{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-discrete-dynamical-system-01", "mko_id": "mko-discrete-dynamical-system", "title": { "zh-Hant": "離散時間動力系統：在七個狀態上迭代 T(x)=(3x+1) mod 7", "en": "Discrete-time dynamical system: iterating T(x)=(3x+1) mod 7 on seven states" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-discrete-dynamical-system-01", "source": { "mko_id": "mko-discrete-dynamical-system", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "83db6a4ce40522862935c10fd58632bc890dbcf7e2d2277368dad51ee0b72837", "english_sha256": "8513c6a6ed28df30295d27ab05a1d1c2a008427aceaf56ba9bfa33819c320e29" }, "mode": "worked_example", "audience_prerequisites": [ "function_def", "division_algorithm", "nat_induction", "subset_def", "image_preimage_def" ], "goal": { "zh-Hant": "狀態空間 X={0,1,2,3,4,5,6}，更新映射 T(x)=(3x+1) mod 7。由 x_0=0 依 x_(n+1)=T(x_n) 算出軌道，證明對所有自然數 n 有 x_(n+6)=x_n，據此求 x_2026；完整枚舉固定點；判定 A={0,1,2,4,5,6} 是正向不變子集而 B={0,1} 不是，並證明從 0 出發的軌道永遠不會到達固定點。", "en": "State space X={0,1,2,3,4,5,6}, update map T(x)=(3x+1) mod 7. From x_0=0 compute the orbit given by x_(n+1)=T(x_n), prove x_(n+6)=x_n for every natural number n and use it to find x_2026; enumerate the fixed points completely; decide that A={0,1,2,4,5,6} is forward invariant while B={0,1} is not, and prove that the orbit from 0 never reaches the fixed point." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "X={0,1,2,3,4,5,6}；時間索引為包含 0 的自然數。", "en": "X={0,1,2,3,4,5,6}; time is indexed by the natural numbers including 0." } }, { "id": "G2", "statement": { "zh-Hant": "T:X→X，T(x)=(3x+1) mod 7，其中 m mod 7 是整數 m 除以 7 的餘數 r（0≤r<7）。", "en": "T:X→X, T(x)=(3x+1) mod 7, where m mod 7 is the remainder r (0≤r<7) of the integer m on division by 7." } }, { "id": "G3", "statement": { "zh-Hant": "初始狀態 x_0=0，對每個自然數 n，x_(n+1)=T(x_n)。", "en": "Initial state x_0=0, and x_(n+1)=T(x_n) for every natural number n." } }, { "id": "G4", "statement": { "zh-Hant": "待檢查的子集 A={0,1,2,4,5,6} 與 B={0,1}。", "en": "Subsets to be tested: A={0,1,2,4,5,6} and B={0,1}." } } ], "steps": [ { "id": "S01", "uses": [ "G1", "G2", "G3" ], "before_tex": "\\text{state space and update rule}", "after_tex": "(X,T),\\ X=\\{0,1,2,3,4,5,6\\},\\ x_{n+1}=T(x_n),\\ x_0=0", "relation": "definition", "rule": "mko_dds", "conditions": [], "explanation": { "zh-Hant": "依 MKO，系統由狀態空間 X 與自映射 T 組成，軌道由初始狀態與 T 共同決定。這裡 X 有 7 個明列的狀態，T 由 G2 給出，時間從 0 開始。只有 X 而沒有 T 不足以決定任何後續狀態（見錯誤控制）。", "en": "By the MKO, a system consists of a state space X and a self-map T, and the orbit is determined by the initial state together with T. Here X has 7 listed states, T is given by G2, and time starts at 0. X without T determines no later state (see the error controls)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "G2", "G3", "mko_dds" ], "explanation": { "zh-Hant": "此步只寫出系統與軌道的定義，不推出新事實；T 是否確實為 X 上的總自映射在 S02 檢查。", "en": "This step only states the system and the orbit and derives no new fact; that T really is a total self-map of X is checked in S02." } }, "leaf_basis": "mko_dds + def_intro" }, { "id": "S02", "uses": [ "G1", "G2" ], "before_tex": "x\\in X", "after_tex": "T(x)=(3x+1)\\bmod 7\\in X", "relation": "implies", "rule": "division_algorithm", "conditions": [ "7>0" ], "explanation": { "zh-Hant": "對每個 x∈X，3x+1 是整數；帶餘除法給出唯一的整數 q、r 使 3x+1=7q+r 且 0≤r<7，於是 T(x)=r 恰是 X 的一個元素。存在性說明 T 對每個狀態都有值（總映射），唯一性說明值是確定的（函數），0≤r<7 說明值落在 X 中（自映射）。這正是 MKO 假設 a1 的要求。", "en": "For each x∈X, 3x+1 is an integer; the division algorithm gives unique integers q, r with 3x+1=7q+r and 0≤r<7, so T(x)=r is exactly one element of X. Existence says T has a value at every state (total), uniqueness says that value is determined (a function), and 0≤r<7 says it lies in X (a self-map). This is what MKO assumption a1 requires." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "division_algorithm", "function_def" ], "explanation": { "zh-Hant": "需要除數 7>0。此步不宣稱 T 是雙射；那件事可由 S03 的表看出，但後面沒有用到。", "en": "It needs the divisor 7>0. The step does not claim that T is a bijection; that can be read off the table in S03 but is not used later." } }, "leaf_basis": "division_algorithm + function_def" }, { "id": "S03", "uses": [ "S02" ], "before_tex": "T(0),T(1),T(2),T(3),T(4),T(5),T(6)", "after_tex": "1,\\ 4,\\ 0,\\ 3,\\ 6,\\ 2,\\ 5", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "完整列出 T 的值表，計算在 S03.1 與 S03.2。因為 X 只有 7 個狀態，這張表就是 T 的全部資料，之後的枚舉都以它為依據。", "en": "List the complete value table of T, computed in S03.1 and S03.2. Since X has only 7 states, this table is all the data of T, and every later enumeration rests on it." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "numeric_evaluation", "division_algorithm" ], "explanation": { "zh-Hant": "每一項都是同一個數的兩種寫法，可以獨立重算；反向即把 r 寫回 3x+1=7q+r。", "en": "Each entry names one number in two ways and can be recomputed independently; backwards, write r back into 3x+1=7q+r." } }, "leaf_basis": "numeric_evaluation + division_algorithm" }, { "id": "S03.1", "uses": [ "G2" ], "before_tex": "3x+1\\ \\text{for }x=0,1,2,3,4,5,6", "after_tex": "1,\\ 4,\\ 7,\\ 10,\\ 13,\\ 16,\\ 19", "relation": "equivalent", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "逐一計算 3x+1：3·0+1=1，3·1+1=4，3·2+1=7，3·3+1=10，3·4+1=13，3·5+1=16，3·6+1=19。", "en": "Compute 3x+1 one by one: 3·0+1=1, 3·1+1=4, 3·2+1=7, 3·3+1=10, 3·4+1=13, 3·5+1=16, 3·6+1=19." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "numeric_evaluation" ], "explanation": { "zh-Hant": "整數算式的精確求值，雙向都是同一個數。", "en": "Exact evaluation of integer expressions; both sides are the same number." } }, "leaf_basis": "numeric_evaluation" }, { "id": "S03.2", "uses": [ "S03.1", "division_algorithm" ], "before_tex": "1,4,7,10,13,16,19\\ \\bmod 7", "after_tex": "1,\\ 4,\\ 0,\\ 3,\\ 6,\\ 2,\\ 5", "relation": "equivalent", "rule": "division_algorithm", "conditions": [ "0\\le r<7" ], "explanation": { "zh-Hant": "寫出帶餘除法：1=7·0+1，4=7·0+4，7=7·1+0，10=7·1+3，13=7·1+6，16=7·2+2，19=7·2+5。每個餘數都在 0 到 6 之間，所以由唯一性它就是 mod 7 的值。", "en": "Write out the division algorithm: 1=7·0+1, 4=7·0+4, 7=7·1+0, 10=7·1+3, 13=7·1+6, 16=7·2+2, 19=7·2+5. Each remainder lies between 0 and 6, so by uniqueness it is the value mod 7." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "division_algorithm", "numeric_evaluation" ], "explanation": { "zh-Hant": "每個等式 7q+r 都可展開回原數；餘數條件 0≤r<7 保證這是唯一的分解。", "en": "Each equation 7q+r expands back to the original number; the condition 0≤r<7 makes this the unique decomposition." } }, "leaf_basis": "division_algorithm + numeric_evaluation" }, { "id": "S04", "uses": [ "G3", "S03" ], "before_tex": "x_0=0", "after_tex": "x_1=1,\\ x_2=4,\\ x_3=6,\\ x_4=5,\\ x_5=2,\\ x_6=0", "relation": "implies", "rule": "eq_replace", "conditions": [], "explanation": { "zh-Hant": "依遞迴式查表：x_1=T(0)=1，x_2=T(1)=4，x_3=T(4)=6，x_4=T(6)=5，x_5=T(5)=2，x_6=T(2)=0。每一步只用到前一個狀態，這就是迭代演算法。", "en": "Apply the recurrence using the table: x_1=T(0)=1, x_2=T(1)=4, x_3=T(4)=6, x_4=T(6)=5, x_5=T(5)=2, x_6=T(2)=0. Each step uses only the previous state; this is the iteration algorithm." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G3", "S03", "eq_replace" ], "explanation": { "zh-Hant": "需要 x_0 與完整的值表；每個值由前一個值唯一決定。由 x_6 不能反推 x_5 是否唯一，這裡也不宣稱。", "en": "It needs x_0 and the complete table; each value is determined by the previous one. Whether x_5 could be recovered uniquely from x_6 is not claimed." } }, "leaf_basis": "eq_replace + numeric_evaluation" }, { "id": "S05", "uses": [ "S04", "G3" ], "before_tex": "x_6=0,\\ x_0=0", "after_tex": "x_6=x_0", "relation": "implies", "rule": "eq_refl_symm_trans", "conditions": [], "explanation": { "zh-Hant": "x_6 與 x_0 都等於 0，依等號的對稱與傳遞得 x_6=x_0：軌道在第 6 步回到起點。", "en": "Both x_6 and x_0 equal 0, so symmetry and transitivity of equality give x_6=x_0: the orbit returns to its start at step 6." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S04", "eq_refl_symm_trans" ], "explanation": { "zh-Hant": "只是一個已算出的等式；它本身尚未說明之後的狀態，那需要 S06 的歸納。", "en": "This is one computed equality; by itself it says nothing about later states, which needs the induction of S06." } }, "leaf_basis": "eq_refl_symm_trans" }, { "id": "S06", "uses": [ "S05", "L_period" ], "before_tex": "x_6=x_0", "after_tex": "\\forall n\\in\\mathbb{N}:\\ x_{n+6}=x_n", "relation": "implies", "rule": "L_period", "conditions": [], "explanation": { "zh-Hant": "依引理 L_period（在 S06.1 到 S06.3 以數學歸納法證明）。關鍵是決定性：下一個狀態只由目前狀態決定，所以兩個時刻狀態相同，之後就永遠同步。這不是由六個數值外推得到的。", "en": "By Lemma L_period (proved by induction in S06.1 to S06.3). The key is determinism: the next state depends only on the current state, so once two times share a state they stay in step forever. This is not an extrapolation from six values." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_period" ], "explanation": { "zh-Hant": "依賴該引理；引理只用到 S05 與 T 是函數。", "en": "It depends on the lemma, which uses only S05 and that T is a function." } }, "leaf_basis": "nat_induction + eq_congruence" }, { "id": "S06.1", "uses": [ "S05" ], "before_tex": "n=0", "after_tex": "x_{0+6}=x_0", "relation": "induction_base", "rule": "nat_induction", "conditions": [], "explanation": { "zh-Hant": "基底：令 P(n) 為 x_(n+6)=x_n。P(0) 就是 x_6=x_0，已由 S05 給出。", "en": "Base: let P(n) be x_(n+6)=x_n. P(0) is x_6=x_0, given by S05." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S05", "nat_induction" ], "explanation": { "zh-Hant": "基底只需要 S05 的計算。", "en": "The base needs only the computation of S05." } }, "leaf_basis": "nat_induction" }, { "id": "S06.2", "uses": [ "G3", "G2" ], "before_tex": "x_{n+6}=x_n", "after_tex": "x_{(n+1)+6}=x_{n+1}", "relation": "induction_step", "rule": "eq_congruence", "conditions": [], "explanation": { "zh-Hant": "歸納步：假設 x_(n+6)=x_n。兩邊同用函數 T：T(x_(n+6))=T(x_n)。依遞迴式左邊是 x_(n+7)，右邊是 x_(n+1)，所以 x_((n+1)+6)=x_(n+1)。", "en": "Inductive step: assume x_(n+6)=x_n. Apply the function T to both sides: T(x_(n+6))=T(x_n). By the recurrence the left side is x_(n+7) and the right side is x_(n+1), so x_((n+1)+6)=x_(n+1)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G3", "eq_congruence", "function_def" ], "explanation": { "zh-Hant": "需要 T 是函數（同一輸入得同一輸出），以及遞迴式對每個 n 成立。不宣稱反向：由 x_(n+7)=x_(n+1) 推回 x_(n+6)=x_n 需要 T 單射，這裡沒有用到。", "en": "It needs T to be a function (equal inputs give equal outputs) and the recurrence for every n. No converse is claimed: going back from x_(n+7)=x_(n+1) to x_(n+6)=x_n would need T to be injective, which is not used." } }, "leaf_basis": "eq_congruence + function_def" }, { "id": "S06.3", "uses": [ "S06.1", "S06.2" ], "before_tex": "P(0),\\ \\forall n\\,(P(n)\\Rightarrow P(n+1))", "after_tex": "\\forall n\\in\\mathbb{N}:\\ x_{n+6}=x_n", "relation": "implies", "rule": "nat_induction", "conditions": [], "explanation": { "zh-Hant": "依數學歸納原理，P(n) 對每個自然數 n 成立。S06.1 到 S06.3 合起來證得引理 L_period。", "en": "By the principle of induction, P(n) holds for every natural number n. S06.1 to S06.3 together prove Lemma L_period." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S06.1", "S06.2", "nat_induction" ], "explanation": { "zh-Hant": "歸納原理的兩個前提分別是 S06.1 與 S06.2，n 在歸納步中是任意的。", "en": "The two premises of induction are S06.1 and S06.2, with n arbitrary in the inductive step." } }, "leaf_basis": "nat_induction" }, { "id": "S07", "uses": [ "S06", "L_reduce" ], "before_tex": "\\forall n:\\ x_{n+6}=x_n", "after_tex": "\\forall q,r\\in\\mathbb{N}:\\ x_{6q+r}=x_r", "relation": "implies", "rule": "L_reduce", "conditions": [], "explanation": { "zh-Hant": "依引理 L_reduce（在 S07.1 到 S07.3 對 q 歸納證明）：時間每加 6，狀態不變，所以可以一次去掉任意多個 6。", "en": "By Lemma L_reduce (proved by induction on q in S07.1 to S07.3): adding 6 to the time leaves the state unchanged, so any number of 6s can be removed at once." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_reduce" ], "explanation": { "zh-Hant": "依賴該引理；引理只用到 L_period。", "en": "It depends on the lemma, which uses only L_period." } }, "leaf_basis": "nat_induction + L_period" }, { "id": "S07.1", "uses": [ "G3" ], "before_tex": "q=0", "after_tex": "x_{6\\cdot 0+r}=x_r", "relation": "induction_base", "rule": "nat_induction", "conditions": [], "explanation": { "zh-Hant": "固定任意自然數 r，令 R(q) 為 x_(6q+r)=x_r。R(0) 成立，因為 6·0+r=r。", "en": "Fix an arbitrary natural number r and let R(q) be x_(6q+r)=x_r. R(0) holds since 6·0+r=r." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "nat_induction", "mul_zero", "add_zero" ], "explanation": { "zh-Hant": "只用到 6·0=0 與 0+r=r。", "en": "It uses only 6·0=0 and 0+r=r." } }, "leaf_basis": "nat_induction + mul_zero + add_zero" }, { "id": "S07.2", "uses": [ "S06" ], "before_tex": "x_{6q+r}=x_r", "after_tex": "x_{6(q+1)+r}=x_r", "relation": "induction_step", "rule": "L_period", "conditions": [], "explanation": { "zh-Hant": "假設 R(q)。因 6(q+1)+r=(6q+r)+6，依 L_period（取 n=6q+r）得 x_(6(q+1)+r)=x_(6q+r)，再由歸納假設等於 x_r。", "en": "Assume R(q). Since 6(q+1)+r=(6q+r)+6, L_period with n=6q+r gives x_(6(q+1)+r)=x_(6q+r), which equals x_r by the inductive hypothesis." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S06", "distributive", "add_assoc", "add_comm", "eq_refl_symm_trans" ], "explanation": { "zh-Hant": "需要指標的改寫 6(q+1)+r=(6q+r)+6（分配律、加法結合與交換）以及 L_period；這裡使用的是另一個已證引理，不是正在證明的 L_reduce。", "en": "It needs the index rewrite 6(q+1)+r=(6q+r)+6 (distributivity, associativity and commutativity of addition) and L_period; that is a different, already proved lemma, not L_reduce itself." } }, "leaf_basis": "L_period + distributive + add_assoc + add_comm + eq_refl_symm_trans" }, { "id": "S07.3", "uses": [ "S07.1", "S07.2" ], "before_tex": "R(0),\\ \\forall q\\,(R(q)\\Rightarrow R(q+1))", "after_tex": "\\forall q,r:\\ x_{6q+r}=x_r", "relation": "implies", "rule": "nat_induction", "conditions": [], "explanation": { "zh-Hant": "依歸納原理 R(q) 對所有 q 成立；r 是任意固定的，所以依全稱推廣對所有 r 也成立。S07.1 到 S07.3 合起來證得引理 L_reduce。", "en": "By induction R(q) holds for every q; r was arbitrary, so by generalization it holds for every r too. S07.1 to S07.3 together prove Lemma L_reduce." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S07.1", "S07.2", "nat_induction", "forall_intro" ], "explanation": { "zh-Hant": "全稱推廣可用，因為證明中對 r 沒有附加任何條件。", "en": "Generalization is allowed because no extra condition was placed on r in the proof." } }, "leaf_basis": "nat_induction + forall_intro" }, { "id": "S08", "uses": [ "division_algorithm" ], "before_tex": "2026", "after_tex": "2026=6\\cdot 337+4,\\ 0\\le 4<6", "relation": "equivalent", "rule": "division_algorithm", "conditions": [ "6>0" ], "explanation": { "zh-Hant": "把時間 2026 除以週期 6：6·337=2022，2026−2022=4，餘數 4 滿足 0≤4<6。", "en": "Divide the time 2026 by the period 6: 6·337=2022 and 2026−2022=4, with remainder 4 satisfying 0≤4<6." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "division_algorithm", "numeric_evaluation" ], "explanation": { "zh-Hant": "6·337+4 可算回 2026；帶餘除法保證商與餘數唯一。", "en": "6·337+4 evaluates back to 2026; the division algorithm makes the quotient and remainder unique." } }, "leaf_basis": "division_algorithm + numeric_evaluation" }, { "id": "S09", "uses": [ "S07", "S08", "S04" ], "before_tex": "x_{2026}", "after_tex": "x_{2026}=x_{6\\cdot 337+4}=x_4=5", "relation": "equivalent", "rule": "L_reduce", "conditions": [], "explanation": { "zh-Hant": "依 S08 改寫指標，再依 L_reduce（q=337、r=4）得 x_2026=x_4，由 S04 得 x_4=5。這個值是證明出來的，不是由前幾項的樣子猜出來的。", "en": "Rewrite the index by S08, then L_reduce with q=337 and r=4 gives x_2026=x_4, and S04 gives x_4=5. This value is proved, not guessed from the look of the first few terms." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_reduce", "S08", "S04", "eq_refl_symm_trans" ], "explanation": { "zh-Hant": "這是一串等式，每個等號都有理由（指標相等、引理、計算），等號對稱，雙向成立。", "en": "This is a chain of equalities, each with a reason (equal indices, the lemma, a computation); equality is symmetric, so the step holds both ways." } }, "leaf_basis": "L_reduce + eq_replace + eq_refl_symm_trans" }, { "id": "S10", "uses": [ "S03", "G1", "mko_dds_props" ], "before_tex": "\\{x\\in X\\mid T(x)=x\\}", "after_tex": "\\{3\\}", "relation": "equivalent", "rule": "case_analysis", "conditions": [], "explanation": { "zh-Hant": "依值表逐一比較：T(0)=1≠0，T(1)=4≠1，T(2)=0≠2，T(3)=3，T(4)=6≠4，T(5)=2≠5，T(6)=5≠6。七種情況就是 X 的全部狀態，所以這個枚舉是完整的，證明了固定點集合恰為 {3}。", "en": "Compare each table entry: T(0)=1≠0, T(1)=4≠1, T(2)=0≠2, T(3)=3, T(4)=6≠4, T(5)=2≠5, T(6)=5≠6. The seven cases are all the states of X, so the enumeration is complete and proves that the set of fixed points is exactly {3}." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "case_analysis", "set_extensionality", "S03" ], "explanation": { "zh-Hant": "集合相等的兩個包含：3 在集合中（T(3)=3）；集合中的元素必是 3（其他六個都不滿足）。完整性來自 X 只有這七個狀態。", "en": "Both inclusions of the set equality: 3 is in the set (T(3)=3), and every element of the set is 3 (the other six fail). Completeness comes from X having exactly these seven states." } }, "leaf_basis": "case_analysis + set_extensionality + set_builder" }, { "id": "S11", "uses": [ "G2", "S10" ], "before_tex": "3x+1\\equiv x\\pmod 7", "after_tex": "x\\equiv 3\\pmod 7", "relation": "equivalent", "rule": "mod7_arith", "conditions": [ "4\\cdot 2\\equiv 1\\pmod 7" ], "explanation": { "zh-Hant": "獨立的代數交叉核對。對 x∈X，x 就是自己的餘數，所以 T(x)=x 等價於 3x+1≡x (mod 7)。兩邊同減 x+1 得 2x≡−1≡6；兩邊同乘 4（因 4·2≡1）得 x≡24≡3。X 中這個同餘類的唯一代表是 3，與 S10 的枚舉一致。", "en": "An independent algebraic cross-check. For x∈X, x is its own remainder, so T(x)=x is equivalent to 3x+1≡x (mod 7). Subtracting x+1 from both sides gives 2x≡−1≡6; multiplying both sides by 4 (since 4·2≡1) gives x≡24≡3. The unique representative of this class in X is 3, agreeing with the enumeration of S10." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mod7_arith", "eq_congruence" ], "explanation": { "zh-Hant": "乘 4 與乘 2 在模 7 下互為反運算（8≡1），減 x+1 與加 x+1 互為反運算，所以每一步雙向成立。", "en": "Multiplying by 4 and by 2 undo each other modulo 7 (8≡1), and subtracting and adding x+1 undo each other, so each move holds both ways." } }, "leaf_basis": "mod7_arith + eq_congruence" }, { "id": "S12", "uses": [ "S10", "G2" ], "before_tex": "y_0=3,\\ y_{n+1}=T(y_n)", "after_tex": "\\forall n:\\ y_n=3", "relation": "implies", "rule": "nat_induction", "conditions": [], "explanation": { "zh-Hant": "從固定點出發的軌道（另記為 y_n）是常數：基底 y_0=3；若 y_n=3，則 y_(n+1)=T(3)=3。依歸納對所有 n 成立。", "en": "The orbit starting at the fixed point (written y_n) is constant: base y_0=3; if y_n=3 then y_(n+1)=T(3)=3. By induction this holds for all n." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S10", "nat_induction", "eq_replace" ], "explanation": { "zh-Hant": "只需要 T(3)=3。這不表示其他軌道會趨向 3（見 S15）。", "en": "It needs only T(3)=3. It does not mean that other orbits approach 3 (see S15)." } }, "leaf_basis": "nat_induction + eq_replace" }, { "id": "S13", "uses": [ "S03", "G4", "mko_dds_props" ], "before_tex": "T(A),\\ A=\\{0,1,2,4,5,6\\}", "after_tex": "T(A)=\\{1,4,0,6,2,5\\}=A\\subseteq A", "relation": "implies", "rule": "mko_dds_props", "conditions": [], "explanation": { "zh-Hant": "逐元素取像：T(0)=1，T(1)=4，T(2)=0，T(4)=6，T(5)=2，T(6)=5，所以 T(A)={1,4,0,6,2,5}，與 A 元素相同。每個像都在 A 中，T(A)⊆A，A 是正向不變子集。", "en": "Take images element by element: T(0)=1, T(1)=4, T(2)=0, T(4)=6, T(5)=2, T(6)=5, so T(A)={1,4,0,6,2,5}, which has the same elements as A. Every image is in A, so T(A)⊆A and A is forward invariant." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "image_preimage_def", "subset_def", "mko_dds_props" ], "explanation": { "zh-Hant": "需要 A 的六個元素全部檢查過，這裡確實全部列出。只檢查部分元素不足以推出包含關係。", "en": "It needs all six elements of A to be checked, and all are listed here. Checking only some elements would not give the inclusion." } }, "leaf_basis": "image_preimage_def + subset_def + set_extensionality" }, { "id": "S14", "uses": [ "S03", "G4", "mko_dds_props" ], "before_tex": "T(1)=4,\\ B=\\{0,1\\}", "after_tex": "4\\notin B\\ \\Rightarrow\\ T(B)\\not\\subseteq B", "relation": "implies", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "正向不變要求 B 的每個元素的像都在 B 中。1∈B 但 T(1)=4∉B，找到一個反例就否定了「對所有元素」，所以 B 不是正向不變的；T(0)=1∈B 不能挽救。", "en": "Forward invariance requires the image of every element of B to lie in B. 1∈B but T(1)=4∉B; one counterexample negates the 'for every element', so B is not forward invariant, and T(0)=1∈B does not rescue it." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "quantifier_negation", "subset_def" ], "explanation": { "zh-Hant": "否定全稱只需要一個反例 x=1；這裡不宣稱 B 中所有元素都會離開 B。", "en": "Negating a universal statement needs one counterexample, x=1; it is not claimed that every element of B leaves B." } }, "leaf_basis": "quantifier_negation + subset_def" }, { "id": "S15", "uses": [ "S13", "G3", "G4" ], "before_tex": "x_0=0\\in A,\\ T(A)\\subseteq A", "after_tex": "\\forall n:\\ x_n\\in A,\\ \\text{hence }x_n\\neq 3", "relation": "implies", "rule": "nat_induction", "conditions": [], "explanation": { "zh-Hant": "基底：x_0=0∈A。歸納步：若 x_n∈A，則 x_(n+1)=T(x_n)∈T(A)⊆A。所以每個 x_n 都在 A 中；而 3∉A，故對每個 n 有 x_n≠3：從 0 出發的軌道永遠不會到達固定點 3。這是對所有 n 的證明，不是觀察有限前綴。", "en": "Base: x_0=0∈A. Step: if x_n∈A then x_(n+1)=T(x_n)∈T(A)⊆A. So every x_n lies in A, and since 3∉A, x_n≠3 for every n: the orbit from 0 never reaches the fixed point 3. This is a proof for all n, not an observation of a finite prefix." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S13", "nat_induction", "image_preimage_def" ], "explanation": { "zh-Hant": "需要初始狀態在 A 中與 A 的正向不變性；兩者缺一，結論都不成立。", "en": "It needs the initial state in A and the forward invariance of A; without either, the conclusion fails." } }, "leaf_basis": "nat_induction + image_preimage_def + subset_def" }, { "id": "S16", "uses": [ "S06", "S09", "S10", "S13", "S14", "S15" ], "before_tex": "\\text{proved facts about }(X,T)", "after_tex": "x_{n+6}=x_n,\\ x_{2026}=5,\\ \\mathrm{Fix}(T)=\\{3\\},\\ T(A)\\subseteq A,\\ T(B)\\not\\subseteq B", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "合併結論：從 0 出發的軌道週期為 6 且 x_2026=5；唯一固定點是 3；A 正向不變而 B 不是；軌道永不到達 3。全部是關於這個有限系統的精確陳述。", "en": "Combine the conclusions: the orbit from 0 repeats with period 6 and x_2026=5; the only fixed point is 3; A is forward invariant and B is not; the orbit never reaches 3. All are exact statements about this finite system." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S06", "S09", "S10", "S13", "S14", "S15", "and_intro_elim" ], "explanation": { "zh-Hant": "目標 G：本稿目標列出的各項判定。足夠條件 H：S06、S09、S10、S13、S14、S15 各自已證（H⇒G 由合取引入）。H 由給定條件正向推出，沒有把 G 當作前提。", "en": "Goal G: the determinations listed in the goal. Sufficient condition H: S06, S09, S10, S13, S14 and S15, each already proved (H⇒G by conjunction introduction). H is derived forwards from the givens; G is never used as a premise." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "對 X={0,1,2,3,4,5,6} 與 T(x)=(3x+1) mod 7：從 0 出發的軌道為 0,1,4,6,5,2,0，並由歸納證明對所有 n 有 x_(n+6)=x_n，因此 x_2026=x_4=5；完整枚舉顯示唯一固定點是 3；A={0,1,2,4,5,6} 正向不變，B={0,1} 因 T(1)=4 而不是；從 0 出發的軌道永遠留在 A 中，不會到達 3。", "en": "For X={0,1,2,3,4,5,6} and T(x)=(3x+1) mod 7: the orbit from 0 is 0,1,4,6,5,2,0, and induction proves x_(n+6)=x_n for all n, so x_2026=x_4=5; complete enumeration shows that the only fixed point is 3; A={0,1,2,4,5,6} is forward invariant, while B={0,1} is not because T(1)=4; the orbit from 0 stays in A forever and never reaches 3." }, "negative_controls": [ { "claim": { "zh-Hant": "x_0 到 x_5 互不相同，所以這條軌道永不重複", "en": "x_0 to x_5 are pairwise distinct, so this orbit never repeats" }, "expected": "reject", "reason": { "zh-Hant": "x_6=T(2)=0=x_0（S05），且之後每 6 步重複（S06）。", "en": "x_6=T(2)=0=x_0 (S05), and from then on it repeats every 6 steps (S06)." } }, { "claim": { "zh-Hant": "固定點 3 吸引其他狀態，從 0 出發的軌道最後會停在 3", "en": "The fixed point 3 attracts the other states, so the orbit from 0 eventually stays at 3" }, "expected": "reject", "reason": { "zh-Hant": "S15 證明對所有 n 有 x_n∈A，而 3∉A，所以軌道永不到達 3。", "en": "S15 proves x_n∈A for all n, and 3∉A, so the orbit never reaches 3." } }, { "claim": { "zh-Hant": "B={0,1} 正向不變，因為 T(0)=1∈B", "en": "B={0,1} is forward invariant because T(0)=1∈B" }, "expected": "reject", "reason": { "zh-Hant": "正向不變要求每個元素；T(1)=4∉B（S14）。", "en": "Forward invariance concerns every element; T(1)=4∉B (S14)." } }, { "claim": { "zh-Hant": "只要知道 X={0,1,2,3,4,5,6} 與 x_0=0，就能決定 x_2026", "en": "Knowing X={0,1,2,3,4,5,6} and x_0=0 is enough to determine x_2026" }, "expected": "reject", "reason": { "zh-Hant": "換成 T′(x)=(x+1) mod 7 時 x_2026=2026 mod 7=3（N11），與本系統的 5 不同。", "en": "With T′(x)=(x+1) mod 7 one gets x_2026=2026 mod 7=3 (N11), unlike 5 for this system." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "function_def", "kind": "prerequisite", "statement": { "zh-Hant": "函數：f:A→B 對每個 a∈A 指定恰好一個 f(a)∈B。兩函數相等 ⇔ 定義域、對應域相同且對每個 a 都有相同值。", "en": "Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a." } }, { "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