tuned proofs -- clarified flow of facts wrt. calculation;
(* Title: HOL/Isar_Examples/Group_Notepad.thy
Author: Makarius
*)
header {* Some algebraic identities derived from group axioms -- proof notepad version *}
theory Group_Notepad
imports Main
begin
notepad
begin
txt {* hypothetical group axiomatization *}
fix prod :: "'a \<Rightarrow> 'a \<Rightarrow> 'a" (infixl "**" 70)
and one :: "'a"
and inverse :: "'a => 'a"
assume assoc: "\<And>x y z. (x ** y) ** z = x ** (y ** z)"
and left_one: "\<And>x. one ** x = x"
and left_inverse: "\<And>x. inverse x ** x = one"
txt {* some consequences *}
have right_inverse: "\<And>x. x ** inverse x = one"
proof -
fix x
have "x ** inverse x = one ** (x ** inverse x)"
by (simp only: left_one)
also have "\<dots> = one ** x ** inverse x"
by (simp only: assoc)
also have "\<dots> = inverse (inverse x) ** inverse x ** x ** inverse x"
by (simp only: left_inverse)
also have "\<dots> = inverse (inverse x) ** (inverse x ** x) ** inverse x"
by (simp only: assoc)
also have "\<dots> = inverse (inverse x) ** one ** inverse x"
by (simp only: left_inverse)
also have "\<dots> = inverse (inverse x) ** (one ** inverse x)"
by (simp only: assoc)
also have "\<dots> = inverse (inverse x) ** inverse x"
by (simp only: left_one)
also have "\<dots> = one"
by (simp only: left_inverse)
finally show "x ** inverse x = one" .
qed
have right_one: "\<And>x. x ** one = x"
proof -
fix x
have "x ** one = x ** (inverse x ** x)"
by (simp only: left_inverse)
also have "\<dots> = x ** inverse x ** x"
by (simp only: assoc)
also have "\<dots> = one ** x"
by (simp only: right_inverse)
also have "\<dots> = x"
by (simp only: left_one)
finally show "x ** one = x" .
qed
have one_equality: "\<And>e x. e ** x = x \<Longrightarrow> one = e"
proof -
fix e x
assume eq: "e ** x = x"
have "one = x ** inverse x"
by (simp only: right_inverse)
also have "\<dots> = (e ** x) ** inverse x"
by (simp only: eq)
also have "\<dots> = e ** (x ** inverse x)"
by (simp only: assoc)
also have "\<dots> = e ** one"
by (simp only: right_inverse)
also have "\<dots> = e"
by (simp only: right_one)
finally show "one = e" .
qed
have inverse_equality: "\<And>x x'. x' ** x = one \<Longrightarrow> inverse x = x'"
proof -
fix x x'
assume eq: "x' ** x = one"
have "inverse x = one ** inverse x"
by (simp only: left_one)
also have "\<dots> = (x' ** x) ** inverse x"
by (simp only: eq)
also have "\<dots> = x' ** (x ** inverse x)"
by (simp only: assoc)
also have "\<dots> = x' ** one"
by (simp only: right_inverse)
also have "\<dots> = x'"
by (simp only: right_one)
finally show "inverse x = x'" .
qed
end
end