{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-counting-principle-01", "mko_id": "mko-counting-principle", "title": { "zh-Hant": "基本計數原理：數字互異的三位偶數有幾個", "en": "Basic counting principles: how many three-digit even numbers have distinct digits" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-counting-principle-01", "source": { "mko_id": "mko-counting-principle", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "0a0619ebbab98b96543b42abc4e66d927aeea705c65ee85b3a9badb6d3b3cf57", "english_sha256": "122425d1403554c17fc2a0e18e7b67e807bc50d412bad0250aa17710f6a34102" }, "mode": "worked_example", "audience_prerequisites": [ "finite_counting_basics", "cartesian_product_def", "injective_surjective_def", "set_operations_def", "numeric_evaluation" ], "goal": { "zh-Hant": "計算 100 到 999 之間、三個數字兩兩相異的偶數共有幾個。要求：只把互斥的類別相加；每次相乘之前，先確認每個階段的可選數對所有先前選擇都相同，不同時先分支。", "en": "Count the even integers from 100 to 999 whose three digits are pairwise distinct. Requirements: add only disjoint classes, and before each multiplication confirm that the number of options at every stage is the same for all earlier choices; when it is not, branch first." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "三位數是滿足 100≤n≤999 的整數；每個這樣的 n 恰有一種寫法 n=100d₁+10d₂+d₃，其中 d₁∈W⁺、d₂,d₃∈W（見 decimal_rep）。", "en": "A three-digit number is an integer n with 100≤n≤999; each such n has exactly one representation n=100d₁+10d₂+d₃ with d₁∈W⁺ and d₂,d₃∈W (see decimal_rep)." } }, { "id": "G2", "statement": { "zh-Hant": "所求集合 T={n∈ℤ ∣ 100≤n≤999，n 是偶數，且 n 的三個數字兩兩相異}。", "en": "The set to count is T={n∈ℤ ∣ 100≤n≤999, n is even, and the three digits of n are pairwise distinct}." } }, { "id": "G3", "statement": { "zh-Hant": "記號：[n]={k∈ℕ ∣ 1≤k≤n}；|S|=n 表示存在雙射 S→[n]。", "en": "Notation: [n]={k∈ℕ ∣ 1≤k≤n}; |S|=n means there is a bijection S→[n]." } } ], "steps": [ { "id": "S01", "uses": [ "G1", "G2", "decimal_rep" ], "before_tex": "T", "after_tex": "|T|=|D|,\\ D=\\{(d_1,d_2,d_3)\\mid d_1\\in W^{+},\\ d_2,d_3\\in W,\\ 100d_1+10d_2+d_3\\ \\text{even},\\ d_1,d_2,d_3\\ \\text{pairwise distinct}\\}", "relation": "implies", "rule": "decimal_rep", "conditions": [], "explanation": { "zh-Hant": "把每個 n∈T 對應到它的數字三元組 (d₁,d₂,d₃)。依 decimal_rep，這個三元組存在且唯一，所以對應是一個函數；反向以 (d₁,d₂,d₃)↦100d₁+10d₂+d₃ 還原 n，所以它是 T 到 D 的雙射。「偶數」與「數字相異」兩個條件原封不動地搬到 D。依 finite_counting_basics，有雙射的兩個有限集合元素個數相同，所以 |T|=|D|。", "en": "Send each n∈T to its digit triple (d₁,d₂,d₃). By decimal_rep this triple exists and is unique, so the assignment is a function; the map (d₁,d₂,d₃)↦100d₁+10d₂+d₃ recovers n, so it is a bijection from T onto D. The conditions 'even' and 'distinct digits' carry over unchanged to D. By finite_counting_basics, finite sets joined by a bijection have the same number of elements, so |T|=|D|." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "decimal_rep", "finite_counting_basics" ], "explanation": { "zh-Hant": "需要表示的存在（對應有定義）與唯一（不會一個數對到兩個三元組）。若表示不唯一，數字三元組的個數就可能多於數的個數。此步只宣稱兩集合等大，不宣稱其他性質。", "en": "It needs existence of the representation (the map is defined) and uniqueness (no number goes to two triples). Without uniqueness there could be more triples than numbers. The step claims only that the two sets have the same size." } }, "leaf_basis": "decimal_rep + injective_surjective_def + finite_counting_basics" }, { "id": "S02", "uses": [ "S01", "even_def", "divides_def" ], "before_tex": "100d_1+10d_2+d_3\\ \\text{even}", "after_tex": "d_3\\in E=\\{0,2,4,6,8\\}", "relation": "equivalent", "rule": "divides_def", "conditions": [ "d_3\\in W" ], "explanation": { "zh-Hant": "令 n=100d₁+10d₂+d₃。因 100d₁+10d₂=2(50d₁+5d₂)（分配律），n−d₃ 是 2 的倍數。若 n=2k，則 d₃=2(k−50d₁−5d₂) 是偶數；反之若 d₃=2j，則 n=2(50d₁+5d₂+j) 是偶數。所以 n 偶 ⇔ d₃ 偶。0 到 9 之中的偶數字恰為 0、2、4、6、8。", "en": "Let n=100d₁+10d₂+d₃. Since 100d₁+10d₂=2(50d₁+5d₂) by distributivity, n−d₃ is a multiple of 2. If n=2k then d₃=2(k−50d₁−5d₂) is even; conversely, if d₃=2j then n=2(50d₁+5d₂+j) is even. So n is even ⇔ d₃ is even, and the even digits from 0 to 9 are exactly 0, 2, 4, 6 and 8." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "divides_def", "even_def", "distributive" ], "explanation": { "zh-Hant": "兩個方向都各自寫出了整數 k 或 j，所以是等價；N1 核對 n−d₃=2(50d₁+5d₂) 這個多項式恆等式。", "en": "Each direction exhibits its own integer k or j, so this is an equivalence; N1 checks the polynomial identity n−d₃=2(50d₁+5d₂)." } }, "leaf_basis": "divides_def + even_def + distributive" }, { "id": "S03", "uses": [ "S01", "S02" ], "before_tex": "D", "after_tex": "D=\\{(d_1,d_2,d_3)\\mid d_1\\in W^{+},\\ d_2\\in W,\\ d_3\\in E,\\ d_1\\neq d_2,\\ d_1\\neq d_3,\\ d_2\\neq d_3\\}", "relation": "equivalent", "rule": "set_extensionality", "conditions": [], "explanation": { "zh-Hant": "把 S01 中的「偶數」條件換成 S02 的等價條件 d₃∈E，並把「兩兩相異」拆成三個不等式。每個三元組在改寫前後滿足的條件同真同假，所以集合不變。", "en": "Replace the condition 'even' in S01 by the equivalent condition d₃∈E from S02, and spell out 'pairwise distinct' as three inequalities. Each triple satisfies the old and new conditions together, so the set is unchanged." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "S02", "set_extensionality" ], "explanation": { "zh-Hant": "條件逐項等價，集合依外延原則雙向相同。", "en": "The conditions are equivalent item by item, so by extensionality the sets agree in both directions." } }, "leaf_basis": "set_extensionality + set_builder" }, { "id": "S04", "uses": [ "S03", "set_operations_def" ], "before_tex": "D", "after_tex": "D=D_0\\cup D_+,\\ D_0\\cap D_+=\\varnothing,\\ D_0=\\{t\\in D\\mid d_3=0\\},\\ D_+=\\{t\\in D\\mid d_3\\in\\{2,4,6,8\\}\\}", "relation": "case_split", "rule": "set_operations_def", "conditions": [], "explanation": { "zh-Hant": "為什麼要分支：若先選個位 d₃，百位 d₁ 的可選數會依 d₃ 而變。d₃=0 時，0 本來就不在 W⁺ 中，d₁ 有 9 種；d₃∈{2,4,6,8} 時，d₃ 是非零數字，必須從 W⁺ 中扣掉，只剩 8 種。MKO 假設 a3 要求每個階段的可選數相對於先前選擇明確給出，可選數不固定時就必須分支。兩類別互斥，因為 d₃ 不能同時是 0 又屬於 {2,4,6,8}；合起來涵蓋 D，因為 E={0}∪{2,4,6,8}。", "en": "Why branch: if the units digit d₃ is chosen first, the number of options for the hundreds digit d₁ depends on d₃. When d₃=0, 0 is not in W⁺ anyway and d₁ has 9 options; when d₃∈{2,4,6,8}, d₃ is a nonzero digit that must be removed from W⁺, leaving 8. MKO assumption a3 requires the number of options at each stage to be stated relative to earlier choices; when it is not constant we must branch. The classes are disjoint because d₃ cannot be both 0 and in {2,4,6,8}; together they cover D because E={0}∪{2,4,6,8}." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "set_operations_def" ], "explanation": { "zh-Hant": "需要兩件事：類別涵蓋全部 D，且互不重疊。前者保證不漏算，後者保證 S05 的加法不重算。", "en": "Two things are needed: the classes cover all of D, and they do not overlap. The first prevents missing elements; the second prevents double counting in the addition of S05." } }, "leaf_basis": "set_operations_def + case_analysis" }, { "id": "S05", "uses": [ "S04", "finite_counting_basics", "mko_counting" ], "before_tex": "D=D_0\\cup D_+,\\ D_0\\cap D_+=\\varnothing", "after_tex": "|D|=|D_0|+|D_+|", "relation": "theorem_application", "rule": "finite_counting_basics", "conditions": [ "D_0\\cap D_+=\\varnothing" ], "explanation": { "zh-Hant": "加法原理（MKO 第一句、finite_counting_basics）：兩個互斥有限集合之聯集的元素個數是兩者個數之和。D₀、D₊ 都是有限集 D 的子集合，所以有限。", "en": "The addition principle (first sentence of the MKO, finite_counting_basics): the number of elements in the union of two disjoint finite sets is the sum of their individual sizes. D₀ and D₊ are subsets of the finite set D, hence finite." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S04", "finite_counting_basics", "finite_subset" ], "explanation": { "zh-Hant": "必須引用 S04 的互斥性；若類別重疊（例如「末位為 0」與「首位為 2」），直接相加會把交集算兩次（見錯誤控制一）。", "en": "The disjointness from S04 must be cited; if the classes overlapped (for example 'ends in 0' and 'starts with 2'), plain addition would count the intersection twice (see error control 1)." } }, "leaf_basis": "finite_counting_basics + finite_subset + mko_counting" }, { "id": "S06", "uses": [ "S04", "L_remove" ], "before_tex": "d_3=0", "after_tex": "|W^{+}\\setminus\\{0\\}|=9,\\ \\ |W\\setminus\\{0,d_1\\}|=8\\ \\text{for every}\\ d_1\\in W^{+}", "relation": "implies", "rule": "L_remove", "conditions": [ "d_1\\neq 0" ], "explanation": { "zh-Hant": "在 D₀ 中 d₃=0。階段一選 d₁：允許集合是 W⁺∖{0}，因 0∉W⁺，它就是 W⁺，有 9 個元素。階段二選 d₂：要與 d₁、d₃ 都不同，允許集合是 W∖{0,d₁}。因 d₁≥1，d₁≠0，所以 {0,d₁}⊆W 恰有 2 個元素；依引理 L_remove，|W∖{0,d₁}|=10−2=8，而且對每個 d₁ 都是 8。", "en": "In D₀ we have d₃=0. Stage one chooses d₁ from W⁺∖{0}; since 0∉W⁺ this is W⁺ itself, with 9 elements. Stage two chooses d₂, which must differ from d₁ and d₃, from W∖{0,d₁}. Since d₁≥1 we have d₁≠0, so {0,d₁}⊆W has exactly 2 elements, and by Lemma L_remove |W∖{0,d₁}|=10−2=8, the same 8 for every d₁." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_remove", "S04" ], "explanation": { "zh-Hant": "關鍵條件是 d₁≠0，它保證被扣掉的集合有 2 個元素而不是 1 個；這正是 a3 要求的「可選數對所有先前選擇相同」。", "en": "The key condition is d₁≠0, which makes the removed set have 2 elements rather than 1; this is exactly the 'same number of options for every earlier choice' that a3 requires." } }, "leaf_basis": "L_remove + finite_counting_basics + numeric_evaluation" }, { "id": "S06.1", "uses": [ "finite_counting_basics", "set_operations_def", "subset_def", "finite_subset" ], "before_tex": "R\\subseteq S,\\ |S|=s,\\ |R|=r", "after_tex": "|S\\setminus R|=s-r", "relation": "implies", "rule": "finite_counting_basics", "conditions": [ "R\\subseteq S" ], "explanation": { "zh-Hant": "S 的每個元素要嘛屬於 R，要嘛屬於 S∖R，不會兩者都是；又因 R⊆S，R 的元素都在 S 中。所以 S=(S∖R)∪R 且 (S∖R)∩R=∅。S∖R 是 S 的子集合，因此有限。依加法原理 s=|S∖R|+r，兩邊同減 r 得 |S∖R|=s−r。此子步證得引理 L_remove。", "en": "Every element of S lies either in R or in S∖R, never in both; and since R⊆S, every element of R lies in S. So S=(S∖R)∪R with (S∖R)∩R=∅. S∖R is a subset of S, hence finite. The addition principle gives s=|S∖R|+r, and subtracting r from both sides gives |S∖R|=s−r. This sub-step proves Lemma L_remove." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "finite_counting_basics", "subset_def", "set_operations_def" ], "explanation": { "zh-Hant": "需要 R⊆S。反例：S={1,2}、R={2,3}，S∖R={1} 有 1 個元素，但 s−r=0。", "en": "R⊆S is needed. Counterexample: S={1,2} and R={2,3} give S∖R={1} with 1 element, but s−r=0." } }, "leaf_basis": "finite_counting_basics + set_operations_def + subset_def + finite_subset + eq_congruence" }, { "id": "S07", "uses": [ "S04", "S06", "L_stage" ], "before_tex": "D_0", "after_tex": "|D_0|=9\\cdot 8=72", "relation": "theorem_application", "rule": "L_stage", "conditions": [ "|W\\setminus\\{0,d_1\\}|=8\\ \\text{for every}\\ d_1\\in W^{+}" ], "explanation": { "zh-Hant": "映射 (d₁,d₂,0)↦(d₁,d₂) 是 D₀ 到 P₀={(d₁,d₂) ∣ d₁∈W⁺, d₂∈W∖{0,d₁}} 的雙射：在 d₃=0 時，d₁≠d₃ 自動成立，d₂≠d₃ 就是 d₂≠0，再加上 d₂≠d₁。取 X=W⁺（m=9）、Y_{d₁}=W∖{0,d₁}（依 S06 每個都是 n=8 個），依引理 L_stage 得 |D₀|=|P₀|=9·8=72。注意 Y_{d₁} 本身依 d₁ 而不同（Y₁ 不含 0、1，Y₂ 不含 0、2），但個數相同，這正是引理所要求的。", "en": "The map (d₁,d₂,0)↦(d₁,d₂) is a bijection from D₀ onto P₀={(d₁,d₂) ∣ d₁∈W⁺, d₂∈W∖{0,d₁}}: with d₃=0, the condition d₁≠d₃ holds automatically, d₂≠d₃ says d₂≠0, and d₂≠d₁ remains. Take X=W⁺ (m=9) and Y_{d₁}=W∖{0,d₁}, each with n=8 elements by S06; Lemma L_stage gives |D₀|=|P₀|=9·8=72. Note that the set Y_{d₁} changes with d₁ (Y₁ omits 0 and 1, Y₂ omits 0 and 2) while its size does not, which is exactly what the lemma requires." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S06", "L_stage", "finite_counting_basics" ], "explanation": { "zh-Hant": "引理需要每個 Y_{d₁} 的個數都等於同一個 n；S06 對所有 d₁∈W⁺ 證明了 n=8。只檢查一個 d₁ 不夠。", "en": "The lemma needs every Y_{d₁} to have the same size n; S06 proves n=8 for all d₁∈W⁺. Checking a single d₁ would not be enough." } }, "leaf_basis": "L_stage + finite_counting_basics + numeric_evaluation" }, { "id": "S07.1", "uses": [ "G3", "finite_counting_basics" ], "before_tex": "|X|=m,\\ |Y_x|=n\\ (x\\in X)", "after_tex": "\\Phi:P\\to X\\times[n],\\ \\Phi(x,y)=(x,\\varphi_x(y))", "relation": "definition", "rule": "def_intro", "conditions": [ "|Y_x|=n\\ \\text{for every}\\ x\\in X" ], "explanation": { "zh-Hant": "依 G3，|Yₓ|=n 表示有雙射 φₓ:Yₓ→[n]；對有限多個 x 各固定一個。定義 Φ(x,y)=(x,φₓ(y))。因 x∈X 且 φₓ(y)∈[n]，Φ 的值確實在 X×[n] 中。", "en": "By G3, |Yₓ|=n means there is a bijection φₓ:Yₓ→[n]; fix one for each of the finitely many x. Define Φ(x,y)=(x,φₓ(y)). Since x∈X and φₓ(y)∈[n], the values of Φ lie in X×[n]." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G3", "def_intro" ], "explanation": { "zh-Hant": "此步只引入記號；它依賴「所有 Yₓ 對應到同一個 [n]」，也就是每個 Yₓ 都恰有 n 個元素。", "en": "This step only introduces notation; it relies on every Yₓ being matched with the same [n], that is, on each Yₓ having exactly n elements." } }, "leaf_basis": "def_intro + function_def" }, { "id": "S07.2", "uses": [ "S07.1", "injective_surjective_def", "cartesian_product_def" ], "before_tex": "\\Phi(x,y)=(x,\\varphi_x(y))", "after_tex": "\\Phi\\ \\text{is a bijection}", "relation": "implies", "rule": "injective_surjective_def", "conditions": [], "explanation": { "zh-Hant": "單射：若 Φ(x,y)=Φ(x′,y′)，依有序對相等得 x=x′ 且 φₓ(y)=φₓ(y′)，φₓ 是單射，所以 y=y′。滿射：任取 (x,k)∈X×[n]，φₓ 是滿射，有 y∈Yₓ 使 φₓ(y)=k，於是 (x,y)∈P 且 Φ(x,y)=(x,k)。可以依 x 而不同的是雙射 φₓ，不能依 x 而不同的是目標 [n]。", "en": "Injective: if Φ(x,y)=Φ(x′,y′), equality of ordered pairs gives x=x′ and φₓ(y)=φₓ(y′), and since φₓ is injective, y=y′. Surjective: for any (x,k)∈X×[n], φₓ is surjective, so some y∈Yₓ has φₓ(y)=k; then (x,y)∈P and Φ(x,y)=(x,k). What may depend on x is the bijection φₓ; what may not is the target [n]." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S07.1", "injective_surjective_def", "cartesian_product_def" ], "explanation": { "zh-Hant": "兩個方向各自用到 φₓ 的單射與滿射性質；不宣稱 Φ 以外的任何對應是雙射。", "en": "The two parts use the injectivity and surjectivity of φₓ respectively; nothing is claimed about maps other than Φ." } }, "leaf_basis": "injective_surjective_def + cartesian_product_def" }, { "id": "S07.3", "uses": [ "S07.2", "finite_counting_basics" ], "before_tex": "\\Phi:P\\to X\\times[n]\\ \\text{bijective}", "after_tex": "|P|=|X\\times[n]|=m\\cdot n", "relation": "implies", "rule": "finite_counting_basics", "conditions": [], "explanation": { "zh-Hant": "依 finite_counting_basics，有雙射的兩個有限集合個數相同，且 |X×[n]|=|X|·|[n]|=m·n。S07.1 到 S07.3 合起來證得引理 L_stage。", "en": "By finite_counting_basics, finite sets joined by a bijection have the same size, and |X×[n]|=|X|·|[n]|=m·n. S07.1 to S07.3 together prove Lemma L_stage." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S07.2", "finite_counting_basics" ], "explanation": { "zh-Hant": "若各 Yₓ 個數不同，就沒有共同的 [n]，乘積公式失效。反例：X={a,b}、|Y_a|=1、|Y_b|=3，則 |P|=4，不等於 2·1 也不等於 2·3。", "en": "If the Yₓ had different sizes there would be no common [n] and the product formula fails. Counterexample: X={a,b}, |Y_a|=1, |Y_b|=3 give |P|=4, which is neither 2·1 nor 2·3." } }, "leaf_basis": "finite_counting_basics" }, { "id": "S08", "uses": [ "S04", "L_remove" ], "before_tex": "d_3\\in\\{2,4,6,8\\}", "after_tex": "|W^{+}\\setminus\\{d_3\\}|=8,\\ \\ |W\\setminus\\{d_1,d_3\\}|=8\\ \\text{for every}\\ d_1\\in W^{+}\\setminus\\{d_3\\}", "relation": "implies", "rule": "L_remove", "conditions": [ "d_3\\in W^{+}", "d_1\\neq d_3" ], "explanation": { "zh-Hant": "在 D₊ 中，d₃∈{2,4,6,8}⊆W⁺，所以 {d₃}⊆W⁺，依 L_remove |W⁺∖{d₃}|=9−1=8，對四個 d₃ 都一樣。d₁ 從 W⁺∖{d₃} 中選，所以 d₁≠d₃，{d₁,d₃}⊆W 恰有 2 個元素，|W∖{d₁,d₃}|=10−2=8，對每一對 (d₃,d₁) 都一樣。", "en": "In D₊, d₃∈{2,4,6,8}⊆W⁺, so {d₃}⊆W⁺ and L_remove gives |W⁺∖{d₃}|=9−1=8, the same for all four values of d₃. Since d₁ is chosen from W⁺∖{d₃}, d₁≠d₃, so {d₁,d₃}⊆W has exactly 2 elements and |W∖{d₁,d₃}|=10−2=8 for every pair (d₃,d₁)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_remove", "S04" ], "explanation": { "zh-Hant": "需要 d₃∈W⁺（被扣掉的確實在 W⁺ 中）與 d₁≠d₃（被扣掉的兩個數不同）。d₃=0 時第一個條件不成立，這正是 S04 分支的原因。", "en": "It needs d₃∈W⁺ (the removed digit really lies in W⁺) and d₁≠d₃ (the two removed digits differ). The first condition fails when d₃=0, which is exactly why S04 branches." } }, "leaf_basis": "L_remove + numeric_evaluation" }, { "id": "S09", "uses": [ "S08", "L_stage" ], "before_tex": "Q=\\{(d_3,d_1)\\mid d_3\\in\\{2,4,6,8\\},\\ d_1\\in W^{+}\\setminus\\{d_3\\}\\}", "after_tex": "|Q|=4\\cdot 8=32", "relation": "theorem_application", "rule": "L_stage", "conditions": [ "|W^{+}\\setminus\\{d_3\\}|=8\\ \\text{for every}\\ d_3\\in\\{2,4,6,8\\}" ], "explanation": { "zh-Hant": "D₊ 的前兩個階段：先選 d₃（4 種），再選 d₁（依 S08 每次 8 種）。依引理 L_stage（m=4、n=8），這些有序對共 32 個。", "en": "The first two stages for D₊: choose d₃ (4 ways), then d₁ (8 ways each time by S08). Lemma L_stage with m=4 and n=8 gives 32 such ordered pairs." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S08", "L_stage" ], "explanation": { "zh-Hant": "引理的條件「每個 d₃ 之後都恰有 8 種 d₁」由 S08 提供。", "en": "The lemma's condition that every d₃ is followed by exactly 8 choices of d₁ is supplied by S08." } }, "leaf_basis": "L_stage + numeric_evaluation" }, { "id": "S10", "uses": [ "S03", "S04", "S08", "S09", "L_stage" ], "before_tex": "D_+", "after_tex": "|D_+|=32\\cdot 8=256", "relation": "theorem_application", "rule": "L_stage", "conditions": [ "|W\\setminus\\{d_1,d_3\\}|=8\\ \\text{for every}\\ (d_3,d_1)\\in Q" ], "explanation": { "zh-Hant": "映射 (d₁,d₂,d₃)↦((d₃,d₁),d₂) 只重排坐標，是 D₊ 到 {(q,d₂) ∣ q=(d₃,d₁)∈Q, d₂∈W∖{d₁,d₃}} 的雙射；兩邊的條件相同：d₃ 為非零偶數字、d₁∈W⁺、d₁≠d₃、d₂≠d₁、d₂≠d₃。再用一次 L_stage（X=Q，m=32；依 S08 每個 q 之後 n=8）得 |D₊|=32·8=256。", "en": "The map (d₁,d₂,d₃)↦((d₃,d₁),d₂) only reorders coordinates and is a bijection from D₊ onto {(q,d₂) ∣ q=(d₃,d₁)∈Q, d₂∈W∖{d₁,d₃}}; both sides impose the same conditions: d₃ a nonzero even digit, d₁∈W⁺, d₁≠d₃, d₂≠d₁, d₂≠d₃. Applying L_stage again (X=Q with m=32, and n=8 after every q by S08) gives |D₊|=32·8=256." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S08", "S09", "L_stage", "finite_counting_basics" ], "explanation": { "zh-Hant": "第三階段的可選數 8 必須對每一對 (d₃,d₁) 都成立，S08 用 d₁≠d₃ 證明了這一點。坐標重排是雙射，所以個數不變。", "en": "The 8 options at the third stage must hold for every pair (d₃,d₁); S08 proves this using d₁≠d₃. Reordering coordinates is a bijection, so the count is unchanged." } }, "leaf_basis": "L_stage + injective_surjective_def + finite_counting_basics + numeric_evaluation" }, { "id": "S11", "uses": [ "S05", "S07", "S10" ], "before_tex": "|D|=|D_0|+|D_+|", "after_tex": "|D|=72+256=328", "relation": "equivalent", "rule": "eq_replace", "conditions": [], "explanation": { "zh-Hant": "把 S07 的 |D₀|=72 與 S10 的 |D₊|=256 代入 S05 的加法式：|D|=72+256=328。", "en": "Substitute |D₀|=72 from S07 and |D₊|=256 from S10 into the addition of S05: |D|=72+256=328." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "S07", "S10", "eq_replace" ], "explanation": { "zh-Hant": "等值代換與精確加法（N8）兩向都成立。", "en": "Replacement of equals and the exact sum (N8) hold in both directions." } }, "leaf_basis": "eq_replace + numeric_evaluation" }, { "id": "S12", "uses": [ "S03", "L_stage", "L_remove", "S05" ], "before_tex": "D=D'_{\\mathrm{odd}}\\cup D'_{\\mathrm{even}}\\ \\text{(branch on}\\ d_1)", "after_tex": "|D|=5\\cdot 5\\cdot 8+4\\cdot 4\\cdot 8=200+128=328", "relation": "implies", "rule": "L_stage", "conditions": [ "d_1\\ \\text{odd: } |E\\setminus\\{d_1\\}|=5", "d_1\\ \\text{even: } |E\\setminus\\{d_1\\}|=4" ], "explanation": { "zh-Hant": "獨立核對：改成先選百位 d₁。此時個位 d₃∈E∖{d₁} 的個數依 d₁ 的奇偶而變，所以依 d₁ 的奇偶分成兩個互斥類別。d₁ 奇（1、3、5、7、9，5 種）：d₁∉E，d₃ 有 5 種，d₂∈W∖{d₁,d₃} 有 8 種，得 5·5·8=200。d₁ 偶（2、4、6、8，4 種）：d₁∈E，d₃ 有 4 種，d₂ 有 8 種，得 4·4·8=128。相加 328，與 S11 一致。", "en": "Independent check: choose the hundreds digit d₁ first. Then the number of choices of d₃∈E∖{d₁} depends on the parity of d₁, so split into two disjoint classes by the parity of d₁. d₁ odd (1, 3, 5, 7, 9; 5 ways): d₁∉E, so d₃ has 5 options and d₂∈W∖{d₁,d₃} has 8, giving 5·5·8=200. d₁ even (2, 4, 6, 8; 4 ways): d₁∈E, so d₃ has 4 options and d₂ has 8, giving 4·4·8=128. The sum is 328, agreeing with S11." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_stage", "L_remove", "S05" ], "explanation": { "zh-Hant": "這是換一個階段順序的第二次計數，同樣先分支再相乘；它與 S11 一致，增加信心，但結論仍由 S01 到 S11 推出。", "en": "This is a second count with a different stage order, again branching before multiplying; its agreement with S11 adds confidence, while the conclusion is still derived by S01 to S11." } }, "leaf_basis": "L_stage + L_remove + finite_counting_basics + numeric_evaluation" }, { "id": "S13", "uses": [ "S01", "S11", "S12" ], "before_tex": "|T|=|D|,\\ |D|=328", "after_tex": "|T|=328", "relation": "implies", "rule": "eq_refl_symm_trans", "conditions": [], "explanation": { "zh-Hant": "由 S01 的 |T|=|D| 與 S11 的 |D|=328，依等號傳遞得 |T|=328：100 到 999 之間數字兩兩相異的偶數共 328 個，其中 72 個以 0 結尾，256 個以 2、4、6 或 8 結尾。", "en": "From |T|=|D| in S01 and |D|=328 in S11, transitivity of equality gives |T|=328: there are 328 even numbers from 100 to 999 with pairwise distinct digits, 72 ending in 0 and 256 ending in 2, 4, 6 or 8." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S01", "S11", "eq_refl_symm_trans" ], "explanation": { "zh-Hant": "目標 G：|T|=328。足夠條件 H：T 與 D 之間有雙射且 |D|=328；H⇒G 由等號傳遞。H 由 S01 與 S11 從已知正向證成，沒有把 G 當作前提；S12 只是獨立核對。", "en": "Goal G: |T|=328. Sufficient condition H: a bijection between T and D, and |D|=328; H⇒G by transitivity of equality. H is proved forwards from the givens in S01 and S11, and G is never used as a premise; S12 is only an independent check." } }, "leaf_basis": "eq_refl_symm_trans" } ], "result": { "zh-Hant": "100 到 999 之間、三個數字兩兩相異的偶數共有 328 個：依個位分成互斥的兩類，個位為 0 的有 9·8=72 個，個位為 2、4、6、8 的有 4·8·8=256 個。改為依百位奇偶分支也得到 200+128=328。", "en": "There are 328 even numbers from 100 to 999 with pairwise distinct digits: splitting by the units digit into two disjoint classes, 9·8=72 end in 0 and 4·8·8=256 end in 2, 4, 6 or 8. Branching instead on the parity of the hundreds digit also gives 200+128=328." }, "negative_controls": [ { "claim": { "zh-Hant": "個位 5 種、百位 8 種、十位 8 種，所以 |T|=5·8·8=320", "en": "Units 5 ways, hundreds 8 ways, tens 8 ways, so |T|=5·8·8=320" }, "expected": "reject", "reason": { "zh-Hant": "個位為 0 時百位有 9 種；正確答案是 328。", "en": "When the units digit is 0 the hundreds digit has 9 options; the correct count is 328." } }, { "claim": { "zh-Hant": "百位 9 種、十位 8 種、個位 5 種，所以 |T|=360", "en": "Hundreds 9 ways, tens 8 ways, units 5 ways, so |T|=360" }, "expected": "reject", "reason": { "zh-Hant": "個位可選的偶數字數依前兩位而變；d₁=2、d₂=4 時只剩 3 種。", "en": "The number of even digits left for the units place depends on the first two digits; with d₁=2 and d₂=4 only 3 remain." } }, { "claim": { "zh-Hant": "T 中個位為 0 或百位為 2 的數有 72+32=104 個", "en": "The numbers in T that end in 0 or start with 2 number 72+32=104" }, "expected": "reject", "reason": { "zh-Hant": "兩類重疊 8 個（210、230 等）；正確個數是 96。", "en": "The classes share 8 numbers (210, 230 and so on); the correct count is 96." } }, { "claim": { "zh-Hant": "D₀ 與 D₊ 的個數都是乘出來的，所以 |D|=72·256", "en": "The sizes of D₀ and D₊ were obtained by multiplying, so |D|=72·256" }, "expected": "reject", "reason": { "zh-Hant": "互斥類別用加法合併；乘法只用於同一流程中先後相接的階段。", "en": "Disjoint classes are combined by addition; multiplication is for successive stages of one process." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "finite_counting_basics", "kind": "prerequisite", "statement": { "zh-Hant": "有限計數：互斥有限集合 A、B 有 |A∪B|=|A|+|B|；|A×B|=|A|·|B|；有限集之間存在雙射 ⇔ 元素個數相同。", "en": "Finite counting: for disjoint finite sets A, B, |A∪B|=|A|+|B|; |A×B|=|A|·|B|; there is a bijection between finite sets ⇔ they have the same number of elements." } }, { "id": "cartesian_product_def", "kind": "prerequisite", "statement": { "zh-Hant": "笛卡兒積：A×B 是所有有序對 (a,b)（a∈A、b∈B）的集合；(a,b)=(c,d) ⇔ a=c 且 b=d。", "en": "Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d." } }, { "id": "injective_surjective_def", "kind": "prerequisite", "statement": { "zh-Hant": "單射：f(a)=f(a′) ⇒ a=a′。滿射：每個 b∈B 都有 a∈A 使 f(a)=b。雙射：既單又滿，此時有唯一反函數 f⁻¹ 使 f⁻¹∘f=id_A、f∘f⁻¹=id_B。", "en": "Injective: f(a)=f(a′) ⇒ a=a′. Surjective: every b∈B has some a∈A with f(a)=b. Bijective: both; then there is a unique inverse f⁻¹ with f⁻¹∘f=id_A and f∘f⁻¹=id_B." } }, { "id": "set_operations_def", "kind": "prerequisite", "statement": { "zh-Hant": "集合運算：x∈A∪B ⇔ x∈A 或 x∈B；x∈A∩B ⇔ x∈A 且 x∈B；x∈A∖B ⇔ x∈A 且 x∉B；在全集 U 中 Aᶜ=U∖A。", "en": "Set operations: x∈A∪B ⇔ x∈A or x∈B; x∈A∩B ⇔ x∈A and x∈B; x∈A∖B ⇔ x∈A and x∉B; in a universe U, Aᶜ=U∖A." } }, { "id": "numeric_evaluation", "kind": "prerequisite", "statement": { "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。", "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable." } }, { "id": "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": "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": "set_extensionality", "kind": "prerequisite", "statement": { "zh-Hant": "集合相等：A=B 當且僅當 A⊆B 且 B⊆A，即兩者有相同元素。", "en": "Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements." } }, { "id": "set_builder", "kind": "prerequisite", "statement": { "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。", "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x)." } }, { "id": "case_analysis", "kind": "prerequisite", "statement": { "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。", "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility." } }, { "id": "subset_def", "kind": "prerequisite", "statement": { "zh-Hant": "子集：A⊆B 表示 A 的每個元素都是 B 的元素，即 ∀x (x∈A⇒x∈B)。", "en": "Subset: A⊆B means every element of A is an element of B, i.e. ∀x (x∈A⇒x∈B)." } }, { "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_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": "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": "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": "eq_refl_symm_trans", "kind": "prerequisite", "statement": { "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。", "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w." } }, { "id": "mko_counting", "kind": "cited", "source": { "mko_id": "mko-counting-principle", "sha256": "0a0619ebbab98b96543b42abc4e66d927aeea705c65ee85b3a9badb6d3b3cf57", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "互斥選擇類別的數量相加；連續選擇階段的可能組合數，在各階段選項數與條件明確時相乘。", "en": "Counts of mutually exclusive choice classes are added. Counts of combinations across successive choice stages are multiplied when the options and conditions at every stage are explicit." } }, { "id": "decimal_rep", "kind": "prerequisite", "statement": { "zh-Hant": "十進位表示：每個滿足 100≤n≤999 的整數 n 恰有一組數字 (d₁,d₂,d₃) 使 n=100d₁+10d₂+d₃，其中 d₁∈{1,2,3,4,5,6,7,8,9}、d₂,d₃∈{0,1,2,3,4,5,6,7,8,9}；反之每組這樣的數字給出一個這樣的 n。", "en": "Decimal representation: every integer n with 100≤n≤999 has exactly one digit triple (d₁,d₂,d₃) with n=100d₁+10d₂+d₃, d₁∈{1,2,3,4,5,6,7,8,9} and d₂,d₃∈{0,1,2,3,4,5,6,7,8,9}; conversely every such triple gives such an n." } }, { "id": "even_def", "kind": "prerequisite", "statement": { "zh-Hant": "整數 n 是偶數，意思是 2∣n，即存在整數 k 使 n=2k。", "en": "An integer n is even when 2∣n, that is, n=2k for some integer k." } }, { "id": "finite_subset", "kind": "prerequisite", "statement": { "zh-Hant": "有限集合的子集合仍是有限集合。", "en": "A subset of a finite set is finite." } }, { "id": "L_remove", "kind": "lemma", "proved_in": [ "S06.1" ], "statement": { "zh-Hant": "若有限集合 S 有 s 個元素，且 R⊆S 有 r 個元素，則 |S∖R|=s−r。", "en": "If a finite set S has s elements and R⊆S has r elements, then |S∖R|=s−r." } }, { "id": "L_stage", "kind": "lemma", "proved_in": [ "S07.1", "S07.2", "S07.3" ], "statement": { "zh-Hant": "設有限集合 X 有 m 個元素，且對每個 x∈X，有限集合 Yₓ 恰有 n 個元素（n 對所有 x 相同；集合 Yₓ 本身可以依 x 而不同）。則 P={(x,y) ∣ x∈X, y∈Yₓ} 有 m·n 個元素。", "en": "Let a finite set X have m elements, and for each x∈X let the finite set Yₓ have exactly n elements (the number n is the same for every x, while the set Yₓ itself may depend on x). Then P={(x,y) ∣ x∈X, y∈Yₓ} has m·n elements." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S02", "vars": [ "a", "b", "c" ], "degree": 1, "lhs": "(100*a+10*b+c)-c", "rhs": "2*(50*a+5*b)" }, { "id": "N2", "step": "S02", "vars": [ "n" ], "range": [ 100, 999 ], "lhs": "n%2", "rhs": "(n%10)%2" }, { "id": "N3", "step": "S06", "lhs": "10-2", "rhs": "8" }, { "id": "N4", "step": "S07", "lhs": "9*8", "rhs": "72" }, { "id": "N5", "step": "S08", "lhs": "9-1", "rhs": "8" }, { "id": "N6", "step": "S09", "lhs": "4*8", "rhs": "32" }, { "id": "N7", "step": "S10", "lhs": "32*8", "rhs": "256" }, { "id": "N8", "step": "S11", "lhs": "72+256", "rhs": "328" }, { "id": "N9", "step": "S12", "lhs": "5*5*8", "rhs": "200" }, { "id": "N10", "step": "S12", "lhs": "4*4*8", "rhs": "128" }, { "id": "N11", "step": "S12", "lhs": "200+128", "rhs": "328" }, { "id": "N12", "lhs": "5*8*8", "relation": "!=", "rhs": "328" }, { "id": "N13", "lhs": "72-1*8*8", "rhs": "8" }, { "id": "N14", "lhs": "9*8*5", "relation": "!=", "rhs": "328" }, { "id": "N15", "lhs": "72+32-8", "rhs": "96" }, { "id": "N16", "step": "S06.1", "lhs": "2-2", "relation": "!=", "rhs": "1" } ] }, "presentation": { "md_scope": { "zh-Hant": "只處理有限集合的計數（MKO 假設 a1）：用互斥類別的加法原理（a2）與逐階段的乘法原理（a3），並由有限計數的基本事實證明本稿使用的「依賴型」乘法引理。不處理一般的排容原理、排列組合公式或無限集合；不改動 canonical MKO 或英文 edition。", "en": "Counts finite sets only (MKO assumption a1), using the addition principle for disjoint classes (a2) and the stage-by-stage multiplication principle (a3); the 'dependent' multiplication lemma used here is proved from basic facts of finite counting. It does not treat general inclusion-exclusion, permutation or combination formulas, or infinite sets, and does not change the canonical MKO or its English edition." }, "md_symbols": [ [ "T", "所求的數的集合", "the set of numbers to be counted" ], [ "d_1,d_2,d_3", "百位、十位、個位數字", "hundreds, tens and units digits" ], [ "W,\\ W^{+}", "W={0,1,2,3,4,5,6,7,8,9}，W⁺=W∖{0}", "W={0,1,2,3,4,5,6,7,8,9} and W⁺=W∖{0}" ], [ "E", "偶數字 {0,2,4,6,8}", "the even digits {0,2,4,6,8}" ], [ "D,\\ D_0,\\ D_+", "數字三元組的集合及其兩個類別", "the set of digit triples and its two classes" ], [ "[n]", "集合 {k∈ℕ ∣ 1≤k≤n}", "the set {k∈ℕ ∣ 1≤k≤n}" ], [ "|S|", "有限集合 S 的元素個數", "the number of elements of a finite set S" ] ], "md_error_controls": [ { "title": "互斥才能直接相加 / Add directly only when the classes are disjoint", "zh": "MKO 指出的第一個常見誤解是對會重疊的案例直接相加。在 T 中，「個位為 0」有 72 個，「百位為 2」有 4·8=32 個，但兩者重疊於 210、230、240、250、260、270、280、290 這 8 個數。直接相加得 104，把重疊部分算了兩次；正確的聯集個數是 72+32−8=96（MKO 假設 a2：重疊需另外修正）。S04 的兩類以 d₃ 是否為 0 來分，不可能重疊，所以 S05 可以直接相加。", "en": "The first common mistake named by the MKO is adding cases that overlap. In T, 72 numbers end in 0 and 4·8=32 start with 2, but the two classes share the 8 numbers 210, 230, 240, 250, 260, 270, 280 and 290. Adding directly gives 104 and counts the overlap twice; the union really has 72+32−8=96 elements (MKO assumption a2: an overlap must be corrected separately). The classes of S04 are split by whether d₃ is 0, so they cannot overlap and S05 may add them directly." }, { "title": "可選數依前一步改變時不能無條件相乘 / Do not multiply blindly when the number of options depends on an earlier choice", "zh": "MKO 指出的第二個誤解：「先選個位 5 種，再選百位 8 種、十位 8 種」得 5·8·8=320。錯在「百位 8 種」只在個位非零時成立；個位為 0 時百位有 9 種。單看 d₃=0 這一類，錯誤算法給 1·8·8=64，實際是 72，正好少了 8 個；320+8=328。正確做法是 S04 的分支，或 S12 的另一種分支。", "en": "The second mistake named by the MKO: 'units digit 5 ways, then hundreds 8 ways and tens 8 ways' gives 5·8·8=320. The claim 'hundreds 8 ways' holds only when the units digit is nonzero; when it is 0 the hundreds digit has 9 options. In the class d₃=0 alone the wrong count gives 1·8·8=64 instead of 72, exactly 8 too few, and 320+8=328. The correct method is the branching of S04, or the other branching of S12." }, { "title": "階段順序可換，但每個順序都要重新核對 / The order of stages may change, but each order must be rechecked", "zh": "先選百位、再選十位、最後選個位時，若寫成 9·8·5=360，就假設了個位永遠剩 5 個偶數字。但百位或十位若用掉偶數字，剩下的會變少，例如 d₁=2、d₂=4 時個位只剩 0、6、8 三種。換順序本身沒有錯，錯在沒有核對每個階段的可選數是否固定；S12 依 d₁ 的奇偶分支後才相乘。", "en": "Choosing hundreds, then tens, then units and writing 9·8·5=360 assumes that 5 even digits always remain for the units place. If the hundreds or tens digit uses an even digit, fewer remain; for example d₁=2 and d₂=4 leave only 0, 6 and 8. Changing the order is not the error; the error is not checking that each stage has a fixed number of options. S12 branches on the parity of d₁ before multiplying." } ], "md_open_items": [ { "zh-Hant": "N2 在 100 到 999 的每個整數上核對「n 的奇偶與個位數字的奇偶相同」，是有限檢查，恰好涵蓋本題用到的全部 900 個數；S02 的論證不依賴它。", "en": "N2 checks, at each integer from 100 to 999, that n has the same parity as its units digit; it is a finite check that covers exactly the 900 numbers used here, and the argument of S02 does not depend on it." }, { "zh-Hant": "十進位表示的存在與唯一（decimal_rep）是本稿的前提，未在本稿中由帶餘除法推出。", "en": "Existence and uniqueness of the decimal representation (decimal_rep) is a premise of this guide; it is not derived here from the division algorithm." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "7f8dccaca5069630be4159d5fe0095ba10e312d6050c11349077668b0fd09b5a", "original_source_sha256": "7f8dccaca5069630be4159d5fe0095ba10e312d6050c11349077668b0fd09b5a", "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." } }