Relación 3

header {* R3: Razonamiento sobre programas en Isabelle/HOL *}

theory R3
imports Main 
begin

text {* --------------------------------------------------------------- 
  Ejercicio 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
  ------------------------------------------------------------------ *}

-- "davoremar juacorvic jeshorcob domcadgom danrodcha carvelcab"
fun sumaImpares :: "nat ⇒ nat" where
  "sumaImpares 0 = 0"
| "sumaImpares (Suc n) = (2*n+1) + sumaImpares n"

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

-- "jeshorcob"
fun sumaImpares3 :: "nat ⇒ nat" where
  "sumaImpares3 n = foldr (λ x y. y + (2 * x) + 1) (upt 0  n) 0"

(* Esta definición en principio es más eficiente pero a la hora de
   demostrar se complica todo. Estaría bien si en clase pudieramos
   explicar si se puede demostrar el lema usando esta definición de
   manera sencilla o explicar un poco cómo iría la prueba *) 

value "sumaImpares 5" -- "= 25"

text {* --------------------------------------------------------------- 
  Ejercicio 2. Demostrar que 
     sumaImpares n = n*n
  ------------------------------------------------------------------- *}

  (* Pedrosrei: lo dejo de menos a más estructurado *)

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

-- "davoremar juacorvic jeshorcob danrodcha carvelcab"
 
lemma "sumaImpares n = n*n"
by (induct n) auto
 
lemma "sumaImpares n = n*n"
proof (induct n)
  show "sumaImpares 0 = 0 * 0" by simp
next
  fix n
  assume HI: "sumaImpares n = n * n"
  thus "sumaImpares (Suc n) = Suc n * Suc n" by simp
qed

-- "davoremar juacorvic marnajgom jeshorcob domcadgom danrodcha carvelcab"

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

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

-- "davoremar juacorvic marnajgom jeshorcob domcadgom danrodcha carvelcab"

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

value "sumaPotenciasDeDosMasUno 3" -- "= 16"

text {* --------------------------------------------------------------- 
  Ejercicio 4. Demostrar que 
     sumaPotenciasDeDosMasUno n = 2^(n+1)
  ------------------------------------------------------------------- *}

-- "davoremar juacorvic jeshorcob danrodcha carvelcab"
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
by (induct n) auto

-- "jeshorcob"
lemma "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)"
  thus "sumaPotenciasDeDosMasUno (Suc n) = 2^(Suc n + 1)" by simp
qed

-- "davoremar juacorvic marnajgom jeshorcob domcadgom danrodcha carvelcab"
lemma "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)"
  have "sumaPotenciasDeDosMasUno (Suc n) = 
        sumaPotenciasDeDosMasUno n + 2^(Suc n)" by simp
  also have "... = 2^(n+1) + 2^(Suc n)" using HI by simp
  also have "... = 2^(Suc n + 1)" by simp
  finally show "sumaPotenciasDeDosMasUno (Suc n) = 2^(Suc n + 1)" 
    by simp
qed

text {* --------------------------------------------------------------- 
  Ejercicio 5. 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]
  ------------------------------------------------------------------ *}

-- "davoremar juacorvic marnajgom jeshorcob domcadgom danrodcha carvelcab"
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 6. 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 ∧
  ----------------------------------------------------------------- *}

-- "davoremar juacorvic marnajgom jeshorcob domcadgom danrodcha carvelcab"
fun todos :: "('a ⇒ bool) ⇒ 'a list ⇒ bool" where
  "todos p []     = True"
| "todos p (x#xs) = ((p x) ∧ (todos p xs))"

(* jeshorcob: me gustaría saber por qué es necesario poner el paréntesis
  en la definición(2) *) 

(* Pedrosrei: porque aunque están predefinidas las prioridades entre la
  suma y el producto en los cuerpos y anillos más comunes, la conjunción
  y la igualdad no tienen definida esa prioridad*) 

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

text {* --------------------------------------------------------------- 
  Ejercicio 7. Demostrar que todos los elementos de (copia n x) son
  iguales a x. 
  ------------------------------------------------------------------- *}

-- "davoremar juacorvic jeshorcob danrodcha carvelcab"
lemma "todos (λy. y=x) (copia n x)"
by (induct n) auto

-- "davoremar marnajgom jeshorcob domcadgom"
lemma "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 "... = todos (λy. y=x) (copia n x)" by simp
  finally show "todos (λy. y=x) (copia (Suc n) x)" using HI by simp
qed

-- "juacorvic jeshorcob"
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
  show "todos (λy. y=x) (copia 0 x)" by simp
  fix n
    assume HI: "todos (λy. y=x) (copia n x)"
    thus  "todos (λy. y=x) (copia (Suc n) x)" by simp
qed

-- "juacorvic"
lemma "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 "...= todos (λy. y=x)[x]" using HI by simp
  finally show "todos (λy. y=x) (copia (Suc n) x)" using HI by simp
qed

-- "danrodcha carvelcab"
lemma "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 only: copia.simps)
  also have "... = (((λy. y=x) x) ∧ (todos (λy. y=x) (x#(copia n x))))" by simp
  also have "... = (todos (λy. y=x) (copia n x))" using HI by simp
  also have "... = True" using HI by simp
  finally show "todos (λy. y=x) (copia (Suc n) x) " by simp
qed

text {* --------------------------------------------------------------- 
  Ejercicio 8. Definir la función
    factR :: nat ⇒ nat
  tal que (factR n) es el factorial de n. Por ejemplo,
    factR 4 = 24
  ------------------------------------------------------------------ *}

-- "davoremar juacorvic marnajgom jeshorcob domcadgom carvelcab"
fun factR :: "nat ⇒ nat" where
  "factR 0 = 1"
| "factR (Suc n) = (Suc n) * factR n"

value "factR 4" -- "= 24"

text {* --------------------------------------------------------------- 
  Ejercicio 9. 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 (Suc n)*x"

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

value "factI 4" -- "= 24"
     
-- "davoremar jeshorcob domcadgom"
lemma fact1: "factI' n x = x* factR n"
by (induct n arbitrary: x) auto 

(* juacorvic: ¿Por que tenemos que indicar arbitrary:x? *)

(* Pedrosrei: no tienes que hacerlo como resulta de mi ejemplo o las 
  correcciones que he indicado en el tuyo. En este caso concreto
  totalmente automatizado es porque posees dos variables y tienes que
  decirle qué hacer con cada una. En una haces la inducción pero la otra
  la dejas sin tocar, o como solemos decir: "sea un x cualquiera..." *)
  
-- "davoremar jeshorcob domcadgom carvelcab"
lemma fact: "factI' n x = x * factR n"
proof (induct n arbitrary: x)
  show "⋀x. factI' 0 x = x * factR 0" by simp
next
  fix n x
  assume HI: "⋀x. factI' n x = x * factR n"
  have "factI' (Suc n) x = factI' n (Suc n)*x" by simp
  also have "... = x * factI' n (Suc n)" by simp
  also have "... = x * ((Suc n) * factR n)" using HI by simp
  also have "... = x * factR (Suc n)" by simp
  finally show "factI' (Suc n) x = x * factR (Suc n)" by simp
qed

(* marnajgom: no entiendo el paso 
      "also have "... = x * factI' n (Suc n)" by simp" 
   ¿Por qué no sacas fuera (Suc n)*x, sólo sacas x? *)

(* Pedrosrei: También podemos prescindir de arbitrary x *)

lemma fact2: "factI' n x = x * factR n"
proof (induct n)
  show "factI' 0 x = x * factR 0" by simp
next
  fix n
  assume HI: "factI' n x = x * factR n"
  show "factI' (Suc n) x = x * factR (Suc n)"
  proof -
    have "factI' (Suc n) x = (factI' n (Suc n))*x" by simp
    also have " ... = ((Suc n) * factR n) * x" using HI fact by simp
    -- "Si usamos auto: finally show ?thesis by auto    "
    also have " ... = x * factR (Suc n)" by simp
    finally show ?thesis .
  qed
qed


(* Pedrosrei: no resuelve finally debido a la conjunción grande que
  emplea isabelle para indicarte un "para todo de cualquier conjunto"
  (en realidad es la conjunción de conjuntos, creo recordar). Si quieres
  poner un para todo recurre a ! ó \forall. Si sólo has puesto lo que te
  ha indicado la máquina, quería indicarte un valor cualquiera, vamos,
  lo que fijas con "arbitrary". 

  Sólo tienes que eliminar los ⋀ de la prueba y añadirle un paso
  intermedio con la regla adecuada (usa find_theorem y el nombre que
  crees que tendrá) entre el also have último y el finally para que vea
  la igualdad de derecha a izquierda y listo. También puedes emplear "by
  arith" para estas cosas tan sencillas y no tener que irte a la regla
  exacta de la aritmética (que no siempre es fácil).

*)
-- "juacorvic" 
(* Se intenta demostración en sentido inverso: 
      x * factR n = factI' n x 
   No resuelve finally *)

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

text {* --------------------------------------------------------------- 
  Ejercicio 10. Demostrar que
     factI n = factR n
  ------------------------------------------------------------------- *}

-- "davoremar juacorvic jeshorcob domcadgom"
corollary "factI n = factR n"
by (simp add: fact)

-- "juacorvic"
corollary "factI n = factR n"
proof - 
  show "factI n = factR n" by (simp add:fact)
qed

-- "davoremar juacorvic jeshorcob carvelcab"
corollary "factI n = factR n"
proof -
  have "factI n = factI' n 1" by simp
  also have "... = 1 * factR n" by (simp add: fact)
  finally show "factI n = factR n" by simp
qed

-- "marnajgom" 
corollary "factI n = factR n"
proof (induct n)
  show "factI 0 = factR 0" by simp
next 
  fix n
  assume HI:"factI n = factR n"
  have "factI (Suc n) = factI' (Suc n) 1" by simp
  also have "... = factI' n (Suc n)*1" by simp
  also have "... = (Suc n)*1 * factR n" using fact by simp
  also have "... = (Suc n) * factR n" by simp
  finally show "factI (Suc n) = factR (Suc n)" by simp
qed

text {* --------------------------------------------------------------- 
  Ejercicio 11. 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]
  ------------------------------------------------------------------ *}

-- "davoremar marnajgom jeshorcob domcadgom carvelcab"
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 12. Demostrar que 
     amplia xs y = xs @ [y]
  ------------------------------------------------------------------- *}

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

-- "juacorvic" (* Añado esquema intermedio *)
lemma "amplia xs y = xs @ [y]"
proof (induct xs)
  show "amplia [] y = [] @ [y]" by simp
next
  fix a xs
  assume HI: "amplia xs y = xs @ [y]"  
  thus "amplia (a # xs) y = (a # xs) @ [y]" by simp
qed

-- "davoremar marnajgom jeshorcob domcadgom carvelcab"
lemma "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 (* Pedrosrei: prescindible *)
  finally show "amplia (x#xs) y = (x#xs) @ [y]" by simp
qed

(* Pedrosrei: las variables son mudas. Da igual 'a' que 'x', del mismo
   modo que da igual en papel el nombre de la incógnita. Puedes usar
   'incognitabonita' y sigue siendo lo mismo *) 

-- "juacorvic" 
(* Demostración idéntica a 'davoremar' pero el sistema me sugirió usar
   'a' en vez de 'x' *) 
lemma "amplia xs y = xs @ [y]"
proof (induct xs)
  show "amplia [] y = [] @ [y]" by simp
next
  fix a xs
  assume HI: "amplia xs y = xs @ [y]"
  have "amplia (a # xs) y  =  a # (amplia xs y) " by simp
  also have "... = a # (xs @ [y])" using HI by simp
  also have "... = (a # xs) @ [y]" by simp
  finally show "amplia (a # xs) y = (a # xs) @ [y]" by simp
qed

end