與 x=5 同真同假；反向直接核對 T(5)=5。",
            "en": "All three sub-steps are two-way equivalences, so T(x)=x and x=5 are true together; backwards, T(5)=5 is checked directly."
          }
        },
        "leaf_basis": "mod6_arith + divides_def + case_analysis"
      },
      {
        "id": "S05.1",
        "uses": [
          "G2",
          "mod6_arith"
        ],
        "before_tex": "T(x)=x",
        "after_tex": "2x+1\\equiv x\\pmod 6",
        "relation": "equivalent",
        "rule": "mod6_arith",
        "conditions": [
          "x\\in X"
        ],
        "explanation": {
          "zh-Hant": "T(x) 與 x 都在 X 中，X 中兩元素相等當且僅當模 6 同餘；而 T(x)≡2x+1 (mod 6)。所以 T(x)=x 等價於 2x+1≡x (mod 6)。",
          "en": "T(x) and x both lie in X, where two elements are equal exactly when they are congruent modulo 6; and T(x)≡2x+1 (mod 6). So T(x)=x is equivalent to 2x+1≡x (mod 6)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mod6_arith"
          ],
          "explanation": {
            "zh-Hant": "「X 中相等 ⇔ 同餘」本身是雙向的。",
            "en": "'Equal in X ⇔ congruent' is itself two-way."
          }
        },
        "leaf_basis": "mod6_arith"
      },
      {
        "id": "S05.2",
        "uses": [
          "S05.1",
          "mod6_arith",
          "divides_def"
        ],
        "before_tex": "2x+1\\equiv x\\pmod 6",
        "after_tex": "6\\mid x+1",
        "relation": "equivalent",
        "rule": "divides_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依定義，2x+1≡x (mod 6) 表示 6∣(2x+1−x)，而 2x+1−x=x+1。",
          "en": "By definition, 2x+1≡x (mod 6) means 6∣(2x+1−x), and 2x+1−x=x+1."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "divides_def",
            "mod6_arith"
          ],
          "explanation": {
            "zh-Hant": "這是同餘定義的展開與一個恆等式，雙向成立。",
            "en": "This unfolds the definition of congruence and uses one identity, so it holds both ways."
          }
        },
        "leaf_basis": "divides_def + mod6_arith + add_inverse"
      },
      {
        "id": "S05.3",
        "uses": [
          "S05.2",
          "G1"
        ],
        "before_tex": "6\\mid x+1,\\ x\\in X",
        "after_tex": "x=5",
        "relation": "equivalent",
        "rule": "case_analysis",
        "conditions": [
          "0\\le x\\le 5"
        ],
        "explanation": {
          "zh-Hant": "x∈X 時 x+1 取 1、2、3、4、5、6 之一；其中唯一的 6 的倍數是 6，所以 x+1=6，即 x=5。反向：x=5 時 x+1=6，6∣6。",
          "en": "For x∈X, x+1 is one of 1, 2, 3, 4, 5, 6; the only multiple of 6 among them is 6, so x+1=6, that is x=5. Backwards: for x=5, x+1=6 and 6∣6."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "case_analysis",
            "divides_def"
          ],
          "explanation": {
            "zh-Hant": "六種情況涵蓋 X 的全部元素，逐一判定，兩個方向都核對了。",
            "en": "The six cases cover every element of X and each is decided, so both directions are checked."
          }
        },
        "leaf_basis": "case_analysis + divides_def + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S02",
          "mod6_arith",
          "composition_def"
        ],
        "before_tex": "T^2(x)=T(T(x))",
        "after_tex": "T^2(x)=(4x+3)\\bmod 6",
        "relation": "equivalent",
        "rule": "mod6_arith",
        "conditions": [],
        "explanation": {
          "zh-Hant": "T²(x)=T(T(x))≡2T(x)+1≡2(2x+1)+1=4x+3 (mod 6)，第二個同餘用到 T(x)≡2x+1 與同餘對乘法、加法的相容。T²(x) 在 X 中，所以它就是 4x+3 除以 6 的餘數。",
          "en": "T²(x)=T(T(x))≡2T(x)+1≡2(2x+1)+1=4x+3 (mod 6); the second congruence uses T(x)≡2x+1 and the compatibility of congruence with multiplication and addition. Since T²(x) lies in X, it is the remainder of 4x+3 on division by 6."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mod6_arith",
            "composition_def"
          ],
          "explanation": {
            "zh-Hant": "兩邊是 X 的同一個元素（同一剩餘類在 X 中唯一的代表），等式雙向成立；N6 對全部六個狀態交叉核對。",
            "en": "Both sides name the same element of X (the unique representative of one residue class), so the equation holds both ways; N6 cross-checks all six states."
          }
        },
        "leaf_basis": "mod6_arith + composition_def + distributive"
      },
      {
        "id": "S07",
        "uses": [
          "S06",
          "mod6_arith"
        ],
        "before_tex": "T^2(x)=x,\\ x\\in X",
        "after_tex": "x\\in\\{1,3,5\\}",
        "relation": "equivalent",
        "rule": "mod6_arith",
        "conditions": [
          "x\\in X"
        ],
        "explanation": {
          "zh-Hant": "與 S05 同法判定所有滿足 T²(x)=x 的狀態，分三個子步。結論：恰為 1、3、5。",
          "en": "As in S05, decide every state with T²(x)=x, in three sub-steps. Conclusion: exactly 1, 3 and 5."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mod6_arith",
            "divides_def"
          ],
          "explanation": {
            "zh-Hant": "三個子步都是等價；反向可直接核對 T²(1)=1、T²(3)=3、T²(5)=5。",
            "en": "All three sub-steps are equivalences; backwards one checks directly T²(1)=1, T²(3)=3 and T²(5)=5."
          }
        },
        "leaf_basis": "mod6_arith + divides_def + case_analysis"
      },
      {
        "id": "S07.1",
        "uses": [
          "S06",
          "mod6_arith",
          "divides_def"
        ],
        "before_tex": "4x+3\\equiv x\\pmod 6",
        "after_tex": "6\\mid 3(x+1)",
        "relation": "equivalent",
        "rule": "divides_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由 S06 與 S05.1 同理，T²(x)=x 等價於 4x+3≡x (mod 6)，即 6∣(4x+3−x)；而 4x+3−x=3x+3=3(x+1)（分配律）。",
          "en": "As in S05.1, by S06 T²(x)=x is equivalent to 4x+3≡x (mod 6), that is 6∣(4x+3−x); and 4x+3−x=3x+3=3(x+1) by distributivity."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "divides_def",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "同餘定義的展開與恆等式 4x+3−x=3(x+1)（N7）都是雙向的。",
            "en": "Unfolding congruence and the identity 4x+3−x=3(x+1) (N7) are both two-way."
          }
        },
        "leaf_basis": "divides_def + distributive + mod6_arith"
      },
      {
        "id": "S07.2",
        "uses": [
          "S07.1",
          "divides_def"
        ],
        "before_tex": "6\\mid 3(x+1)",
        "after_tex": "2\\mid x+1",
        "relation": "equivalent",
        "rule": "divides_def",
        "conditions": [
          "3\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "6∣3(x+1) 表示有整數 k 使 3(x+1)=6k=3·2k；兩邊同除以 3（3≠0）得 x+1=2k，即 2∣x+1。反向：x+1=2k 兩邊乘 3 得 3(x+1)=6k。",
          "en": "6∣3(x+1) means 3(x+1)=6k=3·2k for some integer k; dividing both sides by 3 (3≠0) gives x+1=2k, that is 2∣x+1. Backwards: multiplying x+1=2k by 3 gives 3(x+1)=6k."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "divides_def",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "乘 3 與除以 3 互為反運算，前提 3≠0；同一個 k 在兩個方向都適用。",
            "en": "Multiplying by 3 and dividing by 3 undo each other given 3≠0; the same k works in both directions."
          }
        },
        "leaf_basis": "divides_def + mul_inverse_nonzero + eq_congruence"
      },
      {
        "id": "S07.3",
        "uses": [
          "S07.2",
          "G1"
        ],
        "before_tex": "2\\mid x+1,\\ x\\in X",
        "after_tex": "x\\in\\{1,3,5\\}",
        "relation": "equivalent",
        "rule": "case_analysis",
        "conditions": [
          "0\\le x\\le 5"
        ],
        "explanation": {
          "zh-Hant": "x+1 取 1 到 6；其中偶數為 2、4、6，對應 x=1、3、5。",
          "en": "x+1 runs through 1 to 6; the even values are 2, 4 and 6, corresponding to x=1, 3 and 5."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "case_analysis"
          ],
          "explanation": {
            "zh-Hant": "六種情況涵蓋整個 X，逐一判定，兩向皆核。",
            "en": "The six cases cover all of X and each is decided, so both directions are checked."
          }
        },
        "leaf_basis": "case_analysis + divides_def"
      },
      {
        "id": "S08",
        "uses": [
          "S03",
          "S05",
          "S07",
          "G4"
        ],
        "before_tex": "x\\in\\{1,3,5\\}",
        "after_tex": "p(5)=1,\\ p(1)=2,\\ p(3)=2",
        "relation": "implies",
        "rule": "case_analysis",
        "conditions": [],
        "explanation": {
          "zh-Hant": "最小週期要排除所有更小的正返回時間，這裡只需排除 1。對 5：T(5)=5，1 是正返回時間且是最小的正整數，所以 p(5)=1；雖然 T²(5)=5，5 的最小週期不是 2。對 1：T(1)=3≠1，所以 1 不是返回時間；T²(1)=1（S07），所以 p(1)=2。對 3：T(3)=1≠3，T²(3)=3，所以 p(3)=2。最小週期恰為 2 的點是 1 與 3，形成循環 1→3→1。",
          "en": "The least period requires excluding every smaller positive return time; here only 1 must be excluded. For 5: T(5)=5, so 1 is a positive return time and the least positive integer, hence p(5)=1; although T²(5)=5, the least period of 5 is not 2. For 1: T(1)=3≠1, so 1 is not a return time; T²(1)=1 (S07), so p(1)=2. For 3: T(3)=1≠3 and T²(3)=3, so p(3)=2. The points of least period exactly 2 are 1 and 3, forming the cycle 1→3→1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S05",
            "S07",
            "G4"
          ],
          "explanation": {
            "zh-Hant": "需要兩件事：T^p(x)=x（S05、S07），以及對每個 1≤n<p 有 T^n(x)≠x（由值表 S03）。只有前者不足以確定最小週期。",
            "en": "Two things are needed: T^p(x)=x (S05, S07), and T^n(x)≠x for every 1≤n<p (from the table S03). The first alone does not determine the least period."
          }
        },
        "leaf_basis": "case_analysis + def_unfold + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "G3",
          "G4",
          "iterate_law"
        ],
        "before_tex": "p=\\min\\{n\\geq 1\\mid T^n(x)=x\\}",
        "after_tex": "\\forall n\\geq 1:\\ T^n(x)=x\\iff p\\mid n",
        "relation": "implies",
        "rule": "L_return",
        "conditions": [
          "x\\ \\text{periodic}"
        ],
        "explanation": {
          "zh-Hant": "證明引理 L_return：最小週期 p 的點，其正返回時間恰為 p 的倍數。這把 MKO 的提醒精確化：任何返回時間都是 p 的倍數，但不一定等於 p。證明分四個子步：兩步歸納、一次帶餘除法、一次由最小性導出矛盾。",
          "en": "Prove Lemma L_return: for a point of least period p, the positive return times are exactly the multiples of p. This makes the MKO's warning precise: every return time is a multiple of p but need not equal p. The proof has four sub-steps: two for an induction, one division with remainder and one contradiction from minimality."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_return",
            "iterate_law",
            "G4"
          ],
          "explanation": {
            "zh-Hant": "引理只需要 p 是最小正返回時間與迭代律；它不依賴 X 有限或 T 的具體公式。",
            "en": "The lemma needs only that p is the least positive return time and the iterate law; it does not depend on X being finite or on the formula for T."
          }
        },
        "leaf_basis": "nat_induction + division_algorithm + contradiction + iterate_law"
      },
      {
        "id": "S09.1",
        "uses": [
          "G3"
        ],
        "before_tex": "q=0",
        "after_tex": "T^{0\\cdot p}(x)=T^0(x)=x",
        "relation": "induction_base",
        "rule": "nat_induction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "要證對所有自然數 q，T^(qp)(x)=x。基底 q=0：0·p=0，而 T^0 是恆等映射，所以 T^0(x)=x。",
          "en": "To prove T^(qp)(x)=x for every natural number q. Base q=0: 0·p=0 and T^0 is the identity, so T^0(x)=x."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "nat_induction"
          ],
          "explanation": {
            "zh-Hant": "只用 T^0 的定義。",
            "en": "Only the definition of T^0 is used."
          }
        },
        "leaf_basis": "nat_induction + def_unfold"
      },
      {
        "id": "S09.2",
        "uses": [
          "S09.1",
          "iterate_law",
          "G4"
        ],
        "before_tex": "T^{qp}(x)=x",
        "after_tex": "T^{(q+1)p}(x)=x",
        "relation": "induction_step",
        "rule": "iterate_law",
        "conditions": [
          "T^p(x)=x"
        ],
        "explanation": {
          "zh-Hant": "設 T^(qp)(x)=x。由迭代律 T^((q+1)p)=T^(p+qp)=T^p∘T^(qp)，所以 T^((q+1)p)(x)=T^p(T^(qp)(x))=T^p(x)=x，最後一步用到 p 是返回時間。依歸納原理，對所有 q 有 T^(qp)(x)=x。",
          "en": "Assume T^(qp)(x)=x. By the iterate law T^((q+1)p)=T^(p+qp)=T^p∘T^(qp), so T^((q+1)p)(x)=T^p(T^(qp)(x))=T^p(x)=x, the last equality because p is a return time. By induction, T^(qp)(x)=x for every q."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "iterate_law",
            "nat_induction",
            "G4"
          ],
          "explanation": {
            "zh-Hant": "需要歸納假設、迭代律與 T^p(x)=x；不宣稱 p 的倍數以外沒有返回時間（那是 S09.4 的工作）。",
            "en": "It needs the induction hypothesis, the iterate law and T^p(x)=x; it does not claim that there are no return times other than multiples of p (that is S09.4's job)."
          }
        },
        "leaf_basis": "iterate_law + nat_induction + composition_def"
      },
      {
        "id": "S09.3",
        "uses": [
          "S09.2",
          "division_algorithm",
          "iterate_law"
        ],
        "before_tex": "n=qp+r,\\ 0\\le r<p",
        "after_tex": "T^n(x)=T^r(x)",
        "relation": "implies",
        "rule": "division_algorithm",
        "conditions": [
          "p\\geq 1"
        ],
        "explanation": {
          "zh-Hant": "對正整數 n 用帶餘除法：n=qp+r，0≤r<p。由迭代律 T^n=T^(r+qp)=T^r∘T^(qp)，再用 S09.2 的 T^(qp)(x)=x，得 T^n(x)=T^r(T^(qp)(x))=T^r(x)。",
          "en": "Divide the positive integer n with remainder: n=qp+r with 0≤r<p. By the iterate law T^n=T^(r+qp)=T^r∘T^(qp), and with T^(qp)(x)=x from S09.2, T^n(x)=T^r(T^(qp)(x))=T^r(x)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "division_algorithm",
            "iterate_law",
            "S09.2"
          ],
          "explanation": {
            "zh-Hant": "需要 p≥1 才能做帶餘除法，以及 S09.2 的歸納結論。",
            "en": "It needs p≥1 for division with remainder, and the induction result of S09.2."
          }
        },
        "leaf_basis": "division_algorithm + iterate_law + eq_replace"
      },
      {
        "id": "S09.4",
        "uses": [
          "S09.2",
          "S09.3",
          "G4"
        ],
        "before_tex": "T^n(x)=x",
        "after_tex": "r=0,\\ p\\mid n",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [
          "0\\le r<p"
        ],
        "explanation": {
          "zh-Hant": "⇒：若 T^n(x)=x，由 S09.3 得 T^r(x)=x。若 r≥1，r 就是小於 p 的正返回時間，與 p 的最小性矛盾；所以 r=0，n=qp，p∣n。⇐：若 p∣n，則 n=qp，由 S09.2 得 T^n(x)=x。S09.1 至 S09.4 合起來證得引理 L_return。",
          "en": "⇒: if T^n(x)=x, S09.3 gives T^r(x)=x. If r≥1, r would be a positive return time smaller than p, contradicting the minimality of p; so r=0, n=qp and p∣n. ⇐: if p∣n, then n=qp and S09.2 gives T^n(x)=x. S09.1 to S09.4 together prove Lemma L_return."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09.2",
            "S09.3",
            "contradiction",
            "G4"
          ],
          "explanation": {
            "zh-Hant": "矛盾只用到 p 的最小性與 0≤r<p；兩個方向都已處理。",
            "en": "The contradiction uses only the minimality of p and 0≤r<p; both directions are handled."
          }
        },
        "leaf_basis": "contradiction + divides_def + and_intro_elim"
      },
      {
        "id": "S10",
        "uses": [
          "S08",
          "S09"
        ],
        "before_tex": "x=1,\\ p(1)=2",
        "after_tex": "T^n(1)=1\\iff 2\\mid n;\\ T^4(1)=1,\\ T^3(1)=3",
        "relation": "implies",
        "rule": "L_return",
        "conditions": [
          "n\\geq 1"
        ],
        "explanation": {
          "zh-Hant": "對 x=1、p=2 用 L_return：1 的正返回時間恰為 2、4、6 等偶數。所以 T^4(1)=1，但 4 只是返回時間，最小週期仍是 2。奇數 n=2q+1 時，由 S09.3 得 T^n(1)=T^1(1)=3，例如 T^3(1)=3、T^7(1)=3。同理，p(5)=1 整除每個正整數，所以 T^n(5)=5 對所有 n≥1 成立，T²(5)=5 不使 5 成為最小週期 2 的點。",
          "en": "Apply L_return with x=1, p=2: the positive return times of 1 are exactly the even numbers 2, 4, 6 and so on. So T^4(1)=1, but 4 is only a return time and the least period is still 2. For odd n=2q+1, S09.3 gives T^n(1)=T^1(1)=3, for example T^3(1)=3 and T^7(1)=3. Likewise p(5)=1 divides every positive integer, so T^n(5)=5 for all n≥1, and T²(5)=5 does not make 5 a point of least period 2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "L_return"
          ],
          "explanation": {
            "zh-Hant": "需要 p(1)=2（S08）與引理；N9、N10 是對 T^4(1)、T^3(1) 的直接計算交叉核對。",
            "en": "It needs p(1)=2 (S08) and the lemma; N9 and N10 cross-check T^4(1) and T^3(1) by direct computation."
          }
        },
        "leaf_basis": "L_return + division_algorithm + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S04",
          "G3"
        ],
        "before_tex": "n\\geq 1",
        "after_tex": "T^n(x)\\in T(X)=\\{1,3,5\\}",
        "relation": "implies",
        "rule": "L_image",
        "conditions": [
          "n\\geq 1"
        ],
        "explanation": {
          "zh-Hant": "證明引理 L_image：至少走一步之後，軌道永遠留在 T 的像集合裡。結合 S04，T^n(x) 對每個 n≥1 都屬於 {1,3,5}。",
          "en": "Prove Lemma L_image: after at least one step, the orbit always stays in the image of T. With S04, T^n(x) belongs to {1,3,5} for every n≥1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_image",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "需要 n≥1；n=0 時 T^0(x)=x 可以不在像集合中，這正是 0、2、4 的情形。",
            "en": "It needs n≥1; for n=0, T^0(x)=x may lie outside the image, which is exactly the case of 0, 2 and 4."
          }
        },
        "leaf_basis": "composition_def + image_preimage_def"
      },
      {
        "id": "S11.1",
        "uses": [
          "G3",
          "composition_def",
          "image_preimage_def"
        ],
        "before_tex": "T^n(x),\\ n\\geq 1",
        "after_tex": "T^n(x)=T(T^{n-1}(x))\\in T(X)",
        "relation": "implies",
        "rule": "composition_def",
        "conditions": [
          "n-1\\in\\mathbb{N}"
        ],
        "explanation": {
          "zh-Hant": "n≥1 時 n−1 是自然數，由 G3 的 T^n=T∘T^(n−1) 得 T^n(x)=T(y)，其中 y=T^(n−1)(x)∈X。T 在 X 某點的值依定義屬於 T(X)。這證得引理 L_image。",
          "en": "For n≥1, n−1 is a natural number, and T^n=T∘T^(n−1) from G3 gives T^n(x)=T(y) with y=T^(n−1)(x)∈X. A value of T at a point of X belongs to T(X) by definition. This proves Lemma L_image."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "composition_def",
            "image_preimage_def"
          ],
          "explanation": {
            "zh-Hant": "只用迭代的遞迴定義與像集合定義，對每個 n≥1 一致成立，不是有限核對。",
            "en": "Only the recursive definition of iterates and the definition of the image are used; the argument holds uniformly for every n≥1 and is not a finite check."
          }
        },
        "leaf_basis": "composition_def + image_preimage_def"
      },
      {
        "id": "S12",
        "uses": [
          "S11",
          "S04"
        ],
        "before_tex": "x\\in\\{0,2,4\\}",
        "after_tex": "\\forall n\\geq 1:\\ T^n(x)\\neq x",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "設 x∈{0,2,4}，若某個 n≥1 使 T^n(x)=x，由 S11 得 x∈{1,3,5}，但 {0,2,4} 與 {1,3,5} 不相交，矛盾。所以 0、2、4 沒有任何正返回時間，不是週期點。這是對所有 n 的證明，不是檢查前幾步的結果。",
          "en": "Let x∈{0,2,4}. If T^n(x)=x for some n≥1, S11 gives x∈{1,3,5}, but {0,2,4} and {1,3,5} are disjoint, a contradiction. So 0, 2 and 4 have no positive return time and are not periodic. This is a proof for every n, not the outcome of checking the first few steps."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "S04",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "需要 x 不在像集合中（S04）與 L_image；不宣稱反向，即不在像中是非週期的充分條件，但不是必要條件。",
            "en": "It needs x outside the image (S04) and L_image; no converse is claimed: lying outside the image is sufficient for non-periodicity, not necessary."
          }
        },
        "leaf_basis": "contradiction + set_operations_def"
      },
      {
        "id": "S13",
        "uses": [
          "S03",
          "S08",
          "S12",
          "G4"
        ],
        "before_tex": "T(0)=1,\\ T(2)=5,\\ T(4)=3",
        "after_tex": "0,2,4\\ \\text{preperiodic, not periodic}",
        "relation": "implies",
        "rule": "def_unfold",
        "conditions": [],
        "explanation": {
          "zh-Hant": "一步之後：T(0)=1、T(4)=3 是最小週期 2 的點，T(2)=5 是固定點（S08）。所以 0 的軌道是 0、1、3、1、3，依此類推；4 的軌道是 4、3、1、3；2 的軌道是 2、5、5。三者都最終進入循環，但由 S12 都不是週期點，依 G4 為前週期點。這正是 MKO 的提醒：循環從起點之後才開始時，起點不是週期點。",
          "en": "After one step: T(0)=1 and T(4)=3 are points of least period 2, and T(2)=5 is the fixed point (S08). So the orbit of 0 is 0, 1, 3, 1, 3 and so on; that of 4 is 4, 3, 1, 3; that of 2 is 2, 5, 5. All three eventually enter a cycle, but by S12 none is periodic, so by G4 they are preperiodic. This is exactly the MKO's warning: when the cycle starts after the starting point, the starting point is not periodic."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "S12",
            "G4"
          ],
          "explanation": {
            "zh-Hant": "前週期需要兩部分：不是週期點（S12）且某個 T^k(x) 是週期點（這裡 k=1，由 S03、S08）。",
            "en": "Preperiodic needs two parts: not periodic (S12) and some T^k(x) periodic (here k=1, by S03 and S08)."
          }
        },
        "leaf_basis": "def_unfold + and_intro_elim"
      },
      {
        "id": "S14",
        "uses": [
          "S05",
          "S08",
          "S12",
          "S13",
          "G1"
        ],
        "before_tex": "X=\\{0,1,2,3,4,5\\}",
        "after_tex": "\\mathrm{Fix}(T)=\\{5\\},\\ \\{x\\mid p(x)=2\\}=\\{1,3\\},\\ \\text{preperiodic}=\\{0,2,4\\}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整：唯一固定點是 5（S05）；最小週期 2 的點是 1、3（S08）；0、2、4 是前週期點（S12、S13）。三個集合兩兩不相交，聯集為 X，所以這是 X 的完整分類；X 恰有六個狀態，沒有遺漏。",
          "en": "Summary: the only fixed point is 5 (S05); the points of least period 2 are 1 and 3 (S08); 0, 2 and 4 are preperiodic (S12, S13). The three sets are pairwise disjoint with union X, so this is a complete classification of X; X has exactly six states and none is missed."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S05",
            "S08",
            "S12",
            "S13"
          ],
          "explanation": {
            "zh-Hant": "目標 G：X 的完整分類。足夠條件 H：S05、S08、S12、S13 的四個結論，加上三集合的聯集是 X；H⇒G 由合取。H 由定義與同餘代數正向證成，沒有把 G 當作前提。",
            "en": "Goal G: the complete classification of X. Sufficient condition H: the four conclusions S05, S08, S12, S13, plus the fact that the union of the three sets is X; H⇒G by conjunction. H is proved forwards from the definitions and congruence algebra; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim + set_operations_def"
      }
    ],
    "result": {
      "zh-Hant": "對 T(x)=(2x+1) mod 6：唯一固定點是 5；1 與 3 構成最小週期 2 的循環；0、2、4 不在 T 的像中，所以永遠不會回到自己，是前週期點（一步後分別進入 1、5、3）。對最小週期 p 的點，T^n(x)=x 當且僅當 p∣n；所以 T^4(1)=1 只是返回時間，1 的最小週期仍是 2。",
      "en": "For T(x)=(2x+1) mod 6: the only fixed point is 5; 1 and 3 form a cycle of least period 2; 0, 2 and 4 are not in the image of T, so they never return to themselves and are preperiodic (after one step they move to 1, 5 and 3). For a point of least period p, T^n(x)=x exactly when p∣n; so T^4(1)=1 is only a return time, and the least period of 1 is still 2."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "T^4(1)=1，所以 1 的最小週期是 4",
          "en": "T^4(1)=1, so the least period of 1 is 4"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "T²(1)=T(3)=1，2 是更小的正返回時間。",
          "en": "T²(1)=T(3)=1, so 2 is a smaller positive return time."
        }
      },
      {
        "claim": {
          "zh-Hant": "0 的軌道最後在 1 與 3 之間循環，所以 0 是週期點",
          "en": "The orbit of 0 ends up cycling between 1 and 3, so 0 is periodic"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "0∉T(X)={1,3,5}，所以對每個 n≥1 都有 T^n(0)≠0（S12）。",
          "en": "0∉T(X)={1,3,5}, so T^n(0)≠0 for every n≥1 (S12)."
        }
      },
      {
        "claim": {
          "zh-Hant": "T²(5)=5，所以 5 是最小週期 2 的週期點",
          "en": "T²(5)=5, so 5 is a periodic point of least period 2"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "T(5)=11 mod 6=5，最小週期是 1。",
          "en": "T(5)=11 mod 6=5, so the least period is 1."
        }
      },
      {
        "claim": {
          "zh-Hant": "在狀態空間為 ℤ 的系統中，某點的軌道前 200 步都沒有回到起點，所以它不是週期點",
          "en": "In a system on ℤ, the orbit of a point does not return in its first 200 steps, so the point is not periodic"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "取 T(x)=x+1（0≤x<200）、T(200)=0、其餘 T(x)=x：0 在前 200 步不回來，但 T^201(0)=0。有限前綴不證非週期（MKO 假設 a2）。",
          "en": "Take T(x)=x+1 for 0≤x<200, T(200)=0 and T(x)=x otherwise: 0 does not return in the first 200 steps, but T^201(0)=0. A finite prefix proves no non-periodicity (MKO assumption a2)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "division_algorithm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "帶餘除法：對整數 a 與正整數 b，存在唯一的整數 q、r 使 a=bq+r 且 0≤r<b。",
          "en": "Division algorithm: for an integer a and a positive integer b there are unique integers q, r with a=bq+r and 0≤r<b."
        }
      },
      {
        "id": "divides_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整除：對整數 a、b，a∣b 表示存在整數 k 使 b=ak。",
          "en": "Divisibility: for integers a, b, a∣b means there is an integer k with b=ak."
        }
      },
      {
        "id": "nat_induction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "數學歸納原理：對從 n₀ 起的命題族 P(n)，若 P(n₀) 成立，且對每個 n≥n₀ 都有 P(n)⇒P(n+1)，則所有 n≥n₀ 都有 P(n)。",
          "en": "Principle of induction: for a family P(n) starting at n₀, if P(n₀) holds and P(n)⇒P(n+1) for every n≥n₀, then P(n) holds for every n≥n₀."
        }
      },
      {
        "id": "composition_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合成：對 f:A→B 與 g:B→C，g∘f:A→C 定義為 (g∘f)(a)=g(f(a))。",
          "en": "Composition: for f:A→B and g:B→C, g∘f:A→C is defined by (g∘f)(a)=g(f(a))."
        }
      },
      {
        "id": "image_preimage_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "像與原像：f(S)={f(a) ∣ a∈S}；f⁻¹(T)={a∈A ∣ f(a)∈T}。原像記號不預設 f 有反函數。",
          "en": "Image and preimage: f(S)={f(a) ∣ a∈S}; f⁻¹(T)={a∈A ∣ f(a)∈T}. The preimage notation does not presuppose that f has an inverse."
        }
      },
      {
        "id": "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": "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": "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": "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": "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": "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": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "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": "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": "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": "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": "mko_orbit_defs",
        "kind": "cited",
        "source": {
          "mko_id": "mko-orbit-fixed-periodic",
          "sha256": "702889d6acab1a9c1c6c8d26447c0b2a0201751460d118874f55c0486060083f",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對自映射 T:X→X，x 的前向軌道為 x、T(x)、T²(x) 等。若 T(x)=x，x 是固定點；若存在最小正整數 p 使 T^p(x)=x，則 x 是最小週期 p 的週期點。最終進入循環不代表起點本身為週期點。",
          "en": "For a self-map T:X→X, the forward orbit of x is x, T(x), T²(x) and so on. A point is fixed when T(x)=x. If there is a least positive integer p such that T^p(x)=x, it is periodic with least period p. Eventually entering a cycle does not make the starting point itself periodic."
        }
      },
      {
        "id": "mod6_arith",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "模 6 同餘：u≡v (mod 6) 表示 6∣(u−v)；同餘與加法、乘法相容；m mod 6 是唯一滿足 m≡r (mod 6) 的 r∈X；X 中兩元素相等 ⇔ 它們模 6 同餘。",
          "en": "Congruence modulo 6: u≡v (mod 6) means 6∣(u−v); congruence is compatible with addition and multiplication; m mod 6 is the unique r∈X with m≡r (mod 6); two elements of X are equal ⇔ they are congruent modulo 6."
        }
      },
      {
        "id": "iterate_law",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "對自映射 T 與所有自然數 a、b：T^(a+b)=T^a∘T^b（由 T^0=id、T^(n+1)=T∘T^n 對 a 歸納可得）。",
          "en": "For a self-map T and all natural numbers a, b: T^(a+b)=T^a∘T^b (it follows from T^0=id and T^(n+1)=T∘T^n by induction on a)."
        }
      },
      {
        "id": "L_return",
        "kind": "lemma",
        "proved_in": [
          "S09.1",
          "S09.2",
          "S09.3",
          "S09.4"
        ],
        "statement": {
          "zh-Hant": "若 x 的最小週期為 p，則對每個正整數 n：T^n(x)=x ⇔ p∣n。",
          "en": "If x has least period p, then for every positive integer n: T^n(x)=x ⇔ p∣n."
        }
      },
      {
        "id": "L_image",
        "kind": "lemma",
        "proved_in": [
          "S11.1"
        ],
        "statement": {
          "zh-Hant": "對每個 x∈X 與每個正整數 n：T^n(x)∈T(X)。",
          "en": "For every x∈X and every positive integer n: T^n(x)∈T(X)."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03.2",
        "lhs": "(2*3+1)%6",
        "rhs": "1"
      },
      {
        "id": "N2",
        "step": "S03.2",
        "lhs": "(2*4+1)%6",
        "rhs": "3"
      },
      {
        "id": "N3",
        "step": "S03.2",
        "lhs": "(2*5+1)%6",
        "rhs": "5"
      },
      {
        "id": "N4",
        "step": "S03",
        "vars": [
          "x"
        ],
        "range": [
          0,
          5
        ],
        "lhs": "(2*x+1)%6",
        "rhs": "2*(x%3)+1"
      },
      {
        "id": "N5",
        "step": "S05",
        "vars": [
          "x"
        ],
        "range": [
          0,
          5
        ],
        "lhs": "min(abs((2*x+1)%6-x),1)",
        "rhs": "min(abs(x-5),1)"
      },
      {
        "id": "N6",
        "step": "S06",
        "vars": [
          "x"
        ],
        "range": [
          0,
          5
        ],
        "lhs": "(2*((2*x+1)%6)+1)%6",
        "rhs": "(4*x+3)%6"
      },
      {
        "id": "N7",
        "step": "S07.1",
        "vars": [
          "x"
        ],
        "degree": 1,
        "lhs": "(4*x+3)-x",
        "rhs": "3*(x+1)"
      },
      {
        "id": "N8",
        "step": "S07",
        "vars": [
          "x"
        ],
        "range": [
          0,
          5
        ],
        "lhs": "min(abs((4*x+3)%6-x),1)",
        "rhs": "(x+1)%2"
      },
      {
        "id": "N9",
        "step": "S10",
        "lhs": "(4*((4*1+3)%6)+3)%6",
        "rhs": "1"
      },
      {
        "id": "N10",
        "step": "S10",
        "lhs": "(2*((4*1+3)%6)+1)%6",
        "rhs": "3"
      },
      {
        "id": "N11",
        "step": "S12",
        "vars": [
          "x"
        ],
        "range": [
          0,
          5
        ],
        "lhs": "((2*x+1)%6)%2",
        "rhs": "1"
      },
      {
        "id": "N12",
        "step": "S10",
        "lhs": "7%2",
        "rhs": "1"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：完整有限剩餘類狀態與仿射自映射，分類只適用這個指定的有限模型（假設 a1、a2）。固定點與週期 2 的判定由同餘代數證明，再以覆蓋全部六個狀態的有限核對交叉檢查。不處理吸引性、穩定性、參數分岔或無限系統；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: complete finite residue-class states and an affine self-map, with the classification valid only for this specified finite model (assumptions a1, a2). Fixed points and period 2 are decided by congruence algebra and cross-checked by finite checks that cover all six states. Attraction, stability, parameter bifurcation and infinite systems are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "X",
        "狀態空間 {0,1,2,3,4,5}",
        "the state space {0,1,2,3,4,5}"
      ],
      [
        "T",
        "自映射 T(x)=(2x+1) mod 6",
        "the self-map T(x)=(2x+1) mod 6"
      ],
      [
        "T^n",
        "T 的 n 次迭代，T^0 為恆等映射",
        "the n-th iterate of T, with T^0 the identity"
      ],
      [
        "p(x)",
        "x 的最小週期",
        "the least period of x"
      ],
      [
        "T(X)",
        "T 的像集合",
        "the image of T"
      ],
      [
        "u\\equiv v\\pmod 6",
        "模 6 同餘",
        "congruence modulo 6"
      ]
    ],
    "md_error_controls": [
      {
        "title": "返回時間不一定是最小週期 / A return time need not be the least period",
        "zh": "MKO 指出的誤解之一是把任何返回時間都叫最小週期。T^4(1)=1，但 T²(1)=1 已經成立，所以 1 的最小週期是 2（S08、S10）；T²(5)=5，但 T(5)=5，所以 5 的最小週期是 1。L_return 說明所有返回時間都是 p 的倍數，判定最小週期時必須排除每個更小的正整數。",
        "en": "One mistake named by the MKO is calling any return time the least period. T^4(1)=1, but already T²(1)=1, so the least period of 1 is 2 (S08, S10); T²(5)=5, but T(5)=5, so the least period of 5 is 1. L_return shows every return time is a multiple of p, and deciding the least period requires excluding every smaller positive integer."
      },
      {
        "title": "最終進入循環不保證起點為週期點 / Eventually entering a cycle does not make the starting point periodic",
        "zh": "0 的軌道 0、1、3、1、3 從第 1 步起循環，但循環不含起點 0。S12 證明 0 永遠回不來：每個 T^n(0)（n≥1）都在 {1,3,5} 中。把「最後進入循環」的起點稱為週期點，就是 MKO 指出的另一個誤解。",
        "en": "The orbit 0, 1, 3, 1, 3 of 0 cycles from step 1 on, but the cycle does not contain the start 0. S12 proves 0 never comes back: every T^n(0) with n≥1 lies in {1,3,5}. Calling a point periodic because its orbit eventually cycles is the other mistake the MKO names."
      },
      {
        "title": "有限核對的範圍 / What the finite checks cover",
        "zh": "N4、N5、N6、N8、N11 是對 X 全部六個狀態的有限核對；X 恰有這六個狀態，所以它們涵蓋本系統的整個狀態空間，而 S05、S07、S12 的結論另有代數證明。這些核對對其他模數、其他映射或無限系統不說明任何事（MKO 假設 a2）。",
        "en": "N4, N5, N6, N8 and N11 are finite checks over all six states of X; X has exactly these six states, so they cover the whole state space of this system, and the conclusions of S05, S07 and S12 also have algebraic proofs. The checks say nothing about other moduli, other maps or infinite systems (MKO assumption a2)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "迭代律 T^(a+b)=T^a∘T^b 在本稿作為前置知識使用，沒有在此重新歸納證明。",
        "en": "The iterate law T^(a+b)=T^a∘T^b is used as a prerequisite here and is not re-proved by induction."
      },
      {
        "zh-Hant": "吸引週期、穩定性與參數分岔屬於 MKO 的 formal next，本稿不處理；本稿的分類也不外推到其他仿射映射 (ax+b) mod m。",
        "en": "Attracting cycles, stability and parameter bifurcation belong to the MKO's formal next step and are not treated; the classification here is not extended to other affine maps (ax+b) mod m."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "790a36d058c22ecf3ef889f75d6af706cd0f9c467bc505d2f3267262dbba5620",
    "original_source_sha256": "790a36d058c22ecf3ef889f75d6af706cd0f9c467bc505d2f3267262dbba5620",
    "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."
  }
}