La semana en Calculemus (16 de marzo de 2024)
Esta semana he publicado en Calculemus las demostraciones con Lean4 de las siguientes propiedades:
- 1. s ∪ (⋂ᵢ Aᵢ) = ⋂ᵢ (Aᵢ ∪ s)
- 2. f⁻¹(u ∩ v) = f⁻¹(u) ∩ f⁻¹(v)
- 3. f(s ∪ t) = f(s) ∪ f(t)
- 4. s ⊆ f⁻¹(f(s))
- 5. f(s) ⊆ u ↔ s ⊆ f⁻¹(u)
A continuación se muestran las soluciones.
1. s ∪ (⋂ᵢ Aᵢ) = ⋂ᵢ (Aᵢ ∪ s)
Demostrar con Lean4 que
\[ s ∪ (⋂_i A_i) = ⋂_i (A_i ∪ s) \]
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Set.Basic import Mathlib.Tactic open Set variable {α : Type} variable (s : Set α) variable (A : ℕ → Set α) example : s ∪ (⋂ i, A i) = ⋂ i, (A i ∪ s) := by sorry |
1.1. Demostración en lenguaje natural
Tenemos que demostrar que para todo \(x\),
\[ x ∈ s ∪ ⋂_i A_i ↔ x ∈ ⋂_i (A i ∪ s) \]
Lo haremos mediante la siguiente cadena de equivalencias
\begin{align}
x ∈ s ∪ ⋂_i A_i &↔ x ∈ s ∨ x ∈ ⋂_i A_i \\
&↔ x ∈ s ∨ (∀ i)[x ∈ A_i] \\
&↔ (∀ i)[x ∈ s ∨ x ∈ A_i] \\
&↔ (∀ i)[x ∈ A_i ∨ x ∈ s] \\
&↔ (∀ i)[x ∈ A_i ∪ s] \\
&↔ x ∈ ⋂_i (A_i ∪ s)
\end{align}
1.2. Demostraciones con Lean4
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import Mathlib.Data.Set.Basic import Mathlib.Tactic open Set variable {α : Type} variable (s : Set α) variable (A : ℕ → Set α) -- 1ª demostración -- =============== example : s ∪ (⋂ i, A i) = ⋂ i, (A i ∪ s) := by ext x -- x : α -- ⊢ x ∈ s ∪ ⋂ (i : ℕ), A i ↔ x ∈ ⋂ (i : ℕ), A i ∪ s calc x ∈ s ∪ ⋂ i, A i ↔ x ∈ s ∨ x ∈ ⋂ i, A i := by simp only [mem_union] _ ↔ x ∈ s ∨ ∀ i, x ∈ A i := by simp only [mem_iInter] _ ↔ ∀ i, x ∈ s ∨ x ∈ A i := by simp only [forall_or_left] _ ↔ ∀ i, x ∈ A i ∨ x ∈ s := by simp only [or_comm] _ ↔ ∀ i, x ∈ A i ∪ s := by simp only [mem_union] _ ↔ x ∈ ⋂ i, A i ∪ s := by simp only [mem_iInter] -- 2ª demostración -- =============== example : s ∪ (⋂ i, A i) = ⋂ i, (A i ∪ s) := by ext x -- x : α -- ⊢ x ∈ s ∪ ⋂ (i : ℕ), A i ↔ x ∈ ⋂ (i : ℕ), A i ∪ s simp only [mem_union, mem_iInter] -- ⊢ (x ∈ s ∨ ∀ (i : ℕ), x ∈ A i) ↔ ∀ (i : ℕ), x ∈ A i ∨ x ∈ s constructor . -- ⊢ (x ∈ s ∨ ∀ (i : ℕ), x ∈ A i) → ∀ (i : ℕ), x ∈ A i ∨ x ∈ s intros h i -- h : x ∈ s ∨ ∀ (i : ℕ), x ∈ A i -- i : ℕ -- ⊢ x ∈ A i ∨ x ∈ s rcases h with (xs | xAi) . -- xs : x ∈ s right -- ⊢ x ∈ s exact xs . -- xAi : ∀ (i : ℕ), x ∈ A i left -- ⊢ x ∈ A i exact xAi i . -- ⊢ (∀ (i : ℕ), x ∈ A i ∨ x ∈ s) → x ∈ s ∨ ∀ (i : ℕ), x ∈ A i intro h -- h : ∀ (i : ℕ), x ∈ A i ∨ x ∈ s -- ⊢ x ∈ s ∨ ∀ (i : ℕ), x ∈ A i by_cases cxs : x ∈ s . -- cxs : x ∈ s left -- ⊢ x ∈ s exact cxs . -- cns : ¬x ∈ s right -- ⊢ ∀ (i : ℕ), x ∈ A i intro i -- i : ℕ -- ⊢ x ∈ A i rcases h i with (xAi | xs) . -- ⊢ x ∈ A i exact xAi . -- xs : x ∈ s exact absurd xs cxs -- 3ª demostración -- =============== example : s ∪ (⋂ i, A i) = ⋂ i, (A i ∪ s) := by ext x -- x : α -- ⊢ x ∈ s ∪ ⋂ (i : ℕ), A i ↔ x ∈ ⋂ (i : ℕ), A i ∪ s simp only [mem_union, mem_iInter] -- ⊢ (x ∈ s ∨ ∀ (i : ℕ), x ∈ A i) ↔ ∀ (i : ℕ), x ∈ A i ∨ x ∈ s constructor . -- ⊢ (x ∈ s ∨ ∀ (i : ℕ), x ∈ A i) → ∀ (i : ℕ), x ∈ A i ∨ x ∈ s rintro (xs | xI) i . -- xs : x ∈ s -- i : ℕ -- ⊢ x ∈ A i ∨ x ∈ s right -- ⊢ x ∈ s exact xs . -- xI : ∀ (i : ℕ), x ∈ A i -- i : ℕ -- ⊢ x ∈ A i ∨ x ∈ s left -- ⊢ x ∈ A i exact xI i . -- ⊢ (∀ (i : ℕ), x ∈ A i ∨ x ∈ s) → x ∈ s ∨ ∀ (i : ℕ), x ∈ A i intro h -- h : ∀ (i : ℕ), x ∈ A i ∨ x ∈ s -- ⊢ x ∈ s ∨ ∀ (i : ℕ), x ∈ A i by_cases cxs : x ∈ s . -- cxs : x ∈ s left -- ⊢ x ∈ s exact cxs . -- cxs : ¬x ∈ s right -- ⊢ ∀ (i : ℕ), x ∈ A i intro i -- i : ℕ -- ⊢ x ∈ A i cases h i . -- h : x ∈ A i assumption . -- h : x ∈ s contradiction -- Lemas usados -- ============ -- variable (x : α) -- variable (s t : Set α) -- variable (a b q : Prop) -- variable (p : ℕ → Prop) -- #check (absurd : a → ¬a → b) -- #check (forall_or_left : (∀ x, q ∨ p x) ↔ q ∨ ∀ x, p x) -- #check (mem_iInter : x ∈ ⋂ i, A i ↔ ∀ i, x ∈ A i) -- #check (mem_union x a b : x ∈ s ∪ t ↔ x ∈ s ∨ x ∈ t) -- #check (or_comm : a ∨ b ↔ b ∨ a) |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
1.3. Demostraciones con Isabelle/HOL
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theory Union_con_interseccion_general imports Main begin (* 1ª demostración *) lemma "s ∪ (⋂ i ∈ I. A i) = (⋂ i ∈ I. A i ∪ s)" proof (rule equalityI) show "s ∪ (⋂ i ∈ I. A i) ⊆ (⋂ i ∈ I. A i ∪ s)" proof (rule subsetI) fix x assume "x ∈ s ∪ (⋂ i ∈ I. A i)" then show "x ∈ (⋂ i ∈ I. A i ∪ s)" proof (rule UnE) assume "x ∈ s" show "x ∈ (⋂ i ∈ I. A i ∪ s)" proof (rule INT_I) fix i assume "i ∈ I" show "x ∈ A i ∪ s" using ‹x ∈ s› by (rule UnI2) qed next assume h1 : "x ∈ (⋂ i ∈ I. A i)" show "x ∈ (⋂ i ∈ I. A i ∪ s)" proof (rule INT_I) fix i assume "i ∈ I" with h1 have "x ∈ A i" by (rule INT_D) then show "x ∈ A i ∪ s" by (rule UnI1) qed qed qed next show "(⋂ i ∈ I. A i ∪ s) ⊆ s ∪ (⋂ i ∈ I. A i)" proof (rule subsetI) fix x assume h2 : "x ∈ (⋂ i ∈ I. A i ∪ s)" show "x ∈ s ∪ (⋂ i ∈ I. A i)" proof (cases "x ∈ s") assume "x ∈ s" then show "x ∈ s ∪ (⋂ i ∈ I. A i)" by (rule UnI1) next assume "x ∉ s" have "x ∈ (⋂ i ∈ I. A i)" proof (rule INT_I) fix i assume "i ∈ I" with h2 have "x ∈ A i ∪ s" by (rule INT_D) then show "x ∈ A i" proof (rule UnE) assume "x ∈ A i" then show "x ∈ A i" by this next assume "x ∈ s" with ‹x ∉ s› show "x ∈ A i" by (rule notE) qed qed then show "x ∈ s ∪ (⋂ i ∈ I. A i)" by (rule UnI2) qed qed qed (* 2ª demostración *) lemma "s ∪ (⋂ i ∈ I. A i) = (⋂ i ∈ I. A i ∪ s)" proof show "s ∪ (⋂ i ∈ I. A i) ⊆ (⋂ i ∈ I. A i ∪ s)" proof fix x assume "x ∈ s ∪ (⋂ i ∈ I. A i)" then show "x ∈ (⋂ i ∈ I. A i ∪ s)" proof assume "x ∈ s" show "x ∈ (⋂ i ∈ I. A i ∪ s)" proof fix i assume "i ∈ I" show "x ∈ A i ∪ s" using ‹x ∈ s› by simp qed next assume h1 : "x ∈ (⋂ i ∈ I. A i)" show "x ∈ (⋂ i ∈ I. A i ∪ s)" proof fix i assume "i ∈ I" with h1 have "x ∈ A i" by simp then show "x ∈ A i ∪ s" by simp qed qed qed next show "(⋂ i ∈ I. A i ∪ s) ⊆ s ∪ (⋂ i ∈ I. A i)" proof fix x assume h2 : "x ∈ (⋂ i ∈ I. A i ∪ s)" show "x ∈ s ∪ (⋂ i ∈ I. A i)" proof (cases "x ∈ s") assume "x ∈ s" then show "x ∈ s ∪ (⋂ i ∈ I. A i)" by simp next assume "x ∉ s" have "x ∈ (⋂ i ∈ I. A i)" proof fix i assume "i ∈ I" with h2 have "x ∈ A i ∪ s" by (rule INT_D) then show "x ∈ A i" proof assume "x ∈ A i" then show "x ∈ A i" by this next assume "x ∈ s" with ‹x ∉ s› show "x ∈ A i" by simp qed qed then show "x ∈ s ∪ (⋂ i ∈ I. A i)" by simp qed qed qed (* 3ª demostración *) lemma "s ∪ (⋂ i ∈ I. A i) = (⋂ i ∈ I. A i ∪ s)" proof show "s ∪ (⋂ i ∈ I. A i) ⊆ (⋂ i ∈ I. A i ∪ s)" proof fix x assume "x ∈ s ∪ (⋂ i ∈ I. A i)" then show "x ∈ (⋂ i ∈ I. A i ∪ s)" proof assume "x ∈ s" then show "x ∈ (⋂ i ∈ I. A i ∪ s)" by simp next assume "x ∈ (⋂ i ∈ I. A i)" then show "x ∈ (⋂ i ∈ I. A i ∪ s)" by simp qed qed next show "(⋂ i ∈ I. A i ∪ s) ⊆ s ∪ (⋂ i ∈ I. A i)" proof fix x assume h2 : "x ∈ (⋂ i ∈ I. A i ∪ s)" show "x ∈ s ∪ (⋂ i ∈ I. A i)" proof (cases "x ∈ s") assume "x ∈ s" then show "x ∈ s ∪ (⋂ i ∈ I. A i)" by simp next assume "x ∉ s" then show "x ∈ s ∪ (⋂ i ∈ I. A i)" using h2 by simp qed qed qed (* 4ª demostración *) lemma "s ∪ (⋂ i ∈ I. A i) = (⋂ i ∈ I. A i ∪ s)" proof show "s ∪ (⋂ i ∈ I. A i) ⊆ (⋂ i ∈ I. A i ∪ s)" proof fix x assume "x ∈ s ∪ (⋂ i ∈ I. A i)" then show "x ∈ (⋂ i ∈ I. A i ∪ s)" proof assume "x ∈ s" then show ?thesis by simp next assume "x ∈ (⋂ i ∈ I. A i)" then show ?thesis by simp qed qed next show "(⋂ i ∈ I. A i ∪ s) ⊆ s ∪ (⋂ i ∈ I. A i)" proof fix x assume h2 : "x ∈ (⋂ i ∈ I. A i ∪ s)" show "x ∈ s ∪ (⋂ i ∈ I. A i)" proof (cases "x ∈ s") case True then show ?thesis by simp next case False then show ?thesis using h2 by simp qed qed qed (* 5ª demostración *) lemma "s ∪ (⋂ i ∈ I. A i) = (⋂ i ∈ I. A i ∪ s)" by auto end |
2. f⁻¹[u ∩ v] = f⁻¹[u] ∩ f⁻¹[v]
En Lean, la imagen inversa de un conjunto s (de elementos de tipo β) por la función f (de tipo α → β) es el conjunto f ⁻¹' s de elementos x (de tipo α) tales que f x ∈ s.
Demostrar con Lean4 que
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1 |
f ⁻¹' (u ∩ v) = f ⁻¹' u ∩ f ⁻¹' v |
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Set.Function variable {α β : Type _} variable (f : α → β) variable (u v : Set β) open Set example : f ⁻¹' (u ∩ v) = f ⁻¹' u ∩ f ⁻¹' v := by sorry |
2.1. Demostración en lenguaje natural
Tenemos que demostrar que, para todo \(x\),
\[ x ∈ f⁻¹[u ∩ v] ↔ x ∈ f⁻¹[u] ∩ f⁻¹[v] \]
Lo haremos mediante la siguiente cadena de equivalencias
\begin{align}
x ∈ f⁻¹[u ∩ v] &↔ f x ∈ u ∩ v \\
&↔ f x ∈ u ∧ f x ∈ v \\
&↔ x ∈ f⁻¹[u] ∧ x ∈ f⁻¹[v] \\
&↔ x ∈ f⁻¹[u] ∩ f⁻¹[v] \\
\end{align}
2.2. Demostraciones con Lean4
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import Mathlib.Data.Set.Function variable {α β : Type _} variable (f : α → β) variable (u v : Set β) open Set -- 1ª demostración -- =============== example : f ⁻¹' (u ∩ v) = f ⁻¹' u ∩ f ⁻¹' v := by ext x -- x : α -- ⊢ x ∈ f ⁻¹' (u ∩ v) ↔ x ∈ f ⁻¹' u ∩ f ⁻¹' v calc x ∈ f ⁻¹' (u ∩ v) ↔ f x ∈ u ∩ v := by simp only [mem_preimage] _ ↔ f x ∈ u ∧ f x ∈ v := by simp only [mem_inter_iff] _ ↔ x ∈ f ⁻¹' u ∧ x ∈ f ⁻¹' v := by simp only [mem_preimage] _ ↔ x ∈ f ⁻¹' u ∩ f ⁻¹' v := by simp only [mem_inter_iff] -- 2ª demostración -- =============== example : f ⁻¹' (u ∩ v) = f ⁻¹' u ∩ f ⁻¹' v := by ext x -- x : α -- ⊢ x ∈ f ⁻¹' (u ∩ v) ↔ x ∈ f ⁻¹' u ∩ f ⁻¹' v constructor . -- ⊢ x ∈ f ⁻¹' (u ∩ v) → x ∈ f ⁻¹' u ∩ f ⁻¹' v intro h -- h : x ∈ f ⁻¹' (u ∩ v) -- ⊢ x ∈ f ⁻¹' u ∩ f ⁻¹' v constructor . -- ⊢ x ∈ f ⁻¹' u apply mem_preimage.mpr -- ⊢ f x ∈ u rw [mem_preimage] at h -- h : f x ∈ u ∩ v exact mem_of_mem_inter_left h . -- ⊢ x ∈ f ⁻¹' v apply mem_preimage.mpr -- ⊢ f x ∈ v rw [mem_preimage] at h -- h : f x ∈ u ∩ v exact mem_of_mem_inter_right h . -- ⊢ x ∈ f ⁻¹' u ∩ f ⁻¹' v → x ∈ f ⁻¹' (u ∩ v) intro h -- h : x ∈ f ⁻¹' u ∩ f ⁻¹' v -- ⊢ x ∈ f ⁻¹' (u ∩ v) apply mem_preimage.mpr -- ⊢ f x ∈ u ∩ v constructor . -- ⊢ f x ∈ u apply mem_preimage.mp -- ⊢ x ∈ f ⁻¹' u exact mem_of_mem_inter_left h . -- ⊢ f x ∈ v apply mem_preimage.mp -- ⊢ x ∈ f ⁻¹' v exact mem_of_mem_inter_right h -- 3ª demostración -- =============== example : f ⁻¹' (u ∩ v) = f ⁻¹' u ∩ f ⁻¹' v := by ext x -- x : α -- ⊢ x ∈ f ⁻¹' (u ∩ v) ↔ x ∈ f ⁻¹' u ∩ f ⁻¹' v constructor . -- ⊢ x ∈ f ⁻¹' (u ∩ v) → x ∈ f ⁻¹' u ∩ f ⁻¹' v intro h -- h : x ∈ f ⁻¹' (u ∩ v) -- ⊢ x ∈ f ⁻¹' u ∩ f ⁻¹' v constructor . -- ⊢ x ∈ f ⁻¹' u simp at * -- h : f x ∈ u ∧ f x ∈ v -- ⊢ f x ∈ u exact h.1 . -- ⊢ x ∈ f ⁻¹' v simp at * -- h : f x ∈ u ∧ f x ∈ v -- ⊢ f x ∈ v exact h.2 . -- ⊢ x ∈ f ⁻¹' u ∩ f ⁻¹' v → x ∈ f ⁻¹' (u ∩ v) intro h -- h : x ∈ f ⁻¹' u ∩ f ⁻¹' v -- ⊢ x ∈ f ⁻¹' (u ∩ v) simp at * -- h : f x ∈ u ∧ f x ∈ v -- ⊢ f x ∈ u ∧ f x ∈ v exact h -- 4ª demostración -- =============== example : f ⁻¹' (u ∩ v) = f ⁻¹' u ∩ f ⁻¹' v := by aesop -- 5ª demostración -- =============== example : f ⁻¹' (u ∩ v) = f ⁻¹' u ∩ f ⁻¹' v := preimage_inter -- 6ª demostración -- =============== example : f ⁻¹' (u ∩ v) = f ⁻¹' u ∩ f ⁻¹' v := rfl -- Lemas usados -- ============ -- variable (x : α) -- variable (s t : Set α) -- #check (mem_of_mem_inter_left : x ∈ s ∩ t → x ∈ s) -- #check (mem_of_mem_inter_right : x ∈ s ∩ t → x ∈ t) -- #check (mem_preimage : x ∈ f ⁻¹' u ↔ f x ∈ u) -- #check (preimage_inter : f ⁻¹' (u ∩ v) = f ⁻¹' u ∩ f ⁻¹' v) |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
2.3. Demostraciones con Isabelle/HOL
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theory Imagen_inversa_de_la_interseccion imports Main begin (* 1ª demostración *) lemma "f -` (u ∩ v) = f -` u ∩ f -` v" proof (rule equalityI) show "f -` (u ∩ v) ⊆ f -` u ∩ f -` v" proof (rule subsetI) fix x assume "x ∈ f -` (u ∩ v)" then have h : "f x ∈ u ∩ v" by (simp only: vimage_eq) have "x ∈ f -` u" proof - have "f x ∈ u" using h by (rule IntD1) then show "x ∈ f -` u" by (rule vimageI2) qed moreover have "x ∈ f -` v" proof - have "f x ∈ v" using h by (rule IntD2) then show "x ∈ f -` v" by (rule vimageI2) qed ultimately show "x ∈ f -` u ∩ f -` v" by (rule IntI) qed next show "f -` u ∩ f -` v ⊆ f -` (u ∩ v)" proof (rule subsetI) fix x assume h2 : "x ∈ f -` u ∩ f -` v" have "f x ∈ u" proof - have "x ∈ f -` u" using h2 by (rule IntD1) then show "f x ∈ u" by (rule vimageD) qed moreover have "f x ∈ v" proof - have "x ∈ f -` v" using h2 by (rule IntD2) then show "f x ∈ v" by (rule vimageD) qed ultimately have "f x ∈ u ∩ v" by (rule IntI) then show "x ∈ f -` (u ∩ v)" by (rule vimageI2) qed qed (* 2ª demostración *) lemma "f -` (u ∩ v) = f -` u ∩ f -` v" proof show "f -` (u ∩ v) ⊆ f -` u ∩ f -` v" proof fix x assume "x ∈ f -` (u ∩ v)" then have h : "f x ∈ u ∩ v" by simp have "x ∈ f -` u" proof - have "f x ∈ u" using h by simp then show "x ∈ f -` u" by simp qed moreover have "x ∈ f -` v" proof - have "f x ∈ v" using h by simp then show "x ∈ f -` v" by simp qed ultimately show "x ∈ f -` u ∩ f -` v" by simp qed next show "f -` u ∩ f -` v ⊆ f -` (u ∩ v)" proof fix x assume h2 : "x ∈ f -` u ∩ f -` v" have "f x ∈ u" proof - have "x ∈ f -` u" using h2 by simp then show "f x ∈ u" by simp qed moreover have "f x ∈ v" proof - have "x ∈ f -` v" using h2 by simp then show "f x ∈ v" by simp qed ultimately have "f x ∈ u ∩ v" by simp then show "x ∈ f -` (u ∩ v)" by simp qed qed (* 3ª demostración *) lemma "f -` (u ∩ v) = f -` u ∩ f -` v" proof show "f -` (u ∩ v) ⊆ f -` u ∩ f -` v" proof fix x assume h1 : "x ∈ f -` (u ∩ v)" have "x ∈ f -` u" using h1 by simp moreover have "x ∈ f -` v" using h1 by simp ultimately show "x ∈ f -` u ∩ f -` v" by simp qed next show "f -` u ∩ f -` v ⊆ f -` (u ∩ v)" proof fix x assume h2 : "x ∈ f -` u ∩ f -` v" have "f x ∈ u" using h2 by simp moreover have "f x ∈ v" using h2 by simp ultimately have "f x ∈ u ∩ v" by simp then show "x ∈ f -` (u ∩ v)" by simp qed qed (* 4ª demostración *) lemma "f -` (u ∩ v) = f -` u ∩ f -` v" by (simp only: vimage_Int) (* 5ª demostración *) lemma "f -` (u ∩ v) = f -` u ∩ f -` v" by auto end |
3. f[s ∪ t] = f[s] ∪ f[t]
En Lean4, la imagen de un conjunto s por una función f se representa por f '' s; es decir,
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1 |
f '' s = {y | ∃ x, x ∈ s ∧ f x = y} |
Demostrar con Lean4 que
|
1 |
f '' (s ∪ t) = f '' s ∪ f '' t |
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Set.Function variable {α β : Type _} variable (f : α → β) variable (s t : Set α) open Set example : f '' (s ∪ t) = f '' s ∪ f '' t := by sorry |
2.1. Demostración en lenguaje natural
Tenemos que demostrar, para todo \(y\), que
\[ y ∈ f[s ∪ t] ↔ y ∈ f[s] ∪ f[t] \]
Lo haremos mediante la siguiente cadena de equivalencias
\begin{align}
y ∈ f[s ∪ t] &↔ (∃x)(x ∈ s ∪ t ∧ f x = y) \\
&↔ (∃x)((x ∈ s ∨ x ∈ t) ∧ f x = y) \\
&↔ (∃x)((x ∈ s ∧ f x = y) ∨ (x ∈ t ∧ f x = y)) \\
&↔ (∃x)(x ∈ s ∧ f x = y) ∨ (∃x)(x ∈ t ∧ f x = y) \\
&↔ y ∈ f[s] ∨ y ∈ f[t] \\
&↔ y ∈ f[s] ∪ f[t]
\end{align}
3.2. Demostraciones con Lean4
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import Mathlib.Data.Set.Function variable {α β : Type _} variable (f : α → β) variable (s t : Set α) open Set -- 1ª demostración -- =============== example : f '' (s ∪ t) = f '' s ∪ f '' t := by ext y -- y : β -- ⊢ y ∈ f '' (s ∪ t) ↔ y ∈ f '' s ∪ f '' t calc y ∈ f '' (s ∪ t) ↔ ∃ x, x ∈ s ∪ t ∧ f x = y := by simp only [mem_image] _ ↔ ∃ x, (x ∈ s ∨ x ∈ t) ∧ f x = y := by simp only [mem_union] _ ↔ ∃ x, (x ∈ s ∧ f x = y) ∨ (x ∈ t ∧ f x = y) := by simp only [or_and_right] _ ↔ (∃ x, x ∈ s ∧ f x = y) ∨ (∃ x, x ∈ t ∧ f x = y) := by simp only [exists_or] _ ↔ y ∈ f '' s ∨ y ∈ f '' t := by simp only [mem_image] _ ↔ y ∈ f '' s ∪ f '' t := by simp only [mem_union] -- 2ª demostración -- =============== example : f '' (s ∪ t) = f '' s ∪ f '' t := by ext y -- y : β -- ⊢ y ∈ f '' (s ∪ t) ↔ y ∈ f '' s ∪ f '' t constructor . -- ⊢ y ∈ f '' (s ∪ t) → y ∈ f '' s ∪ f '' t intro h -- h : y ∈ f '' (s ∪ t) -- ⊢ y ∈ f '' s ∪ f '' t rw [mem_image] at h -- h : ∃ x, x ∈ s ∪ t ∧ f x = y rcases h with ⟨x, hx⟩ -- x : α -- hx : x ∈ s ∪ t ∧ f x = y rcases hx with ⟨xst, fxy⟩ -- xst : x ∈ s ∪ t -- fxy : f x = y rw [←fxy] -- ⊢ f x ∈ f '' s ∪ f '' t rw [mem_union] at xst -- xst : x ∈ s ∨ x ∈ t rcases xst with (xs | xt) . -- xs : x ∈ s apply mem_union_left -- ⊢ f x ∈ f '' s apply mem_image_of_mem -- ⊢ x ∈ s exact xs . -- xt : x ∈ t apply mem_union_right -- ⊢ f x ∈ f '' t apply mem_image_of_mem -- ⊢ x ∈ t exact xt . -- ⊢ y ∈ f '' s ∪ f '' t → y ∈ f '' (s ∪ t) intro h -- h : y ∈ f '' s ∪ f '' t -- ⊢ y ∈ f '' (s ∪ t) rw [mem_union] at h -- h : y ∈ f '' s ∨ y ∈ f '' t rcases h with (yfs | yft) . -- yfs : y ∈ f '' s rw [mem_image] -- ⊢ ∃ x, x ∈ s ∪ t ∧ f x = y rw [mem_image] at yfs -- yfs : ∃ x, x ∈ s ∧ f x = y rcases yfs with ⟨x, hx⟩ -- x : α -- hx : x ∈ s ∧ f x = y rcases hx with ⟨xs, fxy⟩ -- xs : x ∈ s -- fxy : f x = y use x -- ⊢ x ∈ s ∪ t ∧ f x = y constructor . -- ⊢ x ∈ s ∪ t apply mem_union_left -- ⊢ x ∈ s exact xs . -- ⊢ f x = y exact fxy . -- yft : y ∈ f '' t rw [mem_image] -- ⊢ ∃ x, x ∈ s ∪ t ∧ f x = y rw [mem_image] at yft -- yft : ∃ x, x ∈ t ∧ f x = y rcases yft with ⟨x, hx⟩ -- x : α -- hx : x ∈ t ∧ f x = y rcases hx with ⟨xt, fxy⟩ -- xt : x ∈ t -- fxy : f x = y use x -- ⊢ x ∈ s ∪ t ∧ f x = y constructor . -- ⊢ x ∈ s ∪ t apply mem_union_right -- ⊢ x ∈ t exact xt . -- ⊢ f x = y exact fxy -- 3ª demostración -- =============== example : f '' (s ∪ t) = f '' s ∪ f '' t := by ext y -- y : β -- ⊢ y ∈ f '' (s ∪ t) ↔ y ∈ f '' s ∪ f '' t constructor . -- ⊢ y ∈ f '' (s ∪ t) → y ∈ f '' s ∪ f '' t rintro ⟨x, xst, rfl⟩ -- x : α -- xst : x ∈ s ∪ t -- ⊢ f x ∈ f '' s ∪ f '' t rcases xst with (xs | xt) . -- xs : x ∈ s left -- ⊢ f x ∈ f '' s exact mem_image_of_mem f xs . -- xt : x ∈ t right -- ⊢ f x ∈ f '' t exact mem_image_of_mem f xt . -- ⊢ y ∈ f '' s ∪ f '' t → y ∈ f '' (s ∪ t) rintro (yfs | yft) . -- yfs : y ∈ f '' s rcases yfs with ⟨x, xs, rfl⟩ -- x : α -- xs : x ∈ s -- ⊢ f x ∈ f '' (s ∪ t) apply mem_image_of_mem -- ⊢ x ∈ s ∪ t left -- ⊢ x ∈ s exact xs . -- yft : y ∈ f '' t rcases yft with ⟨x, xt, rfl⟩ -- x : α -- xs : x ∈ s -- ⊢ f x ∈ f '' (s ∪ t) apply mem_image_of_mem -- ⊢ x ∈ s ∪ t right -- ⊢ x ∈ t exact xt -- 4ª demostración -- =============== example : f '' (s ∪ t) = f '' s ∪ f '' t := by ext y -- y : β -- ⊢ y ∈ f '' (s ∪ t) ↔ y ∈ f '' s ∪ f '' t constructor . -- ⊢ y ∈ f '' (s ∪ t) → y ∈ f '' s ∪ f '' t rintro ⟨x, xst, rfl⟩ -- x : α -- xst : x ∈ s ∪ t -- ⊢ f x ∈ f '' s ∪ f '' t rcases xst with (xs | xt) . -- xs : x ∈ s left -- ⊢ f x ∈ f '' s use x, xs . -- xt : x ∈ t right -- ⊢ f x ∈ f '' t use x, xt . rintro (yfs | yft) . -- yfs : y ∈ f '' s rcases yfs with ⟨x, xs, rfl⟩ -- x : α -- xs : x ∈ s -- ⊢ f x ∈ f '' (s ∪ t) use x, Or.inl xs . -- yft : y ∈ f '' t rcases yft with ⟨x, xt, rfl⟩ -- x : α -- xt : x ∈ t -- ⊢ f x ∈ f '' (s ∪ t) use x, Or.inr xt -- 5ª demostración -- =============== example : f '' (s ∪ t) = f '' s ∪ f '' t := by ext y -- y : β -- ⊢ y ∈ f '' (s ∪ t) ↔ y ∈ f '' s ∪ f '' t constructor . -- ⊢ y ∈ f '' (s ∪ t) → y ∈ f '' s ∪ f '' t rintro ⟨x, xs | xt, rfl⟩ . -- x : α -- xs : x ∈ s -- ⊢ f x ∈ f '' s ∪ f '' t left -- ⊢ f x ∈ f '' s use x, xs . -- x : α -- xt : x ∈ t -- ⊢ f x ∈ f '' s ∪ f '' t right -- ⊢ f x ∈ f '' t use x, xt . -- ⊢ y ∈ f '' s ∪ f '' t → y ∈ f '' (s ∪ t) rintro (⟨x, xs, rfl⟩ | ⟨x, xt, rfl⟩) . -- x : α -- xs : x ∈ s -- ⊢ f x ∈ f '' (s ∪ t) use x, Or.inl xs . -- x : α -- xt : x ∈ t -- ⊢ f x ∈ f '' (s ∪ t) use x, Or.inr xt -- 6ª demostración -- =============== example : f '' (s ∪ t) = f '' s ∪ f '' t := by ext y -- y : β -- ⊢ y ∈ f '' (s ∪ t) ↔ y ∈ f '' s ∪ f '' t constructor . -- ⊢ y ∈ f '' (s ∪ t) → y ∈ f '' s ∪ f '' t aesop . -- ⊢ y ∈ f '' s ∪ f '' t → y ∈ f '' (s ∪ t) aesop -- 7ª demostración -- =============== example : f '' (s ∪ t) = f '' s ∪ f '' t := by ext y constructor <;> aesop -- 8ª demostración -- =============== example : f '' (s ∪ t) = f '' s ∪ f '' t := by ext y -- y : β -- ⊢ y ∈ f '' (s ∪ t) ↔ y ∈ f '' s ∪ f '' t rw [iff_def] -- ⊢ (y ∈ f '' (s ∪ t) → y ∈ f '' s ∪ f '' t) ∧ (y ∈ f '' s ∪ f '' t → y ∈ f '' (s ∪ t)) aesop -- 9ª demostración -- =============== example : f '' (s ∪ t) = f '' s ∪ f '' t := image_union f s t -- Lemas usados -- ============ -- variable (x : α) -- variable (y : β) -- variable (a b c : Prop) -- variable (p q : α → Prop) -- #check (Or.inl : a → a ∨ b) -- #check (Or.inr : b → a ∨ b) -- #check (exists_or : (∃ x, p x ∨ q x) ↔ (∃ x, p x) ∨ ∃ x, q x) -- #check (iff_def : (a ↔ b) ↔ (a → b) ∧ (b → a)) -- #check (image_union f s t : f '' (s ∪ t) = f '' s ∪ f '' t) -- #check (mem_image f s y : (y ∈ f '' s ↔ ∃ (x : α), x ∈ s ∧ f x = y)) -- #check (mem_image_of_mem f : x ∈ s → f x ∈ f '' s) -- #check (mem_union x s t : x ∈ s ∪ t ↔ x ∈ s ∨ x ∈ t) -- #check (mem_union_left t : x ∈ s → x ∈ s ∪ t) -- #check (mem_union_right s : x ∈ t → x ∈ s ∪ t) -- #check (or_and_right : (a ∨ b) ∧ c ↔ a ∧ c ∨ b ∧ c) |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
3.3. Demostraciones con Isabelle/HOL
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theory Imagen_de_la_union imports Main begin (* 1ª demostración *) lemma "f ` (s ∪ t) = f ` s ∪ f ` t" proof (rule equalityI) show "f ` (s ∪ t) ⊆ f ` s ∪ f ` t" proof (rule subsetI) fix y assume "y ∈ f ` (s ∪ t)" then show "y ∈ f ` s ∪ f ` t" proof (rule imageE) fix x assume "y = f x" assume "x ∈ s ∪ t" then show "y ∈ f ` s ∪ f ` t" proof (rule UnE) assume "x ∈ s" with ‹y = f x› have "y ∈ f ` s" by (simp only: image_eqI) then show "y ∈ f ` s ∪ f ` t" by (rule UnI1) next assume "x ∈ t" with ‹y = f x› have "y ∈ f ` t" by (simp only: image_eqI) then show "y ∈ f ` s ∪ f ` t" by (rule UnI2) qed qed qed next show "f ` s ∪ f ` t ⊆ f ` (s ∪ t)" proof (rule subsetI) fix y assume "y ∈ f ` s ∪ f ` t" then show "y ∈ f ` (s ∪ t)" proof (rule UnE) assume "y ∈ f ` s" then show "y ∈ f ` (s ∪ t)" proof (rule imageE) fix x assume "y = f x" assume "x ∈ s" then have "x ∈ s ∪ t" by (rule UnI1) with ‹y = f x› show "y ∈ f ` (s ∪ t)" by (simp only: image_eqI) qed next assume "y ∈ f ` t" then show "y ∈ f ` (s ∪ t)" proof (rule imageE) fix x assume "y = f x" assume "x ∈ t" then have "x ∈ s ∪ t" by (rule UnI2) with ‹y = f x› show "y ∈ f ` (s ∪ t)" by (simp only: image_eqI) qed qed qed qed (* 2ª demostración *) lemma "f ` (s ∪ t) = f ` s ∪ f ` t" proof show "f ` (s ∪ t) ⊆ f ` s ∪ f ` t" proof fix y assume "y ∈ f ` (s ∪ t)" then show "y ∈ f ` s ∪ f ` t" proof fix x assume "y = f x" assume "x ∈ s ∪ t" then show "y ∈ f ` s ∪ f ` t" proof assume "x ∈ s" with ‹y = f x› have "y ∈ f ` s" by simp then show "y ∈ f ` s ∪ f ` t" by simp next assume "x ∈ t" with ‹y = f x› have "y ∈ f ` t" by simp then show "y ∈ f ` s ∪ f ` t" by simp qed qed qed next show "f ` s ∪ f ` t ⊆ f ` (s ∪ t)" proof fix y assume "y ∈ f ` s ∪ f ` t" then show "y ∈ f ` (s ∪ t)" proof assume "y ∈ f ` s" then show "y ∈ f ` (s ∪ t)" proof fix x assume "y = f x" assume "x ∈ s" then have "x ∈ s ∪ t" by simp with ‹y = f x› show "y ∈ f ` (s ∪ t)" by simp qed next assume "y ∈ f ` t" then show "y ∈ f ` (s ∪ t)" proof fix x assume "y = f x" assume "x ∈ t" then have "x ∈ s ∪ t" by simp with ‹y = f x› show "y ∈ f ` (s ∪ t)" by simp qed qed qed qed (* 3ª demostración *) lemma "f ` (s ∪ t) = f ` s ∪ f ` t" by (simp only: image_Un) (* 4ª demostración *) lemma "f ` (s ∪ t) = f ` s ∪ f ` t" by auto end |
4. s ⊆ f⁻¹[f[s]]
Demostrar que si \(s\) es un subconjunto del dominio de la función \(f\), entonces \(s\) está contenido en la imagen inversa de la imagen de \(s\) por \(f\); es decir,
\[ s ⊆ f⁻¹[f[s]] \]
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Set.Function open Set variable {α β : Type _} variable (f : α → β) variable (s : Set α) example : s ⊆ f ⁻¹' (f '' s) := by sorry |
4.1. Demostración en lenguaje natural
Se demuestra mediante la siguiente cadena de implicaciones
\begin{align}
x ∈ s &⟹ f(x) ∈ f[s] \\
&⟹ x ∈ f⁻¹[f[s]]
\end{align}
4.2. Demostraciones con Lean4
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import Mathlib.Data.Set.Function open Set variable {α β : Type _} variable (f : α → β) variable (s : Set α) -- 1ª demostración -- =============== example : s ⊆ f ⁻¹' (f '' s) := by intros x xs -- x : α -- xs : x ∈ s -- ⊢ x ∈ f ⁻¹' (f '' s) have h1 : f x ∈ f '' s := mem_image_of_mem f xs show x ∈ f ⁻¹' (f '' s) exact mem_preimage.mp h1 -- 2ª demostración -- =============== example : s ⊆ f ⁻¹' (f '' s) := by intros x xs -- x : α -- xs : x ∈ s -- ⊢ x ∈ f ⁻¹' (f '' s) apply mem_preimage.mpr -- ⊢ f x ∈ f '' s apply mem_image_of_mem -- ⊢ x ∈ s exact xs -- 3ª demostración -- =============== example : s ⊆ f ⁻¹' (f '' s) := by intros x xs -- x : α -- xs : x ∈ s -- ⊢ x ∈ f ⁻¹' (f '' s) apply mem_image_of_mem -- ⊢ x ∈ s exact xs -- 4ª demostración -- =============== example : s ⊆ f ⁻¹' (f '' s) := fun _ ↦ mem_image_of_mem f -- 5ª demostración -- =============== example : s ⊆ f ⁻¹' (f '' s) := by intros x xs -- x : α -- xs : x ∈ s -- ⊢ x ∈ f ⁻¹' (f '' s) show f x ∈ f '' s use x, xs -- 6ª demostración -- =============== example : s ⊆ f ⁻¹' (f '' s) := by intros x xs -- x : α -- xs : x ∈ s -- ⊢ x ∈ f ⁻¹' (f '' s) use x, xs -- 7ª demostración -- =============== example : s ⊆ f ⁻¹' (f '' s) := subset_preimage_image f s -- Lemas usados -- ============ -- variable (x : α) -- variable (t : Set β) -- #check (mem_preimage : x ∈ f ⁻¹' t ↔ f x ∈ t) -- #check (mem_image_of_mem f : x ∈ s → f x ∈ f '' s) -- #check (subset_preimage_image f s : s ⊆ f ⁻¹' (f '' s)) |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
4.3. Demostraciones con Isabelle/HOL
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theory Imagen_inversa_de_la_imagen imports Main begin (* 1ª demostración *) lemma "s ⊆ f -` (f ` s)" proof (rule subsetI) fix x assume "x ∈ s" then have "f x ∈ f ` s" by (simp only: imageI) then show "x ∈ f -` (f ` s)" by (simp only: vimageI) qed (* 2ª demostración *) lemma "s ⊆ f -` (f ` s)" proof fix x assume "x ∈ s" then have "f x ∈ f ` s" by simp then show "x ∈ f -` (f ` s)" by simp qed (* 3ª demostración *) lemma "s ⊆ f -` (f ` s)" by auto end |
5. f[s] ⊆ u ↔ s ⊆ f⁻¹[u]
Demostrar con Lean4 que
\[ f[s] ⊆ u ↔ s ⊆ f⁻¹[u] \]
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Set.Function open Set variable {α β : Type _} variable (f : α → β) variable (s : Set α) variable (u : Set β) example : f '' s ⊆ u ↔ s ⊆ f ⁻¹' u := by sorry |
5.1. Demostración en lenguaje natural
Los demostraremos probando las dos implicaciones.
(⟹) Supongamos que
\[ f[s] ⊆ u \tag{1} \]
y tenemos que demostrar que
\[ s ⊆ f⁻¹[u] \]
Se prueba mediante las siguientes implicaciones
\begin{align}
x ∈ s &⟹ f(x) ∈ f[s] \\
&⟹ f(x) ∈ u &&\text{[por (1)]} \\
&⟹ x ∈ f⁻¹[u]
\end{align}
(⟸) Supongamos que
\[ s ⊆ f⁻¹[u] \tag{2} \]
y tenemos que demostrar que
\[ f[s] ⊆ u \]
Para ello, sea \(y ∈ f[s]\). Entonces, existe un
\[ x ∈ s \tag{3} \]
tal que
\[ y = f(x) \tag{4} \]
Entonces,
\begin{align}
&x ∈ f⁻¹[u] &&\text{[por (2) y (3)]} \\
⟹ &f(x) ∈ u \\
⟹ &y ∈ u &&\text{[por (4)]}
\end{align}
5.2. Demostraciones con Lean4
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import Mathlib.Data.Set.Function open Set variable {α β : Type _} variable (f : α → β) variable (s : Set α) variable (u : Set β) -- 1ª demostración -- =============== example : f '' s ⊆ u ↔ s ⊆ f ⁻¹' u := calc f '' s ⊆ u ↔ ∀ y, y ∈ f '' s → y ∈ u := by simp only [subset_def] _ ↔ ∀ y, (∃ x, x ∈ s ∧ f x = y) → y ∈ u := by simp only [mem_image] _ ↔ ∀ x, x ∈ s → f x ∈ u := by constructor . -- (∀ y, (∃ x, x ∈ s ∧ f x = y) → y ∈ u) → (∀ x, x ∈ s → f x ∈ u) intro h x xs -- h : ∀ (y : β), (∃ x, x ∈ s ∧ f x = y) → y ∈ u -- x : α -- xs : x ∈ s -- ⊢ f x ∈ u exact h (f x) (by use x, xs) . -- (∀ x, x ∈ s → f x ∈ u) → (∀ y, (∃ x, x ∈ s ∧ f x = y) → y ∈ u) intro h y hy -- h : ∀ (x : α), x ∈ s → f x ∈ u -- y : β -- hy : ∃ x, x ∈ s ∧ f x = y -- ⊢ y ∈ u obtain ⟨x, hx⟩ := hy -- x : α -- hx : x ∈ s ∧ f x = y have h1 : y = f x := hx.2.symm have h2 : f x ∈ u := h x hx.1 show y ∈ u exact mem_of_eq_of_mem h1 h2 _ ↔ ∀ x, x ∈ s → x ∈ f ⁻¹' u := by simp only [mem_preimage] _ ↔ s ⊆ f ⁻¹' u := by simp only [subset_def] -- 2ª demostración -- =============== example : f '' s ⊆ u ↔ s ⊆ f ⁻¹' u := calc f '' s ⊆ u ↔ ∀ y, y ∈ f '' s → y ∈ u := by simp only [subset_def] _ ↔ ∀ y, (∃ x, x ∈ s ∧ f x = y) → y ∈ u := by simp only [mem_image] _ ↔ ∀ x, x ∈ s → f x ∈ u := by constructor . -- (∀ y, (∃ x, x ∈ s ∧ f x = y) → y ∈ u) → (∀ x, x ∈ s → f x ∈ u) intro h x xs -- h : ∀ (y : β), (∃ x, x ∈ s ∧ f x = y) → y ∈ u -- x : α -- xs : x ∈ s -- ⊢ f x ∈ u apply h (f x) -- ⊢ ∃ x_1, x_1 ∈ s ∧ f x_1 = f x use x, xs . -- (∀ x, x ∈ s → f x ∈ u) → (∀ y, (∃ x, x ∈ s ∧ f x = y) → y ∈ u) intro h y hy -- h : ∀ (x : α), x ∈ s → f x ∈ u -- y : β -- hy : ∃ x, x ∈ s ∧ f x = y -- ⊢ y ∈ u obtain ⟨x, hx⟩ := hy -- x : α -- hx : x ∈ s ∧ f x = y rw [←hx.2] -- ⊢ f x ∈ u apply h x -- ⊢ x ∈ s exact hx.1 _ ↔ ∀ x, x ∈ s → x ∈ f ⁻¹' u := by simp only [mem_preimage] _ ↔ s ⊆ f ⁻¹' u := by simp only [subset_def] -- 3ª demostración -- =============== example : f '' s ⊆ u ↔ s ⊆ f ⁻¹' u := by constructor . -- ⊢ f '' s ⊆ u → s ⊆ f ⁻¹' u intros h x xs -- h : f '' s ⊆ u -- x : α -- xs : x ∈ s -- ⊢ x ∈ f ⁻¹' u apply mem_preimage.mpr -- ⊢ f x ∈ u apply h -- ⊢ f x ∈ f '' s apply mem_image_of_mem -- ⊢ x ∈ s exact xs . -- ⊢ s ⊆ f ⁻¹' u → f '' s ⊆ u intros h y hy -- h : s ⊆ f ⁻¹' u -- y : β -- hy : y ∈ f '' s -- ⊢ y ∈ u rcases hy with ⟨x, xs, fxy⟩ -- x : α -- xs : x ∈ s -- fxy : f x = y rw [←fxy] -- ⊢ f x ∈ u exact h xs -- 4ª demostración -- =============== example : f '' s ⊆ u ↔ s ⊆ f ⁻¹' u := by constructor . -- ⊢ f '' s ⊆ u → s ⊆ f ⁻¹' u intros h x xs -- h : f '' s ⊆ u -- x : α -- xs : x ∈ s -- ⊢ x ∈ f ⁻¹' u apply h -- ⊢ f x ∈ f '' s apply mem_image_of_mem -- ⊢ x ∈ s exact xs . -- ⊢ s ⊆ f ⁻¹' u → f '' s ⊆ u rintro h y ⟨x, xs, rfl⟩ -- h : s ⊆ f ⁻¹' u -- x : α -- xs : x ∈ s -- ⊢ f x ∈ u exact h xs -- 5ª demostración -- =============== example : f '' s ⊆ u ↔ s ⊆ f ⁻¹' u := image_subset_iff -- 4ª demostración -- =============== example : f '' s ⊆ u ↔ s ⊆ f ⁻¹' u := by simp -- Lemas usados -- ============ -- variable (x y : α) -- #check (image_subset_iff : f '' s ⊆ u ↔ s ⊆ f ⁻¹' u) -- #check (mem_image_of_mem f : x ∈ s → f x ∈ f '' s) -- #check (mem_of_eq_of_mem : x = y → y ∈ s → x ∈ s) -- #check (mem_preimage : x ∈ f ⁻¹' u ↔ f x ∈ u) |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
5.3. Demostraciones con Isabelle/HOL
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theory Subconjunto_de_la_imagen_inversa imports Main begin (* 1ª demostración *) lemma "f ` s ⊆ u ⟷ s ⊆ f -` u" proof (rule iffI) assume "f ` s ⊆ u" show "s ⊆ f -` u" proof (rule subsetI) fix x assume "x ∈ s" then have "f x ∈ f ` s" by (simp only: imageI) then have "f x ∈ u" using ‹f ` s ⊆ u› by (rule set_rev_mp) then show "x ∈ f -` u" by (simp only: vimageI) qed next assume "s ⊆ f -` u" show "f ` s ⊆ u" proof (rule subsetI) fix y assume "y ∈ f ` s" then show "y ∈ u" proof fix x assume "y = f x" assume "x ∈ s" then have "x ∈ f -` u" using ‹s ⊆ f -` u› by (rule set_rev_mp) then have "f x ∈ u" by (rule vimageD) with ‹y = f x› show "y ∈ u" by (rule ssubst) qed qed qed (* 2ª demostración *) lemma "f ` s ⊆ u ⟷ s ⊆ f -` u" proof assume "f ` s ⊆ u" show "s ⊆ f -` u" proof fix x assume "x ∈ s" then have "f x ∈ f ` s" by simp then have "f x ∈ u" using ‹f ` s ⊆ u› by (simp add: set_rev_mp) then show "x ∈ f -` u" by simp qed next assume "s ⊆ f -` u" show "f ` s ⊆ u" proof fix y assume "y ∈ f ` s" then show "y ∈ u" proof fix x assume "y = f x" assume "x ∈ s" then have "x ∈ f -` u" using ‹s ⊆ f -` u› by (simp only: set_rev_mp) then have "f x ∈ u" by simp with ‹y = f x› show "y ∈ u" by simp qed qed qed (* 3ª demostración *) lemma "f ` s ⊆ u ⟷ s ⊆ f -` u" by (simp only: image_subset_iff_subset_vimage) (* 4ª demostración *) lemma "f ` s ⊆ u ⟷ s ⊆ f -` u" by auto end |