Intersección con la imagen

Demostrar con Lean4 que
$f[s] ∩ v = f[s ∩ f⁻¹[v]]$

Para ello, completar la siguiente teoría de Lean4:

import Mathlib.Data.Set.Function
import Mathlib.Tactic

open Set

variable {α β : Type _}
variable (f : α → β)
variable (s : Set α)
variable (v : Set β)

example : (f '' s) ∩ v = f '' (s ∩ f ⁻¹' v) :=
by sorry

import Mathlib.Data.Set.Function

import Mathlib.Tactic

open Set

variable {α β : Type _}

variable (f : α → β)

variable (s : Set α)

variable (v : Set β)

example : (f '' s) ∩ v = f '' (s ∩ f ⁻¹' v) :=

by sorry

1. Demostración en lenguaje natural

Tenemmos que demostrar que, para todo $y$ ,
$y ∈ f[s] ∩ v ↔ y ∈ f[s ∩ f⁻¹[v]]$
Lo haremos demostrando las dos implicaciones.

(⟹) Supongamos que $y ∈ f[s] ∩ v$ . Entonces,
$\begin{align} &y ∈ f[s] \tag{1} \\ &y ∈ v \tag{2} \end{align}$
Por (1), existe un $x$ tal que
$\begin{align} &x ∈ s \tag{3} \\ &f(x) = y \tag{4} \end{align}$
De (2) y (4), se tiene que
$f(x) ∈ v$
y, por tanto,
$x ∈ f⁻¹[v] \tag{5}$
De (3) y (5), se tiene que
$x ∈ s ∩ f⁻¹[v]$
Por tanto,
$f(x) ∈ f[s ∩ f⁻¹[v]]$
y, por (4),
$y ∈ f[s ∩ f⁻¹[v]]$

(⟸) Supongamos que $y ∈ f[s ∩ f⁻¹[v]]$ . Entonces, existe un $x$ tal que
$\begin{align} &x ∈ s ∩ f⁻¹[v] \tag{6} \\ &f(x) = y \tag{7} \end{align}$
Por (6), se tiene que
$\begin{align} &x ∈ s \tag{8} \\ &x ∈ f⁻¹[v] \tag{9} \end{align}$
Por (8), se tiene que
$f(x) ∈ f[s]$
y, por (7),
$y ∈ f[s] \tag{10}$
Por (9),
$f(x) ∈ v$
y, por (7),
$y ∈ v \tag{11}$
Por (10) y (11),
$y ∈ f[s] ∩ v$

2. Demostraciones con Lean4

import Mathlib.Data.Set.Function
import Mathlib.Tactic

open Set

variable {α β : Type _}
variable (f : α → β)
variable (s : Set α)
variable (v : Set β)

-- 1ª demostración
-- ===============

example : (f '' s) ∩ v = f '' (s ∩ f ⁻¹' v) :=
by
  ext y
  -- y : β
  -- ⊢ y ∈ f '' s ∩ v ↔ y ∈ f '' (s ∩ f ⁻¹' v)
  constructor
  . -- ⊢ y ∈ f '' s ∩ v → y ∈ f '' (s ∩ f ⁻¹' v)
    intro hy
    -- hy : y ∈ f '' s ∩ v
    -- ⊢ y ∈ f '' (s ∩ f ⁻¹' v)
    cases' hy with hyfs yv
    -- hyfs : y ∈ f '' s
    -- yv : y ∈ v
    cases' hyfs with x hx
    -- x : α
    -- hx : x ∈ s ∧ f x = y
    cases' hx with xs fxy
    -- xs : x ∈ s
    -- fxy : f x = y
    have h1 : f x ∈ v := by rwa [←fxy] at yv
    have h3 : x ∈ s ∩ f ⁻¹' v := mem_inter xs h1
    have h4 : f x ∈ f '' (s ∩ f ⁻¹' v) := mem_image_of_mem f h3
    show y ∈ f '' (s ∩ f ⁻¹' v)
    rwa [fxy] at h4
  . -- ⊢ y ∈ f '' (s ∩ f ⁻¹' v) → y ∈ f '' s ∩ v
    intro hy
    -- hy : y ∈ f '' (s ∩ f ⁻¹' v)
    -- ⊢ y ∈ f '' s ∩ v
    cases' hy with x hx
    -- x : α
    -- hx : x ∈ s ∩ f ⁻¹' v ∧ f x = y
    cases' hx with hx1 fxy
    -- hx1 : x ∈ s ∩ f ⁻¹' v
    -- fxy : f x = y
    cases' hx1 with xs xfv
    -- xs : x ∈ s
    -- xfv : x ∈ f ⁻¹' v
    have h5 : f x ∈ f '' s := mem_image_of_mem f xs
    have h6 : y ∈ f '' s := by rwa [fxy] at h5
    have h7 : f x ∈ v := mem_preimage.mp xfv
    have h8 : y ∈ v := by rwa [fxy] at h7
    show y ∈ f '' s ∩ v
    exact mem_inter h6 h8

-- 2ª demostración
-- ===============

example : (f '' s) ∩ v = f '' (s ∩ f ⁻¹' v) :=
by
  ext y
  -- y : β
  -- ⊢ y ∈ f '' s ∩ v ↔ y ∈ f '' (s ∩ f ⁻¹' v)
  constructor
  . -- ⊢ y ∈ f '' s ∩ v → y ∈ f '' (s ∩ f ⁻¹' v)
    intro hy
    -- hy : y ∈ f '' s ∩ v
    -- ⊢ y ∈ f '' (s ∩ f ⁻¹' v)
    cases' hy with hyfs yv
    -- hyfs : y ∈ f '' s
    -- yv : y ∈ v
    cases' hyfs with x hx
    -- x : α
    -- hx : x ∈ s ∧ f x = y
    cases' hx with xs fxy
    -- xs : x ∈ s
    -- fxy : f x = y
    use x
    -- ⊢ x ∈ s ∩ f ⁻¹' v ∧ f x = y
    constructor
    . -- ⊢ x ∈ s ∩ f ⁻¹' v
      constructor
      . -- ⊢ x ∈ s
        exact xs
      . -- ⊢ x ∈ f ⁻¹' v
        rw [mem_preimage]
        -- ⊢ f x ∈ v
        rw [fxy]
        -- ⊢ y ∈ v
        exact yv
    . -- ⊢ f x = y
      exact fxy
  . -- ⊢ y ∈ f '' (s ∩ f ⁻¹' v) → y ∈ f '' s ∩ v
    intro hy
    -- hy : y ∈ f '' (s ∩ f ⁻¹' v)
    -- ⊢ y ∈ f '' s ∩ v
    cases' hy with x hx
    -- x : α
    -- hx : x ∈ s ∩ f ⁻¹' v ∧ f x = y
    constructor
    . -- ⊢ y ∈ f '' s
      use x
      -- ⊢ x ∈ s ∧ f x = y
      constructor
      . -- ⊢ x ∈ s
        exact hx.1.1
      . -- ⊢ f x = y
        exact hx.2
    . -- ⊢ y ∈ v
      cases' hx with hx1 fxy
      -- hx1 : x ∈ s ∩ f ⁻¹' v
      -- fxy : f x = y
      rw [←fxy]
      -- ⊢ f x ∈ v
      rw [←mem_preimage]
      -- ⊢ x ∈ f ⁻¹' v
      exact hx1.2

-- 3ª demostración
-- ===============

example : (f '' s) ∩ v = f '' (s ∩ f ⁻¹' v) :=
by
  ext y
  -- y : β
  -- ⊢ y ∈ f '' s ∩ v ↔ y ∈ f '' (s ∩ f ⁻¹' v)
  constructor
  . -- ⊢ y ∈ f '' s ∩ v → y ∈ f '' (s ∩ f ⁻¹' v)
    rintro ⟨⟨x, xs, fxy⟩, yv⟩
    -- yv : y ∈ v
    -- x : α
    -- xs : x ∈ s
    -- fxy : f x = y
    -- ⊢ y ∈ f '' (s ∩ f ⁻¹' v)
    use x
    -- ⊢ x ∈ s ∩ f ⁻¹' v ∧ f x = y
    constructor
    . -- ⊢ x ∈ s ∩ f ⁻¹' v
      constructor
      . -- ⊢ x ∈ s
        exact xs
      . -- ⊢ x ∈ f ⁻¹' v
        rw [mem_preimage]
        -- ⊢ f x ∈ v
        rw [fxy]
        -- ⊢ y ∈ v
        exact yv
    . exact fxy
  . rintro ⟨x, ⟨xs, xv⟩, fxy⟩
    -- x : α
    -- fxy : f x = y
    -- xs : x ∈ s
    -- xv : x ∈ f ⁻¹' v
    -- ⊢ y ∈ f '' s ∩ v
    constructor
    . -- ⊢ y ∈ f '' s
      use x, xs
      -- ⊢ f x = y
      exact fxy
    . -- ⊢ y ∈ v
      rw [←fxy]
      -- ⊢ f x ∈ v
      rw [←mem_preimage]
      -- ⊢ x ∈ f ⁻¹' v
      exact xv

-- 4ª demostración
-- ===============

example : (f '' s) ∩ v = f '' (s ∩ f ⁻¹' v) :=
by
  ext y
  -- y : β
  -- ⊢ y ∈ f '' s ∩ v ↔ y ∈ f '' (s ∩ f ⁻¹' v)
  constructor
  . -- ⊢ y ∈ f '' s ∩ v → y ∈ f '' (s ∩ f ⁻¹' v)
    rintro ⟨⟨x, xs, fxy⟩, yv⟩
    -- yv : y ∈ v
    -- x : α
    -- xs : x ∈ s
    -- fxy : f x = y
    -- ⊢ y ∈ f '' (s ∩ f ⁻¹' v)
    aesop
  . -- ⊢ y ∈ f '' (s ∩ f ⁻¹' v) → y ∈ f '' s ∩ v
    rintro ⟨x, ⟨xs, xv⟩, fxy⟩
    -- x : α
    -- fxy : f x = y
    -- xs : x ∈ s
    -- xv : x ∈ f ⁻¹' v
    -- ⊢ y ∈ f '' s ∩ v
    aesop

-- 5ª demostración
-- ===============

example : (f '' s) ∩ v = f '' (s ∩ f ⁻¹' v) :=
by ext ; constructor <;> aesop

-- 6ª demostración
-- ===============

example : (f '' s) ∩ v = f '' (s ∩ f ⁻¹' v) :=
(image_inter_preimage f s v).symm

-- Lemas usados
-- ============

-- variable (x : α)
-- variable (a b : Set α)
-- #check (image_inter_preimage f s v : f '' (s ∩ f ⁻¹' v) = f '' s ∩ v)
-- #check (mem_image_of_mem  f : x ∈ a → f x ∈ f '' a)
-- #check (mem_inter : x ∈ a → x ∈ b → x ∈ a ∩ b)
-- #check (mem_preimage : x ∈ f ⁻¹' v ↔ f x ∈ v)

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import Mathlib.Data.Set.Function

import Mathlib.Tactic

open Set

variable {α β : Type _}

variable (f : α → β)

variable (s : Set α)

variable (v : Set β)

-- 1ª demostración

-- ===============

example : (f '' s) ∩ v = f '' (s ∩ f ⁻¹' v) :=

ext y

-- y : β

-- ⊢ y ∈ f '' s ∩ v ↔ y ∈ f '' (s ∩ f ⁻¹' v)

constructor

. -- ⊢ y ∈ f '' s ∩ v → y ∈ f '' (s ∩ f ⁻¹' v)

intro hy

-- hy : y ∈ f '' s ∩ v

-- ⊢ y ∈ f '' (s ∩ f ⁻¹' v)

cases' hy with hyfs yv

-- hyfs : y ∈ f '' s

-- yv : y ∈ v

cases' hyfs with x hx

-- x : α

-- hx : x ∈ s ∧ f x = y

cases' hx with xs fxy

-- xs : x ∈ s

-- fxy : f x = y

have h1 : f x ∈ v := by rwa [←fxy] at yv

have h3 : x ∈ s ∩ f ⁻¹' v := mem_inter xs h1

have h4 : f x ∈ f '' (s ∩ f ⁻¹' v) := mem_image_of_mem f h3

show y ∈ f '' (s ∩ f ⁻¹' v)

rwa [fxy] at h4

. -- ⊢ y ∈ f '' (s ∩ f ⁻¹' v) → y ∈ f '' s ∩ v

intro hy

-- hy : y ∈ f '' (s ∩ f ⁻¹' v)

-- ⊢ y ∈ f '' s ∩ v

cases' hy with x hx

-- x : α

-- hx : x ∈ s ∩ f ⁻¹' v ∧ f x = y

cases' hx with hx1 fxy

-- hx1 : x ∈ s ∩ f ⁻¹' v

-- fxy : f x = y

cases' hx1 with xs xfv

-- xs : x ∈ s

-- xfv : x ∈ f ⁻¹' v

have h5 : f x ∈ f '' s := mem_image_of_mem f xs

have h6 : y ∈ f '' s := by rwa [fxy] at h5

have h7 : f x ∈ v := mem_preimage.mp xfv

have h8 : y ∈ v := by rwa [fxy] at h7

show y ∈ f '' s ∩ v

exact mem_inter h6 h8

-- 2ª demostración

-- ===============

example : (f '' s) ∩ v = f '' (s ∩ f ⁻¹' v) :=

ext y

-- y : β

-- ⊢ y ∈ f '' s ∩ v ↔ y ∈ f '' (s ∩ f ⁻¹' v)

constructor

. -- ⊢ y ∈ f '' s ∩ v → y ∈ f '' (s ∩ f ⁻¹' v)

intro hy

-- hy : y ∈ f '' s ∩ v

-- ⊢ y ∈ f '' (s ∩ f ⁻¹' v)

cases' hy with hyfs yv

-- hyfs : y ∈ f '' s

-- yv : y ∈ v

cases' hyfs with x hx

-- x : α

-- hx : x ∈ s ∧ f x = y

cases' hx with xs fxy

-- xs : x ∈ s

-- fxy : f x = y

use x

-- ⊢ x ∈ s ∩ f ⁻¹' v ∧ f x = y

constructor

. -- ⊢ x ∈ s ∩ f ⁻¹' v

constructor

. -- ⊢ x ∈ s

exact xs

. -- ⊢ x ∈ f ⁻¹' v

rw [mem_preimage]

-- ⊢ f x ∈ v

rw [fxy]

-- ⊢ y ∈ v

exact yv

. -- ⊢ f x = y

exact fxy

. -- ⊢ y ∈ f '' (s ∩ f ⁻¹' v) → y ∈ f '' s ∩ v

intro hy

-- hy : y ∈ f '' (s ∩ f ⁻¹' v)

-- ⊢ y ∈ f '' s ∩ v

cases' hy with x hx

-- x : α

-- hx : x ∈ s ∩ f ⁻¹' v ∧ f x = y

constructor

. -- ⊢ y ∈ f '' s

use x

-- ⊢ x ∈ s ∧ f x = y

constructor

. -- ⊢ x ∈ s

exact hx.1.1

. -- ⊢ f x = y

exact hx.2

. -- ⊢ y ∈ v

cases' hx with hx1 fxy

-- hx1 : x ∈ s ∩ f ⁻¹' v

-- fxy : f x = y

rw [←fxy]

-- ⊢ f x ∈ v

rw [←mem_preimage]

-- ⊢ x ∈ f ⁻¹' v

exact hx1.2

-- 3ª demostración

-- ===============

example : (f '' s) ∩ v = f '' (s ∩ f ⁻¹' v) :=

ext y

-- y : β

-- ⊢ y ∈ f '' s ∩ v ↔ y ∈ f '' (s ∩ f ⁻¹' v)

constructor

. -- ⊢ y ∈ f '' s ∩ v → y ∈ f '' (s ∩ f ⁻¹' v)

rintro ⟨⟨x, xs, fxy⟩, yv⟩

-- yv : y ∈ v

-- x : α

-- xs : x ∈ s

-- fxy : f x = y

-- ⊢ y ∈ f '' (s ∩ f ⁻¹' v)

use x

-- ⊢ x ∈ s ∩ f ⁻¹' v ∧ f x = y

constructor

. -- ⊢ x ∈ s ∩ f ⁻¹' v

constructor

. -- ⊢ x ∈ s

exact xs

. -- ⊢ x ∈ f ⁻¹' v

rw [mem_preimage]

-- ⊢ f x ∈ v

rw [fxy]

-- ⊢ y ∈ v

exact yv

. exact fxy

. rintro ⟨x, ⟨xs, xv⟩, fxy⟩

-- x : α

-- fxy : f x = y

-- xs : x ∈ s

-- xv : x ∈ f ⁻¹' v

-- ⊢ y ∈ f '' s ∩ v

constructor

. -- ⊢ y ∈ f '' s

use x, xs

-- ⊢ f x = y

exact fxy

. -- ⊢ y ∈ v

rw [←fxy]

-- ⊢ f x ∈ v

rw [←mem_preimage]

-- ⊢ x ∈ f ⁻¹' v

exact xv

-- 4ª demostración

-- ===============

example : (f '' s) ∩ v = f '' (s ∩ f ⁻¹' v) :=

ext y

-- y : β

-- ⊢ y ∈ f '' s ∩ v ↔ y ∈ f '' (s ∩ f ⁻¹' v)

constructor

. -- ⊢ y ∈ f '' s ∩ v → y ∈ f '' (s ∩ f ⁻¹' v)

rintro ⟨⟨x, xs, fxy⟩, yv⟩

-- yv : y ∈ v

-- x : α

-- xs : x ∈ s

-- fxy : f x = y

-- ⊢ y ∈ f '' (s ∩ f ⁻¹' v)

aesop

. -- ⊢ y ∈ f '' (s ∩ f ⁻¹' v) → y ∈ f '' s ∩ v

rintro ⟨x, ⟨xs, xv⟩, fxy⟩

-- x : α

-- fxy : f x = y

-- xs : x ∈ s

-- xv : x ∈ f ⁻¹' v

-- ⊢ y ∈ f '' s ∩ v

aesop

-- 5ª demostración

-- ===============

example : (f '' s) ∩ v = f '' (s ∩ f ⁻¹' v) :=

by ext ; constructor <;> aesop

-- 6ª demostración

-- ===============

example : (f '' s) ∩ v = f '' (s ∩ f ⁻¹' v) :=

(image_inter_preimage f s v).symm

-- Lemas usados

-- ============

-- variable (x : α)

-- variable (a b : Set α)

-- #check (image_inter_preimage f s v : f '' (s ∩ f ⁻¹' v) = f '' s ∩ v)

-- #check (mem_image_of_mem f : x ∈ a → f x ∈ f '' a)

-- #check (mem_inter : x ∈ a → x ∈ b → x ∈ a ∩ b)

-- #check (mem_preimage : x ∈ f ⁻¹' v ↔ f x ∈ v)

Se puede interactuar con las demostraciones anteriores en Lean 4 Web.

3. Demostraciones con Isabelle/HOL

theory Interseccion_con_la_imagen_inversa
imports Main
begin

(* 1ª demostración *)

lemma "(f ` s) ∩ v = f ` (s ∩ f -` v)"
proof (rule equalityI)
  show "(f ` s) ∩ v ⊆ f ` (s ∩ f -` v)"
  proof (rule subsetI)
    fix y
    assume "y ∈ (f ` s) ∩ v"
    then show "y ∈ f ` (s ∩ f -` v)"
    proof (rule IntE)
      assume "y ∈ v"
      assume "y ∈ f ` s"
      then show "y ∈ f ` (s ∩ f -` v)"
      proof (rule imageE)
        fix x
        assume "x ∈ s"
        assume "y = f x"
        then have "f x ∈ v"
          using ‹y ∈ v› by (rule subst)
        then have "x ∈ f -` v"
          by (rule vimageI2)
        with ‹x ∈ s› have "x ∈ s ∩ f -` v"
          by (rule IntI)
        then have "f x ∈ f ` (s ∩ f -` v)"
          by (rule imageI)
        with ‹y = f x› show "y ∈ f ` (s ∩ f -` v)"
          by (rule ssubst)
      qed
    qed
  qed
next
  show "f ` (s ∩ f -` v) ⊆ (f ` s) ∩ v"
  proof (rule subsetI)
    fix y
    assume "y ∈ f ` (s ∩ f -` v)"
    then show "y ∈ (f ` s) ∩ v"
    proof (rule imageE)
      fix x
      assume "y = f x"
      assume hx : "x ∈ s ∩ f -` v"
      have "y ∈ f ` s"
      proof -
        have "x ∈ s"
          using hx by (rule IntD1)
        then have "f x ∈ f ` s"
          by (rule imageI)
        with ‹y = f x› show "y ∈ f ` s"
          by (rule ssubst)
      qed
      moreover
      have "y ∈ v"
      proof -
        have "x ∈ f -` v"
          using hx by (rule IntD2)
        then have "f x ∈ v"
          by (rule vimageD)
        with ‹y = f x› show "y ∈ v"
          by (rule ssubst)
      qed
      ultimately show "y ∈ (f ` s) ∩ v"
        by (rule IntI)
    qed
  qed
qed

(* 2ª demostración *)

lemma "(f ` s) ∩ v = f ` (s ∩ f -` v)"
proof
  show "(f ` s) ∩ v ⊆ f ` (s ∩ f -` v)"
  proof
    fix y
    assume "y ∈ (f ` s) ∩ v"
    then show "y ∈ f ` (s ∩ f -` v)"
    proof
      assume "y ∈ v"
      assume "y ∈ f ` s"
      then show "y ∈ f ` (s ∩ f -` v)"
      proof
        fix x
        assume "x ∈ s"
        assume "y = f x"
        then have "f x ∈ v" using ‹y ∈ v› by simp
        then have "x ∈ f -` v" by simp
        with ‹x ∈ s› have "x ∈ s ∩ f -` v" by simp
        then have "f x ∈ f ` (s ∩ f -` v)" by simp
        with ‹y = f x› show "y ∈ f ` (s ∩ f -` v)" by simp
      qed
    qed
  qed
next
  show "f ` (s ∩ f -` v) ⊆ (f ` s) ∩ v"
  proof
    fix y
    assume "y ∈ f ` (s ∩ f -` v)"
    then show "y ∈ (f ` s) ∩ v"
    proof
      fix x
      assume "y = f x"
      assume hx : "x ∈ s ∩ f -` v"
      have "y ∈ f ` s"
      proof -
        have "x ∈ s" using hx by simp
        then have "f x ∈ f ` s" by simp
        with ‹y = f x› show "y ∈ f ` s" by simp
      qed
      moreover
      have "y ∈ v"
      proof -
        have "x ∈ f -` v" using hx by simp
        then have "f x ∈ v" by simp
        with ‹y = f x› show "y ∈ v" by simp
      qed
      ultimately show "y ∈ (f ` s) ∩ v" by simp
    qed
  qed
qed

(* 2ª demostración *)

lemma "(f ` s) ∩ v = f ` (s ∩ f -` v)"
proof
  show "(f ` s) ∩ v ⊆ f ` (s ∩ f -` v)"
  proof
    fix y
    assume "y ∈ (f ` s) ∩ v"
    then show "y ∈ f ` (s ∩ f -` v)"
    proof
      assume "y ∈ v"
      assume "y ∈ f ` s"
      then show "y ∈ f ` (s ∩ f -` v)"
      proof
        fix x
        assume "x ∈ s"
        assume "y = f x"
        then show "y ∈ f ` (s ∩ f -` v)"
          using ‹x ∈ s› ‹y ∈ v› by simp
      qed
    qed
  qed
next
  show "f ` (s ∩ f -` v) ⊆ (f ` s) ∩ v"
  proof
    fix y
    assume "y ∈ f ` (s ∩ f -` v)"
    then show "y ∈ (f ` s) ∩ v"
    proof
      fix x
      assume "y = f x"
      assume hx : "x ∈ s ∩ f -` v"
      then have "y ∈ f ` s" using ‹y = f x› by simp
      moreover
      have "y ∈ v" using hx ‹y = f x› by simp
      ultimately show "y ∈ (f ` s) ∩ v" by simp
    qed
  qed
qed

(* 4ª demostración *)

lemma "(f ` s) ∩ v = f ` (s ∩ f -` v)"
  by auto

end

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theory Interseccion_con_la_imagen_inversa

imports Main

begin

(* 1ª demostración *)

lemma "(f ` s) ∩ v = f ` (s ∩ f -` v)"

proof (rule equalityI)

show "(f ` s) ∩ v ⊆ f ` (s ∩ f -` v)"

proof (rule subsetI)

fix y

assume "y ∈ (f ` s) ∩ v"

then show "y ∈ f ` (s ∩ f -` v)"

proof (rule IntE)

assume "y ∈ v"

assume "y ∈ f ` s"

then show "y ∈ f ` (s ∩ f -` v)"

proof (rule imageE)

fix x

assume "x ∈ s"

assume "y = f x"

then have "f x ∈ v"

using ‹y ∈ v› by (rule subst)

then have "x ∈ f -` v"

by (rule vimageI2)

with ‹x ∈ s› have "x ∈ s ∩ f -` v"

by (rule IntI)

then have "f x ∈ f ` (s ∩ f -` v)"

by (rule imageI)

with ‹y = f x› show "y ∈ f ` (s ∩ f -` v)"

by (rule ssubst)

qed

show "f ` (s ∩ f -` v) ⊆ (f ` s) ∩ v"

proof (rule subsetI)

fix y

assume "y ∈ f ` (s ∩ f -` v)"

then show "y ∈ (f ` s) ∩ v"

proof (rule imageE)

fix x

assume "y = f x"

assume hx : "x ∈ s ∩ f -` v"

have "y ∈ f ` s"

proof -

have "x ∈ s"

using hx by (rule IntD1)

then have "f x ∈ f ` s"

by (rule imageI)

with ‹y = f x› show "y ∈ f ` s"

by (rule ssubst)

qed

moreover

have "y ∈ v"

proof -

have "x ∈ f -` v"

using hx by (rule IntD2)

then have "f x ∈ v"

by (rule vimageD)

with ‹y = f x› show "y ∈ v"

by (rule ssubst)

qed

ultimately show "y ∈ (f ` s) ∩ v"

by (rule IntI)

qed

(* 2ª demostración *)

lemma "(f ` s) ∩ v = f ` (s ∩ f -` v)"

proof

show "(f ` s) ∩ v ⊆ f ` (s ∩ f -` v)"

proof

fix y

assume "y ∈ (f ` s) ∩ v"

then show "y ∈ f ` (s ∩ f -` v)"

proof

assume "y ∈ v"

assume "y ∈ f ` s"

then show "y ∈ f ` (s ∩ f -` v)"

proof

fix x

assume "x ∈ s"

assume "y = f x"

then have "f x ∈ v" using ‹y ∈ v› by simp

then have "x ∈ f -` v" by simp

with ‹x ∈ s› have "x ∈ s ∩ f -` v" by simp

then have "f x ∈ f ` (s ∩ f -` v)" by simp

with ‹y = f x› show "y ∈ f ` (s ∩ f -` v)" by simp

qed

show "f ` (s ∩ f -` v) ⊆ (f ` s) ∩ v"

proof

fix y

assume "y ∈ f ` (s ∩ f -` v)"

then show "y ∈ (f ` s) ∩ v"

proof

fix x

assume "y = f x"

assume hx : "x ∈ s ∩ f -` v"

have "y ∈ f ` s"

proof -

have "x ∈ s" using hx by simp

then have "f x ∈ f ` s" by simp

with ‹y = f x› show "y ∈ f ` s" by simp

qed

moreover

have "y ∈ v"

proof -

have "x ∈ f -` v" using hx by simp

then have "f x ∈ v" by simp

with ‹y = f x› show "y ∈ v" by simp

qed

ultimately show "y ∈ (f ` s) ∩ v" by simp

qed

(* 2ª demostración *)

lemma "(f ` s) ∩ v = f ` (s ∩ f -` v)"

proof

show "(f ` s) ∩ v ⊆ f ` (s ∩ f -` v)"

proof

fix y

assume "y ∈ (f ` s) ∩ v"

then show "y ∈ f ` (s ∩ f -` v)"

proof

assume "y ∈ v"

assume "y ∈ f ` s"

then show "y ∈ f ` (s ∩ f -` v)"

proof

fix x

assume "x ∈ s"

assume "y = f x"

then show "y ∈ f ` (s ∩ f -` v)"

using ‹x ∈ s› ‹y ∈ v› by simp

qed

show "f ` (s ∩ f -` v) ⊆ (f ` s) ∩ v"

proof

fix y

assume "y ∈ f ` (s ∩ f -` v)"

then show "y ∈ (f ` s) ∩ v"

proof

fix x

assume "y = f x"

assume hx : "x ∈ s ∩ f -` v"

then have "y ∈ f ` s" using ‹y = f x› by simp

moreover

have "y ∈ v" using hx ‹y = f x› by simp

ultimately show "y ∈ (f ` s) ∩ v" by simp

qed

(* 4ª demostración *)

lemma "(f ` s) ∩ v = f ` (s ∩ f -` v)"

by auto

end

1. Demostración en lenguaje natural

2. Demostraciones con Lean4

3. Demostraciones con Isabelle/HOL

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