En ℝ, si x² = 1 entonces x = 1 ó x = -1
Demostrar con Lean4 que en \(ℝ\), si \(x² = 1\) entonces \(x = 1\) ó \(x = -1\).
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Data.Real.Basic variable (x y : ℝ) example (h : x^2 = 1) : x = 1 ∨ x = -1 := by sorry |
Demostración en lenguaje natural
Usaremos los siguientes lemas
\begin{align}
&(∀ x ∈ ℝ)[x – x = 0] \tag{L1} \\
&(∀ x, y ∈ ℝ)[xy = 0 → x = 0 ∨ y = 0] \tag{L2} \\
&(∀ x, y ∈ ℝ)[x – y = 0 ↔ x = y] \tag{L3} \\
&(∀ x, y ∈ ℝ)[x + y = 0 → x = -y] \tag{L4}
\end{align}
Se tiene que
\begin{align}
(x – 1)(x + 1) &= x² – 1 \\
&= 1 – 1 &&\text{[por la hipótesis]} \\
&= 0 &&\text{[por L1]}
\end{align}
y, por el lema L2, se tiene que
\[ x – 1 = 0 ∨ x + 1 = 0 \]
Acabaremos la demostración por casos.
Primer caso:
\begin{align}
x – 1 = 0 &⟹ x = 1 &&\text{[por L3]} \\
&⟹ x = 1 ∨ x = -1
\end{align}
Segundo caso:
\begin{align}
x + 1 = 0 &⟹ x = -1 &&\text{[por L4]} \\
&⟹ x = 1 ∨ x = -1
\end{align}
Demostraciones con Lean4
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import Mathlib.Data.Real.Basic variable (x y : ℝ) -- 1ª demostración -- =============== example (h : x^2 = 1) : x = 1 ∨ x = -1 := by have h1 : (x - 1) * (x + 1) = 0 := by calc (x - 1) * (x + 1) = x^2 - 1 := by ring _ = 1 - 1 := by rw [h] _ = 0 := sub_self 1 have h2 : x - 1 = 0 ∨ x + 1 = 0 := by apply eq_zero_or_eq_zero_of_mul_eq_zero h1 rcases h2 with h3 | h4 . -- h3 : x - 1 = 0 left -- ⊢ x = 1 exact sub_eq_zero.mp h3 . -- h4 : x + 1 = 0 right -- ⊢ x = -1 exact eq_neg_of_add_eq_zero_left h4 -- 2ª demostración -- =============== example (h : x^2 = 1) : x = 1 ∨ x = -1 := by have h1 : (x - 1) * (x + 1) = 0 := by nlinarith have h2 : x - 1 = 0 ∨ x + 1 = 0 := by aesop rcases h2 with h3 | h4 . -- h3 : x - 1 = 0 left -- ⊢ x = 1 linarith . -- h4 : x + 1 = 0 right -- ⊢ x = -1 linarith -- 3ª demostración -- =============== example (h : x^2 = 1) : x = 1 ∨ x = -1 := sq_eq_one_iff.mp h -- 3ª demostración -- =============== example (h : x^2 = 1) : x = 1 ∨ x = -1 := by aesop -- Lemas usados -- ============ -- #check (eq_neg_of_add_eq_zero_left : x + y = 0 → x = -y) -- #check (eq_zero_or_eq_zero_of_mul_eq_zero : x * y = 0 → x = 0 ∨ y = 0) -- #check (sq_eq_one_iff : x ^ 2 = 1 ↔ x = 1 ∨ x = -1) -- #check (sub_eq_zero : x - y = 0 ↔ x = y) -- #check (sub_self x : x - x = 0) |
Demostraciones interactivas
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
Referencias
- J. Avigad y P. Massot. Mathematics in Lean, p. 39.
Demostraciones con Isabelle/HOL
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theory Cuadrado_igual_a_uno imports Main HOL.Real begin (* 1ª demostración *) lemma fixes x :: real assumes "x^2 = 1" shows "x = 1 ∨ x = -1" proof - have "(x - 1) * (x + 1) = x^2 - 1" by algebra also have "... = 0" using assms by simp finally have "(x - 1) * (x + 1) = 0" . moreover { assume "(x - 1) = 0" then have "x = 1" by simp } moreover { assume "(x + 1) = 0" then have "x = -1" by simp } ultimately show "x = 1 ∨ x = -1" by auto qed (* 2ª demostración *) lemma fixes x :: real assumes "x^2 = 1" shows "x = 1 ∨ x = -1" proof - have "(x - 1) * (x + 1) = x^2 - 1" by algebra also have "... = 0" using assms by simp finally have "(x - 1) * (x + 1) = 0" . then show "x = 1 ∨ x = -1" by auto qed (* 3ª demostración *) lemma fixes x :: real assumes "x^2 = 1" shows "x = 1 ∨ x = -1" proof - have "(x - 1) * (x + 1) = 0" proof - have "(x - 1) * (x + 1) = x^2 - 1" by algebra also have "… = 0" by (simp add: assms) finally show ?thesis . qed then show "x = 1 ∨ x = -1" by auto qed (* 4ª demostración *) lemma fixes x :: real assumes "x^2 = 1" shows "x = 1 ∨ x = -1" using assms power2_eq_1_iff by blast end |
