{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-relation-01", "mko_id": "mko-relation", "title": { "zh-Hant": "二元關係：定義域、關係像與逆關係的逐步計算", "en": "Binary relations: computing the domain, images and the converse step by step" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-relation-01", "source": { "mko_id": "mko-relation", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "3788215401bc9a5610691985883d04b506880cc9bd5e2ce0a98054cd5a5f0774", "english_sha256": "7d0700fb988776905f5d6ce3bf7399bc7a370ea89a38fe4e6b4c9e8509acdfdb" }, "mode": "definition_walkthrough", "audience_prerequisites": [ "cartesian_product_def", "subset_def", "set_builder", "set_extensionality", "exists_intro", "roster_membership" ], "goal": { "zh-Hant": "設 A={1,2,3,4}、B={2,3,4,5,6}、R={(1,2),(1,3),(3,3),(4,6)}、C={1,3}。核對 R⊆A×B；計算 Dom(R)、R[C]、R[∅]、R⁻¹ 與 Dom(R⁻¹)；判定 R 是否為函數 A→B、R⁻¹ 是否為函數 B→A。", "en": "Let A={1,2,3,4}, B={2,3,4,5,6}, R={(1,2),(1,3),(3,3),(4,6)} and C={1,3}. Check R⊆A×B; compute Dom(R), R[C], R[∅], R⁻¹ and Dom(R⁻¹); decide whether R is a function A→B and whether R⁻¹ is a function B→A." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "來源 A={1,2,3,4}，目標 B={2,3,4,5,6}，都以列舉明示。", "en": "Source A={1,2,3,4} and target B={2,3,4,5,6}, both given explicitly by listing." } }, { "id": "G2", "statement": { "zh-Hant": "R={(1,2),(1,3),(3,3),(4,6)}，以列舉完整給出，宣告為從 A 到 B 的關係；有序對保留方向（MKO 假設 a2）。", "en": "R={(1,2),(1,3),(3,3),(4,6)}, given completely by listing and declared as a relation from A to B; ordered pairs keep their direction (MKO assumption a2)." } }, { "id": "G3", "statement": { "zh-Hant": "C={1,3}⊆A（MKO 假設 a1）。", "en": "C={1,3}⊆A (MKO assumption a1)." } } ], "steps": [ { "id": "S01", "uses": [ "G1", "G2", "subset_def", "cartesian_product_def", "roster_membership" ], "before_tex": "R=\\{(1,2),(1,3),(3,3),(4,6)\\}", "after_tex": "R\\subseteq A\\times B", "relation": "implies", "rule": "subset_def", "conditions": [], "explanation": { "zh-Hant": "依子集定義，要檢查 R 的每個元素都在 A×B 中。R 以列舉給出四個有序對；每一對的第一分量要在 A、第二分量要在 B：(1,2)：1∈A、2∈B；(1,3)：1∈A、3∈B；(3,3)：3∈A、3∈B；(4,6)：4∈A、6∈B。四對都合格，所以 R⊆A×B，R 確實是從 A 到 B 的二元關係。", "en": "By the definition of subset, every element of R must lie in A×B. R is given by listing four ordered pairs; for each pair the first coordinate must be in A and the second in B: (1,2): 1∈A, 2∈B; (1,3): 1∈A, 3∈B; (3,3): 3∈A, 3∈B; (4,6): 4∈A, 6∈B. All four qualify, so R⊆A×B and R really is a binary relation from A to B." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "subset_def", "cartesian_product_def", "roster_membership" ], "explanation": { "zh-Hant": "需要逐對檢查，且分量順序不能交換：(4,6) 的 4 要在 A、6 要在 B；(6,4) 就不在 A×B 中，因為 6∉A。不宣稱 R 等於 A×B（A×B 有 20 個元素，N1）。", "en": "Each pair must be checked, and the order of coordinates may not be swapped: in (4,6), 4 must be in A and 6 in B; (6,4) would not be in A×B, because 6∉A. It is not claimed that R equals A×B (which has 20 elements, N1)." } }, "leaf_basis": "subset_def + cartesian_product_def + roster_membership" }, { "id": "S02", "uses": [ "mko_relation_def", "S01" ], "before_tex": "x\\in\\mathrm{Dom}(R)", "after_tex": "x\\in A\\ \\land\\ \\exists y\\in B:\\ (x,y)\\in R", "relation": "equivalent", "rule": "mko_relation_def", "conditions": [], "explanation": { "zh-Hant": "依 MKO，Dom(R)={x∈A ∣ 存在 y∈B 使 (x,y)∈R}。由集合建構式，x∈Dom(R) 當且僅當 x∈A 且有某個 y∈B 使 (x,y)∈R。Dom(R) 由 R 的資料決定，不是宣告來源 A 的別名。", "en": "By the MKO, Dom(R)={x∈A ∣ there exists y∈B with (x,y)∈R}. By set-builder notation, x∈Dom(R) exactly when x∈A and some y∈B has (x,y)∈R. Dom(R) is determined by the data of R; it is not another name for the declared source A." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mko_relation_def", "set_builder" ], "explanation": { "zh-Hant": "展開定義與集合建構式都是雙向的。", "en": "Unfolding the definition and set-builder notation are both two-way." } }, "leaf_basis": "mko_relation_def + set_builder" }, { "id": "S03", "uses": [ "S02", "G2", "exists_intro" ], "before_tex": "(1,2)\\in R,\\ (3,3)\\in R,\\ (4,6)\\in R", "after_tex": "1,3,4\\in\\mathrm{Dom}(R)", "relation": "implies", "rule": "exists_intro", "conditions": [], "explanation": { "zh-Hant": "找見證：(1,2)∈R 給 x=1 的見證 y=2；(3,3)∈R 給 x=3 的見證 y=3；(4,6)∈R 給 x=4 的見證 y=6。見證都在 B 中，1、3、4 都在 A 中，依 ∃引入與 S02，1、3、4∈Dom(R)。1 還有第二個見證 y=3，但一個就夠。", "en": "Find witnesses: (1,2)∈R gives the witness y=2 for x=1; (3,3)∈R gives y=3 for x=3; (4,6)∈R gives y=6 for x=4. The witnesses are in B and 1, 3, 4 are in A, so by ∃-introduction and S02, 1, 3, 4∈Dom(R). 1 has a second witness y=3, but one is enough." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "exists_intro" ], "explanation": { "zh-Hant": "每個 x 需要一個在 B 中的見證；不宣稱見證唯一。", "en": "Each x needs one witness in B; uniqueness of the witness is not claimed." } }, "leaf_basis": "exists_intro + set_builder" }, { "id": "S04", "uses": [ "G2", "roster_membership", "cartesian_product_def", "forall_intro" ], "before_tex": "y\\in B,\\quad (2,y)\\in R\\ ?", "after_tex": "\\forall y\\in B:\\ (2,y)\\notin R", "relation": "implies", "rule": "cartesian_product_def", "conditions": [], "explanation": { "zh-Hant": "任取 y∈B。由列舉，(2,y)∈R 當且僅當 (2,y) 等於 (1,2)、(1,3)、(3,3)、(4,6) 之一。有序對相等需要第一分量相等，即 2=1、2=1、2=3 或 2=4，四者皆假（N2 以乘積 (2−1)(2−1)(2−3)(2−4)≠0 核對）。所以 (2,y)∉R；y 是任意的，依 ∀引入得結論。", "en": "Take any y∈B. By the listing, (2,y)∈R exactly when (2,y) equals one of (1,2), (1,3), (3,3), (4,6). Equal ordered pairs need equal first coordinates, i.e. 2=1, 2=1, 2=3 or 2=4, all false (N2 checks this with the product (2−1)(2−1)(2−3)(2−4)≠0). So (2,y)∉R; since y is arbitrary, ∀-introduction gives the conclusion." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "roster_membership", "cartesian_product_def", "forall_intro" ], "explanation": { "zh-Hant": "需要 R 的列舉是完整的（G2）；若 R 還有沒列出的對，此步就不成立。", "en": "It needs the listing of R to be complete (G2); if R had unlisted pairs, this step would fail." } }, "leaf_basis": "roster_membership + cartesian_product_def + forall_intro + case_analysis" }, { "id": "S05", "uses": [ "S02", "S04", "quantifier_negation" ], "before_tex": "\\forall y\\in B:\\ (2,y)\\notin R", "after_tex": "2\\notin\\mathrm{Dom}(R)", "relation": "implies", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "依量詞否定律，∀y∈B (2,y)∉R 等價於 ¬∃y∈B (2,y)∈R。由 S02，2∈Dom(R) 需要這樣的 y，所以 2∉Dom(R)。2 屬於宣告來源 A，卻沒有任何出邊。", "en": "By quantifier negation, ∀y∈B (2,y)∉R is equivalent to ¬∃y∈B (2,y)∈R. By S02, 2∈Dom(R) would need such a y, so 2∉Dom(R). 2 belongs to the declared source A but has no outgoing pair at all." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "S04", "quantifier_negation" ], "explanation": { "zh-Hant": "需要 S04 對 B 的每個 y 都成立；只檢查幾個 y 不夠。", "en": "S04 must hold for every y in B; checking a few values of y would not be enough." } }, "leaf_basis": "quantifier_negation + set_builder" }, { "id": "S06", "uses": [ "S03", "S05", "G1", "set_extensionality" ], "before_tex": "1,3,4\\in\\mathrm{Dom}(R),\\ 2\\notin\\mathrm{Dom}(R),\\ \\mathrm{Dom}(R)\\subseteq A", "after_tex": "\\mathrm{Dom}(R)=\\{1,3,4\\}\\neq A", "relation": "implies", "rule": "set_extensionality", "conditions": [], "explanation": { "zh-Hant": "⊇：S03。⊆：Dom(R)⊆A={1,2,3,4}，而 2 已被 S05 排除，所以 Dom(R) 的元素只能是 1、3、4。依集合相等，Dom(R)={1,3,4}。因 2∈A∖Dom(R)，Dom(R)≠A。", "en": "⊇: S03. ⊆: Dom(R)⊆A={1,2,3,4}, and 2 is excluded by S05, so the elements of Dom(R) can only be 1, 3, 4. By set equality, Dom(R)={1,3,4}. Since 2∈A∖Dom(R), Dom(R)≠A." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "S05", "set_extensionality" ], "explanation": { "zh-Hant": "需要兩個方向的包含；A 的四個元素都已處理（N3：3+1=4）。", "en": "Both inclusions are needed; all four elements of A are accounted for (N3: 3+1=4)." } }, "leaf_basis": "set_extensionality + subset_def + roster_membership" }, { "id": "S07", "uses": [ "S05", "G2", "function_graph" ], "before_tex": "2\\notin\\mathrm{Dom}(R),\\ (1,2)\\in R,\\ (1,3)\\in R", "after_tex": "R\\ \\text{is not a function}\\ A\\to B", "relation": "implies", "rule": "function_graph", "conditions": [], "explanation": { "zh-Hant": "函數 A→B 要求每個 x∈A 恰有一個 y。R 有兩處不符：2 沒有任何 y（S05，不是總定義）；1 有兩個 y，即 2 與 3，且 2≠3（不是單值）。任一處就足以否定。MKO 第一段指出關係不必是總函數。", "en": "A function A→B requires exactly one y for each x∈A. R fails in two places: 2 has no y at all (S05, not total), and 1 has two values, 2 and 3, with 2≠3 (not single-valued). Either failure alone is enough. MKO paragraph one notes that a relation need not be a total function." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S05", "function_graph" ], "explanation": { "zh-Hant": "只需一個失敗點（N6 核對 2≠3）；不宣稱 R 經修補後會成為函數。", "en": "One failure suffices (N6 checks 2≠3); it is not claimed that R becomes a function after any repair." } }, "leaf_basis": "function_graph + numeric_evaluation" }, { "id": "S08", "uses": [ "mko_relation_def", "G2", "G3", "exists_intro" ], "before_tex": "y\\in R[C],\\quad C=\\{1,3\\}", "after_tex": "2\\in R[C],\\ 3\\in R[C]", "relation": "implies", "rule": "mko_relation_def", "conditions": [], "explanation": { "zh-Hant": "依 MKO，y∈R[C] 當且僅當 y∈B 且有某個 x∈C 使 (x,y)∈R。第一分量在 C={1,3} 的對是 (1,2)、(1,3)、(3,3)，給出輸出 2、3、3。依 ∃引入，2∈R[C]（見證 x=1），3∈R[C]（見證 x=1 或 x=3）。", "en": "By the MKO, y∈R[C] exactly when y∈B and some x∈C has (x,y)∈R. The pairs with first coordinate in C={1,3} are (1,2), (1,3), (3,3), giving the outputs 2, 3, 3. By ∃-introduction, 2∈R[C] (witness x=1) and 3∈R[C] (witness x=1 or x=3)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_relation_def", "exists_intro" ], "explanation": { "zh-Hant": "每個輸出需要一個在 C 中的見證；此步只證 ⊇ 方向。", "en": "Each output needs a witness in C; this step proves only the ⊇ direction." } }, "leaf_basis": "mko_relation_def + exists_intro + set_builder" }, { "id": "S09", "uses": [ "S08", "G2", "G3", "set_extensionality" ], "before_tex": "2,3\\in R[C]", "after_tex": "R[C]=\\{2,3\\}", "relation": "implies", "rule": "set_extensionality", "conditions": [], "explanation": { "zh-Hant": "⊆：B 的其餘元素是 4、5、6。R 中沒有第二分量為 4 或 5 的對；第二分量為 6 的只有 (4,6)，其第一分量 4∉C（N4：(4−1)(4−3)≠0）。所以 4、5、6∉R[C]。合併 S08，R[C]={2,3}。輸出 3 出現兩次（來自 (1,3) 與 (3,3)），在集合中只列一次。", "en": "⊆: the other elements of B are 4, 5, 6. No pair of R has second coordinate 4 or 5; the only pair with second coordinate 6 is (4,6), whose first coordinate 4∉C (N4: (4−1)(4−3)≠0). So 4, 5, 6∉R[C]. With S08, R[C]={2,3}. The output 3 occurs twice (from (1,3) and (3,3)) but is listed once in the set." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S08", "set_extensionality", "roster_membership" ], "explanation": { "zh-Hant": "⊆ 方向需要排除 B 的每個其餘元素；兩個方向合起來才得集合相等。", "en": "The ⊆ direction must exclude every remaining element of B; only both directions together give set equality." } }, "leaf_basis": "set_extensionality + roster_membership + cartesian_product_def" }, { "id": "S10", "uses": [ "mko_relation_def", "empty_set_def" ], "before_tex": "R[\\varnothing]", "after_tex": "R[\\varnothing]=\\varnothing", "relation": "implies", "rule": "mko_relation_def", "conditions": [], "explanation": { "zh-Hant": "y∈R[∅] 需要某個 x∈∅ 使 (x,y)∈R；∅ 沒有元素，所以對每個 y 這都不成立，R[∅]=∅。這正是 MKO 第二段所說：空輸入集合的像為空。", "en": "y∈R[∅] would need some x∈∅ with (x,y)∈R; ∅ has no elements, so this fails for every y, and R[∅]=∅. This is what MKO paragraph two says: the image of the empty input set is empty." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_relation_def", "empty_set_def", "quantifier_negation" ], "explanation": { "zh-Hant": "這對任何關係都成立，不依賴 R 的具體資料。", "en": "This holds for any relation and does not depend on the data of R." } }, "leaf_basis": "mko_relation_def + empty_set_def + quantifier_negation + set_extensionality" }, { "id": "S11", "uses": [ "mko_relation_def", "G2", "L_converse_member" ], "before_tex": "R=\\{(1,2),(1,3),(3,3),(4,6)\\}", "after_tex": "R^{-1}=\\{(2,1),(3,1),(3,3),(6,4)\\}\\subseteq B\\times A", "relation": "implies", "rule": "mko_relation_def", "conditions": [], "explanation": { "zh-Hant": "依 MKO，R⁻¹={(y,x) ∣ (x,y)∈R}：把每一對交換方向。(1,2)→(2,1)、(1,3)→(3,1)、(3,3)→(3,3)、(4,6)→(6,4)。各對第一分量 2、3、3、6 在 B 中，第二分量 1、1、3、4 在 A 中，所以 R⁻¹⊆B×A：來源與目標也一起交換。由 L_converse_member，R⁻¹ 恰好由這四對組成。", "en": "By the MKO, R⁻¹={(y,x) ∣ (x,y)∈R}: reverse each pair. (1,2)→(2,1), (1,3)→(3,1), (3,3)→(3,3), (4,6)→(6,4). The first coordinates 2, 3, 3, 6 are in B and the second coordinates 1, 1, 3, 4 are in A, so R⁻¹⊆B×A: source and target are swapped as well. By L_converse_member, R⁻¹ consists of exactly these four pairs." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_relation_def", "L_converse_member", "cartesian_product_def" ], "explanation": { "zh-Hant": "需要 R 的列舉完整；(3,3) 交換後不變，但仍是 R⁻¹ 的元素。", "en": "It needs the listing of R to be complete; (3,3) is unchanged by the swap but is still an element of R⁻¹." } }, "leaf_basis": "mko_relation_def + cartesian_product_def + set_extensionality" }, { "id": "S11.1", "uses": [ "mko_relation_def", "cartesian_product_def" ], "before_tex": "(y,x)\\in S^{-1}", "after_tex": "(x,y)\\in S", "relation": "equivalent", "rule": "cartesian_product_def", "conditions": [ "S\\subseteq X\\times Y" ], "explanation": { "zh-Hant": "對任何關係 S⊆X×Y。⇐：若 (x,y)∈S，依定義 (y,x)∈S⁻¹。⇒：若 (y,x)∈S⁻¹，則 (y,x)=(y′,x′) 對某個 (x′,y′)∈S；有序對相等給出 y=y′、x=x′，所以 (x,y)=(x′,y′)∈S。這證得引理 L_converse_member。", "en": "Let S⊆X×Y be any relation. ⇐: if (x,y)∈S, then (y,x)∈S⁻¹ by definition. ⇒: if (y,x)∈S⁻¹, then (y,x)=(y′,x′) for some (x′,y′)∈S; equality of ordered pairs gives y=y′ and x=x′, so (x,y)=(x′,y′)∈S. This proves Lemma L_converse_member." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mko_relation_def", "cartesian_product_def", "eq_replace" ], "explanation": { "zh-Hant": "兩個方向分別由逆關係的定義與有序對相等條件給出。", "en": "The two directions come from the definition of the converse and from the equality condition for ordered pairs." } }, "leaf_basis": "mko_relation_def + cartesian_product_def + eq_replace" }, { "id": "S12", "uses": [ "S01", "S11", "set_extensionality" ], "before_tex": "(1,2)\\in R,\\ (2,1)\\in R^{-1}", "after_tex": "(2,1)\\notin R,\\ \\text{so}\\ R^{-1}\\neq R", "relation": "implies", "rule": "set_extensionality", "conditions": [], "explanation": { "zh-Hant": "有序對保留方向：(2,1)∈R⁻¹，但 (2,1)∉R，因為 R⊆A×B（S01）而第二分量 1∉B。依集合相等，R⁻¹≠R。雖然 A 與 B 有共同標籤 2、3、4，(1,2) 與 (2,1) 仍是不同的有序對。", "en": "Ordered pairs keep their direction: (2,1)∈R⁻¹ but (2,1)∉R, because R⊆A×B (S01) while the second coordinate 1∉B. By set equality, R⁻¹≠R. Although A and B share the labels 2, 3, 4, (1,2) and (2,1) are still different ordered pairs." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S01", "S11", "set_extensionality", "cartesian_product_def" ], "explanation": { "zh-Hant": "一個屬於一邊而不屬於另一邊的元素就足以說明兩集合不等。", "en": "One element lying in one set but not the other suffices to show the sets differ." } }, "leaf_basis": "set_extensionality + cartesian_product_def + subset_def" }, { "id": "S13", "uses": [ "S11", "L_dom_converse" ], "before_tex": "\\mathrm{Dom}(R^{-1})", "after_tex": "\\mathrm{Dom}(R^{-1})=R[A]=\\{2,3,6\\}", "relation": "theorem_application", "rule": "L_dom_converse", "conditions": [], "explanation": { "zh-Hant": "依引理 L_dom_converse，Dom(R⁻¹)=R[A]。計算 R[A]：R 的四對第一分量都在 A 中，所以全部計入，輸出為 2、3、3、6，集合為 {2,3,6}。直接核對：R⁻¹ 各對的第一分量是 2、3、3、6，一致。B 中的 4、5 不在其中。", "en": "By Lemma L_dom_converse, Dom(R⁻¹)=R[A]. Compute R[A]: all four pairs of R have first coordinate in A, so all count, with outputs 2, 3, 3, 6, giving the set {2,3,6}. Direct check: the first coordinates of the pairs of R⁻¹ are 2, 3, 3, 6, in agreement. The elements 4 and 5 of B are not among them." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_dom_converse", "S11" ], "explanation": { "zh-Hant": "引理在 S13.1 證明；R[A] 的計算方法與 S08、S09 相同。", "en": "The lemma is proved in S13.1; R[A] is computed by the same method as in S08 and S09." } }, "leaf_basis": "mko_relation_def + set_extensionality + exists_intro" }, { "id": "S13.1", "uses": [ "mko_relation_def", "L_converse_member", "set_extensionality" ], "before_tex": "y\\in\\mathrm{Dom}(R^{-1})", "after_tex": "y\\in R[A]", "relation": "equivalent", "rule": "L_converse_member", "conditions": [], "explanation": { "zh-Hant": "對 R⁻¹⊆B×A 套用 MKO 的定義域定義：y∈Dom(R⁻¹) ⇔ y∈B 且 ∃x∈A (y,x)∈R⁻¹。由 L_converse_member，(y,x)∈R⁻¹ ⇔ (x,y)∈R，所以上式 ⇔ y∈B 且 ∃x∈A (x,y)∈R ⇔ y∈R[A]（取 C=A）。依集合相等，Dom(R⁻¹)=R[A]。這證得引理 L_dom_converse。", "en": "Apply the MKO's definition of domain to R⁻¹⊆B×A: y∈Dom(R⁻¹) ⇔ y∈B and ∃x∈A (y,x)∈R⁻¹. By L_converse_member, (y,x)∈R⁻¹ ⇔ (x,y)∈R, so this is ⇔ y∈B and ∃x∈A (x,y)∈R ⇔ y∈R[A] (with C=A). By set equality, Dom(R⁻¹)=R[A]. This proves Lemma L_dom_converse." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mko_relation_def", "L_converse_member", "logic_chain" ], "explanation": { "zh-Hant": "鏈中每一環都是等價，所以整條鏈雙向成立。", "en": "Every link of the chain is an equivalence, so the whole chain holds both ways." } }, "leaf_basis": "mko_relation_def + set_builder + set_extensionality + logic_chain" }, { "id": "S14", "uses": [ "S11", "S13", "function_graph" ], "before_tex": "R^{-1}=\\{(2,1),(3,1),(3,3),(6,4)\\}", "after_tex": "R^{-1}\\ \\text{is not a function}\\ B\\to A", "relation": "implies", "rule": "function_graph", "conditions": [], "explanation": { "zh-Hant": "函數 B→A 要求每個 y∈B 恰有一個 x。4∈B 不在 Dom(R⁻¹)={2,3,6} 中（N5），沒有任何 x；3 有兩個 x，即 1 與 3。所以 R⁻¹ 不是函數。逆關係總是存在，但它不自動是反函數，也不提供單射、滿射或可逆性的證明（MKO 第三段）。", "en": "A function B→A requires exactly one x for each y∈B. 4∈B is not in Dom(R⁻¹)={2,3,6} (N5) and has no x at all; 3 has two values of x, namely 1 and 3. So R⁻¹ is not a function. The converse relation always exists, but it is not automatically an inverse function and gives no proof of injectivity, surjectivity or invertibility (MKO paragraph three)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S13", "function_graph" ], "explanation": { "zh-Hant": "一個失敗點就足以否定；不宣稱任何關係的逆關係都不是函數。", "en": "One failure suffices; it is not claimed that the converse of a relation is never a function." } }, "leaf_basis": "function_graph + roster_membership" }, { "id": "S15", "uses": [ "S11", "L_converse_twice" ], "before_tex": "(R^{-1})^{-1}", "after_tex": "(R^{-1})^{-1}=R", "relation": "theorem_application", "rule": "L_converse_twice", "conditions": [], "explanation": { "zh-Hant": "依引理 L_converse_twice，再交換一次就回到 R：(2,1)→(1,2)、(3,1)→(1,3)、(3,3)→(3,3)、(6,4)→(4,6)，正是 R 的四對。取逆關係這個運算在關係層次可以還原，這與 R⁻¹ 是否為函數無關。", "en": "By Lemma L_converse_twice, swapping again returns R: (2,1)→(1,2), (3,1)→(1,3), (3,3)→(3,3), (6,4)→(4,6), exactly the four pairs of R. Taking the converse can be undone at the level of relations, regardless of whether R⁻¹ is a function." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_converse_twice" ], "explanation": { "zh-Hant": "這是關係層次的事實；不宣稱 R 或 R⁻¹ 有反函數。", "en": "This is a fact about relations; it does not claim that R or R⁻¹ has an inverse function." } }, "leaf_basis": "mko_relation_def + set_extensionality" }, { "id": "S15.1", "uses": [ "L_converse_member", "set_extensionality" ], "before_tex": "(x,y)\\in(R^{-1})^{-1}", "after_tex": "(x,y)\\in R", "relation": "equivalent", "rule": "L_converse_member", "conditions": [], "explanation": { "zh-Hant": "對關係 R⁻¹ 套用 L_converse_member：(x,y)∈(R⁻¹)⁻¹ ⇔ (y,x)∈R⁻¹；再對 R 套用一次：(y,x)∈R⁻¹ ⇔ (x,y)∈R。兩集合的元素相同，依集合相等 (R⁻¹)⁻¹=R。這證得引理 L_converse_twice。", "en": "Apply L_converse_member to the relation R⁻¹: (x,y)∈(R⁻¹)⁻¹ ⇔ (y,x)∈R⁻¹; apply it once more to R: (y,x)∈R⁻¹ ⇔ (x,y)∈R. The two sets have the same elements, so (R⁻¹)⁻¹=R by set equality. This proves Lemma L_converse_twice." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_converse_member", "logic_chain" ], "explanation": { "zh-Hant": "兩次等價串接仍是等價。", "en": "Chaining two equivalences gives an equivalence." } }, "leaf_basis": "set_extensionality + logic_chain + cartesian_product_def" }, { "id": "S16", "uses": [ "S01", "S06", "S07", "S09", "S10", "S11", "S13", "S14" ], "before_tex": "\\text{S01, S06, S07, S09, S10, S11, S13, S14}", "after_tex": "\\mathrm{Dom}(R)=\\{1,3,4\\},\\ R[C]=\\{2,3\\},\\ R[\\varnothing]=\\varnothing,\\ \\mathrm{Dom}(R^{-1})=\\{2,3,6\\}", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "彙整：R⊆A×B（S01）；Dom(R)={1,3,4}≠A（S06）；R 不是函數 A→B（S07）；R[C]={2,3}（S09）；R[∅]=∅（S10）；R⁻¹={(2,1),(3,1),(3,3),(6,4)}⊆B×A（S11）；Dom(R⁻¹)=R[A]={2,3,6}（S13）；R⁻¹ 不是函數 B→A（S14）。", "en": "Summary: R⊆A×B (S01); Dom(R)={1,3,4}≠A (S06); R is not a function A→B (S07); R[C]={2,3} (S09); R[∅]=∅ (S10); R⁻¹={(2,1),(3,1),(3,3),(6,4)}⊆B×A (S11); Dom(R⁻¹)=R[A]={2,3,6} (S13); R⁻¹ is not a function B→A (S14)." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S01", "S06", "S07", "S09", "S10", "S11", "S13", "S14", "and_intro_elim" ], "explanation": { "zh-Hant": "目標 G：題目要求的各項計算與判定。足夠條件 H：所列各步的結論；H⇒G 只是合取。H 的每一部分都由 G1 到 G3 正向證成，沒有把 G 當作前提。", "en": "Goal G: the computations and decisions the task asks for. Sufficient condition H: the conclusions of the listed steps; H⇒G is just conjunction. Each part of H is proved forwards from G1 to G3, and G is never used as a premise." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "因 R⊆A×B，R 是從 A 到 B 的關係。Dom(R)={1,3,4}，與宣告來源 A 不同（2 沒有出邊）；R[C]={2,3}，R[∅]=∅；R⁻¹={(2,1),(3,1),(3,3),(6,4)}⊆B×A，Dom(R⁻¹)=R[A]={2,3,6}，且 (R⁻¹)⁻¹=R。R 不是函數 A→B（2 無輸出、1 有兩個輸出），R⁻¹ 也不是函數 B→A（4 無輸出、3 有兩個輸出），所以逆關係並不給出反函數。", "en": "Since R⊆A×B, R is a relation from A to B. Dom(R)={1,3,4}, which differs from the declared source A (2 has no outgoing pair); R[C]={2,3} and R[∅]=∅; R⁻¹={(2,1),(3,1),(3,3),(6,4)}⊆B×A, Dom(R⁻¹)=R[A]={2,3,6}, and (R⁻¹)⁻¹=R. R is not a function A→B (2 has no output, 1 has two), and R⁻¹ is not a function B→A (4 has no output, 3 has two), so the converse relation does not give an inverse function." }, "negative_controls": [ { "claim": { "zh-Hant": "R 宣告為從 A 到 B 的關係，所以 Dom(R)=A", "en": "R is declared as a relation from A to B, so Dom(R)=A" }, "expected": "reject", "reason": { "zh-Hant": "2∈A 沒有任何 (2,y)∈R；Dom(R)={1,3,4}。", "en": "2∈A has no (2,y)∈R; Dom(R)={1,3,4}." } }, { "claim": { "zh-Hant": "R⁻¹ 是 R 的反函數", "en": "R⁻¹ is the inverse function of R" }, "expected": "reject", "reason": { "zh-Hant": "R⁻¹ 不是函數 B→A：4 沒有輸出，3 有 1 與 3 兩個輸出；R 本身也不是函數。", "en": "R⁻¹ is not a function B→A: 4 has no output and 3 has the two outputs 1 and 3; R itself is not a function either." } }, { "claim": { "zh-Hant": "R[C]={2,3,3}，有三個元素", "en": "R[C]={2,3,3}, which has three elements" }, "expected": "reject", "reason": { "zh-Hant": "集合中重複值只算一次；R[C]={2,3} 有兩個元素。", "en": "Repeated values count once in a set; R[C]={2,3} has two elements." } }, { "claim": { "zh-Hant": "(1,2)∈R，所以 (2,1)∈R", "en": "(1,2)∈R, so (2,1)∈R" }, "expected": "reject", "reason": { "zh-Hant": "R⊆A×B 而 1∉B，所以 (2,1)∉R；有序對保留方向。", "en": "R⊆A×B while 1∉B, so (2,1)∉R; ordered pairs keep their direction." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "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": "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": "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": "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": "exists_intro", "kind": "prerequisite", "statement": { "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。", "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)." } }, { "id": "forall_intro", "kind": "prerequisite", "statement": { "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。", "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain." } }, { "id": "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": "quantifier_negation", "kind": "prerequisite", "statement": { "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。", "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)." } }, { "id": "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": "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": "logic_chain", "kind": "prerequisite", "statement": { "zh-Hant": "推論的傳遞：由 P⇒Q 與 Q⇒R 得 P⇒R；⇔ 同理。", "en": "Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔." } }, { "id": "and_intro_elim", "kind": "prerequisite", "statement": { "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。", "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q." } }, { "id": "mko_relation_def", "kind": "cited", "source": { "mko_id": "mko-relation", "sha256": "3788215401bc9a5610691985883d04b506880cc9bd5e2ce0a98054cd5a5f0774", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "對集合 A、B，從 A 到 B 的二元關係 R 是 A×B 的子集合；xRy 表示 (x,y)∈R。關係定義域 Dom(R)={x∈A:存在 y∈B，使 (x,y)∈R}，關係像 R[C]={y∈B:存在 x∈C，使 (x,y)∈R}，逆關係 R⁻¹={(y,x):(x,y)∈R}⊆B×A。", "en": "For sets A and B, a binary relation R from A to B is a subset of A×B; xRy means (x,y)∈R. Its domain is Dom(R)={x∈A:there exists y∈B with (x,y)∈R}; its image of C is R[C]={y∈B:there exists x∈C with (x,y)∈R}; its converse is R⁻¹={(y,x):(x,y)∈R}⊆B×A." } }, { "id": "roster_membership", "kind": "prerequisite", "statement": { "zh-Hant": "列舉集合的成員關係：對以列舉給出的有限集合 S={s_i ∣ 1≤i≤n}，z∈S 當且僅當存在 i（1≤i≤n）使 z=s_i。", "en": "Membership in a listed set: for a finite set given by listing, S={s_i ∣ 1≤i≤n}, z∈S holds exactly when z=s_i for some i with 1≤i≤n." } }, { "id": "empty_set_def", "kind": "prerequisite", "statement": { "zh-Hant": "空集合：對每個 x，x∉∅。", "en": "Empty set: for every x, x∉∅." } }, { "id": "function_graph", "kind": "prerequisite", "statement": { "zh-Hant": "關係 F⊆X×Y 是總函數 X→Y（的圖形），當且僅當每個 x∈X 恰有一個 y∈Y 使 (x,y)∈F：至少一個（總定義）且至多一個（單值）。", "en": "A relation F⊆X×Y is (the graph of) a total function X→Y exactly when each x∈X has exactly one y∈Y with (x,y)∈F: at least one (total) and at most one (single-valued)." } }, { "id": "L_converse_member", "kind": "lemma", "proved_in": [ "S11.1" ], "statement": { "zh-Hant": "對任何關係 S⊆X×Y 與任何 x、y：(y,x)∈S⁻¹ ⇔ (x,y)∈S。", "en": "For any relation S⊆X×Y and any x, y: (y,x)∈S⁻¹ ⇔ (x,y)∈S." } }, { "id": "L_dom_converse", "kind": "lemma", "proved_in": [ "S13.1" ], "statement": { "zh-Hant": "對任何關係 R⊆A×B：Dom(R⁻¹)=R[A]，其中 Dom(R⁻¹)={y∈B ∣ 存在 x∈A 使 (y,x)∈R⁻¹}。", "en": "For any relation R⊆A×B: Dom(R⁻¹)=R[A], where Dom(R⁻¹)={y∈B ∣ there exists x∈A with (y,x)∈R⁻¹}." } }, { "id": "L_converse_twice", "kind": "lemma", "proved_in": [ "S15.1" ], "statement": { "zh-Hant": "對任何關係 R⊆A×B：(R⁻¹)⁻¹=R。", "en": "For any relation R⊆A×B: (R⁻¹)⁻¹=R." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S01", "lhs": "4*5", "rhs": "20" }, { "id": "N2", "step": "S04", "lhs": "(2-1)*(2-1)*(2-3)*(2-4)", "relation": "!=", "rhs": "0" }, { "id": "N3", "step": "S06", "lhs": "3+1", "rhs": "4" }, { "id": "N4", "step": "S09", "lhs": "(4-1)*(4-3)", "relation": "!=", "rhs": "0" }, { "id": "N5", "step": "S14", "lhs": "(4-2)*(4-3)*(4-6)", "relation": "!=", "rhs": "0" }, { "id": "N6", "step": "S07", "lhs": "2-3", "relation": "!=", "rhs": "0" } ] }, "presentation": { "md_scope": { "zh-Hant": "在 MKO 的範圍內工作：明示的有限整數集合與有序對資料。依 MKO 的定義逐步計算定義域、關係像與逆關係，並證明三個一般小引理（逆關係的成員條件、Dom(R⁻¹)=R[A]、(R⁻¹)⁻¹=R）。函數條件只作為額外限制來判定；不處理關係合成、等價關係或序關係；不改動 canonical MKO 或英文 edition。", "en": "Works inside the MKO's scope: explicit finite integer sets and ordered-pair data. Computes the domain, images and the converse step by step from the MKO's definitions, and proves three small general lemmas (membership in the converse, Dom(R⁻¹)=R[A], (R⁻¹)⁻¹=R). The function condition is checked only as an extra restriction; composition, equivalence relations and order relations are not treated; the canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "A,\\ B", "來源 {1,2,3,4} 與目標 {2,3,4,5,6}", "source {1,2,3,4} and target {2,3,4,5,6}" ], [ "R", "從 A 到 B 的關係，即 A×B 的子集合", "the relation from A to B, a subset of A×B" ], [ "\\mathrm{Dom}(R)", "關係定義域：實際有出邊的第一分量", "domain of the relation: first coordinates that actually occur" ], [ "R[C]", "C 在 R 下的關係像", "the image of C under R" ], [ "R^{-1}", "逆關係 {(y,x) ∣ (x,y)∈R}⊆B×A", "the converse {(y,x) ∣ (x,y)∈R}⊆B×A" ] ], "md_error_controls": [ { "title": "宣告來源不是定義域 / The declared source is not the domain", "zh": "MKO 指出的誤解之一是把宣告來源集合當成實際關係定義域。R 宣告為從 A={1,2,3,4} 出發，但 2 沒有任何對（S04、S05），所以 Dom(R)={1,3,4}。若直接把 A 叫作定義域，就漏掉了 2 這個差異。", "en": "One misconception named by the MKO is taking the declared source set as the actual domain. R is declared from A={1,2,3,4}, but 2 has no pair (S04, S05), so Dom(R)={1,3,4}. Calling A the domain would miss exactly the difference at 2." }, { "title": "逆關係不是反函數 / The converse is not an inverse function", "zh": "MKO 指出的另一個誤解是把逆關係當成已存在的反函數。R⁻¹ 對任何關係都存在，但在本例中 4∈B 沒有輸出、3 有兩個輸出（S14），所以 R⁻¹ 不是函數 B→A，更不是 R 的反函數；R 本身也不是函數（S07）。", "en": "The other misconception named by the MKO is treating the converse as an inverse function that already exists. R⁻¹ exists for every relation, but here 4∈B has no output and 3 has two (S14), so R⁻¹ is not a function B→A, let alone an inverse of R; R itself is not a function either (S07)." }, { "title": "有序對保留方向，像只列一次 / Pairs keep direction; images list each value once", "zh": "(1,2)∈R 不推出 (2,1)∈R：R⊆A×B 而 1∉B（S12）。A、B 共用標籤不使順序變得無關。計算 R[C] 時，輸出 3 來自兩個對，但在集合中只列一次，所以 R[C] 有 2 個元素而不是 3 個（S09）。", "en": "(1,2)∈R does not give (2,1)∈R: R⊆A×B while 1∉B (S12). Shared labels in A and B do not make the order irrelevant. In computing R[C], the output 3 comes from two pairs but is listed once in the set, so R[C] has 2 elements, not 3 (S09)." } ], "md_open_items": [ { "zh-Hant": "N2、N4、N5 以「差的乘積非零」表示某數不等於每一個列出的值（零乘積性質），只是對這些具體數字的重算；論證本身在對應步驟中逐項說明。", "en": "N2, N4 and N5 encode 'a number equals none of the listed values' as a nonzero product of differences (zero-product property); they only recompute these particular numbers, and the argument itself is given item by item in the corresponding steps." }, { "zh-Hant": "本稿不處理關係合成、等價關係、序關係或無限關係。", "en": "This guide does not treat composition of relations, equivalence relations, order relations or infinite relations." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "43c26b506fc0b915b03ad39f9c5dcf722f8dfaf1e7be2295ec1d0d519436062d", "original_source_sha256": "43c26b506fc0b915b03ad39f9c5dcf722f8dfaf1e7be2295ec1d0d519436062d", "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." } }