RA2019: Razonamiento estructurado sobre programas con Isabelle/HOL

En la clase de hoy del curso de Razonamiento automático se ha presentado cómo se puede demostrar propiedades de programas funcionales con Isabelle/HOL de forma estructurada con Isar.

Para ello, se ha visto cómo representar en Isabelle/HOL las demostraciones de propiedades de programas estudiadas en el tema 8 del curso de Informática.

Los métodos de demostración utilizados son razonamiento ecuacional, inducción sobre los números naturales, inducción sobre listas e inducción sobre esquemas correspondientes a definiciones recursivas.

Se ha usado Isabelle para analizar detalladamente las demostraciones
poniendo de relieve los lemas usados en cada paso. Para encontralos se
ha empleado la búsqueda de teoremas (con find_theorems) y la traza
(con using [[simp_trace]]).

La teoría con los ejemplos presentados en la clase es la siguiente:

chapter ‹Tema 3: Razonamiento estructurado sobre programas›

theory T3_Razonamiento_estructurado_sobre_programas
imports Main 
begin

text ‹En este tema se demuestra con Isabelle las propiedades de los
  programas funcionales como se expone en el tema 2a y se demostraron
  automáticamente en el tema 2b. A diferencia del tema 2b, ahora
  nos fijamos no sólo en el método de demostración sino en la estructura
  de la prueba resaltando su semejanza con las del tema 2a.›

declare [[names_short]]

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 [a,c,d] = 3
  -------------------------------------------------------------------›

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

text ‹--------------------------------------------------------------- 
  Ejemplo 2. Demostrar que 
     longitud [a,c,d] = 3
  -------------------------------------------------------------------›

(* declare [[show_types]]*)

lemma "longitud [a,c,d] = 3"
proof -
  have "longitud [a,c,d] = 1 + longitud [c,d]"
    by (simp only: longitud.simps(2))
  also have "… = 1 + (1 + longitud [d])"
    by (simp only: longitud.simps(2))
  also have "… = 1 + (1 + (1 + longitud ([] :: 'a list)))"
    by (simp only: longitud.simps(2))
  also have "… = 1 + (1 + (1 + 0))"
    by (simp only: longitud.simps(1))
  also have "… = 3"
    by (simp only: )
  finally show "longitud [a,c,d] = 3" 
    by this
  qed

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)
  -------------------------------------------------------------------›

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 this
qed

text ‹Notas sobre el lenguaje: En la demostración anterior se ha usado
  · "proof" para iniciar la prueba,
  · "-" (después de "proof") para no usar el método por defecto,
  · "have" para establecer un paso,
  · "by (simp only: intercambia.simps)" para indicar que sólo se usa
    como regla de escritura la correspondiente a la definición de
    intercambia,
  · "also" para encadenar pasos ecuacionales,
  · "..." para representar la derecha de la igualdad anterior en un
    razonamiento ecuacional,
  · "finally" para indicar el último pasa de un razonamiento ecuacional,
  · "show" para establecer la conclusión.
  · "by simp" para indicar el método de demostración por simplificación y 
  · "qed" para terminar la pruebas,›

(* Demostración declarativa simplificada *)
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

text ‹
  Nota: La diferencia entre las dos demostraciones es que en los dos
  primeros pasos no se explicita la regla de simplificación. ›

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 declarativa es *)
lemma "inversa [x] = [x]"
proof -
  have "inversa [x] = (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 this
qed

(* La demostración declarativa simplificada es *)
lemma "inversa [x] = [x]"
proof -
  have "inversa [x] = (inversa []) @ [x]" by simp
  also have "… = [] @ [x]" by simp
  also have "… = [x]" by simp 
  finally show "inversa [x] = [x]" by simp
qed

section ‹Razonamiento por inducción sobre los naturales›

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
  -------------------------------------------------------------------›

(* declare [[show_types]] *)

(* La demostración detallada es *)
lemma "longitud (repite n x) = n"
proof (induct n)
  have "longitud (repite 0 x) = longitud ([] :: 'a list)" 
    by (simp only: repite.simps(1))
  also have "… = 0"
    by (simp only: longitud.simps(1))
  finally show "longitud (repite 0 x) = 0" 
    by this
next 
  fix n
  assume HI: "longitud (repite n x) = n"
  have "longitud (repite (Suc n) x) = longitud (x # (repite n x))" 
    by (simp only: repite.simps(2))
  also have "… = 1 + longitud (repite n x)" 
    by (simp only: longitud.simps(2))
  also have "… = 1 + n" 
    by (simp only: HI)
  also have "… = Suc n" 
    (* find_theorems "Suc _ = _" *)
    by (simp only: Suc_eq_plus1)
  finally show "longitud (repite (Suc n) x) = Suc n" 
    by this
qed

(* La demostración simplificada 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

text ‹Comentarios sobre la demostración anterior:
  · A la derecha de proof se indica el método de la demostración.
  · (induct n) indica que la demostración se hará por inducción en n.
  · Se generan dos subobjetivos correspondientes a la base y el paso de
    inducción:
    1. longitud (repite 0 x) = 0
    2. ⋀n. longitud (repite n x) = n ⟹ longitud (repite (Suc n) x) = Suc n
    donde ⋀n se lee "para todo n".  
  · "next" indica el siguiente subobjetivo.
  · "fix n" indica "sea n un número natural cualquiera"
  · assume HI: "longitud (repite n x) = n" indica «supongamos que 
    "longitud (repite n x) = n" y sea HI la etiqueta de este supuesto».
  · "using HI" usando la propiedad etiquetada con HI. ›

section ‹Razonamiento por inducción sobre listas›

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 detallada es *)
lemma "conc xs (conc ys zs) = conc (conc xs ys) zs"
proof (induct xs)
  have "conc [] (conc ys zs) = conc ys zs" 
    by (simp only: conc.simps(1))
  also have "… = conc (conc [] ys) zs"
    by (simp only: conc.simps(1))
  finally show "conc [] (conc ys zs) = conc (conc [] ys) zs" 
    by this 
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 only: conc.simps(2))
  also have "… = x # (conc (conc xs ys) zs)" 
    by (simp only: HI)
  also have "… = conc (conc (x # xs) ys) zs" 
    by (simp only: conc.simps(2))
  finally show "conc (x # xs) (conc ys zs) = conc (conc (x # xs) ys) zs" 
    by this
qed

(* La demostración simplificada 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

text ‹Comentario sobre la demostración anterior
  · (induct xs) genera dos subobjetivos:
    1. conc [] (conc ys zs) = conc (conc [] ys) zs
    2. ⋀a xs. conc xs (conc ys zs) = conc (conc xs ys) zs ⟹
              conc (a#xs) (conc ys zs) = conc (conc (a#xs) ys) zs›

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 = [a2]
  ys = [a1]›

text ‹--------------------------------------------------------------- 
  Ejemplo 12. (p. 28) Demostrar que 
     conc xs [] = xs
  -------------------------------------------------------------------›

(* La demostración detallada es *)
lemma "conc xs [] = xs"
proof (induct xs)
  show "conc [] [] = []" 
    by (simp only: conc.simps(1))
next 
  fix x xs
  assume HI: "conc xs [] = xs" 
  have "conc (x # xs) [] = x # (conc xs [])" 
    by (simp only: conc.simps(2))
  also have "… = x # xs" 
    by (simp only: HI)
  finally show "conc (x # xs) [] = x # xs" 
    by this
qed

(* La demostración simplificada 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

text ‹--------------------------------------------------------------- 
  Ejemplo 13. (p. 30) Demostrar que 
     longitud (conc xs ys) = longitud xs + longitud ys
  -------------------------------------------------------------------›

(* La demostración detallada es *)
lemma "longitud (conc xs ys) = longitud xs + longitud ys"
proof (induct xs)
  have "longitud (conc [] ys) = longitud ys" 
    by (simp only: conc.simps(1))
  also have "… = 0 + longitud ys"
    by (simp only: add_0)
  also  have "… = longitud [] + longitud ys"
    by (simp only: longitud.simps(1))
  finally show "longitud (conc [] ys) = longitud [] + longitud ys" 
    by this
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 only: conc.simps(2))
  also have "… = 1 + longitud (conc xs ys)" 
    by (simp only: longitud.simps(2))
  also have "… = 1 + longitud xs + longitud ys" 
    by (simp only: HI)
  also have "… = longitud (x # xs) + longitud ys" 
    by (simp only: longitud.simps(2))
  finally show "longitud (conc (x # xs) ys) = 
                longitud (x # xs) + longitud ys" 
    by this
qed

(* La demostración simplificada 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

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 ‹--------------------------------------------------------------- 
  Ejemplo 16. (p. 35) Demostrar que 
     conc (coge n xs) (elimina n xs) = xs
  -------------------------------------------------------------------›

(* La demostración detallada es *)
lemma "conc (coge n xs) (elimina n xs) = xs"
proof (induct rule: coge.induct)
  fix n
  have "conc (coge n []) (elimina n []) = conc [] (elimina n [])"
    by (simp only: coge.simps(1))
  also have "… = elimina n []"
    by (simp only: conc.simps(1))
  also have "… = []"
    by (simp only: elimina.simps(1))
  finally show "conc (coge n []) (elimina n []) = []" 
    by this
next
  fix x xs
  have "conc (coge 0 (x#xs)) (elimina 0 (x#xs)) = 
        conc [] (elimina 0 (x#xs))" 
    by (simp only: coge.simps(2))
  also have "… = elimina 0 (x#xs)"
    by  (simp only: conc.simps(1))
  also have "… = x # xs"
    by  (simp only: elimina.simps(2))
  finally show "conc (coge 0 (x#xs)) (elimina 0 (x#xs)) = x#xs" 
    by this 
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 only: coge.simps(3) 
                   elimina.simps(3))
  also have "… = x # (conc (coge n xs) (elimina n xs))" 
    by (simp only: conc.simps(2))
  also have "… = x#xs" 
    by (simp only: HI)  
  finally show "conc (coge (Suc n) (x#xs)) (elimina (Suc n) (x#xs)) = 
                x#xs"
    by this
qed

(* La demostración simplificada 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

text ‹Comentario sobre la demostración anterior:
  · (induct rule: coge.induct) indica que el método de demostración es
    por el esquema de inducción correspondiente a la definición de la
    función coge.
  · Se generan 3 subobjetivos:
    · 1. ⋀n. conc (coge n []) (elimina n []) = []
    · 2. ⋀x xs. conc (coge 0 (x#xs)) (elimina 0 (x#xs)) = x#xs
    · 3. ⋀n x xs. 
            conc (coge n xs) (elimina n xs) = xs ⟹
            conc (coge (Suc n) (x#xs)) (elimina (Suc n) (x#xs)) = x#xs›

section ‹Razonamiento por casos›

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 [a] = 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 = []"
  then show "esVacia xs = esVacia (conc xs xs)" by simp
next
  fix y ys
  assume "xs = y#ys"
  then show "esVacia xs = esVacia (conc xs xs)" by simp
qed

text ‹Comentarios sobre la demostración anterior:
  · "(cases xs)" es el método de demostración por casos según xs.
  · Se generan dos subobjetivos  correspondientes a los dos
    constructores de listas:
    · 1. xs = [] ⟹ esVacia xs = esVacia (conc xs xs)
    · 2. ⋀y ys. xs = y#ys ⟹ esVacia xs = esVacia (conc xs xs)
  · "then" indica "usando la propiedad anterior"›

(* La demostración estructurada simplificada es *)
lemma "esVacia xs = esVacia (conc xs xs)"
proof (cases xs)
  case Nil
  then show "esVacia xs = esVacia (conc xs xs)" by simp
next
  case Cons
  then show "esVacia xs = esVacia (conc xs xs)" by simp
qed

text ‹
  Comentarios sobre la demostración anterior:
  · "case Nil" es una abreviatura de "assume xs = []"
  · "case Cons" es una abreviatura de "fix y ys assume xs = y#ys"›

(* La demostración con el patrón sugerido es *)
lemma "esVacia xs = esVacia (conc xs xs)"
proof (cases xs)
  case Nil
  then show ?thesis by simp
next
  case (Cons x xs)
  then show ?thesis by simp
qed

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 detallada es *)
lemma inversaAcAux_es_inversa:
  "inversaAcAux xs ys = (inversa xs) @ ys"
proof (induct xs arbitrary: ys)
  fix ys
  have "inversaAcAux [] ys = ys"
    by (simp only: inversaAcAux.simps(1))
  also have "… = [] @ ys"
    by (simp only: append.simps(1))
  also have "… = inversa [] @ ys"
    by (simp only: inversa.simps(1))
  finally show "inversaAcAux [] ys = inversa [] @ ys" 
    by this
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 only: inversaAcAux.simps(2))
    also have "… = inversa xs@(a#ys)" 
      by (simp only: HI)
    also have "… = inversa xs @ ([a] @ ys)"
      by (simp only: append.simps) 
    also have "… = (inversa xs @ [a]) @ ys"
      by (simp only: append_assoc) 
    also have "… = inversa (a # xs) @ ys"
      by (simp only: inversa.simps(2)) 
    finally show "inversaAcAux (a#xs) ys = inversa (a#xs)@ys" 
      by this
  qed
qed

(* La demostración simplificada 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

text ‹Comentarios sobre la demostración anterior:
  · "(induct xs arbitrary: ys)" es el método de demostración por
    inducción sobre xs usando ys como variable arbitraria.
  · Se generan dos subobjetivos:
    · 1. ⋀ys. inversaAcAux [] ys = inversa [] @ ys
    · 2. ⋀a xs ys. (⋀ys. inversaAcAux xs ys = inversa xs @ ys) ⟹
                    inversaAcAux (a # xs) ys = inversa (a # xs) @ ys
  · Dentro de una demostración se pueden incluir otras demostraciones.
  · Para demostrar la propiedad universal "⋀ys. P(ys)" se elige una
    lista arbitraria (con "fix ys") y se demuestra "P(ys)". ›

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)

text ‹  Comentario de la demostración anterior:
  · "(simp add: inversaAcAux_es_inversa)" es el método de demostración
    por simplificación usando como regla de simplificación la propiedad
    inversaAcAux_es_inversa. ›

section ‹Demostración por inducción para funciones de orden superior›

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

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

value "suma [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 ((*) 2) [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 ((*) 2) [3::nat,2,5] = [6,4,10]"

text ‹--------------------------------------------------------------- 
  Ejemplo 24. (p. 45) Demostrar que 
     suma (map ((*) 2) xs) = 2 * (suma xs)
  -------------------------------------------------------------------›

(* La demostración detallada es *)
lemma "suma (map ((*) 2) xs) = 2 * (suma xs)"
proof (induct xs)
  have "suma (map ((*) 2) []) = suma []"
    by (simp only: map.simps(1))
  also have "… = 0"
    by (simp only: suma.simps(1))
  also have "… = 2 * 0"
    by  (simp only: mult_0_right)
  also have "… = 2 * suma []"
    by (simp only: suma.simps(1))
  finally show "suma (map ((*) 2) []) = 2 * suma []" 
    by this 
next
  fix a xs
  assume HI: "suma (map ((*) 2) xs) = 2 * suma xs"
  have "suma (map ((*) 2) (a#xs)) = suma ((2*a)#(map ((*) 2) xs))" 
    by (simp only: map.simps(2))
  also have "… = 2*a + suma (map ((*) 2) xs)" 
    by (simp only: suma.simps(2))
  also have "… = 2*a + 2 * suma xs" 
    by (simp only: HI)
  also have "… = 2 * (a + suma xs)" 
    by (simp only: add_mult_distrib2)
  also have "… = 2 * suma (a#xs)" 
    by (simp only: suma.simps(2))
  finally show "suma (map ((*) 2) (a#xs)) = 2 * suma (a#xs)" 
    by this
qed

(* La demostración simplificada es *)
lemma "sum (map ((*) 2) xs) = 2 * (sum xs)"
proof (induct xs)
  show "sum (map ((*) 2) []) = 2 * (sum [])" 
    by simp
next
  fix a xs
  assume HI: "sum (map ((*) 2) xs) = 2 * (sum xs)"
  have "sum (map ((*) 2) (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 ((*) 2) (a#xs)) = 2*(sum (a#xs))" 
    by simp
qed

text ‹--------------------------------------------------------------- 
  Ejemplo 25. (p. 48) Demostrar que 
     longitud (map f xs) = longitud xs
  -------------------------------------------------------------------›

(* declare [[show_types]] *)

(* La demostración detallada 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 only: map.simps(2))
  also have "… = 1 + longitud (map f xs)" 
    by (simp only: longitud.simps(2))
  also have "… = 1 + longitud xs" 
    by (simp only: HI)
  also have "… = longitud (a#xs)" 
    by (simp only: longitud.simps(2))
  finally show "longitud (map f (a#xs)) = longitud (a#xs)" 
    by this
qed

(* La demostración simplificada 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

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

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chapter ‹Tema 3: Razonamiento estructurado sobre programas›

theory T3_Razonamiento_estructurado_sobre_programas

imports Main

begin

text ‹En este tema se demuestra con Isabelle las propiedades de los

programas funcionales como se expone en el tema 2a y se demostraron

automáticamente en el tema 2b. A diferencia del tema 2b, ahora

nos fijamos no sólo en el método de demostración sino en la estructura

de la prueba resaltando su semejanza con las del tema 2a.›

declare [[names_short]]

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 [a,c,d] = 3

-------------------------------------------------------------------›

fun longitud :: "'a list ⇒ nat" where

"longitud [] = 0"

| "longitud (x#xs) = 1 + longitud xs"

value "longitud [a,c,d] = 3"

text ‹---------------------------------------------------------------

Ejemplo 2. Demostrar que

longitud [a,c,d] = 3

-------------------------------------------------------------------›

(* declare [[show_types]]*)

lemma "longitud [a,c,d] = 3"

proof -

have "longitud [a,c,d] = 1 + longitud [c,d]"

by (simp only: longitud.simps(2))

also have "… = 1 + (1 + longitud [d])"

by (simp only: longitud.simps(2))

also have "… = 1 + (1 + (1 + longitud ([] :: 'a list)))"

by (simp only: longitud.simps(2))

also have "… = 1 + (1 + (1 + 0))"

by (simp only: longitud.simps(1))

also have "… = 3"

by (simp only: )

finally show "longitud [a,c,d] = 3"

by this

qed

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)

-------------------------------------------------------------------›

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 this

qed

text ‹Notas sobre el lenguaje: En la demostración anterior se ha usado

· "proof" para iniciar la prueba,

· "-" (después de "proof") para no usar el método por defecto,

· "have" para establecer un paso,

· "by (simp only: intercambia.simps)" para indicar que sólo se usa

como regla de escritura la correspondiente a la definición de

intercambia,

· "also" para encadenar pasos ecuacionales,

· "..." para representar la derecha de la igualdad anterior en un

razonamiento ecuacional,

· "finally" para indicar el último pasa de un razonamiento ecuacional,

· "show" para establecer la conclusión.

· "by simp" para indicar el método de demostración por simplificación y

· "qed" para terminar la pruebas,›

(* Demostración declarativa simplificada *)

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

text ‹

Nota: La diferencia entre las dos demostraciones es que en los dos

primeros pasos no se explicita la regla de simplificación. ›

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 declarativa es *)

lemma "inversa [x] = [x]"

proof -

have "inversa [x] = (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 this

qed

(* La demostración declarativa simplificada es *)

lemma "inversa [x] = [x]"

proof -

have "inversa [x] = (inversa []) @ [x]" by simp

also have "… = [] @ [x]" by simp

also have "… = [x]" by simp

finally show "inversa [x] = [x]" by simp

qed

section ‹Razonamiento por inducción sobre los naturales›

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

-------------------------------------------------------------------›

(* declare [[show_types]] *)

(* La demostración detallada es *)

lemma "longitud (repite n x) = n"

proof (induct n)

have "longitud (repite 0 x) = longitud ([] :: 'a list)"

by (simp only: repite.simps(1))

also have "… = 0"

by (simp only: longitud.simps(1))

finally show "longitud (repite 0 x) = 0"

by this

fix n

assume HI: "longitud (repite n x) = n"

have "longitud (repite (Suc n) x) = longitud (x # (repite n x))"

by (simp only: repite.simps(2))

also have "… = 1 + longitud (repite n x)"

by (simp only: longitud.simps(2))

also have "… = 1 + n"

by (simp only: HI)

also have "… = Suc n"

(* find_theorems "Suc _ = _" *)

by (simp only: Suc_eq_plus1)

finally show "longitud (repite (Suc n) x) = Suc n"

by this

qed

(* La demostración simplificada es *)

lemma "longitud (repite n x) = n"

proof (induct n)

show "longitud (repite 0 x) = 0"

by simp

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

text ‹Comentarios sobre la demostración anterior:

· A la derecha de proof se indica el método de la demostración.

· (induct n) indica que la demostración se hará por inducción en n.

· Se generan dos subobjetivos correspondientes a la base y el paso de

inducción:

1. longitud (repite 0 x) = 0

2. ⋀n. longitud (repite n x) = n ⟹ longitud (repite (Suc n) x) = Suc n

donde ⋀n se lee "para todo n".

· "next" indica el siguiente subobjetivo.

· "fix n" indica "sea n un número natural cualquiera"

· assume HI: "longitud (repite n x) = n" indica «supongamos que

"longitud (repite n x) = n" y sea HI la etiqueta de este supuesto».

· "using HI" usando la propiedad etiquetada con HI. ›

section ‹Razonamiento por inducción sobre listas›

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 detallada es *)

lemma "conc xs (conc ys zs) = conc (conc xs ys) zs"

proof (induct xs)

have "conc [] (conc ys zs) = conc ys zs"

by (simp only: conc.simps(1))

also have "… = conc (conc [] ys) zs"

by (simp only: conc.simps(1))

finally show "conc [] (conc ys zs) = conc (conc [] ys) zs"

by this

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 only: conc.simps(2))

also have "… = x # (conc (conc xs ys) zs)"

by (simp only: HI)

also have "… = conc (conc (x # xs) ys) zs"

by (simp only: conc.simps(2))

finally show "conc (x # xs) (conc ys zs) = conc (conc (x # xs) ys) zs"

by this

qed

(* La demostración simplificada 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

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

text ‹Comentario sobre la demostración anterior

· (induct xs) genera dos subobjetivos:

1. conc [] (conc ys zs) = conc (conc [] ys) zs

2. ⋀a xs. conc xs (conc ys zs) = conc (conc xs ys) zs ⟹

conc (a#xs) (conc ys zs) = conc (conc (a#xs) ys) zs›

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 = [a2]

ys = [a1]›

text ‹---------------------------------------------------------------

Ejemplo 12. (p. 28) Demostrar que

conc xs [] = xs

-------------------------------------------------------------------›

(* La demostración detallada es *)

lemma "conc xs [] = xs"

proof (induct xs)

show "conc [] [] = []"

by (simp only: conc.simps(1))

fix x xs

assume HI: "conc xs [] = xs"

have "conc (x # xs) [] = x # (conc xs [])"

by (simp only: conc.simps(2))

also have "… = x # xs"

by (simp only: HI)

finally show "conc (x # xs) [] = x # xs"

by this

qed

(* La demostración simplificada es *)

lemma "conc xs [] = xs"

proof (induct xs)

show "conc [] [] = []" by simp

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

text ‹---------------------------------------------------------------

Ejemplo 13. (p. 30) Demostrar que

longitud (conc xs ys) = longitud xs + longitud ys

-------------------------------------------------------------------›

(* La demostración detallada es *)

lemma "longitud (conc xs ys) = longitud xs + longitud ys"

proof (induct xs)

have "longitud (conc [] ys) = longitud ys"

by (simp only: conc.simps(1))

also have "… = 0 + longitud ys"

by (simp only: add_0)

also have "… = longitud [] + longitud ys"

by (simp only: longitud.simps(1))

finally show "longitud (conc [] ys) = longitud [] + longitud ys"

by this

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 only: conc.simps(2))

also have "… = 1 + longitud (conc xs ys)"

by (simp only: longitud.simps(2))

also have "… = 1 + longitud xs + longitud ys"

by (simp only: HI)

also have "… = longitud (x # xs) + longitud ys"

by (simp only: longitud.simps(2))

finally show "longitud (conc (x # xs) ys) =

longitud (x # xs) + longitud ys"

by this

qed

(* La demostración simplificada es *)

lemma "longitud (conc xs ys) = longitud xs + longitud ys"

proof (induct xs)

show "longitud (conc [] ys) = longitud [] + longitud ys" by simp

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

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 ‹---------------------------------------------------------------

Ejemplo 16. (p. 35) Demostrar que

conc (coge n xs) (elimina n xs) = xs

-------------------------------------------------------------------›

(* La demostración detallada es *)

lemma "conc (coge n xs) (elimina n xs) = xs"

proof (induct rule: coge.induct)

fix n

have "conc (coge n []) (elimina n []) = conc [] (elimina n [])"

by (simp only: coge.simps(1))

also have "… = elimina n []"

by (simp only: conc.simps(1))

also have "… = []"

by (simp only: elimina.simps(1))

finally show "conc (coge n []) (elimina n []) = []"

by this

fix x xs

have "conc (coge 0 (x#xs)) (elimina 0 (x#xs)) =

conc [] (elimina 0 (x#xs))"

by (simp only: coge.simps(2))

also have "… = elimina 0 (x#xs)"

by (simp only: conc.simps(1))

also have "… = x # xs"

by (simp only: elimina.simps(2))

finally show "conc (coge 0 (x#xs)) (elimina 0 (x#xs)) = x#xs"

by this

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 only: coge.simps(3)

elimina.simps(3))

also have "… = x # (conc (coge n xs) (elimina n xs))"

by (simp only: conc.simps(2))

also have "… = x#xs"

by (simp only: HI)

finally show "conc (coge (Suc n) (x#xs)) (elimina (Suc n) (x#xs)) =

x#xs"

by this

qed

(* La demostración simplificada 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

fix x xs

show "conc (coge 0 (x#xs)) (elimina 0 (x#xs)) = x#xs" by simp

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

text ‹Comentario sobre la demostración anterior:

· (induct rule: coge.induct) indica que el método de demostración es

por el esquema de inducción correspondiente a la definición de la

función coge.

· Se generan 3 subobjetivos:

· 1. ⋀n. conc (coge n []) (elimina n []) = []

· 2. ⋀x xs. conc (coge 0 (x#xs)) (elimina 0 (x#xs)) = x#xs

· 3. ⋀n x xs.

conc (coge n xs) (elimina n xs) = xs ⟹

conc (coge (Suc n) (x#xs)) (elimina (Suc n) (x#xs)) = x#xs›

section ‹Razonamiento por casos›

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 [a] = 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 = []"

then show "esVacia xs = esVacia (conc xs xs)" by simp

fix y ys

assume "xs = y#ys"

then show "esVacia xs = esVacia (conc xs xs)" by simp

qed

text ‹Comentarios sobre la demostración anterior:

· "(cases xs)" es el método de demostración por casos según xs.

· Se generan dos subobjetivos correspondientes a los dos

constructores de listas:

· 1. xs = [] ⟹ esVacia xs = esVacia (conc xs xs)

· 2. ⋀y ys. xs = y#ys ⟹ esVacia xs = esVacia (conc xs xs)

· "then" indica "usando la propiedad anterior"›

(* La demostración estructurada simplificada es *)

lemma "esVacia xs = esVacia (conc xs xs)"

proof (cases xs)

case Nil

then show "esVacia xs = esVacia (conc xs xs)" by simp

case Cons

then show "esVacia xs = esVacia (conc xs xs)" by simp

qed

text ‹

Comentarios sobre la demostración anterior:

· "case Nil" es una abreviatura de "assume xs = []"

· "case Cons" es una abreviatura de "fix y ys assume xs = y#ys"›

(* La demostración con el patrón sugerido es *)

lemma "esVacia xs = esVacia (conc xs xs)"

proof (cases xs)

case Nil

then show ?thesis by simp

case (Cons x xs)

then show ?thesis by simp

qed

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 detallada es *)

lemma inversaAcAux_es_inversa:

"inversaAcAux xs ys = (inversa xs) @ ys"

proof (induct xs arbitrary: ys)

fix ys

have "inversaAcAux [] ys = ys"

by (simp only: inversaAcAux.simps(1))

also have "… = [] @ ys"

by (simp only: append.simps(1))

also have "… = inversa [] @ ys"

by (simp only: inversa.simps(1))

finally show "inversaAcAux [] ys = inversa [] @ ys"

by this

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 only: inversaAcAux.simps(2))

also have "… = inversa xs@(a#ys)"

by (simp only: HI)

also have "… = inversa xs @ ([a] @ ys)"

by (simp only: append.simps)

also have "… = (inversa xs @ [a]) @ ys"

by (simp only: append_assoc)

also have "… = inversa (a # xs) @ ys"

by (simp only: inversa.simps(2))

finally show "inversaAcAux (a#xs) ys = inversa (a#xs)@ys"

by this

qed

(* La demostración simplificada es *)

lemma inversaAcAux_es_inversa:

"inversaAcAux xs ys = (inversa xs) @ ys"

proof (induct xs arbitrary: ys)

show "⋀ys. inversaAcAux [] ys = inversa [] @ ys" by simp

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

text ‹Comentarios sobre la demostración anterior:

· "(induct xs arbitrary: ys)" es el método de demostración por

inducción sobre xs usando ys como variable arbitraria.

· Se generan dos subobjetivos:

· 1. ⋀ys. inversaAcAux [] ys = inversa [] @ ys

· 2. ⋀a xs ys. (⋀ys. inversaAcAux xs ys = inversa xs @ ys) ⟹

inversaAcAux (a # xs) ys = inversa (a # xs) @ ys

· Dentro de una demostración se pueden incluir otras demostraciones.

· Para demostrar la propiedad universal "⋀ys. P(ys)" se elige una

lista arbitraria (con "fix ys") y se demuestra "P(ys)". ›

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)

text ‹ Comentario de la demostración anterior:

· "(simp add: inversaAcAux_es_inversa)" es el método de demostración

por simplificación usando como regla de simplificación la propiedad

inversaAcAux_es_inversa. ›

section ‹Demostración por inducción para funciones de orden superior›

text ‹---------------------------------------------------------------

Ejemplo 22. Definir la función

suma :: nat list ⇒ nat

tal que (suma xs) es la suma de los elementos de xs. Por ejemplo,

suma [3,2,5] = 10

------------------------------------------------------------------›

fun suma :: "nat list ⇒ nat" where

"suma [] = 0"

| "suma (x#xs) = x + suma xs"

value "suma [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 ((*) 2) [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 ((*) 2) [3::nat,2,5] = [6,4,10]"

text ‹---------------------------------------------------------------

Ejemplo 24. (p. 45) Demostrar que

suma (map ((*) 2) xs) = 2 * (suma xs)

-------------------------------------------------------------------›

(* La demostración detallada es *)

lemma "suma (map ((*) 2) xs) = 2 * (suma xs)"

proof (induct xs)

have "suma (map ((*) 2) []) = suma []"

by (simp only: map.simps(1))

also have "… = 0"

by (simp only: suma.simps(1))

also have "… = 2 * 0"

by (simp only: mult_0_right)

also have "… = 2 * suma []"

by (simp only: suma.simps(1))

finally show "suma (map ((*) 2) []) = 2 * suma []"

by this

fix a xs

assume HI: "suma (map ((*) 2) xs) = 2 * suma xs"

have "suma (map ((*) 2) (a#xs)) = suma ((2*a)#(map ((*) 2) xs))"

by (simp only: map.simps(2))

also have "… = 2*a + suma (map ((*) 2) xs)"

by (simp only: suma.simps(2))

also have "… = 2*a + 2 * suma xs"

by (simp only: HI)

also have "… = 2 * (a + suma xs)"

by (simp only: add_mult_distrib2)

also have "… = 2 * suma (a#xs)"

by (simp only: suma.simps(2))

finally show "suma (map ((*) 2) (a#xs)) = 2 * suma (a#xs)"

by this

qed

(* La demostración simplificada es *)

lemma "sum (map ((*) 2) xs) = 2 * (sum xs)"

proof (induct xs)

show "sum (map ((*) 2) []) = 2 * (sum [])"

by simp

fix a xs

assume HI: "sum (map ((*) 2) xs) = 2 * (sum xs)"

have "sum (map ((*) 2) (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 ((*) 2) (a#xs)) = 2*(sum (a#xs))"

by simp

qed

text ‹---------------------------------------------------------------

Ejemplo 25. (p. 48) Demostrar que

longitud (map f xs) = longitud xs

-------------------------------------------------------------------›

(* declare [[show_types]] *)

(* La demostración detallada es *)

lemma "longitud (map f xs) = longitud xs"

proof (induct xs)

show "longitud (map f []) = longitud []"

by simp

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 only: map.simps(2))

also have "… = 1 + longitud (map f xs)"

by (simp only: longitud.simps(2))

also have "… = 1 + longitud xs"

by (simp only: HI)

also have "… = longitud (a#xs)"

by (simp only: longitud.simps(2))

finally show "longitud (map f (a#xs)) = longitud (a#xs)"

by this

qed

(* La demostración simplificada es *)

lemma "longitud (map f xs) = longitud xs"

proof (induct xs)

show "longitud (map f []) = longitud []"

by simp

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

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

Como tarea para la próxima clase se propuso la resolución de los ejercicios de la 3ª relación

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