Tema 5: Razonamiento sobre programas

header {* Tema 5: Razonamiento sobre programas *}

theory T5
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://goo.gl/Imvyt *}

section {* Razonamiento ecuacional *}

text {* ----------------------------------------------------------------
  Ejemplo 1. Definir, por recursión, la función
     longitud :: 'a list ⇒ nat
  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 {* --------------------------------------------------------------- 
  Ejemplo 2. Demostrar que 
     longitud [4,2,5] = 3
  ------------------------------------------------------------------- *}

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

text {* --------------------------------------------------------------- 
  Ejemplo 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 (u,v) = (v,u)
  ------------------------------------------------------------------ *}

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

value "intercambia (u,v)" -- "= (v,u)"

text {* --------------------------------------------------------------- 
  Ejemplo 4. (p.6) Demostrar que 
     intercambia (intercambia (x,y)) = (x,y)
  ------------------------------------------------------------------- *}

-- "La demostración detallada es"
lemma "intercambia (intercambia (x,y)) = (x,y)"
proof -
  have "intercambia (intercambia (x,y)) = intercambia (y,x)"  
    by (simp only: intercambia.simps)
  also have "... = (x,y)" 
    by (simp only: intercambia.simps)
  finally show "intercambia (intercambia (x,y)) = (x,y)" by simp
qed

text {* 
  El razonamiento ecuacional se realiza usando la combinación de "also"
  (además) y "finally" (finalmente). *}

-- "La demostración estructurada es"
lemma "intercambia (intercambia (x,y)) = (x,y)"
proof -
  have "intercambia (intercambia (x,y)) = intercambia (y,x)"  by simp
  also have "... = (x,y)" by simp 
  finally show "intercambia (intercambia (x,y)) = (x,y)" by simp
qed

-- "La demostración automática es"
lemma "intercambia (intercambia (x,y)) = (x,y)"
by simp

text {* --------------------------------------------------------------- 
  Ejemplo 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 [a,d,c] = [c,d,a]
  ------------------------------------------------------------------ *}

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

value "inversa [a,d,c]" -- "= [c,d,a]"

text {* --------------------------------------------------------------- 
  Ejemplo 6. (p. 9) Demostrar que 
     inversa [x] = [x]
  ------------------------------------------------------------------- *}

-- "La demostración detallada es"
lemma "inversa [x] = [x]"
proof -
  have "inversa [x] = inversa (x#[])" by simp
  also have "... = (inversa []) @ [x]" by (simp only: inversa.simps(2))
  also have "... = [] @ [x]" by (simp only: inversa.simps(1))
  also have "... = [x]" by (simp only: append_Nil) 
  finally show "inversa [x] = [x]" by simp
qed

-- "La demostración estructurada es"
lemma "inversa [x] = [x]"
proof -
  have "inversa [x] = inversa (x#[])" by simp
  also have "... = (inversa []) @ [x]" by simp
  also have "... = [] @ [x]" by simp
  also have "... = [x]" by simp 
  finally show "inversa [x] = [x]" by simp
qed

-- "La demostración automática es"
lemma "inversa [x] = [x]"
by simp

section {* Razonamiento por inducción sobre los naturales *}

text {*
  [Principio de inducción sobre los naturales] Para demostrar una
  propiedad P para todos los números naturales basta probar que el 0
  tiene la propiedad P y que si n tiene la propiedad P, entonces n+1
  también la tiene.  
     ⟦P 0; ⋀n. P n ⟹ P (Suc n)⟧ ⟹ P m

  En Isabelle el principio de inducción sobre los naturales está
  formalizado en el teorema nat.induct y puede verse con
     thm nat.induct
*}

text {* --------------------------------------------------------------- 
  Ejemplo 7. Definir la función
     repite :: nat ⇒ 'a ⇒ 'a list
  tal que (repite n x) es la lista formada por n copias del elemento
  x. Por ejemplo, 
     repite 3 a = [a,a,a]
  ------------------------------------------------------------------ *}

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

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

text {* --------------------------------------------------------------- 
  Ejemplo 8. (p. 18) Demostrar que 
     longitud (repite n x) = n
  ------------------------------------------------------------------- *}

-- "La demostración estructurada es"
lemma "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"
  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
  finally show "longitud (repite (Suc n) x) = Suc n" by simp
qed

-- "La demostración automática es"
lemma "longitud (repite n x) = n"
by (induct n) auto

section {* Razonamiento por inducción sobre listas *}

text {*
  Para demostrar una propiedad para todas las listas basta demostrar
  que la lista vacía tiene la propiedad y que al añadir un elemento a una
  lista que tiene la propiedad se obtiene otra lista que también tiene la
  propiedad. 

  En Isabelle el principio de inducción sobre listas está formalizado
  mediante el teorema list.induct que puede verse con 
     thm list.induct
*}

text {* --------------------------------------------------------------- 
  Ejemplo 9. Definir la función
     conc :: 'a list ⇒ 'a list ⇒ 'a list
  tal que (conc xs ys) es la concatención de las listas xs e ys. Por
  ejemplo, 
     conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
  ------------------------------------------------------------------ *}

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

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* --------------------------------------------------------------- 
  Ejemplo 10. (p. 24) Demostrar que 
     conc xs (conc ys zs) = (conc xs ys) zs
  ------------------------------------------------------------------- *}

-- "La demostración estructurada es"
lemma "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" 
  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 (conc (x # xs) ys) zs" by simp
  finally show "conc (x # xs) (conc ys zs) = conc (conc (x # xs) ys) zs" by simp
qed

-- "La demostración automática es"
lemma "conc xs (conc ys zs) = conc (conc xs ys) zs"
by (induct xs) auto

text {* --------------------------------------------------------------- 
  Ejemplo 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 {* --------------------------------------------------------------- 
  Ejemplo 12. (p. 28) Demostrar que 
     conc xs [] = xs
  ------------------------------------------------------------------- *}

-- "La demostración estructurada es"
lemma "conc xs [] = xs"
proof (induct xs)
  show "conc [] [] = []" by simp
next 
  fix x xs
  assume HI: "conc xs [] = xs" 
  have "conc (x # xs) [] = x # (conc xs [])" by simp
  also have "... = x # xs" using HI by simp
  finally show "conc (x # xs) [] = x # xs" by simp
qed

-- "La demostración automática es"
lemma "conc xs [] = xs"
by (induct xs) auto

text {* --------------------------------------------------------------- 
  Ejemplo 13. (p. 30) Demostrar que 
     longitud (conc xs ys) = longitud xs + longitud ys
  ------------------------------------------------------------------- *}

-- "La demostración automática es"
lemma "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"
  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 "... = longitud (x # xs) + longitud ys" by simp
  finally show "longitud (conc (x # xs) ys) = 
                longitud (x # xs) + longitud ys" by simp
qed

-- "La demostración automática es"
lemma "longitud (conc xs ys) = longitud xs + longitud ys"
by (induct xs) auto

section {* Inducción correspondiente a la definición recursiva *}

text {* --------------------------------------------------------------- 
  Ejemplo 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 [a,c,d,b,e] = [a,c]
  ------------------------------------------------------------------ *}

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 [a,c,d,b,e]" -- "= [a,c]"

text {* --------------------------------------------------------------- 
  Ejemplo 15. Definir la función
     elimina :: nat ⇒ 'a list ⇒ 'a list
  tal que (elimina n xs) es la lista obtenida eliminando los n primeros
  elementos de xs. Por ejemplo, 
     elimina 2 [a,c,d,b,e] = [d,b,e]
  ------------------------------------------------------------------ *}

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 [a,c,d,b,e]" -- "= [d,b,e]"

text {* 
  La definición coge genera el esquema de inducción coge.induct:
     ⟦⋀n. P n []; 
      ⋀x xs. P 0 (x#xs); 
      ⋀n x xs. P n xs ⟹ P (Suc n) (x#xs)⟧
     ⟹ P n x

  Puede verse usando "thm coge.induct". *}

text {* --------------------------------------------------------------- 
  Ejemplo 16. (p. 35) Demostrar que 
     conc (coge n xs) (elimina n xs) = xs
  ------------------------------------------------------------------- *}

-- "La demostración estructurada es"
lemma "conc (coge n xs) (elimina n xs) = xs"
proof (induct rule: coge.induct)
  fix n
  show "conc (coge n []) (elimina n []) = []" by simp
next
  fix x xs
  show "conc (coge 0 (x#xs)) (elimina 0 (x#xs)) = x#xs" by simp
next
  fix n x xs
  assume HI: "conc (coge n xs) (elimina n xs) = xs"
  have "conc (coge (Suc n) (x#xs)) (elimina (Suc n) (x#xs)) = 
        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 "conc (coge (Suc n) (x#xs)) (elimina (Suc n) (x#xs)) = x#xs"
    by simp
qed

-- "La demostración automática es"
lemma "conc (coge n xs) (elimina n xs) = xs"
by (induct rule: coge.induct) auto

section {* Razonamiento por casos *}

text {*
  Distinción de casos sobre listas:
  · El método de distinción de casos se activa con (cases xs) donde xs
    es del tipo lista. 
  · "case Nil" es una abreviatura de 
       "assume Nil: xs =[]".
  · "case Cons" es una abreviatura de 
       "fix ? ?? assume Cons: xs = ? # ??"
    donde ? y ?? son variables anónimas. *}

text {* --------------------------------------------------------------- 
  Ejemplo 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 {* --------------------------------------------------------------- 
  Ejemplo 18 (p. 39) . Demostrar que 
     esVacia xs = esVacia (conc xs xs)
  ------------------------------------------------------------------- *}

-- "La demostración estructurada es"
lemma "esVacia xs = esVacia (conc xs xs)"
proof (cases xs)
  assume "xs = []"
  thus "esVacia xs = esVacia (conc xs xs)" by simp
next
  fix y ys
  assume "xs = y#ys"
  thus "esVacia xs = esVacia (conc xs xs)" by simp
qed

-- "La demostración estructurada simplificad es"
lemma "esVacia xs = esVacia (conc xs xs)"
proof (cases xs)
  case Nil
  thus "esVacia xs = esVacia (conc xs xs)" by simp
next
  case Cons
  thus "esVacia xs = esVacia (conc xs xs)" by simp
qed

-- "La demostración automática es"
lemma "esVacia xs = esVacia (conc xs xs)"
by (cases xs) auto

section {* Heurística de generalización *}

text {* 
  Heurística de generalización: Cuando se use demostración estructural,
  cuantificar universalmente las variables libres (o, equivalentemente,
  considerar las variables libres como variables arbitrarias). *}

text {* --------------------------------------------------------------- 
  Ejemplo 19. Definir la función
     inversaAc :: 'a list ⇒ 'a list
  tal que (inversaAc xs) es a inversa de xs calculada usando
  acumuladores. Por ejemplo, 
     inversaAc [a,c,b,e] = [e,b,c,a]
  ------------------------------------------------------------------ *}

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 [a,c,b,e]" -- "= [e,b,c,a]"

text {* --------------------------------------------------------------- 
  Ejemplo 20. (p. 44) Demostrar que 
     inversaAcAux xs ys = (inversa xs) @ ys
  ------------------------------------------------------------------- *}

-- "La demostración estructurada es"
lemma inversaAcAux_es_inversa:
  "inversaAcAux xs ys = (inversa xs) @ ys"
proof (induct xs arbitrary: ys)
  show "⋀ys. inversaAcAux [] ys = inversa [] @ ys" by simp
next
  fix a xs 
  assume HI: "⋀ys. inversaAcAux xs ys = inversa xs@ys"
  show "⋀ys. inversaAcAux (a#xs) ys = inversa (a#xs)@ys"
  proof -
    fix ys
    have "inversaAcAux (a#xs) ys = inversaAcAux xs (a#ys)" by simp
    also have "… = inversa xs@(a#ys)" using HI by simp
    also have "… = inversa (a#xs)@ys" by simp 
    finally show "inversaAcAux (a#xs) ys = inversa (a#xs)@ys" by simp
  qed
qed

-- "La demostración automática es"
lemma "inversaAcAux xs ys = (inversa xs)@ys"
by (induct xs arbitrary: ys) auto

text {* --------------------------------------------------------------- 
  Ejemplo 21. (p. 43) Demostrar que 
     inversaAc xs = inversa xs
  ------------------------------------------------------------------- *}

-- "La demostración automática es"
corollary "inversaAc xs = inversa xs"
by (simp add: inversaAcAux_es_inversa)

section {* Demostración por inducción para funciones de orden superior *}

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

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

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

text {* --------------------------------------------------------------- 
  Ejemplo 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::nat,2,5]" -- "= [6,4,10]"

text {* --------------------------------------------------------------- 
  Ejemplo 24. (p. 45) Demostrar que 
     sum (map (λx. 2*x) xs) = 2 * (sum xs)
  ------------------------------------------------------------------- *}

-- "La demostración estructurada es"
lemma "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)"
  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 "sum (map (λx. 2*x) (a#xs)) = 2*(sum (a#xs))" by simp
qed

-- "La demostración automática es"
lemma "sum (map (λx. 2*x) xs) = 2 * (sum xs)"
by (induct xs) auto

text {* --------------------------------------------------------------- 
  Ejemplo 25. (p. 48) Demostrar que 
     longitud (map f xs) = longitud xs
  ------------------------------------------------------------------- *}

-- "La demostración estructurada es"
lemma "longitud (map f xs) = longitud xs"
proof (induct xs)
  show "longitud (map f []) = longitud []" by simp
next
  fix a xs
  assume HI: "longitud (map f xs) = longitud xs"
  have "longitud (map f (a#xs)) = longitud (f a # (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 (a#xs)" by simp
  finally show "longitud (map f (a#xs)) = longitud (a#xs)" by simp
qed

-- "La demostración automática es"
lemma "longitud (map f xs) = longitud xs"
by (induct xs) auto

section {* Referencias *}

text {*
  · J.A. Alonso. "Razonamiento sobre programas" http://goo.gl/R06O3
  · G. Hutton. "Programming in Haskell". Cap. 13 "Reasoning about
    programms". 
  · S. Thompson. "Haskell: the Craft of Functional Programming, 3rd
    Edition. Cap. 8 "Reasoning about programms". 
  · L. Paulson. "ML for the Working Programmer, 2nd Edition". Cap. 6. 
    "Reasoning about functional programs". 
*}

end