RA2016: Ejercicios de eliminación de duplicados en Isabelle/HOL
En la segunda parte de la clase de hoy del curso de Razonamiento automático se han comentado las soluciones de la 5ª relación de ejercicios sobre eliminación de elementos duplicados de listas en Isabelle/HOL.
La teoría con las soluciones de los ejercicios es la siguiente
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chapter {* R5: Eliminación de duplicados *} theory R5 imports Main begin text {* --------------------------------------------------------------------- Ejercicio 1. Definir la funcion primitiva recursiva estaEn :: 'a ⇒ 'a list ⇒ bool tal que (estaEn x xs) se verifica si el elemento x está en la lista xs. Por ejemplo, estaEn (2::nat) [3,2,4] = True estaEn (1::nat) [3,2,4] = False --------------------------------------------------------------------- *} fun estaEn :: "'a ⇒ 'a list ⇒ bool" where "estaEn x [] = False" | "estaEn x (a#xs) = (x=a ∨ estaEn x xs)" value "estaEn (2::nat) [3,2,4] = True" value "estaEn (1::nat) [3,2,4] = False" text {* --------------------------------------------------------------------- Ejercicio 2. Definir la función primitiva recursiva sinDuplicados :: 'a list ⇒ bool tal que (sinDuplicados xs) se verifica si la lista xs no contiene duplicados. Por ejemplo, sinDuplicados [1::nat,4,2] = True sinDuplicados [1::nat,4,2,4] = False --------------------------------------------------------------------- *} fun sinDuplicados :: "'a list ⇒ bool" where "sinDuplicados [] = True" | "sinDuplicados (a#xs) = ((¬ estaEn a xs) ∧ sinDuplicados xs)" value "sinDuplicados [1::nat,4,2] = True" value "sinDuplicados [1::nat,4,2,4] = False" text {* --------------------------------------------------------------------- Ejercicio 3. Definir la función primitiva recursiva borraDuplicados :: 'a list ⇒ bool tal que (borraDuplicados xs) es la lista obtenida eliminando los elementos duplicados de la lista xs. Por ejemplo, borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3] Nota: La función borraDuplicados es equivalente a la predefinida remdups. --------------------------------------------------------------------- *} fun borraDuplicados :: "'a list ⇒ 'a list" where "borraDuplicados [] = []" | "borraDuplicados (a#xs) = (if estaEn a xs then borraDuplicados xs else (a#borraDuplicados xs))" value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]" text {* --------------------------------------------------------------------- Ejercicio 4.1. Demostrar o refutar automáticamente length (borraDuplicados xs) ≤ length xs --------------------------------------------------------------------- *} -- "La demostración automática es" lemma length_borraDuplicados: "length (borraDuplicados xs) ≤ length xs" by (induct xs) simp_all text {* --------------------------------------------------------------------- Ejercicio 4.2. Demostrar o refutar detalladamente length (borraDuplicados xs) ≤ length xs --------------------------------------------------------------------- *} -- "La demostración estructurada es" lemma length_borraDuplicados_2: "length (borraDuplicados xs) ≤ length xs" proof (induct xs) show "length (borraDuplicados []) ≤ length []" by simp next fix a xs assume HI: "length (borraDuplicados xs) ≤ length xs" thus "length (borraDuplicados (a#xs)) ≤ length (a#xs)" proof (cases) assume "estaEn a xs" thus "length (borraDuplicados (a#xs)) ≤ length (a#xs)" using HI by auto next assume "(¬ estaEn a xs)" thus "length (borraDuplicados (a#xs)) ≤ length (a#xs)" using HI by auto qed qed -- "La demostración detallada es" lemma length_borraDuplicados_3: "length (borraDuplicados xs) ≤ length xs" proof (induct xs) show "length (borraDuplicados []) ≤ length []" by simp next fix a xs assume HI: "length (borraDuplicados xs) ≤ length xs" show "length (borraDuplicados (a#xs)) ≤ length (a#xs)" proof (cases) assume "estaEn a xs" hence "length (borraDuplicados (a#xs)) = length (borraDuplicados xs)" by simp also have "… ≤ length xs" using HI by simp also have "… ≤ length (a#xs)" by simp finally show ?thesis . next assume "(¬ estaEn a xs)" hence "length (borraDuplicados (a#xs)) = length (a#(borraDuplicados xs))" by simp also have "… = 1 + length (borraDuplicados xs)" by simp also have "… ≤ 1 + length xs" using HI by simp also have "… = length (a#xs)" by simp finally show ?thesis . qed qed text {* --------------------------------------------------------------------- Ejercicio 5.1. Demostrar o refutar automáticamente estaEn a (borraDuplicados xs) = estaEn a xs --------------------------------------------------------------------- *} -- "La demostración automática es" lemma estaEn_borraDuplicados: "estaEn a (borraDuplicados xs) = estaEn a xs" by (induct xs) auto text {* --------------------------------------------------------------------- Ejercicio 5.2. Demostrar o refutar detalladamente estaEn a (borraDuplicados xs) = estaEn a xs Nota: Para la demostración de la equivalencia se puede usar proof (rule iffI) La regla iffI es ⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q --------------------------------------------------------------------- *} -- "La demostración estructurada es" lemma estaEn_borraDuplicados_2: "estaEn a (borraDuplicados xs) = estaEn a xs" proof (induct xs) show "estaEn a (borraDuplicados []) = estaEn a []" by simp next fix b xs assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs" show "estaEn a (borraDuplicados (b#xs)) = estaEn a (b#xs)" proof (rule iffI) assume c1: "estaEn a (borraDuplicados (b#xs))" show "estaEn a (b#xs)" proof (cases) assume "estaEn b xs" thus "estaEn a (b#xs)" using c1 HI by auto next assume "¬ estaEn b xs" thus "estaEn a (b#xs)" using c1 HI by auto qed next assume c2: "estaEn a (b#xs)" show "estaEn a (borraDuplicados (b#xs))" proof (cases) assume "a=b" thus "estaEn a (borraDuplicados (b#xs))" using HI by auto next assume "a≠b" thus "estaEn a (borraDuplicados (b#xs))" using `a≠b` c2 HI by auto qed qed qed -- "La demostración detallada es" lemma estaEn_borraDuplicados_3: "estaEn a (borraDuplicados xs) = estaEn a xs" proof (induct xs) show "estaEn a (borraDuplicados []) = estaEn a []" by simp next fix b xs assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs" show "estaEn a (borraDuplicados (b#xs)) = estaEn a (b#xs)" proof (rule iffI) assume c1: "estaEn a (borraDuplicados (b#xs))" show "estaEn a (b#xs)" proof (cases) assume "estaEn b xs" hence "estaEn a (borraDuplicados xs)" using c1 by simp hence "estaEn a xs" using HI by simp thus "estaEn a (b#xs)" by simp next assume "¬ estaEn b xs" hence "estaEn a (b#(borraDuplicados xs))" using c1 by simp hence "a=b ∨ (estaEn a (borraDuplicados xs))" by simp hence "a=b ∨ (estaEn a xs)" using HI by simp thus "estaEn a (b#xs)" by simp qed next assume c2: "estaEn a (b#xs)" show "estaEn a (borraDuplicados (b#xs))" proof (cases) assume "a=b" thus "estaEn a (borraDuplicados (b#xs))" using HI by auto next assume "a≠b" hence "estaEn a xs" using c2 by simp hence "estaEn a (borraDuplicados xs)" using HI by simp thus "estaEn a (borraDuplicados (b#xs))" using `a≠b` by simp qed qed qed text {* --------------------------------------------------------------------- Ejercicio 6.1. Demostrar o refutar automáticamente sinDuplicados (borraDuplicados xs) --------------------------------------------------------------------- *} -- "La demostración automática" lemma sinDuplicados_borraDuplicados: "sinDuplicados (borraDuplicados xs)" by (induct xs) (auto simp add: estaEn_borraDuplicados) text {* --------------------------------------------------------------------- Ejercicio 6.2. Demostrar o refutar detalladamente sinDuplicados (borraDuplicados xs) --------------------------------------------------------------------- *} -- "La demostración estructurada es" lemma sinDuplicados_borraDuplicados_2: "sinDuplicados (borraDuplicados xs)" proof (induct xs) show "sinDuplicados (borraDuplicados [])" by simp next fix a xs assume HI: "sinDuplicados (borraDuplicados xs)" show "sinDuplicados (borraDuplicados (a#xs))" proof (cases) assume "estaEn a xs" thus "sinDuplicados (borraDuplicados (a#xs))" using HI by simp next assume "¬ estaEn a xs" thus "sinDuplicados (borraDuplicados (a#xs))" using `¬ estaEn a xs` HI by (auto simp add: estaEn_borraDuplicados) qed qed -- "La demostración detallada es" lemma sinDuplicados_borraDuplicados_3: "sinDuplicados (borraDuplicados xs)" proof (induct xs) show "sinDuplicados (borraDuplicados [])" by simp next fix a xs assume HI: "sinDuplicados (borraDuplicados xs)" show "sinDuplicados (borraDuplicados (a#xs))" proof (cases) assume "estaEn a xs" thus "sinDuplicados (borraDuplicados (a#xs))" using HI by simp next assume "¬ estaEn a xs" hence "¬ (estaEn a xs) ∧ sinDuplicados (borraDuplicados xs)" using HI by simp hence "¬ estaEn a (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs)" by (simp add: estaEn_borraDuplicados) hence "sinDuplicados (a#borraDuplicados xs)" by simp thus "sinDuplicados (borraDuplicados (a#xs))" using `¬ estaEn a xs` by simp qed qed text {* --------------------------------------------------------------------- Ejercicio 7. Demostrar o refutar: borraDuplicados (rev xs) = rev (borraDuplicados xs) --------------------------------------------------------------------- *} -- "Se busca un contraejemplo con" lemma "borraDuplicados (rev xs) = rev (borraDuplicados xs)" quickcheck oops text {* El contraejemplo encontrado es xs = [3, 2, 3] En efecto, borraDuplicados (rev xs) = borraDuplicados (rev [3,2,3]) = [2,3] rev (borraDuplicados xs) = rev (borraDuplicados [3,2,3]) = [3,2] *} end |