Razonamiento automático (2018-19) Razonamiento automático (2018-19)

chapter {* R3: Razonamiento sobre programas *}

theory R3_Razonamiento_sobre_programas
imports Main 
begin

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
  ------------------------------------------------------------------ *}

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

text {* --------------------------------------------------------------- 
  Ejercicio 1.2. Escribir la demostración detallada de 
     sumaImpares n = n*n
  ------------------------------------------------------------------- *}

(* "La demostración detallada es" *)

(* benber *)
lemma "sumaImpares n = n*n"
proof (induct n)
  show "sumaImpares 0 = 0*0" by (simp only: sumaImpares.simps(1))
next
  fix n
  assume HI: "sumaImpares n = n*n"
  have "sumaImpares (Suc n) = sumaImpares n + (2*n+1)" by (simp only: sumaImpares.simps(2))
  also have "... = n*n + (2*n+1)" by (simp only: HI)
  also have "... = n*n + (n+n+1)" by (simp only:)
  also have "... = n*n + n + n + 1" by (simp only:)
  also have "... = n*n + n*1 + n + 1" by (simp only:)
  also have "... = n*(n+1) + n + 1" by (simp only: Nat.add_mult_distrib2)
  also have "... = n*(n+1) + (n+1)" by (simp only:)
  also have "... = (n+1)*n + (n+1)" by (simp only: Groups.ab_semigroup_mult_class.mult.commute)
  also have "... = (n+1)*n + (n+1)*1" by (simp only: Nat.nat_mult_1_right)
  also have "... = (n+1)*(n+1)" by (simp only: Nat.add_mult_distrib2)
  also have "... = (Suc n) * (Suc n)" by (simp only: Nat.Suc_eq_plus1)
  finally show "sumaImpares (Suc n) = (Suc n) * (Suc n)" .
qed

(* pabalagon josgomrom4 hugrubsan *)
lemma "sumaImpares n = n*n"
proof (induct n)
  show "sumaImpares 0 = 0*0" by (simp only: sumaImpares.simps(1))
next
  fix n
  assume HI: "sumaImpares n = n*n"
  have "sumaImpares (Suc n) = sumaImpares n + 2 * n + 1" by (simp only: sumaImpares.simps(2))
  also have "... = n*n + 2*n + 1" using HI by simp
  also have "... = n*n + n + n + 1" by (simp only: Nat.add_mult_distrib)
  also have "... = (n+1)*n + (n + 1)" by (simp only: Nat.add_mult_distrib)
  also have "... = (n+1)*n + (n + 1)*1" by (simp only: Nat.nat_mult_1_right)
  also have "... = (n+1)*(n+1)" by (simp only: Nat.add_mult_distrib2)
  finally show "sumaImpares (Suc n) = (Suc n)*(Suc n)" 
    by (simp only: Nat.Suc_eq_plus1)
qed

(*marfruman1 raffergon2 aribatval alfmarcua cammonagu*)
lemma "sumaImpares n = n*n"
proof (induct n)
  show "sumaImpares 0 = 0*0" by (simp only: sumaImpares.simps(1))
next
  fix n
  assume HI: "sumaImpares n = n*n"
  have "sumaImpares (Suc n) = sumaImpares n + 2*n +1" by (simp only: sumaImpares.simps(2))
  also have "... = n*n + 2*n + 1" using HI by simp
  also have "... = (Suc n)*(Suc n)" by simp
  finally show "sumaImpares (Suc n) = (Suc n)*(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
  ------------------------------------------------------------------ *}

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

text {* --------------------------------------------------------------- 
  Ejercicio 2.2. Escribir la demostración detallada de 
     sumaPotenciasDeDosMasUno n = 2^(n+1)
  ------------------------------------------------------------------- *}

(* benber josgomrom4 *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
  have "sumaPotenciasDeDosMasUno 0 = 2" by (simp only: sumaPotenciasDeDosMasUno.simps(1))
  also have "... = 2^(0+1)" by simp
  finally show "sumaPotenciasDeDosMasUno 0 = 2^(0+1)" .
next
  fix n
  assume HI: "sumaPotenciasDeDosMasUno n = 2^(n+1)"
  have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1)"
    by (simp only: sumaPotenciasDeDosMasUno.simps(2))
  also have "... = 2^(n+1) + 2^(n+1)" by (simp only: HI)
  also have "... = 2 * 2^(n+1)" by (simp only:)
  also have "... = 2 * 2^(Suc n)" by (simp only: Nat.Suc_eq_plus1)
  also have "... = 2^(Suc (Suc n))" by (simp only: Power.power_class.power.power_Suc)
  also have "... = 2^((Suc n)+1)" by (simp only: Nat.Suc_eq_plus1)
  finally show "sumaPotenciasDeDosMasUno (Suc n) = 2^((Suc n)+1)" .
qed

(*marfruman1 alfmarcua hugrubsan *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
  have "sumaPotenciasDeDosMasUno 0 = 2" by (simp only: sumaPotenciasDeDosMasUno.simps(1))
  also have "... = 2^(0+1)" by simp
  finally 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^(n+1)"
    by (simp only: sumaPotenciasDeDosMasUno.simps(2))
  also have "... = 2^(n+1)+2^(n+1)" using HI by simp
  also have "... =2*2^(n+1)" by simp
  also have "... = 2^((Suc n)+1)" by simp
  finally show "sumaPotenciasDeDosMasUno (Suc n) = 2^((Suc n)+1)" 
        by simp
qed

(* pabalagon raffergon2 aribatval cammonagu hugrubsan *)
lemma "sumaPotenciasDeDosMasUno n = 2^(n+1)"
proof (induct n)
  have "sumaPotenciasDeDosMasUno 0 = 2" 
    by (simp only: sumaPotenciasDeDosMasUno.simps(1))
  also have "... = 2^1" by (simp only: Power.power_one_right)
  finally show "sumaPotenciasDeDosMasUno 0 = 2^(0+1)" by (simp only: Nat.add_0)
next
  fix n
  assume HI: "sumaPotenciasDeDosMasUno n = 2^(n+1)"
  have "sumaPotenciasDeDosMasUno (Suc n) = sumaPotenciasDeDosMasUno n + 2^(n+1)"
    by (simp only: sumaPotenciasDeDosMasUno.simps(2))
  also have "... = 2^(n+1) + 2^(n+1)" by (simp only: HI)
  also have "... = 2*2^(n+1)" by (simp only: Nat.add_mult_distrib)
  also have "... = 2*2^(Suc n)" by (simp only: Nat.Suc_eq_plus1)
  also have "... = 2^(Suc n)*2^1" by (simp only: Power.power_one_right)
  finally show "sumaPotenciasDeDosMasUno (Suc n) = 2^(Suc n + 1)" by (simp only: Power.power_add)
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]
  ------------------------------------------------------------------ *}

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

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

text {* --------------------------------------------------------------- 
  Ejercicio 3.2. Demostrar detalladamente que todos los elementos de
  (copia n x) son iguales a x. 
  ------------------------------------------------------------------- *}

(* benber raffergon2 aribatval josgomrom4 hugrubsan *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
  fix x :: 'a
  have "todos (λy. y=x) (copia 0 x) = todos (λy. y=x) []" by (simp only: copia.simps(1))
  also have "..." by (simp only: todos.simps(1))
  finally show "todos (λy. y = x) (copia 0 x)" .
next
  fix x :: 'a
  fix n :: nat
  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(2))
  also have "... = ( ( (λy. y = x) x ) ∧ todos (λy. y = x) (copia n x) )"
    by (simp only: todos.simps(2))
  also have "..." by (simp only: HI)
  finally show "todos (λy. y = x) (copia (Suc n) x)" .
qed

(* pabalagon *)
lemma "todos (λy. y=x) (copia n x)"
proof (induction n)
  fix x:: 'a
  have "todos (λy. y=x) (copia 0 x) = todos (λy. y=x) []" by simp
  also have "..." by simp
  finally show "todos (λy. y=x) (copia 0 x)" .
next
  fix x:: 'a
  fix n:: nat
  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 "... = ((λy. y=x) x ∧ todos (λy. y=x) (copia n x))" by simp
  finally show "todos (λy. y=x) (copia (Suc n) x)" using HI by simp
qed

(* alfmarcua *)
lemma "todos (λy. y=x) (copia n x)"
proof (induct n)
  have "todos (λy. y = x) (copia 0 x) = todos (λy. y = x) []" by (simp only: copia.simps(1))
  also have "..." by (simp only: todos.simps(1))
  finally show "todos (λy. y = x) (copia 0 x)" .
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(2))
  also have "... = ( ( (λy. y = x) x ) ∧ todos (λy. y = x) (copia n x) )"
    by (simp only: todos.simps(2))
  also have "..." by (simp only: HI)
  finally show "todos (λy. y = x) (copia (Suc n) x)" .
qed

text {* --------------------------------------------------------------- 
  Ejercicio 4.1. Definir la función
    factR :: nat ⇒ nat
  tal que (factR n) es el factorial de n. Por ejemplo,
    factR 4 = 24
  ------------------------------------------------------------------ *}

fun factR :: "nat ⇒ nat" where
  "factR 0       = 1"
| "factR (Suc n) = Suc n * factR n"

text {* --------------------------------------------------------------- 
  Ejercicio 4.2. 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
  Indicación: La propiedad mult_Suc es 
     (Suc m) * n = n + m * n
  Puede que se necesite desactivarla en un paso con 
     (simp del: mult_Suc)
  ------------------------------------------------------------------- *}

fun factI' :: "nat ⇒ nat ⇒ nat" where
  "factI' 0       x = x"
| "factI' (Suc n) x = factI' n (x * Suc n)"

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

(* benber *)
lemma fact: "factI' n x = x * factR n"
proof -
  (* Reformulación para aplicar la hipótesis inductiva con un valor distinto de x. *)
  have "∀x. factI' n x = x * factR n"
  proof (induct n)
    show "∀x. factI' 0 x = x * factR 0"
    proof
      fix x
      have "factI' 0 x = x" by (simp only: factI'.simps(1))
      also have "... = x * 1" by (simp only:)
      also have "... = x * factR 0" by (simp only: factR.simps(1))
      finally show "factI' 0 x = x * factR 0" .
    qed
  next
    show "⋀n. ∀x. factI' n x = x * factR n ⟹ ∀x. factI' (Suc n) x = x * factR (Suc n)"
    proof
      fix x
      fix n
      assume HI: "∀y. factI' n y = y * factR n"
      have "factI' (Suc n) x =  factI' n (x * Suc n)" by (simp only: factI'.simps(2))
      also have "... = (x * Suc n) * factR n" by (simp only: HI)
      also have "... = x * factR (Suc n)" by (simp only: factR.simps(2))
      finally show "factI' (Suc n) x = x * factR (Suc n)" .
    qed
  qed
  thus "factI' n x = x * factR n" by simp
qed

(* pabalagon raffergon2 aribatval josgomrom4 alfmarcua hugrubsan *)
lemma fact: "factI' n x = x * factR n"
proof (induction n arbitrary: x)
  fix x
  have "factI' 0 x = x" by simp
  also have "... = x * 1" by simp
  finally show "factI' 0 x = x * factR 0" by simp
next
  fix n
  assume HI: "⋀x. factI' n x = x * factR n"
  show "⋀x. factI' (Suc n) x = x * factR (Suc n)"
  proof -
    fix x
    have "factI' (Suc n) x = factI' n (x * Suc n)" by (simp only: factI'.simps(2))
    also have "... = x * Suc n * factR n" using HI by simp
    finally show "factI' (Suc n) x = x*factR (Suc n)" by (simp only: factR.simps(2))
  qed
qed

text {* --------------------------------------------------------------- 
  Ejercicio 4.3. Escribir la demostración detallada de
     factI n = factR n
  ------------------------------------------------------------------- *}

(* benber *)
corollary "factI n = factR n"
proof -
  fix n
  have "factI n = factI' n 1" by (simp only: factI.simps(1))
  also have "... = 1 * factR n" by (simp only: fact)
  also have "... = factR n" by (simp only: Groups.mult_1)
  finally show "factI n = factR n" .
qed

(* pabalagon raffergon2 josgomrom4 alfmarcua hugrubsan *)
corollary "factI n = factR n"
proof -
  have "factI n = factI' n 1" by (simp only: factI.simps(1))
  also have "... = 1*factR n" using fact by simp
  finally show "factI n = factR n" by (simp only: mult_1)
qed

text {* --------------------------------------------------------------- 
  Ejercicio 5.1. 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]
  ------------------------------------------------------------------ *}

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

text {* --------------------------------------------------------------- 
  Ejercicio 5.2. Escribir la demostración detallada de
     amplia xs y = xs @ [y]
  ------------------------------------------------------------------- *}

(* benber aribatval alfmarcua *)
lemma "amplia xs y = xs @ [y]"
proof (induct xs)
  have "amplia [] y = [y]" by (simp only: amplia.simps(1))
  also have "... = [] @ [y]" by (simp only: List.append.left_neutral)
  finally show "amplia [] y = [] @ [y]" .
next
  fix x
  fix xs
  assume HI: "amplia xs y = xs @ [y]"
  have "amplia (x#xs) y = x # amplia xs y" by (simp only: amplia.simps(2))
  also have "... = x # (xs @ [y])" by (simp only: HI)
  also have "... = (x#xs) @ [y]" by (simp only: List.append.append_Cons)
  finally show "amplia (x#xs) y = (x#xs) @ [y]" .
qed

(*marfruman1 raffergon2*)
lemma "amplia xs y = xs @ [y]"
proof (induct xs)
  have "amplia [] y = [y]" by (simp only: amplia.simps(1))
  also have "... = [] @ [y]" by simp
  finally 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 only: amplia.simps(2))
  also have "... = x# (xs @ [y])" using HI by simp
  also have "... = (x#xs) @ [y]" by simp
  finally show "amplia (x#xs) y =(x#xs) @ [y]" by simp
qed

(* pabalagon josgomrom4 hugrubsan *)
lemma "amplia xs y = xs @ [y]"
proof (induction xs)
  have "amplia [] y = [y]" by (simp only: amplia.simps(1))
  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 only: amplia.simps(2))
  also have "... = a # (xs @ [y])" using HI by simp
  finally show "amplia (a#xs) y = (a#xs) @ [y]" by simp
qed

end