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Diferencia entre revisiones de «Tema 15»

De Lógica matemática y fundamentos (2012-13)

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 +
<source lang = "Isar">
 +
header {* Tema 15: Razonamiento sobre programas en Isabelle *}
  
 +
theory T15
 +
imports Main
 +
begin
 +
 +
text {*
 +
  En este tema se demuestra con Isabelle las propiedades de los
 +
  programas funcionales como se expone en el tema 8 del curso
 +
  "Informática" que puede leerse en
 +
  http://www.cs.us.es/~jalonso/cursos/i1m/temas/tema-8t.pdf
 +
*}
 +
 +
section {* Razonamiento ecuacional *}
 +
 +
text {* ----------------------------------------------------------------
 +
  Ejercicio 1. Definir, por recursión, la función
 +
    longitud :: "'a list ⇒ nat" where
 +
  tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
 +
    longitud [4,2,5] = 3
 +
  ------------------------------------------------------------------- *}
 +
 +
fun longitud :: "'a list ⇒ nat" where
 +
  "longitud []    = 0"
 +
| "longitud (x#xs) = 1 + longitud xs"
 +
 
 +
value "longitud [4,2,5]" -- "= 3"
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 2. Demostrar que
 +
    longitud [4,2,5] = 3
 +
  ------------------------------------------------------------------- *}
 +
 +
lemma "longitud [4,2,5] = 3"
 +
by simp
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 3. Definir la función
 +
    fun intercambia :: "'a × 'b ⇒ 'b × 'a"
 +
  tal que (intercambia p) es el par obtenido intercambiando las
 +
  componentes del par p. Por ejemplo,
 +
    "intercambia (2,3) = (3,2)
 +
  ------------------------------------------------------------------ *}
 +
 +
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
 +
  "intercambia (x,y) = (y,x)"
 +
 +
value "intercambia (2,3)" -- "= (3,2)"
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 4. Demostrar que
 +
    intercambia (intercambia (x,y)) = (x,y)
 +
  ------------------------------------------------------------------- *}
 +
 +
lemma "intercambia (intercambia (x,y)) = (x,y)"
 +
by simp
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 5. Definir, por recursión, la función
 +
    inversa :: "'a list ⇒ 'a list"
 +
  tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
 +
  elementos de xs. Por ejemplo,
 +
    inversa [3,2,5] = [5,2,3]
 +
  ------------------------------------------------------------------ *}
 +
 +
fun inversa :: "'a list ⇒ 'a list" where
 +
  "inversa [] = []"
 +
| "inversa (x#xs) = inversa xs @ [x]"
 +
 +
value "inversa [3,2,5]" -- "= [5,2,3]"
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 6. Demostrar que
 +
    inversa [x] = [x]
 +
  ------------------------------------------------------------------- *}
 +
 +
lemma "inversa [x] = [x]"
 +
by simp
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 7. Definir la función
 +
    repite :: "nat ⇒ 'a ⇒ 'a list" where
 +
  tal que . Por ejemplo,
 +
    repite 3 5 = [5,5,5]
 +
  ------------------------------------------------------------------ *}
 +
 +
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
 +
  "repite 0 x = []"
 +
| "repite (Suc n) x = x # (repite n x)"
 +
 +
value "repite 3 5" -- "= [5,5,5]"
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 8. Demostrar que
 +
    longitud (repite n x) = n
 +
  ------------------------------------------------------------------- *}
 +
 +
lemma "longitud (repite n x) = n"
 +
apply (induct_tac n)
 +
apply (auto)
 +
done
 +
 +
 +
lemma "longitud (repite n x) = n"
 +
by (induct n) auto
 +
 +
lemma longitud_repite:
 +
  "longitud (repite n x) = n"
 +
proof (induct n)
 +
  show "longitud (repite 0 x) = 0" by simp
 +
next
 +
  fix n
 +
  assume hi: "longitud (repite n x) = n"
 +
  show "longitud (repite (Suc n) x) = Suc n"
 +
  proof -
 +
    have "longitud (repite (Suc n) x) = longitud (x # (repite n x))" by simp
 +
    also have "... = 1 +  longitud (repite n x)" by simp
 +
    also have "... = 1 + n" using hi by simp
 +
    also have "... = Suc n" by simp
 +
    finally show ?thesis .
 +
  qed
 +
qed
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 9. Definir la función
 +
    fun conc :: "'a list ⇒ 'a list ⇒ 'a list"
 +
  tal que . Por ejemplo,
 +
    conc [2,3] [4,3,5] = [2,3,4,3,5]
 +
  ------------------------------------------------------------------ *}
 +
 +
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
 +
  "conc []    ys = ys"
 +
| "conc (x#xs) ys = x # (conc xs ys)"
 +
 +
value "conc [2,3] [4,3,5]" -- "= [2,3,4,3,5]"
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 10. Demostrar que
 +
    conc xs (conc ys zs) = (conc xs ys) zs
 +
  ------------------------------------------------------------------- *}
 +
 +
lemma "conc xs (conc ys zs) = conc (conc xs ys) zs"
 +
by (induct xs) auto
 +
 +
lemma conc_asociativa:
 +
  "conc xs (conc ys zs) = conc (conc xs ys) zs"
 +
proof (induct xs)
 +
  show "conc [] (conc ys zs) = conc (conc [] ys) zs" by simp
 +
next
 +
  fix x xs
 +
  assume hi: "conc xs (conc ys zs) = conc (conc xs ys) zs"
 +
  show "conc (x # xs) (conc ys zs) = conc (conc (x # xs) ys) zs"
 +
  proof -
 +
    have "conc (x # xs) (conc ys zs) = x # (conc xs (conc ys zs))" by simp
 +
    also have "... = x # (conc (conc xs ys) zs)" using hi by simp
 +
    also have "... = conc (x#(conc xs ys)) zs" by simp
 +
    also have "... = conc (conc (x # xs) ys) zs" by simp
 +
    finally show ?thesis .
 +
  qed
 +
qed
 +
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 11. Refutar que
 +
    conc xs ys = conc ys xs
 +
  ------------------------------------------------------------------- *}
 +
 +
lemma "conc xs ys = conc ys xs"
 +
quickcheck
 +
oops
 +
 +
text {* Encuentra el contraejemplo,
 +
  xs = [a\<^isub>2]
 +
  ys = [a\<^isub>1] *}
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 12. Demostrar que
 +
    conc xs [] = xs
 +
  ------------------------------------------------------------------- *}
 +
 +
lemma "conc xs [] = xs"
 +
by (induct xs) auto
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 13. Demostrar que
 +
    longitud (conc xs ys) = longitud xs + longitud ys
 +
  ------------------------------------------------------------------- *}
 +
 +
lemma "longitud (conc xs ys) = longitud xs + longitud ys"
 +
by (induct xs) auto
 +
 +
lemma long_conc:
 +
  "longitud (conc xs ys) = longitud xs + longitud ys"
 +
proof (induct xs)
 +
  show "longitud (conc [] ys) = longitud [] + longitud ys" by simp
 +
next
 +
  fix x xs
 +
  assume hi: "longitud (conc xs ys) = longitud xs + longitud ys"
 +
  show "longitud (conc (x # xs) ys) = longitud (x # xs) + longitud ys"
 +
  proof -
 +
    have "longitud (conc (x # xs) ys) = longitud (x # (conc xs ys))" by simp
 +
    also have "... = 1 + longitud (conc xs ys)" by simp
 +
    also have "... = 1 + (longitud xs + longitud ys)" using hi by simp
 +
    also have "... = (1+ longitud xs) + longitud ys" by simp
 +
    also have "... = longitud (x # xs) + longitud ys" by simp
 +
    finally show ?thesis .
 +
  qed
 +
qed
 +
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 14. Definir la función
 +
    coge :: "nat ⇒ 'a list ⇒ 'a list"
 +
  tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
 +
  ejemplo,
 +
    coge 2 [3,7,5,4] = [3,7]
 +
  ------------------------------------------------------------------ *}
 +
 +
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
 +
  "coge n [] = []"
 +
| "coge 0 xs = []"
 +
| "coge (Suc n) (x#xs) = x # (coge n xs)"
 +
 +
value "coge 2 [3,7,5,4]" -- "= [3,7]"
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 15. Definir la función
 +
    elimina :: "nat ⇒ 'a list ⇒ 'a list"
 +
  tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
 +
  ejemplo,
 +
    elimina 2 [3,7,5,4] = [5,4]
 +
  ------------------------------------------------------------------ *}
 +
 +
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
 +
  "elimina n [] = []"
 +
| "elimina 0 xs = xs"
 +
| "elimina (Suc n) (x#xs) = elimina n xs"
 +
 +
value "elimina 2 [3,7,5,4]" -- "= [5,4]"
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 16. Demostrar que
 +
    conc (coge n xs) (elimina n xs) = xs
 +
  ------------------------------------------------------------------- *}
 +
 +
lemma "conc (coge n xs) (elimina n xs) = xs"
 +
by (induct rule: coge.induct) auto
 +
 +
text {* coge.induct es el esquema de inducción asociado a la definición
 +
  de la función coge. Puede verse como sigue: *}
 +
 +
thm coge.induct
 +
 +
lemma "conc (coge n xs) (elimina n xs) = xs"
 +
by (induct rule: elimina.induct) auto
 +
 +
thm elimina.induct
 +
 +
lemma conc_coge_elimina:
 +
  "conc (coge n xs) (elimina n xs) = xs"
 +
proof (induct rule: coge.induct)
 +
  show "⋀n. conc (coge n []) (elimina n []) = []" by simp
 +
next
 +
  show "⋀v va. conc (coge 0 (v # va)) (elimina 0 (v # va)) = v # va" by simp
 +
next
 +
  fix x xs n
 +
  assume hi: "conc (coge n xs) (elimina n xs) = xs"
 +
  show "conc (coge (Suc n) (x # xs)) (elimina (Suc n) (x # xs)) = x # xs"
 +
  proof -
 +
    have "conc (coge (Suc n) (x # xs)) (elimina (Suc n) (x # xs)) = conc (coge (Suc n) (x # xs)) (elimina n xs)" by simp
 +
    also have "... = conc (x# (coge n xs)) (elimina n xs)" by simp
 +
    also have "... = x#(conc (coge n xs) (elimina n xs))" by simp
 +
    also have "... = (x#xs)" using hi by simp
 +
    finally show ?thesis .
 +
  qed
 +
qed
 +
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 17. Definir la función
 +
    esVacia :: "'a list ⇒ bool"
 +
  tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
 +
    esVacia []  = True
 +
    esVacia [1] = False
 +
  ------------------------------------------------------------------ *}
 +
 +
fun esVacia :: "'a list ⇒ bool" where
 +
  "esVacia []    = True"
 +
| "esVacia (x#xs) = False"
 +
 +
value "esVacia []"  -- "= True"
 +
value "esVacia [1]" -- "= False"
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 18. Demostrar que
 +
    esVacia xs = esVacia (conc xs xs)
 +
  ------------------------------------------------------------------- *}
 +
 +
lemma "esVacia xs = esVacia (conc xs xs)"
 +
by (induct xs) auto
 +
 +
lemma vacia_conc:
 +
  "esVacia xs = esVacia (conc xs xs)"
 +
proof (induct xs)
 +
  show "esVacia [] = esVacia (conc [] [])" by simp
 +
  next
 +
    fix x xs
 +
    assume hi: "esVacia xs = esVacia (conc xs xs)"
 +
    show "esVacia (x # xs) = esVacia (conc (x # xs) (x # xs))"
 +
    proof -
 +
      have "esVacia (conc (x # xs) (x # xs)) = esVacia (x# (conc xs (x#xs)))" by simp
 +
      also have "... = esVacia (x # xs)" by simp
 +
      finally show ?thesis by simp
 +
    qed
 +
qed
 +
 +
lemma vacia_conc':
 +
  "esVacia xs = esVacia (conc xs xs)"
 +
proof (cases xs)
 +
  case Nil thus ?thesis by simp
 +
next
 +
  case Cons thus ?thesis by simp
 +
qed
 +
 +
 +
lemma vacia_conc'':
 +
  "esVacia xs = esVacia (conc xs xs)"
 +
by (cases xs) simp_all
 +
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 19. Definir la función
 +
    inversaAc :: "'a list ⇒ 'a list"
 +
  tal que (inversaAc xs) es a inversa de xs calculada usando
 +
  acumuladores. Por ejemplo,
 +
    inversaAc [3,2,5] = [5,2,3]
 +
  ------------------------------------------------------------------ *}
 +
 +
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
 +
  "inversaAcAux [] ys = ys"
 +
| "inversaAcAux (x#xs) ys = inversaAcAux xs (x#ys)"
 +
 +
fun inversaAc :: "'a list ⇒ 'a list" where
 +
  "inversaAc xs = inversaAcAux xs []"
 +
 +
value "inversaAc [3,2,5]" -- "= [5,2,3]"
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 20. Demostrar que
 +
    inversaAcAux xs ys = (inversa xs)@ys
 +
  ------------------------------------------------------------------- *}
 +
 +
lemma inversaAcAux_es_inversa:
 +
  "inversaAcAux xs ys = (inversa xs)@ys"
 +
by (induct xs arbitrary: ys) auto
 +
 +
 +
lemma inversaAcAux_es_inversa_b:
 +
  "inversaAcAux xs ys = (inversa xs)@ys"
 +
proof (induct xs arbitrary: ys)
 +
  show "⋀ys. inversaAcAux [] ys = inversa [] @ ys" by simp
 +
next
 +
  fix x xs zs
 +
  assume hi: "⋀ys. inversaAcAux xs ys = inversa xs @ ys"
 +
  show "inversaAcAux (x#xs) zs = inversa (x#xs) @ zs"
 +
  proof -
 +
    have "inversaAcAux (x#xs) zs = inversaAcAux xs (x#zs)" by simp
 +
    also have "... = inversa xs @ (x#zs)" using hi by simp
 +
    also have "... = inversa xs @ [x] @ zs" by simp
 +
    also have "... = inversa (x#xs) @ zs" by simp
 +
    finally show ?thesis .
 +
  qed
 +
qed
 +
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 21. Demostrar que
 +
    inversaAc xs = inversa xs
 +
  ------------------------------------------------------------------- *}
 +
 +
corollary "inversaAc xs = inversa xs"
 +
by (simp add: inversaAcAux_es_inversa)
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 22. Definir la función
 +
    sum :: "int list ⇒ int"
 +
  tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
 +
    sum [3,2,5] = 10
 +
  ------------------------------------------------------------------ *}
 +
 +
fun sum :: "int list ⇒ int" where
 +
  "sum []    = 0"
 +
| "sum (x#xs) = x + sum xs"
 +
 +
value "sum [3,2,5]" -- "= 10"
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 23. Definir la función
 +
    map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
 +
  tal que (map f xs) es la lista obtenida aplicando la función f a los
 +
  elementos de xs. Por ejemplo,
 +
    map (λx. 2*x) [3,2,5] = [6,4,10]
 +
  ------------------------------------------------------------------ *}
 +
 +
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
 +
  "map f []    = []"
 +
| "map f (x#xs) = (f x) # map f xs"
 +
 +
value "map (λx. 2*x) [3::int,2,5]" -- "= [6,4,10]"
 +
 +
value "map (λx. 6*x) [3::int,2,5]" -- "= [18,12,30]"
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 24. Demostrar que
 +
    sum (map (λx. 2*x) xs) = 2 * (sum xs)
 +
  ------------------------------------------------------------------- *}
 +
 +
lemma "sum (map (λx. 2*x) xs) = 2 * (sum xs)"
 +
by (induct xs) auto
 +
 +
 +
lemma sum_map:
 +
  "sum (map (λx. 2*x) xs) = 2 * (sum xs)"
 +
proof (induct xs)
 +
  show "sum (map (λx. 2*x) []) = 2 * (sum [])" by simp
 +
next
 +
  fix a xs
 +
  assume hi: "sum (map (λx. 2*x) xs) = 2 * (sum xs)"
 +
  show "sum (map (λx. 2*x) (a#xs)) = 2 * (sum (a#xs))"
 +
  proof -
 +
    have "sum (map (λx. 2*x) (a#xs)) = sum ((2*a)#(map (λx. 2*x) xs))" by simp
 +
    also have "... = 2*a + sum (map (λx. 2*x) xs)" by simp
 +
    also have "... = 2*a + 2 * (sum xs)" using hi by simp
 +
    also have "... = 2*(a + sum xs)" by simp
 +
    also have "... = 2 * (sum (a#xs))"by simp
 +
    finally show ?thesis .
 +
  qed
 +
qed
 +
 +
 +
text {* ---------------------------------------------------------------
 +
  Ejercicio 25. Demostrar que
 +
    longitud (map f xs) = longitud xs
 +
  ------------------------------------------------------------------- *}
 +
 +
lemma "longitud (map f xs) = longitud xs"
 +
by (induct xs) auto
 +
 +
 +
lemma long_map:
 +
  "longitud (map f xs) = longitud xs"
 +
proof (induct xs)
 +
  show "longitud (map f []) = longitud []" by simp
 +
next
 +
  fix x xs
 +
  assume hi: "longitud (map f xs) = longitud xs"
 +
  show "longitud (map f (x#xs)) = longitud (x#xs)"
 +
  proof -
 +
    have "longitud (map f (x#xs)) = longitud ((f x)#(map f xs))" by simp
 +
    also have "... = 1 + longitud (map f xs)" by simp
 +
    also have "... = 1 + longitud xs" using hi by simp
 +
    also have "... = longitud (x#xs)" by simp
 +
    finally show ?thesis .
 +
  qed
 +
qed
 +
 +
 +
end
 +
</source>

Revisión actual del 09:10 27 may 2013

header {* Tema 15: Razonamiento sobre programas en Isabelle *}

theory T15
imports Main
begin

text {* 
  En este tema se demuestra con Isabelle las propiedades de los
  programas funcionales como se expone en el tema 8 del curso
  "Informática" que puede leerse en
  http://www.cs.us.es/~jalonso/cursos/i1m/temas/tema-8t.pdf
*}

section {* Razonamiento ecuacional *}

text {* ----------------------------------------------------------------
  Ejercicio 1. Definir, por recursión, la función
     longitud :: "'a list ⇒ nat" where
  tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
     longitud [4,2,5] = 3
  ------------------------------------------------------------------- *}

fun longitud :: "'a list ⇒ nat" where
  "longitud []     = 0"
| "longitud (x#xs) = 1 + longitud xs"
   
value "longitud [4,2,5]" -- "= 3"

text {* --------------------------------------------------------------- 
  Ejercicio 2. Demostrar que 
     longitud [4,2,5] = 3
  ------------------------------------------------------------------- *}

lemma "longitud [4,2,5] = 3"
by simp

text {* --------------------------------------------------------------- 
  Ejercicio 3. Definir la función
     fun intercambia :: "'a × 'b ⇒ 'b × 'a"
  tal que (intercambia p) es el par obtenido intercambiando las
  componentes del par p. Por ejemplo,
     "intercambia (2,3) = (3,2)
  ------------------------------------------------------------------ *}

fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
  "intercambia (x,y) = (y,x)"

value "intercambia (2,3)" -- "= (3,2)"

text {* --------------------------------------------------------------- 
  Ejercicio 4. Demostrar que 
     intercambia (intercambia (x,y)) = (x,y)
  ------------------------------------------------------------------- *}

lemma "intercambia (intercambia (x,y)) = (x,y)"
by simp

text {* --------------------------------------------------------------- 
  Ejercicio 5. Definir, por recursión, la función
     inversa :: "'a list ⇒ 'a list"
  tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
  elementos de xs. Por ejemplo,
     inversa [3,2,5] = [5,2,3]
  ------------------------------------------------------------------ *}

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

value "inversa [3,2,5]" -- "= [5,2,3]"

text {* --------------------------------------------------------------- 
  Ejercicio 6. Demostrar que 
     inversa [x] = [x]
  ------------------------------------------------------------------- *}

lemma "inversa [x] = [x]"
by simp

text {* --------------------------------------------------------------- 
  Ejercicio 7. Definir la función
     repite :: "nat ⇒ 'a ⇒ 'a list" where
  tal que . Por ejemplo,
     repite 3 5 = [5,5,5]
  ------------------------------------------------------------------ *}

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

value "repite 3 5" -- "= [5,5,5]"

text {* --------------------------------------------------------------- 
  Ejercicio 8. Demostrar que 
     longitud (repite n x) = n
  ------------------------------------------------------------------- *}

lemma "longitud (repite n x) = n"
apply (induct_tac n) 
apply (auto)
done


lemma "longitud (repite n x) = n"
by (induct n) auto

lemma longitud_repite:
  "longitud (repite n x) = n"
proof (induct n)
  show "longitud (repite 0 x) = 0" by simp
next
  fix n
  assume hi: "longitud (repite n x) = n"
  show "longitud (repite (Suc n) x) = Suc n"
  proof -
    have "longitud (repite (Suc n) x) = longitud (x # (repite n x))" by simp
    also have "... = 1 +  longitud (repite n x)" by simp
    also have "... = 1 + n" using hi by simp
    also have "... = Suc n" by simp
    finally show ?thesis .
  qed
qed

text {* --------------------------------------------------------------- 
  Ejercicio 9. Definir la función
     fun conc :: "'a list ⇒ 'a list ⇒ 'a list"
  tal que . Por ejemplo,
     conc [2,3] [4,3,5] = [2,3,4,3,5]
  ------------------------------------------------------------------ *}

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

value "conc [2,3] [4,3,5]" -- "= [2,3,4,3,5]"

text {* --------------------------------------------------------------- 
  Ejercicio 10. Demostrar que 
     conc xs (conc ys zs) = (conc xs ys) zs
  ------------------------------------------------------------------- *}

lemma "conc xs (conc ys zs) = conc (conc xs ys) zs"
by (induct xs) auto

lemma conc_asociativa:
  "conc xs (conc ys zs) = conc (conc xs ys) zs"
proof (induct xs)
  show "conc [] (conc ys zs) = conc (conc [] ys) zs" by simp
next
  fix x xs
  assume hi: "conc xs (conc ys zs) = conc (conc xs ys) zs"
  show "conc (x # xs) (conc ys zs) = conc (conc (x # xs) ys) zs"
  proof -
    have "conc (x # xs) (conc ys zs) = x # (conc xs (conc ys zs))" by simp
    also have "... = x # (conc (conc xs ys) zs)" using hi by simp
    also have "... = conc (x#(conc xs ys)) zs" by simp
    also have "... = conc (conc (x # xs) ys) zs" by simp
    finally show ?thesis .
  qed
qed


text {* --------------------------------------------------------------- 
  Ejercicio 11. Refutar que 
     conc xs ys = conc ys xs
  ------------------------------------------------------------------- *}

lemma "conc xs ys = conc ys xs"
quickcheck
oops

text {* Encuentra el contraejemplo, 
  xs = [a\<^isub>2]
  ys = [a\<^isub>1] *}

text {* --------------------------------------------------------------- 
  Ejercicio 12. Demostrar que 
     conc xs [] = xs
  ------------------------------------------------------------------- *}

lemma "conc xs [] = xs"
by (induct xs) auto

text {* --------------------------------------------------------------- 
  Ejercicio 13. Demostrar que 
     longitud (conc xs ys) = longitud xs + longitud ys
  ------------------------------------------------------------------- *}

lemma "longitud (conc xs ys) = longitud xs + longitud ys"
by (induct xs) auto

lemma long_conc:
  "longitud (conc xs ys) = longitud xs + longitud ys"
proof (induct xs)
  show "longitud (conc [] ys) = longitud [] + longitud ys" by simp
next
  fix x xs
  assume hi: "longitud (conc xs ys) = longitud xs + longitud ys"
  show "longitud (conc (x # xs) ys) = longitud (x # xs) + longitud ys"
  proof -
    have "longitud (conc (x # xs) ys) = longitud (x # (conc xs ys))" by simp
    also have "... = 1 + longitud (conc xs ys)" by simp
    also have "... = 1 + (longitud xs + longitud ys)" using hi by simp
    also have "... = (1+ longitud xs) + longitud ys" by simp
    also have "... = longitud (x # xs) + longitud ys" by simp
    finally show ?thesis .
  qed
qed


text {* --------------------------------------------------------------- 
  Ejercicio 14. Definir la función
     coge :: "nat ⇒ 'a list ⇒ 'a list"
  tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
  ejemplo, 
     coge 2 [3,7,5,4] = [3,7]
  ------------------------------------------------------------------ *}

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

value "coge 2 [3,7,5,4]" -- "= [3,7]"

text {* --------------------------------------------------------------- 
  Ejercicio 15. Definir la función
     elimina :: "nat ⇒ 'a list ⇒ 'a list"
  tal que (coge n xs) es la lista de los n primeros elementos de xs. Por
  ejemplo, 
     elimina 2 [3,7,5,4] = [5,4]
  ------------------------------------------------------------------ *}

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

value "elimina 2 [3,7,5,4]" -- "= [5,4]"

text {* --------------------------------------------------------------- 
  Ejercicio 16. Demostrar que 
     conc (coge n xs) (elimina n xs) = xs
  ------------------------------------------------------------------- *}

lemma "conc (coge n xs) (elimina n xs) = xs"
by (induct rule: coge.induct) auto

text {* coge.induct es el esquema de inducción asociado a la definición
  de la función coge. Puede verse como sigue: *}

thm coge.induct

lemma "conc (coge n xs) (elimina n xs) = xs"
by (induct rule: elimina.induct) auto

thm elimina.induct

lemma conc_coge_elimina:
  "conc (coge n xs) (elimina n xs) = xs"
proof (induct rule: coge.induct) 
  show "⋀n. conc (coge n []) (elimina n []) = []" by simp
next
  show "⋀v va. conc (coge 0 (v # va)) (elimina 0 (v # va)) = v # va" by simp
next
  fix x xs n 
  assume hi: "conc (coge n xs) (elimina n xs) = xs"
  show "conc (coge (Suc n) (x # xs)) (elimina (Suc n) (x # xs)) = x # xs"
  proof -
    have "conc (coge (Suc n) (x # xs)) (elimina (Suc n) (x # xs)) = conc (coge (Suc n) (x # xs)) (elimina n xs)" by simp
    also have "... = conc (x# (coge n xs)) (elimina n xs)" by simp
    also have "... = x#(conc (coge n xs) (elimina n xs))" by simp
    also have "... = (x#xs)" using hi by simp
    finally show ?thesis .
  qed
qed


text {* --------------------------------------------------------------- 
  Ejercicio 17. Definir la función
     esVacia :: "'a list ⇒ bool"
  tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo, 
     esVacia []  = True
     esVacia [1] = False
  ------------------------------------------------------------------ *}

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

value "esVacia []"  -- "= True"
value "esVacia [1]" -- "= False"

text {* --------------------------------------------------------------- 
  Ejercicio 18. Demostrar que 
     esVacia xs = esVacia (conc xs xs)
  ------------------------------------------------------------------- *}

lemma "esVacia xs = esVacia (conc xs xs)"
by (induct xs) auto

lemma vacia_conc:
  "esVacia xs = esVacia (conc xs xs)"
proof (induct xs)
  show "esVacia [] = esVacia (conc [] [])" by simp
  next
    fix x xs
    assume hi: "esVacia xs = esVacia (conc xs xs)"
    show "esVacia (x # xs) = esVacia (conc (x # xs) (x # xs))"
    proof -
      have "esVacia (conc (x # xs) (x # xs)) = esVacia (x# (conc xs (x#xs)))" by simp
      also have "... = esVacia (x # xs)" by simp
      finally show ?thesis by simp
    qed
qed

lemma vacia_conc':
  "esVacia xs = esVacia (conc xs xs)"
proof (cases xs)
  case Nil thus ?thesis by simp
next
  case Cons thus ?thesis by simp
qed


lemma vacia_conc'':
  "esVacia xs = esVacia (conc xs xs)"
by (cases xs) simp_all


text {* --------------------------------------------------------------- 
  Ejercicio 19. Definir la función
     inversaAc :: "'a list ⇒ 'a list"
  tal que (inversaAc xs) es a inversa de xs calculada usando
  acumuladores. Por ejemplo, 
     inversaAc [3,2,5] = [5,2,3]
  ------------------------------------------------------------------ *}

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

fun inversaAc :: "'a list ⇒ 'a list" where
  "inversaAc xs = inversaAcAux xs []"

value "inversaAc [3,2,5]" -- "= [5,2,3]"

text {* --------------------------------------------------------------- 
  Ejercicio 20. Demostrar que 
     inversaAcAux xs ys = (inversa xs)@ys
  ------------------------------------------------------------------- *}

lemma inversaAcAux_es_inversa:
  "inversaAcAux xs ys = (inversa xs)@ys"
by (induct xs arbitrary: ys) auto


lemma inversaAcAux_es_inversa_b:
  "inversaAcAux xs ys = (inversa xs)@ys"
proof (induct xs arbitrary: ys) 
  show "⋀ys. inversaAcAux [] ys = inversa [] @ ys" by simp
next
  fix x xs zs
  assume hi: "⋀ys. inversaAcAux xs ys = inversa xs @ ys"
  show "inversaAcAux (x#xs) zs = inversa (x#xs) @ zs"
  proof -
    have "inversaAcAux (x#xs) zs = inversaAcAux xs (x#zs)" by simp
    also have "... = inversa xs @ (x#zs)" using hi by simp
    also have "... = inversa xs @ [x] @ zs" by simp
    also have "... = inversa (x#xs) @ zs" by simp
    finally show ?thesis .
  qed
qed


text {* --------------------------------------------------------------- 
  Ejercicio 21. Demostrar que 
     inversaAc xs = inversa xs
  ------------------------------------------------------------------- *}

corollary "inversaAc xs = inversa xs"
by (simp add: inversaAcAux_es_inversa)

text {* --------------------------------------------------------------- 
  Ejercicio 22. Definir la función
     sum :: "int list ⇒ int" 
  tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
     sum [3,2,5] = 10
  ------------------------------------------------------------------ *}

fun sum :: "int list ⇒ int" where
  "sum []     = 0"
| "sum (x#xs) = x + sum xs"

value "sum [3,2,5]" -- "= 10"

text {* --------------------------------------------------------------- 
  Ejercicio 23. Definir la función
     map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
  tal que (map f xs) es la lista obtenida aplicando la función f a los
  elementos de xs. Por ejemplo,
     map (λx. 2*x) [3,2,5] = [6,4,10]
  ------------------------------------------------------------------ *}

fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
  "map f []     = []"
| "map f (x#xs) = (f x) # map f xs"

value "map (λx. 2*x) [3::int,2,5]" -- "= [6,4,10]"

value "map (λx. 6*x) [3::int,2,5]" -- "= [18,12,30]"

text {* --------------------------------------------------------------- 
  Ejercicio 24. Demostrar que 
     sum (map (λx. 2*x) xs) = 2 * (sum xs)
  ------------------------------------------------------------------- *}

lemma "sum (map (λx. 2*x) xs) = 2 * (sum xs)"
by (induct xs) auto


lemma sum_map:
  "sum (map (λx. 2*x) xs) = 2 * (sum xs)"
proof (induct xs)
  show "sum (map (λx. 2*x) []) = 2 * (sum [])" by simp
next
  fix a xs
  assume hi: "sum (map (λx. 2*x) xs) = 2 * (sum xs)"
  show "sum (map (λx. 2*x) (a#xs)) = 2 * (sum (a#xs))"
  proof -
    have "sum (map (λx. 2*x) (a#xs)) = sum ((2*a)#(map (λx. 2*x) xs))" by simp
    also have "... = 2*a + sum (map (λx. 2*x) xs)" by simp
    also have "... = 2*a + 2 * (sum xs)" using hi by simp
    also have "... = 2*(a + sum xs)" by simp
    also have "... = 2 * (sum (a#xs))"by simp
    finally show ?thesis .
  qed
qed


text {* --------------------------------------------------------------- 
  Ejercicio 25. Demostrar que 
     longitud (map f xs) = longitud xs
  ------------------------------------------------------------------- *}

lemma "longitud (map f xs) = longitud xs"
by (induct xs) auto


lemma long_map:
  "longitud (map f xs) = longitud xs"
proof (induct xs) 
  show "longitud (map f []) = longitud []" by simp
next
  fix x xs
  assume hi: "longitud (map f xs) = longitud xs"
  show "longitud (map f (x#xs)) = longitud (x#xs)"
  proof -
    have "longitud (map f (x#xs)) = longitud ((f x)#(map f xs))" by simp
    also have "... = 1 + longitud (map f xs)" by simp
    also have "... = 1 + longitud xs" using hi by simp
    also have "... = longitud (x#xs)" by simp
    finally show ?thesis .
  qed
qed


end
Obtenido de «https://www.glc.us.es/~jalonso/LMF2013/index.php?title=Tema_15&oldid=544»