(*

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,

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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