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