La semana en Calculemus (24 de febrero de 2024)
Estas 3 últimas semanas he publicado en Calculemus las demostraciones con Lean4 de las siguientes propiedades:
- 1. Si la sucesión u converge a a y la v a b, entonces u+v converge a a+b
- 2. Unicidad del límite de las sucesiones convergentes
- 3. Si el límite de la sucesión uₙ es a y c ∈ ℝ, entonces el límite de uₙ+c es a+c
- 4. Si el límite de la sucesión uₙ es a y c ∈ ℝ, entonces el límite de cuₙ es ca
- 5. El límite de uₙ es a syss el de uₙ-a es 0
- 6. Si uₙ y vₙ convergen a 0, entonces uₙvₙ converge a 0
- 7. Teorema del emparedado
- 8. Si s ⊆ t, entonces s ∩ u ⊆ t ∩ u
- 9. s ∩ (t ∪ u) ⊆ (s ∩ t) ∪ (s ∩ u)
- 10. (s \ t) \ u ⊆ s \ (t ∪ u)
- 11. (s ∩ t) ∪ (s ∩ u) ⊆ s ∩ (t ∪ u)
A continuación se muestran las soluciones.
1. Si la sucesión u converge a a y la v a b, entonces u+v converge a a+b
Demostrar con Lean4 que si la sucesión \(u\) converge a \(a\) y la \(v\) a \(b\), entonces \(u+v\) converge a \(a+b\).
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Real.Basic variable {s t : ℕ → ℝ} {a b c : ℝ} def limite (s : ℕ → ℝ) (a : ℝ) := ∀ ε > 0, ∃ N, ∀ n ≥ N, |s n - a| < ε example (hu : limite u a) (hv : limite v b) : limite (u + v) (a + b) := by sorry |
1.1. Demostración en lenguaje natural
En la demostración usaremos los siguientes lemas
\begin{align}
&(∀ a ∈ ℝ)\left[a > 0 → \frac{a}{2} > 0\right] \tag{L1} \\
&(∀ a, b, c ∈ ℝ)[\max(a, b) ≤ c → a ≤ c] \tag{L2} \\
&(∀ a, b, c ∈ ℝ)[\max(a, b) ≤ c → b ≤ c] \tag{L3} \\
&(∀ a, b ∈ ℝ)[|a + b| ≤ |a| + |b|] \tag{L4} \\
&(∀ a ∈ ℝ)\left[\frac{a}{2} + \frac{a}{2} = a\right] \tag{L5}
\end{align}
Tenemos que probar que para todo \(ε ∈ ℝ\), si
\[ ε > 0 \tag{1} \]
entonces
\[ (∃N ∈ ℕ)(∀n ∈ ℕ)[n ≥ N → |(u + v)(n) – (a + b)| < ε] \tag{2} \]
Por (1) y el lema L1, se tiene que
\[ \frac{ε}{2} > 0 \tag{3} \]
Por (3) y porque el límite de \(u\) es \(a\), se tiene que
\[ (∃N ∈ ℕ)(∀n ∈ ℕ)\left[n ≥ N → |u(n) – a| < \frac{ε}{2}\right] \]
Sea \(N₁ ∈ ℕ\) tal que
\[ (∀n ∈ ℕ)\left[n ≥ N₁ → |u(n) – a| < \frac{ε}{2}\right] \tag{4} \]
Por (3) y porque el límite de \(v\) es \(b\), se tiene que
\[ (∃N ∈ ℕ)(∀n ∈ ℕ)\left[n ≥ N → |v(n) – b| < \frac{ε}{2}\right] \]
Sea \(N₂ ∈ ℕ\) tal que
\[ (∀n ∈ ℕ)\left[n ≥ N₂ → |v(n) – b| < \frac{ε}{2}\right] \tag{5} \]
Sea \(N = \max(N₁, N₂)\). Veamos que verifica la condición (1). Para ello, sea \(n ∈ ℕ\) tal que \(n ≥ N\). Entonces, \(n ≥ N₁\) (por L2) y \(n ≥ N₂\) (por L3). Por tanto, usando las propiedades (4) y (5) se tiene que
\begin{align}
|u(n) – a| &< \frac{ε}{2} \tag{6} \\
|v(n) – b| &< \frac{ε}{2} \tag{7}
\end{align}
Finalmente,
\begin{align}
|(u + v)(n) – (a + b)| &= |(u(n) + v(n)) – (a + b)| \\
&= |(u(n) – a) + (v(n) – b)| \\
&≤ |u(n) – a| + |v(n) – b| &&\text{[por L4]}\\
&< \frac{ε}{2} + \frac{ε}{2} &&\text{[por (6) y (7)]}\\
&= ε &&\text{[por L5]}
\end{align}
1.2. Demostraciones con Lean4
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import Mathlib.Data.Real.Basic variable {s t : ℕ → ℝ} {a b c : ℝ} def limite (s : ℕ → ℝ) (a : ℝ) := ∀ ε > 0, ∃ N, ∀ n ≥ N, |s n - a| < ε -- 1ª demostración -- =============== example (hu : limite u a) (hv : limite v b) : limite (u + v) (a + b) := by intros ε hε -- ε : ℝ -- hε : ε > 0 -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |(u + v) n - (a + b)| < ε have hε2 : 0 < ε / 2 := half_pos hε cases' hu (ε / 2) hε2 with Nu hNu -- Nu : ℕ -- hNu : ∀ (n : ℕ), n ≥ Nu → |u n - a| < ε / 2 cases' hv (ε / 2) hε2 with Nv hNv -- Nv : ℕ -- hNv : ∀ (n : ℕ), n ≥ Nv → |v n - b| < ε / 2 clear hu hv hε2 hε let N := max Nu Nv use N -- ⊢ ∀ (n : ℕ), n ≥ N → |(s + t) n - (a + b)| < ε intros n hn -- n : ℕ -- hn : n ≥ N have nNu : n ≥ Nu := le_of_max_le_left hn specialize hNu n nNu -- hNu : |u n - a| < ε / 2 have nNv : n ≥ Nv := le_of_max_le_right hn specialize hNv n nNv -- hNv : |v n - b| < ε / 2 clear hn nNu nNv calc |(u + v) n - (a + b)| = |(u n + v n) - (a + b)| := rfl _ = |(u n - a) + (v n - b)| := by { congr; ring } _ ≤ |u n - a| + |v n - b| := by apply abs_add _ < ε / 2 + ε / 2 := by linarith [hNu, hNv] _ = ε := by apply add_halves -- 2ª demostración -- =============== example (hu : limite u a) (hv : limite v b) : limite (u + v) (a + b) := by intros ε hε cases' hu (ε/2) (by linarith) with Nu hNu cases' hv (ε/2) (by linarith) with Nv hNv use max Nu Nv intros n hn have hn₁ : n ≥ Nu := le_of_max_le_left hn specialize hNu n hn₁ have hn₂ : n ≥ Nv := le_of_max_le_right hn specialize hNv n hn₂ calc |(u + v) n - (a + b)| = |(u n + v n) - (a + b)| := by rfl _ = |(u n - a) + (v n - b)| := by {congr; ring} _ ≤ |u n - a| + |v n - b| := by apply abs_add _ < ε / 2 + ε / 2 := by linarith _ = ε := by apply add_halves -- 3ª demostración -- =============== lemma max_ge_iff {α : Type _} [LinearOrder α] {p q r : α} : r ≥ max p q ↔ r ≥ p ∧ r ≥ q := max_le_iff example (hu : limite u a) (hv : limite v b) : limite (u + v) (a + b) := by intros ε hε cases' hu (ε/2) (by linarith) with Nu hNu cases' hv (ε/2) (by linarith) with Nv hNv use max Nu Nv intros n hn cases' max_ge_iff.mp hn with hn₁ hn₂ have cota₁ : |u n - a| < ε/2 := hNu n hn₁ have cota₂ : |v n - b| < ε/2 := hNv n hn₂ calc |(u + v) n - (a + b)| = |(u n + v n) - (a + b)| := by rfl _ = |(u n - a) + (v n - b)| := by { congr; ring } _ ≤ |u n - a| + |v n - b| := by apply abs_add _ < ε := by linarith -- 4ª demostración -- =============== example (hu : limite u a) (hv : limite v b) : limite (u + v) (a + b) := by intros ε hε cases' hu (ε/2) (by linarith) with Nu hNu cases' hv (ε/2) (by linarith) with Nv hNv use max Nu Nv intros n hn cases' max_ge_iff.mp hn with hn₁ hn₂ calc |(u + v) n - (a + b)| = |u n + v n - (a + b)| := by rfl _ = |(u n - a) + (v n - b)| := by { congr; ring } _ ≤ |u n - a| + |v n - b| := by apply abs_add _ < ε/2 + ε/2 := add_lt_add (hNu n hn₁) (hNv n hn₂) _ = ε := by simp -- 5ª demostración -- =============== example (hu : limite u a) (hv : limite v b) : limite (u + v) (a + b) := by intros ε hε cases' hu (ε/2) (by linarith) with Nu hNu cases' hv (ε/2) (by linarith) with Nv hNv use max Nu Nv intros n hn rw [max_ge_iff] at hn calc |(u + v) n - (a + b)| = |u n + v n - (a + b)| := by rfl _ = |(u n - a) + (v n - b)| := by { congr; ring } _ ≤ |u n - a| + |v n - b| := by apply abs_add _ < ε := by linarith [hNu n (by linarith), hNv n (by linarith)] -- 6ª demostración -- =============== example (hu : limite u a) (hv : limite v b) : limite (u + v) (a + b) := by intros ε Hε cases' hu (ε/2) (by linarith) with L HL cases' hv (ε/2) (by linarith) with M HM set N := max L M with _hN use N have HLN : N ≥ L := le_max_left _ _ have HMN : N ≥ M := le_max_right _ _ intros n Hn have H3 : |u n - a| < ε/2 := HL n (by linarith) have H4 : |v n - b| < ε/2 := HM n (by linarith) calc |(u + v) n - (a + b)| = |(u n + v n) - (a + b)| := by rfl _ = |(u n - a) + (v n - b)| := by {congr; ring } _ ≤ |(u n - a)| + |(v n - b)| := by apply abs_add _ < ε/2 + ε/2 := by linarith _ = ε := by ring -- Lemas usados -- ============ -- variable (d : ℝ) -- #check (abs_add a b : |a + b| ≤ |a| + |b|) -- #check (add_halves a : a / 2 + a / 2 = a) -- #check (add_lt_add : a < b → c < d → a + c < b + d) -- #check (half_pos : a > 0 → a / 2 > 0) -- #check (le_max_left a b : a ≤ max a b) -- #check (le_max_right a b : b ≤ max a b) -- #check (le_of_max_le_left : max a b ≤ c → a ≤ c) -- #check (le_of_max_le_right : max a b ≤ c → b ≤ c) -- #check (max_le_iff : max a b ≤ c ↔ a ≤ c ∧ b ≤ c) |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
1.3. Demostraciones con Isabelle/HOL
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theory Limite_de_sucesiones_constantes 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)" (* 1ª demostración *) lemma "limite (λ n. c) c" proof (unfold limite_def) show "∀ε>0. ∃k::nat. ∀n≥k. ¦c - c¦ < ε" proof (intro allI impI) fix ε :: real assume "0 < ε" have "∀n≥0::nat. ¦c - c¦ < ε" proof (intro allI impI) fix n :: nat assume "0 ≤ n" have "c - c = 0" by (simp only: diff_self) then have "¦c - c¦ = 0" by (simp only: abs_eq_0_iff) also have "… < ε" by (simp only: ‹0 < ε›) finally show "¦c - c¦ < ε" by this qed then show "∃k::nat. ∀n≥k. ¦c - c¦ < ε" by (rule exI) qed qed (* 2ª demostración *) lemma "limite (λ n. c) c" proof (unfold limite_def) show "∀ε>0. ∃k::nat. ∀n≥k. ¦c - c¦ < ε" proof (intro allI impI) fix ε :: real assume "0 < ε" have "∀n≥0::nat. ¦c - c¦ < ε" by (simp add: ‹0 < ε›) then show "∃k::nat. ∀n≥k. ¦c - c¦ < ε" by (rule exI) qed qed (* 3ª demostración *) lemma "limite (λ n. c) c" unfolding limite_def by simp (* 4ª demostración *) lemma "limite (λ n. c) c" by (simp add: limite_def) end |
2. Unicidad del límite de las sucesiones convergentes
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
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def limite : (ℕ → ℝ) → ℝ → Prop := fun u c ↦ ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - c| < ε |
Demostrar con Lean4 que cada sucesión tiene como máximo un límite.
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Real.Basic variable {u : ℕ → ℝ} variable {a b : ℝ} def limite : (ℕ → ℝ) → ℝ → Prop := fun u c ↦ ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - c| < ε example (ha : limite u a) (hb : limite u b) : a = b := by sorry |
2.1. Demostración en lenguaje natural
Tenemos que demostrar que si \(u\) es una sucesión y \(a\) y \(b\) son límites de \(u\), entonces \(a = b\). Para ello, basta demostrar que \(a ≤ b\) y \(b ≤ a\).
Demostraremos que \(b ≤ a\) por reducción al absurdo. Supongamos que \(b ≰ a\). Sea \(ε = b – a\). Entonces, ε/2 > 0 y, puesto que \(a\) es un límite de \(u\), existe un \(A ∈ ℕ\) tal que
\[ (∀n ∈ ℕ)\left[n ≥ A → |u(n) – a| < \frac{ε}{2}\right] \tag{1} \]
y, puesto que \(b\) también es un límite de \(u\), existe un \(B ∈ ℕ\) tal que
\[ (∀n ∈ ℕ)\left[n ≥ B → |u(n) – b| < \frac{ε}{2}\right] \tag{2} \]
Sea \(N = máx(A, B)\). Entonces, \(N ≥ A\) y \(N ≥ B\) y, por (2) y (3), se tiene
\begin{align}
|u(N) – a| &< \frac{ε}{2} \tag{3} \\
|u(N) – b| &< \frac{ε}{2} \tag{4}
\end{align}
Para obtener una contradicción basta probar que \(ε < ε\). Su prueba es
\begin{align}
ε &= b – a \\
&= |b – a| \\
&= |(b – a) + (u(N) – u(N))| \\
&= |(u(N) – a) + (b – u(N))| \\
&≤ |u(N) – a| + |b – u(N)| \\
&= |u(N) – a| + |u(N) – b| \\
&< \frac{ε}{2} + \frac{ε}{2} && \text{[por (3) y (4)]} \\
&= ε
\end{align}
La demostración de \(a ≤ b\) es análoga a la anterior.
2.2. Demostraciones con Lean4
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import Mathlib.Data.Real.Basic variable {u : ℕ → ℝ} variable {a b : ℝ} def limite : (ℕ → ℝ) → ℝ → Prop := fun u c ↦ ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - c| < ε -- 1ª demostración del lema auxiliar -- ================================= example (ha : limite u a) (hb : limite u b) : b ≤ a := by by_contra h -- h : ¬b ≤ a -- ⊢ False let ε := b - a have hε : ε > 0 := sub_pos.mpr (not_le.mp h) have hε2 : ε / 2 > 0 := half_pos hε cases' ha (ε/2) hε2 with A hA -- A : ℕ -- hA : ∀ (n : ℕ), n ≥ A → |u n - a| < ε / 2 cases' hb (ε/2) hε2 with B hB -- B : ℕ -- hB : ∀ (n : ℕ), n ≥ B → |u n - b| < ε / 2 let N := max A B have hAN : A ≤ N := le_max_left A B have hBN : B ≤ N := le_max_right A B specialize hA N hAN -- hA : |u N - a| < ε / 2 specialize hB N hBN -- hB : |u N - b| < ε / 2 have h2 : ε < ε := by calc ε = b - a := rfl _ = |b - a| := (abs_of_pos hε).symm _ = |(b - a) + 0| := by {congr ; exact (add_zero (b - a)).symm} _ = |(b - a) + (u N - u N)| := by {congr ; exact (sub_self (u N)).symm} _ = |(u N - a) + (b - u N)| := congrArg (fun x => |x|) (by ring) _ ≤ |u N - a| + |b - u N| := abs_add (u N - a) (b - u N) _ = |u N - a| + |u N - b| := congrArg (|u N - a| + .) (abs_sub_comm b (u N)) _ < ε / 2 + ε / 2 := add_lt_add hA hB _ = ε := add_halves ε have h3 : ¬(ε < ε) := lt_irrefl ε show False exact h3 h2 -- 2ª demostración del lema auxiliar -- ================================= lemma aux (ha : limite u a) (hb : limite u b) : b ≤ a := by by_contra h -- h : ¬b ≤ a -- ⊢ False let ε := b - a cases' ha (ε/2) (by linarith) with A hA -- A : ℕ -- hA : ∀ (n : ℕ), n ≥ A → |u n - a| < ε / 2 cases' hb (ε/2) (by linarith) with B hB -- B : ℕ -- hB : ∀ (n : ℕ), n ≥ B → |u n - b| < ε / 2 let N := max A B have hAN : A ≤ N := le_max_left A B have hBN : B ≤ N := le_max_right A B specialize hA N hAN -- hA : |u N - a| < ε / 2 specialize hB N hBN -- hB : |u N - b| < ε / 2 rw [abs_lt] at hA hB -- hA : -(ε / 2) < u N - a ∧ u N - a < ε / 2 -- hB : -(ε / 2) < u N - b ∧ u N - b < ε / 2 linarith -- 1ª demostración -- =============== example (ha : limite u a) (hb : limite u b) : a = b := le_antisymm (aux hb ha) (aux ha hb) -- Lemas usados -- ============ -- variable (c d : ℝ) -- #check (not_le : ¬a ≤ b ↔ b < a) -- #check (sub_pos : 0 < a - b ↔ b < a) -- #check (half_pos : a > 0 → a / 2 > 0) -- #check (le_max_left a b : a ≤ max a b) -- #check (le_max_right a b : b ≤ max a b) -- #check (abs_lt : |a| < b ↔ -b < a ∧ a < b) -- #check (abs_of_pos : 0 < a → |a| = a) -- #check (add_zero a : a + 0 = a) -- #check (sub_self a : a - a = 0) -- #check (abs_add a b : |a + b| ≤ |a| + |b|) -- #check (abs_sub_comm a b : |a - b| = |b - a|) -- #check (add_lt_add : a < b → c < d → a + c < b + d) -- #check (add_halves a : a / 2 + a / 2 = a) -- #check (lt_irrefl a : ¬a < a) -- #check (le_antisymm : a ≤ b → b ≤ a → a = b) |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
2.3. Demostraciones con Isabelle/HOL
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theory Unicidad_del_limite_de_las_sucesiones_convergentes 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)" lemma aux : assumes "limite u a" "limite u b" shows "b ≤ a" proof (rule ccontr) assume "¬ b ≤ a" let ?ε = "b - a" have "0 < ?ε/2" using ‹¬ b ≤ a› by auto obtain A where hA : "∀n≥A. ¦u n - a¦ < ?ε/2" using assms(1) limite_def ‹0 < ?ε/2› by blast obtain B where hB : "∀n≥B. ¦u n - b¦ < ?ε/2" using assms(2) limite_def ‹0 < ?ε/2› by blast let ?C = "max A B" have hCa : "∀n≥?C. ¦u n - a¦ < ?ε/2" using hA by simp have hCb : "∀n≥?C. ¦u n - b¦ < ?ε/2" using hB by simp have "∀n≥?C. ¦a - b¦ < ?ε" proof (intro allI impI) fix n assume "n ≥ ?C" have "¦a - b¦ = ¦(a - u n) + (u n - b)¦" by simp also have "… ≤ ¦u n - a¦ + ¦u n - b¦" by simp finally show "¦a - b¦ < b - a" using hCa hCb ‹n ≥ ?C› by fastforce qed then show False by fastforce qed theorem assumes "limite u a" "limite u b" shows "a = b" proof (rule antisym) show "a ≤ b" using assms(2) assms(1) by (rule aux) next show "b ≤ a" using assms(1) assms(2) by (rule aux) qed end |
3. Si el límite de la sucesión uₙ es a y c ∈ ℝ, entonces el límite de uₙ+c es a+c
Demostrar con Lean4 que si el límite de la sucesión \(uₙ\) es \(a\) y \(c ∈ ℝ\), entonces el límite de \(uₙ+c\) es \(a+c\).
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Real.Basic import Mathlib.Tactic variable {u : ℕ → ℝ} variable {a c : ℝ} def limite : (ℕ → ℝ) → ℝ → Prop := fun u c ↦ ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - c| < ε example (h : limite u a) : limite (fun i ↦ u i + c) (a + c) := by sorry |
3.1. Demostración en lenguaje natural
Sea \(ε ∈ ℝ\) tal que \(ε > 0\). Tenemos que demostrar que
\[ (∃ N)(∀ n ≥ N)[|(u(n) + c) – (a + c)| < ε] \tag{1} \]
Puesto que el límite de la sucesión \(u\) es \(a\), existe un \(k\) tal que
\[ (∀ n ≥ k)[|u(n) – a| < ε] \tag{2} \]
Veamos que con k se verifica (1); es decir, que
\[ (∀ n ≥ k)[|(u(n) + c) – (a + c)| < ε] \]
Sea \(n ≥ k\). Entonces, por (2),
\[ |u(n) – a| < ε \tag{3} \]
y, por consiguiente,
\begin{align}
|(u(n) + c) – (a + c)| &= |u(n) – a| \\
&< ε &&\text{[por (3)]}
\end{align}
3.2. Demostraciones con Lean4
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import Mathlib.Data.Real.Basic import Mathlib.Tactic variable {u : ℕ → ℝ} variable {a c : ℝ} def limite : (ℕ → ℝ) → ℝ → Prop := fun u c ↦ ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - c| < ε -- 1ª demostración -- =============== example (h : limite u a) : limite (fun i ↦ u i + c) (a + c) := by intros ε hε -- ε : ℝ -- hε : ε > 0 -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |(fun i => u i + c) n - (a + c)| < ε dsimp -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |u n + c - (a + c)| < ε cases' h ε hε with k hk -- k : ℕ -- hk : ∀ (n : ℕ), n ≥ k → |u n - a| < ε use k -- ⊢ ∀ (n : ℕ), n ≥ k → |u n + c - (a + c)| < ε intros n hn -- n : ℕ -- hn : n ≥ k calc |u n + c - (a + c)| = |u n - a| := by norm_num _ < ε := hk n hn -- 2ª demostración -- =============== example (h : limite u a) : limite (fun i ↦ u i + c) (a + c) := by intros ε hε -- ε : ℝ -- hε : ε > 0 -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |(fun i => u i + c) n - (a + c)| < ε dsimp -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |u n + c - (a + c)| < ε cases' h ε hε with k hk -- k : ℕ -- hk : ∀ (n : ℕ), n ≥ k → |u n - a| < ε use k -- ⊢ ∀ (n : ℕ), n ≥ k → |u n + c - (a + c)| < ε intros n hn -- n : ℕ -- hn : n ≥ k -- ⊢ |u n + c - (a + c)| < ε convert hk n hn using 2 -- ⊢ u n + c - (a + c) = u n - a ring -- 3ª demostración -- =============== example (h : limite u a) : limite (fun i ↦ u i + c) (a + c) := by intros ε hε dsimp convert h ε hε using 6 ring -- 4ª demostración -- =============== example (h : limite u a) : limite (fun i ↦ u i + c) (a + c) := fun ε hε ↦ (by convert h ε hε using 6; ring) |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
3.3. Demostraciones con Isabelle/HOL
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theory Limite_cuando_se_suma_una_constante 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)" (* 1ª demostración *) lemma assumes "limite u a" shows "limite (λ i. u i + c) (a + c)" proof (unfold limite_def) show "∀ε>0. ∃k. ∀n≥k. ¦(u n + c) - (a + c)¦ < ε" proof (intro allI impI) fix ε :: real assume "0 < ε" then have "∃k. ∀n≥k. ¦u n - a¦ < ε" using assms limite_def by simp then obtain k where "∀n≥k. ¦u n - a¦ < ε" by (rule exE) then have "∀n≥k. ¦(u n + c) - (a + c)¦ < ε" by simp then show "∃k. ∀n≥k. ¦(u n + c) - (a + c)¦ < ε" by (rule exI) qed qed (* 2ª demostración *) lemma assumes "limite u a" shows "limite (λ i. u i + c) (a + c)" using assms limite_def by simp end |
4. Si el límite de la sucesión uₙ es a y c ∈ ℝ, entonces el límite de cuₙ es ca
Demostrar con Lean4 que si el límite de la sucesión \(uₙ\) es \(a\) y \(c ∈ ℝ\), entonces el límite de \(cuₙ\) es \(ca\).
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Real.Basic import Mathlib.Tactic variable (u v : ℕ → ℝ) variable (a c : ℝ) def limite : (ℕ → ℝ) → ℝ → Prop := fun u c ↦ ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - c| < ε example (h : limite u a) : limite (fun n ↦ c * (u n)) (c * a) := by |
4.1. Demostración en lenguaje natural
Sea \(ε ∈ ℝ\) tal que \(ε > 0\). Tenemos que demostrar que
\[ (∃ N ∈ ℕ)(∀ n ≥ N)[|cuₙ – ca| < ε] \tag{1}\]
Distinguiremos dos casos según sea \(c = 0\) o no.
Primer caso: Supongamos que \(c = 0\). Entonces, (1) se reduce a
\[ (∃ N ∈ ℕ)(∀ n ≥ N)[|0·uₙ – 0·a| < ε] \]
es decir,
\[ (∃ N ∈ ℕ)(∀ n ≥ N)[0 < ε] \]
que se verifica para cualquier número \(N\), ya que \(ε > 0\).
Segundo caso: Supongamos que \(c ≠ 0\). Entonces, \(\dfrac{ε}{|c|}\) > 0 y, puesto que el límite de \(uₙ\) es \(a\), existe un \(k ∈ ℕ\) tal que
\[ (∀ n ≥ k)[|uₙ – a| < \frac{ε}{|c|}] \tag{2} \]
Veamos que con \(k\) se cumple (1). En efecto, sea \(n ≥ k\). Entonces,
\begin{align}
|cuₙ – ca| &= |c(uₙ – a)| \\
&= |c||uₙ – a| \\
&< |c|\frac{ε}{|c|} &&\text{[por (2)]} \\
&= ε
\end{align}
4.2. Demostraciones con Lean4
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import Mathlib.Data.Real.Basic import Mathlib.Tactic variable (u v : ℕ → ℝ) variable (a c : ℝ) def limite : (ℕ → ℝ) → ℝ → Prop := fun u c ↦ ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - c| < ε -- 1ª demostración -- =============== example (h : limite u a) : limite (fun n ↦ c * (u n)) (c * a) := by by_cases hc : c = 0 . -- hc : c = 0 subst hc -- ⊢ limite (fun n => 0 * u n) (0 * a) intros ε hε -- ε : ℝ -- hε : ε > 0 -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |(fun n => 0 * u n) n - 0 * a| < ε aesop . -- hc : ¬c = 0 intros ε hε -- ε : ℝ -- hε : ε > 0 -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |(fun n => c * u n) n - c * a| < ε have hc' : 0 < |c| := abs_pos.mpr hc have hεc : 0 < ε / |c| := div_pos hε hc' specialize h (ε/|c|) hεc -- h : ∃ N, ∀ (n : ℕ), n ≥ N → |u n - a| < ε / |c| cases' h with N hN -- N : ℕ -- hN : ∀ (n : ℕ), n ≥ N → |u n - a| < ε / |c| use N -- ⊢ ∀ (n : ℕ), n ≥ N → |(fun n => c * u n) n - c * a| < ε intros n hn -- n : ℕ -- hn : n ≥ N -- ⊢ |(fun n => c * u n) n - c * a| < ε specialize hN n hn -- hN : |u n - a| < ε / |c| dsimp only calc |c * u n - c * a| = |c * (u n - a)| := congr_arg abs (mul_sub c (u n) a).symm _ = |c| * |u n - a| := abs_mul c (u n - a) _ < |c| * (ε / |c|) := (mul_lt_mul_left hc').mpr hN _ = ε := mul_div_cancel' ε (ne_of_gt hc') -- 2ª demostración -- =============== example (h : limite u a) : limite (fun n ↦ c * (u n)) (c * a) := by by_cases hc : c = 0 . -- hc : c = 0 subst hc -- ⊢ limite (fun n => 0 * u n) (0 * a) intros ε hε -- ε : ℝ -- hε : ε > 0 -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |(fun n => 0 * u n) n - 0 * a| < ε aesop . -- hc : ¬c = 0 intros ε hε -- ε : ℝ -- hε : ε > 0 -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |(fun n => c * u n) n - c * a| < ε have hc' : 0 < |c| := abs_pos.mpr hc have hεc : 0 < ε / |c| := div_pos hε hc' specialize h (ε/|c|) hεc -- h : ∃ N, ∀ (n : ℕ), n ≥ N → |u n - a| < ε / |c| cases' h with N hN -- N : ℕ -- hN : ∀ (n : ℕ), n ≥ N → |u n - a| < ε / |c| use N -- ⊢ ∀ (n : ℕ), n ≥ N → |(fun n => c * u n) n - c * a| < ε intros n hn -- n : ℕ -- hn : n ≥ N -- ⊢ |(fun n => c * u n) n - c * a| < ε specialize hN n hn -- hN : |u n - a| < ε / |c| dsimp only -- ⊢ |c * u n - c * a| < ε rw [← mul_sub] -- ⊢ |c * (u n - a)| < ε rw [abs_mul] -- ⊢ |c| * |u n - a| < ε rw [← lt_div_iff' hc'] -- ⊢ |u n - a| < ε / |c| exact hN -- 3ª demostración -- =============== example (h : limite u a) : limite (fun n ↦ c * (u n)) (c * a) := by by_cases hc : c = 0 . subst hc intros ε hε aesop . intros ε hε have hc' : 0 < |c| := by aesop have hεc : 0 < ε / |c| := div_pos hε hc' cases' h (ε/|c|) hεc with N hN use N intros n hn specialize hN n hn dsimp only rw [← mul_sub, abs_mul, ← lt_div_iff' hc'] exact hN -- Lemas usados -- ============ -- variable (b c : ℝ) -- #check (abs_mul a b : |a * b| = |a| * |b|) -- #check (abs_pos.mpr : a ≠ 0 → 0 < |a|) -- #check (div_pos : 0 < a → 0 < b → 0 < a / b) -- #check (lt_div_iff' : 0 < c → (a < b / c ↔ c * a < b)) -- #check (mul_div_cancel' a : b ≠ 0 → b * (a / b) = a) -- #check (mul_lt_mul_left : 0 < a → (a * b < a * c ↔ b < c)) -- #check (mul_sub a b c : a * (b - c) = a * b - a * c) |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
4.3. Demostraciones con Isabelle/HOL
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theory Limite_multiplicado_por_una_constante 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)" lemma assumes "limite u a" shows "limite (λ n. c * u n) (c * a)" proof (unfold limite_def) show "∀ε>0. ∃k. ∀n≥k. ¦c * u n - c * a¦ < ε" proof (intro allI impI) fix ε :: real assume "0 < ε" show "∃k. ∀n≥k. ¦c * u n - c * a¦ < ε" proof (cases "c = 0") assume "c = 0" then show "∃k. ∀n≥k. ¦c * u n - c * a¦ < ε" by (simp add: ‹0 < ε›) next assume "c ≠ 0" then have "0 < ¦c¦" by simp then have "0 < ε/¦c¦" by (simp add: ‹0 < ε›) then obtain N where hN : "∀n≥N. ¦u n - a¦ < ε/¦c¦" using assms limite_def by auto have "∀n≥N. ¦c * u n - c * a¦ < ε" proof (intro allI impI) fix n assume "n ≥ N" have "¦c * u n - c * a¦ = ¦c * (u n - a)¦" by argo also have "… = ¦c¦ * ¦u n - a¦" by (simp only: abs_mult) also have "… < ¦c¦ * (ε/¦c¦)" using hN ‹n ≥ N› ‹0 < ¦c¦› by (simp only: mult_strict_left_mono) finally show "¦c * u n - c * a¦ < ε" using ‹0 < ¦c¦› by auto qed then show "∃k. ∀n≥k. ¦c * u n - c * a¦ < ε" by (rule exI) qed qed qed end |
5. El límite de uₙ es a syss el de uₙ-a es 0
Demostrar con Lean4 que el límite de \(uₙ\) es \(a\) si, y sólo si, el de \(uₙ-a\) es \(0\).
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Real.Basic import Mathlib.Tactic variable {u : ℕ → ℝ} variable {a c x : ℝ} def limite : (ℕ → ℝ) → ℝ → Prop := fun u c ↦ ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - c| < ε example : limite u a ↔ limite (fun i ↦ u i - a) 0 := by sorry |
5.1. Demostración en lenguaje natural
Se prueba por la siguiente cadena de equivalencias
\begin{align}
&\text{el límite de \(uₙ\) es \(a\)} \\
&↔ (∀ε>0)(∃N)(∀n≥N)[|u(n) – a| < ε] \\
&↔ (∀ε>0)(∃N)(∀n≥N)[|(u(n) – a) – 0| < ε] \\
&↔ \text{el límite de \(uₙ-a\) es \(0\)}
\end{align}
5.2. Demostraciones con Lean4
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import Mathlib.Data.Real.Basic import Mathlib.Tactic variable {u : ℕ → ℝ} variable {a c x : ℝ} def limite : (ℕ → ℝ) → ℝ → Prop := fun u c ↦ ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - c| < ε -- 1ª demostración -- =============== example : limite u a ↔ limite (fun i ↦ u i - a) 0 := by rw [iff_eq_eq] calc limite u a = ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - a| < ε := rfl _ = ∀ ε > 0, ∃ N, ∀ n ≥ N, |(u n - a) - 0| < ε := by simp _ = limite (fun i ↦ u i - a) 0 := rfl -- 2ª demostración -- =============== example : limite u a ↔ limite (fun i ↦ u i - a) 0 := by constructor . -- ⊢ limite u a → limite (fun i => u i - a) 0 intros h ε hε -- h : limite u a -- ε : ℝ -- hε : ε > 0 -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |(fun i => u i - a) n - 0| < ε convert h ε hε using 2 -- x : ℕ -- ⊢ (∀ (n : ℕ), n ≥ x → |(fun i => u i - a) n - 0| < ε) ↔ ∀ (n : ℕ), n ≥ x → |u n - a| < ε norm_num . -- ⊢ limite (fun i => u i - a) 0 → limite u a intros h ε hε -- h : limite (fun i => u i - a) 0 -- ε : ℝ -- hε : ε > 0 -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |u n - a| < ε convert h ε hε using 2 -- x : ℕ -- ⊢ (∀ (n : ℕ), n ≥ x → |u n - a| < ε) ↔ ∀ (n : ℕ), n ≥ x → |(fun i => u i - a) n - 0| < ε norm_num -- 3ª demostración -- =============== example : limite u a ↔ limite (fun i ↦ u i - a) 0 := by constructor <;> { intros h ε hε convert h ε hε using 2 norm_num } -- 4ª demostración -- =============== lemma limite_con_suma (c : ℝ) (h : limite u a) : limite (fun i ↦ u i + c) (a + c) := fun ε hε ↦ (by convert h ε hε using 2; norm_num) lemma CNS_limite_con_suma (c : ℝ) : limite u a ↔ limite (fun i ↦ u i + c) (a + c) := by constructor . -- ⊢ limite u a → limite (fun i => u i + c) (a + c) apply limite_con_suma . -- ⊢ limite (fun i => u i + c) (a + c) → limite u a intro h -- h : limite (fun i => u i + c) (a + c) -- ⊢ limite u a convert limite_con_suma (-c) h using 2 . -- ⊢ u x = u x + c + -c simp . -- ⊢ a = a + c + -c simp example (u : ℕ → ℝ) (a : ℝ) : limite u a ↔ limite (fun i ↦ u i - a) 0 := by convert CNS_limite_con_suma (-a) using 2 -- ⊢ 0 = a + -a simp -- Lemas usados -- ============ -- variable (p q : Prop) -- #check (iff_eq_eq : (p ↔ q) = (p = q)) |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
5.3. Demostraciones con Isabelle/HOL
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theory "El_limite_de_u_es_a_syss_el_de_u-a_es_0" 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)" (* 1ª demostración *) lemma "limite u a ⟷ limite (λ i. u i - a) 0" proof - have "limite u a ⟷ (∀ε>0. ∃k::nat. ∀n≥k. ¦u n - a¦ < ε)" by (rule limite_def) also have "… ⟷ (∀ε>0. ∃k::nat. ∀n≥k. ¦(u n - a) - 0¦ < ε)" by simp also have "… ⟷ limite (λ i. u i - a) 0" by (rule limite_def[symmetric]) finally show "limite u a ⟷ limite (λ i. u i - a) 0" by this qed (* 2ª demostración *) lemma "limite u a ⟷ limite (λ i. u i - a) 0" proof - have "limite u a ⟷ (∀ε>0. ∃k::nat. ∀n≥k. ¦u n - a¦ < ε)" by (simp only: limite_def) also have "… ⟷ (∀ε>0. ∃k::nat. ∀n≥k. ¦(u n - a) - 0¦ < ε)" by simp also have "… ⟷ limite (λ i. u i - a) 0" by (simp only: limite_def) finally show "limite u a ⟷ limite (λ i. u i - a) 0" by this qed (* 3ª demostración *) lemma "limite u a ⟷ limite (λ i. u i - a) 0" using limite_def by simp end |
6. Si uₙ y vₙ convergen a 0, entonces uₙvₙ converge a 0
Demostrar con Lean4 que si \(uₙ\) y \(vₙ\) convergen a \(0\), entonces \(uₙvₙ\) converge a \(0).
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Real.Basic import Mathlib.Tactic variable {u v : ℕ → ℝ} def limite : (ℕ → ℝ) → ℝ → Prop := fun u c ↦ ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - c| < ε example (hu : limite u 0) (hv : limite v 0) : limite (u * v) 0 := by sorry |
6.1. Demostración en lenguaje natural
Sea \(ε ∈ ℝ\) tal que \(ε > 0\). Tenemos que demostrar que
\[ (∃N ∈ ℕ)(∀n ≥ N)[|(u·v)(n) – 0| < ε] \tag{1} \]
Puesto que el límite de \(uₙ\) es \(0\), existe un \(U ∈ ℕ\) tal que
\[ (∀n ≥ U)[|u(n) – 0| < ε] \tag{2} \]
y, puesto que el límite de \(vₙ\) es \(0\), existe un \(V ∈ ℕ\) tal que
\[ (∀n ≥ V)[|v(n) – 0| < 1] \tag{3} \]
Entonces, \(N = \text{máx}(U, V)\) cumple (1). En efecto, sea \(n ≥ N\). Entonces,
\(n ≥ U\) y \(n ≥ V\) y, aplicando (2) y (3), se tiene
\begin{align}
&|u(n) – 0| < ε \tag{4} \\
&|v(n) – 0| < 1 \tag{5}
\end{align}
Por tanto,
\begin{align}
|(u·v)(n) – 0| &= |u(n)·v(n)| \\
&= |u(n)|·|v n| \\
&< ε·1 &&\text{[por (4) y (5)]} \\
&= ε
\end{align}
6.2. Demostraciones con Lean4
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import Mathlib.Data.Real.Basic import Mathlib.Tactic variable {u v : ℕ → ℝ} def limite : (ℕ → ℝ) → ℝ → Prop := fun u c ↦ ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - c| < ε -- 1ª demostración -- =============== example (hu : limite u 0) (hv : limite v 0) : limite (u * v) 0 := by intros ε hε -- ε : ℝ -- hε : ε > 0 -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |(u * v) n - 0| < ε cases' hu ε hε with U hU -- U : ℕ -- hU : ∀ (n : ℕ), n ≥ U → |u n - 0| < ε cases' hv 1 zero_lt_one with V hV -- V : ℕ -- hV : ∀ (n : ℕ), n ≥ V → |v n - 0| < 1 let N := max U V use N -- ⊢ ∀ (n : ℕ), n ≥ N → |(u * v) n - 0| < ε intros n hn -- n : ℕ -- hn : n ≥ N -- ⊢ |(u * v) n - 0| < ε specialize hU n (le_of_max_le_left hn) -- hU : |u n - 0| < ε specialize hV n (le_of_max_le_right hn) -- hV : |v n - 0| < 1 rw [sub_zero] at * -- hU : |u n - 0| < ε -- hV : |v n - 0| < 1 -- ⊢ |(u * v) n - 0| < ε calc |(u * v) n| = |u n * v n| := rfl _ = |u n| * |v n| := abs_mul (u n) (v n) _ < ε * 1 := mul_lt_mul'' hU hV (abs_nonneg (u n)) (abs_nonneg (v n)) _ = ε := mul_one ε -- 2ª demostración -- =============== example (hu : limite u 0) (hv : limite v 0) : limite (u * v) 0 := by intros ε hε -- ε : ℝ -- hε : ε > 0 -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |(u * v) n - 0| < ε cases' hu ε hε with U hU -- U : ℕ -- hU : ∀ (n : ℕ), n ≥ U → |u n - 0| < ε cases' hv 1 (by linarith) with V hV -- V : ℕ -- hV : ∀ (n : ℕ), n ≥ V → |v n - 0| < 1 let N := max U V use N -- ⊢ ∀ (n : ℕ), n ≥ N → |(u * v) n - 0| < ε intros n hn -- n : ℕ -- hn : n ≥ N -- ⊢ |(u * v) n - 0| < ε specialize hU n (le_of_max_le_left hn) -- hU : |u n - 0| < ε specialize hV n (le_of_max_le_right hn) -- hV : |v n - 0| < 1 rw [sub_zero] at * -- hU : |u n| < ε -- hV : |v n| < 1 -- ⊢ |(u * v) n| < ε calc |(u * v) n| = |u n * v n| := rfl _ = |u n| * |v n| := abs_mul (u n) (v n) _ < ε * 1 := by { apply mul_lt_mul'' hU hV <;> simp [abs_nonneg] } _ = ε := mul_one ε -- 3ª demostración -- =============== example (hu : limite u 0) (hv : limite v 0) : limite (u * v) 0 := by intros ε hε -- ε : ℝ -- hε : ε > 0 -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |(u * v) n - 0| < ε cases' hu ε hε with U hU -- U : ℕ -- hU : ∀ (n : ℕ), n ≥ U → |u n - 0| < ε cases' hv 1 (by linarith) with V hV -- V : ℕ -- hV : ∀ (n : ℕ), n ≥ V → |v n - 0| < 1 let N := max U V use N -- ⊢ ∀ (n : ℕ), n ≥ N → |(u * v) n - 0| < ε intros n hn -- n : ℕ -- hn : n ≥ N -- ⊢ |(u * v) n - 0| < ε have hUN : U ≤ N := le_max_left U V have hVN : V ≤ N := le_max_right U V specialize hU n (by linarith) -- hU : |u n - 0| < ε specialize hV n (by linarith) -- hV : |v n - 0| < 1 rw [sub_zero] at * -- hU : |u n| < ε -- hV : |v n| < 1 -- ⊢ |(u * v) n| < ε rw [Pi.mul_apply] -- ⊢ |u n * v n| < ε rw [abs_mul] -- ⊢ |u n| * |v n| < ε convert mul_lt_mul'' hU hV _ _ using 2 <;> simp -- Lemas usados -- ============ -- variable (a b c d : ℝ) -- variable (I : Type _) -- variable (f : I → Type _) -- #check (zero_lt_one : 0 < 1) -- #check (le_of_max_le_left : max a b ≤ c → a ≤ c) -- #check (le_of_max_le_right : max a b ≤ c → b ≤ c) -- #check (sub_zero a : a - 0 = a) -- #check (abs_mul a b : |a * b| = |a| * |b|) -- #check (mul_lt_mul'' : a < c → b < d → 0 ≤ a → 0 ≤ b → a * b < c * d) -- #check (abs_nonneg a : 0 ≤ |a|) -- #check (mul_one a : a * 1 = a) |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
6.3. Demostraciones con Isabelle/HOL
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theory Producto_de_sucesiones_convergentes_a_cero 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)" lemma assumes "limite u 0" "limite v 0" shows "limite (λ n. u n * v n) 0" proof (unfold limite_def; intro allI impI) fix ε :: real assume hε : "0 < ε" then obtain U where hU : "∀n≥U. ¦u n - 0¦ < ε" using assms(1) limite_def by auto obtain V where hV : "∀n≥V. ¦v n - 0¦ < 1" using hε assms(2) limite_def by fastforce have "∀n≥max U V. ¦u n * v n - 0¦ < ε" proof (intro allI impI) fix n assume hn : "max U V ≤ n" then have "U ≤ n" by simp then have "¦u n - 0¦ < ε" using hU by blast have hnV : "V ≤ n" using hn by simp then have "¦v n - 0¦ < 1" using hV by blast have "¦u n * v n - 0¦ = ¦(u n - 0) * (v n - 0)¦" by simp also have "… = ¦u n - 0¦ * ¦v n - 0¦" by (simp add: abs_mult) also have "… < ε * 1" using ‹¦u n - 0¦ < ε› ‹¦v n - 0¦ < 1› by (rule abs_mult_less) also have "… = ε" by simp finally show "¦u n * v n - 0¦ < ε" by this qed then show "∃k. ∀n≥k. ¦u n * v n - 0¦ < ε" by (rule exI) qed end |
7. Teorema del emparedado
Demostrar con Lean4 el teorema del emparedado.
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Real.Basic variable (u v w : ℕ → ℝ) variable (a : ℝ) def limite : (ℕ → ℝ) → ℝ → Prop := fun u c ↦ ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - c| ≤ ε example (hu : limite u a) (hw : limite w a) (h1 : ∀ n, u n ≤ v n) (h2 : ∀ n, v n ≤ w n) : limite v a := by sorry |
7.1. Demostración en lenguaje natural
Tenemos que demostrar que para cada \(ε > 0\), existe un \(N ∈ ℕ\) tal que
\[ (∀ n ≥ N)[|v(n) – a| ≤ ε] \tag{1} \]
Puesto que el límite de \(u\) es \(a\), existe un \(U ∈ ℕ\) tal que
\[ (∀ n ≥ U)[|u(n) – a| ≤ ε] \tag{2} \]
y, puesto que el límite de \(w\) es \(a\), existe un \(W ∈ ℕ\) tal que
\[ (∀ n ≥ W)[|w(n) – a| ≤ ε] \tag{3} \]
Sea \(N = \text{máx}(U, W)\). Veamos que se verifica (1). Para ello, sea \(n ≥ N\). Entonces, \(n ≥ U\), \(n ≥ W\) y, por (2) y (3), se tiene que
\begin{align}
|u(n) – a| &≤ ε \tag{4} \\
|w(n) – a| &≤ ε \tag{5}
\end{align}
Para demostrar que
\[ |v(n) – a| ≤ ε \]
basta demostrar las siguientes desigualdades
\begin{align}
&-ε ≤ v(n) – a \tag{6} \\
&v(n) – a ≤ ε \tag{7}
\end{align}
La demostración de (6) es
\begin{align}
-ε &≤ u(n) – a &&\text{[por (4)]} \\
&≤ v(n) – a &&\text{[por hipótesis]}
\end{align}
La demostración de (7) es
\begin{align}
v(n) – a &≤ w(n) – a &&\text{[por hipótesis]} \\
&≤ ε &&\text{[por (5)]}
\end{align}
7.2. Demostraciones con Lean4
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import Mathlib.Data.Real.Basic variable (u v w : ℕ → ℝ) variable (a : ℝ) def limite : (ℕ → ℝ) → ℝ → Prop := fun u c ↦ ∀ ε > 0, ∃ N, ∀ n ≥ N, |u n - c| ≤ ε -- Nota. En la demostración se usará el siguiente lema: lemma max_ge_iff {p q r : ℕ} : r ≥ max p q ↔ r ≥ p ∧ r ≥ q := max_le_iff -- 1ª demostración -- =============== example (hu : limite u a) (hw : limite w a) (h1 : ∀ n, u n ≤ v n) (h2 : ∀ n, v n ≤ w n) : limite v a := by intros ε hε -- ε : ℝ -- hε : ε > 0 -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |v n - a| ≤ ε rcases hu ε hε with ⟨U, hU⟩ -- U : ℕ -- hU : ∀ (n : ℕ), n ≥ U → |u n - a| ≤ ε clear hu rcases hw ε hε with ⟨W, hW⟩ -- W : ℕ -- hW : ∀ (n : ℕ), n ≥ W → |w n - a| ≤ ε clear hw hε use max U W intros n hn -- n : ℕ -- hn : n ≥ max U W -- ⊢ |v n - a| ≤ ε rw [max_ge_iff] at hn -- hn : n ≥ U ∧ n ≥ W specialize hU n hn.1 -- hU : |u n - a| ≤ ε specialize hW n hn.2 -- hW : |w n - a| ≤ ε specialize h1 n -- h1 : u n ≤ v n specialize h2 n -- h2 : v n ≤ w n clear hn rw [abs_le] at * -- ⊢ -ε ≤ v n - a ∧ v n - a ≤ ε constructor . -- ⊢ -ε ≤ v n - a calc -ε ≤ u n - a := hU.1 _ ≤ v n - a := by linarith . -- ⊢ v n - a ≤ ε calc v n - a ≤ w n - a := by linarith _ ≤ ε := hW.2 -- 2ª demostración example (hu : limite u a) (hw : limite w a) (h1 : ∀ n, u n ≤ v n) (h2 : ∀ n, v n ≤ w n) : limite v a := by intros ε hε -- ε : ℝ -- hε : ε > 0 -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |v n - a| ≤ ε rcases hu ε hε with ⟨U, hU⟩ -- U : ℕ -- hU : ∀ (n : ℕ), n ≥ U → |u n - a| ≤ ε clear hu rcases hw ε hε with ⟨W, hW⟩ -- W : ℕ -- hW : ∀ (n : ℕ), n ≥ W → |w n - a| ≤ ε clear hw hε use max U W intros n hn -- n : ℕ -- hn : n ≥ max U W rw [max_ge_iff] at hn -- hn : n ≥ U ∧ n ≥ W specialize hU n (by linarith) -- hU : |u n - a| ≤ ε specialize hW n (by linarith) -- hW : |w n - a| ≤ ε specialize h1 n -- h1 : u n ≤ v n specialize h2 n -- h2 : v n ≤ w n rw [abs_le] at * -- ⊢ -ε ≤ v n - a ∧ v n - a ≤ ε constructor . -- ⊢ -ε ≤ v n - a linarith . -- ⊢ v n - a ≤ ε linarith -- 3ª demostración example (hu : limite u a) (hw : limite w a) (h1 : ∀ n, u n ≤ v n) (h2 : ∀ n, v n ≤ w n) : limite v a := by intros ε hε -- ε : ℝ -- hε : ε > 0 -- ⊢ ∃ N, ∀ (n : ℕ), n ≥ N → |v n - a| ≤ ε rcases hu ε hε with ⟨U, hU⟩ -- U : ℕ -- hU : ∀ (n : ℕ), n ≥ U → |u n - a| ≤ ε clear hu rcases hw ε hε with ⟨W, hW⟩ -- W : ℕ -- hW : ∀ (n : ℕ), n ≥ W → |w n - a| ≤ ε clear hw hε use max U W intros n hn -- n : ℕ -- hn : n ≥ max U W -- ⊢ |v n - a| ≤ ε rw [max_ge_iff] at hn -- hn : n ≥ U ∧ n ≥ W specialize hU n (by linarith) -- hU : |u n - a| ≤ ε specialize hW n (by linarith) -- hW : |w n - a| ≤ ε specialize h1 n -- h1 : u n ≤ v n specialize h2 n -- h2 : v n ≤ w n rw [abs_le] at * -- hU : -ε ≤ u n - a ∧ u n - a ≤ ε -- hW : -ε ≤ w n - a ∧ w n - a ≤ ε -- ⊢ -ε ≤ v n - a ∧ v n - a ≤ ε constructor <;> linarith |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
7.3. Demostraciones con Isabelle/HOL
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theory Teorema_del_emparedado 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)" lemma assumes "limite u a" "limite w a" "∀n. u n ≤ v n" "∀n. v n ≤ w n" shows "limite v a" proof (unfold limite_def; intro allI impI) fix ε :: real assume hε : "0 < ε" obtain N where hN : "∀n≥N. ¦u n - a¦ < ε" using assms(1) hε limite_def by auto obtain N' where hN' : "∀n≥N'. ¦w n - a¦ < ε" using assms(2) hε limite_def by auto have "∀n≥max N N'. ¦v n - a¦ < ε" proof (intro allI impI) fix n assume hn : "n≥max N N'" have "v n - a < ε" proof - have "v n - a ≤ w n - a" using assms(4) by simp also have "… ≤ ¦w n - a¦" by simp also have "… < ε" using hN' hn by auto finally show "v n - a < ε" . qed moreover have "-(v n - a) < ε" proof - have "-(v n - a) ≤ -(u n - a)" using assms(3) by auto also have "… ≤ ¦u n - a¦" by simp also have "… < ε" using hN hn by auto finally show "-(v n - a) < ε" . qed ultimately show "¦v n - a¦ < ε" by (simp only: abs_less_iff) qed then show "∃k. ∀n≥k. ¦v n - a¦ < ε" by (rule exI) qed end |
8. Si s ⊆ t, entonces s ∩ u ⊆ t ∩ u
Demostrar con Lean4 que “Si \(s ⊆ t\), entonces \(s ∩ u ⊆ t ∩ u\)”.
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 t u : Set α) example (h : s ⊆ t) : s ∩ u ⊆ t ∩ u := by sorry |
8.1. Demostración en lenguaje natural
Sea \(x ∈ s ∩ u\). Entonces, se tiene que
\begin{align}
&x ∈ s \tag{1} \\
&x ∈ u \tag{2}
\end{align}
De (1) y \(s ⊆ t\), se tiene que
\[ x ∈ t \tag{3} \]
De (3) y (2) se tiene que
\[ x ∈ t ∩ u \]
que es lo que teníamos que demostrar.
8.2. Demostraciones con Lean4
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import Mathlib.Data.Set.Basic import Mathlib.Tactic open Set variable {α : Type} variable (s t u : Set α) -- 1ª demostración -- =============== example (h : s ⊆ t) : s ∩ u ⊆ t ∩ u := by rw [subset_def] -- ⊢ ∀ (x : α), x ∈ s ∩ u → x ∈ t ∩ u intros x h1 -- x : α -- h1 : x ∈ s ∩ u -- ⊢ x ∈ t ∩ u rcases h1 with ⟨xs, xu⟩ -- xs : x ∈ s -- xu : x ∈ u constructor . -- ⊢ x ∈ t rw [subset_def] at h -- h : ∀ (x : α), x ∈ s → x ∈ t apply h -- ⊢ x ∈ s exact xs . -- ⊢ x ∈ u exact xu -- 2ª demostración -- =============== example (h : s ⊆ t) : s ∩ u ⊆ t ∩ u := by rw [subset_def] -- ⊢ ∀ (x : α), x ∈ s ∩ u → x ∈ t ∩ u rintro x ⟨xs, xu⟩ -- x : α -- xs : x ∈ s -- xu : x ∈ u rw [subset_def] at h -- h : ∀ (x : α), x ∈ s → x ∈ t exact ⟨h x xs, xu⟩ -- 3ª demostración -- =============== example (h : s ⊆ t) : s ∩ u ⊆ t ∩ u := by simp only [subset_def] -- ⊢ ∀ (x : α), x ∈ s ∩ u → x ∈ t ∩ u rintro x ⟨xs, xu⟩ -- x : α -- xs : x ∈ s -- xu : x ∈ u rw [subset_def] at h -- h : ∀ (x : α), x ∈ s → x ∈ t exact ⟨h _ xs, xu⟩ -- 4ª demostración -- =============== example (h : s ⊆ t) : s ∩ u ⊆ t ∩ u := by intros x xsu -- x : α -- xsu : x ∈ s ∩ u -- ⊢ x ∈ t ∩ u exact ⟨h xsu.1, xsu.2⟩ -- 5ª demostración -- =============== example (h : s ⊆ t) : s ∩ u ⊆ t ∩ u := by rintro x ⟨xs, xu⟩ -- xs : x ∈ s -- xu : x ∈ u -- ⊢ x ∈ t ∩ u exact ⟨h xs, xu⟩ -- 6ª demostración -- =============== example (h : s ⊆ t) : s ∩ u ⊆ t ∩ u := fun _ ⟨xs, xu⟩ ↦ ⟨h xs, xu⟩ -- 7ª demostración -- =============== example (h : s ⊆ t) : s ∩ u ⊆ t ∩ u := inter_subset_inter_left u h -- Lema usado -- ========== -- #check (inter_subset_inter_left u : s ⊆ t → s ∩ u ⊆ t ∩ u) |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
8.3. Demostraciones con Isabelle/HOL
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theory Propiedad_de_monotonia_de_la_interseccion imports Main begin (* 1ª solución *) lemma assumes "s ⊆ t" shows "s ∩ u ⊆ t ∩ u" proof (rule subsetI) fix x assume hx: "x ∈ s ∩ u" have xs: "x ∈ s" using hx by (simp only: IntD1) then have xt: "x ∈ t" using assms by (simp only: subset_eq) have xu: "x ∈ u" using hx by (simp only: IntD2) show "x ∈ t ∩ u" using xt xu by (simp only: Int_iff) qed (* 2 solución *) lemma assumes "s ⊆ t" shows "s ∩ u ⊆ t ∩ u" proof fix x assume hx: "x ∈ s ∩ u" have xs: "x ∈ s" using hx by simp then have xt: "x ∈ t" using assms by auto have xu: "x ∈ u" using hx by simp show "x ∈ t ∩ u" using xt xu by simp qed (* 3ª solución *) lemma assumes "s ⊆ t" shows "s ∩ u ⊆ t ∩ u" using assms by auto (* 4ª solución *) lemma "s ⊆ t ⟹ s ∩ u ⊆ t ∩ u" by auto end |
9. s ∩ (t ∪ u) ⊆ (s ∩ t) ∪ (s ∩ u)
Demostrar con Lean4 que \(s ∩ (t ∪ u) ⊆ (s ∩ t) ∪ (s ∩ u)\).
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 t u : Set α) example : s ∩ (t ∪ u) ⊆ (s ∩ t) ∪ (s ∩ u) := by sorry |
9.1. Demostración en lenguaje natural
Sea \(x ∈ s ∩ (t ∪ u)\). Entonces se tiene que
\begin{align}
&x ∈ s \tag{1} \\
&x ∈ t ∪ u \tag{2}
\end{align}
La relación (2) da lugar a dos casos.
Caso 1: Supongamos que \(x ∈ t\). Entonces, por (1), \(x ∈ s ∩ t\) y, por tanto, \(x ∈ (s ∩ t) ∪ (s ∩ u)\).
Caso 2: Supongamos que \(x ∈ u\). Entonces, por (1), \(x ∈ s ∩ u\) y, por tanto, \(x ∈ (s ∩ t) ∪ (s ∩ u)\).
9.2. Demostraciones con Lean4
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import Mathlib.Data.Set.Basic import Mathlib.Tactic open Set variable {α : Type} variable (s t u : Set α) -- 1ª demostración -- =============== example : s ∩ (t ∪ u) ⊆ (s ∩ t) ∪ (s ∩ u) := by intros x hx -- x : α -- hx : x ∈ s ∩ (t ∪ u) -- ⊢ x ∈ s ∩ t ∪ s ∩ u rcases hx with ⟨hxs, hxtu⟩ -- hxs : x ∈ s -- hxtu : x ∈ t ∪ u rcases hxtu with (hxt | hxu) . -- hxt : x ∈ t left -- ⊢ x ∈ s ∩ t constructor . -- ⊢ x ∈ s exact hxs . -- hxt : x ∈ t exact hxt . -- hxu : x ∈ u right -- ⊢ x ∈ s ∩ u constructor . -- ⊢ x ∈ s exact hxs . -- ⊢ x ∈ u exact hxu -- 2ª demostración -- =============== example : s ∩ (t ∪ u) ⊆ (s ∩ t) ∪ (s ∩ u) := by rintro x ⟨hxs, hxt | hxu⟩ -- x : α -- hxs : x ∈ s -- ⊢ x ∈ s ∩ t ∪ s ∩ u . -- hxt : x ∈ t left -- ⊢ x ∈ s ∩ t exact ⟨hxs, hxt⟩ . -- hxu : x ∈ u right -- ⊢ x ∈ s ∩ u exact ⟨hxs, hxu⟩ -- 3ª demostración -- =============== example : s ∩ (t ∪ u) ⊆ (s ∩ t) ∪ (s ∩ u) := by rintro x ⟨hxs, hxt | hxu⟩ -- x : α -- hxs : x ∈ s -- ⊢ x ∈ s ∩ t ∪ s ∩ u . -- hxt : x ∈ t exact Or.inl ⟨hxs, hxt⟩ . -- hxu : x ∈ u exact Or.inr ⟨hxs, hxu⟩ -- 4ª demostración -- =============== example : s ∩ (t ∪ u) ⊆ (s ∩ t) ∪ (s ∩ u) := by intro x hx -- x : α -- hx : x ∈ s ∩ (t ∪ u) -- ⊢ x ∈ s ∩ t ∪ s ∩ u aesop -- 5ª demostración -- =============== example : s ∩ (t ∪ u) ⊆ (s ∩ t) ∪ (s ∩ u) := by rw [inter_union_distrib_left] |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
9.3. Demostraciones con Isabelle/HOL
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theory Propiedad_semidistributiva_de_la_interseccion_sobre_la_union imports Main begin (* 1ª demostración *) lemma "s ∩ (t ∪ u) ⊆ (s ∩ t) ∪ (s ∩ u)" proof (rule subsetI) fix x assume hx : "x ∈ s ∩ (t ∪ u)" then have xs : "x ∈ s" by (simp only: IntD1) have xtu: "x ∈ t ∪ u" using hx by (simp only: IntD2) then have "x ∈ t ∨ x ∈ u" by (simp only: Un_iff) then show " x ∈ s ∩ t ∪ s ∩ u" proof (rule disjE) assume xt : "x ∈ t" have xst : "x ∈ s ∩ t" using xs xt by (simp only: Int_iff) then show "x ∈ (s ∩ t) ∪ (s ∩ u)" by (simp only: UnI1) next assume xu : "x ∈ u" have xst : "x ∈ s ∩ u" using xs xu by (simp only: Int_iff) then show "x ∈ (s ∩ t) ∪ (s ∩ u)" by (simp only: UnI2) qed qed (* 2ª demostración *) lemma "s ∩ (t ∪ u) ⊆ (s ∩ t) ∪ (s ∩ u)" proof fix x assume hx : "x ∈ s ∩ (t ∪ u)" then have xs : "x ∈ s" by simp have xtu: "x ∈ t ∪ u" using hx by simp then have "x ∈ t ∨ x ∈ u" by simp then show " x ∈ s ∩ t ∪ s ∩ u" proof assume xt : "x ∈ t" have xst : "x ∈ s ∩ t" using xs xt by simp then show "x ∈ (s ∩ t) ∪ (s ∩ u)" by simp next assume xu : "x ∈ u" have xst : "x ∈ s ∩ u" using xs xu by simp then show "x ∈ (s ∩ t) ∪ (s ∩ u)" by simp qed qed (* 3ª demostración *) lemma "s ∩ (t ∪ u) ⊆ (s ∩ t) ∪ (s ∩ u)" proof (rule subsetI) fix x assume hx : "x ∈ s ∩ (t ∪ u)" then have xs : "x ∈ s" by (simp only: IntD1) have xtu: "x ∈ t ∪ u" using hx by (simp only: IntD2) then show " x ∈ s ∩ t ∪ s ∩ u" proof (rule UnE) assume xt : "x ∈ t" have xst : "x ∈ s ∩ t" using xs xt by (simp only: Int_iff) then show "x ∈ (s ∩ t) ∪ (s ∩ u)" by (simp only: UnI1) next assume xu : "x ∈ u" have xst : "x ∈ s ∩ u" using xs xu by (simp only: Int_iff) then show "x ∈ (s ∩ t) ∪ (s ∩ u)" by (simp only: UnI2) qed qed (* 4ª demostración *) lemma "s ∩ (t ∪ u) ⊆ (s ∩ t) ∪ (s ∩ u)" proof fix x assume hx : "x ∈ s ∩ (t ∪ u)" then have xs : "x ∈ s" by simp have xtu: "x ∈ t ∪ u" using hx by simp then show " x ∈ s ∩ t ∪ s ∩ u" proof (rule UnE) assume xt : "x ∈ t" have xst : "x ∈ s ∩ t" using xs xt by simp then show "x ∈ (s ∩ t) ∪ (s ∩ u)" by simp next assume xu : "x ∈ u" have xst : "x ∈ s ∩ u" using xs xu by simp then show "x ∈ (s ∩ t) ∪ (s ∩ u)" by simp qed qed (* 5ª demostración *) lemma "s ∩ (t ∪ u) ⊆ (s ∩ t) ∪ (s ∩ u)" by (simp only: Int_Un_distrib) (* 6ª demostración *) lemma "s ∩ (t ∪ u) ⊆ (s ∩ t) ∪ (s ∩ u)" by auto end |
10. (s \ t) \ u ⊆ s \ (t ∪ u)
Demostrar con Lean4 que
\[ (s \setminus t) \setminus u ⊆ s \setminus (t ∪ u) \]
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Set.Basic open Set variable {α : Type} variable (s t u : Set α) example : (s \ t) \ u ⊆ s \ (t ∪ u) := by sorry |
10.1. Demostración en lenguaje natural
Sea \(x ∈ (s \setminus t) \setminus u\). Entonces, se tiene que
\begin{align}
&x ∈ s \tag{1} \\
&x ∉ t \tag{2} \\
&x ∉ u \tag{3}
\end{align}
Tenemos que demostrar que
\[ x ∈ s \setminus (t ∪ u) \]
pero, por (1), se reduce a
\[ x ∉ t ∪ u \]
que se verifica por (2) y (3).
10.2. Demostraciones con Lean4
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import Mathlib.Data.Set.Basic open Set variable {α : Type} variable (s t u : Set α) -- 1ª demostración -- =============== example : (s \ t) \ u ⊆ s \ (t ∪ u) := by intros x hx -- x : α -- hx : x ∈ (s \ t) \ u -- ⊢ x ∈ s \ (t ∪ u) rcases hx with ⟨hxst, hxnu⟩ -- hxst : x ∈ s \ t -- hxnu : ¬x ∈ u rcases hxst with ⟨hxs, hxnt⟩ -- hxs : x ∈ s -- hxnt : ¬x ∈ t constructor . -- ⊢ x ∈ s exact hxs . -- ⊢ ¬x ∈ t ∪ u by_contra hxtu -- hxtu : x ∈ t ∪ u -- ⊢ False rcases hxtu with (hxt | hxu) . -- hxt : x ∈ t apply hxnt -- ⊢ x ∈ t exact hxt . -- hxu : x ∈ u apply hxnu -- ⊢ x ∈ u exact hxu -- 2ª demostración -- =============== example : (s \ t) \ u ⊆ s \ (t ∪ u) := by rintro x ⟨⟨hxs, hxnt⟩, hxnu⟩ -- x : α -- hxnu : ¬x ∈ u -- hxs : x ∈ s -- hxnt : ¬x ∈ t -- ⊢ x ∈ s \ (t ∪ u) constructor . -- ⊢ x ∈ s exact hxs . -- ⊢ ¬x ∈ t ∪ u by_contra hxtu -- hxtu : x ∈ t ∪ u -- ⊢ False rcases hxtu with (hxt | hxu) . -- hxt : x ∈ t exact hxnt hxt . -- hxu : x ∈ u exact hxnu hxu -- 3ª demostración -- =============== example : (s \ t) \ u ⊆ s \ (t ∪ u) := by rintro x ⟨⟨xs, xnt⟩, xnu⟩ -- x : α -- xnu : ¬x ∈ u -- xs : x ∈ s -- xnt : ¬x ∈ t -- ⊢ x ∈ s \ (t ∪ u) use xs -- ⊢ ¬x ∈ t ∪ u rintro (xt | xu) . -- xt : x ∈ t -- ⊢ False contradiction . -- xu : x ∈ u -- ⊢ False contradiction -- 4ª demostración -- =============== example : (s \ t) \ u ⊆ s \ (t ∪ u) := by rintro x ⟨⟨xs, xnt⟩, xnu⟩ -- x : α -- xnu : ¬x ∈ u -- xs : x ∈ s -- xnt : ¬x ∈ t -- ⊢ x ∈ s \ (t ∪ u) use xs -- ⊢ ¬x ∈ t ∪ u rintro (xt | xu) <;> contradiction -- 5ª demostración -- =============== example : (s \ t) \ u ⊆ s \ (t ∪ u) := by intro x xstu -- x : α -- xstu : x ∈ (s \ t) \ u -- ⊢ x ∈ s \ (t ∪ u) simp at * -- ⊢ x ∈ s ∧ ¬(x ∈ t ∨ x ∈ u) aesop -- 6ª demostración -- =============== example : (s \ t) \ u ⊆ s \ (t ∪ u) := by intro x xstu -- x : α -- xstu : x ∈ (s \ t) \ u -- ⊢ x ∈ s \ (t ∪ u) aesop -- 7ª demostración -- =============== example : (s \ t) \ u ⊆ s \ (t ∪ u) := by rw [diff_diff] -- Lema usado -- ========== -- #check (diff_diff : (s \ t) \ u = s \ (t ∪ u)) |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
10.3. Demostraciones con Isabelle/HOL
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theory Diferencia_de_diferencia_de_conjuntos imports Main begin (* 1ª demostración *) lemma "(s - t) - u ⊆ s - (t ∪ u)" proof (rule subsetI) fix x assume hx : "x ∈ (s - t) - u" then show "x ∈ s - (t ∪ u)" proof (rule DiffE) assume xst : "x ∈ s - t" assume xnu : "x ∉ u" note xst then show "x ∈ s - (t ∪ u)" proof (rule DiffE) assume xs : "x ∈ s" assume xnt : "x ∉ t" have xntu : "x ∉ t ∪ u" proof (rule notI) assume xtu : "x ∈ t ∪ u" then show False proof (rule UnE) assume xt : "x ∈ t" with xnt show False by (rule notE) next assume xu : "x ∈ u" with xnu show False by (rule notE) qed qed show "x ∈ s - (t ∪ u)" using xs xntu by (rule DiffI) qed qed qed (* 2ª demostración *) lemma "(s - t) - u ⊆ s - (t ∪ u)" proof fix x assume hx : "x ∈ (s - t) - u" then have xst : "x ∈ (s - t)" by simp then have xs : "x ∈ s" by simp have xnt : "x ∉ t" using xst by simp have xnu : "x ∉ u" using hx by simp have xntu : "x ∉ t ∪ u" using xnt xnu by simp then show "x ∈ s - (t ∪ u)" using xs by simp qed (* 3ª demostración *) lemma "(s - t) - u ⊆ s - (t ∪ u)" proof fix x assume "x ∈ (s - t) - u" then show "x ∈ s - (t ∪ u)" by simp qed (* 4ª demostración *) lemma "(s - t) - u ⊆ s - (t ∪ u)" by auto end |
11. (s ∩ t) ∪ (s ∩ u) ⊆ s ∩ (t ∪ u)
Demostrar con Lean4 que
\[ (s ∩ t) ∪ (s ∩ u) ⊆ s ∩ (t ∪ u) \]
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Set.Basic open Set variable {α : Type} variable (s t u : Set α) example : (s ∩ t) ∪ (s ∩ u) ⊆ s ∩ (t ∪ u):= by sorry |
11.1. Demostración en lenguaje natural
Sea \(x ∈ (s ∩ t) ∪ (s ∩ u)\). Entonces son posibles dos casos.
1º caso: Supongamos que \(x ∈ s ∩ t\). Entonces, \(x ∈ s\) y \(x ∈ t\) (y, por tanto, \(x ∈ t ∪ u\)). Luego, \(x ∈ s ∩ (t ∪ u)\).
2º caso: Supongamos que \(x ∈ s ∩ u\). Entonces, \(x ∈ s\) y \(x ∈ u\) (y, por tanto, \(x ∈ t ∪ u\)). Luego, \(x ∈ s ∩ (t ∪ u)\).
11.2. Demostraciones con Lean4
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import Mathlib.Data.Set.Basic open Set variable {α : Type} variable (s t u : Set α) -- 1ª demostración -- =============== example : (s ∩ t) ∪ (s ∩ u) ⊆ s ∩ (t ∪ u):= by intros x hx -- x : α -- hx : x ∈ s ∩ t ∪ s ∩ u -- ⊢ x ∈ s ∩ (t ∪ u) rcases hx with (xst | xsu) . -- xst : x ∈ s ∩ t constructor . -- ⊢ x ∈ s exact xst.1 . -- ⊢ x ∈ t ∪ u left -- ⊢ x ∈ t exact xst.2 . -- xsu : x ∈ s ∩ u constructor . -- ⊢ x ∈ s exact xsu.1 . -- ⊢ x ∈ t ∪ u right -- ⊢ x ∈ u exact xsu.2 -- 2ª demostración -- =============== example : (s ∩ t) ∪ (s ∩ u) ⊆ s ∩ (t ∪ u):= by rintro x (⟨xs, xt⟩ | ⟨xs, xu⟩) . -- x : α -- xs : x ∈ s -- xt : x ∈ t -- ⊢ x ∈ s ∩ (t ∪ u) use xs -- ⊢ x ∈ t ∪ u left -- ⊢ x ∈ t exact xt . -- x : α -- xs : x ∈ s -- xu : x ∈ u -- ⊢ x ∈ s ∩ (t ∪ u) use xs -- ⊢ x ∈ t ∪ u right -- ⊢ x ∈ u exact xu -- 3ª demostración -- =============== example : (s ∩ t) ∪ (s ∩ u) ⊆ s ∩ (t ∪ u):= by rw [inter_distrib_left s t u] -- 4ª demostración -- =============== example : (s ∩ t) ∪ (s ∩ u) ⊆ s ∩ (t ∪ u):= by intros x hx -- x : α -- hx : x ∈ s ∩ t ∪ s ∩ u -- ⊢ x ∈ s ∩ (t ∪ u) aesop |
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
11.3. Demostraciones con Isabelle/HOL
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theory Propiedad_semidistributiva_de_la_interseccion_sobre_la_union_2 imports Main begin (* 1ª demostración *) lemma "(s ∩ t) ∪ (s ∩ u) ⊆ s ∩ (t ∪ u)" proof (rule subsetI) fix x assume "x ∈ (s ∩ t) ∪ (s ∩ u)" then show "x ∈ s ∩ (t ∪ u)" proof (rule UnE) assume xst : "x ∈ s ∩ t" then have xs : "x ∈ s" by (simp only: IntD1) have xt : "x ∈ t" using xst by (simp only: IntD2) then have xtu : "x ∈ t ∪ u" by (simp only: UnI1) show "x ∈ s ∩ (t ∪ u)" using xs xtu by (simp only: IntI) next assume xsu : "x ∈ s ∩ u" then have xs : "x ∈ s" by (simp only: IntD1) have xt : "x ∈ u" using xsu by (simp only: IntD2) then have xtu : "x ∈ t ∪ u" by (simp only: UnI2) show "x ∈ s ∩ (t ∪ u)" using xs xtu by (simp only: IntI) qed qed (* 2ª demostración *) lemma "(s ∩ t) ∪ (s ∩ u) ⊆ s ∩ (t ∪ u)" proof fix x assume "x ∈ (s ∩ t) ∪ (s ∩ u)" then show "x ∈ s ∩ (t ∪ u)" proof assume xst : "x ∈ s ∩ t" then have xs : "x ∈ s" by simp have xt : "x ∈ t" using xst by simp then have xtu : "x ∈ t ∪ u" by simp show "x ∈ s ∩ (t ∪ u)" using xs xtu by simp next assume xsu : "x ∈ s ∩ u" then have xs : "x ∈ s" by (simp only: IntD1) have xt : "x ∈ u" using xsu by simp then have xtu : "x ∈ t ∪ u" by simp show "x ∈ s ∩ (t ∪ u)" using xs xtu by simp qed qed (* 3ª demostración *) lemma "(s ∩ t) ∪ (s ∩ u) ⊆ s ∩ (t ∪ u)" proof fix x assume "x ∈ (s ∩ t) ∪ (s ∩ u)" then show "x ∈ s ∩ (t ∪ u)" proof assume "x ∈ s ∩ t" then show "x ∈ s ∩ (t ∪ u)" by simp next assume "x ∈ s ∩ u" then show "x ∈ s ∩ (t ∪ u)" by simp qed qed (* 4ª demostración *) lemma "(s ∩ t) ∪ (s ∩ u) ⊆ s ∩ (t ∪ u)" proof fix x assume "x ∈ (s ∩ t) ∪ (s ∩ u)" then show "x ∈ s ∩ (t ∪ u)" by auto qed (* 5ª demostración *) lemma "(s ∩ t) ∪ (s ∩ u) ⊆ s ∩ (t ∪ u)" by auto (* 6ª demostración *) lemma "(s ∩ t) ∪ (s ∩ u) ⊆ s ∩ (t ∪ u)" by (simp only: distrib_inf_le) end |