https://www.glc.us.es/~jalonso/RA2010/api.php?action=feedcontributions&feedformat=atom&user=Jgiraldez 2026-07-17T14:51:39Z Contribuciones del usuario MediaWiki 1.31.14 https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_3&diff=136 2011-02-14T16:25:39Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 3ª relación de ejercicios *}<br /> <br /> theory Relacion_3<br /> imports Main <br /> begin<br /> <br /> section {* Cons inverso *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 1. Definir recursivamente la función <br /> snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot;<br /> tal que (snoc xs a) es la lista obtenida al añadir el elemento a al<br /> final de la lista xs. Por ejemplo, <br /> value &quot;snoc [2,5] (3::int)&quot; == [2,5,3]<br /> <br /> Nota: No usar @.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> primrec snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot; where<br /> &quot;snoc [] y = [y]&quot;<br /> | &quot;snoc (x#xs) y = x#(snoc xs y)&quot;<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 2. Demostrar el siguiente teorema <br /> snoc xs a = xs @ [a]<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del lema es&quot;<br /> lemma &quot;snoc xs a = xs @ [a]&quot;<br /> by (induct xs) auto<br /> <br /> -- &quot;La demostración estructurada del lema es&quot;<br /> lemma snoc_append: &quot;snoc xs a = xs @ [a]&quot;<br /> proof (induct xs)<br /> show &quot;snoc [] a = [] @ [a]&quot; by simp<br /> next <br /> fix b xs <br /> assume HI: &quot;snoc xs a = xs @ [a]&quot;<br /> thus &quot;snoc (b # xs) a = (b # xs) @ [a]&quot;<br /> proof -<br /> have &quot;snoc (b # xs) a = b#(snoc xs a)&quot; by simp<br /> also have &quot;... = b#(xs @ [a])&quot; using HI by simp<br /> also have &quot;... = (b#xs) @ [a]&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 3. Demostrar el siguiente teorema <br /> rev (x # xs) = snoc (rev xs) x&quot;<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del teorema es&quot;<br /> theorem rev_cons_auto: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> by (simp add:snoc_append)<br /> <br /> -- &quot;La demostración estructurada del teorema es&quot;<br /> theorem rev_cons: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> proof -<br /> have &quot;rev(x#xs)=(rev xs) @ [x]&quot; by simp <br /> also have &quot;... = snoc (rev xs) x&quot; by (simp add:snoc_append)<br /> finally show ?thesis . <br /> qed<br /> <br /> <br /> <br /> <br /> <br /> section {* Cuantificadores sobre listas *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 4. Definir la función <br /> todos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (todos p xs) se verifica si todos los elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> todos (λx. 1&lt;length x) [[2,1,4],[1,3]]<br /> ¬ todos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> <br /> Nota: La función todos es equivalente a la predefinida list_all. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec todos :: &quot;(&#039;a ⇒ bool) ⇒ (&#039;a list ⇒ bool)&quot; where<br /> &quot;todos P [] = True&quot;<br /> |&quot;todos P (x#xs) = (P x ∧ todos P xs)&quot;<br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 5. Definir la función <br /> algunos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (algunos p xs) se verifica si algunos elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> algunos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> ¬ algunos (λx. 1&lt;length x) [[],[3]]&quot;<br /> <br /> Nota: La función algunos es equivalente a la predefinida list_ex. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec algunos :: &quot;(&#039;a =&gt; bool) =&gt; &#039;a list =&gt; bool&quot; where<br /> &quot;algunos P [] = False&quot;<br /> | &quot;algunos P (x#xs) = (P x ∨ algunos P xs)&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 6. Demostrar o refutar: <br /> todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> proof (induct xs)<br /> show &quot;todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])&quot; by simp<br /> next<br /> fix a xs<br /> assume 1: &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> thus &quot;todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))&quot;<br /> proof -<br /> have 2: &quot;todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos (λx. P x ∧ Q x) xs))&quot; by simp<br /> also have 3: &quot;... = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))&quot; using 1 by simp<br /> also have 4: &quot;... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))&quot; by blast<br /> also have 5: &quot;... = ((todos P (a # xs) ∧ todos Q (a # xs)))&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 7. Demostrar o refutar: <br /> todos P (x @ y) = (todos P x ∧ todos P y)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> by (induct x) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> proof (induct x)<br /> show &quot;todos P ([] @ y) = (todos P [] ∧ todos P y)&quot; by simp<br /> next<br /> fix a x y<br /> assume 1: &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> thus &quot;todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)&quot;<br /> proof -<br /> have &quot;(todos P ((a # x) @ y)) = (P a ∧ todos P (x @ y))&quot; by simp<br /> also have &quot;... = (P a ∧ todos P x ∧ todos P y)&quot; using 1 by simp<br /> also have &quot;... = (todos P (a #x) ∧ todos P y)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 8. Demostrar o refutar: <br /> todos P (rev xs) = todos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> proof (induct xs)<br /> show &quot;todos P (rev []) = todos P []&quot; by simp<br /> next<br /> fix a xs<br /> assume 1: &quot;todos P (rev xs) = todos P xs&quot;<br /> thus &quot;todos P (rev (a # xs)) = todos P (a # xs)&quot;<br /> proof -<br /> have &quot;todos P (rev (a # xs)) = todos P (rev (xs) @ [a])&quot; by simp<br /> also have &quot;... = (todos P (rev (xs)) ∧ todos P [a])&quot; by simp<br /> also have &quot;... = (todos P xs ∧ todos P [a])&quot; using 1 by simp<br /> also have &quot;... = (todos P [a] ∧ todos P xs)&quot; by blast<br /> also have &quot;... = todos P ([a] @ xs)&quot; by simp<br /> also have &quot;... = todos P (a # xs)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 9. Demostrar o refutar:<br /> algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;Se busca un contraejemplo con nitpick&quot;<br /> lemma &quot;algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> El contraejemplo encontrado es<br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 10. Demostrar o refutar: <br /> algunos P (map f xs) = algunos (P ∘ f) xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P o f) xs&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P ∘ f) xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 11. Demostrar o refutar: <br /> algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> by (induct xs) auto<br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma algunos_append: <br /> &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> proof (induct xs)<br /> show &quot;algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)&quot; by simp<br /> next<br /> fix a xs<br /> assume 1: &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> thus &quot;algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)&quot;<br /> proof -<br /> have &quot;algunos P ((a # xs) @ ys) = (P a ∨ algunos P (xs @ ys))&quot; by simp<br /> also have &quot;... = (P a ∨ (algunos P xs ∨ algunos P ys))&quot; using 1 by simp<br /> also have &quot;... = ((P a ∨ algunos P xs) ∨ algunos P ys)&quot; by blast<br /> also have &quot;... = (algunos P (a#xs) ∨ algunos P ys)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 12. Demostrar o refutar: <br /> algunos P (rev xs) = algunos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> proof (induct xs)<br /> show &quot; algunos P (rev []) = algunos P []&quot; by simp<br /> next<br /> fix a xs<br /> assume 1: &quot;algunos P (rev xs) = algunos P xs&quot;<br /> thus &quot;algunos P (rev (a # xs)) = algunos P (a # xs)&quot;<br /> proof -<br /> have &quot;algunos P (rev (a # xs)) = algunos P (rev (xs) @ [a])&quot; by simp<br /> also have &quot;... = (algunos P (rev (xs)) ∨ algunos P ([a]))&quot; by (simp add:algunos_append)<br /> also have &quot;... = (algunos P xs ∨ algunos P [a])&quot; using 1 by simp<br /> also have &quot;... = (algunos P [a] ∨ algunos P xs)&quot; by blast<br /> also have &quot;... = algunos P (a # xs)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 13. Encontrar un término no trivial Z tal que sea cierta la <br /> siguiente ecuación:<br /> algunos (λx. P x ∨ Q x) xs = Z<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La ecuación se verifica eligiendo como Z el término <br /> algunos P xs ∨ algunos Q xs<br /> En efecto,<br /> *}<br /> <br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;De forma estructurada&quot;<br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 14. Demostrar o refutar:<br /> algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 15. Definir la funcion primitiva recursiva <br /> estaEn :: &#039;a ⇒ &#039;a list ⇒ bool<br /> tal que (estaEn x xs) se verifica si el elemento x está en la lista<br /> xs. Por ejemplo, <br /> estaEn (2::nat) [3,2,4] = True<br /> estaEn (1::nat) [3,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec estaEn :: &quot;&#039;a ⇒ &#039;a list ⇒ bool&quot; where<br /> &quot;estaEn x [] = False&quot;<br /> | &quot;estaEn x (a#xs) = (x=a ∨ estaEn x xs)&quot; <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 16. Expresar la relación existente entre estaEn y algunos. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La relación es <br /> <br /> En efecto, <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 17. Definir la función primitiva recursiva <br /> sinDuplicados :: &#039;a list ⇒ bool<br /> tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br /> duplicados. Por ejemplo, <br /> sinDuplicados [1::nat,4,2] = True<br /> sinDuplicados [1::nat,4,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> <br /> primrec sinDuplicados :: &quot;&#039;a list ⇒ bool&quot; where<br /> &quot;sinDuplicados [] = True&quot;<br /> | &quot;sinDuplicados (a#xs) = ((¬ (estaEn a xs)) ∧ sinDuplicados xs)&quot;<br /> <br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 18. Definir la función primitiva recursiva <br /> borraDuplicados :: &#039;a list ⇒ bool<br /> tal que (borraDuplicados xs) es la lista obtenida eliminando los<br /> elementos duplicados de la lista xs. Por ejemplo, <br /> borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br /> <br /> Nota: La función borraDuplicados es equivalente a la predefinida remdups. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> <br /> primrec borraDuplicados :: &quot;&#039;a list ⇒ &#039;a list&quot; where<br /> &quot;borraDuplicados [] = []&quot;<br /> | &quot;borraDuplicados (a#xs) = <br /> (if estaEn a xs <br /> then borraDuplicados xs <br /> else (a#borraDuplicados xs))&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 19. Demostrar o refutar:<br /> length (borraDuplicados xs) ≤ length xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma length_borraDuplicados: <br /> &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 20. Demostrar o refutar: <br /> estaEn a (borraDuplicados xs) = estaEn a xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma estaEn_borraDuplicados_auto: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma estaEn_borraDuplicados: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 21. Demostrar o refutar: <br /> sinDuplicados (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática&quot;<br /> lemma &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma sinDuplicados_borraDuplicados:<br /> &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 22. Demostrar o refutar:<br /> borraDuplicados (rev xs) = rev (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma &quot;borraDuplicados (rev xs) = rev (borraDuplicados xs)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_3&diff=135 2011-02-14T16:13:09Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 3ª relación de ejercicios *}<br /> <br /> theory Relacion_3<br /> imports Main <br /> begin<br /> <br /> section {* Cons inverso *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 1. Definir recursivamente la función <br /> snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot;<br /> tal que (snoc xs a) es la lista obtenida al añadir el elemento a al<br /> final de la lista xs. Por ejemplo, <br /> value &quot;snoc [2,5] (3::int)&quot; == [2,5,3]<br /> <br /> Nota: No usar @.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> primrec snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot; where<br /> &quot;snoc [] y = [y]&quot;<br /> | &quot;snoc (x#xs) y = x#(snoc xs y)&quot;<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 2. Demostrar el siguiente teorema <br /> snoc xs a = xs @ [a]<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del lema es&quot;<br /> lemma &quot;snoc xs a = xs @ [a]&quot;<br /> by (induct xs) auto<br /> <br /> -- &quot;La demostración estructurada del lema es&quot;<br /> lemma snoc_append: &quot;snoc xs a = xs @ [a]&quot;<br /> proof (induct xs)<br /> show &quot;snoc [] a = [] @ [a]&quot; by simp<br /> next <br /> fix b xs <br /> assume HI: &quot;snoc xs a = xs @ [a]&quot;<br /> thus &quot;snoc (b # xs) a = (b # xs) @ [a]&quot;<br /> proof -<br /> have &quot;snoc (b # xs) a = b#(snoc xs a)&quot; by simp<br /> also have &quot;... = b#(xs @ [a])&quot; using HI by simp<br /> also have &quot;... = (b#xs) @ [a]&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 3. Demostrar el siguiente teorema <br /> rev (x # xs) = snoc (rev xs) x&quot;<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del teorema es&quot;<br /> theorem rev_cons_auto: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> by (simp add:snoc_append)<br /> <br /> -- &quot;La demostración estructurada del teorema es&quot;<br /> theorem rev_cons: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> proof -<br /> have &quot;rev(x#xs)=(rev xs) @ [x]&quot; by simp <br /> also have &quot;... = snoc (rev xs) x&quot; by (simp add:snoc_append)<br /> finally show ?thesis . <br /> qed<br /> <br /> <br /> <br /> <br /> <br /> section {* Cuantificadores sobre listas *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 4. Definir la función <br /> todos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (todos p xs) se verifica si todos los elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> todos (λx. 1&lt;length x) [[2,1,4],[1,3]]<br /> ¬ todos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> <br /> Nota: La función todos es equivalente a la predefinida list_all. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec todos :: &quot;(&#039;a ⇒ bool) ⇒ (&#039;a list ⇒ bool)&quot; where<br /> &quot;todos P [] = True&quot;<br /> |&quot;todos P (x#xs) = (P x ∧ todos P xs)&quot;<br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 5. Definir la función <br /> algunos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (algunos p xs) se verifica si algunos elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> algunos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> ¬ algunos (λx. 1&lt;length x) [[],[3]]&quot;<br /> <br /> Nota: La función algunos es equivalente a la predefinida list_ex. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec algunos :: &quot;(&#039;a =&gt; bool) =&gt; &#039;a list =&gt; bool&quot; where<br /> &quot;algunos P [] = False&quot;<br /> | &quot;algunos P (x#xs) = (P x ∨ algunos P xs)&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 6. Demostrar o refutar: <br /> todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> proof (induct xs)<br /> show &quot;todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])&quot; by simp<br /> next<br /> fix a xs<br /> assume 1: &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> thus &quot;todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))&quot;<br /> proof -<br /> have 2: &quot;todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos (λx. P x ∧ Q x) xs))&quot; by simp<br /> also have 3: &quot;... = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))&quot; using 1 by simp<br /> also have 4: &quot;... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))&quot; by blast<br /> also have 5: &quot;... = ((todos P (a # xs) ∧ todos Q (a # xs)))&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 7. Demostrar o refutar: <br /> todos P (x @ y) = (todos P x ∧ todos P y)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> by (induct x) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> proof (induct x)<br /> show &quot;todos P ([] @ y) = (todos P [] ∧ todos P y)&quot; by simp<br /> next<br /> fix a x y<br /> assume 1: &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> thus &quot;todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)&quot;<br /> proof -<br /> have &quot;(todos P ((a # x) @ y)) = (P a ∧ todos P (x @ y))&quot; by simp<br /> also have &quot;... = (P a ∧ todos P x ∧ todos P y)&quot; using 1 by simp<br /> also have &quot;... = (todos P (a #x) ∧ todos P y)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 8. Demostrar o refutar: <br /> todos P (rev xs) = todos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> proof (induct xs)<br /> show &quot;todos P (rev []) = todos P []&quot; by simp<br /> next<br /> fix a xs<br /> assume 1: &quot;todos P (rev xs) = todos P xs&quot;<br /> thus &quot;todos P (rev (a # xs)) = todos P (a # xs)&quot;<br /> proof -<br /> have &quot;todos P (rev (a # xs)) = todos P (rev (xs) @ [a])&quot; by simp<br /> also have &quot;... = (todos P (rev (xs)) ∧ todos P [a])&quot; by simp<br /> also have &quot;... = (todos P xs ∧ todos P [a])&quot; using 1 by simp<br /> also have &quot;... = (todos P [a] ∧ todos P xs)&quot; by blast<br /> also have &quot;... = todos P ([a] @ xs)&quot; by simp<br /> also have &quot;... = todos P (a # xs)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 9. Demostrar o refutar:<br /> algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;Se busca un contraejemplo con nitpick&quot;<br /> lemma &quot;algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> El contraejemplo encontrado es<br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 10. Demostrar o refutar: <br /> algunos P (map f xs) = algunos (P ∘ f) xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P o f) xs&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P ∘ f) xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 11. Demostrar o refutar: <br /> algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> by (induct xs) auto<br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma algunos_append: <br /> &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> proof (induct xs)<br /> show &quot;algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)&quot; by simp<br /> next<br /> fix a xs<br /> assume 1: &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> thus &quot;algunos P ((a # xs) @ ys) = (algunos P (a # xs) ∨ algunos P ys)&quot;<br /> proof -<br /> have &quot;algunos P ((a # xs) @ ys) = (P a ∨ algunos P (xs @ ys))&quot; by simp<br /> also have &quot;... = (P a ∨ (algunos P xs ∨ algunos P ys))&quot; using 1 by simp<br /> also have &quot;... = ((P a ∨ algunos P xs) ∨ algunos P ys)&quot; by blast<br /> also have &quot;... = (algunos P (a#xs) ∨ algunos P ys)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 12. Demostrar o refutar: <br /> algunos P (rev xs) = algunos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 13. Encontrar un término no trivial Z tal que sea cierta la <br /> siguiente ecuación:<br /> algunos (λx. P x ∨ Q x) xs = Z<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La ecuación se verifica eligiendo como Z el término <br /> algunos P xs ∨ algunos Q xs<br /> En efecto,<br /> *}<br /> <br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;De forma estructurada&quot;<br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 14. Demostrar o refutar:<br /> algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 15. Definir la funcion primitiva recursiva <br /> estaEn :: &#039;a ⇒ &#039;a list ⇒ bool<br /> tal que (estaEn x xs) se verifica si el elemento x está en la lista<br /> xs. Por ejemplo, <br /> estaEn (2::nat) [3,2,4] = True<br /> estaEn (1::nat) [3,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec estaEn :: &quot;&#039;a ⇒ &#039;a list ⇒ bool&quot; where<br /> &quot;estaEn x [] = False&quot;<br /> | &quot;estaEn x (a#xs) = (x=a ∨ estaEn x xs)&quot; <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 16. Expresar la relación existente entre estaEn y algunos. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La relación es <br /> <br /> En efecto, <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 17. Definir la función primitiva recursiva <br /> sinDuplicados :: &#039;a list ⇒ bool<br /> tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br /> duplicados. Por ejemplo, <br /> sinDuplicados [1::nat,4,2] = True<br /> sinDuplicados [1::nat,4,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> <br /> primrec sinDuplicados :: &quot;&#039;a list ⇒ bool&quot; where<br /> &quot;sinDuplicados [] = True&quot;<br /> | &quot;sinDuplicados (a#xs) = ((¬ (estaEn a xs)) ∧ sinDuplicados xs)&quot;<br /> <br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 18. Definir la función primitiva recursiva <br /> borraDuplicados :: &#039;a list ⇒ bool<br /> tal que (borraDuplicados xs) es la lista obtenida eliminando los<br /> elementos duplicados de la lista xs. Por ejemplo, <br /> borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br /> <br /> Nota: La función borraDuplicados es equivalente a la predefinida remdups. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> <br /> primrec borraDuplicados :: &quot;&#039;a list ⇒ &#039;a list&quot; where<br /> &quot;borraDuplicados [] = []&quot;<br /> | &quot;borraDuplicados (a#xs) = <br /> (if estaEn a xs <br /> then borraDuplicados xs <br /> else (a#borraDuplicados xs))&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 19. Demostrar o refutar:<br /> length (borraDuplicados xs) ≤ length xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma length_borraDuplicados: <br /> &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 20. Demostrar o refutar: <br /> estaEn a (borraDuplicados xs) = estaEn a xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma estaEn_borraDuplicados_auto: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma estaEn_borraDuplicados: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 21. Demostrar o refutar: <br /> sinDuplicados (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática&quot;<br /> lemma &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma sinDuplicados_borraDuplicados:<br /> &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 22. Demostrar o refutar:<br /> borraDuplicados (rev xs) = rev (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma &quot;borraDuplicados (rev xs) = rev (borraDuplicados xs)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_3&diff=134 2011-02-14T16:07:13Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 3ª relación de ejercicios *}<br /> <br /> theory Relacion_3<br /> imports Main <br /> begin<br /> <br /> section {* Cons inverso *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 1. Definir recursivamente la función <br /> snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot;<br /> tal que (snoc xs a) es la lista obtenida al añadir el elemento a al<br /> final de la lista xs. Por ejemplo, <br /> value &quot;snoc [2,5] (3::int)&quot; == [2,5,3]<br /> <br /> Nota: No usar @.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> primrec snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot; where<br /> &quot;snoc [] y = [y]&quot;<br /> | &quot;snoc (x#xs) y = x#(snoc xs y)&quot;<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 2. Demostrar el siguiente teorema <br /> snoc xs a = xs @ [a]<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del lema es&quot;<br /> lemma &quot;snoc xs a = xs @ [a]&quot;<br /> by (induct xs) auto<br /> <br /> -- &quot;La demostración estructurada del lema es&quot;<br /> lemma snoc_append: &quot;snoc xs a = xs @ [a]&quot;<br /> proof (induct xs)<br /> show &quot;snoc [] a = [] @ [a]&quot; by simp<br /> next <br /> fix b xs <br /> assume HI: &quot;snoc xs a = xs @ [a]&quot;<br /> thus &quot;snoc (b # xs) a = (b # xs) @ [a]&quot;<br /> proof -<br /> have &quot;snoc (b # xs) a = b#(snoc xs a)&quot; by simp<br /> also have &quot;... = b#(xs @ [a])&quot; using HI by simp<br /> also have &quot;... = (b#xs) @ [a]&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 3. Demostrar el siguiente teorema <br /> rev (x # xs) = snoc (rev xs) x&quot;<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del teorema es&quot;<br /> theorem rev_cons_auto: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> by (simp add:snoc_append)<br /> <br /> -- &quot;La demostración estructurada del teorema es&quot;<br /> theorem rev_cons: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> proof -<br /> have &quot;rev(x#xs)=(rev xs) @ [x]&quot; by simp <br /> also have &quot;... = snoc (rev xs) x&quot; by (simp add:snoc_append)<br /> finally show ?thesis . <br /> qed<br /> <br /> <br /> <br /> <br /> <br /> section {* Cuantificadores sobre listas *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 4. Definir la función <br /> todos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (todos p xs) se verifica si todos los elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> todos (λx. 1&lt;length x) [[2,1,4],[1,3]]<br /> ¬ todos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> <br /> Nota: La función todos es equivalente a la predefinida list_all. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec todos :: &quot;(&#039;a ⇒ bool) ⇒ (&#039;a list ⇒ bool)&quot; where<br /> &quot;todos P [] = True&quot;<br /> |&quot;todos P (x#xs) = (P x ∧ todos P xs)&quot;<br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 5. Definir la función <br /> algunos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (algunos p xs) se verifica si algunos elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> algunos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> ¬ algunos (λx. 1&lt;length x) [[],[3]]&quot;<br /> <br /> Nota: La función algunos es equivalente a la predefinida list_ex. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec algunos :: &quot;(&#039;a =&gt; bool) =&gt; &#039;a list =&gt; bool&quot; where<br /> &quot;algunos P [] = False&quot;<br /> | &quot;algunos P (x#xs) = (P x ∨ algunos P xs)&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 6. Demostrar o refutar: <br /> todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> proof (induct xs)<br /> show &quot;todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])&quot; by simp<br /> next<br /> fix a xs<br /> assume 1: &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> thus &quot;todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))&quot;<br /> proof -<br /> have 2: &quot;todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos (λx. P x ∧ Q x) xs))&quot; by simp<br /> also have 3: &quot;... = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))&quot; using 1 by simp<br /> also have 4: &quot;... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))&quot; by blast<br /> also have 5: &quot;... = ((todos P (a # xs) ∧ todos Q (a # xs)))&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 7. Demostrar o refutar: <br /> todos P (x @ y) = (todos P x ∧ todos P y)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> by (induct x) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> proof (induct x)<br /> show &quot;todos P ([] @ y) = (todos P [] ∧ todos P y)&quot; by simp<br /> next<br /> fix a x y<br /> assume 1: &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> thus &quot;todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)&quot;<br /> proof -<br /> have &quot;(todos P ((a # x) @ y)) = (P a ∧ todos P (x @ y))&quot; by simp<br /> also have &quot;... = (P a ∧ todos P x ∧ todos P y)&quot; using 1 by simp<br /> also have &quot;... = (todos P (a #x) ∧ todos P y)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 8. Demostrar o refutar: <br /> todos P (rev xs) = todos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> proof (induct xs)<br /> show &quot;todos P (rev []) = todos P []&quot; by simp<br /> next<br /> fix a xs<br /> assume 1: &quot;todos P (rev xs) = todos P xs&quot;<br /> thus &quot;todos P (rev (a # xs)) = todos P (a # xs)&quot;<br /> proof -<br /> have &quot;todos P (rev (a # xs)) = todos P (rev (xs) @ [a])&quot; by simp<br /> also have &quot;... = (todos P (rev (xs)) ∧ todos P [a])&quot; by simp<br /> also have &quot;... = (todos P xs ∧ todos P [a])&quot; using 1 by simp<br /> also have &quot;... = (todos P [a] ∧ todos P xs)&quot; by blast<br /> also have &quot;... = todos P ([a] @ xs)&quot; by simp<br /> also have &quot;... = todos P (a # xs)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 9. Demostrar o refutar:<br /> algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;Se busca un contraejemplo con nitpick&quot;<br /> lemma &quot;algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> El contraejemplo encontrado es<br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 10. Demostrar o refutar: <br /> algunos P (map f xs) = algunos (P ∘ f) xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P o f) xs&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P ∘ f) xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 11. Demostrar o refutar: <br /> algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> by (induct xs) auto<br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma algunos_append: <br /> &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 12. Demostrar o refutar: <br /> algunos P (rev xs) = algunos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 13. Encontrar un término no trivial Z tal que sea cierta la <br /> siguiente ecuación:<br /> algunos (λx. P x ∨ Q x) xs = Z<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La ecuación se verifica eligiendo como Z el término <br /> algunos P xs ∨ algunos Q xs<br /> En efecto,<br /> *}<br /> <br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;De forma estructurada&quot;<br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 14. Demostrar o refutar:<br /> algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 15. Definir la funcion primitiva recursiva <br /> estaEn :: &#039;a ⇒ &#039;a list ⇒ bool<br /> tal que (estaEn x xs) se verifica si el elemento x está en la lista<br /> xs. Por ejemplo, <br /> estaEn (2::nat) [3,2,4] = True<br /> estaEn (1::nat) [3,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec estaEn :: &quot;&#039;a ⇒ &#039;a list ⇒ bool&quot; where<br /> &quot;estaEn x [] = False&quot;<br /> | &quot;estaEn x (a#xs) = (x=a ∨ estaEn x xs)&quot; <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 16. Expresar la relación existente entre estaEn y algunos. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La relación es <br /> <br /> En efecto, <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 17. Definir la función primitiva recursiva <br /> sinDuplicados :: &#039;a list ⇒ bool<br /> tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br /> duplicados. Por ejemplo, <br /> sinDuplicados [1::nat,4,2] = True<br /> sinDuplicados [1::nat,4,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> <br /> primrec sinDuplicados :: &quot;&#039;a list ⇒ bool&quot; where<br /> &quot;sinDuplicados [] = True&quot;<br /> | &quot;sinDuplicados (a#xs) = ((¬ (estaEn a xs)) ∧ sinDuplicados xs)&quot;<br /> <br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 18. Definir la función primitiva recursiva <br /> borraDuplicados :: &#039;a list ⇒ bool<br /> tal que (borraDuplicados xs) es la lista obtenida eliminando los<br /> elementos duplicados de la lista xs. Por ejemplo, <br /> borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br /> <br /> Nota: La función borraDuplicados es equivalente a la predefinida remdups. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> <br /> primrec borraDuplicados :: &quot;&#039;a list ⇒ &#039;a list&quot; where<br /> &quot;borraDuplicados [] = []&quot;<br /> | &quot;borraDuplicados (a#xs) = <br /> (if estaEn a xs <br /> then borraDuplicados xs <br /> else (a#borraDuplicados xs))&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 19. Demostrar o refutar:<br /> length (borraDuplicados xs) ≤ length xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma length_borraDuplicados: <br /> &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 20. Demostrar o refutar: <br /> estaEn a (borraDuplicados xs) = estaEn a xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma estaEn_borraDuplicados_auto: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma estaEn_borraDuplicados: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 21. Demostrar o refutar: <br /> sinDuplicados (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática&quot;<br /> lemma &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma sinDuplicados_borraDuplicados:<br /> &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 22. Demostrar o refutar:<br /> borraDuplicados (rev xs) = rev (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma &quot;borraDuplicados (rev xs) = rev (borraDuplicados xs)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_3&diff=133 2011-02-14T15:55:51Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 3ª relación de ejercicios *}<br /> <br /> theory Relacion_3<br /> imports Main <br /> begin<br /> <br /> section {* Cons inverso *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 1. Definir recursivamente la función <br /> snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot;<br /> tal que (snoc xs a) es la lista obtenida al añadir el elemento a al<br /> final de la lista xs. Por ejemplo, <br /> value &quot;snoc [2,5] (3::int)&quot; == [2,5,3]<br /> <br /> Nota: No usar @.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> primrec snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot; where<br /> &quot;snoc [] y = [y]&quot;<br /> | &quot;snoc (x#xs) y = x#(snoc xs y)&quot;<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 2. Demostrar el siguiente teorema <br /> snoc xs a = xs @ [a]<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del lema es&quot;<br /> lemma &quot;snoc xs a = xs @ [a]&quot;<br /> by (induct xs) auto<br /> <br /> -- &quot;La demostración estructurada del lema es&quot;<br /> lemma snoc_append: &quot;snoc xs a = xs @ [a]&quot;<br /> proof (induct xs)<br /> show &quot;snoc [] a = [] @ [a]&quot; by simp<br /> next <br /> fix b xs <br /> assume HI: &quot;snoc xs a = xs @ [a]&quot;<br /> thus &quot;snoc (b # xs) a = (b # xs) @ [a]&quot;<br /> proof -<br /> have &quot;snoc (b # xs) a = b#(snoc xs a)&quot; by simp<br /> also have &quot;... = b#(xs @ [a])&quot; using HI by simp<br /> also have &quot;... = (b#xs) @ [a]&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 3. Demostrar el siguiente teorema <br /> rev (x # xs) = snoc (rev xs) x&quot;<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del teorema es&quot;<br /> theorem rev_cons_auto: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> by (simp add:snoc_append)<br /> <br /> -- &quot;La demostración estructurada del teorema es&quot;<br /> theorem rev_cons: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> proof -<br /> have &quot;rev(x#xs)=(rev xs) @ [x]&quot; by simp <br /> also have &quot;... = snoc (rev xs) x&quot; by (simp add:snoc_append)<br /> finally show ?thesis . <br /> qed<br /> <br /> <br /> <br /> <br /> <br /> section {* Cuantificadores sobre listas *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 4. Definir la función <br /> todos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (todos p xs) se verifica si todos los elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> todos (λx. 1&lt;length x) [[2,1,4],[1,3]]<br /> ¬ todos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> <br /> Nota: La función todos es equivalente a la predefinida list_all. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec todos :: &quot;(&#039;a ⇒ bool) ⇒ (&#039;a list ⇒ bool)&quot; where<br /> &quot;todos P [] = True&quot;<br /> |&quot;todos P (x#xs) = (P x ∧ todos P xs)&quot;<br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 5. Definir la función <br /> algunos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (algunos p xs) se verifica si algunos elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> algunos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> ¬ algunos (λx. 1&lt;length x) [[],[3]]&quot;<br /> <br /> Nota: La función algunos es equivalente a la predefinida list_ex. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec algunos :: &quot;(&#039;a =&gt; bool) =&gt; &#039;a list =&gt; bool&quot; where<br /> &quot;algunos P [] = False&quot;<br /> | &quot;algunos P (x#xs) = (P x ∨ algunos P xs)&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 6. Demostrar o refutar: <br /> todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> proof (induct xs)<br /> show &quot;todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])&quot; by simp<br /> next<br /> fix a xs<br /> assume 1: &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> thus &quot;todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))&quot;<br /> proof -<br /> have 2: &quot;todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos (λx. P x ∧ Q x) xs))&quot; by simp<br /> also have 3: &quot;... = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))&quot; using 1 by simp<br /> also have 4: &quot;... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))&quot; by blast<br /> also have 5: &quot;... = ((todos P (a # xs) ∧ todos Q (a # xs)))&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 7. Demostrar o refutar: <br /> todos P (x @ y) = (todos P x ∧ todos P y)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> by (induct x) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> proof (induct x)<br /> show &quot;todos P ([] @ y) = (todos P [] ∧ todos P y)&quot; by simp<br /> next<br /> fix a x y<br /> assume 1: &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> thus &quot;todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)&quot;<br /> proof -<br /> have &quot;(todos P ((a # x) @ y)) = (P a ∧ todos P (x @ y))&quot; by simp<br /> also have &quot;... = (P a ∧ todos P x ∧ todos P y)&quot; using 1 by simp<br /> also have &quot;... = (todos P (a #x) ∧ todos P y)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 8. Demostrar o refutar: <br /> todos P (rev xs) = todos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> proof (induct xs)<br /> show &quot;todos P (rev []) = todos P []&quot; by simp<br /> next<br /> fix a xs<br /> assume 1: &quot;todos P (rev xs) = todos P xs&quot;<br /> thus &quot;todos P (rev (a # xs)) = todos P (a # xs)&quot;<br /> proof -<br /> have &quot;todos P (rev (a # xs)) = todos P (rev (xs) @ [a])&quot; by simp<br /> also have &quot;... = (todos P (rev (xs)) ∧ todos P [a])&quot; by simp<br /> also have &quot;... = (todos P xs ∧ todos P [a])&quot; using 1 by simp<br /> also have &quot;... = (todos P [a] ∧ todos P xs)&quot; by blast<br /> also have &quot;... = todos P ([a] @ xs)&quot; by simp<br /> also have &quot;... = todos P (a # xs)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 9. Demostrar o refutar:<br /> algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;Se busca un contraejemplo con nitpick&quot;<br /> lemma &quot;algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> El contraejemplo encontrado es<br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 10. Demostrar o refutar: <br /> algunos P (map f xs) = algunos (P ∘ f) xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P o f) xs&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P ∘ f) xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 11. Demostrar o refutar: <br /> algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma algunos_append: <br /> &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 12. Demostrar o refutar: <br /> algunos P (rev xs) = algunos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 13. Encontrar un término no trivial Z tal que sea cierta la <br /> siguiente ecuación:<br /> algunos (λx. P x ∨ Q x) xs = Z<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La ecuación se verifica eligiendo como Z el término <br /> algunos P xs ∨ algunos Q xs<br /> En efecto,<br /> *}<br /> <br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;De forma estructurada&quot;<br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 14. Demostrar o refutar:<br /> algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 15. Definir la funcion primitiva recursiva <br /> estaEn :: &#039;a ⇒ &#039;a list ⇒ bool<br /> tal que (estaEn x xs) se verifica si el elemento x está en la lista<br /> xs. Por ejemplo, <br /> estaEn (2::nat) [3,2,4] = True<br /> estaEn (1::nat) [3,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec estaEn :: &quot;&#039;a ⇒ &#039;a list ⇒ bool&quot; where<br /> &quot;estaEn x [] = False&quot;<br /> | &quot;estaEn x (a#xs) = (x=a ∨ estaEn x xs)&quot; <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 16. Expresar la relación existente entre estaEn y algunos. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La relación es <br /> <br /> En efecto, <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 17. Definir la función primitiva recursiva <br /> sinDuplicados :: &#039;a list ⇒ bool<br /> tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br /> duplicados. Por ejemplo, <br /> sinDuplicados [1::nat,4,2] = True<br /> sinDuplicados [1::nat,4,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> <br /> primrec sinDuplicados :: &quot;&#039;a list ⇒ bool&quot; where<br /> &quot;sinDuplicados [] = True&quot;<br /> | &quot;sinDuplicados (a#xs) = ((¬ (estaEn a xs)) ∧ sinDuplicados xs)&quot;<br /> <br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 18. Definir la función primitiva recursiva <br /> borraDuplicados :: &#039;a list ⇒ bool<br /> tal que (borraDuplicados xs) es la lista obtenida eliminando los<br /> elementos duplicados de la lista xs. Por ejemplo, <br /> borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br /> <br /> Nota: La función borraDuplicados es equivalente a la predefinida remdups. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> <br /> primrec borraDuplicados :: &quot;&#039;a list ⇒ &#039;a list&quot; where<br /> &quot;borraDuplicados [] = []&quot;<br /> | &quot;borraDuplicados (a#xs) = <br /> (if estaEn a xs <br /> then borraDuplicados xs <br /> else (a#borraDuplicados xs))&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 19. Demostrar o refutar:<br /> length (borraDuplicados xs) ≤ length xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma length_borraDuplicados: <br /> &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 20. Demostrar o refutar: <br /> estaEn a (borraDuplicados xs) = estaEn a xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma estaEn_borraDuplicados_auto: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma estaEn_borraDuplicados: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 21. Demostrar o refutar: <br /> sinDuplicados (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática&quot;<br /> lemma &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma sinDuplicados_borraDuplicados:<br /> &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 22. Demostrar o refutar:<br /> borraDuplicados (rev xs) = rev (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma &quot;borraDuplicados (rev xs) = rev (borraDuplicados xs)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_3&diff=132 2011-02-14T15:53:58Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 3ª relación de ejercicios *}<br /> <br /> theory Relacion_3<br /> imports Main <br /> begin<br /> <br /> section {* Cons inverso *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 1. Definir recursivamente la función <br /> snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot;<br /> tal que (snoc xs a) es la lista obtenida al añadir el elemento a al<br /> final de la lista xs. Por ejemplo, <br /> value &quot;snoc [2,5] (3::int)&quot; == [2,5,3]<br /> <br /> Nota: No usar @.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> primrec snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot; where<br /> &quot;snoc [] y = [y]&quot;<br /> | &quot;snoc (x#xs) y = x#(snoc xs y)&quot;<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 2. Demostrar el siguiente teorema <br /> snoc xs a = xs @ [a]<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del lema es&quot;<br /> lemma &quot;snoc xs a = xs @ [a]&quot;<br /> by (induct xs) auto<br /> <br /> -- &quot;La demostración estructurada del lema es&quot;<br /> lemma snoc_append: &quot;snoc xs a = xs @ [a]&quot;<br /> proof (induct xs)<br /> show &quot;snoc [] a = [] @ [a]&quot; by simp<br /> next <br /> fix b xs <br /> assume HI: &quot;snoc xs a = xs @ [a]&quot;<br /> thus &quot;snoc (b # xs) a = (b # xs) @ [a]&quot;<br /> proof -<br /> have &quot;snoc (b # xs) a = b#(snoc xs a)&quot; by simp<br /> also have &quot;... = b#(xs @ [a])&quot; using HI by simp<br /> also have &quot;... = (b#xs) @ [a]&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 3. Demostrar el siguiente teorema <br /> rev (x # xs) = snoc (rev xs) x&quot;<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del teorema es&quot;<br /> theorem rev_cons_auto: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> by (simp add:snoc_append)<br /> <br /> -- &quot;La demostración estructurada del teorema es&quot;<br /> theorem rev_cons: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> proof -<br /> have &quot;rev(x#xs)=(rev xs) @ [x]&quot; by simp <br /> also have &quot;... = snoc (rev xs) x&quot; by (simp add:snoc_append)<br /> finally show ?thesis . <br /> qed<br /> <br /> <br /> <br /> <br /> <br /> section {* Cuantificadores sobre listas *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 4. Definir la función <br /> todos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (todos p xs) se verifica si todos los elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> todos (λx. 1&lt;length x) [[2,1,4],[1,3]]<br /> ¬ todos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> <br /> Nota: La función todos es equivalente a la predefinida list_all. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec todos :: &quot;(&#039;a ⇒ bool) ⇒ (&#039;a list ⇒ bool)&quot; where<br /> &quot;todos P [] = True&quot;<br /> |&quot;todos P (x#xs) = (P x ∧ todos P xs)&quot;<br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 5. Definir la función <br /> algunos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (algunos p xs) se verifica si algunos elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> algunos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> ¬ algunos (λx. 1&lt;length x) [[],[3]]&quot;<br /> <br /> Nota: La función algunos es equivalente a la predefinida list_ex. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec algunos :: &quot;(&#039;a =&gt; bool) =&gt; &#039;a list =&gt; bool&quot; where<br /> &quot;algunos P [] = False&quot;<br /> | &quot;algunos P (x#xs) = (P x ∨ algunos P xs)&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 6. Demostrar o refutar: <br /> todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> proof (induct xs)<br /> show &quot;todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])&quot; by simp<br /> next<br /> fix a xs<br /> assume 1: &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> thus &quot;todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))&quot;<br /> proof -<br /> have 2: &quot;todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos (λx. P x ∧ Q x) xs))&quot; by simp<br /> also have 3: &quot;... = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))&quot; using 1 by simp<br /> also have 4: &quot;... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))&quot; by blast<br /> also have 5: &quot;... = ((todos P (a # xs) ∧ todos Q (a # xs)))&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 7. Demostrar o refutar: <br /> todos P (x @ y) = (todos P x ∧ todos P y)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> by (induct x) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> proof (induct x)<br /> show &quot;todos P ([] @ y) = (todos P [] ∧ todos P y)&quot; by simp<br /> next<br /> fix a x y<br /> assume 1: &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> thus &quot;todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)&quot;<br /> proof -<br /> have &quot;(todos P ((a # x) @ y)) = (P a ∧ todos P (x @ y))&quot; by simp<br /> also have &quot;... = (P a ∧ todos P x ∧ todos P y)&quot; using 1 by simp<br /> also have &quot;... = (todos P (a #x) ∧ todos P y)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 8. Demostrar o refutar: <br /> todos P (rev xs) = todos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> proof (induct xs)<br /> show &quot;todos P (rev []) = todos P []&quot; by simp<br /> next<br /> fix a xs<br /> assume 1: &quot;todos P (rev xs) = todos P xs&quot;<br /> thus &quot;todos P (rev (a # xs)) = todos P (a # xs)&quot;<br /> proof -<br /> have &quot;todos P (rev (a # xs)) = todos P (rev (xs) @ [a])&quot; by simp<br /> also have &quot;... = (todos P (rev (xs)) ∧ todos P [a])&quot; by simp<br /> also have &quot;... = (todos P xs ∧ todos P [a])&quot; using 1 by simp<br /> also have &quot;... = (todos P [a] ∧ todos P xs)&quot; by blast<br /> also have &quot;... = todos P ([a] @ xs)&quot; by simp<br /> also have &quot;... = todos P (a # xs)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 9. Demostrar o refutar:<br /> algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;Se busca un contraejemplo con nitpick&quot;<br /> lemma &quot;algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> El contraejemplo encontrado es<br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 10. Demostrar o refutar: <br /> algunos P (map f xs) = algunos (P ∘ f) xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P o f) xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P ∘ f) xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 11. Demostrar o refutar: <br /> algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma algunos_append: <br /> &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 12. Demostrar o refutar: <br /> algunos P (rev xs) = algunos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 13. Encontrar un término no trivial Z tal que sea cierta la <br /> siguiente ecuación:<br /> algunos (λx. P x ∨ Q x) xs = Z<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La ecuación se verifica eligiendo como Z el término <br /> algunos P xs ∨ algunos Q xs<br /> En efecto,<br /> *}<br /> <br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;De forma estructurada&quot;<br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 14. Demostrar o refutar:<br /> algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 15. Definir la funcion primitiva recursiva <br /> estaEn :: &#039;a ⇒ &#039;a list ⇒ bool<br /> tal que (estaEn x xs) se verifica si el elemento x está en la lista<br /> xs. Por ejemplo, <br /> estaEn (2::nat) [3,2,4] = True<br /> estaEn (1::nat) [3,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec estaEn :: &quot;&#039;a ⇒ &#039;a list ⇒ bool&quot; where<br /> &quot;estaEn x [] = False&quot;<br /> | &quot;estaEn x (a#xs) = (x=a ∨ estaEn x xs)&quot; <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 16. Expresar la relación existente entre estaEn y algunos. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La relación es <br /> <br /> En efecto, <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 17. Definir la función primitiva recursiva <br /> sinDuplicados :: &#039;a list ⇒ bool<br /> tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br /> duplicados. Por ejemplo, <br /> sinDuplicados [1::nat,4,2] = True<br /> sinDuplicados [1::nat,4,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> <br /> primrec sinDuplicados :: &quot;&#039;a list ⇒ bool&quot; where<br /> &quot;sinDuplicados [] = True&quot;<br /> | &quot;sinDuplicados (a#xs) = ((¬ (estaEn a xs)) ∧ sinDuplicados xs)&quot;<br /> <br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 18. Definir la función primitiva recursiva <br /> borraDuplicados :: &#039;a list ⇒ bool<br /> tal que (borraDuplicados xs) es la lista obtenida eliminando los<br /> elementos duplicados de la lista xs. Por ejemplo, <br /> borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br /> <br /> Nota: La función borraDuplicados es equivalente a la predefinida remdups. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> <br /> primrec borraDuplicados :: &quot;&#039;a list ⇒ &#039;a list&quot; where<br /> &quot;borraDuplicados [] = []&quot;<br /> | &quot;borraDuplicados (a#xs) = <br /> (if estaEn a xs <br /> then borraDuplicados xs <br /> else (a#borraDuplicados xs))&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 19. Demostrar o refutar:<br /> length (borraDuplicados xs) ≤ length xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma length_borraDuplicados: <br /> &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 20. Demostrar o refutar: <br /> estaEn a (borraDuplicados xs) = estaEn a xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma estaEn_borraDuplicados_auto: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma estaEn_borraDuplicados: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 21. Demostrar o refutar: <br /> sinDuplicados (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática&quot;<br /> lemma &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma sinDuplicados_borraDuplicados:<br /> &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 22. Demostrar o refutar:<br /> borraDuplicados (rev xs) = rev (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma &quot;borraDuplicados (rev xs) = rev (borraDuplicados xs)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_3&diff=131 2011-02-14T15:52:47Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 3ª relación de ejercicios *}<br /> <br /> theory Relacion_3<br /> imports Main <br /> begin<br /> <br /> section {* Cons inverso *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 1. Definir recursivamente la función <br /> snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot;<br /> tal que (snoc xs a) es la lista obtenida al añadir el elemento a al<br /> final de la lista xs. Por ejemplo, <br /> value &quot;snoc [2,5] (3::int)&quot; == [2,5,3]<br /> <br /> Nota: No usar @.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> primrec snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot; where<br /> &quot;snoc [] y = [y]&quot;<br /> | &quot;snoc (x#xs) y = x#(snoc xs y)&quot;<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 2. Demostrar el siguiente teorema <br /> snoc xs a = xs @ [a]<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del lema es&quot;<br /> lemma &quot;snoc xs a = xs @ [a]&quot;<br /> by (induct xs) auto<br /> <br /> -- &quot;La demostración estructurada del lema es&quot;<br /> lemma snoc_append: &quot;snoc xs a = xs @ [a]&quot;<br /> proof (induct xs)<br /> show &quot;snoc [] a = [] @ [a]&quot; by simp<br /> next <br /> fix b xs <br /> assume HI: &quot;snoc xs a = xs @ [a]&quot;<br /> thus &quot;snoc (b # xs) a = (b # xs) @ [a]&quot;<br /> proof -<br /> have &quot;snoc (b # xs) a = b#(snoc xs a)&quot; by simp<br /> also have &quot;... = b#(xs @ [a])&quot; using HI by simp<br /> also have &quot;... = (b#xs) @ [a]&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 3. Demostrar el siguiente teorema <br /> rev (x # xs) = snoc (rev xs) x&quot;<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del teorema es&quot;<br /> theorem rev_cons_auto: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> by (simp add:snoc_append)<br /> <br /> -- &quot;La demostración estructurada del teorema es&quot;<br /> theorem rev_cons: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> proof -<br /> have &quot;rev(x#xs)=(rev xs) @ [x]&quot; by simp <br /> also have &quot;... = snoc (rev xs) x&quot; by (simp add:snoc_append)<br /> finally show ?thesis . <br /> qed<br /> <br /> <br /> <br /> <br /> <br /> section {* Cuantificadores sobre listas *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 4. Definir la función <br /> todos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (todos p xs) se verifica si todos los elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> todos (λx. 1&lt;length x) [[2,1,4],[1,3]]<br /> ¬ todos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> <br /> Nota: La función todos es equivalente a la predefinida list_all. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec todos :: &quot;(&#039;a ⇒ bool) ⇒ (&#039;a list ⇒ bool)&quot; where<br /> &quot;todos P [] = True&quot;<br /> |&quot;todos P (x#xs) = (P x ∧ todos P xs)&quot;<br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 5. Definir la función <br /> algunos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (algunos p xs) se verifica si algunos elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> algunos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> ¬ algunos (λx. 1&lt;length x) [[],[3]]&quot;<br /> <br /> Nota: La función algunos es equivalente a la predefinida list_ex. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec algunos :: &quot;(&#039;a =&gt; bool) =&gt; &#039;a list =&gt; bool&quot; where<br /> &quot;algunos P [] = False&quot;<br /> | &quot;algunos P (x#xs) = (P x ∨ algunos P xs)&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 6. Demostrar o refutar: <br /> todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> proof (induct xs)<br /> show &quot;todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])&quot; by simp<br /> next<br /> fix a xs<br /> assume 1: &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> thus &quot;todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))&quot;<br /> proof -<br /> have 2: &quot;todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos (λx. P x ∧ Q x) xs))&quot; by simp<br /> also have 3: &quot;... = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))&quot; using 1 by simp<br /> also have 4: &quot;... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))&quot; by blast<br /> also have 5: &quot;... = ((todos P (a # xs) ∧ todos Q (a # xs)))&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 7. Demostrar o refutar: <br /> todos P (x @ y) = (todos P x ∧ todos P y)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> by (induct x) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> proof (induct x)<br /> show &quot;todos P ([] @ y) = (todos P [] ∧ todos P y)&quot; by simp<br /> next<br /> fix a x y<br /> assume 1: &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> thus &quot;todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)&quot;<br /> proof -<br /> have &quot;(todos P ((a # x) @ y)) = (P a ∧ todos P (x @ y))&quot; by simp<br /> also have &quot;... = (P a ∧ todos P x ∧ todos P y)&quot; using 1 by simp<br /> also have &quot;... = (todos P (a #x) ∧ todos P y)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 8. Demostrar o refutar: <br /> todos P (rev xs) = todos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 9. Demostrar o refutar:<br /> algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;Se busca un contraejemplo con nitpick&quot;<br /> lemma &quot;algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> El contraejemplo encontrado es<br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 10. Demostrar o refutar: <br /> algunos P (map f xs) = algunos (P ∘ f) xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P o f) xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P ∘ f) xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 11. Demostrar o refutar: <br /> algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma algunos_append: <br /> &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 12. Demostrar o refutar: <br /> algunos P (rev xs) = algunos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 13. Encontrar un término no trivial Z tal que sea cierta la <br /> siguiente ecuación:<br /> algunos (λx. P x ∨ Q x) xs = Z<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La ecuación se verifica eligiendo como Z el término <br /> algunos P xs ∨ algunos Q xs<br /> En efecto,<br /> *}<br /> <br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;De forma estructurada&quot;<br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 14. Demostrar o refutar:<br /> algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 15. Definir la funcion primitiva recursiva <br /> estaEn :: &#039;a ⇒ &#039;a list ⇒ bool<br /> tal que (estaEn x xs) se verifica si el elemento x está en la lista<br /> xs. Por ejemplo, <br /> estaEn (2::nat) [3,2,4] = True<br /> estaEn (1::nat) [3,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec estaEn :: &quot;&#039;a ⇒ &#039;a list ⇒ bool&quot; where<br /> &quot;estaEn x [] = False&quot;<br /> | &quot;estaEn x (a#xs) = (x=a ∨ estaEn x xs)&quot; <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 16. Expresar la relación existente entre estaEn y algunos. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La relación es <br /> <br /> En efecto, <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 17. Definir la función primitiva recursiva <br /> sinDuplicados :: &#039;a list ⇒ bool<br /> tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br /> duplicados. Por ejemplo, <br /> sinDuplicados [1::nat,4,2] = True<br /> sinDuplicados [1::nat,4,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> <br /> primrec sinDuplicados :: &quot;&#039;a list ⇒ bool&quot; where<br /> &quot;sinDuplicados [] = True&quot;<br /> | &quot;sinDuplicados (a#xs) = ((¬ (estaEn a xs)) ∧ sinDuplicados xs)&quot;<br /> <br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 18. Definir la función primitiva recursiva <br /> borraDuplicados :: &#039;a list ⇒ bool<br /> tal que (borraDuplicados xs) es la lista obtenida eliminando los<br /> elementos duplicados de la lista xs. Por ejemplo, <br /> borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br /> <br /> Nota: La función borraDuplicados es equivalente a la predefinida remdups. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> <br /> primrec borraDuplicados :: &quot;&#039;a list ⇒ &#039;a list&quot; where<br /> &quot;borraDuplicados [] = []&quot;<br /> | &quot;borraDuplicados (a#xs) = <br /> (if estaEn a xs <br /> then borraDuplicados xs <br /> else (a#borraDuplicados xs))&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 19. Demostrar o refutar:<br /> length (borraDuplicados xs) ≤ length xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma length_borraDuplicados: <br /> &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 20. Demostrar o refutar: <br /> estaEn a (borraDuplicados xs) = estaEn a xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma estaEn_borraDuplicados_auto: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma estaEn_borraDuplicados: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 21. Demostrar o refutar: <br /> sinDuplicados (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática&quot;<br /> lemma &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma sinDuplicados_borraDuplicados:<br /> &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 22. Demostrar o refutar:<br /> borraDuplicados (rev xs) = rev (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma &quot;borraDuplicados (rev xs) = rev (borraDuplicados xs)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_3&diff=130 2011-02-14T15:51:41Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 3ª relación de ejercicios *}<br /> <br /> theory Relacion_3<br /> imports Main <br /> begin<br /> <br /> section {* Cons inverso *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 1. Definir recursivamente la función <br /> snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot;<br /> tal que (snoc xs a) es la lista obtenida al añadir el elemento a al<br /> final de la lista xs. Por ejemplo, <br /> value &quot;snoc [2,5] (3::int)&quot; == [2,5,3]<br /> <br /> Nota: No usar @.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> primrec snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot; where<br /> &quot;snoc [] y = [y]&quot;<br /> | &quot;snoc (x#xs) y = x#(snoc xs y)&quot;<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 2. Demostrar el siguiente teorema <br /> snoc xs a = xs @ [a]<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del lema es&quot;<br /> lemma &quot;snoc xs a = xs @ [a]&quot;<br /> by (induct xs) auto<br /> <br /> -- &quot;La demostración estructurada del lema es&quot;<br /> lemma snoc_append: &quot;snoc xs a = xs @ [a]&quot;<br /> proof (induct xs)<br /> show &quot;snoc [] a = [] @ [a]&quot; by simp<br /> next <br /> fix b xs <br /> assume HI: &quot;snoc xs a = xs @ [a]&quot;<br /> thus &quot;snoc (b # xs) a = (b # xs) @ [a]&quot;<br /> proof -<br /> have &quot;snoc (b # xs) a = b#(snoc xs a)&quot; by simp<br /> also have &quot;... = b#(xs @ [a])&quot; using HI by simp<br /> also have &quot;... = (b#xs) @ [a]&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 3. Demostrar el siguiente teorema <br /> rev (x # xs) = snoc (rev xs) x&quot;<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del teorema es&quot;<br /> theorem rev_cons_auto: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> by (simp add:snoc_append)<br /> <br /> -- &quot;La demostración estructurada del teorema es&quot;<br /> theorem rev_cons: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> proof -<br /> have &quot;rev(x#xs)=(rev xs) @ [x]&quot; by simp <br /> also have &quot;... = snoc (rev xs) x&quot; by (simp add:snoc_append)<br /> finally show ?thesis . <br /> qed<br /> <br /> <br /> <br /> <br /> <br /> section {* Cuantificadores sobre listas *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 4. Definir la función <br /> todos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (todos p xs) se verifica si todos los elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> todos (λx. 1&lt;length x) [[2,1,4],[1,3]]<br /> ¬ todos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> <br /> Nota: La función todos es equivalente a la predefinida list_all. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec todos :: &quot;(&#039;a ⇒ bool) ⇒ (&#039;a list ⇒ bool)&quot; where<br /> &quot;todos P [] = True&quot;<br /> |&quot;todos P (x#xs) = (P x ∧ todos P xs)&quot;<br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 5. Definir la función <br /> algunos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (algunos p xs) se verifica si algunos elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> algunos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> ¬ algunos (λx. 1&lt;length x) [[],[3]]&quot;<br /> <br /> Nota: La función algunos es equivalente a la predefinida list_ex. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec algunos :: &quot;(&#039;a =&gt; bool) =&gt; &#039;a list =&gt; bool&quot; where<br /> &quot;algunos P [] = False&quot;<br /> | &quot;algunos P (x#xs) = (P x ∨ algunos P xs)&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 6. Demostrar o refutar: <br /> todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> proof (induct xs)<br /> show &quot;todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])&quot; by simp<br /> next<br /> fix a xs<br /> assume 1: &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> thus &quot;todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))&quot;<br /> proof -<br /> have 2: &quot;todos (λx. P x ∧ Q x) (a # xs) = ((P a ∧ Q a) ∧ (todos (λx. P x ∧ Q x) xs))&quot; by simp<br /> also have 3: &quot;... = ((P a ∧ Q a) ∧ (todos P xs ∧ todos Q xs))&quot; using 1 by simp<br /> also have 4: &quot;... = ((P a ∧ todos P xs) ∧ (Q a ∧ todos Q xs))&quot; by blast<br /> also have 5: &quot;... = ((todos P (a # xs) ∧ todos Q (a # xs)))&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 7. Demostrar o refutar: <br /> todos P (x @ y) = (todos P x ∧ todos P y)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> by (induct x) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> proof (induct x)<br /> show &quot;todos P ([] @ y) = (todos P [] ∧ todos P y)&quot; by simp<br /> next<br /> fix a x y<br /> assume 1: &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> thus &quot;todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)&quot;<br /> proof -<br /> have &quot;(todos P ((a # x) @ y)) = (P a ∧ todos P (x @ y))&quot; by simp<br /> also have &quot;... = (P a ∧ todos P x ∧ todos P y)&quot; using 1 by simp<br /> also have &quot;... = (todos P (a #x) ∧ todos P y)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 8. Demostrar o refutar: <br /> todos P (rev xs) = todos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 9. Demostrar o refutar:<br /> algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;Se busca un contraejemplo con nitpick&quot;<br /> lemma &quot;algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> El contraejemplo encontrado es<br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 10. Demostrar o refutar: <br /> algunos P (map f xs) = algunos (P ∘ f) xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P o f) xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P ∘ f) xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 11. Demostrar o refutar: <br /> algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma algunos_append: <br /> &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 12. Demostrar o refutar: <br /> algunos P (rev xs) = algunos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 13. Encontrar un término no trivial Z tal que sea cierta la <br /> siguiente ecuación:<br /> algunos (λx. P x ∨ Q x) xs = Z<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La ecuación se verifica eligiendo como Z el término <br /> algunos P xs ∨ algunos Q xs<br /> En efecto,<br /> *}<br /> <br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;De forma estructurada&quot;<br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 14. Demostrar o refutar:<br /> algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 15. Definir la funcion primitiva recursiva <br /> estaEn :: &#039;a ⇒ &#039;a list ⇒ bool<br /> tal que (estaEn x xs) se verifica si el elemento x está en la lista<br /> xs. Por ejemplo, <br /> estaEn (2::nat) [3,2,4] = True<br /> estaEn (1::nat) [3,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec estaEn :: &quot;&#039;a ⇒ &#039;a list ⇒ bool&quot; where<br /> &quot;estaEn x [] = False&quot;<br /> | &quot;estaEn x (a#xs) = (x=a ∨ estaEn x xs)&quot; <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 16. Expresar la relación existente entre estaEn y algunos. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La relación es <br /> <br /> En efecto, <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 17. Definir la función primitiva recursiva <br /> sinDuplicados :: &#039;a list ⇒ bool<br /> tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br /> duplicados. Por ejemplo, <br /> sinDuplicados [1::nat,4,2] = True<br /> sinDuplicados [1::nat,4,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> <br /> primrec sinDuplicados :: &quot;&#039;a list ⇒ bool&quot; where<br /> &quot;sinDuplicados [] = True&quot;<br /> | &quot;sinDuplicados (a#xs) = ((¬ (estaEn a xs)) ∧ sinDuplicados xs)&quot;<br /> <br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 18. Definir la función primitiva recursiva <br /> borraDuplicados :: &#039;a list ⇒ bool<br /> tal que (borraDuplicados xs) es la lista obtenida eliminando los<br /> elementos duplicados de la lista xs. Por ejemplo, <br /> borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br /> <br /> Nota: La función borraDuplicados es equivalente a la predefinida remdups. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> <br /> primrec borraDuplicados :: &quot;&#039;a list ⇒ &#039;a list&quot; where<br /> &quot;borraDuplicados [] = []&quot;<br /> | &quot;borraDuplicados (a#xs) = <br /> (if estaEn a xs <br /> then borraDuplicados xs <br /> else (a#borraDuplicados xs))&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 19. Demostrar o refutar:<br /> length (borraDuplicados xs) ≤ length xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma length_borraDuplicados: <br /> &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 20. Demostrar o refutar: <br /> estaEn a (borraDuplicados xs) = estaEn a xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma estaEn_borraDuplicados_auto: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma estaEn_borraDuplicados: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 21. Demostrar o refutar: <br /> sinDuplicados (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática&quot;<br /> lemma &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma sinDuplicados_borraDuplicados:<br /> &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 22. Demostrar o refutar:<br /> borraDuplicados (rev xs) = rev (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma &quot;borraDuplicados (rev xs) = rev (borraDuplicados xs)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_3&diff=126 2011-02-14T15:10:31Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 3ª relación de ejercicios *}<br /> <br /> theory Relacion_3<br /> imports Main <br /> begin<br /> <br /> section {* Cons inverso *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 1. Definir recursivamente la función <br /> snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot;<br /> tal que (snoc xs a) es la lista obtenida al añadir el elemento a al<br /> final de la lista xs. Por ejemplo, <br /> value &quot;snoc [2,5] (3::int)&quot; == [2,5,3]<br /> <br /> Nota: No usar @.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> primrec snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot; where<br /> &quot;snoc [] y = [y]&quot;<br /> | &quot;snoc (x#xs) y = x#(snoc xs y)&quot;<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 2. Demostrar el siguiente teorema <br /> snoc xs a = xs @ [a]<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del lema es&quot;<br /> lemma &quot;snoc xs a = xs @ [a]&quot;<br /> by (induct xs) auto<br /> <br /> -- &quot;La demostración estructurada del lema es&quot;<br /> lemma snoc_append: &quot;snoc xs a = xs @ [a]&quot;<br /> proof (induct xs)<br /> show &quot;snoc [] a = [] @ [a]&quot; by simp<br /> next <br /> fix b xs <br /> assume HI: &quot;snoc xs a = xs @ [a]&quot;<br /> thus &quot;snoc (b # xs) a = (b # xs) @ [a]&quot;<br /> proof -<br /> have &quot;snoc (b # xs) a = b#(snoc xs a)&quot; by simp<br /> also have &quot;... = b#(xs @ [a])&quot; using HI by simp<br /> also have &quot;... = (b#xs) @ [a]&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 3. Demostrar el siguiente teorema <br /> rev (x # xs) = snoc (rev xs) x&quot;<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del teorema es&quot;<br /> theorem rev_cons_auto: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> by (simp add:snoc_append)<br /> <br /> -- &quot;La demostración estructurada del teorema es&quot;<br /> theorem rev_cons: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> proof -<br /> have &quot;rev(x#xs)=(rev xs) @ [x]&quot; by simp <br /> also have &quot;... = snoc (rev xs) x&quot; by (simp add:snoc_append)<br /> finally show ?thesis . <br /> qed<br /> <br /> <br /> <br /> <br /> <br /> section {* Cuantificadores sobre listas *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 4. Definir la función <br /> todos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (todos p xs) se verifica si todos los elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> todos (λx. 1&lt;length x) [[2,1,4],[1,3]]<br /> ¬ todos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> <br /> Nota: La función todos es equivalente a la predefinida list_all. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec todos :: &quot;(&#039;a ⇒ bool) ⇒ (&#039;a list ⇒ bool)&quot; where<br /> &quot;todos P [] = True&quot;<br /> |&quot;todos P (x#xs) = (P x ∧ todos P xs)&quot;<br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 5. Definir la función <br /> algunos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (algunos p xs) se verifica si algunos elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> algunos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> ¬ algunos (λx. 1&lt;length x) [[],[3]]&quot;<br /> <br /> Nota: La función algunos es equivalente a la predefinida list_ex. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec algunos :: &quot;(&#039;a =&gt; bool) =&gt; &#039;a list =&gt; bool&quot; where<br /> &quot;algunos P [] = False&quot;<br /> | &quot;algunos P (x#xs) = (P x ∨ algunos P xs)&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 6. Demostrar o refutar: <br /> todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 7. Demostrar o refutar: <br /> todos P (x @ y) = (todos P x ∧ todos P y)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> by (induct x) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> proof (induct x)<br /> show &quot;todos P ([] @ y) = (todos P [] ∧ todos P y)&quot; by simp<br /> next<br /> fix a x y<br /> assume 1: &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> thus &quot;todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)&quot;<br /> proof -<br /> have &quot;(todos P ((a # x) @ y)) = (P a ∧ todos P (x @ y))&quot; by simp<br /> also have &quot;... = (P a ∧ todos P x ∧ todos P y)&quot; using 1 by simp<br /> also have &quot;... = (todos P (a #x) ∧ todos P y)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 8. Demostrar o refutar: <br /> todos P (rev xs) = todos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 9. Demostrar o refutar:<br /> algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;Se busca un contraejemplo con nitpick&quot;<br /> lemma &quot;algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> El contraejemplo encontrado es<br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 10. Demostrar o refutar: <br /> algunos P (map f xs) = algunos (P ∘ f) xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P o f) xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P ∘ f) xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 11. Demostrar o refutar: <br /> algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma algunos_append: <br /> &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 12. Demostrar o refutar: <br /> algunos P (rev xs) = algunos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 13. Encontrar un término no trivial Z tal que sea cierta la <br /> siguiente ecuación:<br /> algunos (λx. P x ∨ Q x) xs = Z<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La ecuación se verifica eligiendo como Z el término <br /> algunos P xs ∨ algunos Q xs<br /> En efecto,<br /> *}<br /> <br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;De forma estructurada&quot;<br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 14. Demostrar o refutar:<br /> algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 15. Definir la funcion primitiva recursiva <br /> estaEn :: &#039;a ⇒ &#039;a list ⇒ bool<br /> tal que (estaEn x xs) se verifica si el elemento x está en la lista<br /> xs. Por ejemplo, <br /> estaEn (2::nat) [3,2,4] = True<br /> estaEn (1::nat) [3,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 16. Expresar la relación existente entre estaEn y algunos. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La relación es <br /> <br /> En efecto, <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 17. Definir la función primitiva recursiva <br /> sinDuplicados :: &#039;a list ⇒ bool<br /> tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br /> duplicados. Por ejemplo, <br /> sinDuplicados [1::nat,4,2] = True<br /> sinDuplicados [1::nat,4,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 18. Definir la función primitiva recursiva <br /> borraDuplicados :: &#039;a list ⇒ bool<br /> tal que (borraDuplicados xs) es la lista obtenida eliminando los<br /> elementos duplicados de la lista xs. Por ejemplo, <br /> borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br /> <br /> Nota: La función borraDuplicados es equivalente a la predefinida remdups. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 19. Demostrar o refutar:<br /> length (borraDuplicados xs) ≤ length xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma length_borraDuplicados: <br /> &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 20. Demostrar o refutar: <br /> estaEn a (borraDuplicados xs) = estaEn a xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma estaEn_borraDuplicados_auto: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma estaEn_borraDuplicados: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 21. Demostrar o refutar: <br /> sinDuplicados (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática&quot;<br /> lemma &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma sinDuplicados_borraDuplicados:<br /> &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 22. Demostrar o refutar:<br /> borraDuplicados (rev xs) = rev (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma &quot;borraDuplicados (rev xs) = rev (borraDuplicados xs)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_3&diff=125 2011-02-14T15:09:24Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 3ª relación de ejercicios *}<br /> <br /> theory Relacion_3<br /> imports Main <br /> begin<br /> <br /> section {* Cons inverso *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 1. Definir recursivamente la función <br /> snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot;<br /> tal que (snoc xs a) es la lista obtenida al añadir el elemento a al<br /> final de la lista xs. Por ejemplo, <br /> value &quot;snoc [2,5] (3::int)&quot; == [2,5,3]<br /> <br /> Nota: No usar @.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> primrec snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot; where<br /> &quot;snoc [] y = [y]&quot;<br /> | &quot;snoc (x#xs) y = x#(snoc xs y)&quot;<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 2. Demostrar el siguiente teorema <br /> snoc xs a = xs @ [a]<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del lema es&quot;<br /> lemma &quot;snoc xs a = xs @ [a]&quot;<br /> by (induct xs) auto<br /> <br /> -- &quot;La demostración estructurada del lema es&quot;<br /> lemma snoc_append: &quot;snoc xs a = xs @ [a]&quot;<br /> proof (induct xs)<br /> show &quot;snoc [] a = [] @ [a]&quot; by simp<br /> next <br /> fix b xs <br /> assume HI: &quot;snoc xs a = xs @ [a]&quot;<br /> thus &quot;snoc (b # xs) a = (b # xs) @ [a]&quot;<br /> proof -<br /> have &quot;snoc (b # xs) a = b#(snoc xs a)&quot; by simp<br /> also have &quot;... = b#(xs @ [a])&quot; using HI by simp<br /> also have &quot;... = (b#xs) @ [a]&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 3. Demostrar el siguiente teorema <br /> rev (x # xs) = snoc (rev xs) x&quot;<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del teorema es&quot;<br /> theorem rev_cons_auto: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> by (simp add:snoc_append)<br /> <br /> -- &quot;La demostración estructurada del teorema es&quot;<br /> theorem rev_cons: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> proof -<br /> have &quot;rev(x#xs)=(rev xs) @ [x]&quot; by simp <br /> also have &quot;... = snoc (rev xs) x&quot; by (simp add:snoc_append)<br /> finally show ?thesis . <br /> qed<br /> <br /> <br /> <br /> <br /> <br /> section {* Cuantificadores sobre listas *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 4. Definir la función <br /> todos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (todos p xs) se verifica si todos los elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> todos (λx. 1&lt;length x) [[2,1,4],[1,3]]<br /> ¬ todos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> <br /> Nota: La función todos es equivalente a la predefinida list_all. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec todos :: &quot;(&#039;a ⇒ bool) ⇒ (&#039;a list ⇒ bool)&quot; where<br /> &quot;todos P [] = True&quot;<br /> |&quot;todos P (x#xs) = (P x ∧ todos P xs)&quot;<br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 5. Definir la función <br /> algunos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (algunos p xs) se verifica si algunos elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> algunos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> ¬ algunos (λx. 1&lt;length x) [[],[3]]&quot;<br /> <br /> Nota: La función algunos es equivalente a la predefinida list_ex. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec algunos :: &quot;(&#039;a =&gt; bool) =&gt; &#039;a list =&gt; bool&quot; where<br /> &quot;algunos P [] = False&quot;<br /> | &quot;algunos P (x#xs) = (P x ∨ algunos P xs)&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 6. Demostrar o refutar: <br /> todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 7. Demostrar o refutar: <br /> todos P (x @ y) = (todos P x ∧ todos P y)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> by (induct x) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> proof (induct x)<br /> show &quot;todos P ([] @ y) = (todos P [] ∧ todos P y)&quot; by simp<br /> next<br /> fix a x y<br /> assume 1: &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> thus &quot;todos P ((a # x) @ y) = (todos P (a # x) ∧ todos P y)&quot;<br /> proof -<br /> have &quot;(todos P ((a # x) @ y)) = (P a ∧ todos P (x @ y))&quot; by simp<br /> also have &quot;... = (P a ∧ todos P x ∧ todos P y)&quot; using 1 by simp<br /> also have &quot;... = (todos P (a #x) ∧ todos P y)&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 8. Demostrar o refutar: <br /> todos P (rev xs) = todos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 9. Demostrar o refutar:<br /> algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;Se busca un contraejemplo con nitpick&quot;<br /> lemma &quot;algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> El contraejemplo encontrado es<br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 10. Demostrar o refutar: <br /> algunos P (map f xs) = algunos (P ∘ f) xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P o f) xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P ∘ f) xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 11. Demostrar o refutar: <br /> algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma algunos_append: <br /> &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 12. Demostrar o refutar: <br /> algunos P (rev xs) = algunos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 13. Encontrar un término no trivial Z tal que sea cierta la <br /> siguiente ecuación:<br /> algunos (λx. P x ∨ Q x) xs = Z<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La ecuación se verifica eligiendo como Z el término <br /> algunos P xs ∨ algunos Q xs<br /> En efecto,<br /> *}<br /> <br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;De forma estructurada&quot;<br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 14. Demostrar o refutar:<br /> algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 15. Definir la funcion primitiva recursiva <br /> estaEn :: &#039;a ⇒ &#039;a list ⇒ bool<br /> tal que (estaEn x xs) se verifica si el elemento x está en la lista<br /> xs. Por ejemplo, <br /> estaEn (2::nat) [3,2,4] = True<br /> estaEn (1::nat) [3,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 16. Expresar la relación existente entre estaEn y algunos. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La relación es <br /> <br /> En efecto, <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 17. Definir la función primitiva recursiva <br /> sinDuplicados :: &#039;a list ⇒ bool<br /> tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br /> duplicados. Por ejemplo, <br /> sinDuplicados [1::nat,4,2] = True<br /> sinDuplicados [1::nat,4,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 18. Definir la función primitiva recursiva <br /> borraDuplicados :: &#039;a list ⇒ bool<br /> tal que (borraDuplicados xs) es la lista obtenida eliminando los<br /> elementos duplicados de la lista xs. Por ejemplo, <br /> borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br /> <br /> Nota: La función borraDuplicados es equivalente a la predefinida remdups. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 19. Demostrar o refutar:<br /> length (borraDuplicados xs) ≤ length xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma length_borraDuplicados: <br /> &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 20. Demostrar o refutar: <br /> estaEn a (borraDuplicados xs) = estaEn a xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma estaEn_borraDuplicados_auto: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma estaEn_borraDuplicados: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 21. Demostrar o refutar: <br /> sinDuplicados (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática&quot;<br /> lemma &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma sinDuplicados_borraDuplicados:<br /> &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 22. Demostrar o refutar:<br /> borraDuplicados (rev xs) = rev (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma &quot;borraDuplicados (rev xs) = rev (borraDuplicados xs)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_3&diff=124 2011-02-14T15:02:27Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 3ª relación de ejercicios *}<br /> <br /> theory Relacion_3<br /> imports Main <br /> begin<br /> <br /> section {* Cons inverso *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 1. Definir recursivamente la función <br /> snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot;<br /> tal que (snoc xs a) es la lista obtenida al añadir el elemento a al<br /> final de la lista xs. Por ejemplo, <br /> value &quot;snoc [2,5] (3::int)&quot; == [2,5,3]<br /> <br /> Nota: No usar @.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> primrec snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot; where<br /> &quot;snoc [] y = [y]&quot;<br /> | &quot;snoc (x#xs) y = x#(snoc xs y)&quot;<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 2. Demostrar el siguiente teorema <br /> snoc xs a = xs @ [a]<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del lema es&quot;<br /> lemma &quot;snoc xs a = xs @ [a]&quot;<br /> by (induct xs) auto<br /> <br /> -- &quot;La demostración estructurada del lema es&quot;<br /> lemma snoc_append: &quot;snoc xs a = xs @ [a]&quot;<br /> proof (induct xs)<br /> show &quot;snoc [] a = [] @ [a]&quot; by simp<br /> next <br /> fix b xs <br /> assume HI: &quot;snoc xs a = xs @ [a]&quot;<br /> thus &quot;snoc (b # xs) a = (b # xs) @ [a]&quot;<br /> proof -<br /> have &quot;snoc (b # xs) a = b#(snoc xs a)&quot; by simp<br /> also have &quot;... = b#(xs @ [a])&quot; using HI by simp<br /> also have &quot;... = (b#xs) @ [a]&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 3. Demostrar el siguiente teorema <br /> rev (x # xs) = snoc (rev xs) x&quot;<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del teorema es&quot;<br /> theorem rev_cons_auto: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> by (simp add:snoc_append)<br /> <br /> -- &quot;La demostración estructurada del teorema es&quot;<br /> theorem rev_cons: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> proof -<br /> have &quot;rev(x#xs)=(rev xs) @ [x]&quot; by simp <br /> also have &quot;... = snoc (rev xs) x&quot; by (simp add:snoc_append)<br /> finally show ?thesis . <br /> qed<br /> <br /> <br /> <br /> <br /> <br /> section {* Cuantificadores sobre listas *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 4. Definir la función <br /> todos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (todos p xs) se verifica si todos los elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> todos (λx. 1&lt;length x) [[2,1,4],[1,3]]<br /> ¬ todos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> <br /> Nota: La función todos es equivalente a la predefinida list_all. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec todos :: &quot;(&#039;a ⇒ bool) ⇒ (&#039;a list ⇒ bool)&quot; where<br /> &quot;todos P [] = True&quot;<br /> |&quot;todos P (x#xs) = (P x ∧ todos P xs)&quot;<br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 5. Definir la función <br /> algunos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (algunos p xs) se verifica si algunos elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> algunos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> ¬ algunos (λx. 1&lt;length x) [[],[3]]&quot;<br /> <br /> Nota: La función algunos es equivalente a la predefinida list_ex. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec algunos :: &quot;(&#039;a =&gt; bool) =&gt; &#039;a list =&gt; bool&quot; where<br /> &quot;algunos P [] = False&quot;<br /> | &quot;algunos P (x#xs) = (P x ∨ algunos P xs)&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 6. Demostrar o refutar: <br /> todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 7. Demostrar o refutar: <br /> todos P (x @ y) = (todos P x ∧ todos P y)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> by (induct x) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 8. Demostrar o refutar: <br /> todos P (rev xs) = todos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 9. Demostrar o refutar:<br /> algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;Se busca un contraejemplo con nitpick&quot;<br /> lemma &quot;algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> El contraejemplo encontrado es<br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 10. Demostrar o refutar: <br /> algunos P (map f xs) = algunos (P ∘ f) xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P o f) xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P ∘ f) xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 11. Demostrar o refutar: <br /> algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma algunos_append: <br /> &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 12. Demostrar o refutar: <br /> algunos P (rev xs) = algunos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 13. Encontrar un término no trivial Z tal que sea cierta la <br /> siguiente ecuación:<br /> algunos (λx. P x ∨ Q x) xs = Z<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La ecuación se verifica eligiendo como Z el término <br /> algunos P xs ∨ algunos Q xs<br /> En efecto,<br /> *}<br /> <br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;De forma estructurada&quot;<br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 14. Demostrar o refutar:<br /> algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 15. Definir la funcion primitiva recursiva <br /> estaEn :: &#039;a ⇒ &#039;a list ⇒ bool<br /> tal que (estaEn x xs) se verifica si el elemento x está en la lista<br /> xs. Por ejemplo, <br /> estaEn (2::nat) [3,2,4] = True<br /> estaEn (1::nat) [3,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 16. Expresar la relación existente entre estaEn y algunos. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La relación es <br /> <br /> En efecto, <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 17. Definir la función primitiva recursiva <br /> sinDuplicados :: &#039;a list ⇒ bool<br /> tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br /> duplicados. Por ejemplo, <br /> sinDuplicados [1::nat,4,2] = True<br /> sinDuplicados [1::nat,4,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 18. Definir la función primitiva recursiva <br /> borraDuplicados :: &#039;a list ⇒ bool<br /> tal que (borraDuplicados xs) es la lista obtenida eliminando los<br /> elementos duplicados de la lista xs. Por ejemplo, <br /> borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br /> <br /> Nota: La función borraDuplicados es equivalente a la predefinida remdups. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 19. Demostrar o refutar:<br /> length (borraDuplicados xs) ≤ length xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma length_borraDuplicados: <br /> &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 20. Demostrar o refutar: <br /> estaEn a (borraDuplicados xs) = estaEn a xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma estaEn_borraDuplicados_auto: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma estaEn_borraDuplicados: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 21. Demostrar o refutar: <br /> sinDuplicados (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática&quot;<br /> lemma &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma sinDuplicados_borraDuplicados:<br /> &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 22. Demostrar o refutar:<br /> borraDuplicados (rev xs) = rev (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma &quot;borraDuplicados (rev xs) = rev (borraDuplicados xs)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_3&diff=123 2011-02-14T13:59:57Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 3ª relación de ejercicios *}<br /> <br /> theory Relacion_3<br /> imports Main <br /> begin<br /> <br /> section {* Cons inverso *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 1. Definir recursivamente la función <br /> snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot;<br /> tal que (snoc xs a) es la lista obtenida al añadir el elemento a al<br /> final de la lista xs. Por ejemplo, <br /> value &quot;snoc [2,5] (3::int)&quot; == [2,5,3]<br /> <br /> Nota: No usar @.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> primrec snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot; where<br /> &quot;snoc [] y = [y]&quot;<br /> | &quot;snoc (x#xs) y = x#(snoc xs y)&quot;<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 2. Demostrar el siguiente teorema <br /> snoc xs a = xs @ [a]<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del lema es&quot;<br /> lemma &quot;snoc xs a = xs @ [a]&quot;<br /> by (induct xs) auto<br /> <br /> -- &quot;La demostración estructurada del lema es&quot;<br /> lemma snoc_append: &quot;snoc xs a = xs @ [a]&quot;<br /> proof (induct xs)<br /> show &quot;snoc [] a = [] @ [a]&quot; by simp<br /> next <br /> fix b xs <br /> assume HI: &quot;snoc xs a = xs @ [a]&quot;<br /> thus &quot;snoc (b # xs) a = (b # xs) @ [a]&quot;<br /> proof -<br /> have &quot;snoc (b # xs) a = b#(snoc xs a)&quot; by simp<br /> also have &quot;... = b#(xs @ [a])&quot; using HI by simp<br /> also have &quot;... = (b#xs) @ [a]&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 3. Demostrar el siguiente teorema <br /> rev (x # xs) = snoc (rev xs) x&quot;<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del teorema es&quot;<br /> theorem rev_cons_auto: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> by (simp add:snoc_append)<br /> <br /> -- &quot;La demostración estructurada del teorema es&quot;<br /> theorem rev_cons: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> proof -<br /> have &quot;rev(x#xs)=(rev xs) @ [x]&quot; by simp <br /> also have &quot;... = snoc (rev xs) x&quot; by (simp add:snoc_append)<br /> finally show ?thesis . <br /> qed<br /> <br /> <br /> <br /> <br /> <br /> section {* Cuantificadores sobre listas *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 4. Definir la función <br /> todos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (todos p xs) se verifica si todos los elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> todos (λx. 1&lt;length x) [[2,1,4],[1,3]]<br /> ¬ todos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> <br /> Nota: La función todos es equivalente a la predefinida list_all. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec todos :: &quot;(&#039;a ⇒ bool) ⇒ (&#039;a list ⇒ bool)&quot; where<br /> &quot;todos P [] = True&quot;<br /> |&quot;todos P (x#xs) = (P x ∧ todos P xs)&quot;<br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 5. Definir la función <br /> algunos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (algunos p xs) se verifica si algunos elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> algunos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> ¬ algunos (λx. 1&lt;length x) [[],[3]]&quot;<br /> <br /> Nota: La función algunos es equivalente a la predefinida list_ex. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec algunos :: &quot;(&#039;a =&gt; bool) =&gt; &#039;a list =&gt; bool&quot; where<br /> &quot;algunos P [] = False&quot;<br /> | &quot;algunos P (x#xs) = (P x ∨ algunos P xs)&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 6. Demostrar o refutar: <br /> todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> by (induct xs) auto<br /> <br /> <br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 7. Demostrar o refutar: <br /> todos P (x @ y) = (todos P x ∧ todos P y)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 8. Demostrar o refutar: <br /> todos P (rev xs) = todos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 9. Demostrar o refutar:<br /> algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;Se busca un contraejemplo con nitpick&quot;<br /> lemma &quot;algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> El contraejemplo encontrado es<br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 10. Demostrar o refutar: <br /> algunos P (map f xs) = algunos (P ∘ f) xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P o f) xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P ∘ f) xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 11. Demostrar o refutar: <br /> algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma algunos_append: <br /> &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 12. Demostrar o refutar: <br /> algunos P (rev xs) = algunos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 13. Encontrar un término no trivial Z tal que sea cierta la <br /> siguiente ecuación:<br /> algunos (λx. P x ∨ Q x) xs = Z<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La ecuación se verifica eligiendo como Z el término <br /> algunos P xs ∨ algunos Q xs<br /> En efecto,<br /> *}<br /> <br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;De forma estructurada&quot;<br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 14. Demostrar o refutar:<br /> algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 15. Definir la funcion primitiva recursiva <br /> estaEn :: &#039;a ⇒ &#039;a list ⇒ bool<br /> tal que (estaEn x xs) se verifica si el elemento x está en la lista<br /> xs. Por ejemplo, <br /> estaEn (2::nat) [3,2,4] = True<br /> estaEn (1::nat) [3,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 16. Expresar la relación existente entre estaEn y algunos. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La relación es <br /> <br /> En efecto, <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 17. Definir la función primitiva recursiva <br /> sinDuplicados :: &#039;a list ⇒ bool<br /> tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br /> duplicados. Por ejemplo, <br /> sinDuplicados [1::nat,4,2] = True<br /> sinDuplicados [1::nat,4,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 18. Definir la función primitiva recursiva <br /> borraDuplicados :: &#039;a list ⇒ bool<br /> tal que (borraDuplicados xs) es la lista obtenida eliminando los<br /> elementos duplicados de la lista xs. Por ejemplo, <br /> borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br /> <br /> Nota: La función borraDuplicados es equivalente a la predefinida remdups. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 19. Demostrar o refutar:<br /> length (borraDuplicados xs) ≤ length xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma length_borraDuplicados: <br /> &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 20. Demostrar o refutar: <br /> estaEn a (borraDuplicados xs) = estaEn a xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma estaEn_borraDuplicados_auto: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma estaEn_borraDuplicados: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 21. Demostrar o refutar: <br /> sinDuplicados (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática&quot;<br /> lemma &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma sinDuplicados_borraDuplicados:<br /> &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 22. Demostrar o refutar:<br /> borraDuplicados (rev xs) = rev (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma &quot;borraDuplicados (rev xs) = rev (borraDuplicados xs)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_3&diff=122 2011-02-14T13:42:21Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 3ª relación de ejercicios *}<br /> <br /> theory Relacion_3<br /> imports Main <br /> begin<br /> <br /> section {* Cons inverso *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 1. Definir recursivamente la función <br /> snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot;<br /> tal que (snoc xs a) es la lista obtenida al añadir el elemento a al<br /> final de la lista xs. Por ejemplo, <br /> value &quot;snoc [2,5] (3::int)&quot; == [2,5,3]<br /> <br /> Nota: No usar @.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> primrec snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot; where<br /> &quot;snoc [] y = [y]&quot;<br /> | &quot;snoc (x#xs) y = x#(snoc xs y)&quot;<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 2. Demostrar el siguiente teorema <br /> snoc xs a = xs @ [a]<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del lema es&quot;<br /> lemma &quot;snoc xs a = xs @ [a]&quot;<br /> by (induct xs) auto<br /> <br /> -- &quot;La demostración estructurada del lema es&quot;<br /> lemma snoc_append: &quot;snoc xs a = xs @ [a]&quot;<br /> proof (induct xs)<br /> show &quot;snoc [] a = [] @ [a]&quot; by simp<br /> next <br /> fix b xs <br /> assume HI: &quot;snoc xs a = xs @ [a]&quot;<br /> thus &quot;snoc (b # xs) a = (b # xs) @ [a]&quot;<br /> proof -<br /> have &quot;snoc (b # xs) a = b#(snoc xs a)&quot; by simp<br /> also have &quot;... = b#(xs @ [a])&quot; using HI by simp<br /> also have &quot;... = (b#xs) @ [a]&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 3. Demostrar el siguiente teorema <br /> rev (x # xs) = snoc (rev xs) x&quot;<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del teorema es&quot;<br /> theorem rev_cons_auto: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> by (simp add:snoc_append)<br /> <br /> -- &quot;La demostración estructurada del teorema es&quot;<br /> theorem rev_cons: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> proof -<br /> have &quot;rev(x#xs)=(rev xs) @ [x]&quot; by simp <br /> also have &quot;... = snoc (rev xs) x&quot; by (simp add:snoc_append)<br /> finally show ?thesis . <br /> qed<br /> <br /> <br /> <br /> <br /> <br /> section {* Cuantificadores sobre listas *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 4. Definir la función <br /> todos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (todos p xs) se verifica si todos los elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> todos (λx. 1&lt;length x) [[2,1,4],[1,3]]<br /> ¬ todos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> <br /> Nota: La función todos es equivalente a la predefinida list_all. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec todos :: &quot;(&#039;a ⇒ bool) ⇒ (&#039;a list ⇒ bool)&quot; where<br /> &quot;todos P [] = True&quot;<br /> |&quot;todos P (x#xs) = (P x ∧ todos P xs)&quot;<br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 5. Definir la función <br /> algunos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (algunos p xs) se verifica si algunos elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> algunos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> ¬ algunos (λx. 1&lt;length x) [[],[3]]&quot;<br /> <br /> Nota: La función algunos es equivalente a la predefinida list_ex. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec algunos :: &quot;(&#039;a =&gt; bool) =&gt; &#039;a list =&gt; bool&quot; where<br /> &quot;algunos P [] = False&quot;<br /> | &quot;algunos P (x#xs) = (P x ∨ algunos P xs)&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 6. Demostrar o refutar: <br /> todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 7. Demostrar o refutar: <br /> todos P (x @ y) = (todos P x ∧ todos P y)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 8. Demostrar o refutar: <br /> todos P (rev xs) = todos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 9. Demostrar o refutar:<br /> algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;Se busca un contraejemplo con nitpick&quot;<br /> lemma &quot;algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> El contraejemplo encontrado es<br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 10. Demostrar o refutar: <br /> algunos P (map f xs) = algunos (P ∘ f) xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P o f) xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P ∘ f) xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 11. Demostrar o refutar: <br /> algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma algunos_append: <br /> &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 12. Demostrar o refutar: <br /> algunos P (rev xs) = algunos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 13. Encontrar un término no trivial Z tal que sea cierta la <br /> siguiente ecuación:<br /> algunos (λx. P x ∨ Q x) xs = Z<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La ecuación se verifica eligiendo como Z el término <br /> algunos P xs ∨ algunos Q xs<br /> En efecto,<br /> *}<br /> <br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;De forma estructurada&quot;<br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 14. Demostrar o refutar:<br /> algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 15. Definir la funcion primitiva recursiva <br /> estaEn :: &#039;a ⇒ &#039;a list ⇒ bool<br /> tal que (estaEn x xs) se verifica si el elemento x está en la lista<br /> xs. Por ejemplo, <br /> estaEn (2::nat) [3,2,4] = True<br /> estaEn (1::nat) [3,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 16. Expresar la relación existente entre estaEn y algunos. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La relación es <br /> <br /> En efecto, <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 17. Definir la función primitiva recursiva <br /> sinDuplicados :: &#039;a list ⇒ bool<br /> tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br /> duplicados. Por ejemplo, <br /> sinDuplicados [1::nat,4,2] = True<br /> sinDuplicados [1::nat,4,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 18. Definir la función primitiva recursiva <br /> borraDuplicados :: &#039;a list ⇒ bool<br /> tal que (borraDuplicados xs) es la lista obtenida eliminando los<br /> elementos duplicados de la lista xs. Por ejemplo, <br /> borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br /> <br /> Nota: La función borraDuplicados es equivalente a la predefinida remdups. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 19. Demostrar o refutar:<br /> length (borraDuplicados xs) ≤ length xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma length_borraDuplicados: <br /> &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 20. Demostrar o refutar: <br /> estaEn a (borraDuplicados xs) = estaEn a xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma estaEn_borraDuplicados_auto: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma estaEn_borraDuplicados: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 21. Demostrar o refutar: <br /> sinDuplicados (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática&quot;<br /> lemma &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma sinDuplicados_borraDuplicados:<br /> &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 22. Demostrar o refutar:<br /> borraDuplicados (rev xs) = rev (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma &quot;borraDuplicados (rev xs) = rev (borraDuplicados xs)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_3&diff=121 2011-02-14T13:40:57Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 3ª relación de ejercicios *}<br /> <br /> theory Relacion_3<br /> imports Main <br /> begin<br /> <br /> section {* Cons inverso *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 1. Definir recursivamente la función <br /> snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot;<br /> tal que (snoc xs a) es la lista obtenida al añadir el elemento a al<br /> final de la lista xs. Por ejemplo, <br /> value &quot;snoc [2,5] (3::int)&quot; == [2,5,3]<br /> <br /> Nota: No usar @.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> primrec snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot; where<br /> &quot;snoc [] y = [y]&quot;<br /> | &quot;snoc (x#xs) y = x#(snoc xs y)&quot;<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 2. Demostrar el siguiente teorema <br /> snoc xs a = xs @ [a]<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del lema es&quot;<br /> lemma &quot;snoc xs a = xs @ [a]&quot;<br /> by (induct xs) auto<br /> <br /> -- &quot;La demostración estructurada del lema es&quot;<br /> lemma snoc_append: &quot;snoc xs a = xs @ [a]&quot;<br /> proof (induct xs)<br /> show &quot;snoc [] a = [] @ [a]&quot; by simp<br /> next <br /> fix b xs <br /> assume HI: &quot;snoc xs a = xs @ [a]&quot;<br /> thus &quot;snoc (b # xs) a = (b # xs) @ [a]&quot;<br /> proof -<br /> have &quot;snoc (b # xs) a = b#(snoc xs a)&quot; by simp<br /> also have &quot;... = b#(xs @ [a])&quot; using HI by simp<br /> also have &quot;... = (b#xs) @ [a]&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 3. Demostrar el siguiente teorema <br /> rev (x # xs) = snoc (rev xs) x&quot;<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del teorema es&quot;<br /> theorem rev_cons_auto: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> by (simp add:snoc_append)<br /> <br /> -- &quot;La demostración estructurada del teorema es&quot;<br /> theorem rev_cons: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> proof -<br /> have &quot;rev(x#xs)=(rev xs) @ [x]&quot; by simp <br /> also have &quot;... = snoc (rev xs) x&quot; by (simp add:snoc_append)<br /> finally show ?thesis . <br /> qed<br /> <br /> <br /> <br /> <br /> <br /> section {* Cuantificadores sobre listas *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 4. Definir la función <br /> todos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (todos p xs) se verifica si todos los elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> todos (λx. 1&lt;length x) [[2,1,4],[1,3]]<br /> ¬ todos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> <br /> Nota: La función todos es equivalente a la predefinida list_all. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec todos :: &quot;(&#039;a ⇒ bool) ⇒ (&#039;a list ⇒ bool)&quot; where<br /> &quot;todos P [] = True&quot;<br /> |&quot;todos P (x#xs) = (P x ∧ todos P xs)&quot;<br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 5. Definir la función <br /> algunos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (algunos p xs) se verifica si algunos elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> algunos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> ¬ algunos (λx. 1&lt;length x) [[],[3]]&quot;<br /> <br /> Nota: La función algunos es equivalente a la predefinida list_ex. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> primrec algunos :: &quot;(&#039;a =&gt; bool) =&gt; &#039;a list =&gt; bool&quot; where<br /> &quot;algunos P [] = False&quot;<br /> | &quot;algunos P (x#xs) = (P x ∨ algunosx P xs)&quot;<br /> <br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 6. Demostrar o refutar: <br /> todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 7. Demostrar o refutar: <br /> todos P (x @ y) = (todos P x ∧ todos P y)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 8. Demostrar o refutar: <br /> todos P (rev xs) = todos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 9. Demostrar o refutar:<br /> algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;Se busca un contraejemplo con nitpick&quot;<br /> lemma &quot;algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> El contraejemplo encontrado es<br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 10. Demostrar o refutar: <br /> algunos P (map f xs) = algunos (P ∘ f) xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P o f) xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P ∘ f) xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 11. Demostrar o refutar: <br /> algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma algunos_append: <br /> &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 12. Demostrar o refutar: <br /> algunos P (rev xs) = algunos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 13. Encontrar un término no trivial Z tal que sea cierta la <br /> siguiente ecuación:<br /> algunos (λx. P x ∨ Q x) xs = Z<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La ecuación se verifica eligiendo como Z el término <br /> algunos P xs ∨ algunos Q xs<br /> En efecto,<br /> *}<br /> <br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;De forma estructurada&quot;<br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 14. Demostrar o refutar:<br /> algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 15. Definir la funcion primitiva recursiva <br /> estaEn :: &#039;a ⇒ &#039;a list ⇒ bool<br /> tal que (estaEn x xs) se verifica si el elemento x está en la lista<br /> xs. Por ejemplo, <br /> estaEn (2::nat) [3,2,4] = True<br /> estaEn (1::nat) [3,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 16. Expresar la relación existente entre estaEn y algunos. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La relación es <br /> <br /> En efecto, <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 17. Definir la función primitiva recursiva <br /> sinDuplicados :: &#039;a list ⇒ bool<br /> tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br /> duplicados. Por ejemplo, <br /> sinDuplicados [1::nat,4,2] = True<br /> sinDuplicados [1::nat,4,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 18. Definir la función primitiva recursiva <br /> borraDuplicados :: &#039;a list ⇒ bool<br /> tal que (borraDuplicados xs) es la lista obtenida eliminando los<br /> elementos duplicados de la lista xs. Por ejemplo, <br /> borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br /> <br /> Nota: La función borraDuplicados es equivalente a la predefinida remdups. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 19. Demostrar o refutar:<br /> length (borraDuplicados xs) ≤ length xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma length_borraDuplicados: <br /> &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 20. Demostrar o refutar: <br /> estaEn a (borraDuplicados xs) = estaEn a xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma estaEn_borraDuplicados_auto: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma estaEn_borraDuplicados: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 21. Demostrar o refutar: <br /> sinDuplicados (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática&quot;<br /> lemma &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma sinDuplicados_borraDuplicados:<br /> &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 22. Demostrar o refutar:<br /> borraDuplicados (rev xs) = rev (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma &quot;borraDuplicados (rev xs) = rev (borraDuplicados xs)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_3&diff=120 2011-02-14T13:39:51Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 3ª relación de ejercicios *}<br /> <br /> theory Relacion_3<br /> imports Main <br /> begin<br /> <br /> section {* Cons inverso *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 1. Definir recursivamente la función <br /> snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot;<br /> tal que (snoc xs a) es la lista obtenida al añadir el elemento a al<br /> final de la lista xs. Por ejemplo, <br /> value &quot;snoc [2,5] (3::int)&quot; == [2,5,3]<br /> <br /> Nota: No usar @.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> primrec snoc :: &quot;&#039;a list ⇒ &#039;a ⇒ &#039;a list&quot; where<br /> &quot;snoc [] y = [y]&quot;<br /> | &quot;snoc (x#xs) y = x#(snoc xs y)&quot;<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 2. Demostrar el siguiente teorema <br /> snoc xs a = xs @ [a]<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del lema es&quot;<br /> lemma &quot;snoc xs a = xs @ [a]&quot;<br /> by (induct xs) auto<br /> <br /> -- &quot;La demostración estructurada del lema es&quot;<br /> lemma snoc_append: &quot;snoc xs a = xs @ [a]&quot;<br /> proof (induct xs)<br /> show &quot;snoc [] a = [] @ [a]&quot; by simp<br /> next <br /> fix b xs <br /> assume HI: &quot;snoc xs a = xs @ [a]&quot;<br /> thus &quot;snoc (b # xs) a = (b # xs) @ [a]&quot;<br /> proof -<br /> have &quot;snoc (b # xs) a = b#(snoc xs a)&quot; by simp<br /> also have &quot;... = b#(xs @ [a])&quot; using HI by simp<br /> also have &quot;... = (b#xs) @ [a]&quot; by simp<br /> finally show ?thesis .<br /> qed<br /> qed<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 3. Demostrar el siguiente teorema <br /> rev (x # xs) = snoc (rev xs) x&quot;<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática del teorema es&quot;<br /> theorem rev_cons_auto: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> by (simp add:snoc_append)<br /> <br /> -- &quot;La demostración estructurada del teorema es&quot;<br /> theorem rev_cons: &quot;rev (x # xs) = snoc (rev xs) x&quot;<br /> proof -<br /> have &quot;rev(x#xs)=(rev xs) @ [x]&quot; by simp <br /> also have &quot;... = snoc (rev xs) x&quot; by (simp add:snoc_append)<br /> finally show ?thesis . <br /> qed<br /> <br /> <br /> <br /> <br /> <br /> section {* Cuantificadores sobre listas *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 4. Definir la función <br /> todos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (todos p xs) se verifica si todos los elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> todos (λx. 1&lt;length x) [[2,1,4],[1,3]]<br /> ¬ todos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> <br /> Nota: La función todos es equivalente a la predefinida list_all. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> <br /> <br /> <br /> primrec todos :: &quot;(&#039;a ⇒ bool) ⇒ (&#039;a list ⇒ bool)&quot; where<br /> &quot;todos P [] = True&quot;<br /> |&quot;todos P (x#xs) = (P x ∧ todos P xs)&quot;<br /> <br /> <br /> <br /> <br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 5. Definir la función <br /> algunos :: (&#039;a ⇒ bool) ⇒ &#039;a list ⇒ bool<br /> tal que (algunos p xs) se verifica si algunos elementos de la lista <br /> xs cumplen la propiedad p. Por ejemplo, se verifica <br /> algunos (λx. 1&lt;length x) [[2,1,4],[3]]<br /> ¬ algunos (λx. 1&lt;length x) [[],[3]]&quot;<br /> <br /> Nota: La función algunos es equivalente a la predefinida list_ex. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 6. Demostrar o refutar: <br /> todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 7. Demostrar o refutar: <br /> todos P (x @ y) = (todos P x ∧ todos P y)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma todos_append [simp]: <br /> &quot;todos P (x @ y) = (todos P x ∧ todos P y)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 8. Demostrar o refutar: <br /> todos P (rev xs) = todos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;todos P (rev xs) = todos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 9. Demostrar o refutar:<br /> algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;Se busca un contraejemplo con nitpick&quot;<br /> lemma &quot;algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> El contraejemplo encontrado es<br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 10. Demostrar o refutar: <br /> algunos P (map f xs) = algunos (P ∘ f) xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P o f) xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (map f xs) = algunos (P ∘ f) xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 11. Demostrar o refutar: <br /> algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma algunos_append: <br /> &quot;algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 12. Demostrar o refutar: <br /> algunos P (rev xs) = algunos P xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P (rev xs) = algunos P xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 13. Encontrar un término no trivial Z tal que sea cierta la <br /> siguiente ecuación:<br /> algunos (λx. P x ∨ Q x) xs = Z<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La ecuación se verifica eligiendo como Z el término <br /> algunos P xs ∨ algunos Q xs<br /> En efecto,<br /> *}<br /> <br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> -- &quot;De forma estructurada&quot;<br /> lemma &quot;algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 14. Demostrar o refutar:<br /> algunos P xs = (¬ todos (λx. (¬ P x)) xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma &quot;algunos P xs = (¬ todos (λx. (¬ P x)) xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 15. Definir la funcion primitiva recursiva <br /> estaEn :: &#039;a ⇒ &#039;a list ⇒ bool<br /> tal que (estaEn x xs) se verifica si el elemento x está en la lista<br /> xs. Por ejemplo, <br /> estaEn (2::nat) [3,2,4] = True<br /> estaEn (1::nat) [3,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 16. Expresar la relación existente entre estaEn y algunos. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> Solución: La relación es <br /> <br /> En efecto, <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 17. Definir la función primitiva recursiva <br /> sinDuplicados :: &#039;a list ⇒ bool<br /> tal que (sinDuplicados xs) se verifica si la lista xs no contiene<br /> duplicados. Por ejemplo, <br /> sinDuplicados [1::nat,4,2] = True<br /> sinDuplicados [1::nat,4,2,4] = False<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {* <br /> --------------------------------------------------------------------- <br /> Ejercicio 18. Definir la función primitiva recursiva <br /> borraDuplicados :: &#039;a list ⇒ bool<br /> tal que (borraDuplicados xs) es la lista obtenida eliminando los<br /> elementos duplicados de la lista xs. Por ejemplo, <br /> borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]<br /> <br /> Nota: La función borraDuplicados es equivalente a la predefinida remdups. <br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 19. Demostrar o refutar:<br /> length (borraDuplicados xs) ≤ length xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma length_borraDuplicados: <br /> &quot;length (borraDuplicados xs) ≤ length xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 20. Demostrar o refutar: <br /> estaEn a (borraDuplicados xs) = estaEn a xs<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática es&quot;<br /> lemma estaEn_borraDuplicados_auto: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma estaEn_borraDuplicados: <br /> &quot;estaEn a (borraDuplicados xs) = estaEn a xs&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 21. Demostrar o refutar: <br /> sinDuplicados (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> -- &quot;La demostración automática&quot;<br /> lemma &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> -- &quot;La demostración estructurada es&quot;<br /> lemma sinDuplicados_borraDuplicados:<br /> &quot;sinDuplicados (borraDuplicados xs)&quot;<br /> oops<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Ejercicio 22. Demostrar o refutar:<br /> borraDuplicados (rev xs) = rev (borraDuplicados xs)<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma &quot;borraDuplicados (rev xs) = rev (borraDuplicados xs)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_2&diff=93 2011-02-07T00:26:13Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 2ª relación de ejercicios *}<br /> <br /> theory Relacion_2<br /> imports Main <br /> begin<br /> <br /> text {*<br /> Demostrar o refutar los siguientes lemas usando sólo las reglas<br /> básicas de deducción natural de la lógica proposicional y de la lógica<br /> de predicados (allI, allE, exI y exE). <br /> *}<br /> <br /> lemma &quot;(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)&quot;<br /> proof (rule impI)<br /> assume 1: &quot;∃x. ∀y. P x y&quot;<br /> show &quot;∀y. ∃x. P x y&quot;<br /> proof (rule allI)<br /> from 1 obtain &quot;x&quot; where 2: &quot;∀y. P x y&quot; by (rule exE)<br /> fix y<br /> from 2 have 3: &quot;P x y&quot; by (rule allE)<br /> from 3 show 4: &quot;∃x. P x y&quot; by (rule exI)<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(∀x. P x ⟶ Q) = ((∃x. P x) ⟶ Q)&quot;<br /> proof<br /> { assume 1: &quot;∀x. P x ⟶ Q&quot;<br /> show &quot;(∃x. P x) ⟶ Q&quot;<br /> proof<br /> assume &quot;∃x. P x&quot;<br /> then obtain a where &quot;P a&quot; by (rule exE)<br /> have &quot;P a ⟶ Q&quot; using 1 by (rule allE)<br /> thus &quot;Q&quot; using `P a` by (rule mp)<br /> qed }<br /> next<br /> { assume 2: &quot;(∃x. P x) ⟶ Q&quot;<br /> show &quot;∀x. P x ⟶ Q&quot;<br /> proof<br /> fix a<br /> show &quot;P a ⟶ Q&quot;<br /> proof<br /> assume &quot;P a&quot;<br /> hence 3: &quot;∃x. P x&quot; by (rule exI)<br /> from 2 and 3 show &quot;Q&quot; by (rule mp)<br /> qed<br /> qed }<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;((∀ x. P x) ∧ (∀ x. Q x)) = (∀ x. (P x ∧ Q x))&quot;<br /> proof<br /> { assume 1: &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> show &quot;(∀ x. (P x ∧ Q x))&quot;<br /> proof (rule allI)<br /> from 1 have 2: &quot;(∀ x. P x)&quot; by (rule conjE)<br /> from 1 have 3: &quot;(∀ x. Q x)&quot; by (rule conjE)<br /> fix a<br /> from 2 have 4: &quot;P a&quot; by (rule allE)<br /> from 3 have 5: &quot;Q a&quot; by (rule allE)<br /> from 4 and 5 show 6: &quot;(P a) ∧ (Q a)&quot; by (rule conjI)<br /> qed }<br /> next<br /> { assume 7: &quot;∀ x. (P x ∧ Q x)&quot;<br /> show &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> proof<br /> { show &quot;(∀ x. P x)&quot;<br /> proof (rule allI)<br /> fix a<br /> from 7 have 8: &quot;P a ∧ Q a&quot; by (rule allE)<br /> from 8 show &quot;P a&quot; by (rule conjE)<br /> qed }<br /> <br /> { show &quot;(∀ x. Q x)&quot;<br /> proof (rule allI)<br /> fix a<br /> from 7 have 9: &quot;P a ∧ Q a&quot; by (rule allE)<br /> from 9 show &quot;Q a&quot; by (rule conjE)<br /> qed }<br /> qed }<br /> qed<br /> <br /> <br /> <br /> <br /> <br /> lemma &quot;((∀ x. P x) ∨ (∀ x. Q x)) = (∀ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> <br /> <br /> <br /> <br /> lemma &quot;((∃ x. P x) ∨ (∃ x. Q x)) = (∃ x. (P x ∨ Q x))&quot;<br /> proof<br /> {assume 1:&quot;((∃ x. P x) ∨ (∃ x. Q x))&quot;<br /> show &quot;(∃ x. (P x ∨ Q x))&quot;<br /> proof -<br /> note 1<br /> moreover<br /> { assume 2: &quot;(∃ x. P x)&quot;<br /> obtain a where 3: &quot;P a&quot; using 2 by (rule exE)<br /> from 3 have 4: &quot;P a ∨ Q a&quot; .. <br /> from 4 have &quot;(∃ x. (P x ∨ Q x))&quot; by (rule exI) }<br /> moreover<br /> { assume 2: &quot;(∃ x. Q x)&quot;<br /> obtain a where 3: &quot;Q a&quot; using 2 by (rule exE)<br /> from 3 have 4: &quot;P a ∨ Q a&quot; .. <br /> from 4 have &quot;(∃ x. (P x ∨ Q x))&quot; by (rule exI) }<br /> ultimately show &quot;(∃ x. (P x ∨ Q x))&quot; by (rule disjE) <br /> qed}<br /> next<br /> {assume 1:&quot;(∃ x. (P x ∨ Q x))&quot;<br /> obtain a where 2: &quot;P a ∨ Q a&quot; using 1 by (rule exE)<br /> show &quot;((∃ x. P x) ∨ (∃ x. Q x))&quot; using 2<br /> proof <br /> { assume 3: &quot;P a&quot;<br /> from 3 have 4: &quot;(∃ x. P x)&quot; by (rule exI)<br /> from 4 show &quot;((∃ x. P x) ∨ (∃ x. Q x))&quot; .. }<br /> { assume 3: &quot;Q a&quot;<br /> from 3 have 4: &quot;(∃ x. Q x)&quot; by (rule exI)<br /> from 4 show &quot;((∃ x. P x) ∨ (∃ x. Q x))&quot; .. }<br /> qed }<br /> qed<br /> <br /> <br /> <br /> <br /> <br /> lemma &quot;(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)&quot;<br /> oops<br /> <br /> lemma &quot;(¬ (∀ x. P x)) = (∃ x. ¬ P x)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_2&diff=92 2011-02-06T23:22:38Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 2ª relación de ejercicios *}<br /> <br /> theory Relacion_2<br /> imports Main <br /> begin<br /> <br /> text {*<br /> Demostrar o refutar los siguientes lemas usando sólo las reglas<br /> básicas de deducción natural de la lógica proposicional y de la lógica<br /> de predicados (allI, allE, exI y exE). <br /> *}<br /> <br /> lemma &quot;(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)&quot;<br /> proof (rule impI)<br /> assume 1: &quot;∃x. ∀y. P x y&quot;<br /> show &quot;∀y. ∃x. P x y&quot;<br /> proof (rule allI)<br /> from 1 obtain &quot;x&quot; where 2: &quot;∀y. P x y&quot; by (rule exE)<br /> fix y<br /> from 2 have 3: &quot;P x y&quot; by (rule allE)<br /> from 3 show 4: &quot;∃x. P x y&quot; by (rule exI)<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(∀x. P x ⟶ Q) = ((∃x. P x) ⟶ Q)&quot;<br /> proof<br /> { assume 1: &quot;∀x. P x ⟶ Q&quot;<br /> show &quot;(∃x. P x) ⟶ Q&quot;<br /> proof<br /> assume &quot;∃x. P x&quot;<br /> then obtain a where &quot;P a&quot; by (rule exE)<br /> have &quot;P a ⟶ Q&quot; using 1 by (rule allE)<br /> thus &quot;Q&quot; using `P a` by (rule mp)<br /> qed }<br /> next<br /> { assume 2: &quot;(∃x. P x) ⟶ Q&quot;<br /> show &quot;∀x. P x ⟶ Q&quot;<br /> proof<br /> fix a<br /> show &quot;P a ⟶ Q&quot;<br /> proof<br /> assume &quot;P a&quot;<br /> hence 3: &quot;∃x. P x&quot; by (rule exI)<br /> from 2 and 3 show &quot;Q&quot; by (rule mp)<br /> qed<br /> qed }<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;((∀ x. P x) ∧ (∀ x. Q x)) = (∀ x. (P x ∧ Q x))&quot;<br /> proof<br /> { assume 1: &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> show &quot;(∀ x. (P x ∧ Q x))&quot;<br /> proof (rule allI)<br /> from 1 have 2: &quot;(∀ x. P x)&quot; by (rule conjE)<br /> from 1 have 3: &quot;(∀ x. Q x)&quot; by (rule conjE)<br /> fix a<br /> from 2 have 4: &quot;P a&quot; by (rule allE)<br /> from 3 have 5: &quot;Q a&quot; by (rule allE)<br /> from 4 and 5 show 6: &quot;(P a) ∧ (Q a)&quot; by (rule conjI)<br /> qed }<br /> next<br /> { assume 7: &quot;∀ x. (P x ∧ Q x)&quot;<br /> show &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> proof<br /> { show &quot;(∀ x. P x)&quot;<br /> proof (rule allI)<br /> fix a<br /> from 7 have 8: &quot;P a ∧ Q a&quot; by (rule allE)<br /> from 8 show &quot;P a&quot; by (rule conjE)<br /> qed }<br /> <br /> { show &quot;(∀ x. Q x)&quot;<br /> proof (rule allI)<br /> fix a<br /> from 7 have 9: &quot;P a ∧ Q a&quot; by (rule allE)<br /> from 9 show &quot;Q a&quot; by (rule conjE)<br /> qed }<br /> qed }<br /> qed<br /> <br /> <br /> <br /> <br /> <br /> lemma &quot;((∀ x. P x) ∨ (∀ x. Q x)) = (∀ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;((∃ x. P x) ∨ (∃ x. Q x)) = (∃ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)&quot;<br /> oops<br /> <br /> lemma &quot;(¬ (∀ x. P x)) = (∃ x. ¬ P x)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_2&diff=91 2011-02-06T23:21:12Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 2ª relación de ejercicios *}<br /> <br /> theory Relacion_2<br /> imports Main <br /> begin<br /> <br /> text {*<br /> Demostrar o refutar los siguientes lemas usando sólo las reglas<br /> básicas de deducción natural de la lógica proposicional y de la lógica<br /> de predicados (allI, allE, exI y exE). <br /> *}<br /> <br /> lemma &quot;(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)&quot;<br /> proof (rule impI)<br /> assume 1: &quot;∃x. ∀y. P x y&quot;<br /> show &quot;∀y. ∃x. P x y&quot;<br /> proof (rule allI)<br /> from 1 obtain &quot;x&quot; where 2: &quot;∀y. P x y&quot; by (rule exE)<br /> fix y<br /> from 2 have 3: &quot;P x y&quot; by (rule allE)<br /> from 3 show 4: &quot;∃x. P x y&quot; by (rule exI)<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(∀x. P x ⟶ Q) = ((∃x. P x) ⟶ Q)&quot;<br /> proof<br /> { assume 1: &quot;∀x. P x ⟶ Q&quot;<br /> show &quot;(∃x. P x) ⟶ Q&quot;<br /> proof<br /> assume &quot;∃x. P x&quot;<br /> then obtain a where &quot;P a&quot; by (rule exE)<br /> have &quot;P a ⟶ Q&quot; using 1 by (rule allE)<br /> thus &quot;Q&quot; using `P a` by (rule mp)<br /> qed }<br /> next<br /> { assume 2: &quot;(∃x. P x) ⟶ Q&quot;<br /> show &quot;∀x. P x ⟶ Q&quot;<br /> proof<br /> fix a<br /> show &quot;P a ⟶ Q&quot;<br /> proof<br /> assume &quot;P a&quot;<br /> hence 3: &quot;∃x. P x&quot; by (rule exI)<br /> from 2 and 3 show &quot;Q&quot; by (rule mp)<br /> qed<br /> qed }<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;((∀ x. P x) ∧ (∀ x. Q x)) = (∀ x. (P x ∧ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;((∀ x. P x) ∨ (∀ x. Q x)) = (∀ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;((∃ x. P x) ∨ (∃ x. Q x)) = (∃ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)&quot;<br /> oops<br /> <br /> lemma &quot;(¬ (∀ x. P x)) = (∃ x. ¬ P x)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_1&diff=90 2011-02-06T23:19:47Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 1ª relación de ejercicios *}<br /> <br /> theory Relacion_1<br /> imports Main <br /> begin<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Demostrar los siguientes lemas usando sólo las reglas básicas de<br /> deducción natural de la lógica proposicional.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma I: &quot;A ⟶ A&quot;<br /> by (rule imp_refl)<br /> <br /> -- &quot;Se puede hacer automáticamente&quot;<br /> lemma I_d2: &quot;A ⟶ A&quot;<br /> by auto<br /> <br /> -- &quot;Otra forma (más explicito)<br /> lemma I_d3: &quot;A ⟶ A&quot;<br /> proof (rule impI)<br /> assume 1: &quot;A&quot;<br /> qed<br /> <br /> <br /> <br /> <br /> lemma <br /> assumes 1:&quot;A ∧ B&quot; <br /> shows &quot;B ∧ A&quot; <br /> proof -<br /> from 1 have 2:&quot;A&quot; by (rule conjunct1)<br /> from 1 have 3:&quot;B&quot; by (rule conjunct2)<br /> from 3 2 show &quot;B ∧ A&quot; by (rule conjI)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;A ∧ B ⟶ A ∨ B&quot;<br /> proof<br /> assume &quot;A ∧ B&quot;<br /> hence &quot;A&quot; ..<br /> thus &quot;A ∨ B&quot; ..<br /> qed<br /> <br /> <br /> <br /> <br /> lemma<br /> assumes 1:&quot;(A ∨ B) ∨ C&quot;<br /> shows &quot;A ∨ ( B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover<br /> { assume 2: &quot;C&quot;<br /> from 2 have 3:&quot;B ∨ C&quot; by (rule disjI2) <br /> from 3 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)}<br /> moreover<br /> { assume 4: &quot;A ∨ B&quot;<br /> from 4 have &quot;(A ∨ B) ∨ C&quot; by (rule disjI1) }<br /> ultimately show &quot;A ∨ (B ∨ C) &quot; by (simp)<br /> qed<br /> <br /> -- Otra forma (más explícita)<br /> lemma version2: &quot;(A ∨ B) ∨ C ⟶ A ∨ (B ∨ C)&quot;<br /> proof<br /> assume 1: &quot;(A ∨ B) ∨ C&quot;<br /> show &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover{<br /> assume 2: &quot;A ∨ B&quot;<br /> have &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 2<br /> moreover{<br /> assume 3: &quot;A&quot;<br /> from 3 have 4:&quot;A ∨ (B ∨ C)&quot; by (rule disjI1)<br /> }<br /> moreover{<br /> assume 5:&quot;B&quot;<br /> from 5 have 6: &quot;B ∨ C&quot; by (rule disjI1)<br /> from 6 have 7: &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> }<br /> moreover{<br /> assume 8: &quot;C&quot;<br /> from 8 have 9: &quot;B ∨ C&quot; by (rule disjI2)<br /> from 9 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma Aux:<br /> assumes &quot;A&quot;<br /> shows &quot;B ⟶ A&quot;<br /> using assms by simp<br /> <br /> lemma K: &quot;A ⟶ (B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;B ⟶ A&quot; by (rule Aux)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(A ⟶ (B ⟶C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;A ⟶ (B ⟶ C)&quot;<br /> show &quot;(A ⟶ B) ⟶ (A ⟶C)&quot;<br /> proof<br /> assume 2:&quot;(A ⟶ B)&quot;<br /> show &quot;A ⟶C&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 2 3 have 4:&quot;B&quot; by (rule mp)<br /> from 1 3 have 5:&quot;B ⟶ C&quot; by (rule mp)<br /> from 5 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;(A ⟶ B)&quot;<br /> show &quot;((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof <br /> assume 2:&quot;(B ⟶ C)&quot;<br /> show &quot;(A ⟶ C)&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 1 3 have 4:&quot;B&quot; by (rule mp)<br /> from 2 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;¬¬A ⟶ A&quot;<br /> proof<br /> assume 1:&quot;¬¬A&quot;<br /> from 1 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> <br /> <br /> lemma &quot;A ⟶ ¬¬A&quot;<br /> proof<br /> assume 1:&quot;A&quot;<br /> from 1 show &quot;¬¬A&quot; by (rule contrapos_pn)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma MT:<br /> assumes 1:&quot;F ⟶ G&quot; and 2:&quot;¬G&quot;<br /> shows &quot;¬F&quot;<br /> proof (rule notI)<br /> assume 3:&quot;F&quot;<br /> from 1 and 3 have 4:&quot;G&quot; by (rule mp)<br /> from 2 and 4 show False by (rule notE)<br /> qed<br /> <br /> lemma &quot;(¬A ⟶ B) ⟶ (¬B ⟶ A)&quot;<br /> proof<br /> assume 1:&quot;(¬A ⟶ B)&quot;<br /> show &quot;(¬B ⟶ A)&quot;<br /> proof<br /> assume 2:&quot;¬B&quot;<br /> from 1 2 have 3:&quot;¬¬A&quot; by (rule MT)<br /> from 3 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;((A ⟶ B) ⟶ A) ⟶ A&quot;<br /> proof<br /> assume 1: &quot;(A ⟶ B) ⟶ A&quot;<br /> have 8: &quot;¬¬A&quot;<br /> proof (rule notI)<br /> assume 2: &quot;¬A&quot;<br /> have 3: &quot;A ⟶ B&quot;<br /> proof (rule impI)<br /> assume 4: &quot;A&quot;<br /> from 2 and 4 show &quot;B&quot; by (rule notE)<br /> qed<br /> from 1 and 3 have 5: &quot;A&quot; by (rule mp)<br /> from 2 and 5 show False by (rule notE)<br /> qed<br /> from 8 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;A ∨ ¬A&quot;<br /> proof cases<br /> assume &quot;A&quot; thus ?thesis ..<br /> next<br /> assume &quot;¬A&quot; thus ?thesis ..<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(¬(A ∧ B)) = (¬A ∨ ¬B)&quot;<br /> proof<br /> {<br /> assume 1: &quot;¬(A ∧ B)&quot;<br /> have 2: &quot;¬A ∨ A&quot; by (rule excluded_middle)<br /> thus &quot;¬A ∨ ¬B&quot;<br /> proof (rule disjE)<br /> { assume 3: &quot;¬A&quot;<br /> thus &quot;¬A ∨ ¬B&quot; by (rule disjI1) }<br /> next<br /> { assume 4: &quot;A&quot;<br /> have 5: &quot;¬B ∨ B&quot; by (rule excluded_middle)<br /> thus &quot;¬A ∨ ¬B&quot;<br /> proof (rule disjE)<br /> {assume 6: &quot;¬B&quot;<br /> thus &quot;¬A ∨ ¬B&quot; by (rule disjI2) }<br /> next<br /> {assume 7: &quot;B&quot;<br /> from 4 and 7 have 8: &quot;A ∧ B&quot; by (rule conjI)<br /> from 1 and 8 show &quot;¬A ∨ ¬B&quot; by (rule notE) }<br /> qed<br /> } <br /> qed<br /> }<br /> next<br /> { assume 1: &quot;¬A ∨ ¬B&quot;<br /> show &quot;¬(A ∧ B)&quot;<br /> proof (rule notI)<br /> assume 2: &quot;A ∧ B&quot;<br /> from 1 have 3:&quot;¬A ∨ ¬B&quot; .<br /> moreover<br /> { assume 4: &quot;¬A&quot;<br /> from 2 have 5:&quot;A&quot; by (rule conjunct1)<br /> from 4 and 5 have False by (rule notE) }<br /> moreover<br /> { assume 6: &quot;¬B&quot;<br /> from 2 have 7: &quot;B&quot; by (rule conjunct2)<br /> from 6 and 7 have False by (rule notE) }<br /> ultimately show False by (rule disjE) <br /> qed<br /> <br /> }<br /> qed<br /> <br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_1&diff=89 2011-02-06T23:19:05Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 1ª relación de ejercicios *}<br /> <br /> theory Relacion_1<br /> imports Main <br /> begin<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Demostrar los siguientes lemas usando sólo las reglas básicas de<br /> deducción natural de la lógica proposicional.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma I: &quot;A ⟶ A&quot;<br /> by (rule imp_refl)<br /> <br /> -- &quot;Se puede hacer automáticamente&quot;<br /> lemma I_d2: &quot;A ⟶ A&quot;<br /> by auto<br /> <br /> -- &quot;Otra forma (más explicito)<br /> lemma I: &quot;A ⟶ A&quot;<br /> proof (rule impI)<br /> assume 1: &quot;A&quot;<br /> qed<br /> <br /> <br /> <br /> <br /> lemma <br /> assumes 1:&quot;A ∧ B&quot; <br /> shows &quot;B ∧ A&quot; <br /> proof -<br /> from 1 have 2:&quot;A&quot; by (rule conjunct1)<br /> from 1 have 3:&quot;B&quot; by (rule conjunct2)<br /> from 3 2 show &quot;B ∧ A&quot; by (rule conjI)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;A ∧ B ⟶ A ∨ B&quot;<br /> proof<br /> assume &quot;A ∧ B&quot;<br /> hence &quot;A&quot; ..<br /> thus &quot;A ∨ B&quot; ..<br /> qed<br /> <br /> <br /> <br /> <br /> lemma<br /> assumes 1:&quot;(A ∨ B) ∨ C&quot;<br /> shows &quot;A ∨ ( B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover<br /> { assume 2: &quot;C&quot;<br /> from 2 have 3:&quot;B ∨ C&quot; by (rule disjI2) <br /> from 3 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)}<br /> moreover<br /> { assume 4: &quot;A ∨ B&quot;<br /> from 4 have &quot;(A ∨ B) ∨ C&quot; by (rule disjI1) }<br /> ultimately show &quot;A ∨ (B ∨ C) &quot; by (simp)<br /> qed<br /> <br /> -- Otra forma (más explícita)<br /> lemma &quot;(A ∨ B) ∨ C ⟶ A ∨ (B ∨ C)&quot;<br /> proof<br /> assume 1: &quot;(A ∨ B) ∨ C&quot;<br /> show &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover{<br /> assume 2: &quot;A ∨ B&quot;<br /> have &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 2<br /> moreover{<br /> assume 3: &quot;A&quot;<br /> from 3 have 4:&quot;A ∨ (B ∨ C)&quot; by (rule disjI1)<br /> }<br /> moreover{<br /> assume 5:&quot;B&quot;<br /> from 5 have 6: &quot;B ∨ C&quot; by (rule disjI1)<br /> from 6 have 7: &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> }<br /> moreover{<br /> assume 8: &quot;C&quot;<br /> from 8 have 9: &quot;B ∨ C&quot; by (rule disjI2)<br /> from 9 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma Aux:<br /> assumes &quot;A&quot;<br /> shows &quot;B ⟶ A&quot;<br /> using assms by simp<br /> <br /> lemma K: &quot;A ⟶ (B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;B ⟶ A&quot; by (rule Aux)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(A ⟶ (B ⟶C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;A ⟶ (B ⟶ C)&quot;<br /> show &quot;(A ⟶ B) ⟶ (A ⟶C)&quot;<br /> proof<br /> assume 2:&quot;(A ⟶ B)&quot;<br /> show &quot;A ⟶C&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 2 3 have 4:&quot;B&quot; by (rule mp)<br /> from 1 3 have 5:&quot;B ⟶ C&quot; by (rule mp)<br /> from 5 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;(A ⟶ B)&quot;<br /> show &quot;((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof <br /> assume 2:&quot;(B ⟶ C)&quot;<br /> show &quot;(A ⟶ C)&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 1 3 have 4:&quot;B&quot; by (rule mp)<br /> from 2 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;¬¬A ⟶ A&quot;<br /> proof<br /> assume 1:&quot;¬¬A&quot;<br /> from 1 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> <br /> <br /> lemma &quot;A ⟶ ¬¬A&quot;<br /> proof<br /> assume 1:&quot;A&quot;<br /> from 1 show &quot;¬¬A&quot; by (rule contrapos_pn)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma MT:<br /> assumes 1:&quot;F ⟶ G&quot; and 2:&quot;¬G&quot;<br /> shows &quot;¬F&quot;<br /> proof (rule notI)<br /> assume 3:&quot;F&quot;<br /> from 1 and 3 have 4:&quot;G&quot; by (rule mp)<br /> from 2 and 4 show False by (rule notE)<br /> qed<br /> <br /> lemma &quot;(¬A ⟶ B) ⟶ (¬B ⟶ A)&quot;<br /> proof<br /> assume 1:&quot;(¬A ⟶ B)&quot;<br /> show &quot;(¬B ⟶ A)&quot;<br /> proof<br /> assume 2:&quot;¬B&quot;<br /> from 1 2 have 3:&quot;¬¬A&quot; by (rule MT)<br /> from 3 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;((A ⟶ B) ⟶ A) ⟶ A&quot;<br /> proof<br /> assume 1: &quot;(A ⟶ B) ⟶ A&quot;<br /> have 8: &quot;¬¬A&quot;<br /> proof (rule notI)<br /> assume 2: &quot;¬A&quot;<br /> have 3: &quot;A ⟶ B&quot;<br /> proof (rule impI)<br /> assume 4: &quot;A&quot;<br /> from 2 and 4 show &quot;B&quot; by (rule notE)<br /> qed<br /> from 1 and 3 have 5: &quot;A&quot; by (rule mp)<br /> from 2 and 5 show False by (rule notE)<br /> qed<br /> from 8 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;A ∨ ¬A&quot;<br /> proof cases<br /> assume &quot;A&quot; thus ?thesis ..<br /> next<br /> assume &quot;¬A&quot; thus ?thesis ..<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(¬(A ∧ B)) = (¬A ∨ ¬B)&quot;<br /> proof<br /> {<br /> assume 1: &quot;¬(A ∧ B)&quot;<br /> have 2: &quot;¬A ∨ A&quot; by (rule excluded_middle)<br /> thus &quot;¬A ∨ ¬B&quot;<br /> proof (rule disjE)<br /> { assume 3: &quot;¬A&quot;<br /> thus &quot;¬A ∨ ¬B&quot; by (rule disjI1) }<br /> next<br /> { assume 4: &quot;A&quot;<br /> have 5: &quot;¬B ∨ B&quot; by (rule excluded_middle)<br /> thus &quot;¬A ∨ ¬B&quot;<br /> proof (rule disjE)<br /> {assume 6: &quot;¬B&quot;<br /> thus &quot;¬A ∨ ¬B&quot; by (rule disjI2) }<br /> next<br /> {assume 7: &quot;B&quot;<br /> from 4 and 7 have 8: &quot;A ∧ B&quot; by (rule conjI)<br /> from 1 and 8 show &quot;¬A ∨ ¬B&quot; by (rule notE) }<br /> qed<br /> } <br /> qed<br /> }<br /> next<br /> { assume 1: &quot;¬A ∨ ¬B&quot;<br /> show &quot;¬(A ∧ B)&quot;<br /> proof (rule notI)<br /> assume 2: &quot;A ∧ B&quot;<br /> from 1 have 3:&quot;¬A ∨ ¬B&quot; .<br /> moreover<br /> { assume 4: &quot;¬A&quot;<br /> from 2 have 5:&quot;A&quot; by (rule conjunct1)<br /> from 4 and 5 have False by (rule notE) }<br /> moreover<br /> { assume 6: &quot;¬B&quot;<br /> from 2 have 7: &quot;B&quot; by (rule conjunct2)<br /> from 6 and 7 have False by (rule notE) }<br /> ultimately show False by (rule disjE) <br /> qed<br /> <br /> }<br /> qed<br /> <br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_1&diff=88 2011-02-06T23:18:30Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 1ª relación de ejercicios *}<br /> <br /> theory Relacion_1<br /> imports Main <br /> begin<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Demostrar los siguientes lemas usando sólo las reglas básicas de<br /> deducción natural de la lógica proposicional.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma I: &quot;A ⟶ A&quot;<br /> by (rule imp_refl)<br /> <br /> -- &quot;Se puede hacer automáticamente&quot;<br /> lemma I_d2: &quot;A ⟶ A&quot;<br /> by auto<br /> <br /> -- &quot;Otra forma (más explicito)<br /> lemma I: &quot;A ⟶ A&quot;<br /> proof (rule impI)<br /> assume 1: &quot;A&quot;<br /> qed<br /> <br /> <br /> <br /> <br /> lemma <br /> assumes 1:&quot;A ∧ B&quot; <br /> shows &quot;B ∧ A&quot; <br /> proof -<br /> from 1 have 2:&quot;A&quot; by (rule conjunct1)<br /> from 1 have 3:&quot;B&quot; by (rule conjunct2)<br /> from 3 2 show &quot;B ∧ A&quot; by (rule conjI)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;A ∧ B ⟶ A ∨ B&quot;<br /> proof<br /> assume &quot;A ∧ B&quot;<br /> hence &quot;A&quot; ..<br /> thus &quot;A ∨ B&quot; ..<br /> qed<br /> <br /> <br /> <br /> <br /> lemma<br /> assumes 1:&quot;(A ∨ B) ∨ C&quot;<br /> shows &quot;A ∨ ( B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover<br /> { assume 2: &quot;C&quot;<br /> from 2 have 3:&quot;B ∨ C&quot; by (rule disjI2) <br /> from 3 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)}<br /> moreover<br /> { assume 4: &quot;A ∨ B&quot;<br /> from 4 have &quot;(A ∨ B) ∨ C&quot; by (rule disjI1) }<br /> ultimately show &quot;A ∨ (B ∨ C) &quot; by (simp)<br /> qed<br /> <br /> -- Otra forma (más explícita)<br /> lemma &quot;(A ∨ B) ∨ C ⟶ A ∨ (B ∨ C)&quot;<br /> proof<br /> assume 1: &quot;(A ∨ B) ∨ C&quot;<br /> show &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover{<br /> assume 2: &quot;A ∨ B&quot;<br /> have &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 2<br /> moreover{<br /> assume 3: &quot;A&quot;<br /> from 3 have 4:&quot;A ∨ (B ∨ C)&quot; by (rule disjI1)<br /> }<br /> moreover{<br /> assume 5:&quot;B&quot;<br /> from 5 have 6: &quot;B ∨ C&quot; by (rule disjI1)<br /> from 6 have 7: &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> }<br /> moreover{<br /> assume 8: &quot;C&quot;<br /> from 8 have 9: &quot;B ∨ C&quot; by (rule disjI2)<br /> from 9 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma Aux:<br /> assumes &quot;A&quot;<br /> shows &quot;B ⟶ A&quot;<br /> using assms by simp<br /> <br /> lemma K: &quot;A ⟶ (B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;B ⟶ A&quot; by (rule Aux)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(A ⟶ (B ⟶C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;A ⟶ (B ⟶ C)&quot;<br /> show &quot;(A ⟶ B) ⟶ (A ⟶C)&quot;<br /> proof<br /> assume 2:&quot;(A ⟶ B)&quot;<br /> show &quot;A ⟶C&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 2 3 have 4:&quot;B&quot; by (rule mp)<br /> from 1 3 have 5:&quot;B ⟶ C&quot; by (rule mp)<br /> from 5 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;(A ⟶ B)&quot;<br /> show &quot;((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof <br /> assume 2:&quot;(B ⟶ C)&quot;<br /> show &quot;(A ⟶ C)&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 1 3 have 4:&quot;B&quot; by (rule mp)<br /> from 2 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;¬¬A ⟶ A&quot;<br /> proof<br /> assume 1:&quot;¬¬A&quot;<br /> from 1 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> <br /> <br /> lemma &quot;A ⟶ ¬¬A&quot;<br /> proof<br /> assume 1:&quot;A&quot;<br /> from 1 show &quot;¬¬A&quot; by (rule contrapos_pn)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma MT:<br /> assumes 1:&quot;F ⟶ G&quot; and 2:&quot;¬G&quot;<br /> shows &quot;¬F&quot;<br /> proof (rule notI)<br /> assume 3:&quot;F&quot;<br /> from 1 and 3 have 4:&quot;G&quot; by (rule mp)<br /> from 2 and 4 show False by (rule notE)<br /> qed<br /> <br /> lemma &quot;(¬A ⟶ B) ⟶ (¬B ⟶ A)&quot;<br /> proof<br /> assume 1:&quot;(¬A ⟶ B)&quot;<br /> show &quot;(¬B ⟶ A)&quot;<br /> proof<br /> assume 2:&quot;¬B&quot;<br /> from 1 2 have 3:&quot;¬¬A&quot; by (rule MT)<br /> from 3 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;((A ⟶ B) ⟶ A) ⟶ A&quot;<br /> proof<br /> assume 1: &quot;(A ⟶ B) ⟶ A&quot;<br /> have 8: &quot;¬¬A&quot;<br /> proof (rule notI)<br /> assume 2: &quot;¬A&quot;<br /> have 3: &quot;A ⟶ B&quot;<br /> proof (rule impI)<br /> assume 4: &quot;A&quot;<br /> from 2 and 4 show &quot;B&quot; by (rule notE)<br /> qed<br /> from 1 and 3 have 5: &quot;A&quot; by (rule mp)<br /> from 2 and 5 show False by (rule notE)<br /> qed<br /> from 8 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;A ∨ ¬A&quot;<br /> proof cases<br /> assume &quot;A&quot; thus ?thesis ..<br /> next<br /> assume &quot;¬A&quot; thus ?thesis ..<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(¬(A ∧ B)) = (¬A ∨ ¬B)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_1&diff=87 2011-02-06T23:18:01Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 1ª relación de ejercicios *}<br /> <br /> theory Relacion_1<br /> imports Main <br /> begin<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Demostrar los siguientes lemas usando sólo las reglas básicas de<br /> deducción natural de la lógica proposicional.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma I: &quot;A ⟶ A&quot;<br /> by (rule imp_refl)<br /> <br /> -- &quot;Se puede hacer automáticamente&quot;<br /> lemma I_d2: &quot;A ⟶ A&quot;<br /> by auto<br /> <br /> -- &quot;Otra forma (más explicito)<br /> lemma I: &quot;A ⟶ A&quot;<br /> proof (rule impI)<br /> assume 1: &quot;A&quot;<br /> qed<br /> <br /> <br /> <br /> <br /> lemma <br /> assumes 1:&quot;A ∧ B&quot; <br /> shows &quot;B ∧ A&quot; <br /> proof -<br /> from 1 have 2:&quot;A&quot; by (rule conjunct1)<br /> from 1 have 3:&quot;B&quot; by (rule conjunct2)<br /> from 3 2 show &quot;B ∧ A&quot; by (rule conjI)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;A ∧ B ⟶ A ∨ B&quot;<br /> proof<br /> assume &quot;A ∧ B&quot;<br /> hence &quot;A&quot; ..<br /> thus &quot;A ∨ B&quot; ..<br /> qed<br /> <br /> <br /> <br /> <br /> lemma<br /> assumes 1:&quot;(A ∨ B) ∨ C&quot;<br /> shows &quot;A ∨ ( B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover<br /> { assume 2: &quot;C&quot;<br /> from 2 have 3:&quot;B ∨ C&quot; by (rule disjI2) <br /> from 3 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)}<br /> moreover<br /> { assume 4: &quot;A ∨ B&quot;<br /> from 4 have &quot;(A ∨ B) ∨ C&quot; by (rule disjI1) }<br /> ultimately show &quot;A ∨ (B ∨ C) &quot; by (simp)<br /> qed<br /> <br /> -- Otra forma (más explícita)<br /> lemma &quot;(A ∨ B) ∨ C ⟶ A ∨ (B ∨ C)&quot;<br /> proof<br /> assume 1: &quot;(A ∨ B) ∨ C&quot;<br /> show &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover{<br /> assume 2: &quot;A ∨ B&quot;<br /> have &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 2<br /> moreover{<br /> assume 3: &quot;A&quot;<br /> from 3 have 4:&quot;A ∨ (B ∨ C)&quot; by (rule disjI1)<br /> }<br /> moreover{<br /> assume 5:&quot;B&quot;<br /> from 5 have 6: &quot;B ∨ C&quot; by (rule disjI1)<br /> from 6 have 7: &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> }<br /> moreover{<br /> assume 8: &quot;C&quot;<br /> from 8 have 9: &quot;B ∨ C&quot; by (rule disjI2)<br /> from 9 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma Aux:<br /> assumes &quot;A&quot;<br /> shows &quot;B ⟶ A&quot;<br /> using assms by simp<br /> <br /> lemma K: &quot;A ⟶ (B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;B ⟶ A&quot; by (rule Aux)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(A ⟶ (B ⟶C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;A ⟶ (B ⟶ C)&quot;<br /> show &quot;(A ⟶ B) ⟶ (A ⟶C)&quot;<br /> proof<br /> assume 2:&quot;(A ⟶ B)&quot;<br /> show &quot;A ⟶C&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 2 3 have 4:&quot;B&quot; by (rule mp)<br /> from 1 3 have 5:&quot;B ⟶ C&quot; by (rule mp)<br /> from 5 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;(A ⟶ B)&quot;<br /> show &quot;((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof <br /> assume 2:&quot;(B ⟶ C)&quot;<br /> show &quot;(A ⟶ C)&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 1 3 have 4:&quot;B&quot; by (rule mp)<br /> from 2 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;¬¬A ⟶ A&quot;<br /> proof<br /> assume 1:&quot;¬¬A&quot;<br /> from 1 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> <br /> <br /> lemma &quot;A ⟶ ¬¬A&quot;<br /> proof<br /> assume 1:&quot;A&quot;<br /> from 1 show &quot;¬¬A&quot; by (rule contrapos_pn)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma MT:<br /> assumes 1:&quot;F ⟶ G&quot; and 2:&quot;¬G&quot;<br /> shows &quot;¬F&quot;<br /> proof (rule notI)<br /> assume 3:&quot;F&quot;<br /> from 1 and 3 have 4:&quot;G&quot; by (rule mp)<br /> from 2 and 4 show False by (rule notE)<br /> qed<br /> <br /> lemma &quot;(¬A ⟶ B) ⟶ (¬B ⟶ A)&quot;<br /> proof<br /> assume 1:&quot;(¬A ⟶ B)&quot;<br /> show &quot;(¬B ⟶ A)&quot;<br /> proof<br /> assume 2:&quot;¬B&quot;<br /> from 1 2 have 3:&quot;¬¬A&quot; by (rule MT)<br /> from 3 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;((A ⟶ B) ⟶ A) ⟶ A&quot;<br /> proof<br /> assume 1: &quot;(A ⟶ B) ⟶ A&quot;<br /> have 8: &quot;¬¬A&quot;<br /> proof (rule notI)<br /> assume 2: &quot;¬A&quot;<br /> have 3: &quot;A ⟶ B&quot;<br /> proof (rule impI)<br /> assume 4: &quot;A&quot;<br /> from 2 and 4 show &quot;B&quot; by (rule notE)<br /> qed<br /> from 1 and 3 have 5: &quot;A&quot; by (rule mp)<br /> from 2 and 5 show False by (rule notE)<br /> qed<br /> from 8 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;A ∨ ¬A&quot;<br /> oops<br /> <br /> lemma &quot;(¬(A ∧ B)) = (¬A ∨ ¬B)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_1&diff=86 2011-02-06T23:16:50Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 1ª relación de ejercicios *}<br /> <br /> theory Relacion_1<br /> imports Main <br /> begin<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Demostrar los siguientes lemas usando sólo las reglas básicas de<br /> deducción natural de la lógica proposicional.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma I: &quot;A ⟶ A&quot;<br /> by (rule imp_refl)<br /> <br /> -- &quot;Se puede hacer automáticamente&quot;<br /> lemma I_d2: &quot;A ⟶ A&quot;<br /> by auto<br /> <br /> -- &quot;Otra forma (más explicito)<br /> lemma I: &quot;A ⟶ A&quot;<br /> proof (rule impI)<br /> assume 1: &quot;A&quot;<br /> qed<br /> <br /> <br /> <br /> <br /> lemma <br /> assumes 1:&quot;A ∧ B&quot; <br /> shows &quot;B ∧ A&quot; <br /> proof -<br /> from 1 have 2:&quot;A&quot; by (rule conjunct1)<br /> from 1 have 3:&quot;B&quot; by (rule conjunct2)<br /> from 3 2 show &quot;B ∧ A&quot; by (rule conjI)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;A ∧ B ⟶ A ∨ B&quot;<br /> proof<br /> assume &quot;A ∧ B&quot;<br /> hence &quot;A&quot; ..<br /> thus &quot;A ∨ B&quot; ..<br /> qed<br /> <br /> <br /> <br /> <br /> lemma<br /> assumes 1:&quot;(A ∨ B) ∨ C&quot;<br /> shows &quot;A ∨ ( B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover<br /> { assume 2: &quot;C&quot;<br /> from 2 have 3:&quot;B ∨ C&quot; by (rule disjI2) <br /> from 3 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)}<br /> moreover<br /> { assume 4: &quot;A ∨ B&quot;<br /> from 4 have &quot;(A ∨ B) ∨ C&quot; by (rule disjI1) }<br /> ultimately show &quot;A ∨ (B ∨ C) &quot; by (simp)<br /> qed<br /> <br /> -- Otra forma (más explícita)<br /> lemma &quot;(A ∨ B) ∨ C ⟶ A ∨ (B ∨ C)&quot;<br /> proof<br /> assume 1: &quot;(A ∨ B) ∨ C&quot;<br /> show &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover{<br /> assume 2: &quot;A ∨ B&quot;<br /> have &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 2<br /> moreover{<br /> assume 3: &quot;A&quot;<br /> from 3 have 4:&quot;A ∨ (B ∨ C)&quot; by (rule disjI1)<br /> }<br /> moreover{<br /> assume 5:&quot;B&quot;<br /> from 5 have 6: &quot;B ∨ C&quot; by (rule disjI1)<br /> from 6 have 7: &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> }<br /> moreover{<br /> assume 8: &quot;C&quot;<br /> from 8 have 9: &quot;B ∨ C&quot; by (rule disjI2)<br /> from 9 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma Aux:<br /> assumes &quot;A&quot;<br /> shows &quot;B ⟶ A&quot;<br /> using assms by simp<br /> <br /> lemma K: &quot;A ⟶ (B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;B ⟶ A&quot; by (rule Aux)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(A ⟶ (B ⟶C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;A ⟶ (B ⟶ C)&quot;<br /> show &quot;(A ⟶ B) ⟶ (A ⟶C)&quot;<br /> proof<br /> assume 2:&quot;(A ⟶ B)&quot;<br /> show &quot;A ⟶C&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 2 3 have 4:&quot;B&quot; by (rule mp)<br /> from 1 3 have 5:&quot;B ⟶ C&quot; by (rule mp)<br /> from 5 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;(A ⟶ B)&quot;<br /> show &quot;((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof <br /> assume 2:&quot;(B ⟶ C)&quot;<br /> show &quot;(A ⟶ C)&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 1 3 have 4:&quot;B&quot; by (rule mp)<br /> from 2 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;¬¬A ⟶ A&quot;<br /> proof<br /> assume 1:&quot;¬¬A&quot;<br /> from 1 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> <br /> <br /> lemma &quot;A ⟶ ¬¬A&quot;<br /> proof<br /> assume 1:&quot;A&quot;<br /> from 1 show &quot;¬¬A&quot; by (rule contrapos_pn)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma MT:<br /> assumes 1:&quot;F ⟶ G&quot; and 2:&quot;¬G&quot;<br /> shows &quot;¬F&quot;<br /> proof (rule notI)<br /> assume 3:&quot;F&quot;<br /> from 1 and 3 have 4:&quot;G&quot; by (rule mp)<br /> from 2 and 4 show False by (rule notE)<br /> qed<br /> <br /> lemma &quot;(¬A ⟶ B) ⟶ (¬B ⟶ A)&quot;<br /> proof<br /> assume 1:&quot;(¬A ⟶ B)&quot;<br /> show &quot;(¬B ⟶ A)&quot;<br /> proof<br /> assume 2:&quot;¬B&quot;<br /> from 1 2 have 3:&quot;¬¬A&quot; by (rule MT)<br /> from 3 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;((A ⟶ B) ⟶ A) ⟶ A&quot;<br /> oops<br /> <br /> lemma &quot;A ∨ ¬A&quot;<br /> oops<br /> <br /> lemma &quot;(¬(A ∧ B)) = (¬A ∨ ¬B)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_1&diff=85 2011-02-06T23:14:44Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 1ª relación de ejercicios *}<br /> <br /> theory Relacion_1<br /> imports Main <br /> begin<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Demostrar los siguientes lemas usando sólo las reglas básicas de<br /> deducción natural de la lógica proposicional.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma I: &quot;A ⟶ A&quot;<br /> by (rule imp_refl)<br /> <br /> -- &quot;Se puede hacer automáticamente&quot;<br /> lemma I_d2: &quot;A ⟶ A&quot;<br /> by auto<br /> <br /> -- &quot;Otra forma (más explicito)<br /> lemma I: &quot;A ⟶ A&quot;<br /> proof (rule impI)<br /> assume 1: &quot;A&quot;<br /> qed<br /> <br /> <br /> <br /> <br /> lemma <br /> assumes 1:&quot;A ∧ B&quot; <br /> shows &quot;B ∧ A&quot; <br /> proof -<br /> from 1 have 2:&quot;A&quot; by (rule conjunct1)<br /> from 1 have 3:&quot;B&quot; by (rule conjunct2)<br /> from 3 2 show &quot;B ∧ A&quot; by (rule conjI)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;A ∧ B ⟶ A ∨ B&quot;<br /> proof<br /> assume &quot;A ∧ B&quot;<br /> hence &quot;A&quot; ..<br /> thus &quot;A ∨ B&quot; ..<br /> qed<br /> <br /> <br /> <br /> <br /> lemma<br /> assumes 1:&quot;(A ∨ B) ∨ C&quot;<br /> shows &quot;A ∨ ( B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover<br /> { assume 2: &quot;C&quot;<br /> from 2 have 3:&quot;B ∨ C&quot; by (rule disjI2) <br /> from 3 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)}<br /> moreover<br /> { assume 4: &quot;A ∨ B&quot;<br /> from 4 have &quot;(A ∨ B) ∨ C&quot; by (rule disjI1) }<br /> ultimately show &quot;A ∨ (B ∨ C) &quot; by (simp)<br /> qed<br /> <br /> -- Otra forma (más explícita)<br /> lemma &quot;(A ∨ B) ∨ C ⟶ A ∨ (B ∨ C)&quot;<br /> proof<br /> assume 1: &quot;(A ∨ B) ∨ C&quot;<br /> show &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover{<br /> assume 2: &quot;A ∨ B&quot;<br /> have &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 2<br /> moreover{<br /> assume 3: &quot;A&quot;<br /> from 3 have 4:&quot;A ∨ (B ∨ C)&quot; by (rule disjI1)<br /> }<br /> moreover{<br /> assume 5:&quot;B&quot;<br /> from 5 have 6: &quot;B ∨ C&quot; by (rule disjI1)<br /> from 6 have 7: &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> }<br /> moreover{<br /> assume 8: &quot;C&quot;<br /> from 8 have 9: &quot;B ∨ C&quot; by (rule disjI2)<br /> from 9 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma Aux:<br /> assumes &quot;A&quot;<br /> shows &quot;B ⟶ A&quot;<br /> using assms by simp<br /> <br /> lemma K: &quot;A ⟶ (B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;B ⟶ A&quot; by (rule Aux)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;(A ⟶ (B ⟶C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;A ⟶ (B ⟶ C)&quot;<br /> show &quot;(A ⟶ B) ⟶ (A ⟶C)&quot;<br /> proof<br /> assume 2:&quot;(A ⟶ B)&quot;<br /> show &quot;A ⟶C&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 2 3 have 4:&quot;B&quot; by (rule mp)<br /> from 1 3 have 5:&quot;B ⟶ C&quot; by (rule mp)<br /> from 5 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> lemma &quot;(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;(A ⟶ B)&quot;<br /> show &quot;((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof <br /> assume 2:&quot;(B ⟶ C)&quot;<br /> show &quot;(A ⟶ C)&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 1 3 have 4:&quot;B&quot; by (rule mp)<br /> from 2 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> lemma &quot;¬¬A ⟶ A&quot;<br /> proof<br /> assume 1:&quot;¬¬A&quot;<br /> from 1 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> lemma &quot;A ⟶ ¬¬A&quot;<br /> proof<br /> assume 1:&quot;A&quot;<br /> from 1 show &quot;¬¬A&quot; by (rule contrapos_pn)<br /> qed<br /> <br /> lemma MT:<br /> assumes 1:&quot;F ⟶ G&quot; and 2:&quot;¬G&quot;<br /> shows &quot;¬F&quot;<br /> proof (rule notI)<br /> assume 3:&quot;F&quot;<br /> from 1 and 3 have 4:&quot;G&quot; by (rule mp)<br /> from 2 and 4 show False by (rule notE)<br /> qed<br /> <br /> lemma &quot;(¬A ⟶ B) ⟶ (¬B ⟶ A)&quot;<br /> proof<br /> assume 1:&quot;(¬A ⟶ B)&quot;<br /> show &quot;(¬B ⟶ A)&quot;<br /> proof<br /> assume 2:&quot;¬B&quot;<br /> from 1 2 have 3:&quot;¬¬A&quot; by (rule MT)<br /> from 3 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> qed<br /> <br /> lemma &quot;((A ⟶ B) ⟶ A) ⟶ A&quot;<br /> oops<br /> <br /> lemma &quot;A ∨ ¬A&quot;<br /> oops<br /> <br /> lemma &quot;(¬(A ∧ B)) = (¬A ∨ ¬B)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_1&diff=84 2011-02-06T23:13:15Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 1ª relación de ejercicios *}<br /> <br /> theory Relacion_1<br /> imports Main <br /> begin<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Demostrar los siguientes lemas usando sólo las reglas básicas de<br /> deducción natural de la lógica proposicional.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma I: &quot;A ⟶ A&quot;<br /> by (rule imp_refl)<br /> <br /> -- &quot;Se puede hacer automáticamente&quot;<br /> lemma I_d2: &quot;A ⟶ A&quot;<br /> by auto<br /> <br /> -- &quot;Otra forma (más explicito)<br /> lemma I: &quot;A ⟶ A&quot;<br /> proof (rule impI)<br /> assume 1: &quot;A&quot;<br /> qed<br /> <br /> <br /> <br /> <br /> lemma <br /> assumes 1:&quot;A ∧ B&quot; <br /> shows &quot;B ∧ A&quot; <br /> proof -<br /> from 1 have 2:&quot;A&quot; by (rule conjunct1)<br /> from 1 have 3:&quot;B&quot; by (rule conjunct2)<br /> from 3 2 show &quot;B ∧ A&quot; by (rule conjI)<br /> qed<br /> <br /> <br /> <br /> <br /> lemma &quot;A ∧ B ⟶ A ∨ B&quot;<br /> proof<br /> assume &quot;A ∧ B&quot;<br /> hence &quot;A&quot; ..<br /> thus &quot;A ∨ B&quot; ..<br /> qed<br /> <br /> <br /> <br /> <br /> lemma<br /> assumes 1:&quot;(A ∨ B) ∨ C&quot;<br /> shows &quot;A ∨ ( B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover<br /> { assume 2: &quot;C&quot;<br /> from 2 have 3:&quot;B ∨ C&quot; by (rule disjI2) <br /> from 3 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)}<br /> moreover<br /> { assume 4: &quot;A ∨ B&quot;<br /> from 4 have &quot;(A ∨ B) ∨ C&quot; by (rule disjI1) }<br /> ultimately show &quot;A ∨ (B ∨ C) &quot; by (simp)<br /> qed<br /> <br /> -- Otra forma (más explícita)<br /> lemma &quot;(A ∨ B) ∨ C ⟶ A ∨ (B ∨ C)&quot;<br /> proof<br /> assume 1: &quot;(A ∨ B) ∨ C&quot;<br /> show &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover{<br /> assume 2: &quot;A ∨ B&quot;<br /> have &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 2<br /> moreover{<br /> assume 3: &quot;A&quot;<br /> from 3 have 4:&quot;A ∨ (B ∨ C)&quot; by (rule disjI1)<br /> }<br /> moreover{<br /> assume 5:&quot;B&quot;<br /> from 5 have 6: &quot;B ∨ C&quot; by (rule disjI1)<br /> from 6 have 7: &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> }<br /> moreover{<br /> assume 8: &quot;C&quot;<br /> from 8 have 9: &quot;B ∨ C&quot; by (rule disjI2)<br /> from 9 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> qed<br /> <br /> <br /> <br /> <br /> lemma K: &quot;A ⟶ (B ⟶ A)&quot;<br /> oops<br /> lemma &quot;(A ⟶ (B ⟶C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;A ⟶ (B ⟶ C)&quot;<br /> show &quot;(A ⟶ B) ⟶ (A ⟶C)&quot;<br /> proof<br /> assume 2:&quot;(A ⟶ B)&quot;<br /> show &quot;A ⟶C&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 2 3 have 4:&quot;B&quot; by (rule mp)<br /> from 1 3 have 5:&quot;B ⟶ C&quot; by (rule mp)<br /> from 5 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> lemma &quot;(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;(A ⟶ B)&quot;<br /> show &quot;((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof <br /> assume 2:&quot;(B ⟶ C)&quot;<br /> show &quot;(A ⟶ C)&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 1 3 have 4:&quot;B&quot; by (rule mp)<br /> from 2 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> lemma &quot;¬¬A ⟶ A&quot;<br /> proof<br /> assume 1:&quot;¬¬A&quot;<br /> from 1 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> lemma &quot;A ⟶ ¬¬A&quot;<br /> proof<br /> assume 1:&quot;A&quot;<br /> from 1 show &quot;¬¬A&quot; by (rule contrapos_pn)<br /> qed<br /> <br /> lemma MT:<br /> assumes 1:&quot;F ⟶ G&quot; and 2:&quot;¬G&quot;<br /> shows &quot;¬F&quot;<br /> proof (rule notI)<br /> assume 3:&quot;F&quot;<br /> from 1 and 3 have 4:&quot;G&quot; by (rule mp)<br /> from 2 and 4 show False by (rule notE)<br /> qed<br /> <br /> lemma &quot;(¬A ⟶ B) ⟶ (¬B ⟶ A)&quot;<br /> proof<br /> assume 1:&quot;(¬A ⟶ B)&quot;<br /> show &quot;(¬B ⟶ A)&quot;<br /> proof<br /> assume 2:&quot;¬B&quot;<br /> from 1 2 have 3:&quot;¬¬A&quot; by (rule MT)<br /> from 3 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> qed<br /> <br /> lemma &quot;((A ⟶ B) ⟶ A) ⟶ A&quot;<br /> oops<br /> <br /> lemma &quot;A ∨ ¬A&quot;<br /> oops<br /> <br /> lemma &quot;(¬(A ∧ B)) = (¬A ∨ ¬B)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Relaci%C3%B3n_1&diff=83 2011-02-06T23:09:19Z

<p>Jgiraldez: </p> <hr /> <div>&lt;source lang=&quot;isar&quot;&gt;<br /> header {* 1ª relación de ejercicios *}<br /> <br /> theory Relacion_1<br /> imports Main <br /> begin<br /> <br /> text {*<br /> --------------------------------------------------------------------- <br /> Demostrar los siguientes lemas usando sólo las reglas básicas de<br /> deducción natural de la lógica proposicional.<br /> --------------------------------------------------------------------- <br /> *}<br /> <br /> lemma I: &quot;A ⟶ A&quot;<br /> by (rule imp_refl)<br /> <br /> -- &quot;Se puede hacer automáticamente&quot;<br /> lemma I_d2: &quot;A ⟶ A&quot;<br /> by auto<br /> <br /> -- &quot;Otra forma (más explicito)<br /> lemma I: &quot;A ⟶ A&quot;<br /> proof (rule impI)<br /> assume 1: &quot;A&quot;<br /> qed<br /> <br /> <br /> <br /> <br /> lemma <br /> assumes 1:&quot;A ∧ B&quot; <br /> shows &quot;B ∧ A&quot; <br /> proof -<br /> from 1 have 2:&quot;A&quot; by (rule conjunct1)<br /> from 1 have 3:&quot;B&quot; by (rule conjunct2)<br /> from 3 2 show &quot;B ∧ A&quot; by (rule conjI)<br /> qed<br /> <br /> lemma &quot;A ∧ B ⟶ A ∨ B&quot;<br /> proof<br /> assume &quot;A ∧ B&quot;<br /> hence &quot;A&quot; ..<br /> thus &quot;A ∨ B&quot; ..<br /> qed <br /> <br /> lemma<br /> assumes 1:&quot;(A ∨ B) ∨ C&quot;<br /> shows &quot;A ∨ ( B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover<br /> { assume 2: &quot;C&quot;<br /> from 2 have 3:&quot;B ∨ C&quot; by (rule disjI2) <br /> from 3 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)}<br /> moreover<br /> { assume 4: &quot;A ∨ B&quot;<br /> from 4 have &quot;(A ∨ B) ∨ C&quot; by (rule disjI1) }<br /> ultimately show &quot;A ∨ (B ∨ C) &quot; by (simp)<br /> qed<br /> <br /> <br /> lemma K: &quot;A ⟶ (B ⟶ A)&quot;<br /> oops<br /> lemma &quot;(A ⟶ (B ⟶C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;A ⟶ (B ⟶ C)&quot;<br /> show &quot;(A ⟶ B) ⟶ (A ⟶C)&quot;<br /> proof<br /> assume 2:&quot;(A ⟶ B)&quot;<br /> show &quot;A ⟶C&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 2 3 have 4:&quot;B&quot; by (rule mp)<br /> from 1 3 have 5:&quot;B ⟶ C&quot; by (rule mp)<br /> from 5 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> lemma &quot;(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1:&quot;(A ⟶ B)&quot;<br /> show &quot;((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof <br /> assume 2:&quot;(B ⟶ C)&quot;<br /> show &quot;(A ⟶ C)&quot;<br /> proof<br /> assume 3:&quot;A&quot;<br /> from 1 3 have 4:&quot;B&quot; by (rule mp)<br /> from 2 4 show &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> lemma &quot;¬¬A ⟶ A&quot;<br /> proof<br /> assume 1:&quot;¬¬A&quot;<br /> from 1 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> lemma &quot;A ⟶ ¬¬A&quot;<br /> proof<br /> assume 1:&quot;A&quot;<br /> from 1 show &quot;¬¬A&quot; by (rule contrapos_pn)<br /> qed<br /> <br /> lemma MT:<br /> assumes 1:&quot;F ⟶ G&quot; and 2:&quot;¬G&quot;<br /> shows &quot;¬F&quot;<br /> proof (rule notI)<br /> assume 3:&quot;F&quot;<br /> from 1 and 3 have 4:&quot;G&quot; by (rule mp)<br /> from 2 and 4 show False by (rule notE)<br /> qed<br /> <br /> lemma &quot;(¬A ⟶ B) ⟶ (¬B ⟶ A)&quot;<br /> proof<br /> assume 1:&quot;(¬A ⟶ B)&quot;<br /> show &quot;(¬B ⟶ A)&quot;<br /> proof<br /> assume 2:&quot;¬B&quot;<br /> from 1 2 have 3:&quot;¬¬A&quot; by (rule MT)<br /> from 3 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> qed<br /> <br /> lemma &quot;((A ⟶ B) ⟶ A) ⟶ A&quot;<br /> oops<br /> <br /> lemma &quot;A ∨ ¬A&quot;<br /> oops<br /> <br /> lemma &quot;(¬(A ∧ B)) = (¬A ∨ ¬B)&quot;<br /> oops<br /> <br /> end<br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario:Jgiraldez&diff=82 2011-02-06T23:05:41Z

<p>Jgiraldez: /* RAZONAMIENTO AUTOMÁTICO 2010/11 */</p> <hr /> <div>==RAZONAMIENTO AUTOMÁTICO 2010/11==<br /> Realizado por Jesús Giráldez Crú<br /> ----</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario_discusi%C3%B3n:Jgiraldez&diff=81 2011-02-06T23:05:21Z

<p>Jgiraldez: Página blanqueada</p> <hr /> <div></div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario_discusi%C3%B3n:Jgiraldez&diff=73 2011-02-03T16:00:35Z

<p>Jgiraldez: /* RA_Relacion_2 */</p> <hr /> <div>== RA: Relacion 1 ==<br /> &lt;source lang=&quot;isar&quot;&gt;<br /> theory Relacion_1<br /> imports Main <br /> begin<br /> <br /> (* Demostrar los siguientes lemas usando sólo las reglas básicas de deducción<br /> natural de la lógica proposicional. *)<br /> <br /> lemma I: &quot;A ⟶ A&quot;<br /> proof <br /> assume 1: &quot;A&quot;<br /> qed<br /> <br /> lemma 0: &quot;A ∧ B ⟶ B ∧ A&quot;<br /> proof <br /> assume 1: &quot;A ∧ B&quot;<br /> from 1 have 2: &quot;A&quot; by (rule conjunct1) <br /> from 1 have 3: &quot;B&quot; by (rule conjunct2)<br /> from 3 and 2 show &quot;B ∧ A&quot; by (rule conjI)<br /> qed<br /> <br /> lemma &quot;A ∧ B ⟶ A ∨ B&quot;<br /> proof<br /> assume 1: &quot;A ∧ B&quot;<br /> from 1 have 2: &quot;A&quot; by (rule conjunct1)<br /> from 1 have 3: &quot;B&quot; by (rule conjunct2)<br /> from 2 show &quot;A ∨ B&quot; by (rule disjI1)<br /> qed<br /> <br /> lemma &quot;(A ∨ B) ∨ C ⟶ A ∨ (B ∨ C)&quot;<br /> proof<br /> assume 1: &quot;(A ∨ B) ∨ C&quot;<br /> show &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover{<br /> assume 2: &quot;A ∨ B&quot;<br /> have &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 2<br /> moreover{<br /> assume 3: &quot;A&quot;<br /> from 3 have 4:&quot;A ∨ (B ∨ C)&quot; by (rule disjI1)<br /> }<br /> moreover{<br /> assume 5:&quot;B&quot;<br /> from 5 have 6: &quot;B ∨ C&quot; by (rule disjI1)<br /> from 6 have 7: &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> }<br /> moreover{<br /> assume 8: &quot;C&quot;<br /> from 8 have 9: &quot;B ∨ C&quot; by (rule disjI2)<br /> from 9 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> qed<br /> <br /> lemma Aux:<br /> assumes &quot;A&quot;<br /> shows &quot;B ⟶ A&quot;<br /> using assms by simp<br /> <br /> lemma K: &quot;A ⟶ (B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;B ⟶ A&quot; by (rule Aux)<br /> qed<br /> <br /> <br /> lemma S: &quot;(A ⟶ (B ⟶ C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1: &quot;(A ⟶ (B ⟶ C))&quot;<br /> show &quot;(A ⟶ B) ⟶ (A ⟶ C)&quot;<br /> proof<br /> assume 2: &quot;A ⟶ B&quot;<br /> show &quot;A ⟶ C&quot;<br /> proof<br /> assume 3: &quot;A&quot;<br /> from 2 and 3 have 4: &quot;B&quot; by (rule mp)<br /> from 1 and 3 have 5: &quot;B ⟶ C&quot; by (rule mp)<br /> from 5 and 4 show 6: &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> lemma &quot;(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1: &quot;A ⟶ B&quot;<br /> show &quot;(B ⟶ C) ⟶ (A ⟶ C)&quot;<br /> proof<br /> assume 2: &quot;B ⟶ C&quot;<br /> show &quot;A ⟶ C&quot;<br /> proof<br /> assume 3: &quot;A&quot;<br /> from 1 and 3 have 4: &quot;B&quot; by (rule mp)<br /> from 2 and 4 show 5: &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> <br /> lemma &quot;¬¬A ⟶ A&quot;<br /> proof<br /> assume 1: &quot;¬¬A&quot;<br /> from 1 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> lemma &quot;A ⟶ ¬¬A&quot;<br /> proof <br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;¬¬A&quot; by (rule contrapos_pn)<br /> qed<br /> <br /> lemma MT:<br /> assumes 1: &quot;F ⟶ G&quot; and<br /> 2: &quot;¬G&quot;<br /> shows &quot;¬F&quot;<br /> proof (rule notI)<br /> assume 3: &quot;F&quot;<br /> from 1 and 3 have 4: &quot;G&quot; by (rule mp)<br /> from 2 and 4 show False by (rule notE)<br /> qed<br /> <br /> lemma &quot;(¬A ⟶ B) ⟶ (¬B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;¬A ⟶ B&quot;<br /> show &quot;¬B ⟶ A&quot;<br /> proof<br /> assume 2: &quot;¬B&quot;<br /> from 1 and 2 have 3:&quot;¬¬A&quot; by (rule MT)<br /> from 3 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> qed<br /> <br /> lemma &quot;((A ⟶ B) ⟶ A) ⟶ A&quot;<br /> proof<br /> assume 1: &quot;(A ⟶ B) ⟶ A&quot;<br /> have 8: &quot;¬¬A&quot;<br /> proof (rule notI)<br /> assume 2: &quot;¬A&quot;<br /> have 3: &quot;A ⟶ B&quot;<br /> proof (rule impI)<br /> assume 4: &quot;A&quot;<br /> from 2 and 4 show &quot;B&quot; by (rule notE)<br /> qed<br /> from 1 and 3 have 5: &quot;A&quot; by (rule mp)<br /> from 2 and 5 show False by (rule notE)<br /> qed<br /> from 8 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> lemma &quot;A ∨ ¬A&quot;<br /> proof cases<br /> assume &quot;A&quot; thus ?thesis ..<br /> next<br /> assume &quot;¬A&quot; thus ?thesis ..<br /> qed<br /> <br /> <br /> lemma &quot;(¬(A ∧ B)) = (¬A ∨ ¬B)&quot;<br /> proof<br /> {<br /> assume 1: &quot;¬(A ∧ B)&quot;<br /> have 2: &quot;¬A ∨ A&quot; by (rule excluded_middle)<br /> thus &quot;¬A ∨ ¬B&quot;<br /> proof (rule disjE)<br /> { assume 3: &quot;¬A&quot;<br /> thus &quot;¬A ∨ ¬B&quot; by (rule disjI1) }<br /> next<br /> { assume 4: &quot;A&quot;<br /> have 5: &quot;¬B ∨ B&quot; by (rule excluded_middle)<br /> thus &quot;¬A ∨ ¬B&quot;<br /> proof (rule disjE)<br /> {assume 6: &quot;¬B&quot;<br /> thus &quot;¬A ∨ ¬B&quot; by (rule disjI2) }<br /> next<br /> {assume 7: &quot;B&quot;<br /> from 4 and 7 have 8: &quot;A ∧ B&quot; by (rule conjI)<br /> from 1 and 8 show &quot;¬A ∨ ¬B&quot; by (rule notE) }<br /> qed<br /> } <br /> qed<br /> }<br /> next<br /> { assume 1: &quot;¬A ∨ ¬B&quot;<br /> show &quot;¬(A ∧ B)&quot;<br /> proof (rule notI)<br /> assume 2: &quot;A ∧ B&quot;<br /> from 1 have 3:&quot;¬A ∨ ¬B&quot; .<br /> moreover<br /> { assume 4: &quot;¬A&quot;<br /> from 2 have 5:&quot;A&quot; by (rule conjunct1)<br /> from 4 and 5 have False by (rule notE) }<br /> moreover<br /> { assume 6: &quot;¬B&quot;<br /> from 2 have 7: &quot;B&quot; by (rule conjunct2)<br /> from 6 and 7 have False by (rule notE) }<br /> ultimately show False by (rule disjE) <br /> qed<br /> <br /> }<br /> qed<br /> <br /> end<br /> <br /> &lt;/source&gt;<br /> <br /> == RA: Relacion 2 ==<br /> &lt;source lang=&quot;isar&quot;&gt;<br /> <br /> theory Relacion_2<br /> imports Main <br /> begin<br /> <br /> (* Demostrar o refutar los siguientes lemas usando sólo las reglas básicas<br /> de deducción natural de la lógica proposicional y de la lógica de <br /> predicados (allI, allE, exI y exE). *)<br /> <br /> <br /> lemma &quot;(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)&quot;<br /> proof (rule impI)<br /> assume 1: &quot;(∃x. ∀y. P x y)&quot;<br /> show &quot;(∀y. ∃x. P x y)&quot;<br /> proof (rule allI)<br /> from 1 obtain x where 2: &quot;∀y. P x y&quot; by (rule exE)<br /> fix y<br /> from 2 have 3: &quot;P x y&quot; by (rule allE)<br /> from 3 show 4: &quot;∃x. P x y&quot; by (rule exI)<br /> qed<br /> qed<br /> <br /> lemma &quot;(∀x. P x ⟶ Q) = ((∃x. P x) ⟶ Q)&quot;<br /> proof<br /> { assume 1: &quot;∀x. P x ⟶ Q&quot;<br /> show &quot;(∃x. P x) ⟶ Q&quot;<br /> proof<br /> assume &quot;∃x. P x&quot;<br /> then obtain a where &quot;P a&quot; by (rule exE)<br /> have &quot;P a ⟶ Q&quot; using 1 by (rule allE)<br /> thus &quot;Q&quot; using `P a` by (rule mp)<br /> qed }<br /> next<br /> { assume 2: &quot;(∃x. P x) ⟶ Q&quot;<br /> show &quot;∀x. P x ⟶ Q&quot;<br /> proof<br /> fix a<br /> show &quot;P a ⟶ Q&quot;<br /> proof<br /> assume &quot;P a&quot;<br /> hence 3: &quot;∃x. P x&quot; by (rule exI)<br /> from 2 and 3 show &quot;Q&quot; by (rule mp)<br /> qed<br /> qed }<br /> qed<br /> <br /> lemma &quot;((∀ x. P x) ∧ (∀ x. Q x)) = (∀ x. (P x ∧ Q x))&quot;<br /> proof<br /> {<br /> assume 1: &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> show &quot;(∀ x. (P x ∧ Q x))&quot;<br /> proof (rule allI)<br /> from 1 have 2: &quot;(∀ x. P x)&quot; by (rule conjE)<br /> from 1 have 3: &quot;(∀ x. Q x)&quot; by (rule conjE)<br /> fix a<br /> from 2 have 4: &quot;P a&quot; by (rule allE)<br /> from 3 have 5: &quot;Q a&quot; by (rule allE)<br /> from 4 and 5 show 6: &quot;(P a) ∧ (Q a)&quot; by (rule conjI)<br /> qed<br /> }<br /> next<br /> {<br /> assume 7: &quot;∀ x. (P x ∧ Q x)&quot;<br /> show &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> proof<br /> fix a<br /> from 7 have 8: &quot;P a ∧ Q a&quot; by (rule allE)<br /> from 8 have 9: &quot;P a&quot; by (rule conjE)<br /> (* seguir... *)<br /> from 8 have 11: &quot;Q a&quot; by (rule allE)<br /> (* seguir... *)<br /> qed<br /> }<br /> qed <br /> <br /> oops<br /> <br /> lemma &quot;((∀ x. P x) ∨ (∀ x. Q x)) = (∀ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;((∃ x. P x) ∨ (∃ x. Q x)) = (∃ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)&quot;<br /> oops<br /> (*<br /> proof (rule impI)<br /> assume 1: &quot;∀x. ∃y. P x y&quot;<br /> show &quot;∃y. ∀x. P x y&quot;<br /> proof (rule exI)<br /> show &quot; ∀x. P x y&quot;<br /> proof (rule allI)<br /> fix x<br /> from 1 have 2: &quot;∃y. P x y&quot; by (rule allE)<br /> from 2 obtain y where 3: &quot;P x y&quot; by (rule exE)<br /> thus ?thesis ..<br /> qed<br /> *)<br /> <br /> lemma &quot;(¬ (∀ x. P x)) = (∃ x. ¬ P x)&quot;<br /> proof<br /> {<br /> assume 1: &quot;¬ (∀ x. P x)&quot; <br /> show &quot;∃ x. ¬ P x&quot;<br /> proof<br /> <br /> <br /> qed<br /> }<br /> next<br /> {<br /> assume 2: &quot;∃ x. ¬ P x&quot;<br /> show &quot;¬ (∀ x. P x)&quot;<br /> proof<br /> <br /> qed<br /> }<br /> qed<br /> <br /> end<br /> <br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario_discusi%C3%B3n:Jgiraldez&diff=72 2011-02-03T16:00:21Z

<p>Jgiraldez: /* RA_Relacion_1 */</p> <hr /> <div>== RA: Relacion 1 ==<br /> &lt;source lang=&quot;isar&quot;&gt;<br /> theory Relacion_1<br /> imports Main <br /> begin<br /> <br /> (* Demostrar los siguientes lemas usando sólo las reglas básicas de deducción<br /> natural de la lógica proposicional. *)<br /> <br /> lemma I: &quot;A ⟶ A&quot;<br /> proof <br /> assume 1: &quot;A&quot;<br /> qed<br /> <br /> lemma 0: &quot;A ∧ B ⟶ B ∧ A&quot;<br /> proof <br /> assume 1: &quot;A ∧ B&quot;<br /> from 1 have 2: &quot;A&quot; by (rule conjunct1) <br /> from 1 have 3: &quot;B&quot; by (rule conjunct2)<br /> from 3 and 2 show &quot;B ∧ A&quot; by (rule conjI)<br /> qed<br /> <br /> lemma &quot;A ∧ B ⟶ A ∨ B&quot;<br /> proof<br /> assume 1: &quot;A ∧ B&quot;<br /> from 1 have 2: &quot;A&quot; by (rule conjunct1)<br /> from 1 have 3: &quot;B&quot; by (rule conjunct2)<br /> from 2 show &quot;A ∨ B&quot; by (rule disjI1)<br /> qed<br /> <br /> lemma &quot;(A ∨ B) ∨ C ⟶ A ∨ (B ∨ C)&quot;<br /> proof<br /> assume 1: &quot;(A ∨ B) ∨ C&quot;<br /> show &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover{<br /> assume 2: &quot;A ∨ B&quot;<br /> have &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 2<br /> moreover{<br /> assume 3: &quot;A&quot;<br /> from 3 have 4:&quot;A ∨ (B ∨ C)&quot; by (rule disjI1)<br /> }<br /> moreover{<br /> assume 5:&quot;B&quot;<br /> from 5 have 6: &quot;B ∨ C&quot; by (rule disjI1)<br /> from 6 have 7: &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> }<br /> moreover{<br /> assume 8: &quot;C&quot;<br /> from 8 have 9: &quot;B ∨ C&quot; by (rule disjI2)<br /> from 9 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> qed<br /> <br /> lemma Aux:<br /> assumes &quot;A&quot;<br /> shows &quot;B ⟶ A&quot;<br /> using assms by simp<br /> <br /> lemma K: &quot;A ⟶ (B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;B ⟶ A&quot; by (rule Aux)<br /> qed<br /> <br /> <br /> lemma S: &quot;(A ⟶ (B ⟶ C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1: &quot;(A ⟶ (B ⟶ C))&quot;<br /> show &quot;(A ⟶ B) ⟶ (A ⟶ C)&quot;<br /> proof<br /> assume 2: &quot;A ⟶ B&quot;<br /> show &quot;A ⟶ C&quot;<br /> proof<br /> assume 3: &quot;A&quot;<br /> from 2 and 3 have 4: &quot;B&quot; by (rule mp)<br /> from 1 and 3 have 5: &quot;B ⟶ C&quot; by (rule mp)<br /> from 5 and 4 show 6: &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> lemma &quot;(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1: &quot;A ⟶ B&quot;<br /> show &quot;(B ⟶ C) ⟶ (A ⟶ C)&quot;<br /> proof<br /> assume 2: &quot;B ⟶ C&quot;<br /> show &quot;A ⟶ C&quot;<br /> proof<br /> assume 3: &quot;A&quot;<br /> from 1 and 3 have 4: &quot;B&quot; by (rule mp)<br /> from 2 and 4 show 5: &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> <br /> lemma &quot;¬¬A ⟶ A&quot;<br /> proof<br /> assume 1: &quot;¬¬A&quot;<br /> from 1 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> lemma &quot;A ⟶ ¬¬A&quot;<br /> proof <br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;¬¬A&quot; by (rule contrapos_pn)<br /> qed<br /> <br /> lemma MT:<br /> assumes 1: &quot;F ⟶ G&quot; and<br /> 2: &quot;¬G&quot;<br /> shows &quot;¬F&quot;<br /> proof (rule notI)<br /> assume 3: &quot;F&quot;<br /> from 1 and 3 have 4: &quot;G&quot; by (rule mp)<br /> from 2 and 4 show False by (rule notE)<br /> qed<br /> <br /> lemma &quot;(¬A ⟶ B) ⟶ (¬B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;¬A ⟶ B&quot;<br /> show &quot;¬B ⟶ A&quot;<br /> proof<br /> assume 2: &quot;¬B&quot;<br /> from 1 and 2 have 3:&quot;¬¬A&quot; by (rule MT)<br /> from 3 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> qed<br /> <br /> lemma &quot;((A ⟶ B) ⟶ A) ⟶ A&quot;<br /> proof<br /> assume 1: &quot;(A ⟶ B) ⟶ A&quot;<br /> have 8: &quot;¬¬A&quot;<br /> proof (rule notI)<br /> assume 2: &quot;¬A&quot;<br /> have 3: &quot;A ⟶ B&quot;<br /> proof (rule impI)<br /> assume 4: &quot;A&quot;<br /> from 2 and 4 show &quot;B&quot; by (rule notE)<br /> qed<br /> from 1 and 3 have 5: &quot;A&quot; by (rule mp)<br /> from 2 and 5 show False by (rule notE)<br /> qed<br /> from 8 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> lemma &quot;A ∨ ¬A&quot;<br /> proof cases<br /> assume &quot;A&quot; thus ?thesis ..<br /> next<br /> assume &quot;¬A&quot; thus ?thesis ..<br /> qed<br /> <br /> <br /> lemma &quot;(¬(A ∧ B)) = (¬A ∨ ¬B)&quot;<br /> proof<br /> {<br /> assume 1: &quot;¬(A ∧ B)&quot;<br /> have 2: &quot;¬A ∨ A&quot; by (rule excluded_middle)<br /> thus &quot;¬A ∨ ¬B&quot;<br /> proof (rule disjE)<br /> { assume 3: &quot;¬A&quot;<br /> thus &quot;¬A ∨ ¬B&quot; by (rule disjI1) }<br /> next<br /> { assume 4: &quot;A&quot;<br /> have 5: &quot;¬B ∨ B&quot; by (rule excluded_middle)<br /> thus &quot;¬A ∨ ¬B&quot;<br /> proof (rule disjE)<br /> {assume 6: &quot;¬B&quot;<br /> thus &quot;¬A ∨ ¬B&quot; by (rule disjI2) }<br /> next<br /> {assume 7: &quot;B&quot;<br /> from 4 and 7 have 8: &quot;A ∧ B&quot; by (rule conjI)<br /> from 1 and 8 show &quot;¬A ∨ ¬B&quot; by (rule notE) }<br /> qed<br /> } <br /> qed<br /> }<br /> next<br /> { assume 1: &quot;¬A ∨ ¬B&quot;<br /> show &quot;¬(A ∧ B)&quot;<br /> proof (rule notI)<br /> assume 2: &quot;A ∧ B&quot;<br /> from 1 have 3:&quot;¬A ∨ ¬B&quot; .<br /> moreover<br /> { assume 4: &quot;¬A&quot;<br /> from 2 have 5:&quot;A&quot; by (rule conjunct1)<br /> from 4 and 5 have False by (rule notE) }<br /> moreover<br /> { assume 6: &quot;¬B&quot;<br /> from 2 have 7: &quot;B&quot; by (rule conjunct2)<br /> from 6 and 7 have False by (rule notE) }<br /> ultimately show False by (rule disjE) <br /> qed<br /> <br /> }<br /> qed<br /> <br /> end<br /> <br /> &lt;/source&gt;<br /> <br /> == RA_Relacion_2 ==<br /> &lt;source lang=&quot;isar&quot;&gt;<br /> <br /> theory Relacion_2<br /> imports Main <br /> begin<br /> <br /> (* Demostrar o refutar los siguientes lemas usando sólo las reglas básicas<br /> de deducción natural de la lógica proposicional y de la lógica de <br /> predicados (allI, allE, exI y exE). *)<br /> <br /> <br /> lemma &quot;(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)&quot;<br /> proof (rule impI)<br /> assume 1: &quot;(∃x. ∀y. P x y)&quot;<br /> show &quot;(∀y. ∃x. P x y)&quot;<br /> proof (rule allI)<br /> from 1 obtain x where 2: &quot;∀y. P x y&quot; by (rule exE)<br /> fix y<br /> from 2 have 3: &quot;P x y&quot; by (rule allE)<br /> from 3 show 4: &quot;∃x. P x y&quot; by (rule exI)<br /> qed<br /> qed<br /> <br /> lemma &quot;(∀x. P x ⟶ Q) = ((∃x. P x) ⟶ Q)&quot;<br /> proof<br /> { assume 1: &quot;∀x. P x ⟶ Q&quot;<br /> show &quot;(∃x. P x) ⟶ Q&quot;<br /> proof<br /> assume &quot;∃x. P x&quot;<br /> then obtain a where &quot;P a&quot; by (rule exE)<br /> have &quot;P a ⟶ Q&quot; using 1 by (rule allE)<br /> thus &quot;Q&quot; using `P a` by (rule mp)<br /> qed }<br /> next<br /> { assume 2: &quot;(∃x. P x) ⟶ Q&quot;<br /> show &quot;∀x. P x ⟶ Q&quot;<br /> proof<br /> fix a<br /> show &quot;P a ⟶ Q&quot;<br /> proof<br /> assume &quot;P a&quot;<br /> hence 3: &quot;∃x. P x&quot; by (rule exI)<br /> from 2 and 3 show &quot;Q&quot; by (rule mp)<br /> qed<br /> qed }<br /> qed<br /> <br /> lemma &quot;((∀ x. P x) ∧ (∀ x. Q x)) = (∀ x. (P x ∧ Q x))&quot;<br /> proof<br /> {<br /> assume 1: &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> show &quot;(∀ x. (P x ∧ Q x))&quot;<br /> proof (rule allI)<br /> from 1 have 2: &quot;(∀ x. P x)&quot; by (rule conjE)<br /> from 1 have 3: &quot;(∀ x. Q x)&quot; by (rule conjE)<br /> fix a<br /> from 2 have 4: &quot;P a&quot; by (rule allE)<br /> from 3 have 5: &quot;Q a&quot; by (rule allE)<br /> from 4 and 5 show 6: &quot;(P a) ∧ (Q a)&quot; by (rule conjI)<br /> qed<br /> }<br /> next<br /> {<br /> assume 7: &quot;∀ x. (P x ∧ Q x)&quot;<br /> show &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> proof<br /> fix a<br /> from 7 have 8: &quot;P a ∧ Q a&quot; by (rule allE)<br /> from 8 have 9: &quot;P a&quot; by (rule conjE)<br /> (* seguir... *)<br /> from 8 have 11: &quot;Q a&quot; by (rule allE)<br /> (* seguir... *)<br /> qed<br /> }<br /> qed <br /> <br /> oops<br /> <br /> lemma &quot;((∀ x. P x) ∨ (∀ x. Q x)) = (∀ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;((∃ x. P x) ∨ (∃ x. Q x)) = (∃ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)&quot;<br /> oops<br /> (*<br /> proof (rule impI)<br /> assume 1: &quot;∀x. ∃y. P x y&quot;<br /> show &quot;∃y. ∀x. P x y&quot;<br /> proof (rule exI)<br /> show &quot; ∀x. P x y&quot;<br /> proof (rule allI)<br /> fix x<br /> from 1 have 2: &quot;∃y. P x y&quot; by (rule allE)<br /> from 2 obtain y where 3: &quot;P x y&quot; by (rule exE)<br /> thus ?thesis ..<br /> qed<br /> *)<br /> <br /> lemma &quot;(¬ (∀ x. P x)) = (∃ x. ¬ P x)&quot;<br /> proof<br /> {<br /> assume 1: &quot;¬ (∀ x. P x)&quot; <br /> show &quot;∃ x. ¬ P x&quot;<br /> proof<br /> <br /> <br /> qed<br /> }<br /> next<br /> {<br /> assume 2: &quot;∃ x. ¬ P x&quot;<br /> show &quot;¬ (∀ x. P x)&quot;<br /> proof<br /> <br /> qed<br /> }<br /> qed<br /> <br /> end<br /> <br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario:Jgiraldez&diff=71 2011-02-03T15:55:49Z

<p>Jgiraldez: /* Relaciones de Ejercicios resueltos: */</p> <hr /> <div>==RAZONAMIENTO AUTOMÁTICO 2010/11==<br /> Realizado por Jesús Giráldez Crú<br /> ----<br /> <br /> ===Relaciones de Ejercicios resueltos:===<br /> *Relación 1: [http://www.glc.us.es/~jalonso/WikiRA2010/index.php5/Usuario_Discusi%C3%B3n:Jgiraldez#RA_Relacion_1 Relacion_1] COMPLETA (Actualizado 03/02/2011)<br /> *Relación 2: [http://www.glc.us.es/~jalonso/WikiRA2010/index.php5/Usuario_Discusi%C3%B3n:Jgiraldez#RA_Relacion_2 Relacion_2] 2/7 ejercicios completados (Actualizado 03/02/2011)</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario_discusi%C3%B3n:Jgiraldez&diff=70 2011-02-03T15:53:26Z

<p>Jgiraldez: /* RA_Relacion_2 */</p> <hr /> <div>== RA_Relacion_1 ==<br /> &lt;source lang=&quot;isar&quot;&gt;<br /> theory Relacion_1<br /> imports Main <br /> begin<br /> <br /> (* Demostrar los siguientes lemas usando sólo las reglas básicas de deducción<br /> natural de la lógica proposicional. *)<br /> <br /> lemma I: &quot;A ⟶ A&quot;<br /> proof <br /> assume 1: &quot;A&quot;<br /> qed<br /> <br /> lemma 0: &quot;A ∧ B ⟶ B ∧ A&quot;<br /> proof <br /> assume 1: &quot;A ∧ B&quot;<br /> from 1 have 2: &quot;A&quot; by (rule conjunct1) <br /> from 1 have 3: &quot;B&quot; by (rule conjunct2)<br /> from 3 and 2 show &quot;B ∧ A&quot; by (rule conjI)<br /> qed<br /> <br /> lemma &quot;A ∧ B ⟶ A ∨ B&quot;<br /> proof<br /> assume 1: &quot;A ∧ B&quot;<br /> from 1 have 2: &quot;A&quot; by (rule conjunct1)<br /> from 1 have 3: &quot;B&quot; by (rule conjunct2)<br /> from 2 show &quot;A ∨ B&quot; by (rule disjI1)<br /> qed<br /> <br /> lemma &quot;(A ∨ B) ∨ C ⟶ A ∨ (B ∨ C)&quot;<br /> proof<br /> assume 1: &quot;(A ∨ B) ∨ C&quot;<br /> show &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover{<br /> assume 2: &quot;A ∨ B&quot;<br /> have &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 2<br /> moreover{<br /> assume 3: &quot;A&quot;<br /> from 3 have 4:&quot;A ∨ (B ∨ C)&quot; by (rule disjI1)<br /> }<br /> moreover{<br /> assume 5:&quot;B&quot;<br /> from 5 have 6: &quot;B ∨ C&quot; by (rule disjI1)<br /> from 6 have 7: &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> }<br /> moreover{<br /> assume 8: &quot;C&quot;<br /> from 8 have 9: &quot;B ∨ C&quot; by (rule disjI2)<br /> from 9 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> qed<br /> <br /> lemma Aux:<br /> assumes &quot;A&quot;<br /> shows &quot;B ⟶ A&quot;<br /> using assms by simp<br /> <br /> lemma K: &quot;A ⟶ (B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;B ⟶ A&quot; by (rule Aux)<br /> qed<br /> <br /> <br /> lemma S: &quot;(A ⟶ (B ⟶ C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1: &quot;(A ⟶ (B ⟶ C))&quot;<br /> show &quot;(A ⟶ B) ⟶ (A ⟶ C)&quot;<br /> proof<br /> assume 2: &quot;A ⟶ B&quot;<br /> show &quot;A ⟶ C&quot;<br /> proof<br /> assume 3: &quot;A&quot;<br /> from 2 and 3 have 4: &quot;B&quot; by (rule mp)<br /> from 1 and 3 have 5: &quot;B ⟶ C&quot; by (rule mp)<br /> from 5 and 4 show 6: &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> lemma &quot;(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1: &quot;A ⟶ B&quot;<br /> show &quot;(B ⟶ C) ⟶ (A ⟶ C)&quot;<br /> proof<br /> assume 2: &quot;B ⟶ C&quot;<br /> show &quot;A ⟶ C&quot;<br /> proof<br /> assume 3: &quot;A&quot;<br /> from 1 and 3 have 4: &quot;B&quot; by (rule mp)<br /> from 2 and 4 show 5: &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> <br /> lemma &quot;¬¬A ⟶ A&quot;<br /> proof<br /> assume 1: &quot;¬¬A&quot;<br /> from 1 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> lemma &quot;A ⟶ ¬¬A&quot;<br /> proof <br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;¬¬A&quot; by (rule contrapos_pn)<br /> qed<br /> <br /> lemma MT:<br /> assumes 1: &quot;F ⟶ G&quot; and<br /> 2: &quot;¬G&quot;<br /> shows &quot;¬F&quot;<br /> proof (rule notI)<br /> assume 3: &quot;F&quot;<br /> from 1 and 3 have 4: &quot;G&quot; by (rule mp)<br /> from 2 and 4 show False by (rule notE)<br /> qed<br /> <br /> lemma &quot;(¬A ⟶ B) ⟶ (¬B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;¬A ⟶ B&quot;<br /> show &quot;¬B ⟶ A&quot;<br /> proof<br /> assume 2: &quot;¬B&quot;<br /> from 1 and 2 have 3:&quot;¬¬A&quot; by (rule MT)<br /> from 3 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> qed<br /> <br /> lemma &quot;((A ⟶ B) ⟶ A) ⟶ A&quot;<br /> proof<br /> assume 1: &quot;(A ⟶ B) ⟶ A&quot;<br /> have 8: &quot;¬¬A&quot;<br /> proof (rule notI)<br /> assume 2: &quot;¬A&quot;<br /> have 3: &quot;A ⟶ B&quot;<br /> proof (rule impI)<br /> assume 4: &quot;A&quot;<br /> from 2 and 4 show &quot;B&quot; by (rule notE)<br /> qed<br /> from 1 and 3 have 5: &quot;A&quot; by (rule mp)<br /> from 2 and 5 show False by (rule notE)<br /> qed<br /> from 8 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> lemma &quot;A ∨ ¬A&quot;<br /> proof cases<br /> assume &quot;A&quot; thus ?thesis ..<br /> next<br /> assume &quot;¬A&quot; thus ?thesis ..<br /> qed<br /> <br /> <br /> lemma &quot;(¬(A ∧ B)) = (¬A ∨ ¬B)&quot;<br /> proof<br /> {<br /> assume 1: &quot;¬(A ∧ B)&quot;<br /> have 2: &quot;¬A ∨ A&quot; by (rule excluded_middle)<br /> thus &quot;¬A ∨ ¬B&quot;<br /> proof (rule disjE)<br /> { assume 3: &quot;¬A&quot;<br /> thus &quot;¬A ∨ ¬B&quot; by (rule disjI1) }<br /> next<br /> { assume 4: &quot;A&quot;<br /> have 5: &quot;¬B ∨ B&quot; by (rule excluded_middle)<br /> thus &quot;¬A ∨ ¬B&quot;<br /> proof (rule disjE)<br /> {assume 6: &quot;¬B&quot;<br /> thus &quot;¬A ∨ ¬B&quot; by (rule disjI2) }<br /> next<br /> {assume 7: &quot;B&quot;<br /> from 4 and 7 have 8: &quot;A ∧ B&quot; by (rule conjI)<br /> from 1 and 8 show &quot;¬A ∨ ¬B&quot; by (rule notE) }<br /> qed<br /> } <br /> qed<br /> }<br /> next<br /> { assume 1: &quot;¬A ∨ ¬B&quot;<br /> show &quot;¬(A ∧ B)&quot;<br /> proof (rule notI)<br /> assume 2: &quot;A ∧ B&quot;<br /> from 1 have 3:&quot;¬A ∨ ¬B&quot; .<br /> moreover<br /> { assume 4: &quot;¬A&quot;<br /> from 2 have 5:&quot;A&quot; by (rule conjunct1)<br /> from 4 and 5 have False by (rule notE) }<br /> moreover<br /> { assume 6: &quot;¬B&quot;<br /> from 2 have 7: &quot;B&quot; by (rule conjunct2)<br /> from 6 and 7 have False by (rule notE) }<br /> ultimately show False by (rule disjE) <br /> qed<br /> <br /> }<br /> qed<br /> <br /> end<br /> <br /> &lt;/source&gt;<br /> <br /> == RA_Relacion_2 ==<br /> &lt;source lang=&quot;isar&quot;&gt;<br /> <br /> theory Relacion_2<br /> imports Main <br /> begin<br /> <br /> (* Demostrar o refutar los siguientes lemas usando sólo las reglas básicas<br /> de deducción natural de la lógica proposicional y de la lógica de <br /> predicados (allI, allE, exI y exE). *)<br /> <br /> <br /> lemma &quot;(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)&quot;<br /> proof (rule impI)<br /> assume 1: &quot;(∃x. ∀y. P x y)&quot;<br /> show &quot;(∀y. ∃x. P x y)&quot;<br /> proof (rule allI)<br /> from 1 obtain x where 2: &quot;∀y. P x y&quot; by (rule exE)<br /> fix y<br /> from 2 have 3: &quot;P x y&quot; by (rule allE)<br /> from 3 show 4: &quot;∃x. P x y&quot; by (rule exI)<br /> qed<br /> qed<br /> <br /> lemma &quot;(∀x. P x ⟶ Q) = ((∃x. P x) ⟶ Q)&quot;<br /> proof<br /> { assume 1: &quot;∀x. P x ⟶ Q&quot;<br /> show &quot;(∃x. P x) ⟶ Q&quot;<br /> proof<br /> assume &quot;∃x. P x&quot;<br /> then obtain a where &quot;P a&quot; by (rule exE)<br /> have &quot;P a ⟶ Q&quot; using 1 by (rule allE)<br /> thus &quot;Q&quot; using `P a` by (rule mp)<br /> qed }<br /> next<br /> { assume 2: &quot;(∃x. P x) ⟶ Q&quot;<br /> show &quot;∀x. P x ⟶ Q&quot;<br /> proof<br /> fix a<br /> show &quot;P a ⟶ Q&quot;<br /> proof<br /> assume &quot;P a&quot;<br /> hence 3: &quot;∃x. P x&quot; by (rule exI)<br /> from 2 and 3 show &quot;Q&quot; by (rule mp)<br /> qed<br /> qed }<br /> qed<br /> <br /> lemma &quot;((∀ x. P x) ∧ (∀ x. Q x)) = (∀ x. (P x ∧ Q x))&quot;<br /> proof<br /> {<br /> assume 1: &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> show &quot;(∀ x. (P x ∧ Q x))&quot;<br /> proof (rule allI)<br /> from 1 have 2: &quot;(∀ x. P x)&quot; by (rule conjE)<br /> from 1 have 3: &quot;(∀ x. Q x)&quot; by (rule conjE)<br /> fix a<br /> from 2 have 4: &quot;P a&quot; by (rule allE)<br /> from 3 have 5: &quot;Q a&quot; by (rule allE)<br /> from 4 and 5 show 6: &quot;(P a) ∧ (Q a)&quot; by (rule conjI)<br /> qed<br /> }<br /> next<br /> {<br /> assume 7: &quot;∀ x. (P x ∧ Q x)&quot;<br /> show &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> proof<br /> fix a<br /> from 7 have 8: &quot;P a ∧ Q a&quot; by (rule allE)<br /> from 8 have 9: &quot;P a&quot; by (rule conjE)<br /> (* seguir... *)<br /> from 8 have 11: &quot;Q a&quot; by (rule allE)<br /> (* seguir... *)<br /> qed<br /> }<br /> qed <br /> <br /> oops<br /> <br /> lemma &quot;((∀ x. P x) ∨ (∀ x. Q x)) = (∀ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;((∃ x. P x) ∨ (∃ x. Q x)) = (∃ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)&quot;<br /> oops<br /> (*<br /> proof (rule impI)<br /> assume 1: &quot;∀x. ∃y. P x y&quot;<br /> show &quot;∃y. ∀x. P x y&quot;<br /> proof (rule exI)<br /> show &quot; ∀x. P x y&quot;<br /> proof (rule allI)<br /> fix x<br /> from 1 have 2: &quot;∃y. P x y&quot; by (rule allE)<br /> from 2 obtain y where 3: &quot;P x y&quot; by (rule exE)<br /> thus ?thesis ..<br /> qed<br /> *)<br /> <br /> lemma &quot;(¬ (∀ x. P x)) = (∃ x. ¬ P x)&quot;<br /> proof<br /> {<br /> assume 1: &quot;¬ (∀ x. P x)&quot; <br /> show &quot;∃ x. ¬ P x&quot;<br /> proof<br /> <br /> <br /> qed<br /> }<br /> next<br /> {<br /> assume 2: &quot;∃ x. ¬ P x&quot;<br /> show &quot;¬ (∀ x. P x)&quot;<br /> proof<br /> <br /> qed<br /> }<br /> qed<br /> <br /> end<br /> <br /> &lt;/source&gt;</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario_discusi%C3%B3n:Jgiraldez&diff=69 2011-02-03T15:53:00Z

<p>Jgiraldez: /* RA_Relacion_1 */</p> <hr /> <div>== RA_Relacion_1 ==<br /> &lt;source lang=&quot;isar&quot;&gt;<br /> theory Relacion_1<br /> imports Main <br /> begin<br /> <br /> (* Demostrar los siguientes lemas usando sólo las reglas básicas de deducción<br /> natural de la lógica proposicional. *)<br /> <br /> lemma I: &quot;A ⟶ A&quot;<br /> proof <br /> assume 1: &quot;A&quot;<br /> qed<br /> <br /> lemma 0: &quot;A ∧ B ⟶ B ∧ A&quot;<br /> proof <br /> assume 1: &quot;A ∧ B&quot;<br /> from 1 have 2: &quot;A&quot; by (rule conjunct1) <br /> from 1 have 3: &quot;B&quot; by (rule conjunct2)<br /> from 3 and 2 show &quot;B ∧ A&quot; by (rule conjI)<br /> qed<br /> <br /> lemma &quot;A ∧ B ⟶ A ∨ B&quot;<br /> proof<br /> assume 1: &quot;A ∧ B&quot;<br /> from 1 have 2: &quot;A&quot; by (rule conjunct1)<br /> from 1 have 3: &quot;B&quot; by (rule conjunct2)<br /> from 2 show &quot;A ∨ B&quot; by (rule disjI1)<br /> qed<br /> <br /> lemma &quot;(A ∨ B) ∨ C ⟶ A ∨ (B ∨ C)&quot;<br /> proof<br /> assume 1: &quot;(A ∨ B) ∨ C&quot;<br /> show &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover{<br /> assume 2: &quot;A ∨ B&quot;<br /> have &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 2<br /> moreover{<br /> assume 3: &quot;A&quot;<br /> from 3 have 4:&quot;A ∨ (B ∨ C)&quot; by (rule disjI1)<br /> }<br /> moreover{<br /> assume 5:&quot;B&quot;<br /> from 5 have 6: &quot;B ∨ C&quot; by (rule disjI1)<br /> from 6 have 7: &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> }<br /> moreover{<br /> assume 8: &quot;C&quot;<br /> from 8 have 9: &quot;B ∨ C&quot; by (rule disjI2)<br /> from 9 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> qed<br /> <br /> lemma Aux:<br /> assumes &quot;A&quot;<br /> shows &quot;B ⟶ A&quot;<br /> using assms by simp<br /> <br /> lemma K: &quot;A ⟶ (B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;B ⟶ A&quot; by (rule Aux)<br /> qed<br /> <br /> <br /> lemma S: &quot;(A ⟶ (B ⟶ C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1: &quot;(A ⟶ (B ⟶ C))&quot;<br /> show &quot;(A ⟶ B) ⟶ (A ⟶ C)&quot;<br /> proof<br /> assume 2: &quot;A ⟶ B&quot;<br /> show &quot;A ⟶ C&quot;<br /> proof<br /> assume 3: &quot;A&quot;<br /> from 2 and 3 have 4: &quot;B&quot; by (rule mp)<br /> from 1 and 3 have 5: &quot;B ⟶ C&quot; by (rule mp)<br /> from 5 and 4 show 6: &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> lemma &quot;(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1: &quot;A ⟶ B&quot;<br /> show &quot;(B ⟶ C) ⟶ (A ⟶ C)&quot;<br /> proof<br /> assume 2: &quot;B ⟶ C&quot;<br /> show &quot;A ⟶ C&quot;<br /> proof<br /> assume 3: &quot;A&quot;<br /> from 1 and 3 have 4: &quot;B&quot; by (rule mp)<br /> from 2 and 4 show 5: &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> <br /> lemma &quot;¬¬A ⟶ A&quot;<br /> proof<br /> assume 1: &quot;¬¬A&quot;<br /> from 1 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> lemma &quot;A ⟶ ¬¬A&quot;<br /> proof <br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;¬¬A&quot; by (rule contrapos_pn)<br /> qed<br /> <br /> lemma MT:<br /> assumes 1: &quot;F ⟶ G&quot; and<br /> 2: &quot;¬G&quot;<br /> shows &quot;¬F&quot;<br /> proof (rule notI)<br /> assume 3: &quot;F&quot;<br /> from 1 and 3 have 4: &quot;G&quot; by (rule mp)<br /> from 2 and 4 show False by (rule notE)<br /> qed<br /> <br /> lemma &quot;(¬A ⟶ B) ⟶ (¬B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;¬A ⟶ B&quot;<br /> show &quot;¬B ⟶ A&quot;<br /> proof<br /> assume 2: &quot;¬B&quot;<br /> from 1 and 2 have 3:&quot;¬¬A&quot; by (rule MT)<br /> from 3 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> qed<br /> <br /> lemma &quot;((A ⟶ B) ⟶ A) ⟶ A&quot;<br /> proof<br /> assume 1: &quot;(A ⟶ B) ⟶ A&quot;<br /> have 8: &quot;¬¬A&quot;<br /> proof (rule notI)<br /> assume 2: &quot;¬A&quot;<br /> have 3: &quot;A ⟶ B&quot;<br /> proof (rule impI)<br /> assume 4: &quot;A&quot;<br /> from 2 and 4 show &quot;B&quot; by (rule notE)<br /> qed<br /> from 1 and 3 have 5: &quot;A&quot; by (rule mp)<br /> from 2 and 5 show False by (rule notE)<br /> qed<br /> from 8 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> lemma &quot;A ∨ ¬A&quot;<br /> proof cases<br /> assume &quot;A&quot; thus ?thesis ..<br /> next<br /> assume &quot;¬A&quot; thus ?thesis ..<br /> qed<br /> <br /> <br /> lemma &quot;(¬(A ∧ B)) = (¬A ∨ ¬B)&quot;<br /> proof<br /> {<br /> assume 1: &quot;¬(A ∧ B)&quot;<br /> have 2: &quot;¬A ∨ A&quot; by (rule excluded_middle)<br /> thus &quot;¬A ∨ ¬B&quot;<br /> proof (rule disjE)<br /> { assume 3: &quot;¬A&quot;<br /> thus &quot;¬A ∨ ¬B&quot; by (rule disjI1) }<br /> next<br /> { assume 4: &quot;A&quot;<br /> have 5: &quot;¬B ∨ B&quot; by (rule excluded_middle)<br /> thus &quot;¬A ∨ ¬B&quot;<br /> proof (rule disjE)<br /> {assume 6: &quot;¬B&quot;<br /> thus &quot;¬A ∨ ¬B&quot; by (rule disjI2) }<br /> next<br /> {assume 7: &quot;B&quot;<br /> from 4 and 7 have 8: &quot;A ∧ B&quot; by (rule conjI)<br /> from 1 and 8 show &quot;¬A ∨ ¬B&quot; by (rule notE) }<br /> qed<br /> } <br /> qed<br /> }<br /> next<br /> { assume 1: &quot;¬A ∨ ¬B&quot;<br /> show &quot;¬(A ∧ B)&quot;<br /> proof (rule notI)<br /> assume 2: &quot;A ∧ B&quot;<br /> from 1 have 3:&quot;¬A ∨ ¬B&quot; .<br /> moreover<br /> { assume 4: &quot;¬A&quot;<br /> from 2 have 5:&quot;A&quot; by (rule conjunct1)<br /> from 4 and 5 have False by (rule notE) }<br /> moreover<br /> { assume 6: &quot;¬B&quot;<br /> from 2 have 7: &quot;B&quot; by (rule conjunct2)<br /> from 6 and 7 have False by (rule notE) }<br /> ultimately show False by (rule disjE) <br /> qed<br /> <br /> }<br /> qed<br /> <br /> end<br /> <br /> &lt;/source&gt;<br /> <br /> == RA_Relacion_2 ==<br /> <br /> theory Relacion_2<br /> imports Main <br /> begin<br /> <br /> (* Demostrar o refutar los siguientes lemas usando sólo las reglas básicas<br /> de deducción natural de la lógica proposicional y de la lógica de <br /> predicados (allI, allE, exI y exE). *)<br /> <br /> <br /> lemma &quot;(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)&quot;<br /> proof (rule impI)<br /> assume 1: &quot;(∃x. ∀y. P x y)&quot;<br /> show &quot;(∀y. ∃x. P x y)&quot;<br /> proof (rule allI)<br /> from 1 obtain x where 2: &quot;∀y. P x y&quot; by (rule exE)<br /> fix y<br /> from 2 have 3: &quot;P x y&quot; by (rule allE)<br /> from 3 show 4: &quot;∃x. P x y&quot; by (rule exI)<br /> qed<br /> qed<br /> <br /> lemma &quot;(∀x. P x ⟶ Q) = ((∃x. P x) ⟶ Q)&quot;<br /> proof<br /> { assume 1: &quot;∀x. P x ⟶ Q&quot;<br /> show &quot;(∃x. P x) ⟶ Q&quot;<br /> proof<br /> assume &quot;∃x. P x&quot;<br /> then obtain a where &quot;P a&quot; by (rule exE)<br /> have &quot;P a ⟶ Q&quot; using 1 by (rule allE)<br /> thus &quot;Q&quot; using `P a` by (rule mp)<br /> qed }<br /> next<br /> { assume 2: &quot;(∃x. P x) ⟶ Q&quot;<br /> show &quot;∀x. P x ⟶ Q&quot;<br /> proof<br /> fix a<br /> show &quot;P a ⟶ Q&quot;<br /> proof<br /> assume &quot;P a&quot;<br /> hence 3: &quot;∃x. P x&quot; by (rule exI)<br /> from 2 and 3 show &quot;Q&quot; by (rule mp)<br /> qed<br /> qed }<br /> qed<br /> <br /> lemma &quot;((∀ x. P x) ∧ (∀ x. Q x)) = (∀ x. (P x ∧ Q x))&quot;<br /> proof<br /> {<br /> assume 1: &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> show &quot;(∀ x. (P x ∧ Q x))&quot;<br /> proof (rule allI)<br /> from 1 have 2: &quot;(∀ x. P x)&quot; by (rule conjE)<br /> from 1 have 3: &quot;(∀ x. Q x)&quot; by (rule conjE)<br /> fix a<br /> from 2 have 4: &quot;P a&quot; by (rule allE)<br /> from 3 have 5: &quot;Q a&quot; by (rule allE)<br /> from 4 and 5 show 6: &quot;(P a) ∧ (Q a)&quot; by (rule conjI)<br /> qed<br /> }<br /> next<br /> {<br /> assume 7: &quot;∀ x. (P x ∧ Q x)&quot;<br /> show &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> proof<br /> fix a<br /> from 7 have 8: &quot;P a ∧ Q a&quot; by (rule allE)<br /> from 8 have 9: &quot;P a&quot; by (rule conjE)<br /> (* seguir... *)<br /> from 8 have 11: &quot;Q a&quot; by (rule allE)<br /> (* seguir... *)<br /> qed<br /> }<br /> qed <br /> <br /> oops<br /> <br /> lemma &quot;((∀ x. P x) ∨ (∀ x. Q x)) = (∀ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;((∃ x. P x) ∨ (∃ x. Q x)) = (∃ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)&quot;<br /> oops<br /> (*<br /> proof (rule impI)<br /> assume 1: &quot;∀x. ∃y. P x y&quot;<br /> show &quot;∃y. ∀x. P x y&quot;<br /> proof (rule exI)<br /> show &quot; ∀x. P x y&quot;<br /> proof (rule allI)<br /> fix x<br /> from 1 have 2: &quot;∃y. P x y&quot; by (rule allE)<br /> from 2 obtain y where 3: &quot;P x y&quot; by (rule exE)<br /> thus ?thesis ..<br /> qed<br /> *)<br /> <br /> lemma &quot;(¬ (∀ x. P x)) = (∃ x. ¬ P x)&quot;<br /> proof<br /> {<br /> assume 1: &quot;¬ (∀ x. P x)&quot; <br /> show &quot;∃ x. ¬ P x&quot;<br /> proof<br /> <br /> <br /> qed<br /> }<br /> next<br /> {<br /> assume 2: &quot;∃ x. ¬ P x&quot;<br /> show &quot;¬ (∀ x. P x)&quot;<br /> proof<br /> <br /> qed<br /> }<br /> qed<br /> <br /> end</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario:Jgiraldez&diff=68 2011-02-03T15:48:30Z

<p>Jgiraldez: /* Relaciones de Ejercicios resueltos: */</p> <hr /> <div>==RAZONAMIENTO AUTOMÁTICO 2010/11==<br /> Realizado por Jesús Giráldez Crú<br /> ----<br /> <br /> ===Relaciones de Ejercicios resueltos:===<br /> *Relación 1: COMPLETA! (Actualizado 03/02/2011)<br /> *Relación 2:</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario:Jgiraldez&diff=67 2011-02-03T15:41:18Z

<p>Jgiraldez: </p> <hr /> <div>==RAZONAMIENTO AUTOMÁTICO 2010/11==<br /> Realizado por Jesús Giráldez Crú<br /> ----<br /> <br /> ===Relaciones de Ejercicios resueltos:===<br /> Relación 1: COMPLETA! (Actualizado 03/02/2011)<br /> <br /> Relación 2:</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario:Jgiraldez&diff=66 2011-02-03T15:41:07Z

<p>Jgiraldez: </p> <hr /> <div>==RAZONAMIENTO AUTOMÁTICO 2010/11==<br /> Realizado por Jesús Giráldez Crú<br /> ----<br /> <br /> ===Relaciones de Ejercicios resueltos:===<br /> Relación 1: COMPLETA! (Actualizado 03/02/2011)<br /> Relación 2:</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario:Jgiraldez&diff=65 2011-02-03T15:40:41Z

<p>Jgiraldez: </p> <hr /> <div>==RAZONAMIENTO AUTOMÁTICO 2010/11==<br /> Realizado por Jesús Giráldez Crú<br /> ----<br /> <br /> <br /> <br /> <br /> <br /> ===Relaciones de Ejercicios resueltos:===<br /> <br /> Relación 1: COMPLETA! (Actualizado 03/02/2011)<br /> <br /> Relación 2:</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario:Jgiraldez&diff=64 2011-02-03T15:40:27Z

<p>Jgiraldez: </p> <hr /> <div>==RAZONAMIENTO AUTOMÁTICO 2010/11==<br /> Realizado por Jesús Giráldez Crú<br /> ----<br /> <br /> <br /> <br /> <br /> <br /> =Relaciones de Ejercicios resueltos:=<br /> <br /> Relación 1: COMPLETA! (Actualizado 03/02/2011)<br /> <br /> Relación 2:</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario_discusi%C3%B3n:Jgiraldez&diff=63 2011-02-03T15:37:27Z

<p>Jgiraldez: /* RA_Relacion_1 */</p> <hr /> <div>== RA_Relacion_1 ==<br /> <br /> theory Relacion_1<br /> imports Main <br /> begin<br /> <br /> (* Demostrar los siguientes lemas usando sólo las reglas básicas de deducción<br /> natural de la lógica proposicional. *)<br /> <br /> lemma I: &quot;A ⟶ A&quot;<br /> proof <br /> assume 1: &quot;A&quot;<br /> qed<br /> <br /> lemma 0: &quot;A ∧ B ⟶ B ∧ A&quot;<br /> proof <br /> assume 1: &quot;A ∧ B&quot;<br /> from 1 have 2: &quot;A&quot; by (rule conjunct1) <br /> from 1 have 3: &quot;B&quot; by (rule conjunct2)<br /> from 3 and 2 show &quot;B ∧ A&quot; by (rule conjI)<br /> qed<br /> <br /> lemma &quot;A ∧ B ⟶ A ∨ B&quot;<br /> proof<br /> assume 1: &quot;A ∧ B&quot;<br /> from 1 have 2: &quot;A&quot; by (rule conjunct1)<br /> from 1 have 3: &quot;B&quot; by (rule conjunct2)<br /> from 2 show &quot;A ∨ B&quot; by (rule disjI1)<br /> qed<br /> <br /> lemma &quot;(A ∨ B) ∨ C ⟶ A ∨ (B ∨ C)&quot;<br /> proof<br /> assume 1: &quot;(A ∨ B) ∨ C&quot;<br /> show &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 1<br /> moreover{<br /> assume 2: &quot;A ∨ B&quot;<br /> have &quot;A ∨ (B ∨ C)&quot;<br /> proof -<br /> note 2<br /> moreover{<br /> assume 3: &quot;A&quot;<br /> from 3 have 4:&quot;A ∨ (B ∨ C)&quot; by (rule disjI1)<br /> }<br /> moreover{<br /> assume 5:&quot;B&quot;<br /> from 5 have 6: &quot;B ∨ C&quot; by (rule disjI1)<br /> from 6 have 7: &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> }<br /> moreover{<br /> assume 8: &quot;C&quot;<br /> from 8 have 9: &quot;B ∨ C&quot; by (rule disjI2)<br /> from 9 have &quot;A ∨ (B ∨ C)&quot; by (rule disjI2)<br /> }<br /> ultimately show &quot;A ∨ (B ∨ C)&quot; by (rule disjE)<br /> qed<br /> qed<br /> <br /> lemma Aux:<br /> assumes &quot;A&quot;<br /> shows &quot;B ⟶ A&quot;<br /> using assms by simp<br /> <br /> lemma K: &quot;A ⟶ (B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;B ⟶ A&quot; by (rule Aux)<br /> qed<br /> <br /> <br /> lemma S: &quot;(A ⟶ (B ⟶ C)) ⟶ ((A ⟶ B) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1: &quot;(A ⟶ (B ⟶ C))&quot;<br /> show &quot;(A ⟶ B) ⟶ (A ⟶ C)&quot;<br /> proof<br /> assume 2: &quot;A ⟶ B&quot;<br /> show &quot;A ⟶ C&quot;<br /> proof<br /> assume 3: &quot;A&quot;<br /> from 2 and 3 have 4: &quot;B&quot; by (rule mp)<br /> from 1 and 3 have 5: &quot;B ⟶ C&quot; by (rule mp)<br /> from 5 and 4 show 6: &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> lemma &quot;(A ⟶ B) ⟶ ((B ⟶ C) ⟶ (A ⟶ C))&quot;<br /> proof<br /> assume 1: &quot;A ⟶ B&quot;<br /> show &quot;(B ⟶ C) ⟶ (A ⟶ C)&quot;<br /> proof<br /> assume 2: &quot;B ⟶ C&quot;<br /> show &quot;A ⟶ C&quot;<br /> proof<br /> assume 3: &quot;A&quot;<br /> from 1 and 3 have 4: &quot;B&quot; by (rule mp)<br /> from 2 and 4 show 5: &quot;C&quot; by (rule mp)<br /> qed<br /> qed<br /> qed<br /> <br /> <br /> lemma &quot;¬¬A ⟶ A&quot;<br /> proof<br /> assume 1: &quot;¬¬A&quot;<br /> from 1 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> lemma &quot;A ⟶ ¬¬A&quot;<br /> proof <br /> assume 1: &quot;A&quot;<br /> from 1 show &quot;¬¬A&quot; by (rule contrapos_pn)<br /> qed<br /> <br /> lemma MT:<br /> assumes 1: &quot;F ⟶ G&quot; and<br /> 2: &quot;¬G&quot;<br /> shows &quot;¬F&quot;<br /> proof (rule notI)<br /> assume 3: &quot;F&quot;<br /> from 1 and 3 have 4: &quot;G&quot; by (rule mp)<br /> from 2 and 4 show False by (rule notE)<br /> qed<br /> <br /> lemma &quot;(¬A ⟶ B) ⟶ (¬B ⟶ A)&quot;<br /> proof<br /> assume 1: &quot;¬A ⟶ B&quot;<br /> show &quot;¬B ⟶ A&quot;<br /> proof<br /> assume 2: &quot;¬B&quot;<br /> from 1 and 2 have 3:&quot;¬¬A&quot; by (rule MT)<br /> from 3 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> qed<br /> <br /> lemma &quot;((A ⟶ B) ⟶ A) ⟶ A&quot;<br /> proof<br /> assume 1: &quot;(A ⟶ B) ⟶ A&quot;<br /> have 8: &quot;¬¬A&quot;<br /> proof (rule notI)<br /> assume 2: &quot;¬A&quot;<br /> have 3: &quot;A ⟶ B&quot;<br /> proof (rule impI)<br /> assume 4: &quot;A&quot;<br /> from 2 and 4 show &quot;B&quot; by (rule notE)<br /> qed<br /> from 1 and 3 have 5: &quot;A&quot; by (rule mp)<br /> from 2 and 5 show False by (rule notE)<br /> qed<br /> from 8 show &quot;A&quot; by (rule notnotD)<br /> qed<br /> <br /> lemma &quot;A ∨ ¬A&quot;<br /> proof cases<br /> assume &quot;A&quot; thus ?thesis ..<br /> next<br /> assume &quot;¬A&quot; thus ?thesis ..<br /> qed<br /> <br /> <br /> lemma &quot;(¬(A ∧ B)) = (¬A ∨ ¬B)&quot;<br /> proof<br /> {<br /> assume 1: &quot;¬(A ∧ B)&quot;<br /> have 2: &quot;¬A ∨ A&quot; by (rule excluded_middle)<br /> thus &quot;¬A ∨ ¬B&quot;<br /> proof (rule disjE)<br /> { assume 3: &quot;¬A&quot;<br /> thus &quot;¬A ∨ ¬B&quot; by (rule disjI1) }<br /> next<br /> { assume 4: &quot;A&quot;<br /> have 5: &quot;¬B ∨ B&quot; by (rule excluded_middle)<br /> thus &quot;¬A ∨ ¬B&quot;<br /> proof (rule disjE)<br /> {assume 6: &quot;¬B&quot;<br /> thus &quot;¬A ∨ ¬B&quot; by (rule disjI2) }<br /> next<br /> {assume 7: &quot;B&quot;<br /> from 4 and 7 have 8: &quot;A ∧ B&quot; by (rule conjI)<br /> from 1 and 8 show &quot;¬A ∨ ¬B&quot; by (rule notE) }<br /> qed<br /> } <br /> qed<br /> }<br /> next<br /> { assume 1: &quot;¬A ∨ ¬B&quot;<br /> show &quot;¬(A ∧ B)&quot;<br /> proof (rule notI)<br /> assume 2: &quot;A ∧ B&quot;<br /> from 1 have 3:&quot;¬A ∨ ¬B&quot; .<br /> moreover<br /> { assume 4: &quot;¬A&quot;<br /> from 2 have 5:&quot;A&quot; by (rule conjunct1)<br /> from 4 and 5 have False by (rule notE) }<br /> moreover<br /> { assume 6: &quot;¬B&quot;<br /> from 2 have 7: &quot;B&quot; by (rule conjunct2)<br /> from 6 and 7 have False by (rule notE) }<br /> ultimately show False by (rule disjE) <br /> qed<br /> <br /> }<br /> qed<br /> <br /> end<br /> <br /> == RA_Relacion_2 ==<br /> <br /> theory Relacion_2<br /> imports Main <br /> begin<br /> <br /> (* Demostrar o refutar los siguientes lemas usando sólo las reglas básicas<br /> de deducción natural de la lógica proposicional y de la lógica de <br /> predicados (allI, allE, exI y exE). *)<br /> <br /> <br /> lemma &quot;(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)&quot;<br /> proof (rule impI)<br /> assume 1: &quot;(∃x. ∀y. P x y)&quot;<br /> show &quot;(∀y. ∃x. P x y)&quot;<br /> proof (rule allI)<br /> from 1 obtain x where 2: &quot;∀y. P x y&quot; by (rule exE)<br /> fix y<br /> from 2 have 3: &quot;P x y&quot; by (rule allE)<br /> from 3 show 4: &quot;∃x. P x y&quot; by (rule exI)<br /> qed<br /> qed<br /> <br /> lemma &quot;(∀x. P x ⟶ Q) = ((∃x. P x) ⟶ Q)&quot;<br /> proof<br /> { assume 1: &quot;∀x. P x ⟶ Q&quot;<br /> show &quot;(∃x. P x) ⟶ Q&quot;<br /> proof<br /> assume &quot;∃x. P x&quot;<br /> then obtain a where &quot;P a&quot; by (rule exE)<br /> have &quot;P a ⟶ Q&quot; using 1 by (rule allE)<br /> thus &quot;Q&quot; using `P a` by (rule mp)<br /> qed }<br /> next<br /> { assume 2: &quot;(∃x. P x) ⟶ Q&quot;<br /> show &quot;∀x. P x ⟶ Q&quot;<br /> proof<br /> fix a<br /> show &quot;P a ⟶ Q&quot;<br /> proof<br /> assume &quot;P a&quot;<br /> hence 3: &quot;∃x. P x&quot; by (rule exI)<br /> from 2 and 3 show &quot;Q&quot; by (rule mp)<br /> qed<br /> qed }<br /> qed<br /> <br /> lemma &quot;((∀ x. P x) ∧ (∀ x. Q x)) = (∀ x. (P x ∧ Q x))&quot;<br /> proof<br /> {<br /> assume 1: &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> show &quot;(∀ x. (P x ∧ Q x))&quot;<br /> proof (rule allI)<br /> from 1 have 2: &quot;(∀ x. P x)&quot; by (rule conjE)<br /> from 1 have 3: &quot;(∀ x. Q x)&quot; by (rule conjE)<br /> fix a<br /> from 2 have 4: &quot;P a&quot; by (rule allE)<br /> from 3 have 5: &quot;Q a&quot; by (rule allE)<br /> from 4 and 5 show 6: &quot;(P a) ∧ (Q a)&quot; by (rule conjI)<br /> qed<br /> }<br /> next<br /> {<br /> assume 7: &quot;∀ x. (P x ∧ Q x)&quot;<br /> show &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> proof<br /> fix a<br /> from 7 have 8: &quot;P a ∧ Q a&quot; by (rule allE)<br /> from 8 have 9: &quot;P a&quot; by (rule conjE)<br /> (* seguir... *)<br /> from 8 have 11: &quot;Q a&quot; by (rule allE)<br /> (* seguir... *)<br /> qed<br /> }<br /> qed <br /> <br /> oops<br /> <br /> lemma &quot;((∀ x. P x) ∨ (∀ x. Q x)) = (∀ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;((∃ x. P x) ∨ (∃ x. Q x)) = (∃ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)&quot;<br /> oops<br /> (*<br /> proof (rule impI)<br /> assume 1: &quot;∀x. ∃y. P x y&quot;<br /> show &quot;∃y. ∀x. P x y&quot;<br /> proof (rule exI)<br /> show &quot; ∀x. P x y&quot;<br /> proof (rule allI)<br /> fix x<br /> from 1 have 2: &quot;∃y. P x y&quot; by (rule allE)<br /> from 2 obtain y where 3: &quot;P x y&quot; by (rule exE)<br /> thus ?thesis ..<br /> qed<br /> *)<br /> <br /> lemma &quot;(¬ (∀ x. P x)) = (∃ x. ¬ P x)&quot;<br /> proof<br /> {<br /> assume 1: &quot;¬ (∀ x. P x)&quot; <br /> show &quot;∃ x. ¬ P x&quot;<br /> proof<br /> <br /> <br /> qed<br /> }<br /> next<br /> {<br /> assume 2: &quot;∃ x. ¬ P x&quot;<br /> show &quot;¬ (∀ x. P x)&quot;<br /> proof<br /> <br /> qed<br /> }<br /> qed<br /> <br /> end</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario_discusi%C3%B3n:Jgiraldez&diff=62 2011-02-03T15:36:55Z

<p>Jgiraldez: Nueva sección: /* RA_Relacion_2 */</p> <hr /> <div>== RA_Relacion_1 ==<br /> <br /> theory Relacion_2<br /> imports Main <br /> begin<br /> <br /> (* Demostrar o refutar los siguientes lemas usando sólo las reglas básicas<br /> de deducción natural de la lógica proposicional y de la lógica de <br /> predicados (allI, allE, exI y exE). *)<br /> <br /> <br /> lemma &quot;(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)&quot;<br /> proof (rule impI)<br /> assume 1: &quot;(∃x. ∀y. P x y)&quot;<br /> show &quot;(∀y. ∃x. P x y)&quot;<br /> proof (rule allI)<br /> from 1 obtain x where 2: &quot;∀y. P x y&quot; by (rule exE)<br /> fix y<br /> from 2 have 3: &quot;P x y&quot; by (rule allE)<br /> from 3 show 4: &quot;∃x. P x y&quot; by (rule exI)<br /> qed<br /> qed<br /> <br /> lemma &quot;(∀x. P x ⟶ Q) = ((∃x. P x) ⟶ Q)&quot;<br /> proof<br /> { assume 1: &quot;∀x. P x ⟶ Q&quot;<br /> show &quot;(∃x. P x) ⟶ Q&quot;<br /> proof<br /> assume &quot;∃x. P x&quot;<br /> then obtain a where &quot;P a&quot; by (rule exE)<br /> have &quot;P a ⟶ Q&quot; using 1 by (rule allE)<br /> thus &quot;Q&quot; using `P a` by (rule mp)<br /> qed }<br /> next<br /> { assume 2: &quot;(∃x. P x) ⟶ Q&quot;<br /> show &quot;∀x. P x ⟶ Q&quot;<br /> proof<br /> fix a<br /> show &quot;P a ⟶ Q&quot;<br /> proof<br /> assume &quot;P a&quot;<br /> hence 3: &quot;∃x. P x&quot; by (rule exI)<br /> from 2 and 3 show &quot;Q&quot; by (rule mp)<br /> qed<br /> qed }<br /> qed<br /> <br /> lemma &quot;((∀ x. P x) ∧ (∀ x. Q x)) = (∀ x. (P x ∧ Q x))&quot;<br /> proof<br /> {<br /> assume 1: &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> show &quot;(∀ x. (P x ∧ Q x))&quot;<br /> proof (rule allI)<br /> from 1 have 2: &quot;(∀ x. P x)&quot; by (rule conjE)<br /> from 1 have 3: &quot;(∀ x. Q x)&quot; by (rule conjE)<br /> fix a<br /> from 2 have 4: &quot;P a&quot; by (rule allE)<br /> from 3 have 5: &quot;Q a&quot; by (rule allE)<br /> from 4 and 5 show 6: &quot;(P a) ∧ (Q a)&quot; by (rule conjI)<br /> qed<br /> }<br /> next<br /> {<br /> assume 7: &quot;∀ x. (P x ∧ Q x)&quot;<br /> show &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> proof<br /> fix a<br /> from 7 have 8: &quot;P a ∧ Q a&quot; by (rule allE)<br /> from 8 have 9: &quot;P a&quot; by (rule conjE)<br /> (* seguir... *)<br /> from 8 have 11: &quot;Q a&quot; by (rule allE)<br /> (* seguir... *)<br /> qed<br /> }<br /> qed <br /> <br /> oops<br /> <br /> lemma &quot;((∀ x. P x) ∨ (∀ x. Q x)) = (∀ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;((∃ x. P x) ∨ (∃ x. Q x)) = (∃ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)&quot;<br /> oops<br /> (*<br /> proof (rule impI)<br /> assume 1: &quot;∀x. ∃y. P x y&quot;<br /> show &quot;∃y. ∀x. P x y&quot;<br /> proof (rule exI)<br /> show &quot; ∀x. P x y&quot;<br /> proof (rule allI)<br /> fix x<br /> from 1 have 2: &quot;∃y. P x y&quot; by (rule allE)<br /> from 2 obtain y where 3: &quot;P x y&quot; by (rule exE)<br /> thus ?thesis ..<br /> qed<br /> *)<br /> <br /> lemma &quot;(¬ (∀ x. P x)) = (∃ x. ¬ P x)&quot;<br /> proof<br /> {<br /> assume 1: &quot;¬ (∀ x. P x)&quot; <br /> show &quot;∃ x. ¬ P x&quot;<br /> proof<br /> <br /> <br /> qed<br /> }<br /> next<br /> {<br /> assume 2: &quot;∃ x. ¬ P x&quot;<br /> show &quot;¬ (∀ x. P x)&quot;<br /> proof<br /> <br /> qed<br /> }<br /> qed<br /> <br /> end<br /> <br /> == RA_Relacion_2 ==<br /> <br /> theory Relacion_2<br /> imports Main <br /> begin<br /> <br /> (* Demostrar o refutar los siguientes lemas usando sólo las reglas básicas<br /> de deducción natural de la lógica proposicional y de la lógica de <br /> predicados (allI, allE, exI y exE). *)<br /> <br /> <br /> lemma &quot;(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)&quot;<br /> proof (rule impI)<br /> assume 1: &quot;(∃x. ∀y. P x y)&quot;<br /> show &quot;(∀y. ∃x. P x y)&quot;<br /> proof (rule allI)<br /> from 1 obtain x where 2: &quot;∀y. P x y&quot; by (rule exE)<br /> fix y<br /> from 2 have 3: &quot;P x y&quot; by (rule allE)<br /> from 3 show 4: &quot;∃x. P x y&quot; by (rule exI)<br /> qed<br /> qed<br /> <br /> lemma &quot;(∀x. P x ⟶ Q) = ((∃x. P x) ⟶ Q)&quot;<br /> proof<br /> { assume 1: &quot;∀x. P x ⟶ Q&quot;<br /> show &quot;(∃x. P x) ⟶ Q&quot;<br /> proof<br /> assume &quot;∃x. P x&quot;<br /> then obtain a where &quot;P a&quot; by (rule exE)<br /> have &quot;P a ⟶ Q&quot; using 1 by (rule allE)<br /> thus &quot;Q&quot; using `P a` by (rule mp)<br /> qed }<br /> next<br /> { assume 2: &quot;(∃x. P x) ⟶ Q&quot;<br /> show &quot;∀x. P x ⟶ Q&quot;<br /> proof<br /> fix a<br /> show &quot;P a ⟶ Q&quot;<br /> proof<br /> assume &quot;P a&quot;<br /> hence 3: &quot;∃x. P x&quot; by (rule exI)<br /> from 2 and 3 show &quot;Q&quot; by (rule mp)<br /> qed<br /> qed }<br /> qed<br /> <br /> lemma &quot;((∀ x. P x) ∧ (∀ x. Q x)) = (∀ x. (P x ∧ Q x))&quot;<br /> proof<br /> {<br /> assume 1: &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> show &quot;(∀ x. (P x ∧ Q x))&quot;<br /> proof (rule allI)<br /> from 1 have 2: &quot;(∀ x. P x)&quot; by (rule conjE)<br /> from 1 have 3: &quot;(∀ x. Q x)&quot; by (rule conjE)<br /> fix a<br /> from 2 have 4: &quot;P a&quot; by (rule allE)<br /> from 3 have 5: &quot;Q a&quot; by (rule allE)<br /> from 4 and 5 show 6: &quot;(P a) ∧ (Q a)&quot; by (rule conjI)<br /> qed<br /> }<br /> next<br /> {<br /> assume 7: &quot;∀ x. (P x ∧ Q x)&quot;<br /> show &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> proof<br /> fix a<br /> from 7 have 8: &quot;P a ∧ Q a&quot; by (rule allE)<br /> from 8 have 9: &quot;P a&quot; by (rule conjE)<br /> (* seguir... *)<br /> from 8 have 11: &quot;Q a&quot; by (rule allE)<br /> (* seguir... *)<br /> qed<br /> }<br /> qed <br /> <br /> oops<br /> <br /> lemma &quot;((∀ x. P x) ∨ (∀ x. Q x)) = (∀ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;((∃ x. P x) ∨ (∃ x. Q x)) = (∃ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)&quot;<br /> oops<br /> (*<br /> proof (rule impI)<br /> assume 1: &quot;∀x. ∃y. P x y&quot;<br /> show &quot;∃y. ∀x. P x y&quot;<br /> proof (rule exI)<br /> show &quot; ∀x. P x y&quot;<br /> proof (rule allI)<br /> fix x<br /> from 1 have 2: &quot;∃y. P x y&quot; by (rule allE)<br /> from 2 obtain y where 3: &quot;P x y&quot; by (rule exE)<br /> thus ?thesis ..<br /> qed<br /> *)<br /> <br /> lemma &quot;(¬ (∀ x. P x)) = (∃ x. ¬ P x)&quot;<br /> proof<br /> {<br /> assume 1: &quot;¬ (∀ x. P x)&quot; <br /> show &quot;∃ x. ¬ P x&quot;<br /> proof<br /> <br /> <br /> qed<br /> }<br /> next<br /> {<br /> assume 2: &quot;∃ x. ¬ P x&quot;<br /> show &quot;¬ (∀ x. P x)&quot;<br /> proof<br /> <br /> qed<br /> }<br /> qed<br /> <br /> end</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario_discusi%C3%B3n:Jgiraldez&diff=61 2011-02-03T15:35:55Z

<p>Jgiraldez: RA_Relacion_1</p> <hr /> <div>== RA_Relacion_1 ==<br /> <br /> theory Relacion_2<br /> imports Main <br /> begin<br /> <br /> (* Demostrar o refutar los siguientes lemas usando sólo las reglas básicas<br /> de deducción natural de la lógica proposicional y de la lógica de <br /> predicados (allI, allE, exI y exE). *)<br /> <br /> <br /> lemma &quot;(∃x. ∀y. P x y) ⟶ (∀y. ∃x. P x y)&quot;<br /> proof (rule impI)<br /> assume 1: &quot;(∃x. ∀y. P x y)&quot;<br /> show &quot;(∀y. ∃x. P x y)&quot;<br /> proof (rule allI)<br /> from 1 obtain x where 2: &quot;∀y. P x y&quot; by (rule exE)<br /> fix y<br /> from 2 have 3: &quot;P x y&quot; by (rule allE)<br /> from 3 show 4: &quot;∃x. P x y&quot; by (rule exI)<br /> qed<br /> qed<br /> <br /> lemma &quot;(∀x. P x ⟶ Q) = ((∃x. P x) ⟶ Q)&quot;<br /> proof<br /> { assume 1: &quot;∀x. P x ⟶ Q&quot;<br /> show &quot;(∃x. P x) ⟶ Q&quot;<br /> proof<br /> assume &quot;∃x. P x&quot;<br /> then obtain a where &quot;P a&quot; by (rule exE)<br /> have &quot;P a ⟶ Q&quot; using 1 by (rule allE)<br /> thus &quot;Q&quot; using `P a` by (rule mp)<br /> qed }<br /> next<br /> { assume 2: &quot;(∃x. P x) ⟶ Q&quot;<br /> show &quot;∀x. P x ⟶ Q&quot;<br /> proof<br /> fix a<br /> show &quot;P a ⟶ Q&quot;<br /> proof<br /> assume &quot;P a&quot;<br /> hence 3: &quot;∃x. P x&quot; by (rule exI)<br /> from 2 and 3 show &quot;Q&quot; by (rule mp)<br /> qed<br /> qed }<br /> qed<br /> <br /> lemma &quot;((∀ x. P x) ∧ (∀ x. Q x)) = (∀ x. (P x ∧ Q x))&quot;<br /> proof<br /> {<br /> assume 1: &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> show &quot;(∀ x. (P x ∧ Q x))&quot;<br /> proof (rule allI)<br /> from 1 have 2: &quot;(∀ x. P x)&quot; by (rule conjE)<br /> from 1 have 3: &quot;(∀ x. Q x)&quot; by (rule conjE)<br /> fix a<br /> from 2 have 4: &quot;P a&quot; by (rule allE)<br /> from 3 have 5: &quot;Q a&quot; by (rule allE)<br /> from 4 and 5 show 6: &quot;(P a) ∧ (Q a)&quot; by (rule conjI)<br /> qed<br /> }<br /> next<br /> {<br /> assume 7: &quot;∀ x. (P x ∧ Q x)&quot;<br /> show &quot;(∀ x. P x) ∧ (∀ x. Q x)&quot;<br /> proof<br /> fix a<br /> from 7 have 8: &quot;P a ∧ Q a&quot; by (rule allE)<br /> from 8 have 9: &quot;P a&quot; by (rule conjE)<br /> (* seguir... *)<br /> from 8 have 11: &quot;Q a&quot; by (rule allE)<br /> (* seguir... *)<br /> qed<br /> }<br /> qed <br /> <br /> oops<br /> <br /> lemma &quot;((∀ x. P x) ∨ (∀ x. Q x)) = (∀ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;((∃ x. P x) ∨ (∃ x. Q x)) = (∃ x. (P x ∨ Q x))&quot;<br /> oops<br /> <br /> lemma &quot;(∀x. ∃y. P x y) ⟶ (∃y. ∀x. P x y)&quot;<br /> oops<br /> (*<br /> proof (rule impI)<br /> assume 1: &quot;∀x. ∃y. P x y&quot;<br /> show &quot;∃y. ∀x. P x y&quot;<br /> proof (rule exI)<br /> show &quot; ∀x. P x y&quot;<br /> proof (rule allI)<br /> fix x<br /> from 1 have 2: &quot;∃y. P x y&quot; by (rule allE)<br /> from 2 obtain y where 3: &quot;P x y&quot; by (rule exE)<br /> thus ?thesis ..<br /> qed<br /> *)<br /> <br /> lemma &quot;(¬ (∀ x. P x)) = (∃ x. ¬ P x)&quot;<br /> proof<br /> {<br /> assume 1: &quot;¬ (∀ x. P x)&quot; <br /> show &quot;∃ x. ¬ P x&quot;<br /> proof<br /> <br /> <br /> qed<br /> }<br /> next<br /> {<br /> assume 2: &quot;∃ x. ¬ P x&quot;<br /> show &quot;¬ (∀ x. P x)&quot;<br /> proof<br /> <br /> qed<br /> }<br /> qed<br /> <br /> end</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario:Jgiraldez&diff=60 2011-02-03T15:33:04Z

<p>Jgiraldez: </p> <hr /> <div>RAZONAMIENTO AUTOMÁTICO 2010/11<br /> <br /> Realizado por Jesús Giráldez Crú<br /> <br /> <br /> Relaciones de Ejercicios resueltos:<br /> <br /> Relación 1: COMPLETA! (Actualizado 03/02/2011)<br /> <br /> Relación 2:</div>

Jgiraldez https://www.glc.us.es/~jalonso/RA2010/index.php?title=Usuario:Jgiraldez&diff=59 2011-02-03T15:30:16Z

<p>Jgiraldez: Página creada con &#039;RAZONAMIENTO AUTOMÁTICO 2010/11 Realizado por Jesús Giráldez Crú Relaciones de Ejercicios resueltos: Relación 1: COMPLETA! (Actualizado 03/02/2011) Relación 2:&#039;</p> <hr /> <div>RAZONAMIENTO AUTOMÁTICO 2010/11<br /> Realizado por Jesús Giráldez Crú<br /> <br /> <br /> Relaciones de Ejercicios resueltos:<br /> Relación 1: COMPLETA! (Actualizado 03/02/2011)<br /> Relación 2:</div>

Jgiraldez