En los monoides, los inversos a la izquierda y a la derecha son iguales
Un monoide es un conjunto junto con una operación binaria que es asociativa y tiene elemento neutro.
En Lean4, está definida la clase de los monoides (como Monoid) y sus propiedades características son
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mul_assoc : (a * b) * c = a * (b * c) one_mul : 1 * a = a mul_one : a * 1 = a |
Demostrar que si \(M\) es un monoide, \(a ∈ M\), \(b\) es un inverso de \(a\) por la izquierda y \(c\) es un inverso de \(a\) por la derecha, entonces \(b = c\).
Para ello, completar la siguiente teoría de Lean4:
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import Mathlib.Algebra.Group.Defs variable {M : Type} [Monoid M] variable {a b c : M} example (hba : b * a = 1) (hac : a * c = 1) : b = c := by sorry |
1. Demostración en lenguaje natural
Por la siguiente cadena de igualdades
\begin{align}
b &= b * 1 &&\text{[por mul_one]} \\
&= b * (a * c) &&\text{[por hipótesis]} \\
&= (b * a) * c &&\text{[por mul_assoc]} \\
&= 1 * c &&\text{[por hipótesis]} \\
&= c &&\text{[por one_mul]} \\
\end{align}
2. Demostraciones con Lean4
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import Mathlib.Algebra.Group.Defs variable {M : Type} [Monoid M] variable {a b c : M} -- 1ª demostración -- =============== example (hba : b * a = 1) (hac : a * c = 1) : b = c := calc b = b * 1 := (mul_one b).symm _ = b * (a * c) := congrArg (b * .) hac.symm _ = (b * a) * c := (mul_assoc b a c).symm _ = 1 * c := congrArg (. * c) hba _ = c := one_mul c -- 2ª demostración -- =============== example (hba : b * a = 1) (hac : a * c = 1) : b = c := calc b = b * 1 := by aesop _ = b * (a * c) := by aesop _ = (b * a) * c := (mul_assoc b a c).symm _ = 1 * c := by aesop _ = c := by aesop -- 1ª demostración -- =============== example (hba : b * a = 1) (hac : a * c = 1) : b = c := by rw [←one_mul c] -- ⊢ b = 1 * c rw [←hba] -- ⊢ b = (b * a) * c rw [mul_assoc] -- ⊢ b = b * (a * c) rw [hac] -- ⊢ b = b * 1 rw [mul_one b] -- 2ª demostración -- =============== example (hba : b * a = 1) (hac : a * c = 1) : b = c := by rw [←one_mul c, ←hba, mul_assoc, hac, mul_one b] -- 5ª demostración -- =============== example (hba : b * a = 1) (hac : a * c = 1) : b = c := left_inv_eq_right_inv hba hac -- Lemas usados -- ============ -- #check (left_inv_eq_right_inv : b * a = 1 → a * c = 1 → b = c) -- #check (mul_assoc a b c : (a * b) * c = a * (b * c)) -- #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
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theory En_los_monoides_los_inversos_a_la_izquierda_y_a_la_derecha_son_iguales imports Main begin context monoid begin (* 1ª demostración *) lemma assumes "b * a = 1" "a * c = 1" shows "b = c" proof - have "b = b * 1" by (simp only: right_neutral) also have "… = b * (a * c)" by (simp only: ‹a * c = 1›) also have "… = (b * a) * c" by (simp only: assoc) also have "… = 1 * c" by (simp only: ‹b * a = 1›) also have "… = c" by (simp only: left_neutral) finally show "b = c" by this qed (* 2ª demostración *) lemma assumes "b * a = 1" "a * c = 1" shows "b = c" proof - have "b = b * 1" by simp also have "… = b * (a * c)" using ‹a * c = 1› by simp also have "… = (b * a) * c" by (simp add: assoc) also have "… = 1 * c" using ‹b * a = 1› by simp also have "… = c" by simp finally show "b = c" by this qed (* 3ª demostración *) lemma assumes "b * a = 1" "a * c = 1" shows "b = c" using assms by (metis assoc left_neutral right_neutral) end end |
