src/HOL/Isar_Examples/Group_Context.thy

tuned;

(*  Title:      HOL/Isar_Examples/Group_Context.thy
    Author:     Makarius
*)

section \<open>Some algebraic identities derived from group axioms -- theory context version\<close>

theory Group_Context
  imports Main
begin

text \<open>hypothetical group axiomatization\<close>

context
  fixes prod :: "'a \<Rightarrow> 'a \<Rightarrow> 'a"  (infixl "\<odot>" 70)
    and one :: "'a"
    and inverse :: "'a \<Rightarrow> 'a"
  assumes assoc: "(x \<odot> y) \<odot> z = x \<odot> (y \<odot> z)"
    and left_one: "one \<odot> x = x"
    and left_inverse: "inverse x \<odot> x = one"
begin

text \<open>some consequences\<close>

lemma right_inverse: "x \<odot> inverse x = one"
proof -
  have "x \<odot> inverse x = one \<odot> (x \<odot> inverse x)"
    by (simp only: left_one)
  also have "\<dots> = one \<odot> x \<odot> inverse x"
    by (simp only: assoc)
  also have "\<dots> = inverse (inverse x) \<odot> inverse x \<odot> x \<odot> inverse x"
    by (simp only: left_inverse)
  also have "\<dots> = inverse (inverse x) \<odot> (inverse x \<odot> x) \<odot> inverse x"
    by (simp only: assoc)
  also have "\<dots> = inverse (inverse x) \<odot> one \<odot> inverse x"
    by (simp only: left_inverse)
  also have "\<dots> = inverse (inverse x) \<odot> (one \<odot> inverse x)"
    by (simp only: assoc)
  also have "\<dots> = inverse (inverse x) \<odot> inverse x"
    by (simp only: left_one)
  also have "\<dots> = one"
    by (simp only: left_inverse)
  finally show ?thesis .
qed

lemma right_one: "x \<odot> one = x"
proof -
  have "x \<odot> one = x \<odot> (inverse x \<odot> x)"
    by (simp only: left_inverse)
  also have "\<dots> = x \<odot> inverse x \<odot> x"
    by (simp only: assoc)
  also have "\<dots> = one \<odot> x"
    by (simp only: right_inverse)
  also have "\<dots> = x"
    by (simp only: left_one)
  finally show ?thesis .
qed

lemma one_equality:
  assumes eq: "e \<odot> x = x"
  shows "one = e"
proof -
  have "one = x \<odot> inverse x"
    by (simp only: right_inverse)
  also have "\<dots> = (e \<odot> x) \<odot> inverse x"
    by (simp only: eq)
  also have "\<dots> = e \<odot> (x \<odot> inverse x)"
    by (simp only: assoc)
  also have "\<dots> = e \<odot> one"
    by (simp only: right_inverse)
  also have "\<dots> = e"
    by (simp only: right_one)
  finally show ?thesis .
qed

lemma inverse_equality:
  assumes eq: "x' \<odot> x = one"
  shows "inverse x = x'"
proof -
  have "inverse x = one \<odot> inverse x"
    by (simp only: left_one)
  also have "\<dots> = (x' \<odot> x) \<odot> inverse x"
    by (simp only: eq)
  also have "\<dots> = x' \<odot> (x \<odot> inverse x)"
    by (simp only: assoc)
  also have "\<dots> = x' \<odot> one"
    by (simp only: right_inverse)
  also have "\<dots> = x'"
    by (simp only: right_one)
  finally show ?thesis .
qed

end

end