Equivalencia de inversos iguales al neutro
Sea \(M\) un monoide y \(a, b ∈ M\) tales que \(ab = 1\). Demostrar con Lean4 que \(a = 1\) si y sólo si \(b = 1\).
Para ello, completar la siguiente teoría de Lean4:
import Mathlib.Algebra.Group.Basic variable {M : Type} [Monoid M] variable {a b : M} example (h : a * b = 1) : a = 1 ↔ b = 1 := by sorry
1. Demostración en lenguaje natural
Demostraremos las dos implicaciones.
(⟹) Supongamos que \(a = 1\). Entonces, \begin{align} b &= 1·b &&\text{[por neutro por la izquierda]} \newline &= a·b &&\text{[por supuesto]} \newline &= 1 &&\text{[por hipótesis]} \end{align}
(⟸) Supongamos que \(b = 1\). Entonces, \begin{align} a &= a·1 &&\text{[por neutro por la derecha]} \newline &= a·b &&\text{[por supuesto]} \newline &= 1 &&\text{[por hipótesis]} \end{align}
2. Demostraciones con Lean4
import Mathlib.Algebra.Group.Basic variable {M : Type} [Monoid M] variable {a b : M} -- 1ª demostración -- =============== example (h : a * b = 1) : a = 1 ↔ b = 1 := by constructor . -- ⊢ a = 1 → b = 1 intro a1 -- a1 : a = 1 -- ⊢ b = 1 calc b = 1 * b := (one_mul b).symm _ = a * b := congrArg (. * b) a1.symm _ = 1 := h . -- ⊢ b = 1 → a = 1 intro b1 -- b1 : b = 1 -- ⊢ a = 1 calc a = a * 1 := (mul_one a).symm _ = a * b := congrArg (a * .) b1.symm _ = 1 := h -- 2ª demostración -- =============== example (h : a * b = 1) : a = 1 ↔ b = 1 := by constructor . -- ⊢ a = 1 → b = 1 intro a1 -- a1 : a = 1 -- ⊢ b = 1 rw [a1] at h -- h : 1 * b = 1 rw [one_mul] at h -- h : b = 1 exact h . -- ⊢ b = 1 → a = 1 intro b1 -- b1 : b = 1 -- ⊢ a = 1 rw [b1] at h -- h : a * 1 = 1 rw [mul_one] at h -- h : a = 1 exact h -- 3ª demostración -- =============== example (h : a * b = 1) : a = 1 ↔ b = 1 := by constructor . -- ⊢ a = 1 → b = 1 rintro rfl -- h : 1 * b = 1 simpa using h . -- ⊢ b = 1 → a = 1 rintro rfl -- h : a * 1 = 1 simpa using h -- 4ª demostración -- =============== example (h : a * b = 1) : a = 1 ↔ b = 1 := by constructor <;> (rintro rfl; simpa using h) -- 5ª demostración -- =============== example (h : a * b = 1) : a = 1 ↔ b = 1 := eq_one_iff_eq_one_of_mul_eq_one h -- Lemas usados -- ============ -- #check (eq_one_iff_eq_one_of_mul_eq_one : a * b = 1 → (a = 1 ↔ b = 1)) -- #check (mul_one a : a * 1 = a) -- #check (one_mul a : 1 * a = a)
Se puede interactuar con las demostraciones anteriores en Lean 4 Web.
3. Demostraciones con Isabelle/HOL
theory Equivalencia_de_inversos_iguales_al_neutro imports Main begin context monoid begin (* 1ª demostración *) lemma assumes "a ❙* b = ❙1" shows "a = ❙1 ⟷ b = ❙1" proof (rule iffI) assume "a = ❙1" have "b = ❙1 ❙* b" by (simp only: left_neutral) also have "… = a ❙* b" by (simp only: ‹a = ❙1›) also have "… = ❙1" by (simp only: ‹a ❙* b = ❙1›) finally show "b = ❙1" by this next assume "b = ❙1" have "a = a ❙* ❙1" by (simp only: right_neutral) also have "… = a ❙* b" by (simp only: ‹b = ❙1›) also have "… = ❙1" by (simp only: ‹a ❙* b = ❙1›) finally show "a = ❙1" by this qed (* 2ª demostración *) lemma assumes "a ❙* b = ❙1" shows "a = ❙1 ⟷ b = ❙1" proof assume "a = ❙1" have "b = ❙1 ❙* b" by simp also have "… = a ❙* b" using ‹a = ❙1› by simp also have "… = ❙1" using ‹a ❙* b = ❙1› by simp finally show "b = ❙1" . next assume "b = ❙1" have "a = a ❙* ❙1" by simp also have "… = a ❙* b" using ‹b = ❙1› by simp also have "… = ❙1" using ‹a ❙* b = ❙1› by simp finally show "a = ❙1" . qed (* 3ª demostración *) lemma assumes "a ❙* b = ❙1" shows "a = ❙1 ⟷ b = ❙1" by (metis assms left_neutral right_neutral) end end