Relación 2

chapter {* R2: Razonamiento sobre programas en Isabelle/HOL *}

theory R2_Razonamiento_automatico_sobre_programas
imports Main 
begin

declare [[names_short]]

text {* --------------------------------------------------------------- 
  Ejercicio 1.1. Definir la función
     sumaImpares :: nat ⇒ nat
  tal que (sumaImpares n) es la suma de los n primeros números
  impares. Por ejemplo,
     sumaImpares 5  =  25
  ------------------------------------------------------------------ *}

(* ivamenjim wilmorort*)

fun sumaImpares :: "nat ⇒ nat" where
  "sumaImpares 0 = 0"
| "sumaImpares n = 2*(n-1) + 1 + sumaImpares (n-1)"

(*wilmorort*)

(*Notar: Por la propiedad de Gauss se puede deducir que:
         la suma de los n números impares es igual a la suma de los 
         n y n-1 números consecutivos*)   
fun suma :: "nat ⇒ nat" where
    "suma 0 = 0"|
    "suma n = n + suma(n-1)"

fun sumaImpares1 :: "nat ⇒ nat" where
   "sumaImpares1 0 = 0" |
   "sumaImpares1 n = suma(n) + suma(n-1)"

value "sumaImpares 5 = 25"

(* fraortmoy *)

fun sumaImpares2 :: "nat ⇒ nat" where
  "sumaImpares2 0 = 0"
| "sumaImpares2 (Suc(n)) = ((Suc(n)*2)-1)+sumaImpares2 n"

value "sumaImpares2 5 = 25"

(* anaprarod, danrodcha crigomgom*)

fun sumaImpares3 :: "nat ⇒ nat" where
  "sumaImpares3 0 = 0"
 |"sumaImpares3 (Suc n) = (2*n +1) + (sumaImpares3 n)"

value "sumaImpares3 5 = 25"

text {* --------------------------------------------------------------- 
  Ejercicio 1.2. Demostrar que 
     sumaImpares n = n*n
  ------------------------------------------------------------------- *}

(* ivamenjim wilmorort anaprarod crigomgom*)

lemma "sumaImpares n = n*n"
apply (induct n)
apply auto
done

(*wilmorort*)
(*Demostración estructurada*)

lemma aux1 : "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)"
apply (induct n)
apply auto
done

lemma aux2: "(n+1) + suma(n) + n + suma(n-1) = (n+1) + n + sumaImpares1 n "
apply (induct n)
apply auto
done

lemma "sumaImpares1 n = n*n"
proof (induct n)
show "sumaImpares1 0 = 0*0" by simp
next 
 fix n
 assume HI:"sumaImpares1 n = n*n"
 have "sumaImpares1 (Suc n) = suma(Suc n) + suma(n)" by simp
 have "suma(Suc n) + suma(n) = (n+1) + suma(n) + n + suma(n-1)"  using
 "aux1" by simp
 also have "... = (n+1) + n + sumaImpares1 n " using "aux2" by simp
 also have "... = 2*n +1 + n*n" using HI by simp
 also have "... = (n+1)*(n+1)" by simp
 finally show "sumaImpares1 (Suc n) = Suc n * Suc n" by simp
 qed

(* fraortmoy *)
(* la Demostración para la función sumaImpares2 es la misma que la de sumaImpares*)

lemma "sumaImpares2 n = n*n"
apply (induct n)
apply auto
done

(* danrodcha *)
lemma "sumaImpares n = n*n"
by (induct n) auto

(* danrodcha *)
lemma "sumaImpares n = n*n" (is "?P n")
proof (induct n)
  show "?P 0" by simp
next
  fix n assume "?P n"
  thus "?P (Suc n)" by simp
qed

text {* --------------------------------------------------------------- 
  Ejercicio 2.1. Definir la función
     sumaPotenciasDeDosMasUno :: nat ⇒ nat
  tal que 
     (sumaPotenciasDeDosMasUno n) = 1 + 2^0 + 2^1 + 2^2 + ... + 2^n. 
  Por ejemplo, 
     sumaPotenciasDeDosMasUno 3  =  16
  ------------------------------------------------------------------ *}

(* wilmorort ivamenjim *)
  
fun sumaPotenciasDeDosMasUno :: "nat ⇒ nat" where
  "sumaPotenciasDeDosMasUno 0 = 2"|
  "sumaPotenciasDeDosMasUno n = 2^(n) + sumaPotenciasDeDosMasUno (n-1) "

value "sumaPotenciasDeDosMasUno 3 = 16"


(* fraortmoy anaprarod danrodcha crigomgom*)
fun sumaPotenciasDeDosMasUno2 :: "nat ⇒ nat" where
  "sumaPotenciasDeDosMasUno2 0 = 2"
| "sumaPotenciasDeDosMasUno2 (Suc(n)) = (2^(Suc n)) + sumaPotenciasDeDosMasUno2 n"

value "sumaPotenciasDeDosMasUno2 3 = 16"


text {* --------------------------------------------------------------- 
  Ejercicio 2.2. Demostrar que 
     sumaPotenciasDeDosMasUno n = 2^(n+1)
  ------------------------------------------------------------------- *}

(* wilmorort ivamenjim fraortmoy anaprarod crigomgom*)
(* esta demostración funciona con sumaPotenciasDeDosMasUno2 *)

lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
apply (induct n)
apply auto
done

(* danrodcha *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
by (induct n) auto

(* danrodcha *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)" (is "?P n")
proof (induct n)
  show "?P 0" by simp
next
  fix n assume "?P n"
  thus "?P (Suc n)"by simp
qed

text {* --------------------------------------------------------------- 
  Ejercicio 3.1. 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]
  ------------------------------------------------------------------ *}
(* ivamenjim wilmorort*)

fun copia :: "nat ⇒ 'a ⇒ 'a list" where
  "copia 0 x = [] " 
| "copia n x = x # (copia (n-1) x)"

(* wilmorort anaprarod *)

fun copia1 :: "nat ⇒ 'a ⇒ 'a list" where
  "copia1 0 x = []" |
  "copia1 (Suc n) x = [x] @ (copia1 n x) "

(* danrodcha crigomgom*)
fun copia2 :: "nat ⇒ 'a ⇒ 'a list" where
  "copia2 0 x = []"
| "copia2 (Suc n) x = x # copia2 n x"

value "copia 3 x = [x,x,x]"


text {* --------------------------------------------------------------- 
  Ejercicio 3.2. 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 ∧
  ----------------------------------------------------------------- *}

(*wilmorort*)

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

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


(* fraortmoy anaprarod danrodcha crigomgom*)

fun todos1 :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
  "todos1 p [] = True"
| "todos1 p (x#xs) = (p x ∧ todos1 p xs )"

value "todos1 (λx. x>(1::nat)) [2,6,4] = True"
value "todos1 (λx. x>(2::nat)) [2,6,4] = False"


text {* --------------------------------------------------------------- 
  Ejercicio 3.3. Demostrar que todos los elementos de (copia n x) son
  iguales a x. 
  ------------------------------------------------------------------- *}
(*wilmorort fraortmoy anaprarod crigomgom*)

lemma "todos (λy. y=x) (copia n x)"
apply (induct n)
apply auto
done

(* danrodcha*)
lemma "todos (λy. y=x) (copia n x)"
by (induct n) auto

(* danrodcha *)
lemma "todos (λy. y=x) (copia n x)" (is "?P n")
proof (induct n)
  show "?P 0" by simp
next
  fix n assume "?P n"
  thus "?P (Suc n)" by simp
qed

text {* --------------------------------------------------------------- 
  Ejercicio 4.1. 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]
  ------------------------------------------------------------------ *}

(*wilmorort fraortmoy anaprarod crigomgom*)

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 4.2. Demostrar que 
     amplia xs y = xs @ [y]
  ------------------------------------------------------------------- *}

(*wilmorort fraortmoy anaprarod crigomgom*)

lemma "amplia xs y = xs @ [y]"
apply (induct xs)
apply auto
done

(* danrodcha *)

lemma "amplia xs y = xs @ [y]"
by (induct xs) auto

(* danrodcha *)
lemma "amplia xs y = xs @ [y]" (is "?P xs")
proof (induct xs)
  show "?P []" by simp
next
  fix a xs assume "?P xs"
  thus "?P (a#xs)" by simp
qed

end