ssh client via regular OpenSSH tools, with authentic use of .ssh/config (notably proxy configuration);
tuned signature;
(* 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