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Inverso_del_producto.thy
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Inverso_del_producto.thy
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(* Inverso_del_producto.thy
-- Si G es un grupo y a, b \<in> G, entonces (ab)⁻¹ = b⁻¹a⁻¹
-- José A. Alonso Jiménez <https://jaalonso.github.io>
-- Sevilla, 14-mayo-2024
-- ------------------------------------------------------------------ *)
(* ---------------------------------------------------------------------
-- Sea G un grupo y a, b \<in> G. Entonces,
-- (a * b)⁻¹ = b⁻¹ * a⁻¹
-- ------------------------------------------------------------------ *)
theory Inverso_del_producto
imports Main
begin
context group
begin
(* 1\<ordfeminine> demostración *)
lemma "inverse (a \<^bold>* b) = inverse b \<^bold>* inverse a"
proof (rule inverse_unique)
have "(a \<^bold>* b) \<^bold>* (inverse b \<^bold>* inverse a) =
((a \<^bold>* b) \<^bold>* inverse b) \<^bold>* inverse a"
by (simp only: assoc)
also have "\<dots> = (a \<^bold>* (b \<^bold>* inverse b)) \<^bold>* inverse a"
by (simp only: assoc)
also have "\<dots> = (a \<^bold>* \<^bold>1) \<^bold>* inverse a"
by (simp only: right_inverse)
also have "\<dots> = a \<^bold>* inverse a"
by (simp only: right_neutral)
also have "\<dots> = \<^bold>1"
by (simp only: right_inverse)
finally show "a \<^bold>* b \<^bold>* (inverse b \<^bold>* inverse a) = \<^bold>1"
by this
qed
(* 2\<ordfeminine> demostración *)
lemma "inverse (a \<^bold>* b) = inverse b \<^bold>* inverse a"
proof (rule inverse_unique)
have "(a \<^bold>* b) \<^bold>* (inverse b \<^bold>* inverse a) =
((a \<^bold>* b) \<^bold>* inverse b) \<^bold>* inverse a"
by (simp only: assoc)
also have "\<dots> = (a \<^bold>* (b \<^bold>* inverse b)) \<^bold>* inverse a"
by (simp only: assoc)
also have "\<dots> = (a \<^bold>* \<^bold>1) \<^bold>* inverse a"
by simp
also have "\<dots> = a \<^bold>* inverse a"
by simp
also have "\<dots> = \<^bold>1"
by simp
finally show "a \<^bold>* b \<^bold>* (inverse b \<^bold>* inverse a) = \<^bold>1"
.
qed
(* 3\<ordfeminine> demostración *)
lemma "inverse (a \<^bold>* b) = inverse b \<^bold>* inverse a"
proof (rule inverse_unique)
have "a \<^bold>* b \<^bold>* (inverse b \<^bold>* inverse a) =
a \<^bold>* (b \<^bold>* inverse b) \<^bold>* inverse a"
by (simp only: assoc)
also have "\<dots> = \<^bold>1"
by simp
finally show "a \<^bold>* b \<^bold>* (inverse b \<^bold>* inverse a) = \<^bold>1" .
qed
(* 4\<ordfeminine> demostración *)
lemma "inverse (a \<^bold>* b) = inverse b \<^bold>* inverse a"
by (simp only: inverse_distrib_swap)
end
end