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Etiqueta: L.simp

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 →

Pruebas de take n xs ++ drop n xs = xs

José A. Alonso-

En Lean están definidas las funciones

   take : nat → list α → nat
   drop : nat → list α → nat
   (++) : list α → list α → list α

tales que

  • (take n xs) es la lista formada por los n primeros elementos de xs. Por ejemplo,
     take 2 [3,5,1,9,7] = [3,5]
  • (drop n xs) es la lista formada eliminando los n primeros elementos de xs. Por ejemplo,
     drop 2 [3,5,1,9,7] = [1,9,7]
  • (xs ++ ys) es la lista obtenida concatenando xs e ys. Por ejemplo.
     [3,5] ++ [1,9,7] = [3,5,1,9,7]

Dichas funciones están caracterizadas por los siguientes lemas:

   take_zero   : take 0 xs = []
   take_nil    : take n [] = []
   take_cons   : take (succ n) (x :: xs) = x :: take n xs
   drop_zero   : drop 0 xs = xs
   drop_nil    : drop n [] = []
   drop_cons   : drop (succ n) (x :: xs) = drop n xs := rfl
   nil_append  : [] ++ ys = ys
   cons_append : (x :: xs) ++ y = x :: (xs ++ ys)

Demostrar que

   take n xs ++ drop n xs = xs

Para ello, completar la siguiente teoría de Lean:

import data.list.basic
import tactic
open list nat
 
variable {α : Type}
variable (x : α)
variable (xs : list α)
variable (n : ℕ)
 
lemma drop_zero : drop 0 xs = xs := rfl
lemma drop_cons : drop (succ n) (x :: xs) = drop n xs := rfl
 
example :
  take n xs ++ drop n xs = xs :=
sorry
Soluciones con Lean
import data.list.basic
import tactic
open list
open nat
 
variable {α : Type}
variable (n : ℕ)
variable (x : α)
variable (xs : list α)
 
lemma drop_zero : drop 0 xs = xs := rfl
lemma drop_cons : drop (succ n) (x :: xs) = drop n xs := rfl
 
-- 1ª demostración
example :
  take n xs ++ drop n xs = xs :=
begin
  induction n with m HI1 generalizing xs,
  { rw take_zero,
    rw drop_zero,
    rw nil_append, },
  { induction xs with a as HI2,
    { rw take_nil,
      rw drop_nil,
      rw nil_append, },
    { rw take_cons,
      rw drop_cons,
      rw cons_append,
      rw (HI1 as), }, },
end
 
-- 2ª demostración
example :
  take n xs ++ drop n xs = xs :=
begin
  induction n with m HI1 generalizing xs,
  { calc take 0 xs ++ drop 0 xs
         = [] ++ drop 0 xs        : by rw take_zero
     ... = [] ++ xs               : by rw drop_zero
     ... = xs                     : by rw nil_append, },
  { induction xs with a as HI2,
    { calc take (succ m) [] ++ drop (succ m) []
           = ([] : list α) ++ drop (succ m) [] : by rw take_nil
       ... = [] ++ []                          : by rw drop_nil
       ... = []                                : by rw nil_append, },
    { calc take (succ m) (a :: as) ++ drop (succ m) (a :: as)
           = (a :: take m as) ++ drop (succ m) (a :: as) : by rw take_cons
       ... = (a :: take m as) ++ drop m as               : by rw drop_cons
       ... = a :: (take m as ++ drop m as)               : by rw cons_append
       ... = a :: as                                     : by rw (HI1 as), }, },
end
 
-- 3ª demostración
example :
  take n xs ++ drop n xs = xs :=
begin
  induction n with m HI1 generalizing xs,
  { simp, },
  { induction xs with a as HI2,
    { simp, },
    { simp [HI1 as], }, },
end
 
-- 4ª demostración
lemma conc_take_drop_1 :
  ∀ (n : ℕ) (xs : list α), take n xs ++ drop n xs = xs
| 0 xs := by calc
    take 0 xs ++ drop 0 xs
        = [] ++ drop 0 xs   : by rw take_zero
    ... = [] ++ xs          : by rw drop_zero
    ... = xs                : by rw nil_append
| (succ m) [] := by calc
    take (succ m) [] ++ drop (succ m) []
        = ([] : list α) ++ drop (succ m) [] : by rw take_nil
    ... = [] ++ []                          : by rw drop_nil
    ... = []                                : by rw nil_append
| (succ m) (a :: as) := by calc
    take (succ m) (a :: as) ++ drop (succ m) (a :: as)
        = (a :: take m as) ++ drop (succ m) (a :: as)    : by rw take_cons
    ... = (a :: take m as) ++ drop m as                  : by rw drop_cons
    ... = a :: (take m as ++ drop m as)                  : by rw cons_append
    ... = a :: as                                        : by rw conc_take_drop_1
 
-- 5ª demostración
lemma conc_take_drop_2 :
  ∀ (n : ℕ) (xs : list α), take n xs ++ drop n xs = xs
| 0        xs        := by simp
| (succ m) []        := by simp
| (succ m) (a :: as) := by simp [conc_take_drop_2]
 
-- 6ª demostración
lemma conc_take_drop_3 :
  ∀ (n : ℕ) (xs : list α), take n xs ++ drop n xs = xs
| 0        xs        := rfl
| (succ m) []        := rfl
| (succ m) (a :: as) := congr_arg (cons a) (conc_take_drop_3 m as)
 
-- 7ª demostración
example : take n xs ++ drop n xs = xs :=
-- by library_search
take_append_drop n xs
 
-- 8ª demostración
example : take n xs ++ drop n xs = xs :=
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 "Pruebas_de_take_n_xs_++_drop_n_xs_Ig_xs"
imports Main
begin
 
fun take' :: "nat ⇒ 'a list ⇒ 'a list" where
  "take' n []           = []"
| "take' 0 xs           = []"
| "take' (Suc n) (x#xs) = x # (take' n xs)"
 
fun drop' :: "nat ⇒ 'a list ⇒ 'a list" where
  "drop' n []           = []"
| "drop' 0 xs           = xs"
| "drop' (Suc n) (x#xs) = drop' n xs"
 
(* 1ª demostración *)
lemma "take' n xs @ drop' n xs = xs"
proof (induct rule: take'.induct)
  fix n
  have "take' n [] @ drop' n [] = [] @ drop' n []"
    by (simp only: take'.simps(1))
  also have "… = drop' n []"
    by (simp only: append.simps(1))
  also have "… = []"
    by (simp only: drop'.simps(1))
  finally show "take' n [] @ drop' n [] = []"
    by this
next
  fix x :: 'a and xs :: "'a list"
  have "take' 0 (x#xs) @ drop' 0 (x#xs) =
        [] @ drop' 0 (x#xs)"
    by (simp only: take'.simps(2))
  also have "… = drop' 0 (x#xs)"
    by  (simp only: append.simps(1))
  also have "… = x # xs"
    by  (simp only: drop'.simps(2))
  finally show "take' 0 (x#xs) @ drop' 0 (x#xs) = x#xs"
    by this
next
  fix n :: nat and x :: 'a and xs :: "'a list"
  assume HI: "take' n xs @ drop' n xs = xs"
  have "take' (Suc n) (x # xs) @ drop' (Suc n) (x # xs) =
        (x # (take' n xs)) @ drop' n xs"
    by (simp only: take'.simps(3)
                   drop'.simps(3))
  also have "… = x # (take' n xs @ drop' n xs)"
    by (simp only: append.simps(2))
  also have "… = x#xs"
    by (simp only: HI)
  finally show "take' (Suc n) (x#xs) @ drop' (Suc n) (x#xs) =
                x#xs"
    by this
qed
 
(* 2ª demostración *)
lemma "take' n xs @ drop' n xs = xs"
proof (induct rule: take'.induct)
case (1 n)
  then show ?case by simp
next
  case (2 x xs)
  then show ?case by simp
next
  case (3 n x xs)
  then show ?case by simp
qed
 
(* 3ª demostración *)
lemma "take' n xs @ drop' n xs = xs"
by (induct rule: take'.induct) simp_all
 
end

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

Soluciones →

Pruebas de length (xs ++ ys) = length xs + length ys

José A. Alonso-

En Lean están definidas las funciones

   length : list α → nat
   (++)   : list α → list α → list α

tales que

  • (length xs) es la longitud de xs. Por ejemplo,
     length [2,3,5,3] = 4
  • (xs ++ ys) es la lista obtenida concatenando xs e ys. Por ejemplo.
     [1,2] ++ [2,3,5,3] = [1,2,2,3,5,3]

Dichas funciones están caracterizadas por los siguientes lemas:

   length_nil  : length [] = 0
   length_cons : length (x :: xs) = length xs + 1
   nil_append  : [] ++ ys = ys
   cons_append : (x :: xs) ++ y = x :: (xs ++ ys)

Demostrar que

   length (xs ++ ys) = length xs + length ys

Para ello, completar la siguiente teoría de Lean:

import tactic
open list
 
variable  {α : Type}
variable  (x : α)
variables (xs ys zs : list α)
 
lemma length_nil  : length ([] : list α) = 0 := rfl
 
example :
  length (xs ++ ys) = length xs + length ys :=
sorry
Soluciones con Lean
import tactic
open list
 
variable  {α : Type}
variable  (x : α)
variables (xs ys zs : list α)
 
lemma length_nil  : length ([] : list α) = 0 := rfl
 
-- 1ª demostración
example :
  length (xs ++ ys) = length xs + length ys :=
begin
  induction xs with a as HI,
  { rw nil_append,
    rw length_nil,
    rw zero_add, },
  { rw cons_append,
    rw length_cons,
    rw HI,
    rw length_cons,
    rw add_assoc,
    rw add_comm (length ys),
    rw add_assoc, },
end
 
-- 2ª demostración
example :
  length (xs ++ ys) = length xs + length ys :=
begin
  induction xs with a as HI,
  { rw nil_append,
    rw length_nil,
    rw zero_add, },
  { rw cons_append,
    rw length_cons,
    rw HI,
    rw length_cons,
    -- library_search,
    exact add_right_comm (length as) (length ys) 1 },
end
 
-- 3ª demostración
example :
  length (xs ++ ys) = length xs + length ys :=
begin
  induction xs with a as HI,
  { rw nil_append,
    rw length_nil,
    rw zero_add, },
  { rw cons_append,
    rw length_cons,
    rw HI,
    rw length_cons,
    -- by hint,
    linarith, },
end
 
-- 4ª demostración
example :
  length (xs ++ ys) = length xs + length ys :=
begin
  induction xs with a as HI,
  { simp, },
  { simp [HI],
    linarith, },
end
 
-- 5ª demostración
example :
  length (xs ++ ys) = length xs + length ys :=
begin
  induction xs with a as HI,
  { simp, },
  { finish [HI],},
end
 
-- 6ª demostración
example :
  length (xs ++ ys) = length xs + length ys :=
by induction xs ; finish [*]
 
-- 7ª demostración
example :
  length (xs ++ ys) = length xs + length ys :=
begin
  induction xs with a as HI,
  { calc length ([] ++ ys)
         = length ys                    : congr_arg length (nil_append ys)
     ... = 0 + length ys                : (zero_add (length ys)).symm
     ... = length [] + length ys        : congr_arg2 (+) length_nil.symm rfl, },
  { calc length ((a :: as) ++ ys)
         = length (a :: (as ++ ys))     : congr_arg length (cons_append a as ys)
     ... = length (as ++ ys) + 1        : length_cons a (as ++ ys)
     ... = (length as + length ys) + 1  : congr_arg2 (+) HI rfl
     ... = (length as + 1) + length ys  : add_right_comm (length as) (length ys) 1
     ... = length (a :: as) + length ys : congr_arg2 (+) (length_cons a as).symm rfl, },
end
 
-- 8ª demostración
example :
  length (xs ++ ys) = length xs + length ys :=
begin
  induction xs with a as HI,
  { calc length ([] ++ ys)
         = length ys                    : by rw nil_append
     ... = 0 + length ys                : (zero_add (length ys)).symm
     ... = length [] + length ys        : by rw length_nil },
  { calc length ((a :: as) ++ ys)
         = length (a :: (as ++ ys))     : by rw cons_append
     ... = length (as ++ ys) + 1        : by rw length_cons
     ... = (length as + length ys) + 1  : by rw HI
     ... = (length as + 1) + length ys  : add_right_comm (length as) (length ys) 1
     ... = length (a :: as) + length ys : by rw length_cons, },
end
 
-- 9ª demostración
example :
  length (xs ++ ys) = length xs + length ys :=
list.rec_on xs
  ( show length ([] ++ ys) = length [] + length ys, from
      calc length ([] ++ ys)
           = length ys                    : by rw nil_append
       ... = 0 + length ys                : by exact (zero_add (length ys)).symm
       ... = length [] + length ys        : by rw length_nil )
  ( assume a as,
    assume HI : length (as ++ ys) = length as + length ys,
    show length ((a :: as) ++ ys) = length (a :: as) + length ys, from
      calc length ((a :: as) ++ ys)
           = length (a :: (as ++ ys))     : by rw cons_append
       ... = length (as ++ ys) + 1        : by rw length_cons
       ... = (length as + length ys) + 1  : by rw HI
       ... = (length as + 1) + length ys  : by exact add_right_comm (length as) (length ys) 1
       ... = length (a :: as) + length ys : by rw length_cons)
 
-- 10ª demostración
example :
  length (xs ++ ys) = length xs + length ys :=
list.rec_on xs
  ( by simp)
  ( λ a as HI, by simp [HI, add_right_comm])
 
-- 11ª demostración
lemma longitud_conc_1 :
  ∀ xs, length (xs ++ ys) = length xs + length ys
| [] := by calc
    length ([] ++ ys)
        = length ys                    : by rw nil_append
    ... = 0 + length ys                : by rw zero_add
    ... = length [] + length ys        : by rw length_nil
| (a :: as) := by calc
    length ((a :: as) ++ ys)
        = length (a :: (as ++ ys))     : by rw cons_append
    ... = length (as ++ ys) + 1        : by rw length_cons
    ... = (length as + length ys) + 1  : by rw longitud_conc_1
    ... = (length as + 1) + length ys  : by exact add_right_comm (length as) (length ys) 1
    ... = length (a :: as) + length ys : by rw length_cons
 
-- 12ª demostración
lemma longitud_conc_2 :
  ∀ xs, length (xs ++ ys) = length xs + length ys
| []        := by simp
| (a :: as) := by simp [longitud_conc_2 as, add_right_comm]
 
-- 13ª demostración
example :
  length (xs ++ ys) = length xs + length ys :=
-- by library_search
length_append xs ys
 
-- 14ª demostración
example :
  length (xs ++ ys) = length xs + length ys :=
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 "Pruebas_de_length(xs_++_ys)_Ig_length_xs+length_ys"
imports Main
begin
 
(* 1ª demostración *)
lemma "length (xs @ ys) = length xs + length ys"
proof (induct xs)
  have "length ([] @ ys) = length ys"
    by (simp only: append_Nil)
  also have "… = 0 + length ys"
    by (rule add_0 [symmetric])
  also have "… = length [] + length ys"
    by (simp only: list.size(3))
  finally show "length ([] @ ys) = length [] + length ys"
    by this
next
  fix x xs
  assume HI : "length (xs @ ys) = length xs + length ys"
  have "length ((x # xs) @ ys) =
        length (x # (xs @ ys))"
    by (simp only: append_Cons)
  also have "… = length (xs @ ys) + 1"
    by (simp only: list.size(4))
  also have "… = (length xs + length ys) + 1"
    by (simp only: HI)
  also have "… = (length xs + 1) + length ys"
    by (simp only: add.assoc add.commute)
  also have "… = length (x # xs) + length ys"
    by (simp only: list.size(4))
  then show "length ((x # xs) @ ys) = length (x # xs) + length ys"
    by simp
qed
 
(* 2ª demostración *)
lemma "length (xs @ ys) = length xs + length ys"
proof (induct xs)
  show "length ([] @ ys) = length [] + length ys"
    by simp
next
  fix x xs
  assume "length (xs @ ys) = length xs + length ys"
  then show "length ((x # xs) @ ys) = length (x # xs) + length ys"
    by simp
qed
 
(* 3ª demostración *)
lemma "length (xs @ ys) = length xs + length ys"
proof (induct xs)
  case Nil
  then show ?case by simp
next
  case (Cons a xs)
  then show ?case by simp
qed
 
(* 4ª demostración *)
lemma "length (xs @ ys) = length xs + length ys"
by (induct xs) simp_all
 
(* 5ª demostración *)
lemma "length (xs @ ys) = length xs + length ys"
by (fact length_append)
 
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 →

Pruebas de length (repeat x n) = n

José A. Alonso-

En Lean están definidas las funciones length y repeat tales que

  • (length xs) es la longitud de la lista xs. Por ejemplo,
     length [1,2,5,2] = 4
  • (repeat x n) es la lista que tiene el elemento x n veces. Por ejemplo,
     repeat 7 3 = [7, 7, 7]

Demostrar que

   length (repeat x n) = n

Para ello, completar la siguiente teoría de Lean:

import data.list.basic
open nat
open list
 
variable {α : Type}
variable (x : α)
variable (n : ℕ)
 
example :
  length (repeat x n) = n :=
sorry
Soluciones con Lean
import data.list.basic
open nat
open list
 
set_option pp.structure_projections false
 
variable {α : Type}
variable (x : α)
variable (n : ℕ)
 
-- 1ª demostración
example :
  length (repeat x n) = n :=
begin
  induction n with n HI,
  { calc length (repeat x 0)
          = length []                : congr_arg length (repeat.equations._eqn_1 x)
      ... = 0                        : length.equations._eqn_1 },
  { calc length (repeat x (succ n))
         = length (x :: repeat x n)  : congr_arg length (repeat.equations._eqn_2 x n)
     ... = length (repeat x n) + 1   : length.equations._eqn_2 x (repeat x n)
     ... = n + 1                     : congr_arg2 (+) HI rfl
     ... = succ n                    : rfl, },
end
 
-- 2ª demostración
example :
  length (repeat x n) = n :=
begin
  induction n with n HI,
  { calc length (repeat x 0)
          = length []                : rfl
      ... = 0                        : rfl },
  { calc length (repeat x (succ n))
         = length (x :: repeat x n)  : rfl
     ... = length (repeat x n) + 1   : rfl
     ... = n + 1                     : by rw HI
     ... = succ n                    : rfl, },
end
 
-- 3ª demostración
example : length (repeat x n) = n :=
begin
  induction n with n HI,
  { refl, },
  { dsimp,
    rw HI, },
end
 
-- 4ª demostración
example : length (repeat x n) = n :=
begin
  induction n with n HI,
  { simp, },
  { simp [HI], },
end
 
-- 5ª demostración
example : length (repeat x n) = n :=
by induction n ; simp [*]
 
-- 6ª demostración
example : length (repeat x n) = n :=
nat.rec_on n
  ( show length (repeat x 0) = 0, from
      calc length (repeat x 0)
           = length []                : rfl
       ... = 0                        : rfl )
  ( assume n,
    assume HI : length (repeat x n) = n,
    show length (repeat x (succ n)) = succ n, from
      calc length (repeat x (succ n))
           = length (x :: repeat x n) : rfl
       ... = length (repeat x n) + 1  : rfl
       ... = n + 1                    : by rw HI
       ... = succ n                   : rfl )
 
-- 7ª demostración
example : length (repeat x n) = n :=
nat.rec_on n
  ( by simp )
  ( λ n HI, by simp [HI])
 
-- 8ª demostración
example : length (repeat x n) = n :=
length_repeat x n
 
-- 9ª demostración
example : length (repeat x n) = n :=
by simp
 
-- 10ª demostración
lemma length_repeat_1 :
  ∀ n, length (repeat x n) = n
| 0 := by calc length (repeat x 0)
               = length ([] : list α)         : rfl
           ... = 0                            : rfl
| (n+1) := by calc length (repeat x (n + 1))
                   = length (x :: repeat x n) : rfl
               ... = length (repeat x n) + 1  : rfl
               ... = n + 1                    : by rw length_repeat_1
 
-- 11ª demostración
lemma length_repeat_2 :
  ∀ n, length (repeat x n) = n
| 0     := by simp
| (n+1) := 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 "Pruebas_de_length_(repeat_x_n)_Ig_n"
imports Main
begin
 
(* 1ª demostración⁾*)
lemma "length (replicate n x) = n"
proof (induct n)
  have "length (replicate 0 x) = length ([] :: 'a list)"
    by (simp only: replicate.simps(1))
  also have "… = 0"
    by (rule list.size(3))
  finally show "length (replicate 0 x) = 0"
    by this
next
  fix n
  assume HI : "length (replicate n x) = n"
  have "length (replicate (Suc n) x) =
        length (x # replicate n x)"
    by (simp only: replicate.simps(2))
  also have "… = length (replicate n x) + 1"
    by (simp only: list.size(4))
  also have "… = Suc n"
    by (simp only: HI)
  finally show "length (replicate (Suc n) x) = Suc n"
    by this
qed
 
(* 2ª demostración⁾*)
lemma "length (replicate n x) = n"
proof (induct n)
  show "length (replicate 0 x) = 0"
    by simp
next
  fix n
  assume "length (replicate n x) = n"
  then show "length (replicate (Suc n) x) = Suc n"
    by simp
qed
 
(* 3ª demostración⁾*)
lemma "length (replicate n x) = n"
proof (induct n)
  case 0
  then show ?case by simp
next
  case (Suc n)
  then show ?case by simp
qed
 
(* 4ª demostración⁾*)
lemma "length (replicate n x) = n"
  by (rule length_replicate)
 
end

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

Soluciones →

Si a es un punto de acumulación de la sucesión de Cauchy u, entonces a es el límite de u

José A. Alonso-

En Lean, una sucesión u₀, u₁, u₂, … se puede representar mediante una función (u : ℕ → ℝ) de forma que u(n) es uₙ.

Para extraer una subsucesión se aplica una función de extracción queconserva el orden; por ejemplo, la subsucesión

   uₒ, u₂, u₄, u₆, ...

se ha obtenido con la función de extracción φ tal que φ(n) = 2*n.

En Lean, se puede definir que φ es una función de extracción por

   def extraccion (φ : ℕ → ℕ) :=
     ∀ n m, n < m → φ n < φ m

que a es un límite de u por

   def limite (u : ℕ → ℝ) (a : ℝ) :=
     ∀ ε > 0, ∃ N, ∀ k ≥ N, |u k - a| < ε

que a es un punto de acumulación de u por

   def punto_acumulacion (u : ℕ → ℝ) (a : ℝ) :=
     ∃ φ, extraccion φ ∧ limite (u ∘ φ) a

que la sucesión u es de Cauchy por

   def suc_cauchy (u : ℕ → ℝ) :=
     ∀ ε > 0, ∃ N, ∀ p ≥ N, ∀ q ≥ N, |u p - u q| < ε

Demostrar que si u es una sucesión de Cauchy y a es un punto de acumulación de u, entonces a es el límite de u.

Para ello, completar la siguiente teoría de Lean:

import data.real.basic
open nat
 
variable  {u : ℕ → ℝ}
variables {a : ℝ}
variable  {φ : ℕ → ℕ}
 
notation `|`x`|` := abs x
 
def extraccion (φ : ℕ → ℕ) :=
  ∀ n m, n < m → φ n < φ m
 
def limite (u : ℕ → ℝ) (l : ℝ) : Prop :=
  ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - l| < ε
 
def punto_acumulacion (u : ℕ → ℝ) (a : ℝ) :=
  ∃ φ, extraccion φ ∧ limite (u ∘ φ) a
 
def suc_cauchy (u : ℕ → ℝ) :=
  ∀ ε > 0, ∃ N, ∀ p ≥ N, ∀ q ≥ N, |u p - u q| < ε
 
example
  (hu : suc_cauchy u)
  (ha : punto_acumulacion u a)
  : limite u a :=
sorry
Soluciones con Lean
import data.real.basic
open nat
 
variable  {u : ℕ → ℝ}
variables {a : ℝ}
variable  {φ : ℕ → ℕ}
 
notation `|`x`|` := abs x
 
def extraccion (φ : ℕ → ℕ) :=
  ∀ n m, n < m → φ n < φ m
 
def limite (u : ℕ → ℝ) (l : ℝ) : Prop :=
  ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - l| < ε
 
def punto_acumulacion (u : ℕ → ℝ) (a : ℝ) :=
  ∃ φ, extraccion φ ∧ limite (u ∘ φ) a
 
def suc_cauchy (u : ℕ → ℝ) :=
  ∀ ε > 0, ∃ N, ∀ p ≥ N, ∀ q ≥ N, |u p - u q| < ε
 
lemma aux1
  (h : extraccion φ)
  : ∀ n, n ≤ φ n :=
begin
  intro n,
  induction n with m HI,
  { exact nat.zero_le (φ 0), },
  { apply nat.succ_le_of_lt,
    calc m ≤ φ m        : HI
       ... < φ (succ m) : h m (m+1) (lt_add_one m), },
end
 
lemma aux2
  (h : extraccion φ)
  : ∀ N N', ∃ n ≥ N', φ n ≥ N :=
λ N N', ⟨max N N', ⟨le_max_right N N',
                    le_trans (le_max_left N N')
                             (aux1 h (max N N'))⟩⟩
 
lemma cerca_acumulacion
  (h : punto_acumulacion u a)
  : ∀ ε > 0, ∀ N, ∃ n ≥ N, |u n - a| < ε :=
begin
  intros ε hε N,
  rcases h with ⟨φ, hφ1, hφ2⟩,
  cases hφ2 ε hε with N' hN',
  rcases aux2 hφ1 N N' with ⟨m, hm, hm'⟩,
  exact ⟨φ m, hm', hN' _ hm⟩,
end
 
-- 1ª demostración
example
  (hu : suc_cauchy u)
  (ha : punto_acumulacion u a)
  : limite u a :=
begin
  unfold limite,
  intros ε hε,
  unfold suc_cauchy at hu,
  cases hu (ε/2) (half_pos hε) with N hN,
  use N,
  have ha' : ∃ N' ≥ N, |u N' - a| < ε/2,
    apply cerca_acumulacion ha (ε/2) (half_pos hε),
  cases ha' with N' h,
  cases h with hNN' hN',
  intros n hn,
  calc   |u n - a|
       = |(u n - u N') + (u N' - a)| : by ring_nf
   ... ≤ |u n - u N'| + |u N' - a|   : abs_add (u n - u N') (u N' - a)
   ... < ε/2 + |u N' - a|            : add_lt_add_right (hN n hn N' hNN') _
   ... < ε/2 + ε/2                   : add_lt_add_left hN' (ε / 2)
   ... = ε                           : add_halves ε
end
 
-- 2ª demostración
example
  (hu : suc_cauchy u)
  (ha : punto_acumulacion u a)
  : limite u a :=
begin
  intros ε hε,
  cases hu (ε/2) (by linarith) with N hN,
  use N,
  have ha' : ∃ N' ≥ N, |u N' - a| < ε/2,
    apply cerca_acumulacion ha (ε/2) (by linarith),
  rcases ha' with ⟨N', hNN', hN'⟩,
  intros n hn,
  calc  |u n - a|
      = |(u n - u N') + (u N' - a)| : by ring_nf
  ... ≤ |u n - u N'| + |u N' - a|   : by simp [abs_add]
  ... < ε                           : by linarith [hN n hn N' hNN'],
end

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 "Si_a_es_un_punto_de_acumulacion_de_la_sucesion_de_Cauchy_u,_entonces_a_es_el_limite_de_u"
imports Main HOL.Real
begin
 
definition extraccion :: "(nat ⇒ nat) ⇒ bool" where
  "extraccion φ ⟷ (∀ n m. n < m ⟶ φ n < φ m)"
 
definition limite :: "(nat ⇒ real) ⇒ real ⇒ bool"
  where "limite u a ⟷ (∀ε>0. ∃N. ∀k≥N. ¦u k - a¦ < ε)"
 
definition punto_acumulacion :: "(nat ⇒ real) ⇒ real ⇒ bool"
  where "punto_acumulacion u a ⟷ (∃φ. extraccion φ ∧ limite (u ∘ φ) a)"
 
definition suc_cauchy :: "(nat ⇒ real) ⇒ bool"
  where "suc_cauchy u ⟷ (∀ε>0. ∃k. ∀m≥k. ∀n≥k. ¦u m - u n¦ < ε)"
 
(* Lemas auxiliares *)
 
lemma aux1 :
  assumes "extraccion φ"
  shows   "n ≤ φ n"
proof (induct n)
  show "0 ≤ φ 0" by simp
next
  fix n assume HI : "n ≤ φ n"
  then show "Suc n ≤ φ (Suc n)"
    using assms extraccion_def
    by (metis Suc_leI lessI order_le_less_subst1)
qed
 
lemma aux2 :
  assumes "extraccion φ"
  shows   "∀ N N'. ∃ k ≥ N'. φ k ≥ N"
proof (intro allI)
  fix N N' :: nat
  have "max N N' ≥ N' ∧ φ (max N N') ≥ N"
    by (meson assms aux1 max.bounded_iff max.cobounded2)
  then show "∃k ≥ N'. φ k ≥ N"
    by blast
qed
 
lemma cerca_acumulacion :
  assumes "punto_acumulacion u a"
  shows   "∀ε>0. ∀ N. ∃k≥N. ¦u k - a¦ < ε"
proof (intro allI impI)
  fix ε :: real and N :: nat
  assume "ε > 0"
  obtain φ where hφ1 : "extraccion φ"
             and hφ2 : "limite (u ∘ φ) a"
    using assms punto_acumulacion_def by blast
  obtain N' where hN' : "∀k≥N'. ¦(u ∘ φ) k - a¦ < ε"
    using hφ2 limite_def ‹ε > 0› by auto
  obtain m where "m ≥ N' ∧ φ m ≥ N"
    using aux2 hφ1 by blast
  then show "∃k≥N. ¦u k - a¦ < ε"
    using hN' by auto
qed
 
(* Demostración *)
lemma
  assumes "suc_cauchy u"
          "punto_acumulacion u a"
  shows   "limite u a"
proof (unfold limite_def; intro allI impI)
  fix ε :: real
  assume "ε > 0"
  then have "ε/2 > 0"
    by simp
  then obtain N where hN : "∀m≥N. ∀n≥N. ¦u m - u n¦ < ε/2"
    using assms(1) suc_cauchy_def
    by blast
  have "∀k≥N. ¦u k - a¦ < ε"
  proof (intro allI impI)
    fix k
    assume hk : "k ≥ N"
    obtain N' where hN'1 : "N' ≥ N" and
                    hN'2 : "¦u N' - a¦ < ε/2"
      using assms(2) cerca_acumulacion ‹ε/2 > 0› by blast
    have "¦u k - a¦ = ¦(u k - u N') + (u N'  - a)¦"
      by simp
    also have "… ≤ ¦u k - u N'¦ + ¦u N'  - a¦"
      by simp
    also have "… < ε/2 + ¦u N'  - a¦"
      using hk hN hN'1 by auto
    also have "… < ε/2 + ε/2"
      using hN'2 by auto
    also have "… = ε"
      by simp
    finally show "¦u k - a¦ < ε" .
  qed
  then show "∃N. ∀k≥N. ¦u k - a¦ < ε"
    by auto
qed
 
end

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

Soluciones →

La suma de los n primeros impares es n^2

José A. Alonso-

En Lean, se puede definir el n-ésimo número primo por

   def impar (n : ℕ) := 2 * n + 1

Además, en la librería finset están definidas las funciones

   range :: ℕ → finset ℕ
   sum :: finset α → (α → β) → β

tales que

+ (range n) es el conjunto de los n primeros números naturales. Por ejemplo, el valor de (range 3) es {0, 1, 2}.
+ (sum A f) es la suma del conjunto obtenido aplicando la función f a los elementos del conjunto finito A. Por ejemplo, el valor de (sum (range 3) impar) es 9.

Demostrar que la suma de los n primeros números impares es n².

Para ello, completar la siguiente teoría de Lean:

import data.finset
import tactic.ring
open nat
 
variable (n : ℕ)
 
def impar (n : ℕ) := 2 * n + 1
 
example :
  finset.sum (finset.range n) impar = n ^ 2 :=
sorry
Soluciones con Lean
import data.finset
import tactic.ring
open nat
 
set_option pp.structure_projections false
 
variable (n : ℕ)
 
def impar (n : ℕ) := 2 * n + 1
 
-- 1ª demostración
example :
  finset.sum (finset.range n) impar = n ^ 2 :=
begin
  induction n with m HI,
  { calc finset.sum (finset.range 0) impar
          = 0
            : by simp
     ...  = 0 ^ 2
            : rfl, },
  { calc finset.sum (finset.range (succ m)) impar
         = finset.sum (finset.range m) impar + impar m
           : finset.sum_range_succ impar m
     ... = m ^ 2 + impar m
           : congr_arg2 (+) HI rfl
     ... = m ^ 2 + 2 * m + 1
           : rfl
     ... = (m + 1) ^ 2
           : by ring_nf
     ... = succ m ^ 2
           : rfl },
end
 
-- 2ª demostración
example :
  finset.sum (finset.range n) impar = n ^ 2 :=
begin
  induction n with d hd,
  { refl, },
  { rw finset.sum_range_succ,
    rw hd,
    change d ^ 2 + (2 * d + 1) = (d + 1) ^ 2,
    ring_nf, },
end

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 "La_suma_de_los_n_primeros_impares_es_n^2"
imports Main
begin
 
definition impar :: "nat ⇒ nat" where
  "impar n ≡ 2 * n + 1"
 
lemma "sum impar {i::nat. i < n} = n⇧2"
proof (induct n)
  show "sum impar {i. i < 0} = 0⇧2"
    by simp
next
  fix n
  assume HI : "sum impar {i. i < n} = n⇧2"
  have "{i. i < Suc n} = {i. i < n} ∪ {n}"
    by auto
  then have "sum impar {i. i < Suc n} =
             sum impar {i. i < n} + impar n"
    by simp
  also have "… = n⇧2 + (2 * n + 1)"
    using HI impar_def by simp
  also have "… = (n + 1)⇧2"
    by algebra
  also have "… = (Suc n)⇧2"
    by simp
  finally show "sum impar {i. i < Suc n} = (Suc n)⇧2" .
qed
 
end

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

Soluciones →

El conjunto de las clases de equivalencia es una partición

José A. Alonso-

Demostrar que si R es una relación de equivalencia en X, entonces las clases de equivalencia de R es una partición de X.

Para ello, completar la siguiente teoría de Lean:

import tactic
 
variable  {X : Type}
variables {x y: X}
variable  {R : X → X → Prop}
 
def clase (R : X → X → Prop) (x : X) :=
  {y : X | R x y}
 
def particion (A : set (set X)) : Prop :=
  (∀ x, (∃ B ∈ A, x ∈ B ∧ ∀ C ∈ A, x ∈ C → B = C)) ∧ ∅ ∉ A
 
example
  (h : equivalence R)
  : particion {a : set X | ∃ s : X, a = clase R s} :=
sorry
Soluciones con Lean
import tactic
 
variable  {X : Type}
variables {x y: X}
variable  {R : X → X → Prop}
 
def clase (R : X → X → Prop) (x : X) :=
  {y : X | R x y}
 
def particion (A : set (set X)) : Prop :=
  (∀ x, (∃ B ∈ A, x ∈ B ∧ ∀ C ∈ A, x ∈ C → B = C)) ∧ ∅ ∉ A
 
lemma aux
  (h : equivalence R)
  (hxy : R x y)
  : clase R y ⊆ clase R x :=
λ z hz, h.2.2 hxy hz
 
-- 1ª demostración
example
  (h : equivalence R)
  : particion {a : set X | ∃ s : X, a = clase R s} :=
begin
  split,
  { simp,
    intro y,
    use (clase R y),
    split,
    { use y, },
    { split,
      { exact h.1 y, },
      { intros x hx,
        apply le_antisymm,
        { exact aux h hx, },
        { exact aux h (h.2.1 hx), }}}},
  { simp,
    intros x hx,
    have h1 : x ∈ clase R x := h.1 x,
    rw ← hx at h1,
    exact set.not_mem_empty x h1, },
end
 
-- 2ª demostración
example
  (h : equivalence R)
  : particion {a : set X | ∃ s : X, a = clase R s} :=
begin
  split,
  { simp,
    intro y,
    use (clase R y),
    split,
    { use y, },
    { split,
      { exact h.1 y, },
      { intros x hx,
        exact le_antisymm (aux h hx) (aux h (h.2.1 hx)), }}},
  { simp,
    intros x hx,
    have h1 : x ∈ clase R x := h.1 x,
    rw ← hx at h1,
    exact set.not_mem_empty x h1, },
end
 
-- 3ª demostración
example
  (h : equivalence R)
  : particion {a : set X | ∃ s : X, a = clase R s} :=
begin
  split,
  { simp,
    intro y,
    use [clase R y,
         ⟨by use y,
          ⟨h.1 y, λ x hx, le_antisymm (aux h hx) (aux h (h.2.1 hx))⟩⟩], },
  { simp,
    intros x hx,
    have h1 : x ∈ clase R x := h.1 x,
    rw ← hx at h1,
    exact set.not_mem_empty x h1, },
end

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 El_conjunto_de_las_clases_de_equivalencia_es_una_particion
imports Main
begin
 
definition clase :: "('a ⇒ 'a ⇒ bool) ⇒ 'a ⇒ 'a set"
  where "clase R x = {y. R x y}"
 
definition particion :: "('a set) set ⇒ bool" where
  "particion P ⟷ (∀x. (∃B∈P. x ∈ B ∧ (∀C∈P. x ∈ C ⟶ B = C))) ∧ {} ∉ P"
 
lemma
  fixes   R :: "'a ⇒ 'a ⇒ bool"
  assumes "equivp R"
  shows   "particion (⋃x. {clase R x})" (is "particion ?P")
proof (unfold particion_def; intro conjI)
  show "(∀x. ∃B∈?P. x ∈ B ∧ (∀C∈?P. x ∈ C ⟶ B = C))"
  proof (intro allI)
    fix x
    have "clase R x ∈ ?P"
      by auto
    moreover have "x ∈ clase R x"
      using assms clase_def equivp_def
      by (metis CollectI)
    moreover have "∀C∈?P. x ∈ C ⟶ clase R x = C"
    proof
      fix C
      assume "C ∈ ?P"
      then obtain y where "C = clase R y"
        by auto
      show "x ∈ C ⟶ clase R x = C"
      proof
        assume "x ∈ C"
        then have "R y x"
          using ‹C = clase R y› assms clase_def
          by (metis CollectD)
        then show "clase R x = C"
          using assms ‹C = clase R y› clase_def equivp_def
          by metis
      qed
    qed
    ultimately show "∃B∈?P. x ∈ B ∧ (∀C∈?P. x ∈ C ⟶ B = C)"
      by blast
  qed
next
  show "{} ∉ ?P"
  proof
    assume "{} ∈ ?P"
    then obtain x where "{} = clase R x"
      by auto
    moreover have "x ∈ clase R x"
      using assms clase_def equivp_def
      by (metis CollectI)
    ultimately show False
      by simp
  qed
qed
 
end

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

Soluciones →

Las sucesiones convergentes son sucesiones de Cauchy

José A. Alonso-

Nota: El problema de hoy lo ha escrito Sara Díaz Real y es uno de los que se encuentran en su Trabajo Fin de Máster Formalización en Lean de problemas de las Olimpiadas Internacionales de Matemáticas (IMO). Concretamente, el problema se encuentra en la página 52 junto con la demostración en lenguaje natural.

En Lean, una sucesión u₀, u₁, u₂, … se puede representar mediante una función (u : ℕ → ℝ) de forma que u(n) es uₙ.

Se define

  • el valor absoluto de x por
     notation `|`x`|` := abs x
  • a es un límite de la sucesión u, por
     def limite (u : ℕ → ℝ) (a : ℝ) :=
       ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - a| ≤ ε
  • la sucesión u es convergente por
     def suc_convergente (u : ℕ → ℝ) :=
       ∃ l, limite u l
  • la sucesión u es de Cauchy por
     def suc_cauchy (u : ℕ → ℝ) :=
       ∀ ε > 0, ∃ N, ∀ p ≥ N, ∀ q ≥ N, |u p - u q| ≤ ε

Demostrar que las sucesiones convergentes son de Cauchy.

Para ello, completar la siguiente teoría de Lean:

import data.real.basic
 
variable {u : ℕ → ℝ }
 
notation `|`x`|` := abs x
 
def limite (u : ℕ → ℝ) (l : ℝ) : Prop :=
  ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - l| ≤ ε
 
def suc_convergente (u : ℕ → ℝ) :=
  ∃ l, limite u l
 
def suc_cauchy (u : ℕ → ℝ) :=
  ∀ ε > 0, ∃ N, ∀ p ≥ N, ∀ q ≥ N, |u p - u q| ≤ ε
 
example
  (h : suc_convergente u)
  : suc_cauchy u :=
sorry
Soluciones con Lean
import data.real.basic
 
variable {u : ℕ → ℝ }
 
notation `|`x`|` := abs x
 
def limite (u : ℕ → ℝ) (l : ℝ) : Prop :=
  ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - l| ≤ ε
 
def suc_convergente (u : ℕ → ℝ) :=
  ∃ l, limite u l
 
def suc_cauchy (u : ℕ → ℝ) :=
  ∀ ε > 0, ∃ N, ∀ p ≥ N, ∀ q ≥ N, |u p - u q| ≤ ε
 
-- 1ª demostración
example
  (h : suc_convergente u)
  : suc_cauchy u :=
begin
  unfold suc_cauchy,
  intros ε hε,
  have hε2 : 0 < ε/2 := half_pos hε,
  cases h with l hl,
  cases hl (ε/2) hε2 with N hN,
  clear hε hl hε2,
  use N,
  intros p hp q hq,
  calc |u p - u q|
       = |(u p - l) + (l - u q)| : by ring_nf
   ... ≤ |u p - l|  + |l - u q|  : abs_add (u p - l) (l - u q)
   ... = |u p - l|  + |u q - l|  : congr_arg2 (+) rfl (abs_sub_comm l (u q))
   ... ≤ ε/2 + ε/2               : add_le_add (hN p hp) (hN q hq)
   ... = ε                       : add_halves ε,
end
 
-- 2ª demostración
example
  (h : suc_convergente u)
  : suc_cauchy u :=
begin
  intros ε hε,
  cases h with l hl,
  cases hl (ε/2) (half_pos hε) with N hN,
  clear hε hl,
  use N,
  intros p hp q hq,
  calc |u p - u q|
       = |(u p - l) + (l - u q)| : by ring_nf
   ... ≤ |u p - l|  + |l - u q|  : abs_add (u p - l) (l - u q)
   ... = |u p - l|  + |u q - l|  : congr_arg2 (+) rfl (abs_sub_comm l (u q))
   ... ≤ ε/2 + ε/2               : add_le_add (hN p hp) (hN q hq)
   ... = ε                       : add_halves ε,
end
 
-- 3ª demostración
example
  (h : suc_convergente u)
  : suc_cauchy u :=
begin
  intros ε hε,
  cases h with l hl,
  cases hl (ε/2) (half_pos hε) with N hN,
  clear hε hl,
  use N,
  intros p hp q hq,
  have cota1 : |u p - l| ≤ ε / 2 := hN p hp,
  have cota2 : |u q - l| ≤ ε / 2 := hN q hq,
  clear hN hp hq,
  calc |u p - u q|
       = |(u p - l) + (l - u q)| : by ring_nf
   ... ≤ |u p - l|  + |l - u q|  : abs_add (u p - l) (l - u q)
   ... = |u p - l|  + |u q - l|  : by rw abs_sub_comm l (u q)
   ... ≤ ε                       : by linarith,
end
 
-- 4ª demostración
example
  (h : suc_convergente u)
  : suc_cauchy u :=
begin
  intros ε hε,
  cases h with l hl,
  cases hl (ε/2) (half_pos hε) with N hN,
  clear hε hl,
  use N,
  intros p hp q hq,
  calc |u p - u q|
       = |(u p - l) + (l - u q)| : by ring_nf
   ... ≤ |u p - l|  + |l - u q|  : abs_add (u p - l) (l - u q)
   ... = |u p - l|  + |u q - l|  : by rw abs_sub_comm l (u q)
   ... ≤ ε                       : by linarith [hN p hp, hN q hq],
end
 
-- 5ª demostración
example
  (h : suc_convergente u)
  : suc_cauchy u :=
begin
  intros ε hε,
  cases h with l hl,
  cases hl (ε/2) (by linarith) with N hN,
  use N,
  intros p hp q hq,
  calc |u p - u q|
       = |(u p - l) + (l - u q)| : by ring_nf
   ... ≤ |u p - l|  + |l - u q|  : by simp [abs_add]
   ... = |u p - l|  + |u q - l|  : by simp [abs_sub_comm]
   ... ≤ ε                       : by linarith [hN p hp, hN q hq],
end

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Soluciones con Isabelle/HOL
theory Las_sucesiones_convergentes_son_sucesiones_de_Cauchy
imports Main HOL.Real
begin
 
definition limite :: "(nat ⇒ real) ⇒ real ⇒ bool"
  where "limite u c ⟷ (∀ε>0. ∃k::nat. ∀n≥k. ¦u n - c¦ < ε)"
 
definition suc_convergente :: "(nat ⇒ real) ⇒ bool"
  where "suc_convergente u ⟷ (∃ l. limite u l)"
 
definition suc_cauchy :: "(nat ⇒ real) ⇒ bool"
  where "suc_cauchy u ⟷ (∀ε>0. ∃k. ∀m≥k. ∀n≥k. ¦u m - u n¦ < ε)"
 
(* 1ª demostración *)
lemma
  assumes "suc_convergente u"
  shows   "suc_cauchy u"
proof (unfold suc_cauchy_def; intro allI impI)
  fix ε :: real
  assume "0 < ε"
  then have "0 < ε/2"
    by simp
  obtain a where "limite u a"
    using assms suc_convergente_def by blast
  then obtain k where hk : "∀n≥k. ¦u n - a¦ < ε/2"
    using ‹0 < ε / 2› limite_def by blast
  have "∀m≥k. ∀n≥k. ¦u m - u n¦ < ε"
  proof (intro allI impI)
    fix p q
    assume hp : "p ≥ k" and hq : "q ≥ k"
    then have hp' : "¦u p - a¦ < ε/2"
      using hk by blast
    have hq' : "¦u q - a¦ < ε/2"
      using hk hq by blast
    have "¦u p - u q¦ = ¦(u p - a) + (a - u q)¦"
      by simp
    also have "… ≤ ¦u p - a¦  + ¦a - u q¦"
      by simp
    also have "… = ¦u p - a¦  + ¦u q - a¦"
      by simp
    also have "… < ε/2 + ε/2"
      using hp' hq' by simp
    also have "… = ε"
      by simp
    finally show "¦u p - u q¦ < ε"
      by this
  qed
  then show "∃k. ∀m≥k. ∀n≥k. ¦u m - u n¦ < ε"
    by (rule exI)
qed
 
(* 2ª demostración *)
lemma
  assumes "suc_convergente u"
  shows   "suc_cauchy u"
proof (unfold suc_cauchy_def; intro allI impI)
  fix ε :: real
  assume "0 < ε"
  then have "0 < ε/2"
    by simp
  obtain a where "limite u a"
    using assms suc_convergente_def by blast
  then obtain k where hk : "∀n≥k. ¦u n - a¦ < ε/2"
    using ‹0 < ε / 2› limite_def by blast
  have "∀m≥k. ∀n≥k. ¦u m - u n¦ < ε"
  proof (intro allI impI)
    fix p q
    assume hp : "p ≥ k" and hq : "q ≥ k"
    then have hp' : "¦u p - a¦ < ε/2"
      using hk by blast
    have hq' : "¦u q - a¦ < ε/2"
      using hk hq by blast
    show "¦u p - u q¦ < ε"
      using hp' hq' by argo
  qed
  then show "∃k. ∀m≥k. ∀n≥k. ¦u m - u n¦ < ε"
    by (rule exI)
qed
 
(* 3ª demostración *)
lemma
  assumes "suc_convergente u"
  shows   "suc_cauchy u"
proof (unfold suc_cauchy_def; intro allI impI)
  fix ε :: real
  assume "0 < ε"
  then have "0 < ε/2"
    by simp
  obtain a where "limite u a"
    using assms suc_convergente_def by blast
  then obtain k where hk : "∀n≥k. ¦u n - a¦ < ε/2"
    using ‹0 < ε / 2› limite_def by blast
  have "∀m≥k. ∀n≥k. ¦u m - u n¦ < ε"
    using hk by (smt (z3) field_sum_of_halves)
  then show "∃k. ∀m≥k. ∀n≥k. ¦u m - u n¦ < ε"
    by (rule exI)
qed
 
(* 3ª demostración *)
lemma
  assumes "suc_convergente u"
  shows   "suc_cauchy u"
proof (unfold suc_cauchy_def; intro allI impI)
  fix ε :: real
  assume "0 < ε"
  then have "0 < ε/2"
    by simp
  obtain a where "limite u a"
    using assms suc_convergente_def by blast
  then obtain k where hk : "∀n≥k. ¦u n - a¦ < ε/2"
    using ‹0 < ε / 2› limite_def by blast
  then show "∃k. ∀m≥k. ∀n≥k. ¦u m - u n¦ < ε"
    by (smt (z3) field_sum_of_halves)
qed
 
end

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

Soluciones →