Theory AbelianGroup_ZF

theory AbelianGroup_ZF
imports Group_ZF
(* 
This file is a part of IsarMathLib -
a library of formalized mathematics for Isabelle/Isar.

Copyright (C) 2005, 2006 Slawomir Kolodynski

This program is free software; Redistribution and use in source and binary forms,
with or without modification, are permitted provided that the
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derived from this software without specific prior written permission.

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*)


header{*\isaheader{AbelianGroup\_ZF.thy}*}

theory AbelianGroup_ZF imports Group_ZF

begin

text{*A group is called ``abelian`` if its operation is commutative, i.e.
$P\langle a,b \rangle = P\langle a,b \rangle$ for all group
elements $a,b$, where $P$ is the group operation. It is customary
to use the additive notation for abelian groups, so this condition
is typically written as $a+b = b+a$. We will be using multiplicative
notation though (in which the commutativity condition
of the operation is written as $a\cdot b = b\cdot a$), just to avoid
the hassle of changing the notation we used for general groups.
*}


section{*Rearrangement formulae*}

text{*This section is not interesting and should not be read.
Here we will prove formulas is which right hand side uses the same
factors as the left hand side, just in different order. These facts
are obvious in informal math sense, but Isabelle prover is not able
to derive them automatically, so we have to prove them by hand.
*}


text{*Proving the facts about associative and commutative operations is
quite tedious in formalized mathematics. To a human the thing is simple:
we can arrange the elements in any order and put parantheses wherever we
want, it is all the same. However, formalizing this statement would be
rather difficult (I think). The next lemma attempts a quasi-algorithmic
approach to this type of problem. To prove that two expressions are equal,
we first strip one from parantheses, then rearrange the elements in proper
order, then put the parantheses where we want them to be. The algorithm for
rearrangement is easy to describe: we keep putting the first element
(from the right) that is in the wrong place at the left-most position
until we get the proper arrangement.
As far removing parantheses is concerned Isabelle does its job
automatically.*}


lemma (in group0) group0_4_L2:
assumes A1:"P {is commutative on} G"
and A2:"a∈G" "b∈G" "c∈G" "d∈G" "E∈G" "F∈G"
shows "(a·b)·(c·d)·(E·F) = (a·(d·F))·(b·(c·E))"
proof -;
from A2 have "(a·b)·(c·d)·(E·F) = a·b·c·d·E·F"
using group_op_closed group_oper_assoc
by simp;
also have "a·b·c·d·E·F = a·d·F·b·c·E"
proof -
from A1 A2 have "a·b·c·d·E·F = F·(a·b·c·d·E)"
using IsCommutative_def group_op_closed
by simp;
also from A2 have "F·(a·b·c·d·E) = F·a·b·c·d·E"
using group_op_closed group_oper_assoc
by simp;
also from A1 A2 have "F·a·b·c·d·E = d·(F·a·b·c)·E"
using IsCommutative_def group_op_closed
by simp;
also from A2 have "d·(F·a·b·c)·E = d·F·a·b·c·E"
using group_op_closed group_oper_assoc
by simp;
also from A1 A2 have " d·F·a·b·c·E = a·(d·F)·b·c·E"
using IsCommutative_def group_op_closed
by simp;
also from A2 have "a·(d·F)·b·c·E = a·d·F·b·c·E"
using group_op_closed group_oper_assoc
by simp;
finally show ?thesis by simp;
qed;
also from A2 have "a·d·F·b·c·E = (a·(d·F))·(b·(c·E))"
using group_op_closed group_oper_assoc
by simp;
finally show ?thesis by simp;
qed;

text{*Another useful rearrangement.*}

lemma (in group0) group0_4_L3:
assumes A1:"P {is commutative on} G"
and A2: "a∈G" "b∈G" and A3: "c∈G" "d∈G" "E∈G" "F∈G"
shows "a·b·((c·d)¯·(E·F)¯) = (a·(E·c)¯)·(b·(F·d)¯)"
proof -;
from A3 have T1:
"c¯∈G" "d¯∈G" "E¯∈G" "F¯∈G" "(c·d)¯∈G" "(E·F)¯∈G"
using inverse_in_group group_op_closed
by auto;
from A2 T1 have
"a·b·((c·d)¯·(E·F)¯) = a·b·(c·d)¯·(E·F)¯"
using group_op_closed group_oper_assoc
by simp;
also from A2 A3 have
"a·b·(c·d)¯·(E·F)¯ = (a·b)·(d¯·c¯)·(F¯·E¯)"
using group_inv_of_two by simp;
also from A1 A2 T1 have
"(a·b)·(d¯·c¯)·(F¯·E¯) = (a·(c¯·E¯))·(b·(d¯·F¯))"
using group0_4_L2 by simp;
also from A2 A3 have
"(a·(c¯·E¯))·(b·(d¯·F¯)) = (a·(E·c)¯)·(b·(F·d)¯)"
using group_inv_of_two by simp;
finally show ?thesis by simp;
qed;

text{*Some useful rearrangements for two elements of a group.*}

lemma (in group0) group0_4_L4:
assumes A1:"P {is commutative on} G"
and A2: "a∈G" "b∈G"
shows
"b¯·a¯ = a¯·b¯"
"(a·b)¯ = a¯·b¯"
"(a·b¯)¯ = a¯·b"
proof -;
from A2 have T1: "b¯∈G" "a¯∈G" using inverse_in_group by auto;
with A1 show "b¯·a¯ = a¯·b¯" using IsCommutative_def by simp;
with A2 show "(a·b)¯ = a¯·b¯" using group_inv_of_two by simp;
from A2 T1 have "(a·b¯)¯ = (b¯)¯·a¯" using group_inv_of_two by simp;
with A1 A2 T1 show "(a·b¯)¯ = a¯·b"
using group_inv_of_inv IsCommutative_def by simp;
qed;

text{*Another bunch of useful rearrangements with three elements.*}

lemma (in group0) group0_4_L4A:
assumes A1: "P {is commutative on} G"
and A2: "a∈G" "b∈G" "c∈G"
shows
"a·b·c = c·a·b"
"a¯·(b¯·c¯)¯ = (a·(b·c)¯)¯"
"a·(b·c)¯ = a·b¯·c¯"
"a·(b·c¯)¯ = a·b¯·c"
"a·b¯·c¯ = a·c¯·b¯"
proof -
from A1 A2 have "a·b·c = c·(a·b)"
using IsCommutative_def group_op_closed
by simp;
with A2 show "a·b·c = c·a·b" using
group_op_closed group_oper_assoc
by simp
from A2 have T:
"b¯∈G" "c¯∈G" "b¯·c¯ ∈ G" "a·b ∈ G"
using inverse_in_group group_op_closed
by auto;
with A1 A2 show "a¯·(b¯·c¯)¯ = (a·(b·c)¯)¯"
using group_inv_of_two IsCommutative_def
by simp;
from A1 A2 T have "a·(b·c)¯ = a·(b¯·c¯)"
using group_inv_of_two IsCommutative_def by simp;
with A2 T show "a·(b·c)¯ = a·b¯·c¯"
using group_oper_assoc by simp;
from A1 A2 T have "a·(b·c¯)¯ = a·(b¯·(c¯)¯)"
using group_inv_of_two IsCommutative_def by simp;
with A2 T show "a·(b·c¯)¯ = a·b¯·c"
using group_oper_assoc group_inv_of_inv by simp;
from A1 A2 T have "a·b¯·c¯ = a·(c¯·b¯)"
using group_oper_assoc IsCommutative_def by simp;
with A2 T show "a·b¯·c¯ = a·c¯·b¯"
using group_oper_assoc by simp
qed;

text{*Another useful rearrangement.*}

lemma (in group0) group0_4_L4B:
assumes "P {is commutative on} G"
and "a∈G" "b∈G" "c∈G"
shows "a·b¯·(b·c¯) = a·c¯"
using assms inverse_in_group group_op_closed
group0_4_L4 group_oper_assoc inv_cancel_two by simp;

text{*A couple of permutations of order for three alements.*}

lemma (in group0) group0_4_L4C:
assumes A1: "P {is commutative on} G"
and A2: "a∈G" "b∈G" "c∈G"
shows
"a·b·c = c·a·b"
"a·b·c = a·(c·b)"
"a·b·c = c·(a·b)"
"a·b·c = c·b·a"
proof -
from A1 A2 show I: "a·b·c = c·a·b"
using group0_4_L4A by simp
also from A1 A2 have "c·a·b = a·c·b"
using IsCommutative_def by simp;
also from A2 have "a·c·b = a·(c·b)"
using group_oper_assoc by simp;
finally show "a·b·c = a·(c·b)" by simp;
from A2 I show "a·b·c = c·(a·b)"
using group_oper_assoc by simp;
also from A1 A2 have "c·(a·b) = c·(b·a)"
using IsCommutative_def by simp;
also from A2 have "c·(b·a) = c·b·a"
using group_oper_assoc by simp;
finally show "a·b·c = c·b·a" by simp;
qed;

text{*Some rearangement with three elements and inverse.*}

lemma (in group0) group0_4_L4D:
assumes A1: "P {is commutative on} G"
and A2: "a∈G" "b∈G" "c∈G"
shows
"a¯·b¯·c = c·a¯·b¯"
"b¯·a¯·c = c·a¯·b¯"
"(a¯·b·c)¯ = a·b¯·c¯"
proof -
from A2 have T:
"a¯ ∈ G" "b¯ ∈ G" "c¯∈G"
using inverse_in_group by auto;
with A1 A2 show
"a¯·b¯·c = c·a¯·b¯"
"b¯·a¯·c = c·a¯·b¯"
using group0_4_L4A by auto
from A1 A2 T show "(a¯·b·c)¯ = a·b¯·c¯"
using group_inv_of_three group_inv_of_inv group0_4_L4C
by simp;
qed;

text{*Another rearrangement lemma with three elements and equation.*}

lemma (in group0) group0_4_L5: assumes A1:"P {is commutative on} G"
and A2: "a∈G" "b∈G" "c∈G"
and A3: "c = a·b¯"
shows "a = b·c"
proof -;
from A2 A3 have "c·(b¯)¯ = a"
using inverse_in_group group0_2_L18
by simp;
with A1 A2 show ?thesis using
group_inv_of_inv IsCommutative_def by simp;
qed;

text{*In abelian groups we can cancel an element with its inverse
even if separated by another element.*}


lemma (in group0) group0_4_L6A: assumes A1: "P {is commutative on} G"
and A2: "a∈G" "b∈G"
shows
"a·b·a¯ = b"
"a¯·b·a = b"
"a¯·(b·a) = b"
"a·(b·a¯) = b"
proof -;
from A1 A2 have
"a·b·a¯ = a¯·a·b"
using inverse_in_group group0_4_L4A by blast
also from A2 have "… = b"
using group0_2_L6 group0_2_L2 by simp;
finally show "a·b·a¯ = b" by simp;
from A1 A2 have
"a¯·b·a = a·a¯·b"
using inverse_in_group group0_4_L4A by blast;
also from A2 have "… = b"
using group0_2_L6 group0_2_L2 by simp;
finally show "a¯·b·a = b" by simp
moreover from A2 have "a¯·b·a = a¯·(b·a)"
using inverse_in_group group_oper_assoc by simp;
ultimately show "a¯·(b·a) = b" by simp;
from A1 A2 show "a·(b·a¯) = b"
using inverse_in_group IsCommutative_def inv_cancel_two
by simp;
qed;

text{*Another lemma about cancelling with two elements.*}

lemma (in group0) group0_4_L6AA:
assumes A1: "P {is commutative on} G" and A2: "a∈G" "b∈G"
shows "a·b¯·a¯ = b¯"
using assms inverse_in_group group0_4_L6A
by auto;

text{*Another lemma about cancelling with two elements.*}

lemma (in group0) group0_4_L6AB:
assumes A1: "P {is commutative on} G" and A2: "a∈G" "b∈G"
shows
"a·(a·b)¯ = b¯"
"a·(b·a¯) = b"
proof -
from A2 have "a·(a·b)¯ = a·(b¯·a¯)"
using group_inv_of_two by simp
also from A2 have "… = a·b¯·a¯"
using inverse_in_group group_oper_assoc by simp;
also from A1 A2 have "… = b¯"
using group0_4_L6AA by simp;
finally show "a·(a·b)¯ = b¯" by simp;
from A1 A2 have "a·(b·a¯) = a·(a¯·b)"
using inverse_in_group IsCommutative_def by simp;
also from A2 have "… = b"
using inverse_in_group group_oper_assoc group0_2_L6 group0_2_L2
by simp;
finally show "a·(b·a¯) = b" by simp;
qed;

text{*Another lemma about cancelling with two elements.*}

lemma (in group0) group0_4_L6AC:
assumes "P {is commutative on} G" and "a∈G" "b∈G"
shows "a·(a·b¯)¯ = b"
using assms inverse_in_group group0_4_L6AB group_inv_of_inv
by simp;


text{*In abelian groups we can cancel an element with its inverse
even if separated by two other elements.*}


lemma (in group0) group0_4_L6B: assumes A1: "P {is commutative on} G"
and A2: "a∈G" "b∈G" "c∈G"
shows
"a·b·c·a¯ = b·c"
"a¯·b·c·a = b·c"
proof -
from A2 have
"a·b·c·a¯ = a·(b·c)·a¯"
"a¯·b·c·a = a¯·(b·c)·a"
using group_op_closed group_oper_assoc inverse_in_group
by auto;
with A1 A2 show
"a·b·c·a¯ = b·c"
"a¯·b·c·a = b·c"
using group_op_closed group0_4_L6A
by auto;
qed;

text{*In abelian groups we can cancel an element with its inverse
even if separated by three other elements.*}


lemma (in group0) group0_4_L6C: assumes A1: "P {is commutative on} G"
and A2: "a∈G" "b∈G" "c∈G" "d∈G"
shows "a·b·c·d·a¯ = b·c·d"
proof -
from A2 have "a·b·c·d·a¯ = a·(b·c·d)·a¯"
using group_op_closed group_oper_assoc
by simp;
with A1 A2 show ?thesis
using group_op_closed group0_4_L6A
by simp;
qed;

text{*Another couple of useful rearrangements of three elements
and cancelling.*}


lemma (in group0) group0_4_L6D:
assumes A1: "P {is commutative on} G"
and A2: "a∈G" "b∈G" "c∈G"
shows
"a·b¯·(a·c¯)¯ = c·b¯"
"(a·c)¯·(b·c) = a¯·b"
"a·(b·(c·a¯·b¯)) = c"
"a·b·c¯·(c·a¯) = b"
proof -
from A2 have T:
"a¯ ∈ G" "b¯ ∈ G" "c¯ ∈ G"
"a·b ∈ G" "a·b¯ ∈ G" "c¯·a¯ ∈ G" "c·a¯ ∈ G"
using inverse_in_group group_op_closed by auto;
with A1 A2 show "a·b¯·(a·c¯)¯ = c·b¯"
using group0_2_L12 group_oper_assoc group0_4_L6B
IsCommutative_def by simp;
from A2 T have "(a·c)¯·(b·c) = c¯·a¯·b·c"
using group_inv_of_two group_oper_assoc by simp;
also from A1 A2 T have "… = a¯·b"
using group0_4_L6B by simp;
finally show "(a·c)¯·(b·c) = a¯·b"
by simp;
from A1 A2 T show "a·(b·(c·a¯·b¯)) = c"
using group_oper_assoc group0_4_L6B group0_4_L6A
by simp;
from T have "a·b·c¯·(c·a¯) = a·b·(c¯·(c·a¯))"
using group_oper_assoc by simp;
also from A1 A2 T have "… = b"
using group_oper_assoc group0_2_L6 group0_2_L2 group0_4_L6A
by simp;
finally show "a·b·c¯·(c·a¯) = b" by simp;
qed;

text{*Another useful rearrangement of three elements
and cancelling.*}


lemma (in group0) group0_4_L6E:
assumes A1: "P {is commutative on} G"
and A2: "a∈G" "b∈G" "c∈G"
shows
"a·b·(a·c)¯ = b·c¯"
proof -
from A2 have T: "b¯ ∈ G" "c¯ ∈ G"
using inverse_in_group by auto;
with A1 A2 have
"a·(b¯)¯·(a·(c¯)¯)¯ = c¯·(b¯)¯"
using group0_4_L6D by simp;
with A1 A2 T show "a·b·(a·c)¯ = b·c¯"
using group_inv_of_inv IsCommutative_def
by simp;
qed;

text{*A rearrangement with two elements and canceelling,
special case of @{text "group0_4_L6D"} when $c=b^{-1}$.*}


lemma (in group0) group0_4_L6F:
assumes A1: "P {is commutative on} G"
and A2: "a∈G" "b∈G"
shows "a·b¯·(a·b)¯ = b¯·b¯"
proof -
from A2 have "b¯ ∈ G"
using inverse_in_group by simp
with A1 A2 have "a·b¯·(a·(b¯)¯)¯ = b¯·b¯"
using group0_4_L6D by simp;
with A2 show "a·b¯·(a·b)¯ = b¯·b¯"
using group_inv_of_inv by simp;
qed;

text{*Some other rearrangements with four elements.
The algorithm for proof as in @{text "group0_4_L2"}
works very well here.*}


lemma (in group0) rearr_ab_gr_4_elemA:
assumes A1: "P {is commutative on} G"
and A2: "a∈G" "b∈G" "c∈G" "d∈G"
shows
"a·b·c·d = a·d·b·c"
"a·b·c·d = a·c·(b·d)"
proof -
from A1 A2 have "a·b·c·d = d·(a·b·c)"
using IsCommutative_def group_op_closed
by simp;
also from A2 have "… = d·a·b·c"
using group_op_closed group_oper_assoc
by simp;
also from A1 A2 have "… = a·d·b·c"
using IsCommutative_def group_op_closed
by simp;
finally show "a·b·c·d = a·d·b·c"
by simp;
from A1 A2 have "a·b·c·d = c·(a·b)·d"
using IsCommutative_def group_op_closed
by simp
also from A2 have "… = c·a·b·d"
using group_op_closed group_oper_assoc
by simp
also from A1 A2 have "… = a·c·b·d"
using IsCommutative_def group_op_closed
by simp
also from A2 have "… = a·c·(b·d)"
using group_op_closed group_oper_assoc
by simp
finally show "a·b·c·d = a·c·(b·d)"
by simp
qed;

text{*Some rearrangements with four elements and inverse
that are applications of @{text "rearr_ab_gr_4_elem"}
*}


lemma (in group0) rearr_ab_gr_4_elemB:
assumes A1: "P {is commutative on} G"
and A2: "a∈G" "b∈G" "c∈G" "d∈G"
shows
"a·b¯·c¯·d¯ = a·d¯·b¯·c¯"
"a·b·c·d¯ = a·d¯·b·c"
"a·b·c¯·d¯ = a·c¯·(b·d¯)"
proof -
from A2 have T: "b¯ ∈ G" "c¯ ∈ G" "d¯ ∈ G"
using inverse_in_group by auto;
with A1 A2 show
"a·b¯·c¯·d¯ = a·d¯·b¯·c¯"
"a·b·c·d¯ = a·d¯·b·c"
"a·b·c¯·d¯ = a·c¯·(b·d¯)"
using rearr_ab_gr_4_elemA by auto;
qed;

text{*Some rearrangement lemmas with four elements.*}

lemma (in group0) group0_4_L7:
assumes A1: "P {is commutative on} G"
and A2: "a∈G" "b∈G" "c∈G" "d∈G"
shows
"a·b·c·d¯ = a·d¯· b·c"
"a·d·(b·d·(c·d))¯ = a·(b·c)¯·d¯"
"a·(b·c)·d = a·b·d·c"
proof -
from A2 have T:
"b·c ∈ G" "d¯ ∈ G" "b¯∈G" "c¯∈G"
"d¯·b ∈ G" "c¯·d ∈ G" "(b·c)¯ ∈ G"
"b·d ∈ G" "b·d·c ∈ G" "(b·d·c)¯ ∈ G"
"a·d ∈ G" "b·c ∈ G"
using group_op_closed inverse_in_group
by auto;
with A1 A2 have "a·b·c·d¯ = a·(d¯·b·c)"
using group_oper_assoc group0_4_L4A by simp;
also from A2 T have "a·(d¯·b·c) = a·d¯·b·c"
using group_oper_assoc by simp;
finally show "a·b·c·d¯ = a·d¯· b·c" by simp;
from A2 T have "a·d·(b·d·(c·d))¯ = a·d·(d¯·(b·d·c)¯)"
using group_oper_assoc group_inv_of_two by simp;
also from A2 T have "… = a·(b·d·c)¯"
using group_oper_assoc inv_cancel_two by simp;
also from A1 A2 have "… = a·(d·(b·c))¯"
using IsCommutative_def group_oper_assoc by simp;
also from A2 T have "… = a·((b·c)¯·d¯)"
using group_inv_of_two by simp;
also from A2 T have "… = a·(b·c)¯·d¯"
using group_oper_assoc by simp;
finally show "a·d·(b·d·(c·d))¯ = a·(b·c)¯·d¯"
by simp;
from A2 have "a·(b·c)·d = a·(b·(c·d))"
using group_op_closed group_oper_assoc by simp;
also from A1 A2 have "… = a·(b·(d·c))"
using IsCommutative_def group_op_closed by simp;
also from A2 have "… = a·b·d·c"
using group_op_closed group_oper_assoc by simp;
finally show "a·(b·c)·d = a·b·d·c" by simp;
qed;

text{*Some other rearrangements with four elements.*}

lemma (in group0) group0_4_L8:
assumes A1: "P {is commutative on} G"
and A2: "a∈G" "b∈G" "c∈G" "d∈G"
shows
"a·(b·c)¯ = (a·d¯·c¯)·(d·b¯)"
"a·b·(c·d) = c·a·(b·d)"
"a·b·(c·d) = a·c·(b·d)"
"a·(b·c¯)·d = a·b·d·c¯"
"(a·b)·(c·d)¯·(b·d¯)¯ = a·c¯"
proof -
from A2 have T:
"b·c ∈ G" "a·b ∈ G" "d¯ ∈ G" "b¯∈G" "c¯∈G"
"d¯·b ∈ G" "c¯·d ∈ G" "(b·c)¯ ∈ G"
"a·b ∈ G" "(c·d)¯ ∈ G" "(b·d¯)¯ ∈ G" "d·b¯ ∈ G"
using group_op_closed inverse_in_group
by auto;
from A2 have "a·(b·c)¯ = a·c¯·b¯" using group0_2_L14A by blast;
moreover from A2 have "a·c¯ = (a·d¯)·(d·c¯)" using group0_2_L14A
by blast;
ultimately have "a·(b·c)¯ = (a·d¯)·(d·c¯)·b¯" by simp;
with A1 A2 T have "a·(b·c)¯= a·d¯·(c¯·d)·b¯"
using IsCommutative_def by simp;
with A2 T show "a·(b·c)¯ = (a·d¯·c¯)·(d·b¯)"
using group_op_closed group_oper_assoc by simp;
from A2 T have "a·b·(c·d) = a·b·c·d"
using group_oper_assoc by simp;
also have "a·b·c·d = c·a·b·d"
proof -;
from A1 A2 have "a·b·c·d = c·(a·b)·d"
using IsCommutative_def group_op_closed
by simp;
also from A2 have "… = c·a·b·d"
using group_op_closed group_oper_assoc
by simp;
finally show ?thesis by simp;
qed;
also from A2 have "c·a·b·d = c·a·(b·d)"
using group_op_closed group_oper_assoc
by simp;
finally show "a·b·(c·d) = c·a·(b·d)" by simp;
with A1 A2 show "a·b·(c·d) = a·c·(b·d)"
using IsCommutative_def by simp
from A1 A2 T show "a·(b·c¯)·d = a·b·d·c¯"
using group0_4_L7 by simp;
from T have "(a·b)·(c·d)¯·(b·d¯)¯ = (a·b)·((c·d)¯·(b·d¯)¯)"
using group_oper_assoc by simp;
also from A1 A2 T have "… = (a·b)·(c¯·d¯·(d·b¯))"
using group_inv_of_two group0_2_L12 IsCommutative_def
by simp;
also from T have "… = (a·b)·(c¯·(d¯·(d·b¯)))"
using group_oper_assoc by simp;
also from A1 A2 T have "… = a·c¯"
using group_oper_assoc group0_2_L6 group0_2_L2 IsCommutative_def
inv_cancel_two by simp;
finally show "(a·b)·(c·d)¯·(b·d¯)¯ = a·c¯"
by simp;
qed;


text{*Some other rearrangements with four elements.*}

lemma (in group0) group0_4_L8A:
assumes A1: "P {is commutative on} G"
and A2: "a∈G" "b∈G" "c∈G" "d∈G"
shows
"a·b¯·(c·d¯) = a·c·(b¯·d¯)"
"a·b¯·(c·d¯) = a·c·b¯·d¯"
proof -
from A2 have
T: "a∈G" "b¯ ∈ G" "c∈G" "d¯ ∈ G"
using inverse_in_group by auto
with A1 show "a·b¯·(c·d¯) = a·c·(b¯·d¯)"
by (rule group0_4_L8);
with A2 T show "a·b¯·(c·d¯) = a·c·b¯·d¯"
using group_op_closed group_oper_assoc
by simp;
qed;

text{*Some rearrangements with an equation.*}

lemma (in group0) group0_4_L9:
assumes A1: "P {is commutative on} G"
and A2: "a∈G" "b∈G" "c∈G" "d∈G"
and A3: "a = b·c¯·d¯"
shows
"d = b·a¯·c¯"
"d = a¯·b·c¯"
"b = a·d·c"
proof -
from A2 have T:
"a¯ ∈ G" "c¯ ∈ G" "d¯ ∈ G" "b·c¯ ∈ G"
using group_op_closed inverse_in_group
by auto
with A2 A3 have "a·(d¯)¯ = b·c¯"
using group0_2_L18 by simp;
with A2 have "b·c¯ = a·d"
using group_inv_of_inv by simp;
with A2 T have I: "a¯·(b·c¯) = d"
using group0_2_L18 by simp;
with A1 A2 T show
"d = b·a¯·c¯"
"d = a¯·b·c¯"
using group_oper_assoc IsCommutative_def by auto;
from A3 have "a·d·c = (b·c¯·d¯)·d·c" by simp;
also from A2 T have "… = b·c¯·(d¯·d)·c"
using group_oper_assoc by simp;
also from A2 T have "… = b·c¯·c"
using group0_2_L6 group0_2_L2 by simp;
also from A2 T have "… = b·(c¯·c)"
using group_oper_assoc by simp;
also from A2 have "… = b"
using group0_2_L6 group0_2_L2 by simp
finally have "a·d·c = b" by simp;
thus "b = a·d·c" by simp;
qed;

end