section ‹More on ordered groups›
theory OrderedGroup_ZF_1 imports OrderedGroup_ZF
begin
text‹In this theory we continue the ‹OrderedGroup_ZF›
theory development.›
subsection‹Absolute value and the triangle inequality›
text‹The goal of this section is to prove the triangle inequality for
ordered groups.›
text‹Absolute value maps $G$ into $G$.›
lemma (in group3) OrderedGroup_ZF_3_L1:
shows "AbsoluteValue(G,P,r) : G→G"
proof -
let ?f = "id(G⇧+)"
let ?g = "restrict(GroupInv(G,P),G-G⇧+)"
have "?f : G⇧+→G⇧+" using id_type by simp
then have "?f : G⇧+→G" using OrderedGroup_ZF_1_L4E fun_weaken_type
by blast
moreover have "?g : G-G⇧+→G"
proof -
from ordGroupAssum have "GroupInv(G,P) : G→G"
using IsAnOrdGroup_def group0_2_T2 by simp
moreover have "G-G⇧+ ⊆ G" by auto
ultimately show ?thesis using restrict_type2 by simp
qed
moreover have "G⇧+∩(G-G⇧+) = 0" by blast
ultimately have "?f ∪ ?g : G⇧+∪(G-G⇧+)→G∪G"
by (rule fun_disjoint_Un)
moreover have "G⇧+∪(G-G⇧+) = G" using OrderedGroup_ZF_1_L4E
by auto
ultimately show "AbsoluteValue(G,P,r) : G→G"
using AbsoluteValue_def by simp
qed
text‹If $a\in G^+$, then $|a| = a$.›
lemma (in group3) OrderedGroup_ZF_3_L2:
assumes A1: "a∈G⇧+" shows "|a| = a"
proof -
from ordGroupAssum have "GroupInv(G,P) : G→G"
using IsAnOrdGroup_def group0_2_T2 by simp
with A1 show ?thesis using
func1_1_L1 OrderedGroup_ZF_1_L4E fun_disjoint_apply1
AbsoluteValue_def id_conv by simp
qed
text‹The absolute value of the unit is the unit. In the
additive totation that would be $|0| = 0$.›
lemma (in group3) OrderedGroup_ZF_3_L2A:
shows "|𝟭| = 𝟭" using OrderedGroup_ZF_1_L3A OrderedGroup_ZF_3_L2
by simp
text‹If $a$ is positive, then $|a| = a$.›
lemma (in group3) OrderedGroup_ZF_3_L2B:
assumes "a∈G⇩+" shows "|a| = a"
using assms PositiveSet_def Nonnegative_def OrderedGroup_ZF_3_L2
by auto
text‹If $a\in G\setminus G^+$, then $|a| = a^{-1}$.›
lemma (in group3) OrderedGroup_ZF_3_L3:
assumes A1: "a ∈ G-G⇧+" shows "|a| = a¯"
proof -
have "domain(id(G⇧+)) = G⇧+"
using id_type func1_1_L1 by auto
with A1 show ?thesis using fun_disjoint_apply2 AbsoluteValue_def
restrict by simp
qed
text‹For elements that not greater than the unit, the absolute value is
the inverse.›
lemma (in group3) OrderedGroup_ZF_3_L3A:
assumes A1: "a\<lsq>𝟭"
shows "|a| = a¯"
proof -
{ assume "a=𝟭" then have "|a| = a¯"
using OrderedGroup_ZF_3_L2A OrderedGroup_ZF_1_L1 group0.group_inv_of_one
by simp }
moreover
{ assume "a≠𝟭"
with A1 have "|a| = a¯" using OrderedGroup_ZF_1_L4C OrderedGroup_ZF_3_L3
by simp }
ultimately show "|a| = a¯" by blast
qed
text‹In linearly ordered groups the absolute value of any element
is in $G^+$.›
lemma (in group3) OrderedGroup_ZF_3_L3B:
assumes A1: "r {is total on} G" and A2: "a∈G"
shows "|a| ∈ G⇧+"
proof -
{ assume "a ∈ G⇧+" then have "|a| ∈ G⇧+"
using OrderedGroup_ZF_3_L2 by simp }
moreover
{ assume "a ∉ G⇧+"
with A1 A2 have "|a| ∈ G⇧+" using OrderedGroup_ZF_3_L3
OrderedGroup_ZF_1_L6 by simp }
ultimately show "|a| ∈ G⇧+" by blast
qed
text‹For linearly ordered groups (where the order is total), the absolute
value maps the group into the positive set.›
lemma (in group3) OrderedGroup_ZF_3_L3C:
assumes A1: "r {is total on} G"
shows "AbsoluteValue(G,P,r) : G→G⇧+"
proof-
have "AbsoluteValue(G,P,r) : G→G" using OrderedGroup_ZF_3_L1
by simp
moreover from A1 have T2:
"∀g∈G. AbsoluteValue(G,P,r)`(g) ∈ G⇧+"
using OrderedGroup_ZF_3_L3B by simp
ultimately show ?thesis by (rule func1_1_L1A)
qed
text‹If the absolute value is the unit, then the elemnent is the unit.›
lemma (in group3) OrderedGroup_ZF_3_L3D:
assumes A1: "a∈G" and A2: "|a| = 𝟭"
shows "a = 𝟭"
proof -
{ assume "a ∈ G⇧+"
with A2 have "a = 𝟭" using OrderedGroup_ZF_3_L2 by simp }
moreover
{ assume "a ∉ G⇧+"
with A1 A2 have "a = 𝟭" using
OrderedGroup_ZF_3_L3 OrderedGroup_ZF_1_L1 group0.group0_2_L8A
by auto }
ultimately show "a = 𝟭" by blast
qed
text‹In linearly ordered groups the unit is not greater than the absolute
value of any element.›
lemma (in group3) OrderedGroup_ZF_3_L3E:
assumes "r {is total on} G" and "a∈G"
shows "𝟭 \<lsq> |a|"
using assms OrderedGroup_ZF_3_L3B OrderedGroup_ZF_1_L2 by simp
text‹If $b$ is greater than both $a$ and $a^{-1}$, then $b$ is greater than
$|a|$.›
lemma (in group3) OrderedGroup_ZF_3_L4:
assumes A1: "a\<lsq>b" and A2: "a¯\<lsq> b"
shows "|a|\<lsq> b"
proof -
{ assume "a∈G⇧+"
with A1 have "|a|\<lsq> b" using OrderedGroup_ZF_3_L2 by simp }
moreover
{ assume "a∉G⇧+"
with A1 A2 have "|a|\<lsq> b"
using OrderedGroup_ZF_1_L4 OrderedGroup_ZF_3_L3 by simp }
ultimately show "|a|\<lsq> b" by blast
qed
text‹In linearly ordered groups $a\leq |a|$.›
lemma (in group3) OrderedGroup_ZF_3_L5:
assumes A1: "r {is total on} G" and A2: "a∈G"
shows "a \<lsq> |a|"
proof -
{ assume "a ∈ G⇧+"
with A2 have "a \<lsq> |a|"
using OrderedGroup_ZF_3_L2 OrderedGroup_ZF_1_L3 by simp }
moreover
{ assume "a ∉ G⇧+"
with A1 A2 have "a \<lsq> |a|"
using OrderedGroup_ZF_3_L3B OrderedGroup_ZF_1_L4B by simp }
ultimately show "a \<lsq> |a|" by blast
qed
text‹$a^{-1}\leq |a|$ (in additive notation it would be $-a\leq |a|$.›
lemma (in group3) OrderedGroup_ZF_3_L6:
assumes A1: "a∈G" shows "a¯ \<lsq> |a|"
proof -
{ assume "a ∈ G⇧+"
then have T1: "𝟭\<lsq>a" and T2: "|a| = a" using OrderedGroup_ZF_1_L2
OrderedGroup_ZF_3_L2 by auto
then have "a¯\<lsq>𝟭¯" using OrderedGroup_ZF_1_L5 by simp
then have T3: "a¯\<lsq>𝟭"
using OrderedGroup_ZF_1_L1 group0.group_inv_of_one by simp
from T3 T1 have "a¯\<lsq>a" by (rule Group_order_transitive)
with T2 have "a¯ \<lsq> |a|" by simp }
moreover
{ assume A2: "a ∉ G⇧+"
from A1 have "|a| ∈ G"
using OrderedGroup_ZF_3_L1 apply_funtype by auto
with ordGroupAssum have "|a| \<lsq> |a|"
using IsAnOrdGroup_def IsPartOrder_def refl_def by simp
with A1 A2 have "a¯ \<lsq> |a|" using OrderedGroup_ZF_3_L3 by simp }
ultimately show "a¯ \<lsq> |a|" by blast
qed
text‹Some inequalities about the product of two elements of a linearly
ordered group and its absolute value.›
lemma (in group3) OrderedGroup_ZF_3_L6A:
assumes "r {is total on} G" and "a∈G" "b∈G"
shows
"a⋅b \<lsq>|a|⋅|b|"
"a⋅b¯ \<lsq>|a|⋅|b|"
"a¯⋅b \<lsq>|a|⋅|b|"
"a¯⋅b¯ \<lsq>|a|⋅|b|"
using assms OrderedGroup_ZF_3_L5 OrderedGroup_ZF_3_L6
OrderedGroup_ZF_1_L5B by auto
text‹$|a^{-1}|\leq |a|$.›
lemma (in group3) OrderedGroup_ZF_3_L7:
assumes "r {is total on} G" and "a∈G"
shows "|a¯|\<lsq>|a|"
using assms OrderedGroup_ZF_3_L5 OrderedGroup_ZF_1_L1 group0.group_inv_of_inv
OrderedGroup_ZF_3_L6 OrderedGroup_ZF_3_L4 by simp
text‹$|a^{-1}| = |a|$.›
lemma (in group3) OrderedGroup_ZF_3_L7A:
assumes A1: "r {is total on} G" and A2: "a∈G"
shows "|a¯| = |a|"
proof -
from A2 have "a¯∈G" using OrderedGroup_ZF_1_L1 group0.inverse_in_group
by simp
with A1 have "|(a¯)¯| \<lsq> |a¯|" using OrderedGroup_ZF_3_L7 by simp
with A1 A2 have "|a¯| \<lsq> |a|" "|a| \<lsq> |a¯|"
using OrderedGroup_ZF_1_L1 group0.group_inv_of_inv OrderedGroup_ZF_3_L7
by auto
then show ?thesis by (rule group_order_antisym)
qed
text‹$|a\cdot b^{-1}| = |b\cdot a^{-1}|$. It doesn't look so strange in the
additive notation: $|a-b| = |b-a|$.›
lemma (in group3) OrderedGroup_ZF_3_L7B:
assumes A1: "r {is total on} G" and A2: "a∈G" "b∈G"
shows "|a⋅b¯| = |b⋅a¯|"
proof -
from A1 A2 have "|(a⋅b¯)¯| = |a⋅b¯|" using
OrderedGroup_ZF_1_L1 group0.inverse_in_group group0.group0_2_L1
monoid0.group0_1_L1 OrderedGroup_ZF_3_L7A by simp
moreover from A2 have "(a⋅b¯)¯ = b⋅a¯"
using OrderedGroup_ZF_1_L1 group0.group0_2_L12 by simp
ultimately show ?thesis by simp
qed
text‹Triangle inequality for linearly ordered abelian groups.
It would be nice to drop commutativity or give an example that shows we
can't do that.›
theorem (in group3) OrdGroup_triangle_ineq:
assumes A1: "P {is commutative on} G"
and A2: "r {is total on} G" and A3: "a∈G" "b∈G"
shows "|a⋅b| \<lsq> |a|⋅|b|"
proof -
from A1 A2 A3 have
"a \<lsq> |a|" "b \<lsq> |b|" "a¯ \<lsq> |a|" "b¯ \<lsq> |b|"
using OrderedGroup_ZF_3_L5 OrderedGroup_ZF_3_L6 by auto
then have "a⋅b \<lsq> |a|⋅|b|" "a¯⋅b¯ \<lsq> |a|⋅|b|"
using OrderedGroup_ZF_1_L5B by auto
with A1 A3 show "|a⋅b| \<lsq> |a|⋅|b|"
using OrderedGroup_ZF_1_L1 group0.group_inv_of_two IsCommutative_def
OrderedGroup_ZF_3_L4 by simp
qed
text‹We can multiply the sides of an inequality with absolute value.›
lemma (in group3) OrderedGroup_ZF_3_L7C:
assumes A1: "P {is commutative on} G"
and A2: "r {is total on} G" and A3: "a∈G" "b∈G"
and A4: "|a| \<lsq> c" "|b| \<lsq> d"
shows "|a⋅b| \<lsq> c⋅d"
proof -
from A1 A2 A3 A4 have "|a⋅b| \<lsq> |a|⋅|b|"
using OrderedGroup_ZF_1_L4 OrdGroup_triangle_ineq
by simp
moreover from A4 have "|a|⋅|b| \<lsq> c⋅d"
using OrderedGroup_ZF_1_L5B by simp
ultimately show ?thesis by (rule Group_order_transitive)
qed
text‹A version of the ‹OrderedGroup_ZF_3_L7C›
but with multiplying by the inverse.›
lemma (in group3) OrderedGroup_ZF_3_L7CA:
assumes "P {is commutative on} G"
and "r {is total on} G" and "a∈G" "b∈G"
and "|a| \<lsq> c" "|b| \<lsq> d"
shows "|a⋅b¯| \<lsq> c⋅d"
using assms OrderedGroup_ZF_1_L1 group0.inverse_in_group
OrderedGroup_ZF_3_L7A OrderedGroup_ZF_3_L7C by simp
text‹Triangle inequality with three integers.›
lemma (in group3) OrdGroup_triangle_ineq3:
assumes A1: "P {is commutative on} G"
and A2: "r {is total on} G" and A3: "a∈G" "b∈G" "c∈G"
shows "|a⋅b⋅c| \<lsq> |a|⋅|b|⋅|c|"
proof -
from A3 have T: "a⋅b ∈ G" "|c| ∈ G"
using OrderedGroup_ZF_1_L1 group0.group_op_closed
OrderedGroup_ZF_3_L1 apply_funtype by auto
with A1 A2 A3 have "|a⋅b⋅c| \<lsq> |a⋅b|⋅|c|"
using OrdGroup_triangle_ineq by simp
moreover from ordGroupAssum A1 A2 A3 T have
"|a⋅b|⋅|c| \<lsq> |a|⋅|b|⋅|c|"
using OrdGroup_triangle_ineq IsAnOrdGroup_def by simp
ultimately show "|a⋅b⋅c| \<lsq> |a|⋅|b|⋅|c|"
by (rule Group_order_transitive)
qed
text‹Some variants of the triangle inequality.›
lemma (in group3) OrderedGroup_ZF_3_L7D:
assumes A1: "P {is commutative on} G"
and A2: "r {is total on} G" and A3: "a∈G" "b∈G"
and A4: "|a⋅b¯| \<lsq> c"
shows
"|a| \<lsq> c⋅|b|"
"|a| \<lsq> |b|⋅c"
"c¯⋅a \<lsq> b"
"a⋅c¯ \<lsq> b"
"a \<lsq> b⋅c"
proof -
from A3 A4 have
T: "a⋅b¯ ∈ G" "|b| ∈ G" "c∈G" "c¯ ∈ G"
using OrderedGroup_ZF_1_L1
group0.inverse_in_group group0.group0_2_L1 monoid0.group0_1_L1
OrderedGroup_ZF_3_L1 apply_funtype OrderedGroup_ZF_1_L4
by auto
from A3 have "|a| = |a⋅b¯⋅b|"
using OrderedGroup_ZF_1_L1 group0.inv_cancel_two
by simp
with A1 A2 A3 T have "|a| \<lsq> |a⋅b¯|⋅|b|"
using OrdGroup_triangle_ineq by simp
with T A4 show "|a| \<lsq> c⋅|b|" using OrderedGroup_ZF_1_L5C
by blast
with T A1 show "|a| \<lsq> |b|⋅c"
using IsCommutative_def by simp
from A2 T have "a⋅b¯ \<lsq> |a⋅b¯|"
using OrderedGroup_ZF_3_L5 by simp
moreover note A4
ultimately have I: "a⋅b¯ \<lsq> c"
by (rule Group_order_transitive)
with A3 show "c¯⋅a \<lsq> b"
using OrderedGroup_ZF_1_L5H by simp
with A1 A3 T show "a⋅c¯ \<lsq> b"
using IsCommutative_def by simp
from A1 A3 T I show "a \<lsq> b⋅c"
using OrderedGroup_ZF_1_L5H IsCommutative_def
by auto
qed
text‹Some more variants of the triangle inequality.›
lemma (in group3) OrderedGroup_ZF_3_L7E:
assumes A1: "P {is commutative on} G"
and A2: "r {is total on} G" and A3: "a∈G" "b∈G"
and A4: "|a⋅b¯| \<lsq> c"
shows "b⋅c¯ \<lsq> a"
proof -
from A3 have "a⋅b¯ ∈ G"
using OrderedGroup_ZF_1_L1
group0.inverse_in_group group0.group_op_closed
by auto
with A2 have "|(a⋅b¯)¯| = |a⋅b¯|"
using OrderedGroup_ZF_3_L7A by simp
moreover from A3 have "(a⋅b¯)¯ = b⋅a¯"
using OrderedGroup_ZF_1_L1 group0.group0_2_L12
by simp
ultimately have "|b⋅a¯| = |a⋅b¯|"
by simp
with A1 A2 A3 A4 show "b⋅c¯ \<lsq> a"
using OrderedGroup_ZF_3_L7D by simp
qed
text‹An application of the triangle inequality with four group
elements.›
lemma (in group3) OrderedGroup_ZF_3_L7F:
assumes A1: "P {is commutative on} G"
and A2: "r {is total on} G" and
A3: "a∈G" "b∈G" "c∈G" "d∈G"
shows "|a⋅c¯| \<lsq> |a⋅b|⋅|c⋅d|⋅|b⋅d¯|"
proof -
from A3 have T:
"a⋅c¯ ∈ G" "a⋅b ∈ G" "c⋅d ∈ G" "b⋅d¯ ∈ G"
"(c⋅d)¯ ∈ G" "(b⋅d¯)¯ ∈ G"
using OrderedGroup_ZF_1_L1
group0.inverse_in_group group0.group_op_closed
by auto
with A1 A2 have "|(a⋅b)⋅(c⋅d)¯⋅(b⋅d¯)¯| \<lsq> |a⋅b|⋅|(c⋅d)¯|⋅|(b⋅d¯)¯|"
using OrdGroup_triangle_ineq3 by simp
moreover from A2 T have "|(c⋅d)¯| =|c⋅d|" and "|(b⋅d¯)¯| = |b⋅d¯|"
using OrderedGroup_ZF_3_L7A by auto
moreover from A1 A3 have "(a⋅b)⋅(c⋅d)¯⋅(b⋅d¯)¯ = a⋅c¯"
using OrderedGroup_ZF_1_L1 group0.group0_4_L8
by simp
ultimately show "|a⋅c¯| \<lsq> |a⋅b|⋅|c⋅d|⋅|b⋅d¯|"
by simp
qed
text‹$|a|\leq L$ implies $L^{-1} \leq a$
(it would be $-L \leq a$ in the additive notation).›
lemma (in group3) OrderedGroup_ZF_3_L8:
assumes A1: "a∈G" and A2: "|a|\<lsq>L"
shows
"L¯\<lsq>a"
proof -
from A1 have I: "a¯ \<lsq> |a|" using OrderedGroup_ZF_3_L6 by simp
from I A2 have "a¯ \<lsq> L" by (rule Group_order_transitive)
then have "L¯\<lsq>(a¯)¯" using OrderedGroup_ZF_1_L5 by simp
with A1 show "L¯\<lsq>a" using OrderedGroup_ZF_1_L1 group0.group_inv_of_inv
by simp
qed
text‹In linearly ordered groups $|a|\leq L$ implies $a \leq L$
(it would be $a \leq L$ in the additive notation).›
lemma (in group3) OrderedGroup_ZF_3_L8A:
assumes A1: "r {is total on} G"
and A2: "a∈G" and A3: "|a|\<lsq>L"
shows
"a\<lsq>L"
"𝟭\<lsq>L"
proof -
from A1 A2 have I: "a \<lsq> |a|" using OrderedGroup_ZF_3_L5 by simp
from I A3 show "a\<lsq>L" by (rule Group_order_transitive)
from A1 A2 A3 have "𝟭 \<lsq> |a|" "|a|\<lsq>L"
using OrderedGroup_ZF_3_L3B OrderedGroup_ZF_1_L2 by auto
then show "𝟭\<lsq>L" by (rule Group_order_transitive)
qed
text‹A somewhat generalized version of the above lemma.›
lemma (in group3) OrderedGroup_ZF_3_L8B:
assumes A1: "a∈G" and A2: "|a|\<lsq>L" and A3: "𝟭\<lsq>c"
shows "(L⋅c)¯ \<lsq> a"
proof -
from A1 A2 A3 have "c¯⋅L¯ \<lsq> 𝟭⋅a"
using OrderedGroup_ZF_3_L8 OrderedGroup_ZF_1_L5AB
OrderedGroup_ZF_1_L5B by simp
with A1 A2 A3 show "(L⋅c)¯ \<lsq> a"
using OrderedGroup_ZF_1_L4 OrderedGroup_ZF_1_L1
group0.group_inv_of_two group0.group0_2_L2
by simp
qed
text‹If $b$ is between $a$ and $a\cdot c$, then $b\cdot a^{-1}\leq c$.›
lemma (in group3) OrderedGroup_ZF_3_L8C:
assumes A1: "a\<lsq>b" and A2: "c∈G" and A3: "b\<lsq>c⋅a"
shows "|b⋅a¯| \<lsq> c"
proof -
from A1 A2 A3 have "b⋅a¯ \<lsq> c"
using OrderedGroup_ZF_1_L9C OrderedGroup_ZF_1_L4
by simp
moreover have "(b⋅a¯)¯ \<lsq> c"
proof -
from A1 have T: "a∈G" "b∈G"
using OrderedGroup_ZF_1_L4 by auto
with A1 have "a⋅b¯ \<lsq> 𝟭"
using OrderedGroup_ZF_1_L9 by blast
moreover
from A1 A3 have "a\<lsq>c⋅a"
by (rule Group_order_transitive)
with ordGroupAssum T have "a⋅a¯ \<lsq> c⋅a⋅a¯"
using OrderedGroup_ZF_1_L1 group0.inverse_in_group
IsAnOrdGroup_def by simp
with T A2 have "𝟭 \<lsq> c"
using OrderedGroup_ZF_1_L1
group0.group0_2_L6 group0.inv_cancel_two
by simp
ultimately have "a⋅b¯ \<lsq> c"
by (rule Group_order_transitive)
with T show "(b⋅a¯)¯ \<lsq> c"
using OrderedGroup_ZF_1_L1 group0.group0_2_L12
by simp
qed
ultimately show "|b⋅a¯| \<lsq> c"
using OrderedGroup_ZF_3_L4 by simp
qed
text‹For linearly ordered groups if the absolute values of elements in a set
are bounded, then the set is bounded.›
lemma (in group3) OrderedGroup_ZF_3_L9:
assumes A1: "r {is total on} G"
and A2: "A⊆G" and A3: "∀a∈A. |a| \<lsq> L"
shows "IsBounded(A,r)"
proof -
from A1 A2 A3 have
"∀a∈A. a\<lsq>L" "∀a∈A. L¯\<lsq>a"
using OrderedGroup_ZF_3_L8 OrderedGroup_ZF_3_L8A by auto
then show "IsBounded(A,r)" using
IsBoundedAbove_def IsBoundedBelow_def IsBounded_def
by auto
qed
text‹A slightly more general version of the previous lemma, stating the same
fact for a set defined by separation.›
lemma (in group3) OrderedGroup_ZF_3_L9A:
assumes A1: "r {is total on} G"
and A2: "∀x∈X. b(x)∈G ∧ |b(x)|\<lsq>L"
shows "IsBounded({b(x). x∈X},r)"
proof -
from A2 have "{b(x). x∈X} ⊆ G" "∀a∈{b(x). x∈X}. |a| \<lsq> L"
by auto
with A1 show ?thesis using OrderedGroup_ZF_3_L9 by blast
qed
text‹A special form of the previous lemma stating a similar fact for an
image of a set by a function with values in a linearly ordered group.›
lemma (in group3) OrderedGroup_ZF_3_L9B:
assumes A1: "r {is total on} G"
and A2: "f:X→G" and A3: "A⊆X"
and A4: "∀x∈A. |f`(x)| \<lsq> L"
shows "IsBounded(f``(A),r)"
proof -
from A2 A3 A4 have "∀x∈A. f`(x) ∈ G ∧ |f`(x)| \<lsq> L"
using apply_funtype by auto
with A1 have "IsBounded({f`(x). x∈A},r)"
by (rule OrderedGroup_ZF_3_L9A)
with A2 A3 show "IsBounded(f``(A),r)"
using func_imagedef by simp
qed
text‹For linearly ordered groups if $l\leq a\leq u$ then
$|a|$ is smaller than the greater of $|l|,|u|$.›
lemma (in group3) OrderedGroup_ZF_3_L10:
assumes A1: "r {is total on} G"
and A2: "l\<lsq>a" "a\<lsq>u"
shows
"|a| \<lsq> GreaterOf(r,|l|,|u|)"
proof -
from A2 have T1: "|l| ∈ G" "|a| ∈ G" "|u| ∈ G"
using OrderedGroup_ZF_1_L4 OrderedGroup_ZF_3_L1 apply_funtype
by auto
{ assume A3: "a∈G⇧+"
with A2 have "𝟭\<lsq>a" "a\<lsq>u"
using OrderedGroup_ZF_1_L2 by auto
then have "𝟭\<lsq>u" by (rule Group_order_transitive)
with A2 A3 have "|a|\<lsq>|u|"
using OrderedGroup_ZF_1_L2 OrderedGroup_ZF_3_L2 by simp
moreover from A1 T1 have "|u| \<lsq> GreaterOf(r,|l|,|u|)"
using Order_ZF_3_L2 by simp
ultimately have "|a| \<lsq> GreaterOf(r,|l|,|u|)"
by (rule Group_order_transitive) }
moreover
{ assume A4: "a∉G⇧+"
with A2 have T2:
"l∈G" "|l| ∈ G" "|a| ∈ G" "|u| ∈ G" "a ∈ G-G⇧+"
using OrderedGroup_ZF_1_L4 OrderedGroup_ZF_3_L1 apply_funtype
by auto
with A2 have "l ∈ G-G⇧+" using OrderedGroup_ZF_1_L4D by fast
with T2 A2 have "|a| \<lsq> |l|"
using OrderedGroup_ZF_3_L3 OrderedGroup_ZF_1_L5
by simp
moreover from A1 T2 have "|l| \<lsq> GreaterOf(r,|l|,|u|)"
using Order_ZF_3_L2 by simp
ultimately have "|a| \<lsq> GreaterOf(r,|l|,|u|)"
by (rule Group_order_transitive) }
ultimately show ?thesis by blast
qed
text‹For linearly ordered groups if a set is bounded then the absolute
values are bounded.›
lemma (in group3) OrderedGroup_ZF_3_L10A:
assumes A1: "r {is total on} G"
and A2: "IsBounded(A,r)"
shows "∃L. ∀a∈A. |a| \<lsq> L"
proof -
{ assume "A = 0" then have ?thesis by auto }
moreover
{ assume A3: "A≠0"
with A2 have "∃u. ∀g∈A. g\<lsq>u" and "∃l.∀g∈A. l\<lsq>g"
using IsBounded_def IsBoundedAbove_def IsBoundedBelow_def
by auto
then obtain u l where "∀g∈A. l\<lsq>g ∧ g\<lsq>u"
by auto
with A1 have "∀a∈A. |a| \<lsq> GreaterOf(r,|l|,|u|)"
using OrderedGroup_ZF_3_L10 by simp
then have ?thesis by auto }
ultimately show ?thesis by blast
qed
text‹A slightly more general version of the previous lemma, stating the same
fact for a set defined by separation.›
lemma (in group3) OrderedGroup_ZF_3_L11:
assumes "r {is total on} G"
and "IsBounded({b(x).x∈X},r)"
shows "∃L. ∀x∈X. |b(x)| \<lsq> L"
using assms OrderedGroup_ZF_3_L10A by blast
text‹Absolute values of elements of a finite image of a nonempty set are
bounded by an element of the group.›
lemma (in group3) OrderedGroup_ZF_3_L11A:
assumes A1: "r {is total on} G"
and A2: "X≠0" and A3: "{b(x). x∈X} ∈ Fin(G)"
shows "∃L∈G. ∀x∈X. |b(x)| \<lsq> L"
proof -
from A1 A3 have "∃L. ∀x∈X. |b(x)| \<lsq> L"
using ord_group_fin_bounded OrderedGroup_ZF_3_L11
by simp
then obtain L where I: "∀x∈X. |b(x)| \<lsq> L"
using OrderedGroup_ZF_3_L11 by auto
from A2 obtain x where "x∈X" by auto
with I show ?thesis using OrderedGroup_ZF_1_L4
by blast
qed
text‹In totally oredered groups the absolute value of a
nonunit element is in ‹G⇩+›.›
lemma (in group3) OrderedGroup_ZF_3_L12:
assumes A1: "r {is total on} G"
and A2: "a∈G" and A3: "a≠𝟭"
shows "|a| ∈ G⇩+"
proof -
from A1 A2 have "|a| ∈ G" "𝟭 \<lsq> |a|"
using OrderedGroup_ZF_3_L1 apply_funtype
OrderedGroup_ZF_3_L3B OrderedGroup_ZF_1_L2
by auto
moreover from A2 A3 have "|a| ≠ 𝟭"
using OrderedGroup_ZF_3_L3D by auto
ultimately show "|a| ∈ G⇩+"
using PositiveSet_def by auto
qed
subsection‹Maximum absolute value of a set›
text‹Quite often when considering inequalities we prefer to talk about
the absolute values instead of raw elements of a set. This section formalizes
some material that is useful for that.›
text‹If a set has a maximum and minimum, then the greater of the
absolute value of the maximum and minimum belongs to the image of the set
by the absolute value function.›
lemma (in group3) OrderedGroup_ZF_4_L1:
assumes "A ⊆ G"
and "HasAmaximum(r,A)" "HasAminimum(r,A)"
and "M = GreaterOf(r,|Minimum(r,A)|,|Maximum(r,A)|)"
shows "M ∈ AbsoluteValue(G,P,r)``(A)"
using ordGroupAssum assms IsAnOrdGroup_def IsPartOrder_def
Order_ZF_4_L3 Order_ZF_4_L4 OrderedGroup_ZF_3_L1
func_imagedef GreaterOf_def by auto
text‹If a set has a maximum and minimum, then the greater of the
absolute value of the maximum and minimum bounds absolute values of all
elements of the set.›
lemma (in group3) OrderedGroup_ZF_4_L2:
assumes A1: "r {is total on} G"
and A2: "HasAmaximum(r,A)" "HasAminimum(r,A)"
and A3: "a∈A"
shows "|a|\<lsq> GreaterOf(r,|Minimum(r,A)|,|Maximum(r,A)|)"
proof -
from ordGroupAssum A2 A3 have
"Minimum(r,A)\<lsq> a" "a\<lsq> Maximum(r,A)"
using IsAnOrdGroup_def IsPartOrder_def Order_ZF_4_L3 Order_ZF_4_L4
by auto
with A1 show ?thesis by (rule OrderedGroup_ZF_3_L10)
qed
text‹If a set has a maximum and minimum, then the greater of the
absolute value of the maximum and minimum bounds absolute values of all
elements of the set. In this lemma the absolute values of ekements of a
set are represented as the elements of the image of the set by the absolute
value function.›
lemma (in group3) OrderedGroup_ZF_4_L3:
assumes "r {is total on} G" and "A ⊆ G"
and "HasAmaximum(r,A)" "HasAminimum(r,A)"
and "b ∈ AbsoluteValue(G,P,r)``(A)"
shows "b\<lsq> GreaterOf(r,|Minimum(r,A)|,|Maximum(r,A)|)"
using assms OrderedGroup_ZF_3_L1 func_imagedef OrderedGroup_ZF_4_L2
by auto
text‹If a set has a maximum and minimum, then the set of absolute values
also has a maximum.›
lemma (in group3) OrderedGroup_ZF_4_L4:
assumes A1: "r {is total on} G" and A2: "A ⊆ G"
and A3: "HasAmaximum(r,A)" "HasAminimum(r,A)"
shows "HasAmaximum(r,AbsoluteValue(G,P,r)``(A))"
proof -
let ?M = "GreaterOf(r,|Minimum(r,A)|,|Maximum(r,A)|)"
from A2 A3 have "?M ∈ AbsoluteValue(G,P,r)``(A)"
using OrderedGroup_ZF_4_L1 by simp
moreover from A1 A2 A3 have
"∀b ∈ AbsoluteValue(G,P,r)``(A). b \<lsq> ?M"
using OrderedGroup_ZF_4_L3 by simp
ultimately show ?thesis using HasAmaximum_def by auto
qed
text‹If a set has a maximum and a minimum, then all absolute values are
bounded by the maximum of the set of absolute values.›
lemma (in group3) OrderedGroup_ZF_4_L5:
assumes A1: "r {is total on} G" and A2: "A ⊆ G"
and A3: "HasAmaximum(r,A)" "HasAminimum(r,A)"
and A4: "a∈A"
shows "|a| \<lsq> Maximum(r,AbsoluteValue(G,P,r)``(A))"
proof -
from A2 A4 have "|a| ∈ AbsoluteValue(G,P,r)``(A)"
using OrderedGroup_ZF_3_L1 func_imagedef by auto
with ordGroupAssum A1 A2 A3 show ?thesis using
IsAnOrdGroup_def IsPartOrder_def OrderedGroup_ZF_4_L4
Order_ZF_4_L3 by simp
qed
subsection‹Alternative definitions›
text‹Sometimes it is usful to define the order by prescibing the set
of positive or nonnegative elements. This section deals with two such
definitions. One takes a subset $H$ of $G$ that is closed under the group
operation, $1\notin H$ and for every $a\in H$ we have either $a\in H$ or
$a^{-1}\in H$. Then the order is defined as $a\leq b$ iff $a=b$ or
$a^{-1}b \in H$. For abelian groups this makes a linearly ordered group.
We will refer to order defined this way in the comments as the order
defined by a positive set. The context used in this section is the
‹group0› context defined in ‹Group_ZF› theory. Recall that
‹f› in that context denotes the group operation (unlike in the
previous sections where the group operation was denoted ‹P›.›
text‹The order defined by a positive set is the same as the order defined by
a nonnegative set.›
lemma (in group0) OrderedGroup_ZF_5_L1:
assumes A1: "r = {p ∈ G×G. fst(p) = snd(p) ∨ fst(p)¯⋅snd(p) ∈ H}"
shows "⟨a,b⟩ ∈ r ⟷ a∈G ∧ b∈G ∧ a¯⋅b ∈ H ∪ {𝟭}"
proof
assume "⟨a,b⟩ ∈ r"
with A1 show "a∈G ∧ b∈G ∧ a¯⋅b ∈ H ∪ {𝟭}"
using group0_2_L6 by auto
next assume "a∈G ∧ b∈G ∧ a¯⋅b ∈ H ∪ {𝟭}"
then have "a∈G ∧ b∈G ∧ b=(a¯)¯ ∨ a∈G ∧ b∈G ∧ a¯⋅b ∈ H"
using inverse_in_group group0_2_L9 by auto
with A1 show "⟨a,b⟩ ∈ r" using group_inv_of_inv
by auto
qed
text‹The relation defined by a positive set is antisymmetric.›
lemma (in group0) OrderedGroup_ZF_5_L2:
assumes A1: "r = {p ∈ G×G. fst(p) = snd(p) ∨ fst(p)¯⋅snd(p) ∈ H}"
and A2: "∀a∈G. a≠𝟭 ⟶ (a∈H) Xor (a¯∈H)"
shows "antisym(r)"
proof -
{ fix a b assume A3: "⟨a,b⟩ ∈ r" "⟨b,a⟩ ∈ r"
with A1 have T: "a∈G" "b∈G" by auto
{ assume A4: "a≠b"
with A1 A3 have "a¯⋅b ∈ G" "a¯⋅b ∈ H" "(a¯⋅b)¯ ∈ H"
using inverse_in_group group0_2_L1 monoid0.group0_1_L1 group0_2_L12
by auto
with A2 have "a¯⋅b = 𝟭" using Xor_def by auto
with T A4 have False using group0_2_L11 by auto
} then have "a=b" by auto
} then show "antisym(r)" by (rule antisymI)
qed
text‹The relation defined by a positive set is transitive.›
lemma (in group0) OrderedGroup_ZF_5_L3:
assumes A1: "r = {p ∈ G×G. fst(p) = snd(p) ∨ fst(p)¯⋅snd(p) ∈ H}"
and A2: "H⊆G" "H {is closed under} P"
shows "trans(r)"
proof -
{ fix a b c assume "⟨a,b⟩ ∈ r" "⟨b,c⟩ ∈ r"
with A1 have
"a∈G ∧ b∈G ∧ a¯⋅b ∈ H ∪ {𝟭}"
"b∈G ∧ c∈G ∧ b¯⋅c ∈ H ∪ {𝟭}"
using OrderedGroup_ZF_5_L1 by auto
with A2 have
I: "a∈G" "b∈G" "c∈G"
and "(a¯⋅b)⋅(b¯⋅c) ∈ H ∪ {𝟭}"
using inverse_in_group group0_2_L17 IsOpClosed_def
by auto
moreover from I have "a¯⋅c = (a¯⋅b)⋅(b¯⋅c)"
by (rule group0_2_L14A)
ultimately have "⟨a,c⟩ ∈ G×G" "a¯⋅c ∈ H ∪ {𝟭}"
by auto
with A1 have "⟨a,c⟩ ∈ r" using OrderedGroup_ZF_5_L1
by auto
} then have "∀ a b c. ⟨a, b⟩ ∈ r ∧ ⟨b, c⟩ ∈ r ⟶ ⟨a, c⟩ ∈ r"
by blast
then show "trans(r)" by (rule Fol1_L2)
qed
text‹The relation defined by a positive set is translation invariant.
With our definition this step requires the group to be abelian.›
lemma (in group0) OrderedGroup_ZF_5_L4:
assumes A1: "r = {p ∈ G×G. fst(p) = snd(p) ∨ fst(p)¯⋅snd(p) ∈ H}"
and A2: "P {is commutative on} G"
and A3: "⟨a,b⟩ ∈ r" and A4: "c∈G"
shows "⟨a⋅c,b⋅c⟩ ∈ r ∧ ⟨c⋅a,c⋅b⟩ ∈ r"
proof
from A1 A3 A4 have
I: "a∈G" "b∈G" "a⋅c ∈ G" "b⋅c ∈ G"
and II: "a¯⋅b ∈ H ∪ {𝟭}"
using OrderedGroup_ZF_5_L1 group_op_closed
by auto
with A2 A4 have "(a⋅c)¯⋅(b⋅c) ∈ H ∪ {𝟭}"
using group0_4_L6D by simp
with A1 I show "⟨a⋅c,b⋅c⟩ ∈ r" using OrderedGroup_ZF_5_L1
by auto
with A2 A4 I show "⟨c⋅a,c⋅b⟩ ∈ r"
using IsCommutative_def by simp
qed
text‹If $H\subseteq G$ is closed under the group operation
$1\notin H$ and for every $a\in H$ we have either $a\in H$ or
$a^{-1}\in H$, then the relation "$\leq$" defined by
$a\leq b \Leftrightarrow a^{-1}b \in H$ orders the group $G$.
In such order $H$ may be the set of positive or nonnegative
elements.›
lemma (in group0) OrderedGroup_ZF_5_L5:
assumes A1: "P {is commutative on} G"
and A2: "H⊆G" "H {is closed under} P"
and A3: "∀a∈G. a≠𝟭 ⟶ (a∈H) Xor (a¯∈H)"
and A4: "r = {p ∈ G×G. fst(p) = snd(p) ∨ fst(p)¯⋅snd(p) ∈ H}"
shows
"IsAnOrdGroup(G,P,r)"
"r {is total on} G"
"Nonnegative(G,P,r) = PositiveSet(G,P,r) ∪ {𝟭}"
proof -
from groupAssum A2 A3 A4 have
"IsAgroup(G,P)" "r ⊆ G×G" "IsPartOrder(G,r)"
using refl_def OrderedGroup_ZF_5_L2 OrderedGroup_ZF_5_L3
IsPartOrder_def by auto
moreover from A1 A4 have
"∀g∈G. ∀a b. ⟨ a,b⟩ ∈ r ⟶ ⟨a⋅g,b⋅g⟩ ∈ r ∧ ⟨g⋅a,g⋅b⟩ ∈ r"
using OrderedGroup_ZF_5_L4 by blast
ultimately show "IsAnOrdGroup(G,P,r)"
using IsAnOrdGroup_def by simp
then show "Nonnegative(G,P,r) = PositiveSet(G,P,r) ∪ {𝟭}"
using group3_def group3.OrderedGroup_ZF_1_L24
by simp
{ fix a b
assume T: "a∈G" "b∈G"
then have T1: "a¯⋅b ∈ G"
using inverse_in_group group_op_closed by simp
{ assume "⟨ a,b⟩ ∉ r"
with A4 T have I: "a≠b" and II: "a¯⋅b ∉ H"
by auto
from A3 T T1 I have "(a¯⋅b ∈ H) Xor ((a¯⋅b)¯ ∈ H)"
using group0_2_L11 by auto
with A4 T II have "⟨ b,a⟩ ∈ r"
using Xor_def group0_2_L12 by simp
} then have "⟨ a,b⟩ ∈ r ∨ ⟨ b,a⟩ ∈ r" by auto
} then show "r {is total on} G" using IsTotal_def
by simp
qed
text‹If the set defined as in ‹OrderedGroup_ZF_5_L4› does not
contain the neutral element, then it is the positive set for the resulting
order.›
lemma (in group0) OrderedGroup_ZF_5_L6:
assumes "P {is commutative on} G"
and "H⊆G" and "𝟭 ∉ H"
and "r = {p ∈ G×G. fst(p) = snd(p) ∨ fst(p)¯⋅snd(p) ∈ H}"
shows "PositiveSet(G,P,r) = H"
using assms group_inv_of_one group0_2_L2 PositiveSet_def
by auto
text‹The next definition describes how we construct an order relation
from the prescribed set of positive elements.›
definition
"OrderFromPosSet(G,P,H) ≡
{p ∈ G×G. fst(p) = snd(p) ∨ P`⟨GroupInv(G,P)`(fst(p)),snd(p)⟩ ∈ H }"
text‹The next theorem rephrases lemmas
‹OrderedGroup_ZF_5_L5› and ‹OrderedGroup_ZF_5_L6›
using the definition of the order from the positive set
‹OrderFromPosSet›. To summarize, this is what it says:
Suppose that $H\subseteq G$ is a set closed under that group operation
such that $1\notin H$ and for every nonunit group element $a$ either $a\in H$
or $a^{-1}\in H$. Define the order as $a\leq b$ iff $a=b$ or
$a^{-1}\cdot b \in H$. Then this order makes $G$ into a linearly ordered
group such $H$ is the set of positive elements (and then of course
$H \cup \{1\}$ is the set of nonnegative elements).›
theorem (in group0) Group_ord_by_positive_set:
assumes "P {is commutative on} G"
and "H⊆G" "H {is closed under} P" "𝟭 ∉ H"
and "∀a∈G. a≠𝟭 ⟶ (a∈H) Xor (a¯∈H)"
shows
"IsAnOrdGroup(G,P,OrderFromPosSet(G,P,H))"
"OrderFromPosSet(G,P,H) {is total on} G"
"PositiveSet(G,P,OrderFromPosSet(G,P,H)) = H"
"Nonnegative(G,P,OrderFromPosSet(G,P,H)) = H ∪ {𝟭}"
using assms OrderFromPosSet_def OrderedGroup_ZF_5_L5 OrderedGroup_ZF_5_L6
by auto
subsection‹Odd Extensions›
text‹In this section we verify properties of odd extensions of functions
defined on $G_+$. An odd extension of a function $f: G_+ \rightarrow G$
is a function $f^\circ : G\rightarrow G$ defined by $f^\circ (x) = f(x)$
if $x\in G_+$, $f(1) = 1$ and $f^\circ (x) = (f(x^{-1}))^{-1}$ for $x < 1$.
Such function is the unique odd function that is equal to $f$ when
restricted to $G_+$.›
text‹The next lemma is just to see the definition of the odd extension
in the notation used in the ‹group1› context.›
lemma (in group3) OrderedGroup_ZF_6_L1:
shows "f° = f ∪ {⟨a, (f`(a¯))¯⟩. a ∈ \<sm>G⇩+} ∪ {⟨𝟭,𝟭⟩}"
using OddExtension_def by simp
text‹A technical lemma that states that from a function defined on
‹G⇩+› with values in $G$ we have $(f(a^{-1}))^{-1}\in G$.›
lemma (in group3) OrderedGroup_ZF_6_L2:
assumes "f: G⇩+→G" and "a∈\<sm>G⇩+"
shows
"f`(a¯) ∈ G"
"(f`(a¯))¯ ∈ G"
using assms OrderedGroup_ZF_1_L27 apply_funtype
OrderedGroup_ZF_1_L1 group0.inverse_in_group
by auto
text‹The main theorem about odd extensions. It basically says that the odd
extension of a function is what we want to to be.›
lemma (in group3) odd_ext_props:
assumes A1: "r {is total on} G" and A2: "f: G⇩+→G"
shows
"f° : G → G"
"∀a∈G⇩+. (f°)`(a) = f`(a)"
"∀a∈(\<sm>G⇩+). (f°)`(a) = (f`(a¯))¯"
"(f°)`(𝟭) = 𝟭"
proof -
from A1 A2 have I:
"f: G⇩+→G"
"∀a∈\<sm>G⇩+. (f`(a¯))¯ ∈ G"
"G⇩+∩(\<sm>G⇩+) = 0"
"𝟭 ∉ G⇩+∪(\<sm>G⇩+)"
"f° = f ∪ {⟨a, (f`(a¯))¯⟩. a ∈ \<sm>G⇩+} ∪ {⟨𝟭,𝟭⟩}"
using OrderedGroup_ZF_6_L2 OrdGroup_decomp2 OrderedGroup_ZF_6_L1
by auto
then have "f°: G⇩+ ∪ (\<sm>G⇩+) ∪ {𝟭} →G∪G∪{𝟭}"
by (rule func1_1_L11E)
moreover from A1 have
"G⇩+ ∪ (\<sm>G⇩+) ∪ {𝟭} = G"
"G∪G∪{𝟭} = G"
using OrdGroup_decomp2 OrderedGroup_ZF_1_L1 group0.group0_2_L2
by auto
ultimately show "f° : G → G" by simp
from I show "∀a∈G⇩+. (f°)`(a) = f`(a)"
by (rule func1_1_L11E)
from I show "∀a∈(\<sm>G⇩+). (f°)`(a) = (f`(a¯))¯"
by (rule func1_1_L11E)
from I show "(f°)`(𝟭) = 𝟭"
by (rule func1_1_L11E)
qed
text‹Odd extensions are odd, of course.›
lemma (in group3) oddext_is_odd:
assumes A1: "r {is total on} G" and A2: "f: G⇩+→G"
and A3: "a∈G"
shows "(f°)`(a¯) = ((f°)`(a))¯"
proof -
from A1 A3 have "a∈G⇩+ ∨ a ∈ (\<sm>G⇩+) ∨ a=𝟭"
using OrdGroup_decomp2 by blast
moreover
{ assume "a∈G⇩+"
with A1 A2 have "a¯ ∈ \<sm>G⇩+" and "(f°)`(a) = f`(a)"
using OrderedGroup_ZF_1_L25 odd_ext_props by auto
with A1 A2 have
"(f°)`(a¯) = (f`((a¯)¯))¯" and "(f`(a))¯ = ((f°)`(a))¯"
using odd_ext_props by auto
with A3 have "(f°)`(a¯) = ((f°)`(a))¯"
using OrderedGroup_ZF_1_L1 group0.group_inv_of_inv
by simp }
moreover
{ assume A4: "a ∈ \<sm>G⇩+"
with A1 A2 have "a¯ ∈ G⇩+" and "(f°)`(a) = (f`(a¯))¯"
using OrderedGroup_ZF_1_L27 odd_ext_props
by auto
with A1 A2 A4 have "(f°)`(a¯) = ((f°)`(a))¯"
using odd_ext_props OrderedGroup_ZF_6_L2
OrderedGroup_ZF_1_L1 group0.group_inv_of_inv
by simp }
moreover
{ assume "a = 𝟭"
with A1 A2 have "(f°)`(a¯) = ((f°)`(a))¯"
using OrderedGroup_ZF_1_L1 group0.group_inv_of_one
odd_ext_props by simp
}
ultimately show "(f°)`(a¯) = ((f°)`(a))¯"
by auto
qed
text‹Another way of saying that odd extensions are odd.›
lemma (in group3) oddext_is_odd_alt:
assumes A1: "r {is total on} G" and A2: "f: G⇩+→G"
and A3: "a∈G"
shows "((f°)`(a¯))¯ = (f°)`(a)"
proof -
from A1 A2 have
"f° : G → G"
"∀a∈G. (f°)`(a¯) = ((f°)`(a))¯"
using odd_ext_props oddext_is_odd by auto
then have "∀a∈G. ((f°)`(a¯))¯ = (f°)`(a)"
using OrderedGroup_ZF_1_L1 group0.group0_6_L2 by simp
with A3 show "((f°)`(a¯))¯ = (f°)`(a)" by simp
qed
subsection‹Functions with infinite limits›
text‹In this section we consider functions $f: G\rightarrow G$ with the
property that for $f(x)$ is arbitrarily large for large enough $x$.
More precisely, for every $a\in G$ there exist $b\in G_+$ such that
for every $x\geq b$ we have $f(x)\geq a$. In a sense this means that
$\lim_{x\rightarrow \infty}f(x) = \infty$, hence the title of this section.
We also prove dual statements for functions such that
$\lim_{x\rightarrow -\infty}f(x) = -\infty$.
›
text‹If an image of a set by a function with infinite positive limit
is bounded above, then the set itself is bounded above.›
lemma (in group3) OrderedGroup_ZF_7_L1:
assumes A1: "r {is total on} G" and A2: "G ≠ {𝟭}" and
A3: "f:G→G" and
A4: "∀a∈G.∃b∈G⇩+.∀x. b\<lsq>x ⟶ a \<lsq> f`(x)" and
A5: "A⊆G" and
A6: "IsBoundedAbove(f``(A),r)"
shows "IsBoundedAbove(A,r)"
proof -
{ assume "¬IsBoundedAbove(A,r)"
then have I: "∀u. ∃x∈A. ¬(x\<lsq>u)"
using IsBoundedAbove_def by auto
have "∀a∈G. ∃y∈f``(A). a\<lsq>y"
proof -
{ fix a assume "a∈G"
with A4 obtain b where
II: "b∈G⇩+" and III: "∀x. b\<lsq>x ⟶ a \<lsq> f`(x)"
by auto
from I obtain x where IV: "x∈A" and "¬(x\<lsq>b)"
by auto
with A1 A5 II have
"r {is total on} G"
"x∈G" "b∈G" "¬(x\<lsq>b)"
using PositiveSet_def by auto
with III have "a \<lsq> f`(x)"
using OrderedGroup_ZF_1_L8 by blast
with A3 A5 IV have "∃y∈f``(A). a\<lsq>y"
using func_imagedef by auto
} thus ?thesis by simp
qed
with A1 A2 A6 have False using OrderedGroup_ZF_2_L2A
by simp
} thus ?thesis by auto
qed
text‹If an image of a set defined by separation
by a function with infinite positive limit
is bounded above, then the set itself is bounded above.›
lemma (in group3) OrderedGroup_ZF_7_L2:
assumes A1: "r {is total on} G" and A2: "G ≠ {𝟭}" and
A3: "X≠0" and A4: "f:G→G" and
A5: "∀a∈G.∃b∈G⇩+.∀y. b\<lsq>y ⟶ a \<lsq> f`(y)" and
A6: "∀x∈X. b(x) ∈ G ∧ f`(b(x)) \<lsq> U"
shows "∃u.∀x∈X. b(x) \<lsq> u"
proof -
let ?A = "{b(x). x∈X}"
from A6 have I: "?A⊆G" by auto
moreover note assms
moreover have "IsBoundedAbove(f``(?A),r)"
proof -
from A4 A6 I have "∀z∈f``(?A). ⟨z,U⟩ ∈ r"
using func_imagedef by simp
then show "IsBoundedAbove(f``(?A),r)"
by (rule Order_ZF_3_L10)
qed
ultimately have "IsBoundedAbove(?A,r)" using OrderedGroup_ZF_7_L1
by simp
with A3 have "∃u.∀y∈?A. y \<lsq> u"
using IsBoundedAbove_def by simp
then show "∃u.∀x∈X. b(x) \<lsq> u" by auto
qed
text‹If the image of a set defined by separation
by a function with infinite negative limit
is bounded below, then the set itself is bounded above.
This is dual to ‹OrderedGroup_ZF_7_L2›.›
lemma (in group3) OrderedGroup_ZF_7_L3:
assumes A1: "r {is total on} G" and A2: "G ≠ {𝟭}" and
A3: "X≠0" and A4: "f:G→G" and
A5: "∀a∈G.∃b∈G⇩+.∀y. b\<lsq>y ⟶ f`(y¯) \<lsq> a" and
A6: "∀x∈X. b(x) ∈ G ∧ L \<lsq> f`(b(x))"
shows "∃l.∀x∈X. l \<lsq> b(x)"
proof -
let ?g = "GroupInv(G,P) O f O GroupInv(G,P)"
from ordGroupAssum have I: "GroupInv(G,P) : G→G"
using IsAnOrdGroup_def group0_2_T2 by simp
with A4 have II: "∀x∈G. ?g`(x) = (f`(x¯))¯"
using func1_1_L18 by simp
note A1 A2 A3
moreover from A4 I have "?g : G→G"
using comp_fun by blast
moreover have "∀a∈G.∃b∈G⇩+.∀y. b\<lsq>y ⟶ a \<lsq> ?g`(y)"
proof -
{ fix a assume A7: "a∈G"
then have "a¯ ∈ G"
using OrderedGroup_ZF_1_L1 group0.inverse_in_group
by simp
with A5 obtain b where
III: "b∈G⇩+" and "∀y. b\<lsq>y ⟶ f`(y¯) \<lsq> a¯"
by auto
with II A7 have "∀y. b\<lsq>y ⟶ a \<lsq> ?g`(y)"
using OrderedGroup_ZF_1_L5AD OrderedGroup_ZF_1_L4
by simp
with III have "∃b∈G⇩+.∀y. b\<lsq>y ⟶ a \<lsq> ?g`(y)"
by auto
} then show "∀a∈G.∃b∈G⇩+.∀y. b\<lsq>y ⟶ a \<lsq> ?g`(y)"
by simp
qed
moreover have "∀x∈X. b(x)¯ ∈ G ∧ ?g`(b(x)¯) \<lsq> L¯"
proof-
{ fix x assume "x∈X"
with A6 have
T: "b(x) ∈ G" "b(x)¯ ∈ G" and "L \<lsq> f`(b(x))"
using OrderedGroup_ZF_1_L1 group0.inverse_in_group
by auto
then have "(f`(b(x)))¯ \<lsq> L¯"
using OrderedGroup_ZF_1_L5 by simp
moreover from II T have "(f`(b(x)))¯ = ?g`(b(x)¯)"
using OrderedGroup_ZF_1_L1 group0.group_inv_of_inv
by simp
ultimately have "?g`(b(x)¯) \<lsq> L¯" by simp
with T have "b(x)¯ ∈ G ∧ ?g`(b(x)¯) \<lsq> L¯"
by simp
} then show "∀x∈X. b(x)¯ ∈ G ∧ ?g`(b(x)¯) \<lsq> L¯"
by simp
qed
ultimately have "∃u.∀x∈X. (b(x))¯ \<lsq> u"
by (rule OrderedGroup_ZF_7_L2)
then have "∃u.∀x∈X. u¯ \<lsq> (b(x)¯)¯"
using OrderedGroup_ZF_1_L5 by auto
with A6 show "∃l.∀x∈X. l \<lsq> b(x)"
using OrderedGroup_ZF_1_L1 group0.group_inv_of_inv
by auto
qed
text‹The next lemma combines ‹OrderedGroup_ZF_7_L2› and
‹OrderedGroup_ZF_7_L3› to show that if an image of a set
defined by separation by a function with infinite limits is bounded,
then the set itself i bounded.›
lemma (in group3) OrderedGroup_ZF_7_L4:
assumes A1: "r {is total on} G" and A2: "G ≠ {𝟭}" and
A3: "X≠0" and A4: "f:G→G" and
A5: "∀a∈G.∃b∈G⇩+.∀y. b\<lsq>y ⟶ a \<lsq> f`(y)" and
A6: "∀a∈G.∃b∈G⇩+.∀y. b\<lsq>y ⟶ f`(y¯) \<lsq> a" and
A7: "∀x∈X. b(x) ∈ G ∧ L \<lsq> f`(b(x)) ∧ f`(b(x)) \<lsq> U"
shows "∃M.∀x∈X. |b(x)| \<lsq> M"
proof -
from A7 have
I: "∀x∈X. b(x) ∈ G ∧ f`(b(x)) \<lsq> U" and
II: "∀x∈X. b(x) ∈ G ∧ L \<lsq> f`(b(x))"
by auto
from A1 A2 A3 A4 A5 I have "∃u.∀x∈X. b(x) \<lsq> u"
by (rule OrderedGroup_ZF_7_L2)
moreover from A1 A2 A3 A4 A6 II have "∃l.∀x∈X. l \<lsq> b(x)"
by (rule OrderedGroup_ZF_7_L3)
ultimately have "∃u l. ∀x∈X. l\<lsq>b(x) ∧ b(x) \<lsq> u"
by auto
with A1 have "∃u l.∀x∈X. |b(x)| \<lsq> GreaterOf(r,|l|,|u|)"
using OrderedGroup_ZF_3_L10 by blast
then show "∃M.∀x∈X. |b(x)| \<lsq> M"
by auto
qed
end