Relación 9

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Relación 9

2017-01-20T11:29:48Z

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Relación 9

2017-01-20T11:28:14Z

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<source lang="isar">
chapter {* R5: Eliminación de duplicados *}

theory R5_Eliminacion_de_duplicados
imports Main
begin

text {*
---------------------------------------------------------------------
Ejercicio 1. Definir la funcion primitiva recursiva
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar bowma wilmorort pablucoto serrodcal anaprarod migtermor paupeddeg fraortmoy marpoldia1*)
fun estaEn :: "'a ⇒ 'a list ⇒ bool" where
"estaEn _ [] = False"
| "estaEn x (a#xs) = ((a = x) ∨ (estaEn x xs))"

value "estaEn (2::nat) [3,2,4] = True"
value "estaEn (1::nat) [3,2,4] = False"

(* ivamenjim, ferrenseg *)
(* Igual que la anterior pero con x en lugar de _ en el caso base *)

fun estaEn1 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn1 x [] = False"
| "estaEn1 x (a#xs) = ((x=a) ∨ estaEn1 x xs)"

value "estaEn1 (2::nat) [3,2,4] = True"
value "estaEn1 (1::nat) [3,2,4] = False"

(* wilmorort *)
(* reutilizando la funcion "algunos" de R4.thy*)
fun estaEn2 :: "'a ⇒ 'a list ⇒ bool" where
"estaEn2 a xs = algunos (λx. x = a) xs"

value "estaEn2 (2::nat) [3,2,4] = True"
value "estaEn2 (1::nat) [3,2,4] = False"

text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función primitiva recursiva
sinDuplicados :: 'a list ⇒ bool
tal que (sinDuplicados xs) se verifica si la lista xs no contiene
duplicados. Por ejemplo,
sinDuplicados [1::nat,4,2] = True
sinDuplicados [1::nat,4,2,4] = False
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar ivamenjim wilmorort bowma pablucoto serrodcal anaprarod migtermor paupeddeg fraortmoy marpoldia1 ferrenseg *)
fun sinDuplicados :: "'a list ⇒ bool" where
"sinDuplicados [] = True"
| "sinDuplicados (x#xs) = (¬ estaEn x xs ∧ sinDuplicados xs)"

value "sinDuplicados [1::nat,4,2] = True"
value "sinDuplicados [1::nat,4,2,4] = False"

(* paupeddeg *)
(* Utilizando la función ∉ de Isabelle *)
fun sinDuplicados2 :: "'a list ⇒ bool" where
"sinDuplicados2 [] = True"
| "sinDuplicados2 (a#xs) = ((a ∉ set xs) ∧ sinDuplicados2 xs ) "

text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función primitiva recursiva
borraDuplicados :: 'a list ⇒ bool
tal que (borraDuplicados xs) es la lista obtenida eliminando los
elementos duplicados de la lista xs. Por ejemplo,
borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]

Nota: La función borraDuplicados es equivalente a la predefinida
remdups.
---------------------------------------------------------------------
*}

(* crigomgom rubgonmar wilmorort bowma pablucoto serrodcal anaprarod migtermor paupeddeg fraortmoy marpoldia1 ferrenseg*)
fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) =( if estaEn x xs then borraDuplicados xs else x#borraDuplicados xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

(* ivamenjim *)
(* Utilizando la negación primero *)

fun borraDuplicados :: "'a list ⇒ 'a list" where
"borraDuplicados [] = []"
| "borraDuplicados (x#xs) = (if ¬(estaEn x xs) then (x#(borraDuplicados xs)) else borraDuplicados xs)"

(* rubgonmar *)
(* Otra forma Sin usar if
Utilizando case aunque se le sacaría más partido con más de 2 casos *)

fun borraDuplicados1 :: "'a list ⇒ 'a list" where
"borraDuplicados1 [] = []" |
"borraDuplicados1 (x#xs) = ( case estaEn x xs of False => x#borraDuplicados1 xs | True => borraDuplicados1 xs )"

(* rubgonmar *)
(*Otra forma utilizando let*)
fun borraDuplicados2 :: "'a list ⇒ 'a list" where
"borraDuplicados2 [] = []" |
"borraDuplicados2 (x#xs) = (let condicion = estaEn x xs::bool in
if condicion then borraDuplicados2 xs else x # borraDuplicados2 xs)"

value "borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]"

text {*
---------------------------------------------------------------------
Ejercicio 4.1. Demostrar o refutar automáticamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(*crigomgom anaprarod ferrenseg*)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs, simp_all)

(* rubgonmar wilmorort pablucoto serrodcal migtermor paupeddeg fraortmoy marpoldia1*)
lemma length_borraDuplicados:
"length ( borraDuplicados xs ) ≤ length xs"
by (induct xs) auto

(* ivamenjim *)
(* Demostrando objetivo a objetivo *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply simp
apply auto
done

(* bowma anaprarod *)
lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
apply (induct xs)
apply (simp, simp) (* creo que es mejor poner aquí simp_all *)
done

(* ferrenseg *)

lemma length_borraDuplicados:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs) simp_all (* Creo que se puede poner simp_all fuera de parentesis *)

text {*
---------------------------------------------------------------------
Ejercicio 4.2. Demostrar o refutar detalladamente
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"
(* crigomgom *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix x xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
show "length (borraDuplicados (x#xs)) ≤ length (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "length (borraDuplicados (x#xs)) = length (borraDuplicados xs)" by simp
also have "... ≤ length xs" using HI by simp
also have "... ≤ length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
next
assume "(¬ estaEn x xs)"
then have "length (borraDuplicados (x#xs)) = length (x#borraDuplicados xs)" by simp
also have "... = 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... = length (x#xs)" by simp
finally show "length (borraDuplicados (x#xs)) ≤ length (x#xs)" by simp
qed
qed

(* ivamenjim wilmorort ferrenseg *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* serrodcal anaprarod *)
lemma length_borraDuplicados_2: "length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a # xs)) ≤ 1+length (borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
finally show "?P (a # xs)" by simp
qed

(* pablucoto *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof(induct xs)
show "length (borraDuplicados []) ≤ length [] " by simp
next
fix a xs
assume HI: " length (borraDuplicados xs) ≤ length xs "
have "length (borraDuplicados (a # xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs) " by simp
qed

(* bowma *)
lemma length_borraDuplicados_3:
"length (borraDuplicados xs) ≤ length xs" (is "?p xs")
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "?p xs"
have "length (a#xs) = 1 + length xs" by simp
have "length(borraDuplicados (a#xs)) ≤ 1 + length(borraDuplicados xs)" by simp
also have "... ≤ 1+length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show "?p (a # xs)" by simp
qed

(* migtermor *)

lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix a xs
assume HI: "?P xs"
have "length (borraDuplicados (a#xs)) ≤ (length (a#xs))"
proof (cases)
assume "(estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
next
assume "¬ (estaEn a xs)"
then have Aux: "length (borraDuplicados (a#xs)) = 1+ length (borraDuplicados xs)" by simp
also have "… ≤ length (a#xs)" using HI by simp
then show "length (borraDuplicados (a#xs)) ≤ (length (a#xs))" using Aux by simp
qed
then show "?P (a#xs)" by simp
qed

(* paupeddeg marpoldia1*)
lemma length_borraDuplicados_4:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length [a] + length (borraDuplicados xs) " by simp
also have "... ≤ 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

(* fraortmoy *)
(* muy parecida a alguna anterior, pero yo dí mas pasos *)
lemma length_borraDuplicados_2:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume H1: "length (borraDuplicados xs) ≤ length xs"
have "length (borraDuplicados (a # xs)) ≤ length(borraDuplicados [a])+length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length (borraDuplicados xs)" by simp
also have "… ≤ 1 + length xs" using H1 by simp
finally show "length (borraDuplicados (a # xs)) ≤ length (a # xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 5.1. Demostrar o refutar automáticamente
estaEn a (borraDuplicados xs) = estaEn a xs
---------------------------------------------------------------------
*}

-- "La demostración automática es"
(* crigomgom rubgonmar wilmorort pablucoto serrodcal bowma migtermor fraortmoy marpoldia1 ferrenseg *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs) auto

(* ivamenjim *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply auto
done

(* anaprarod *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs, simp_all, blast)

(* anaprarod *)
lemma estaEn_borraDuplicados':
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply (cases "estaEn x xs")
apply (simp_all)
apply blast
done

(* bowma *)
lemma estaEn_borraDuplicados:
"estaEn a (borraDuplicados xs) = estaEn a xs"
apply (induct xs)
apply simp
apply (simp, blast)
done

text {*
---------------------------------------------------------------------
Ejercicio 5.2. Demostrar o refutar detalladamente
estaEn a (borraDuplicados xs) = estaEn a xs
Nota: Para la demostración de la equivalencia se puede usar
proof (rule iffI)
La regla iffI es
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
---------------------------------------------------------------------
*}

-- "La demostración estructurada es"

(* wilmorort *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix b xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (b#xs)) = estaEn a (b#xs)"
proof (rule iffI)
assume H1: "estaEn a (borraDuplicados (b#xs))"
show "estaEn a (b#xs)"
proof (cases)
assume "estaEn b xs"
then have "estaEn a (borraDuplicados xs)" using H1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (b#xs)" by simp
next
assume "¬ estaEn b xs"
then have "estaEn a (b#(borraDuplicados xs))" using H1 by simp
then have "a=b ∨ (estaEn a (borraDuplicados xs))" by simp
then have " a=b ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (b#xs)" by simp
qed
next
assume H2: "estaEn a (b#xs)"
show "estaEn a (borraDuplicados (b#xs))"
proof (cases)
assume "a=b"
then have "estaEn b (borraDuplicados xs) = estaEn b xs" using HI by simp
then have "(estaEn b xs ⟶ estaEn b (borraDuplicados xs)) ∧
(¬ estaEn b xs ⟶ estaEn b (b # borraDuplicados xs))" by simp
then have "estaEn b (borraDuplicados (b#xs))" by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a=b` by simp
next
assume "a≠b"
then have "estaEn a (b#xs)" using H2 by simp
then have "a = b ∨ estaEn a xs" by simp
then have "False ∨ estaEn a xs " using `a≠b` by simp
then have "estaEn a xs" by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (b#xs))" using `a≠b` by simp
qed
qed
qed

(* anaprarod marpoldia1 ferrenseg *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "...= estaEn a xs" using HI by simp
finally show "?P (x#xs)" by auto
next
assume "¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "...= (x = a ∨ estaEn a (borraDuplicados xs))" by simp
finally show "?P (x#xs)" using HI by simp
qed
qed

(* migtermor *)

lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix aa xs
assume HI: "?P xs"
have P1: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)"
proof (cases)
assume C1: "(estaEn aa xs)"
have "estaEn a (borraDuplicados (aa#xs)) = estaEn a (borraDuplicados xs)"
using C1 by simp
also have P3: "… = estaEn a xs" using HI by simp
also have "… = estaEn a (aa#xs)"
proof (cases)
assume "(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" using C1 by simp
next
assume "¬(a=aa)"
then show "estaEn a xs = estaEn a (aa#xs)" by simp
qed
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P3 by simp
next
assume C2: "¬(estaEn aa xs)"
then show "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using HI by simp
qed
also have Conc: "estaEn a (borraDuplicados (aa#xs)) = estaEn a (aa#xs)" using P1 by simp
finally show "?P (aa#xs)" using Conc by simp
qed

(* crigomgom *)
-- "La demostración estructurada es"
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI)
assume a1: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then have "estaEn a (borraDuplicados xs)" using a1 by simp
then have "estaEn a xs" using HI by simp
then show "estaEn a (x#xs)" by simp
next
assume "¬ estaEn x xs"
then have "estaEn a (x#(borraDuplicados xs))" using a1 by simp
then have " x=a ∨ (estaEn a (borraDuplicados xs))" by simp
then have " x=a ∨ (estaEn a xs)" using HI by simp
then show "estaEn a (x#xs)" by simp
qed
next
assume a2: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by simp
next
assume b1: "a≠x"
then have "estaEn a (x#xs)" using a2 by simp
then have "x = a ∨ estaEn a xs" by simp
then have "estaEn a xs " using b1 by simp
then have "estaEn a (borraDuplicados xs)" using HI by simp
then show "estaEn a (borraDuplicados (x#xs))" using b1 by simp
qed
qed
qed

(* rubgonmar *)
lemma estaEn_borraDuplicados_2:
"estaEn a ( borraDuplicados xs ) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix x
fix xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#xs)"
proof (rule iffI) (* usamos proof de la regla dada iffI*)
assume cprim: "estaEn a (borraDuplicados (x#xs))"
show "estaEn a (x#xs)"
proof (cases)
assume "estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
next
assume "¬ estaEn x xs"
then show "estaEn a (x#xs)" using cprim HI by simp
qed
next
assume cseg: "estaEn a (x#xs)"
show "estaEn a (borraDuplicados (x#xs))"
proof (cases)
assume "a=x"
then show "estaEn a (borraDuplicados (x#xs))" using HI by auto
next
assume "a≠x"
thus "estaEn a (borraDuplicados (x#xs))" using `a≠x` cseg HI by simp
qed
qed
qed

(*
Aplico la regla iffI:
⟦P ⟹ Q ; Q ⟹ P⟧ ⟹ P = Q
Así:
[estaEn a (borraDuplicados (x # xs)) ⟹ estaEn a (x # xs); estaEn a (x # xs) ⟹ estaEn a (borraDuplicados (x # xs))]
⟹ estaEn a (borraDuplicados (x # xs)) = estaEn a (x # xs)
*)

(* bowma *)
lemma estaEn_borraDuplicados_2:
"estaEn a (borraDuplicados xs) = estaEn a xs" (is "?p xs")
proof (induct xs)
show "?p []" by simp
next
fix x xs
assume HI: "?p xs"
show "?p (x#xs)"
proof (cases)
assume H1:"estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (borraDuplicados xs)" by simp
also have "... = estaEn a xs" using HI by simp
also have "... = estaEn a (x#xs)"
proof(cases)
assume "x=a"
then show "estaEn a xs = estaEn a (x#xs)" using H1 by simp
next
assume "x≠a"
then show "estaEn a xs = estaEn a (x#xs)" by simp
qed
finally show "?p (x#xs)" by simp
next
assume H2:"¬estaEn x xs"
then have "estaEn a (borraDuplicados (x#xs)) = estaEn a (x#borraDuplicados xs)" by simp
also have "... = ((x=a) ∨ estaEn a (borraDuplicados xs))" by simp
also have "... = ((x=a) ∨ estaEn a xs)" using HI by simp
also have "... = estaEn a (x#xs)" by simp
finally show "?p (x#xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 6.1. Demostrar o refutar automáticamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

-- "La demostración automática"

(* ivamenjim wilmorort serrodcal crigomgom anaprarod fraortmoy marpoldia1*)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs) (auto simp add: estaEn_borraDuplicados)

(* migtermor *)
lemma sinDuplicados_borraDuplicados:
"sinDuplicados (borraDuplicados xs)"
by (induct xs, simp_all add: estaEn_borraDuplicados_2)

text {*
---------------------------------------------------------------------
Ejercicio 6.2. Demostrar o refutar detalladamente
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*wilmorort*)

-- "La demostración estructurada es"
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show " sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a#xs))" using HI by simp
next
assume"¬ estaEn a xs"
then have "¬ (estaEn a xs) ∧ sinDuplicados (borraDuplicados xs)" using HI by simp
then have "¬ estaEn a (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs)"
by (simp add: estaEn_borraDuplicados)
then have " sinDuplicados (a#borraDuplicados xs)" by simp
then show " sinDuplicados (borraDuplicados(a #xs))" by simp
qed
qed

(* ivamenjim migtermor crigomgom rubgonmar fraortmoy *)

lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show "sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a # xs))"
proof (cases)
assume "estaEn a xs"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by simp
next
assume "¬(estaEn a xs)"
then show "sinDuplicados (borraDuplicados (a # xs))" using HI by (simp add: estaEn_borraDuplicados)
qed
qed

(* anaprarod *)
lemma sinDuplicados_borraDuplicados_2:
"sinDuplicados (borraDuplicados xs)" (is "?P xs")
proof (induct xs)
show "?P []" by simp
next
fix x xs
assume HI: "?P xs"
show "?P (x#xs)"
proof (cases)
assume c1: "estaEn x xs"
then show "sinDuplicados (borraDuplicados (x#xs))" using HI by simp
next
assume c2: "¬ estaEn x xs"
then have "sinDuplicados (borraDuplicados (x#xs)) =sinDuplicados (x#borraDuplicados xs)" by simp
also have "…= (¬estaEn x (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs))" by simp
also have "… = (¬estaEn x (borraDuplicados xs))" using HI by simp
also have "… = (¬(estaEn x xs))" by (simp add:estaEn_borraDuplicados)
also have "… = True" using c2 by simp
finally show "?P (x#xs)" by simp
qed
qed

text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar o refutar:
borraDuplicados (rev xs) = rev (borraDuplicados xs)
---------------------------------------------------------------------
*}

(*crigomgom rubgonmar ivamenjim wilmorort pablucoto migtermor anaprarod fraortmoy ferrenseg *)
lemma "borraDuplicados (rev xs) = rev (borraDuplicados xs)"
quickcheck
oops

(* ivamenjim: Quickcheck encuentra el siguiente contraejemplo:
xs = [a1, a2, a1]
Por lo que:
· "borraDuplicados (rev xs) = [a2, a1]"
· "rev (borraDuplicados xs) = [a1, a2]" *)

end
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