IH.add.commute – Calculemus

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

PorJosé A. Alonso 9 septiembre 20215 septiembre 2021

En Lean están definidas las funciones

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

1 2	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

1	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]

1	[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)

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

1	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

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

[expand title=»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

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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>
[/expand]

[expand title=»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

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

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

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

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>
[/expand]

Las sucesiones divergentes positivas no tienen límites finitos

PorJosé A. Alonso 5 septiembre 20211 septiembre 2021

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 que a es el límite de la sucesión u, por

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

1 2	def limite (u: ℕ → ℝ) (a: ℝ) := ∀ ε > 0, ∃ N, ∀ n ≥ N, \|u n - a\| < ε

donde se usa la notación |x| para el valor absoluto de x

   notation `|`x`|` := abs x

1	notation `\|`x`\|` := abs x

La sucesión u diverge positivamente cuando, para cada número real A, se puede encontrar un número natural m tal que, para n > m , se tenga u(n) > A. En Lean se puede definir por

   def diverge_positivamente (u : ℕ → ℝ) :=
     ∀ A, ∃ m, ∀ n ≥ m, u n > A

1 2	def diverge_positivamente (u : ℕ → ℝ) := ∀ A, ∃ m, ∀ n ≥ m, u n > A

Demostrar que si u diverge positivamente, entonces ningún número real es límite de u.

Para ello, completar la siguiente teoría de Lean:

import data.real.basic
import tactic

variable  {u : ℕ → ℝ}

notation `|`x`|` := abs x

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

def diverge_positivamente (u : ℕ → ℝ) :=
  ∀ A, ∃ m, ∀ n ≥ m, u n > A

example
  (h : diverge_positivamente u)
  : ¬(∃ a, limite u a) :=
sorry

import data.real.basic

import tactic

variable {u : ℕ → ℝ}

notation `|`x`|` := abs x

def limite (u : ℕ → ℝ) (a : ℝ) :=

∀ ε > 0, ∃ m, ∀ n ≥ m, |u n - a| < ε

def diverge_positivamente (u : ℕ → ℝ) :=

∀ A, ∃ m, ∀ n ≥ m, u n > A

example

(h : diverge_positivamente u)

: ¬(∃ a, limite u a) :=

sorry

[expand title=»Soluciones con Lean»]

import data.real.basic
import tactic

variable  {u : ℕ → ℝ}

notation `|`x`|` := abs x

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

def diverge_positivamente (u : ℕ → ℝ) :=
  ∀ A, ∃ m, ∀ n ≥ m, u n > A

-- 1ª demostración
example
  (h : diverge_positivamente u)
  : ¬(∃ a, limite u a) :=
begin
  push_neg,
  intros a ha,
  cases ha 1 zero_lt_one with m1 hm1,
  cases h (a+1) with m2 hm2,
  let m := max m1 m2,
  specialize hm1 m (le_max_left _ _),
  specialize hm2 m (le_max_right _ _),
  replace hm1 : u m - a < 1 := lt_of_abs_lt hm1,
  replace hm2 : 1 < u m - a := lt_sub_iff_add_lt'.mpr hm2,
  apply lt_irrefl (u m),
  calc u m < a + 1 : sub_lt_iff_lt_add'.mp hm1
       ... < u m   : lt_sub_iff_add_lt'.mp hm2,
end

-- 2ª demostración
example
  (h : diverge_positivamente u)
  : ¬(∃ a, limite u a) :=
begin
  push_neg,
  intros a ha,
  cases ha 1 (by linarith) with m1 hm1,
  cases h (a+1) with m2 hm2,
  let m := max m1 m2,
  replace hm1 : |u m - a| < 1 := by finish,
  replace hm1 : u m - a < 1   := lt_of_abs_lt hm1,
  replace hm2 : u m > a + 1   := by finish,
  replace hm2 : 1 < u m - a   := lt_sub_iff_add_lt'.mpr hm2,
  apply lt_irrefl (u m),
  calc u m < a + 1 : by linarith
       ... < u m   : by linarith
end

-- 3ª demostración
example
  (h : diverge_positivamente u)
  : ¬(∃ a, limite u a) :=
begin
  push_neg,
  intros a ha,
  cases ha 1 (by linarith) with m1 hm1,
  cases h (a+1) with m2 hm2,
  let m := max m1 m2,
  specialize hm1 m (le_max_left _ _),
  specialize hm2 m (le_max_right _ _),
  rw abs_lt at hm1,
  linarith,
end

import data.real.basic

import tactic

variable {u : ℕ → ℝ}

notation `|`x`|` := abs x

def limite (u : ℕ → ℝ) (a : ℝ) :=

∀ ε > 0, ∃ m, ∀ n ≥ m, |u n - a| < ε

def diverge_positivamente (u : ℕ → ℝ) :=

∀ A, ∃ m, ∀ n ≥ m, u n > A

-- 1ª demostración

example

(h : diverge_positivamente u)

: ¬(∃ a, limite u a) :=

begin

push_neg,

intros a ha,

cases ha 1 zero_lt_one with m1 hm1,

cases h (a+1) with m2 hm2,

let m := max m1 m2,

specialize hm1 m (le_max_left _ _),

specialize hm2 m (le_max_right _ _),

replace hm1 : u m - a < 1 := lt_of_abs_lt hm1,

replace hm2 : 1 < u m - a := lt_sub_iff_add_lt'.mpr hm2,

apply lt_irrefl (u m),

calc u m < a + 1 : sub_lt_iff_lt_add'.mp hm1

... < u m : lt_sub_iff_add_lt'.mp hm2,

end

-- 2ª demostración

example

(h : diverge_positivamente u)

: ¬(∃ a, limite u a) :=

begin

push_neg,

intros a ha,

cases ha 1 (by linarith) with m1 hm1,

cases h (a+1) with m2 hm2,

let m := max m1 m2,

replace hm1 : |u m - a| < 1 := by finish,

replace hm1 : u m - a < 1 := lt_of_abs_lt hm1,

replace hm2 : u m > a + 1 := by finish,

replace hm2 : 1 < u m - a := lt_sub_iff_add_lt'.mpr hm2,

apply lt_irrefl (u m),

calc u m < a + 1 : by linarith

... < u m : by linarith

end

-- 3ª demostración

example

(h : diverge_positivamente u)

: ¬(∃ a, limite u a) :=

begin

push_neg,

intros a ha,

cases ha 1 (by linarith) with m1 hm1,

cases h (a+1) with m2 hm2,

let m := max m1 m2,

specialize hm1 m (le_max_left _ _),

specialize hm2 m (le_max_right _ _),

rw abs_lt at hm1,

linarith,

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>
[/expand]

[expand title=»Soluciones con Isabelle/HOL»]

theory Las_sucesiones_divergentes_positivas_no_tienen_limites_finitos
imports Main HOL.Real
begin

definition limite :: "(nat ⇒ real) ⇒ real ⇒ bool"
  where "limite u a ⟷ (∀ε>0. ∃N. ∀k≥N. ¦u k - a¦ < ε)"

definition diverge_positivamente :: "(nat ⇒ real) ⇒ bool"
  where "diverge_positivamente u ⟷ (∀A. ∃m. ∀n≥m. u n > A)"

(* 1ª demostración *)
lemma
  assumes "diverge_positivamente u"
  shows   "∄a. limite u a"
proof (rule notI)
  assume "∃a. limite u a"
  then obtain a where "limite u a" try
    by auto
  then obtain m1 where hm1 : "∀n≥m1. ¦u n - a¦ < 1"
    using limite_def by fastforce
  obtain m2 where hm2 : "∀n≥m2. u n > a + 1"
    using assms diverge_positivamente_def by blast
  let ?m = "max m1 m2"
  have "u ?m < u ?m" using hm1 hm2
  proof -
    have "?m ≥ m1"
      by (rule max.cobounded1)
    have "?m ≥ m2"
      by (rule max.cobounded2)
    have "u ?m - a < 1"
      using hm1 ‹?m ≥ m1› by fastforce
    moreover have "u ?m > a + 1"
      using hm2 ‹?m ≥ m2› by simp
    ultimately show "u ?m < u ?m"
      by simp
  qed
  then show False
    by auto
qed

(* 2ª demostración *)
lemma
  assumes "diverge_positivamente u"
  shows   "∄a. limite u a"
proof (rule notI)
  assume "∃a. limite u a"
  then obtain a where "limite u a" try
    by auto
  then obtain m1 where hm1 : "∀n≥m1. ¦u n - a¦ < 1"
    using limite_def by fastforce
  obtain m2 where hm2 : "∀n≥m2. u n > a + 1"
    using assms diverge_positivamente_def by blast
  let ?m = "max m1 m2"
  have "1 < 1"
  proof -
    have "1 < u ?m - a"
      using hm2
      by (metis add.commute less_diff_eq max.cobounded2)
    also have "… < 1"
      using hm1
      by (metis abs_less_iff max_def order_refl)
    finally show "1 < 1" .
  qed
  then show False
    by auto
qed

end

theory Las_sucesiones_divergentes_positivas_no_tienen_limites_finitos

imports Main HOL.Real

begin

definition limite :: "(nat ⇒ real) ⇒ real ⇒ bool"

where "limite u a ⟷ (∀ε>0. ∃N. ∀k≥N. ¦u k - a¦ < ε)"

definition diverge_positivamente :: "(nat ⇒ real) ⇒ bool"

where "diverge_positivamente u ⟷ (∀A. ∃m. ∀n≥m. u n > A)"

(* 1ª demostración *)

lemma

assumes "diverge_positivamente u"

shows "∄a. limite u a"

proof (rule notI)

assume "∃a. limite u a"

then obtain a where "limite u a" try

by auto

then obtain m1 where hm1 : "∀n≥m1. ¦u n - a¦ < 1"

using limite_def by fastforce

obtain m2 where hm2 : "∀n≥m2. u n > a + 1"

using assms diverge_positivamente_def by blast

let ?m = "max m1 m2"

have "u ?m < u ?m" using hm1 hm2

proof -

have "?m ≥ m1"

by (rule max.cobounded1)

have "?m ≥ m2"

by (rule max.cobounded2)

have "u ?m - a < 1"

using hm1 ‹?m ≥ m1› by fastforce

moreover have "u ?m > a + 1"

using hm2 ‹?m ≥ m2› by simp

ultimately show "u ?m < u ?m"

by simp

qed

then show False

by auto

qed

(* 2ª demostración *)

lemma

assumes "diverge_positivamente u"

shows "∄a. limite u a"

proof (rule notI)

assume "∃a. limite u a"

then obtain a where "limite u a" try

by auto

then obtain m1 where hm1 : "∀n≥m1. ¦u n - a¦ < 1"

using limite_def by fastforce

obtain m2 where hm2 : "∀n≥m2. u n > a + 1"

using assms diverge_positivamente_def by blast

let ?m = "max m1 m2"

have "1 < 1"

proof -

have "1 < u ?m - a"

using hm2

by (metis add.commute less_diff_eq max.cobounded2)

also have "… < 1"

using hm1

by (metis abs_less_iff max_def order_refl)

finally show "1 < 1" .

qed

then show False

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>
[/expand]