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Etiqueta: L.append_assoc

Pruebas de equivalencia de definiciones de inversa

José A. Alonso-

En Lean, está definida la función

   reverse : list α → list α

tal que (reverse xs) es la lista obtenida invirtiendo el orden de los elementos de xs. Por ejemplo,

   reverse  [3,2,5,1] = [1,5,2,3]

Su definición es

   def reverse_core : list α → list α → list α
   | []     r := r
   | (a::l) r := reverse_core l (a::r)
 
   def reverse : list α → list α :=
   λ l, reverse_core l []

Una definición alternativa es

   def inversa : list α → list α
   | []        := []
   | (x :: xs) := inversa xs ++ [x]

Demostrar que las dos definiciones son equivalentes; es decir,

   reverse xs = inversa xs

Para ello, completar la siguiente teoría de Lean:

import data.list.basic
open list
 
variable  {α : Type*}
variable  (x : α)
variables (xs ys : list α)
 
def inversa : list α → list α
| []        := []
| (x :: xs) := inversa xs ++ [x]
 
example : reverse xs = inversa xs :=
sorry
Soluciones con Lean
import data.list.basic
open list
 
variable  {α : Type*}
variable  (x : α)
variables (xs ys : list α)
 
-- Definición y reglas de simplificación de inversa
-- ================================================
 
def inversa : list α → list α
| []        := []
| (x :: xs) := inversa xs ++ [x]
 
@[simp]
lemma inversa_nil :
  inversa ([] : list α) = [] :=
rfl
 
@[simp]
lemma inversa_cons :
  inversa (x :: xs) = inversa xs ++ [x] :=
rfl
 
-- Reglas de simplificación de reverse_core
-- ========================================
 
@[simp]
lemma reverse_core_nil :
  reverse_core [] ys = ys :=
rfl
 
@[simp]
lemma reverse_core_cons :
  reverse_core (x :: xs) ys = reverse_core xs (x :: ys) :=
rfl
 
-- Lema auxiliar: reverse_core xs ys = (inversa xs) ++ ys
-- ======================================================
 
-- 1ª demostración del lema auxiliar
example :
  reverse_core xs ys = (inversa xs) ++ ys :=
begin
  induction xs with a as HI generalizing ys,
  { calc reverse_core [] ys
         = ys                        : reverse_core_nil ys
     ... = [] ++ ys                  : (nil_append ys).symm
     ... = inversa [] ++ ys          : congr_arg2 (++) inversa_nil.symm rfl, },
  { calc reverse_core (a :: as) ys
         = reverse_core as (a :: ys) : reverse_core_cons a as ys
     ... = inversa as ++ (a :: ys)   : (HI (a :: ys))
     ... = inversa as ++ ([a] ++ ys) : congr_arg2 (++) rfl singleton_append
     ... = (inversa as ++ [a]) ++ ys : (append_assoc (inversa as) [a] ys).symm
     ... = inversa (a :: as) ++ ys   : congr_arg2 (++) (inversa_cons a as).symm rfl},
end
 
-- 2ª demostración del lema auxiliar
example :
  reverse_core xs ys = (inversa xs) ++ ys :=
begin
  induction xs with a as HI generalizing ys,
  { calc reverse_core [] ys
         = ys                        : by rw reverse_core_nil
     ... = [] ++ ys                  : by rw nil_append
     ... = inversa [] ++ ys          : by rw inversa_nil },
  { calc reverse_core (a :: as) ys
         = reverse_core as (a :: ys) : by rw reverse_core_cons
     ... = inversa as ++ (a :: ys)   : by rw (HI (a :: ys))
     ... = inversa as ++ ([a] ++ ys) : by rw singleton_append
     ... = (inversa as ++ [a]) ++ ys : by rw append_assoc
     ... = inversa (a :: as) ++ ys   : by rw inversa_cons },
end
 
-- 3ª demostración del lema auxiliar
example :
  reverse_core xs ys = (inversa xs) ++ ys :=
begin
  induction xs with a as HI generalizing ys,
  { calc reverse_core [] ys
         = ys                        : rfl
     ... = [] ++ ys                  : rfl
     ... = inversa [] ++ ys          : rfl },
  { calc reverse_core (a :: as) ys
         = reverse_core as (a :: ys) : rfl
     ... = inversa as ++ (a :: ys)   : (HI (a :: ys))
     ... = inversa as ++ ([a] ++ ys) : rfl
     ... = (inversa as ++ [a]) ++ ys : by rw append_assoc
     ... = inversa (a :: as) ++ ys   : rfl },
end
 
-- 3ª demostración del lema auxiliar
example :
  reverse_core xs ys = (inversa xs) ++ ys :=
begin
  induction xs with a as HI generalizing ys,
  { calc reverse_core [] ys
         = ys                        : by simp
     ... = [] ++ ys                  : by simp
     ... = inversa [] ++ ys          : by simp },
  { calc reverse_core (a :: as) ys
         = reverse_core as (a :: ys) : by simp
     ... = inversa as ++ (a :: ys)   : (HI (a :: ys))
     ... = inversa as ++ ([a] ++ ys) : by simp
     ... = (inversa as ++ [a]) ++ ys : by simp
     ... = inversa (a :: as) ++ ys   : by simp },
end
 
-- 4ª demostración del lema auxiliar
example :
  reverse_core xs ys = (inversa xs) ++ ys :=
begin
  induction xs with a as HI generalizing ys,
  { by simp, },
  { calc reverse_core (a :: as) ys
         = reverse_core as (a :: ys) : by simp
     ... = inversa as ++ (a :: ys)   : (HI (a :: ys))
     ... = inversa (a :: as) ++ ys   : by simp },
end
 
-- 5ª demostración del lema auxiliar
example :
  reverse_core xs ys = (inversa xs) ++ ys :=
begin
  induction xs with a as HI generalizing ys,
  { simp, },
  { simp [HI (a :: ys)], },
end
 
-- 6ª demostración del lema auxiliar
example :
  reverse_core xs ys = (inversa xs) ++ ys :=
by induction xs generalizing ys ; simp [*]
 
-- 7ª demostración del lema auxiliar
example :
  reverse_core xs ys = (inversa xs) ++ ys :=
begin
  induction xs with a as HI generalizing ys,
  { rw reverse_core_nil,
    rw inversa_nil,
    rw nil_append, },
  { rw reverse_core_cons,
    rw (HI (a :: ys)),
    rw inversa_cons,
    rw append_assoc,
    rw singleton_append, },
end
 
-- 8ª demostración  del lema auxiliar
@[simp]
lemma inversa_equiv :
  ∀ xs : list α, ∀ ys, reverse_core xs ys = (inversa xs) ++ ys
| []         := by simp
| (a :: as)  := by simp [inversa_equiv as]
 
-- Demostraciones del lema principal
-- =================================
 
-- 1ª demostración
example : reverse xs = inversa xs :=
calc reverse xs
     = reverse_core xs [] : rfl
 ... = inversa xs ++ []   : by rw inversa_equiv
 ... = inversa xs         : by rw append_nil
 
-- 2ª demostración
example : reverse xs = inversa xs :=
by simp [inversa_equiv, reverse]
 
-- 3ª demostración
example : reverse xs = inversa xs :=
by simp [reverse]

Se puede interactuar con la prueba anterior en esta sesión con Lean.

En los comentarios se pueden escribir otras soluciones, escribiendo el código entre una línea con <pre lang="lean"> y otra con </pre>

Soluciones con Isabelle/HOL
theory Pruebas_de_equivalencia_de_definiciones_de_inversa
imports Main
begin
 
(* Definición alternativa *)
(* ====================== *)
 
fun inversa_aux :: "'a list ⇒ 'a list ⇒ 'a list" where
  "inversa_aux [] ys     = ys"
| "inversa_aux (x#xs) ys = inversa_aux xs (x#ys)"
 
fun inversa :: "'a list ⇒ 'a list" where
  "inversa xs = inversa_aux xs []"
 
(* Lema auxiliar: inversa_aux xs ys = (rev xs) @ ys *)
(* ================================================ *)
 
(* 1ª demostración del lema auxiliar *)
lemma
  "inversa_aux xs ys = (rev xs) @ ys"
proof (induct xs arbitrary: ys)
  fix ys :: "'a list"
  have "inversa_aux [] ys = ys"
    by (simp only: inversa_aux.simps(1))
  also have "… = [] @ ys"
    by (simp only: append.simps(1))
  also have "… = rev [] @ ys"
    by (simp only: rev.simps(1))
  finally show "inversa_aux [] ys = rev [] @ ys"
    by this
next
  fix a ::'a and xs :: "'a list"
  assume HI: "⋀ys. inversa_aux xs ys = rev xs@ys"
  show "⋀ys. inversa_aux (a#xs) ys = rev (a#xs)@ys"
  proof -
    fix ys
    have "inversa_aux (a#xs) ys = inversa_aux xs (a#ys)"
      by (simp only: inversa_aux.simps(2))
    also have "… = rev xs@(a#ys)"
      by (simp only: HI)
    also have "… = rev xs @ ([a] @ ys)"
      by (simp only: append.simps)
    also have "… = (rev xs @ [a]) @ ys"
      by (simp only: append_assoc)
    also have "… = rev (a # xs) @ ys"
      by (simp only: rev.simps(2))
    finally show "inversa_aux (a#xs) ys = rev (a#xs)@ys"
      by this
  qed
qed
 
(* 2ª demostración del lema auxiliar *)
lemma
  "inversa_aux xs ys = (rev xs) @ ys"
proof (induct xs arbitrary: ys)
  fix ys :: "'a list"
  have "inversa_aux [] ys = ys" by simp
  also have "… = [] @ ys" by simp
  also have "… = rev [] @ ys" by simp
  finally show "inversa_aux [] ys = rev [] @ ys" .
next
  fix a ::'a and xs :: "'a list"
  assume HI: "⋀ys. inversa_aux xs ys = rev xs@ys"
  show "⋀ys. inversa_aux (a#xs) ys = rev (a#xs)@ys"
  proof -
    fix ys
    have "inversa_aux (a#xs) ys = inversa_aux xs (a#ys)" by simp
    also have "… = rev xs@(a#ys)" using HI by simp
    also have "… = rev xs @ ([a] @ ys)" by simp
    also have "… = (rev xs @ [a]) @ ys" by simp
    also have "… = rev (a # xs) @ ys" by simp
    finally show "inversa_aux (a#xs) ys = rev (a#xs)@ys" .
  qed
qed
 
(* 3ª demostración del lema auxiliar *)
lemma
  "inversa_aux xs ys = (rev xs) @ ys"
proof (induct xs arbitrary: ys)
  fix ys :: "'a list"
  show "inversa_aux [] ys = rev [] @ ys" by simp
next
  fix a ::'a and xs :: "'a list"
  assume HI: "⋀ys. inversa_aux xs ys = rev xs@ys"
  show "⋀ys. inversa_aux (a#xs) ys = rev (a#xs)@ys"
  proof -
    fix ys
    have "inversa_aux (a#xs) ys = rev xs@(a#ys)" using HI by simp
    also have "… = rev (a # xs) @ ys" by simp
    finally show "inversa_aux (a#xs) ys = rev (a#xs)@ys" .
  qed
qed
 
(* 4ª demostración del lema auxiliar *)
lemma
  "inversa_aux xs ys = (rev xs) @ ys"
proof (induct xs arbitrary: ys)
  show "⋀ys. inversa_aux [] ys = rev [] @ ys" by simp
next
  fix a ::'a and xs :: "'a list"
  assume "⋀ys. inversa_aux xs ys = rev xs@ys"
  then show "⋀ys. inversa_aux (a#xs) ys = rev (a#xs)@ys" by simp
qed
 
(* 5ª demostración del lema auxiliar *)
lemma
  "inversa_aux xs ys = (rev xs) @ ys"
proof (induct xs arbitrary: ys)
  case Nil
  then show ?case by simp
next
  case (Cons a xs)
  then show ?case by simp
qed
 
(* 6ª demostración del lema auxiliar *)
lemma inversa_equiv:
  "inversa_aux xs ys = (rev xs) @ ys"
by (induct xs arbitrary: ys) simp_all
 
(* Demostraciones del lema principal *)
(* ================================= *)
 
(* 1ª demostración *)
lemma "inversa xs = rev xs"
proof -
  have "inversa xs = inversa_aux xs []"
    by (rule inversa.simps)
  also have "… = (rev xs) @ []"
    by (rule inversa_equiv)
  also have "… = rev xs"
    by (rule append.right_neutral)
  finally show "inversa xs = rev xs"
    by this
qed
 
(* 2ª demostración *)
lemma "inversa xs = rev xs"
proof -
  have "inversa xs = inversa_aux xs []"  by simp
  also have "… = (rev xs) @ []" by (rule inversa_equiv)
  also have "… = rev xs" by simp
  finally show "inversa xs = rev xs" .
qed
 
(* 3ª demostración *)
lemma "inversa xs = rev xs"
by (simp add: inversa_equiv)
 
end

En los comentarios se pueden escribir otras soluciones, escribiendo el código entre una línea con <pre lang="isar"> y otra con </pre>

Soluciones →

Asociatividad de la concatenación de listas

José A. Alonso-

En Lean la operación de concatenación de listas se representa por (++) y está caracterizada por los siguientes lemas

   nil_append  : [] ++ ys = ys
   cons_append : (x :: xs) ++ y = x :: (xs ++ ys)

Demostrar que la concatenación es asociativa; es decir,

   xs ++ (ys ++ zs) = (xs ++ ys) ++ zs

Para ello, completar la siguiente teoría de Lean:

import data.list.basic
import tactic
open list
 
variable  {α : Type}
variable  (x : α)
variables (xs ys zs : list α)
 
example :
  xs ++ (ys ++ zs) = (xs ++ ys) ++ zs :=
sorry
Soluciones con Lean
import data.list.basic
import tactic
open list
 
variable  {α : Type}
variable  (x : α)
variables (xs ys zs : list α)
 
-- 1ª demostración
example :
  xs ++ (ys ++ zs) = (xs ++ ys) ++ zs :=
begin
  induction xs with a as HI,
  { calc [] ++ (ys ++ zs)
         = ys ++ zs                : append.equations._eqn_1 (ys ++ zs)
     ... = ([] ++ ys) ++ zs        : congr_arg2 (++) (append.equations._eqn_1 ys) rfl, },
  { calc (a :: as) ++ (ys ++ zs)
         = a :: (as ++ (ys ++ zs)) : append.equations._eqn_2 a as (ys ++ zs)
     ... = a :: ((as ++ ys) ++ zs) : congr_arg2 (::) rfl HI
     ... = (a :: (as ++ ys)) ++ zs : (append.equations._eqn_2 a (as ++ ys) zs).symm
     ... = ((a :: as) ++ ys) ++ zs : congr_arg2 (++) (append.equations._eqn_2 a as ys).symm rfl, },
end
 
-- 2ª demostración
example :
  xs ++ (ys ++ zs) = (xs ++ ys) ++ zs :=
begin
  induction xs with a as HI,
  { calc [] ++ (ys ++ zs)
         = ys ++ zs                : nil_append (ys ++ zs)
     ... = ([] ++ ys) ++ zs        : congr_arg2 (++) (nil_append ys) rfl, },
  { calc (a :: as) ++ (ys ++ zs)
         = a :: (as ++ (ys ++ zs)) : cons_append a as (ys ++ zs)
     ... = a :: ((as ++ ys) ++ zs) : congr_arg2 (::) rfl HI
     ... = (a :: (as ++ ys)) ++ zs : (cons_append a (as ++ ys) zs).symm
     ... = ((a :: as) ++ ys) ++ zs : congr_arg2 (++) (cons_append a as ys).symm rfl, },
end
 
-- 3ª demostración
example :
  xs ++ (ys ++ zs) = (xs ++ ys) ++ zs :=
begin
  induction xs with a as HI,
  { calc [] ++ (ys ++ zs)
         = ys ++ zs                : by rw nil_append
     ... = ([] ++ ys) ++ zs        : by rw nil_append, },
  { calc (a :: as) ++ (ys ++ zs)
         = a :: (as ++ (ys ++ zs)) : by rw cons_append
     ... = a :: ((as ++ ys) ++ zs) : by rw HI
     ... = (a :: (as ++ ys)) ++ zs : by rw cons_append
     ... = ((a :: as) ++ ys) ++ zs : by rw ← cons_append, },
end
 
-- 4ª demostración
example :
  xs ++ (ys ++ zs) = (xs ++ ys) ++ zs :=
begin
  induction xs with a as HI,
  { calc [] ++ (ys ++ zs)
         = ys ++ zs                : rfl
     ... = ([] ++ ys) ++ zs        : rfl, },
  { calc (a :: as) ++ (ys ++ zs)
         = a :: (as ++ (ys ++ zs)) : rfl
     ... = a :: ((as ++ ys) ++ zs) : by rw HI
     ... = (a :: (as ++ ys)) ++ zs : rfl
     ... = ((a :: as) ++ ys) ++ zs : rfl, },
end
 
-- 5ª demostración
example :
  xs ++ (ys ++ zs) = (xs ++ ys) ++ zs :=
begin
  induction xs with a as HI,
  { calc [] ++ (ys ++ zs)
         = ys ++ zs                : by simp
     ... = ([] ++ ys) ++ zs        : by simp, },
  { calc (a :: as) ++ (ys ++ zs)
         = a :: (as ++ (ys ++ zs)) : by simp
     ... = a :: ((as ++ ys) ++ zs) : congr_arg (cons a) HI
     ... = (a :: (as ++ ys)) ++ zs : by simp
     ... = ((a :: as) ++ ys) ++ zs : by simp, },
end
 
-- 6ª demostración
example :
  xs ++ (ys ++ zs) = (xs ++ ys) ++ zs :=
begin
  induction xs with a as HI,
  { by simp, },
  { by exact (cons_inj a).mpr HI, },
end
 
-- 7ª demostración
example :
  xs ++ (ys ++ zs) = (xs ++ ys) ++ zs :=
begin
  induction xs with a as HI,
  { rw nil_append,
    rw nil_append, },
  { rw cons_append,
    rw HI,
    rw cons_append,
    rw cons_append, },
end
 
-- 8ª demostración
example :
  xs ++ (ys ++ zs) = (xs ++ ys) ++ zs :=
list.rec_on xs
  ( show [] ++ (ys ++ zs) = ([] ++ ys) ++ zs,
      from calc
        [] ++ (ys ++ zs)
            = ys ++ zs         : by rw nil_append
        ... = ([] ++ ys) ++ zs : by rw nil_append )
  ( assume a as,
    assume HI : as ++ (ys ++ zs) = (as ++ ys) ++ zs,
    show (a :: as) ++ (ys  ++ zs) = ((a :: as) ++ ys) ++ zs,
      from calc
        (a :: as) ++ (ys ++ zs)
            = a :: (as ++ (ys ++ zs)) : by rw cons_append
        ... = a :: ((as ++ ys) ++ zs) : by rw HI
        ... = (a :: (as ++ ys)) ++ zs : by rw cons_append
        ... = ((a :: as) ++ ys) ++ zs : by rw ← cons_append)
 
-- 9ª demostración
example :
  xs ++ (ys ++ zs) = (xs ++ ys) ++ zs :=
list.rec_on xs
  (by simp)
  (by simp [*])
 
-- 10ª demostración
lemma conc_asoc_1 :
  ∀ xs, xs ++ (ys ++ zs) = (xs ++ ys) ++ zs
| [] := by calc
    [] ++ (ys ++ zs)
        = ys ++ zs         : by rw nil_append
    ... = ([] ++ ys) ++ zs : by rw nil_append
| (a :: as) := by calc
    (a :: as) ++ (ys ++ zs)
        = a :: (as ++ (ys ++ zs)) : by rw cons_append
    ... = a :: ((as ++ ys) ++ zs) : by rw conc_asoc_1
    ... = (a :: (as ++ ys)) ++ zs : by rw cons_append
    ... = ((a :: as) ++ ys) ++ zs : by rw ← cons_append
 
-- 11ª demostración
example :
  (xs ++ ys) ++ zs = xs ++ (ys ++ zs) :=
-- by library_search
append_assoc xs ys zs
 
-- 12ª demostración
example :
  (xs ++ ys) ++ zs = xs ++ (ys ++ zs) :=
by induction xs ; simp [*]
 
-- 13ª demostración
example :
  (xs ++ ys) ++ zs = xs ++ (ys ++ zs) :=
by simp

Se puede interactuar con la prueba anterior en esta sesión con Lean.

En los comentarios se pueden escribir otras soluciones, escribiendo el código entre una línea con <pre lang="lean"> y otra con </pre>

Soluciones con Isabelle/HOL
theory Asociatividad_de_la_concatenacion_de_listas
imports Main
begin
 
(* 1ª demostración *)
lemma "xs @ (ys @ zs) = (xs @ ys) @ zs"
proof (induct xs)
  have "[] @ (ys @ zs) = ys @ zs"
    by (simp only: append_Nil)
  also have "… = ([] @ ys) @ zs"
    by (simp only: append_Nil)
  finally show "[] @ (ys @ zs) = ([] @ ys) @ zs"
    by this
next
  fix x xs
  assume HI : "xs @ (ys @ zs) = (xs @ ys) @ zs"
  have "(x # xs) @ (ys @ zs) = x # (xs @ (ys @ zs))"
    by (simp only: append_Cons)
  also have "… = x # ((xs @ ys) @ zs)"
    by (simp only: HI)
  also have "… = (x # (xs @ ys)) @ zs"
    by (simp only: append_Cons)
  also have "… = ((x # xs) @ ys) @ zs"
    by (simp only: append_Cons)
  finally show "(x # xs) @ (ys @ zs) = ((x # xs) @ ys) @ zs"
    by this
qed
 
(* 2ª demostración *)
lemma "xs @ (ys @ zs) = (xs @ ys) @ zs"
proof (induct xs)
  have "[] @ (ys @ zs) = ys @ zs" by simp
  also have "… = ([] @ ys) @ zs" by simp
  finally show "[] @ (ys @ zs) = ([] @ ys) @ zs" .
next
  fix x xs
  assume HI : "xs @ (ys @ zs) = (xs @ ys) @ zs"
  have "(x # xs) @ (ys @ zs) = x # (xs @ (ys @ zs))" by simp
  also have "… = x # ((xs @ ys) @ zs)" by simp
  also have "… = (x # (xs @ ys)) @ zs" by simp
  also have "… = ((x # xs) @ ys) @ zs" by simp
  finally show "(x # xs) @ (ys @ zs) = ((x # xs) @ ys) @ zs" .
qed
 
(* 3ª demostración *)
lemma "xs @ (ys @ zs) = (xs @ ys) @ zs"
proof (induct xs)
  show "[] @ (ys @ zs) = ([] @ ys) @ zs" by simp
next
  fix x xs
  assume "xs @ (ys @ zs) = (xs @ ys) @ zs"
  then show "(x # xs) @ (ys @ zs) = ((x # xs) @ ys) @ zs" by simp
qed
 
(* 4ª demostración *)
lemma "xs @ (ys @ zs) = (xs @ ys) @ zs"
proof (induct xs)
  case Nil
  then show ?case by simp
next
  case (Cons a xs)
  then show ?case by simp
qed
 
(* 5ª demostración *)
lemma "xs @ (ys @ zs) = (xs @ ys) @ zs"
  by (rule append_assoc [symmetric])
 
(* 6ª demostración *)
lemma "xs @ (ys @ zs) = (xs @ ys) @ zs"
  by (induct xs) simp_all
 
(* 7ª demostración *)
lemma "xs @ (ys @ zs) = (xs @ ys) @ zs"
  by simp
 
end

En los comentarios se pueden escribir otras soluciones, escribiendo el código entre una línea con <pre lang="isar"> y otra con </pre>

Soluciones →