{"schema_version":"ocme-site-results-v1","locale":"zh-Hant","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"page":"4"},"kind":"guides","total":299,"pages":13,"page":4,"entries":[{"id":"derivation-mko-derivative-rules-01","title":{"zh-Hant":"微分法則：以展開求導核對乘積與鏈式規則","en":"Derivative rules: checking the product and chain rules against differentiating the expansion"},"summary":{"zh-Hant":"以升冪係數列表 f=[2,0,1]、g=[1,3]（即 f(x)=2+x²、g(x)=1+3x）：先由導數定義證明三次以下多項式的求導公式；再以兩條獨立路徑求 (fg)' 與 (f∘g)'——展開後求導，以及使用乘積、鏈式規則——並確認兩者對每個 x 相同；同時核對和與純量倍的規則，指出 f'g'、漏乘內導數、外導數取錯點、反讀係數等錯誤，最後求 x=1 的值。","en":"With the ascending coefficient lists f=[2,0,1] and g=[1,3] (that is f(x)=2+x² and g(x)=1+3x): first prove from the definition of the derivative the differentiation formula for polynomials of degree at most three; then find (fg)' and (f∘g)' along two independent routes, differentiating the expansion and applying the product and chain rules, and confirm they agree at every x; also check the sum and scalar rules, expose the errors f'g', a missing inner derivative, an outer derivative at the wrong point and reversed coefficient reading, and evaluate at x=1."},"mko_ids":["mko-derivative-rules"]},{"id":"derivation-mko-antiderivative-01","title":{"zh-Hant":"原函數：多項式代表原函數、常數族與候選驗收","en":"Antiderivatives: the representative polynomial antiderivative, the constant family and checking candidates"},"summary":{"zh-Hant":"對 f=[3,−4,6]（f(x)=3−4x+6x²）：由定義證明四次以下多項式的求導公式；構造代表原函數 F0=[0,3,−2,2] 並驗證 F0'=f；證明 F0+C 對每個實數 C 都是原函數，且在至多五個係數的多項式中每個原函數都是 F0+C；驗收候選 G=[7,3,−2,2]（接受）、H=[0,3,−4,2]（拒絕）、K=[0,3,−3,8/3]（兩個測點相符仍拒絕）；求 F0(2)+C（C=−5），並說明原函數不是反函數。","en":"For f=[3,−4,6] (f(x)=3−4x+6x²): prove from the definition the derivative formula for polynomials of degree at most four; build the representative antiderivative F0=[0,3,−2,2] and verify F0'=f; prove that F0+C is an antiderivative for every real C and that, among polynomials with at most five coefficients, every antiderivative is F0+C; check the candidates G=[7,3,−2,2] (accept), H=[0,3,−4,2] (reject) and K=[0,3,−3,8/3] (reject although it matches at two test points); evaluate F0(2)+C with C=−5; and explain why an antiderivative is not an inverse function."},"mko_ids":["mko-antiderivative"]},{"id":"derivation-mko-definite-integral-01","title":{"zh-Hant":"定積分：以左端和的極限與矩公式計算 ∫₀³(x²−4)dx","en":"Definite integral: computing ∫₀³(x²−4)dx as a limit of left sums and by the moment formula"},"summary":{"zh-Hant":"設 f(x)=x²−4、a=0、b=3。(1) 只用 Riemann 定義（所有合法標記分割和在網格趨 0 時的共同極限）與 f 可積的前提，由 N 等分左端和 L_N 的極限求出 I(f,0,3)；(2) 用本 MKO 明列的矩公式獨立再算一次並核對；(3) 求反向端點的 I(f,3,0)；(4) 以 N=3、N=50 說明有限左端和不等於積分，並說明有號積分不是非負面積。","en":"Let f(x)=x²−4, a=0, b=3. (1) Using only the Riemann definition (the common limit of all valid tagged-partition sums as the mesh tends to zero) and the premise that f is integrable, find I(f,0,3) as the limit of the N-piece left sums L_N; (2) recompute it independently with the moment formula listed in this MKO and compare; (3) find the reversed integral I(f,3,0); (4) use N=3 and N=50 to show that a finite left sum is not the integral, and explain why the signed integral is not a nonnegative area."},"mko_ids":["mko-definite-integral"]},{"id":"derivation-mko-fundamental-theorem-calculus-01","title":{"zh-Hant":"微積分基本定理：由連續性證明 G′=f，再推出 ∫_a^b f=F(b)−F(a)","en":"Fundamental theorem of calculus: proving G′=f from continuity, then ∫_a^b f=F(b)−F(a)"},"summary":{"zh-Hant":"設 f:ℝ→ℝ 在全 ℝ 連續，a∈ℝ，G(x)=∫_a^x f(t)dt 為有向積分。(1) 證明 G 在每一點可微且 G′(x)=f(x)；(2) 若 F:ℝ→ℝ 可微且 F′=f，證明對任意實數 a、b 有 ∫_a^b f(t)dt=F(b)−F(a)；(3) 以 f(t)=6t²−2t+1 與候選 F(x)=2x³−x²+x+1/2 算出 ∫_2^{−1} f(t)dt，並說明 G(2)=0 不推出 G′(2)=0、上下限在 x=0 重合也不推出導數為 0。","en":"Let f:ℝ→ℝ be continuous on all of ℝ, a∈ℝ, and G(x)=∫_a^x f(t)dt an oriented integral. (1) Prove that G is differentiable at every point with G′(x)=f(x); (2) if F:ℝ→ℝ is differentiable with F′=f, prove ∫_a^b f(t)dt=F(b)−F(a) for arbitrary real a, b; (3) with f(t)=6t²−2t+1 and the candidate F(x)=2x³−x²+x+1/2, compute ∫_2^{−1} f(t)dt, and explain why G(2)=0 does not give G′(2)=0 and why bounds coinciding at x=0 do not give a zero derivative."},"mko_ids":["mko-fundamental-theorem-calculus"]},{"id":"derivation-mko-tree-01","title":{"zh-Hant":"樹：判定一個 6 頂點圖是樹，求葉、唯一道路與加邊後的唯一環","en":"Trees: deciding that a 6-vertex graph is a tree, with its leaves, a unique path and the unique cycle after adding an edge"},"summary":{"zh-Hant":"設 G=(V,E)，V={1,2,3,4,5,6}，E={{1,2},{2,3},{2,4},{4,5},{4,6}}。(1) 依 MKO 定義證明 G 是樹（V 非空、連通、無簡單環）；(2) 列出葉；(3) 求從 3 到 6 的道路並證明它唯一；(4) 加入邊 {3,6}，寫出所形成的簡單環並證明它是唯一的簡單環；(5) 對照：三角形加一個孤立點雖有 |E|=|V|−1 卻不是樹；單點無邊圖是樹但沒有葉。","en":"Let G=(V,E) with V={1,2,3,4,5,6} and E={{1,2},{2,3},{2,4},{4,5},{4,6}}. (1) Prove from the MKO's definition that G is a tree (V nonempty, connected, no simple cycle); (2) list its leaves; (3) find the path from 3 to 6 and prove it is unique; (4) add the edge {3,6}, write the simple cycle it creates and prove it is the only simple cycle; (5) contrast: a triangle plus an isolated vertex has |E|=|V|−1 but is not a tree; a single vertex without edges is a tree with no leaf."},"mko_ids":["mko-tree"]},{"id":"derivation-mko-topological-space-01","title":{"zh-Hant":"拓撲空間：完整驗證三點集合上的一個鏈狀開集族，並辨認非拓撲","en":"Topological space: fully verifying a chain of open sets on a three-point set, and recognising non-topologies"},"summary":{"zh-Hant":"設 X={a,b,c}（a、b、c 相異）。(1) 證明 τ₁={∅,{a},{a,b},X} 是 X 上的拓撲，涵蓋全部 16 個子族（含空族）；(2) 證明 τ₂={∅,{a},{b},X} 不是拓撲，並指出它通過了哪些檢查；(3) 比較 τ₁ 與不可分拓撲、離散拓撲的粗細；(4) 在 ℕ 上，證明 τ_fin={ℕ 的有限子集}∪{ℕ} 對兩兩聯集與有限交集封閉，卻不是拓撲。","en":"Let X={a,b,c} with a, b, c distinct. (1) Prove that τ₁={∅,{a},{a,b},X} is a topology on X, covering all 16 subfamilies (including the empty one); (2) prove that τ₂={∅,{a},{b},X} is not a topology, and say which checks it does pass; (3) compare τ₁ with the indiscrete and discrete topologies; (4) on ℕ, prove that τ_fin={finite subsets of ℕ}∪{ℕ} is closed under pairwise unions and finite intersections but is not a topology."},"mko_ids":["mko-topological-space"]},{"id":"derivation-mko-natural-divisibility-01","title":{"zh-Hant":"自然數整除：以乘數見證判定、處理 0 的情形並證明遞移性","en":"Natural-number divisibility: deciding by a multiplier witness, handling zero, and proving transitivity"},"summary":{"zh-Hant":"依 MKO 定義（a∣b ⇔ 存在自然數 k 使 b=a·k）：(1) 證明 6∣84；(2) 證明 6∤87；(3) 證明 4∤10，雖然 10/4=5/2 是一個數；(4) 證明每個 a 都整除 0；(5) 證明 0∣b ⇔ b=0，並說明 0∣0 的見證不唯一、0∤5；(6) 證明 a>0 時見證至多一個（若 a∣b，則既有見證唯一）；(7) 證明整除的自反性與遞移性，並由 3∣6、6∣84 求出 3∣84 的見證。","en":"From the MKO's definition (a∣b ⇔ there is a natural k with b=a·k): (1) prove 6∣84; (2) prove 6∤87; (3) prove 4∤10, although 10/4=5/2 is a number; (4) prove that every a divides 0; (5) prove 0∣b ⇔ b=0, and explain why the witness for 0∣0 is not unique and why 0∤5; (6) prove that for a>0 there is at most one witness (and when a∣b that existing witness is unique); (7) prove reflexivity and transitivity of divisibility, and from 3∣6 and 6∣84 find a witness for 3∣84."},"mko_ids":["mko-natural-divisibility"]},{"id":"derivation-mko-gcd-coprime-01","title":{"zh-Hant":"最大公因數與互質：以整除刻畫驗證歐幾里得算法","en":"Greatest common divisor and coprimality: checking Euclid's algorithm against the divisibility characterization"},"summary":{"zh-Hant":"依 MKO 的整除刻畫（d∣a、d∣b，且每個公因數 c 都整除 d）：(1) 證明滿足刻畫的 d 至多一個；(2) 證明歐幾里得算法每一步保持公因數集合且必會停止，並以它算出 gcd(84,36)=12，再直接驗證 12 滿足刻畫；(3) 說明公因數 6 不是 gcd；(4) 算出 gcd(8,15)=1，說明 8 與 15 互質但都不是質數；(5) 處理含 0 的情形 gcd(0,1)=1、gcd(0,6)=6、gcd(0,0)=0。","en":"From the MKO's divisibility characterization (d∣a, d∣b, and every common divisor c divides d): (1) prove that at most one d satisfies it; (2) prove that each step of Euclid's algorithm preserves the set of common divisors and that the algorithm stops, use it to compute gcd(84,36)=12, and verify directly that 12 satisfies the characterization; (3) explain why the common divisor 6 is not the gcd; (4) compute gcd(8,15)=1 and explain that 8 and 15 are coprime although neither is prime; (5) treat the zero cases gcd(0,1)=1, gcd(0,6)=6 and gcd(0,0)=0."},"mko_ids":["mko-gcd-coprime"]},{"id":"derivation-mko-prime-number-01","title":{"zh-Hant":"自然數質數：證明 97 是質數、91 是合數，並排除 0 與 1","en":"Prime natural numbers: proving 97 prime and 91 composite, and excluding 0 and 1"},"summary":{"zh-Hant":"依 MKO 定義（p 為質數 ⇔ 2≤p 且 p 的每個自然數因數 d 都滿足 d=1 或 d=p）：(1) 說明 0 與 1 不是質數；(2) 證明 2 是質數；(3) 以見證 7 證明 91 是合數，並說明只測 2、3、5 會誤判；(4) 證明「若 n=dk 且 d、k≥2，則 min(d,k)²≤n」，據此只需檢查 2 到 9，證明 97 是質數；(5) 證明 2 是唯一的偶質數。","en":"From the MKO's definition (p is prime ⇔ 2≤p and every natural divisor d of p satisfies d=1 or d=p): (1) explain why 0 and 1 are not prime; (2) prove that 2 is prime; (3) prove that 91 is composite with the witness 7, and show that testing only 2, 3, 5 misjudges it; (4) prove 'if n=dk with d, k≥2 then min(d,k)²≤n', and use it to prove that 97 is prime by checking only 2 to 9; (5) prove that 2 is the only even prime."},"mko_ids":["mko-prime-number"]},{"id":"derivation-mko-modular-congruence-01","title":{"zh-Hant":"自然數模同餘：餘數定義、見證形式、運算保持與取消的限制","en":"Congruence modulo a positive modulus: the remainder definition, the witness form, preservation by operations and the limits of cancellation"},"summary":{"zh-Hant":"依 MKO 定義（n>0，a≡b (mod n) ⇔ a mod n=b mod n）：(1) 判定 100≡23 (mod 7)；(2) 證明它等價於見證形式「存在自然數 k、l 使 a+nk=b+nl」，並給出 100 與 23 的見證；(3) 說明同餘是等價關係；(4) 證明加法與乘法保持同餘，並由 15≡1 (mod 7) 得 115≡24、1500≡23 (mod 7)；(5) 以 2·3≡2·0 (mod 6) 說明不能無條件取消公因子；(6) 說明不同模數不能串接，並說明模數 0 被排除。","en":"From the MKO's definition (n>0, a≡b (mod n) ⇔ a mod n=b mod n): (1) decide 100≡23 (mod 7); (2) prove that this is equivalent to the witness form 'there are naturals k, l with a+nk=b+nl', and give witnesses for 100 and 23; (3) explain why congruence is an equivalence relation; (4) prove that addition and multiplication preserve congruence, and from 15≡1 (mod 7) obtain 115≡24 and 1500≡23 (mod 7); (5) use 2·3≡2·0 (mod 6) to show that a common factor cannot be cancelled unconditionally; (6) explain why different moduli cannot be chained, and why the modulus 0 is excluded."},"mko_ids":["mko-modular-congruence"]},{"id":"derivation-mko-decision-problem-01","title":{"zh-Hant":"決定問題：在明示有限輸入集上證明試除程序是 decider","en":"Decision problem: proving that a trial-division procedure is a decider on an explicit finite input set"},"summary":{"zh-Hant":"取輸入集合 X={0,1,2,3,4,5,6,7,8,9}、規格 P(x)：「x 是質數」與 yes-instance 集合 Y={x∈X ∣ P(x)}。在明示的 while 程式計算模型 W 中，證明試除程序 D1 對每個 x∈X 都終止，且 D1(x)=true 當且僅當 x∈Y，因此 D1 是 (X,Y) 的 decider；再說明只在 yes-instance 上停止的 D2 與只看奇偶的 D3 都不是 decider。","en":"Take the input set X={0,1,2,3,4,5,6,7,8,9}, the specification P(x): 'x is prime', and the yes-instance set Y={x∈X ∣ P(x)}. In the explicit while-program model W, prove that the trial-division procedure D1 terminates on every x∈X and that D1(x)=true exactly when x∈Y, so D1 is a decider for (X,Y); then show that D2, which halts only on yes-instances, and D3, which tests only parity, are not deciders."},"mko_ids":["mko-decision-problem"]},{"id":"derivation-mko-lean-formal-statement-01","title":{"zh-Hant":"證明助理中的形式陳述：逐層讀懂一個 Lean 陳述並區分「是陳述」與「已證明」","en":"Formal statements in a proof assistant: reading a Lean statement layer by layer and separating 'is a statement' from 'is proved'"},"summary":{"zh-Hant":"以源碼字串 s：`∀ (a b : Nat), a ≤ b → a < b + 1` 為教學對象，在一個明示固定但未實際執行的環境 E 中逐層說明：它的約束變數與型別、假設與結論、若 elaboration 成功應得到的核心項 S、為何 S 的型別會是 Prop、一個證明項必須建立什麼；再給出其數學內容的非正式證明，並以 s′：`∀ (a : Nat), a - 1 < a`（是 Prop 但為假）與 s″：`∀ (a : Int), a - 1 < a`（為真）對照。全程未執行 Lean，不宣稱任何內容可解析、可 elaboration、已編譯或已被 Lean 證明。","en":"Take the source string s: `∀ (a b : Nat), a ≤ b → a < b + 1` as a teaching object and, in an environment E that is fixed explicitly but never executed, explain layer by layer: its bound variables and their types, its hypothesis and conclusion, the core term S that elaboration would have to produce if it succeeded, why S would have type Prop, and what a proof term would have to establish; then give an informal proof of its mathematical content and contrast it with s′: `∀ (a : Nat), a - 1 < a` (a Prop, but false) and s″: `∀ (a : Int), a - 1 < a` (true). Lean is not run at any point, and nothing here claims that any text parses, elaborates, compiles or is proved by Lean."},"mko_ids":["mko-lean-formal-statement"]},{"id":"derivation-mko-open-closed-set-01","title":{"zh-Hant":"開集與閉集：在四點拓撲中完整分類十六個子集","en":"Open and closed sets: classifying all sixteen subsets of a four-point topology"},"summary":{"zh-Hant":"設 X={1,2,3,4}，τ={∅,{1},{1,2},{3,4},{1,3,4},X}。先驗證 τ 是 X 上的拓撲，再列出全部閉集，並把 X 的十六個子集分成「既開又閉」「只開」「只閉」「既不開也不閉」四類；最後證明在任何拓撲空間中 ∅ 與 X 都既開又閉，以及兩個閉集的聯集仍是閉集。","en":"Let X={1,2,3,4} and τ={∅,{1},{1,2},{3,4},{1,3,4},X}. First verify that τ is a topology on X, then list all closed sets and sort the sixteen subsets of X into four classes: clopen, open only, closed only, and neither; finally prove that in every topological space ∅ and X are clopen, and that the union of two closed sets is closed."},"mko_ids":["mko-open-closed-set"]},{"id":"derivation-mko-neighborhood-01","title":{"zh-Hant":"鄰域：在四點拓撲中搜尋開集見證並列出全部鄰域","en":"Neighborhood: searching for open witnesses and listing all neighborhoods in a four-point topology"},"summary":{"zh-Hant":"設 X={1,2,3,4}，τ={∅,{1},{1,2},{3,4},{1,3,4},X}。判定 N1={1,2,3} 與 N2={2,3,4} 是否為點 2 的鄰域；證明在有限拓撲中，點 x 的鄰域恰是包含 O_x（含 x 的全部開集之交）的子集，據此列出點 1、2、3、4 的全部鄰域；證明鄰域向上封閉；並說明同一集合在離散拓撲下的判定可能不同。","en":"Let X={1,2,3,4} and τ={∅,{1},{1,2},{3,4},{1,3,4},X}. Decide whether N1={1,2,3} and N2={2,3,4} are neighborhoods of the point 2; prove that in a finite topology the neighborhoods of x are exactly the subsets containing O_x (the intersection of all open sets containing x), and use this to list all neighborhoods of the points 1, 2, 3 and 4; prove that neighborhoods are closed upwards; and show that the verdict for the same set can change under the discrete topology."},"mko_ids":["mko-neighborhood"]},{"id":"derivation-mko-topological-continuity-01","title":{"zh-Hant":"拓撲連續映射：在有限空間中用逆像判定連續，並與開映射區分","en":"Continuous maps between topological spaces: deciding continuity by preimages in finite spaces and separating it from openness"},"summary":{"zh-Hant":"設 X={1,2,3}，τX={∅,{1},{1,2},X}；Y={a,b}，τY={∅,{a},Y}。對三個以完整表格給出的映射 f（1↦a、2↦a、3↦b）、g（1↦a、2↦b、3↦a）、c（全部 ↦b），用逆像條件判定是否連續，並判定是否為開映射；證明連續映射的合成仍連續並以 q:Y→Z 驗證；最後數出 X→Y 的全部連續映射。","en":"Let X={1,2,3} with τX={∅,{1},{1,2},X}, and Y={a,b} with τY={∅,{a},Y}. For three maps given by complete tables, f (1↦a, 2↦a, 3↦b), g (1↦a, 2↦b, 3↦a) and c (everything ↦b), decide continuity by the preimage condition and decide whether each is an open map; prove that a composite of continuous maps is continuous and check it with q:Y→Z; finally count all continuous maps X→Y."},"mko_ids":["mko-topological-continuity"]},{"id":"derivation-mko-computational-model-01","title":{"zh-Hant":"抽象計算模型：在七狀態模型中截斷停止邊、計算可達集並區分停止、卡住與不終止","en":"Abstract computational model: truncating halt edges, computing reachability and separating halting, stuck and non-terminating runs in a seven-state model"},"summary":{"zh-Hant":"給定有限模型 (S,Init,Step,Halt)：S={s0,s1,s2,s3,s4,s5,s6}，Init={s0}，Halt={s3}，原始邊資料 {(s0,s1),(s0,s2),(s1,s3),(s1,s5),(s2,s4),(s4,s2),(s3,s6)}。依 MKO 規則截斷停止狀態的外出邊，以不動點迭代算出可達集，找出停止狀態、卡住狀態與可達循環，給出合法的停止、卡住與無窮執行，排除三條不合法的 trace，並說明為何尚未窮盡可達圖的有限深度探索，不能僅憑未見反例就斷言所有極大執行終止。","en":"Given the finite model (S,Init,Step,Halt) with S={s0,s1,s2,s3,s4,s5,s6}, Init={s0}, Halt={s3} and raw edge data {(s0,s1),(s0,s2),(s1,s3),(s1,s5),(s2,s4),(s4,s2),(s3,s6)}: truncate edges out of halt states as the MKO requires, compute the reachable set by fixpoint iteration, identify halt states, stuck states and reachable cycles, exhibit legal halting, stuck and infinite executions, rule out three illegal traces, and explain why depth-bounded exploration that has not exhausted the reachable graph cannot certify termination of all maximal executions merely from the absence of an observed counterexample."},"mko_ids":["mko-computational-model"]},{"id":"derivation-mko-algorithm-correctness-01","title":{"zh-Hant":"算法正確性：非確定性相鄰交換排序的部分正確、終止與總正確","en":"Algorithm correctness: partial correctness, termination and total correctness of nondeterministic adjacent-swap sorting"},"summary":{"zh-Hant":"在有限狀態模型 M 中，狀態是 (1,2,3) 的六個排列，初始狀態 (3,2,1)，每步可非確定性地交換任一個相鄰逆序對，Halt={(1,2,3)}。以後置條件「遞增排列且為輸入的重排」證明 M 的部分正確性、終止性與總正確性：用排名函數 inv（逆序數）證明每條執行至多 3 步，且每條極大執行停在 Halt。再以變體 M′（允許交換任何相鄰對）說明一條成功路徑與部分正確都不足以得到總正確，以變體 M″（錯誤的停止條件）說明終止不足以得到正確。","en":"In a finite-state model M whose states are the six permutations of (1,2,3), with initial state (3,2,1), each step may nondeterministically swap any adjacent inverted pair, and Halt={(1,2,3)}. With the postcondition 'increasing, and a rearrangement of the input', prove partial correctness, termination and total correctness of M, using the ranking function inv (the number of inversions) to show every execution takes at most 3 steps and every maximal execution stops in Halt. Then use the variant M′ (any adjacent pair may be swapped) to show that one successful path and partial correctness do not give total correctness, and the variant M″ (a wrong halting condition) to show that termination does not give correctness."},"mko_ids":["mko-algorithm-correctness"]},{"id":"derivation-mko-asymptotic-resource-bound-01","title":{"zh-Hant":"漸近資源界：證明 3n²+5n+2∈Θ(n²) 並排除 Θ(n)","en":"Asymptotic resource bounds: proving 3n²+5n+2∈Θ(n²) and ruling out Θ(n)"},"summary":{"zh-Hant":"設成本函數 f(n)=3n²+5n+2（n 從 0 起），比較函數 g(n)=n² 與 h(n)=n。以明示的全尾段 witness 證明 f∈O(g)（c=10、N=1）、f∈Ω(g)（c=3、N=0），故 f∈Θ(g)；說明 N=0 對 O 不可行、c=3 對 O 不可行、另一組 witness c=4、N=6 也可行；再證明 f∉O(h)，因此 f∈Ω(h) 但 f∉Θ(h)。","en":"Let the cost function be f(n)=3n²+5n+2 (n starting at 0), with comparison functions g(n)=n² and h(n)=n. With explicit full-tail witnesses prove f∈O(g) (c=10, N=1) and f∈Ω(g) (c=3, N=0), hence f∈Θ(g); show that N=0 cannot work for O, that c=3 cannot work for O, and that another witness c=4, N=6 also works; then prove f∉O(h), so f∈Ω(h) but f∉Θ(h)."},"mko_ids":["mko-asymptotic-resource-bound"]},{"id":"derivation-mko-lean-proof-term-01","title":{"zh-Hant":"證明項：逐部分解讀 λh. ⟨snd h, fst h⟩ 為何是 P∧Q→Q∧P 的證明項","en":"Proof term: reading λh. ⟨snd h, fst h⟩ part by part as a proof term for P∧Q→Q∧P"},"summary":{"zh-Hant":"在一個明示的合成小型命題型別系統 ST 中（對應 MKO 假設 a2），取相異命題原子 P、Q：逐部分說明項 t=λh:P∧Q. ⟨snd h, fst h⟩ 的每一部分建立了什麼、具有什麼型別，推導判斷 ⊢ t : P∧Q→Q∧P，並說明 t 的型別為什麼與命題 S=P∧Q→Q∧P 相符；再證明外觀相近的 t′=λh:P∧Q. ⟨fst h, snd h⟩ 不是 S 的證明項。最後區分「ST 中的型別推導」「Lean kernel 接受」與「已保存的 proof-term receipt」三者。","en":"In an explicit synthetic small propositional type system ST (matching MKO assumption a2), with distinct propositional atoms P and Q: explain part by part what each piece of the term t=λh:P∧Q. ⟨snd h, fst h⟩ establishes and what type it has, derive the judgment ⊢ t : P∧Q→Q∧P, and explain why the type of t matches the proposition S=P∧Q→Q∧P; then prove that the similar-looking t′=λh:P∧Q. ⟨fst h, snd h⟩ is not a proof term for S. Finally separate three things: a typing derivation in ST, acceptance by the Lean kernel, and a saved proof-term receipt."},"mko_ids":["mko-lean-proof-term"]},{"id":"derivation-mko-discrete-dynamical-system-01","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"},"summary":{"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."},"mko_ids":["mko-discrete-dynamical-system"]},{"id":"derivation-mko-ordinary-differential-equation-01","title":{"zh-Hant":"常微分方程：判定階數並核對候選解 y=x² 與 y=x²+x","en":"Ordinary differential equation: finding the order and checking the candidates y=x² and y=x²+x"},"summary":{"zh-Hant":"在區間 I=ℝ 上考慮常微分方程 y″−2y′+y=x²−4x+2。(1) 判定它的階數；(2) 核對 y₁(x)=x² 在整個 ℝ 上是解（殘差恆為零）；(3) 證明 y₂(x)=x²+x 的殘差是 x−2，只在 x=2 為零，所以它不是 ℝ 上的解；(4) 核對初值 y(0)=0、y′(0)=0，並引用二階線性方程的存在唯一定理（含其假設）判定 y₁ 是此初值問題的唯一解；(5) 說明沒有初值時解不唯一。","en":"On the interval I=ℝ consider the ordinary differential equation y″−2y′+y=x²−4x+2. (1) Determine its order; (2) check that y₁(x)=x² is a solution on all of ℝ (its residual is identically zero); (3) prove that y₂(x)=x²+x has residual x−2, which vanishes only at x=2, so it is not a solution on ℝ; (4) check the initial values y(0)=0, y′(0)=0 and, citing the existence and uniqueness theorem for second-order linear equations with its hypotheses, conclude that y₁ is the unique solution of this initial value problem; (5) show that without initial values the solution is not unique."},"mko_ids":["mko-ordinary-differential-equation"]},{"id":"derivation-mko-numerical-approximation-error-01","title":{"zh-Hant":"數值近似與誤差：以 0.33 近似 1/3 的絕對誤差、相對誤差與容許範圍","en":"Numerical approximation and error: absolute error, relative error and tolerance for 0.33 approximating 1/3"},"summary":{"zh-Hant":"以精確有理數計算：近似值 a=33/100 相對參照值 x=1/3 的絕對誤差與相對誤差；判定 a 是否在容許誤差 τ₁=1/250、τ₂=1/400、τ₃=1/300 之內（τ₃ 為端點情形）；證明 |a−x|≤τ ⇔ x−τ≤a≤x+τ；比較放大 1000 倍後的誤差；說明參照值為 0 時相對誤差沒有定義，以及誤差數值本身不能說明誤差來源。","en":"Using exact rationals: compute the absolute and relative error of the approximation a=33/100 to the reference x=1/3; decide whether a is within the tolerances τ₁=1/250, τ₂=1/400 and τ₃=1/300 (τ₃ is the endpoint case); prove |a−x|≤τ ⇔ x−τ≤a≤x+τ; compare the errors after scaling by 1000; explain why relative error is undefined for a zero reference and why an error value alone does not reveal the source of the error."},"mko_ids":["mko-numerical-approximation-error"]},{"id":"derivation-mko-optimization-problem-01","title":{"zh-Hant":"最佳化問題：在九個整數候選中先判可行、再比較目標值","en":"Optimization problem: feasibility first, then objective values, over nine integer candidates"},"summary":{"zh-Hant":"候選集合 C={0,1,2}² 共九點，限制 x+2y≥3，目標函數 f(x,y)=x+y。以「先判定可行性、再逐一比較」的掃描演算法求最小值，並以不變量證明結果：最優值 m=2，argmin={(0,2),(1,1)}（兩個平手）；說明 (0,0) 與 (2,0) 雖然目標值不大於 2 卻不能入選；求最大值；並說明同一限制在實數非負象限上的最小值為 3/2，其達到點 (0,3/2) 不在宣告的有限候選集合中。","en":"The candidate set C={0,1,2}² has nine points, the constraint is x+2y≥3 and the objective is f(x,y)=x+y. Minimize with a scan algorithm that first decides feasibility and then compares values one by one, and prove the result with an invariant: optimal value m=2, argmin={(0,2),(1,1)} (a tie of two); explain why (0,0) and (2,0) are excluded although their objective values do not exceed 2; find the maximum; and show that under the same constraint over the real nonnegative quadrant, the minimum is 3/2, attained at (0,3/2), which is not in the declared finite candidate set."},"mko_ids":["mko-optimization-problem"]},{"id":"derivation-mko-dimensional-analysis-01","title":{"zh-Hant":"量綱分析：以指數向量核對 ½mv²=mgh 並排除 v=2gh","en":"Dimensional analysis: checking ½mv²=mgh with exponent vectors and ruling out v=2gh"},"summary":{"zh-Hant":"以基本量綱 L、M、T 的整數指數向量 d(Q)=(a,b,c)：算出速度、加速度與兩種能量式的量綱；判定 ½mv²=mgh 量綱一致、v=2gh 不一致且無法以純數常數修正、v²=4gh 量綱一致卻在等加速度模型中錯誤；判定 m+v 不能相加；由 E=C·m^α v^β 解出 α=1、β=2；並說明量綱相同不代表單位換算因子相同。","en":"With integer exponent vectors d(Q)=(a,b,c) over the base dimensions L, M, T: compute the dimensions of speed, acceleration and two energy expressions; decide that ½mv²=mgh is dimensionally consistent, that v=2gh is not and cannot be repaired by a pure-number constant, and that v²=4gh is consistent but wrong in the constant-acceleration model; decide that m+v cannot be added; solve E=C·m^α v^β for α=1 and β=2; and explain that equal dimensions do not mean equal unit conversion factors."},"mko_ids":["mko-dimensional-analysis"]}],"title":"概念推導導讀"}