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Propiedad_cancelativa_en_grupos.thy
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Propiedad_cancelativa_en_grupos.thy
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(* Propiedad_cancelativa_en_grupos.thy
-- Si G es un grupo y a, b, c ∈ G tales que a·b = a·c, entonces b = c.
-- José A. Alonso Jiménez <https://jaalonso.github.io>
-- Sevilla, 16-mayo-2024
-- ------------------------------------------------------------------ *)
(* ---------------------------------------------------------------------
-- Sea G un grupo y a,b,c \<in> G. Demostrar que si a * b = a* c, entonces
-- b = c.
-- ------------------------------------------------------------------ *)
theory Propiedad_cancelativa_en_grupos
imports Main
begin
context group
begin
(* 1\<ordfeminine> demostración *)
lemma
assumes "a \<^bold>* b = a \<^bold>* c"
shows "b = c"
proof -
have "b = \<^bold>1 \<^bold>* b" by (simp only: left_neutral)
also have "\<dots> = (inverse a \<^bold>* a) \<^bold>* b" by (simp only: left_inverse)
also have "\<dots> = inverse a \<^bold>* (a \<^bold>* b)" by (simp only: assoc)
also have "\<dots> = inverse a \<^bold>* (a \<^bold>* c)" by (simp only: \<open>a \<^bold>* b = a \<^bold>* c\<close>)
also have "\<dots> = (inverse a \<^bold>* a) \<^bold>* c" by (simp only: assoc)
also have "\<dots> = \<^bold>1 \<^bold>* c" by (simp only: left_inverse)
also have "\<dots> = c" by (simp only: left_neutral)
finally show "b = c" by this
qed
(* 2\<ordfeminine> demostración *)
lemma
assumes "a \<^bold>* b = a \<^bold>* c"
shows "b = c"
proof -
have "b = \<^bold>1 \<^bold>* b" by simp
also have "\<dots> = (inverse a \<^bold>* a) \<^bold>* b" by simp
also have "\<dots> = inverse a \<^bold>* (a \<^bold>* b)" by (simp only: assoc)
also have "\<dots> = inverse a \<^bold>* (a \<^bold>* c)" using \<open>a \<^bold>* b = a \<^bold>* c\<close> by simp
also have "\<dots> = (inverse a \<^bold>* a) \<^bold>* c" by (simp only: assoc)
also have "\<dots> = \<^bold>1 \<^bold>* c" by simp
finally show "b = c" by simp
qed
(* 3\<ordfeminine> demostración *)
lemma
assumes "a \<^bold>* b = a \<^bold>* c"
shows "b = c"
proof -
have "b = (inverse a \<^bold>* a) \<^bold>* b" by simp
also have "\<dots> = inverse a \<^bold>* (a \<^bold>* b)" by (simp only: assoc)
also have "\<dots> = inverse a \<^bold>* (a \<^bold>* c)" using \<open>a \<^bold>* b = a \<^bold>* c\<close> by simp
also have "\<dots> = (inverse a \<^bold>* a) \<^bold>* c" by (simp only: assoc)
finally show "b = c" by simp
qed
(* 4\<ordfeminine> demostración *)
lemma
assumes "a \<^bold>* b = a \<^bold>* c"
shows "b = c"
proof -
have "inverse a \<^bold>* (a \<^bold>* b) = inverse a \<^bold>* (a \<^bold>* c)"
by (simp only: \<open>a \<^bold>* b = a \<^bold>* c\<close>)
then have "(inverse a \<^bold>* a) \<^bold>* b = (inverse a \<^bold>* a) \<^bold>* c"
by (simp only: assoc)
then have "\<^bold>1 \<^bold>* b = \<^bold>1 \<^bold>* c"
by (simp only: left_inverse)
then show "b = c"
by (simp only: left_neutral)
qed
(* 5\<ordfeminine> demostración *)
lemma
assumes "a \<^bold>* b = a \<^bold>* c"
shows "b = c"
proof -
have "inverse a \<^bold>* (a \<^bold>* b) = inverse a \<^bold>* (a \<^bold>* c)"
by (simp only: \<open>a \<^bold>* b = a \<^bold>* c\<close>)
then have "(inverse a \<^bold>* a) \<^bold>* b = (inverse a \<^bold>* a) \<^bold>* c"
by (simp only: assoc)
then have "\<^bold>1 \<^bold>* b = \<^bold>1 \<^bold>* c"
by (simp only: left_inverse)
then show "b = c"
by (simp only: left_neutral)
qed
(* 6\<ordfeminine> demostración *)
lemma
assumes "a \<^bold>* b = a \<^bold>* c"
shows "b = c"
proof -
have "inverse a \<^bold>* (a \<^bold>* b) = inverse a \<^bold>* (a \<^bold>* c)"
using \<open>a \<^bold>* b = a \<^bold>* c\<close> by simp
then have "(inverse a \<^bold>* a) \<^bold>* b = (inverse a \<^bold>* a) \<^bold>* c"
by (simp only: assoc)
then have "\<^bold>1 \<^bold>* b = \<^bold>1 \<^bold>* c"
by simp
then show "b = c"
by simp
qed
(* 7\<ordfeminine> demostración *)
lemma
assumes "a \<^bold>* b = a \<^bold>* c"
shows "b = c"
using assms
by (simp only: left_cancel)
end
end