Mjoseh en 15:56 26 jun 2019

2019-06-26T15:56:15Z

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<source lang = "isabelle">

chapter {* R11: Razonamiento sobre programas en Isabelle/HOL *}

theory R11_sol
imports Main
begin

text {* ---------------------------------------------------------------
En toda la relación de ejercicios las demostraciones han de realizarse
de las formas siguientes:
(*) detallada
(*) estructurada
(*) aplicativa
(*) automática
------------------------------------------------------------------ *}

text {* ---------------------------------------------------------------
Ejercicio 1. Definir la función
sumaPotenciasDeDosMasUno :: nat ⇒ nat
tal que
(sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n.
Por ejemplo,
sumaPotenciasDeDosMasUno 3 = 16
------------------------------------------------------------------ *}

fun sumaPotenciasDeDosMasUno :: "nat ⇒ nat" where
"sumaPotenciasDeDosMasUno 0 = 2"
| "sumaPotenciasDeDosMasUno (Suc n) =
sumaPotenciasDeDosMasUno n + 2^(n+1)"

value "sumaPotenciasDeDosMasUno 3" ― ‹= 16›

text {* ---------------------------------------------------------------
Ejercicio 2. Demostrar que
sumaPotenciasDeDosMasUno n = 2^(n+1)
------------------------------------------------------------------- *}

lemma sumaPotenciasDeDosMasUno_d:
"sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
show "sumaPotenciasDeDosMasUno 0 = 2^(0+1)" by simp
next
fix n
assume HI: "sumaPotenciasDeDosMasUno n = 2 ^ (n + 1)"
show "sumaPotenciasDeDosMasUno (Suc n) = 2 ^ (Suc n + 1)"
proof -
have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1)" by simp
also have "... = 2 ^ (n + 1) + 2^(n+1)" using HI by simp
also have "... = 2 ^ (Suc n + 1)" by simp
finally show ?thesis .
qed
qed

lemma sumaPotenciasDeDosMasUno_e:
"sumaPotenciasDeDosMasUno n = 2^(n+1)"
oops

lemma sumaPotenciasDeDosMasUno_a:
"sumaPotenciasDeDosMasUno n = 2^(n+1)"
apply (induct n)
apply simp_all
done

lemma sumaPotenciasDeDosMasUno_auto:
"sumaPotenciasDeDosMasUno n = 2^(n+1)"
by (induct n) simp_all

text {* ---------------------------------------------------------------
Ejercicio 3. Definir la función
copia :: nat ⇒ 'a ⇒ 'a list
tal que (copia n x) es la lista formado por n copias del elemento
x. Por ejemplo,
copia 3 x = [x,x,x]
------------------------------------------------------------------ *}

fun copia :: "nat ⇒ 'a ⇒ 'a list" where
"copia 0 x = []"
| "copia (Suc n) x = x # copia n x"

value "copia 3 x" ― ‹= [x,x,x]›

text {* ---------------------------------------------------------------
Ejercicio 4. Definir la función
todos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (todos p xs) se verifica si todos los elementos de xs cumplen
la propiedad p. Por ejemplo,
todos (λx. x>(1::nat)) [2,6,4] = True
todos (λx. x>(2::nat)) [2,6,4] = False
Nota: La conjunción se representa por ∧
----------------------------------------------------------------- *}

fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
"todos p [] = True"
| "todos p (x#xs) = (p x ∧ todos p xs)"

value "todos (λx. x>(1::nat)) [2,6,4]" ― ‹= True›
value "todos (λx. x>(2::nat)) [2,6,4]" ― ‹= False›

text {* ---------------------------------------------------------------
Ejercicio 5. Demostrar que todos los elementos de (copia n x) son
iguales a x.
------------------------------------------------------------------- *}

lemma todos_copia_d: "todos (λy. y=x) (copia n x)"
proof (induct n)
show "todos (λy. y = x) (copia 0 x)" by simp
next
fix n
assume HI: "todos (λy. y = x) (copia n x)"
have "todos (λy. y = x) (copia (Suc n) x) =
todos (λy. y = x) (x # copia n x)" by simp
also have "... = (x = x ∧ todos (λy. y = x) (copia n x))"
by simp
also have "... = True" using HI by simp
finally show "todos (λy. y = x) (copia (Suc n) x)" by simp
qed

lemma todos_copia_e: "todos (λy. y=x) (copia n x)"
oops

lemma todos_copia_a: "todos (λy. y=x) (copia n x)"
apply (induct n)
apply simp_all
done

lemma todos_copia_auto: "todos (λy. y=x) (copia n x)"
by (induct n) simp_all

text {* ---------------------------------------------------------------
Ejercicio 6. Definir la función
factR :: nat ⇒ nat
tal que (factR n) es el factorial de n. Por ejemplo,
factR 4 = 24
------------------------------------------------------------------ *}

fun factR :: "nat ⇒ nat" where
"factR 0 = 1"
| "factR (Suc n) = Suc n * factR n"

value "factR 4" ― ‹= 24›

text {* ---------------------------------------------------------------
Ejercicio 7. Se considera la siguiente definición iterativa de la
función factorial
factI :: "nat ⇒ nat" where
factI n = factI' n 1

factI' :: nat ⇒ nat ⇒ nat" where
factI' 0 x = x
factI' (Suc n) x = factI' n (Suc n)*x
Demostrar que, para todo n y todo x, se tiene
factI' n x = x * factR n
------------------------------------------------------------------- *}

fun factI' :: "nat ⇒ nat ⇒ nat" where
"factI' 0 x = x"
| "factI' (Suc n) x = factI' n (x* (Suc n))"

fun factI :: "nat ⇒ nat" where
"factI n = factI' n 1"

lemma fact_d: "factI' n x = x * factR n"
proof (induct n arbitrary: x)
show "⋀x. factI' 0 x = x * factR 0" by simp
next
fix n
assume HI: "⋀x. factI' n x = x * factR n"
show "⋀x. factI' (Suc n) x = x * factR (Suc n)"
proof -
fix x
have "factI' (Suc n) x = factI' n (x * (Suc n))" by simp
also have "... = (x * (Suc n)) * factR n" using HI by simp
also have "... = x * ((Suc n) * factR n)" by (simp only: mult.assoc)
also have "... = x * factR (Suc n)" by simp
finally show "factI' (Suc n) x = x * factR (Suc n)" by simp
qed
qed

lemma fact_e: "factI' n x = x * factR n"
oops

thm mult_Suc
thm mult.assoc

lemma fact_a: "factI' n x = x * factR n"
apply (induct n arbitrary: x)
apply simp
apply (simp del: mult_Suc)
done

lemma fact_auto: "factI' n x = x * factR n"
by (induct n arbitrary: x) (auto simp del: mult_Suc)

text {* ---------------------------------------------------------------
Ejercicio 8. Demostrar que
factI n = factR n
------------------------------------------------------------------- *}

corollary fact_equiv: "factI n = factR n"
by (simp add: fact_a)

text {* ---------------------------------------------------------------
Ejercicio 9. Definir, recursivamente y sin usar (@), la función
amplia :: 'a list ⇒ 'a ⇒ 'a list
tal que (amplia xs y) es la lista obtenida añadiendo el elemento y al
final de la lista xs. Por ejemplo,
amplia [d,a] t = [d,a,t]
------------------------------------------------------------------ *}

fun amplia :: "'a list ⇒ 'a ⇒ 'a list" where
"amplia [] y = [y]"
| "amplia (x#xs) y = x # amplia xs y"

value "amplia [d,a] t" ― ‹= [d,a,t]›

text {* ---------------------------------------------------------------
Ejercicio 10. Demostrar que
amplia xs y = xs @ [y]
------------------------------------------------------------------- *}

lemma amplia_append_d: "amplia xs y = xs @ [y]"
proof (induct xs)
show "amplia [] y = [] @ [y]" by simp
next
fix x xs
assume HI: "amplia xs y = xs @ [y]"
have "amplia (x # xs) y = x # amplia xs y" by simp
also have "... = x # (xs @ [y])" using HI by simp
also have "... = (x # xs) @ [y]" by simp
finally show "amplia (x # xs) y = (x # xs) @ [y]" by simp
qed

lemma amplia_append_e: "amplia xs y = xs @ [y]"
oops

lemma amplia_append_a: "amplia xs y = xs @ [y]"
apply (induct xs)
apply simp_all
done

lemma amplia_append_auto: "amplia xs y = xs @ [y]"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 11. Definir la función
algunos :: ('a ⇒ bool) ⇒ 'a list ⇒ bool
tal que (algunos p xs) se verifica si algunos elementos de la lista
xs cumplen la propiedad p. Por ejemplo, se verifica
algunos (λx. 1<length x) [[2,1,4],[3]]
¬algunos (λx. 1<length x) [[],[3]]"

Nota: La función algunos es equivalente a la predefinida list_ex.
---------------------------------------------------------------------
*}

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

text {*
---------------------------------------------------------------------
Ejercicio 12. Demostrar o refutar
todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)
---------------------------------------------------------------------
*}

lemma conjCommuta4: "a ∧ b ∧ c ∧ d ⟷ a ∧ c ∧ b ∧ d"
apply (rule iffI)
apply (rule conjI)
apply (erule conjunct1)
apply (rule conjI)
apply (drule conjunct2)
apply (drule conjunct2)
apply (erule conjunct1)
apply (drule conjunct2)
apply (frule conjunct1)
apply (drule conjunct2)+
apply (rule conjI, assumption+)
apply auto
done

lemma todos_conj_d: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
oops

lemma todos_conj_e: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
proof (induct xs)
show "todos (λx. P x ∧ Q x) [] = (todos P [] ∧ todos Q [])" by simp
next
fix a xs
assume HI: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
thus "todos (λx. P x ∧ Q x) (a # xs) = (todos P (a # xs) ∧ todos Q (a # xs))" by auto
qed

thm mult_Suc
thm mult.assoc

lemma todos_conj_a: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
apply (induct xs)
apply simp
apply simp
apply (simp add: conjCommuta4)
done

lemma todos_conj_auto: "todos (λx. P x ∧ Q x) xs = (todos P xs ∧ todos Q xs)"
by (induct xs) auto

text {*
---------------------------------------------------------------------
Ejercicio 13. Demostrar o refutar
todos P (x @ y) = (todos P x ∧ todos P y)
---------------------------------------------------------------------
*}

lemma todos_append_d: "todos P (xs @ ys) = (todos P xs ∧ todos P ys)"
proof (induct xs)
show "todos P ([] @ ys) = (todos P [] ∧ todos P ys)" by simp
next
fix a xs
assume HI:"todos P (xs @ ys) = (todos P xs ∧ todos P ys)"
show "todos P ((a # xs) @ ys) = (todos P (a # xs) ∧ todos P ys)"
proof -
have "todos P ((a#xs) @ ys) = ((P a) ∧ todos P (xs@ys))" by simp
also have "... = ((P a) ∧ todos P xs ∧ todos P ys)" using HI by simp
also have "... = (todos P (a#xs) ∧ todos P ys)" by simp
finally show ?thesis .
qed
qed

lemma todos_append_e: "todos P (xs @ ys) = (todos P xs ∧ todos P ys)"
proof (induct xs)
show "todos P ([] @ ys) = (todos P [] ∧ todos P ys)" by simp
next
fix a xs
assume "todos P (xs @ ys) = (todos P xs ∧ todos P ys)"
thus "todos P ((a # xs) @ ys) = (todos P (a # xs) ∧ todos P ys)" by auto
qed

lemma todos_append_a: "todos P (xs @ ys) = (todos P xs ∧ todos P ys)"
apply (induct xs)
apply simp_all
done

lemma todos_append_auto: "todos P (x @ y) = (todos P x ∧ todos P y)"
by (induct x) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 14. Demostrar o refutar
todos P (rev xs) = todos P xs
---------------------------------------------------------------------
*}

lemma todos_rev_d: "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume HI:"todos P (rev xs) = todos P xs"
show "todos P (rev (a # xs)) = todos P (a # xs)"
proof -
have "todos P (rev (a#xs)) = todos P ((rev xs)@[a])" by simp
also have "... = (todos P (rev xs) ∧ todos P [a])"
by (simp add: todos_append_a)
also have "... = (todos P xs ∧ todos P [a])" using HI by simp
also have "... = (todos P [a] ∧ todos P xs)" by auto
also have "... = (P a ∧ todos P xs)" by simp
also have "... = todos P (a#xs)" by simp
finally show ?thesis .
qed
qed

lemma todos_rev_e: "todos P (rev xs) = todos P xs"
proof (induct xs)
show "todos P (rev []) = todos P []" by simp
next
fix a xs
assume "todos P (rev xs) = todos P xs"
thus "todos P (rev (a#xs)) = todos P (a#xs)"
by (auto simp add: todos_append_a)
qed

lemma todos_rev_a: "todos P (rev xs) = todos P xs"
oops

lemma todos_rev_auto: "todos P (rev xs) = todos P xs"
by (induct xs) (auto simp add: todos_append_a)

text {*
---------------------------------------------------------------------
Ejercicio 15. Demostrar o refutar:
algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)
---------------------------------------------------------------------
*}

lemma "algunos (λx. P x ∧ Q x) xs = (algunos P xs ∧ algunos Q xs)"
oops

text {*
---------------------------------------------------------------------
Ejercicio 16. Demostrar o refutar
algunos P (map f xs) = algunos (P ∘ f) xs
---------------------------------------------------------------------
*}

lemma algunos_map_d: "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P o f) []" by simp
next
fix a xs
assume HI: "algunos P (map f xs) = algunos (P o f) xs"
show "algunos P (map f (a#xs)) = algunos (P o f) (a#xs)"
proof -
have "algunos P (map f (a#xs)) = algunos P ((f a)#(map f xs))" by simp
also have "... = ((P (f a)) ∨ (algunos P (map f xs)))" by simp
also have "... = (((P o f) a) ∨ (algunos (P o f) xs))" using HI by simp
also have "... = algunos (P o f) (a#xs)" by simp
finally show ?thesis .
qed
qed

lemma algunos_map_e: "algunos P (map f xs) = algunos (P o f) xs"
proof (induct xs)
show "algunos P (map f []) = algunos (P o f) []" by simp
next
fix a xs
assume "algunos P (map f xs) = algunos (P o f) xs"
thus "algunos P (map f (a#xs)) = algunos (P o f) (a#xs)" by auto
qed

thm algunos.simps

lemma algunos_map_a: "algunos P (map f xs) = algunos (P o f) xs"
apply (induct xs)
apply simp_all
done

thm conj_commute

lemma algunos_map_auto: "algunos P (map f xs) = algunos (P o f) xs"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 17. Demostrar o refutar
algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)
---------------------------------------------------------------------
*}

lemma algunos_append_d: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume HI: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
show "algunos P ((a#xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)"
proof -
have "algunos P ((a#xs) @ ys) = algunos P (a#(xs @ ys))" by simp
also have "... = ((P a) ∨ algunos P (xs @ ys))" by simp
also have "... = ((P a) ∨ algunos P xs ∨ algunos P ys)" using HI by simp
also have "... = (algunos P (a#xs) ∨ algunos P ys)" by simp
finally show ?thesis .
qed
qed

thm Fun.comp_apply

lemma algunos_append_e: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
proof (induct xs)
show "algunos P ([] @ ys) = (algunos P [] ∨ algunos P ys)" by simp
next
fix a xs
assume "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
thus "algunos P ((a#xs) @ ys) = (algunos P (a#xs) ∨ algunos P ys)"
by auto
qed

lemma algunos_append_a: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
oops

lemma algunos_append_auto: "algunos P (xs @ ys) = (algunos P xs ∨ algunos P ys)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 18. Demostrar o refutar
algunos P (rev xs) = algunos P xs
---------------------------------------------------------------------
*}

lemma algunos_rev_d: "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix a xs
assume HI: "algunos P (rev xs) = algunos P xs"
show "algunos P (rev (a#xs)) = algunos P (a#xs)"
proof -
have "algunos P (rev (a#xs)) = algunos P ((rev xs) @ [a])" by simp
also have "... = (algunos P (rev xs) ∨ algunos P [a])"
by (simp add: algunos_append_auto)
also have "... = (algunos P xs ∨ algunos P [a])" using HI by simp
also have "... = (algunos P xs ∨ P a)" by simp
also have "... = (P a ∨ algunos P xs)" by (rule disj_commute)
also have "... = algunos P (a#xs)" by simp
finally show ?thesis .
qed
qed

lemma algunos_rev_e: "algunos P (rev xs) = algunos P xs"
proof (induct xs)
show "algunos P (rev []) = algunos P []" by simp
next
fix a xs
assume "algunos P (rev xs) = algunos P xs"
thus "algunos P (rev (a#xs)) = algunos P (a#xs)"
by (auto simp add: algunos_append_auto)
qed

lemma algunos_rev_a: "algunos P (rev xs) = algunos P xs"
apply (induct xs)
apply simp
apply (simp add: algunos_append_auto)
apply (simp add: disj_commute)
done

lemma algunos_rev_auto: "algunos P (rev xs) = algunos P xs"
by (induct xs) (auto simp add: algunos_append_auto)

text {*
---------------------------------------------------------------------
Ejercicio 19. Encontrar un término no trivial Z tal que sea cierta la
siguiente ecuación:
algunos (λx. P x ∨ Q x) xs = Z
y demostrar la equivalencia de forma automática y detallada.
---------------------------------------------------------------------
*}

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
by (induct xs) auto

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨ algunos Q xs)"
apply (induct xs)
apply simp
apply simp
by auto

lemma "algunos (λx. P x ∨ Q x) xs = (algunos P xs ∨algunos Q xs)"
proof (induct xs)
show "algunos (λx. P x ∨ Q x) [] = (algunos P []∨ algunos Q [])" by simp
next
fix a xs
assume "algunos (λx. (P x ∨ Q x)) xs = (algunos P xs ∨ algunos Q xs)"
thus "algunos (λx. P x ∨ Q x) (a#xs) = (algunos P(a#xs) ∨ algunos Q (a#xs))"
by auto
qed

text {*
---------------------------------------------------------------------
Ejercicio 20. Demostrar o refutar
algunos P xs = (¬ todos (λx. (¬ P x)) xs)
---------------------------------------------------------------------
*}

lemma algunos_no_todos_d: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
proof (induct xs)
show "algunos P [] = (¬ todos (λx. ¬ P x) [])" by simp
next
fix a xs
assume HI: "algunos P xs = (¬ todos (λx. ¬ P x) xs)"
show "algunos P (a # xs) = (¬ todos (λx. ¬ P x) (a # xs))"
proof -
have "algunos P (a # xs) = ((P a) ∨ (algunos P xs))" by simp
also have "... = ((P a) ∨ (¬ todos (λx. ¬ P x) xs))" using HI by simp
also have "... = (¬ (¬ (P a) ∧ todos (λx. (¬ P x)) xs))" by simp
also have "... = (¬ todos (λx. (¬ P x)) (a#xs))" by simp
finally show ?thesis .
qed
qed

lemma algunos_no_todos_e: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
oops

lemma algunos_no_todos_a: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
oops

lemma algunos_no_todos_auto: "algunos P xs = (¬ todos (λx. (¬ P x)) xs)"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 21. Definir la función
estaEn :: 'a ⇒ 'a list ⇒ bool
tal que (estaEn x xs) se verifica si el elemento x está en la lista
xs. Por ejemplo,
estaEn (2::nat) [3,2,4] = True
estaEn (1::nat) [3,2,4] = False
---------------------------------------------------------------------
*}

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

text {*
---------------------------------------------------------------------
Ejercicio 22. Expresar la relación existente entre estaEn y algunos.
Demostrar dicha relación de forma automática y detallada.
---------------------------------------------------------------------
*}

lemma estaEn_algunos_auto:
"estaEn y xs = algunos (λx. y=x) xs"
by (induct xs) simp_all

lemma estaEn_algunos_a:
"estaEn y xs = algunos (λx. y=x) xs"
apply (induct xs)
apply simp
apply simp
done

lemma estaEn_algunos_e:
"estaEn y xs = algunos (λ x. y=x) xs"
proof (induct xs)
show "estaEn y [] = algunos (λ x. y=x) []" by simp
next
fix a xs
assume "estaEn y xs = algunos (λ x. y=x) xs"
thus "estaEn y (a#xs) = algunos (λ x. y=x) (a#xs)" by simp
qed

text {*
---------------------------------------------------------------------
Ejercicio 23. Definir la función
sinDuplicados :: 'a list ⇒ bool
tal que (sinDuplicados xs) se verifica si la lista xs no contiene
duplicados. Por ejemplo,
sinDuplicados [1::nat,4,2] = True
sinDuplicados [1::nat,4,2,4] = False
---------------------------------------------------------------------
*}

fun sinDuplicados :: "'a list ⇒ bool" where
"sinDuplicados [] = True"
| "sinDuplicados (a#xs) = ((¬ estaEn a xs) ∧ sinDuplicados xs)"

text {*
---------------------------------------------------------------------
Ejercicio 24. Definir la función
borraDuplicados :: 'a list ⇒ bool
tal que (borraDuplicados xs) es la lista obtenida eliminando los
elementos duplicados de la lista xs. Por ejemplo,
borraDuplicados [1::nat,2,4,2,3] = [1,4,2,3]

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

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

text {*
---------------------------------------------------------------------
Ejercicio 25. Demostrar o refutar
length (borraDuplicados xs) ≤ length xs
---------------------------------------------------------------------
*}

lemma length_borraDuplicados_d:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
show "length (borraDuplicados (a#xs)) ≤ length (a#xs)"
proof (cases)
assume "estaEn a xs"
hence "length (borraDuplicados (a#xs)) = length (borraDuplicados xs)"
by simp
also have "... ≤ length xs" using HI by simp
also have "... ≤ length (a#xs)" by simp
finally show ?thesis .
next
assume "(¬ estaEn a xs)"
hence "length (borraDuplicados (a#xs)) = length (a#(borraDuplicados xs))"
by simp
also have "... = 1 + length (borraDuplicados xs)" by simp
also have "... ≤ 1 + length xs" using HI by simp
also have "... = length (a#xs)" by simp
finally show ?thesis .
qed
qed

lemma length_borraDuplicados_e:
"length (borraDuplicados xs) ≤ length xs"
proof (induct xs)
show "length (borraDuplicados []) ≤ length []" by simp
next
fix a xs
assume HI: "length (borraDuplicados xs) ≤ length xs"
thus "length (borraDuplicados (a#xs)) ≤ length (a#xs)"
proof (cases)
assume "estaEn a xs"
thus "length (borraDuplicados (a#xs)) ≤ length (a#xs)"
using HI by auto
next
assume "(¬ estaEn a xs)"
thus "length (borraDuplicados (a#xs)) ≤ length (a#xs)"
using HI by auto
qed
qed

lemma length_borraDuplicados_a:
"length (borraDuplicados xs) ≤ length xs"
oops

lemma length_borraDuplicados_auto:
"length (borraDuplicados xs) ≤ length xs"
by (induct xs) simp_all

text {*
---------------------------------------------------------------------
Ejercicio 26. Demostrar o refutar
estaEn a (borraDuplicados xs) = estaEn a xs
---------------------------------------------------------------------
*}

lemma estaEn_borraDuplicados_d:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix b xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (b#xs)) = estaEn a (b#xs)"
proof (rule iffI)
assume c1: "estaEn a (borraDuplicados (b#xs))"
show "estaEn a (b#xs)"
proof (cases)
assume "estaEn b xs"
thus "estaEn a (b#xs)" using c1 HI by auto
next
assume "¬ estaEn b xs"
thus "estaEn a (b#xs)" using c1 HI by auto
qed
next
assume c2: "estaEn a (b#xs)"
show "estaEn a (borraDuplicados (b#xs))"
proof (cases)
assume "a=b"
thus "estaEn a (borraDuplicados (b#xs))" using HI by auto
next
assume "a ≠ b"
thus "estaEn a (borraDuplicados (b#xs))" using `a ≠ b` c2 HI by auto
qed
qed
qed

lemma estaEn_borraDuplicados_e:
"estaEn a (borraDuplicados xs) = estaEn a xs"
proof (induct xs)
show "estaEn a (borraDuplicados []) = estaEn a []" by simp
next
fix b xs
assume HI: "estaEn a (borraDuplicados xs) = estaEn a xs"
show "estaEn a (borraDuplicados (b#xs)) = estaEn a (b#xs)"
proof (rule iffI)
assume c1: "estaEn a (borraDuplicados (b#xs))"
show "estaEn a (b#xs)"
proof (cases)
assume "estaEn b xs"
thus "estaEn a (b#xs)" using c1 HI by auto
next
assume "¬ estaEn b xs"
thus "estaEn a (b#xs)" using c1 HI by auto
qed
next
assume c2: "estaEn a (b#xs)"
show "estaEn a (borraDuplicados (b#xs))"
proof (cases)
assume "a=b"
thus "estaEn a (borraDuplicados (b#xs))" using HI by auto
next
assume "a ≠ b"
thus "estaEn a (borraDuplicados (b#xs))" using `a ≠ b` c2 HI by auto
qed
qed
qed

lemma estaEn_borraDuplicados_a:
"estaEn a (borraDuplicados xs) = estaEn a xs"
oops

lemma estaEn_borraDuplicados_auto:
"estaEn a (borraDuplicados xs) = estaEn a xs"
by (induct xs) auto

text {*
---------------------------------------------------------------------
Ejercicio 27. Demostrar o refutar
sinDuplicados (borraDuplicados xs)
---------------------------------------------------------------------
*}

lemma sinDuplicados_borraDuplicados_d:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show "sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a#xs))"
proof (cases)
assume "estaEn a xs"
thus "sinDuplicados (borraDuplicados (a#xs))" using HI by simp
next
assume "¬ estaEn a xs"
hence "¬ (estaEn a xs) ∧ sinDuplicados (borraDuplicados xs)"
using HI by simp
hence "¬ estaEn a (borraDuplicados xs) ∧ sinDuplicados (borraDuplicados xs)"
by (simp add: estaEn_borraDuplicados_auto)
hence "sinDuplicados (a#borraDuplicados xs)" by simp
thus "sinDuplicados (borraDuplicados (a#xs))"
using `¬ estaEn a xs` by simp
qed
qed

lemma sinDuplicados_borraDuplicados_e:
"sinDuplicados (borraDuplicados xs)"
proof (induct xs)
show "sinDuplicados (borraDuplicados [])" by simp
next
fix a xs
assume HI: "sinDuplicados (borraDuplicados xs)"
show "sinDuplicados (borraDuplicados (a#xs))"
proof (cases)
assume "estaEn a xs"
thus "sinDuplicados (borraDuplicados (a#xs))" using HI by simp
next
assume "¬ estaEn a xs"
thus "sinDuplicados (borraDuplicados (a#xs))"
using `¬ estaEn a xs` HI by (auto simp add: estaEn_borraDuplicados_auto)
qed
qed

lemma sinDuplicados_borraDuplicados_a:
"sinDuplicados (borraDuplicados xs)"
oops

lemma sinDuplicados_borraDuplicados_auto:
"sinDuplicados (borraDuplicados xs)"
by (induct xs) (auto simp add: estaEn_borraDuplicados_auto)

text {*
---------------------------------------------------------------------
Ejercicio 28. Demostrar o refutar:
borraDuplicados (rev xs) = rev (borraDuplicados xs)
---------------------------------------------------------------------
*}

lemma "borraDuplicados (rev xs) = rev (borraDuplicados xs)"
oops

end

</source>

Sol 11 - Historial de revisiones

Mjoseh en 15:59 26 jun 2019

Mjoseh en 15:56 26 jun 2019

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