(*

This file is a part of IsarMathLib -

a library of formalized mathematics for Isabelle/Isar.

Copyright (C) 2005, 2006 Slawomir Kolodynski

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*)

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

theory OrderedRing_ZF imports Ring_ZF OrderedGroup_ZF_1

begin

text{*In this theory file we consider ordered rings.*}

section{*Definition and notation*}

text{*This section defines ordered rings and sets up appriopriate notation.*}

text{*We define ordered ring as a commutative ring with linear order

that is preserved by

translations and such that the set of nonnegative elements is closed

under multiplication. Note that this definition does not guarantee

that there are no zero divisors in the ring.*}

definition

"IsAnOrdRing(R,A,M,r) ≡

( IsAring(R,A,M) ∧ (M {is commutative on} R) ∧

r⊆R×R ∧ IsLinOrder(R,r) ∧

(∀a b. ∀ c∈R. ⟨ a,b⟩ ∈ r --> ⟨A`⟨ a,c⟩,A`⟨ b,c⟩⟩ ∈ r) ∧

(Nonnegative(R,A,r) {is closed under} M))"

text{*The next context (locale) defines notation used for ordered rings.

We do that by extending the notation defined in the

@{text "ring0"} locale and adding some assumptions to make sure we are

talking about ordered rings in this context.*}

locale ring1 = ring0 +

assumes mult_commut: "M {is commutative on} R"

fixes r

assumes ordincl: "r ⊆ R×R"

assumes linord: "IsLinOrder(R,r)"

fixes lesseq (infix "\<lsq>" 68)

defines lesseq_def [simp]: "a \<lsq> b ≡ ⟨ a,b⟩ ∈ r"

fixes sless (infix "\<ls>" 68)

defines sless_def [simp]: "a \<ls> b ≡ a\<lsq>b ∧ a≠b"

assumes ordgroup: "∀a b. ∀ c∈R. a\<lsq>b --> a\<ra>c \<lsq> b\<ra>c"

assumes pos_mult_closed: "Nonnegative(R,A,r) {is closed under} M"

fixes abs ("| _ |")

defines abs_def [simp]: "|a| ≡ AbsoluteValue(R,A,r)`(a)"

fixes positiveset ("R⇩_{+}")

defines positiveset_def [simp]: "R⇩_{+}≡ PositiveSet(R,A,r)"

text{*The next lemma assures us that we are talking about ordered rings

in the @{text "ring1"} context.*}

lemma (in ring1) OrdRing_ZF_1_L1: shows "IsAnOrdRing(R,A,M,r)"

using ring0_def ringAssum mult_commut ordincl linord ordgroup

pos_mult_closed IsAnOrdRing_def by simp;

text{*We can use theorems proven in the @{text "ring1"} context whenever we

talk about an ordered ring.*}

lemma OrdRing_ZF_1_L2: assumes "IsAnOrdRing(R,A,M,r)"

shows "ring1(R,A,M,r)"

using assms IsAnOrdRing_def ring1_axioms.intro ring0_def ring1_def

by simp;

text{*In the @{text "ring1"} context $a\leq b$ implies that $a,b$ are

elements of the ring.*}

lemma (in ring1) OrdRing_ZF_1_L3: assumes "a\<lsq>b"

shows "a∈R" "b∈R"

using assms ordincl by auto;

text{*Ordered ring is an ordered group, hence we can use theorems

proven in the @{text "group3"} context.*}

lemma (in ring1) OrdRing_ZF_1_L4: shows

"IsAnOrdGroup(R,A,r)"

"r {is total on} R"

"A {is commutative on} R"

"group3(R,A,r)"

proof -;

{ fix a b g assume A1: "g∈R" and A2: "a\<lsq>b"

with ordgroup have "a\<ra>g \<lsq> b\<ra>g"

by simp;

moreover from ringAssum A1 A2 have

"a\<ra>g = g\<ra>a" "b\<ra>g = g\<ra>b"

using OrdRing_ZF_1_L3 IsAring_def IsCommutative_def by auto;

ultimately have

"a\<ra>g \<lsq> b\<ra>g" "g\<ra>a \<lsq> g\<ra>b"

by auto;

} hence

"∀g∈R. ∀a b. a\<lsq>b --> a\<ra>g \<lsq> b\<ra>g ∧ g\<ra>a \<lsq> g\<ra>b"

by simp;

with ringAssum ordincl linord show

"IsAnOrdGroup(R,A,r)"

"group3(R,A,r)"

"r {is total on} R"

"A {is commutative on} R"

using IsAring_def Order_ZF_1_L2 IsAnOrdGroup_def group3_def IsLinOrder_def

by auto;

qed;

text{*The order relation in rings is transitive.*}

lemma (in ring1) ring_ord_transitive: assumes A1: "a\<lsq>b" "b\<lsq>c"

shows "a\<lsq>c"

proof -

from A1 have

"group3(R,A,r)" "⟨a,b⟩ ∈ r" "⟨b,c⟩ ∈ r"

using OrdRing_ZF_1_L4 by auto;

then have "⟨a,c⟩ ∈ r" by (rule group3.Group_order_transitive);

then show "a\<lsq>c" by simp

qed

text{*Transitivity for the strict order: if $a<b$ and $b\leq c$, then $a<c$.

Property of ordered groups.*}

lemma (in ring1) ring_strict_ord_trans:

assumes A1: "a\<ls>b" and A2: "b\<lsq>c"

shows "a\<ls>c"

proof -

from A1 A2 have

"group3(R,A,r)"

"⟨a,b⟩ ∈ r ∧ a≠b" "⟨b,c⟩ ∈ r"

using OrdRing_ZF_1_L4 by auto

then have "⟨a,c⟩ ∈ r ∧ a≠c" by (rule group3.OrderedGroup_ZF_1_L4A)

then show "a\<ls>c" by simp

qed;

text{*Another version of transitivity for the strict order:

if $a\leq b$ and $b<c$, then $a<c$. Property of ordered groups.*}

lemma (in ring1) ring_strict_ord_transit:

assumes A1: "a\<lsq>b" and A2: "b\<ls>c"

shows "a\<ls>c"

proof -

from A1 A2 have

"group3(R,A,r)"

"⟨a,b⟩ ∈ r" "⟨b,c⟩ ∈ r ∧ b≠c"

using OrdRing_ZF_1_L4 by auto

then have "⟨a,c⟩ ∈ r ∧ a≠c" by (rule group3.group_strict_ord_transit)

then show "a\<ls>c" by simp

qed;

text{*The next lemma shows what happens when one element of an ordered

ring is not greater or equal than another.*}

lemma (in ring1) OrdRing_ZF_1_L4A: assumes A1: "a∈R" "b∈R"

and A2: "¬(a\<lsq>b)"

shows "b \<lsq> a" "(\<rm>a) \<lsq> (\<rm>b)" "a≠b"

proof -

from A1 A2 have I:

"group3(R,A,r)"

"r {is total on} R"

"a ∈ R" "b ∈ R" "⟨a, b⟩ ∉ r"

using OrdRing_ZF_1_L4 by auto;

then have "⟨b,a⟩ ∈ r" by (rule group3.OrderedGroup_ZF_1_L8);

then show "b \<lsq> a" by simp;

from I have "⟨GroupInv(R,A)`(a),GroupInv(R,A)`(b)⟩ ∈ r"

by (rule group3.OrderedGroup_ZF_1_L8);

then show "(\<rm>a) \<lsq> (\<rm>b)" by simp;

from I show "a≠b" by (rule group3.OrderedGroup_ZF_1_L8);

qed;

text{*A special case of @{text "OrdRing_ZF_1_L4A"} when one of the

constants is $0$. This is useful for many proofs by cases.*}

corollary (in ring1) ord_ring_split2: assumes A1: "a∈R"

shows "a\<lsq>\<zero> ∨ (\<zero>\<lsq>a ∧ a≠\<zero>)"

proof -

{ from A1 have I: "a∈R" "\<zero>∈R"

using Ring_ZF_1_L2 by auto;

moreover assume A2: "¬(a\<lsq>\<zero>)"

ultimately have "\<zero>\<lsq>a" by (rule OrdRing_ZF_1_L4A);

moreover from I A2 have "a≠\<zero>" by (rule OrdRing_ZF_1_L4A);

ultimately have "\<zero>\<lsq>a ∧ a≠\<zero>" by simp}

then show ?thesis by auto

qed;

text{*Taking minus on both sides reverses an inequality.*}

lemma (in ring1) OrdRing_ZF_1_L4B: assumes "a\<lsq>b"

shows "(\<rm>b) \<lsq> (\<rm>a)"

using assms OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L5

by simp;

text{*The next lemma just expands the condition that requires the set

of nonnegative elements to be closed with respect to multiplication.

These are properties of totally ordered groups.*}

lemma (in ring1) OrdRing_ZF_1_L5:

assumes "\<zero>\<lsq>a" "\<zero>\<lsq>b"

shows "\<zero> \<lsq> a·b"

using pos_mult_closed assms OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L2

IsOpClosed_def by simp;

text{*Double nonnegative is nonnegative.*}

lemma (in ring1) OrdRing_ZF_1_L5A: assumes A1: "\<zero>\<lsq>a"

shows "\<zero>\<lsq>\<two>·a"

using assms OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L5G

OrdRing_ZF_1_L3 Ring_ZF_1_L3 by simp;

text{*A sufficient (somewhat redundant) condition for a structure to be an

ordered ring. It says that a commutative ring that is a totally ordered

group with respect to the additive operation such that set of nonnegative

elements is closed under multiplication, is an ordered ring.*}

lemma OrdRing_ZF_1_L6:

assumes

"IsAring(R,A,M)"

"M {is commutative on} R"

"Nonnegative(R,A,r) {is closed under} M"

"IsAnOrdGroup(R,A,r)"

"r {is total on} R"

shows "IsAnOrdRing(R,A,M,r)"

using assms IsAnOrdGroup_def Order_ZF_1_L3 IsAnOrdRing_def

by simp;

text{*$a\leq b$ iff $a-b\leq 0$. This is a fact from

@{text "OrderedGroup.thy"}, where it is stated in multiplicative notation.*}

lemma (in ring1) OrdRing_ZF_1_L7:

assumes "a∈R" "b∈R"

shows "a\<lsq>b <-> a\<rs>b \<lsq> \<zero>"

using assms OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L9

by simp;

text{*Negative times positive is negative.*}

lemma (in ring1) OrdRing_ZF_1_L8:

assumes A1: "a\<lsq>\<zero>" and A2: "\<zero>\<lsq>b"

shows "a·b \<lsq> \<zero>"

proof -

from A1 A2 have T1: "a∈R" "b∈R" "a·b ∈ R"

using OrdRing_ZF_1_L3 Ring_ZF_1_L4 by auto;

from A1 A2 have "\<zero>\<lsq>(\<rm>a)·b"

using OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L5A OrdRing_ZF_1_L5

by simp;

with T1 show "a·b \<lsq> \<zero>"

using Ring_ZF_1_L7 OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L5AA

by simp;

qed;

text{*We can multiply both sides of an inequality by a nonnegative ring

element. This property is sometimes (not here) used to define

ordered rings. *}

lemma (in ring1) OrdRing_ZF_1_L9:

assumes A1: "a\<lsq>b" and A2: "\<zero>\<lsq>c"

shows

"a·c \<lsq> b·c"

"c·a \<lsq> c·b"

proof -

from A1 A2 have T1:

"a∈R" "b∈R" "c∈R" "a·c ∈ R" "b·c ∈ R"

using OrdRing_ZF_1_L3 Ring_ZF_1_L4 by auto;

with A1 A2 have "(a\<rs>b)·c \<lsq> \<zero>"

using OrdRing_ZF_1_L7 OrdRing_ZF_1_L8 by simp;

with T1 show "a·c \<lsq> b·c"

using Ring_ZF_1_L8 OrdRing_ZF_1_L7 by simp;

with mult_commut T1 show "c·a \<lsq> c·b"

using IsCommutative_def by simp;

qed;

text{*A special case of @{text "OrdRing_ZF_1_L9"}: we can multiply

an inequality by a positive ring element.*}

lemma (in ring1) OrdRing_ZF_1_L9A:

assumes A1: "a\<lsq>b" and A2: "c∈R⇩_{+}"

shows

"a·c \<lsq> b·c"

"c·a \<lsq> c·b"

proof -

from A2 have "\<zero> \<lsq> c" using PositiveSet_def

by simp;

with A1 show "a·c \<lsq> b·c" "c·a \<lsq> c·b"

using OrdRing_ZF_1_L9 by auto;

qed;

text{*A square is nonnegative.*}

lemma (in ring1) OrdRing_ZF_1_L10:

assumes A1: "a∈R" shows "\<zero>\<lsq>(a⇧^{2})"

proof -

{ assume "\<zero>\<lsq>a"

then have "\<zero>\<lsq>(a⇧^{2})" using OrdRing_ZF_1_L5 by simp}

moreover

{ assume "¬(\<zero>\<lsq>a)"

with A1 have "\<zero>\<lsq>((\<rm>a)⇧^{2})"

using OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L8A

OrdRing_ZF_1_L5 by simp;

with A1 have "\<zero>\<lsq>(a⇧^{2})" using Ring_ZF_1_L14 by simp }

ultimately show ?thesis by blast

qed;

text{*$1$ is nonnegative.*}

corollary (in ring1) ordring_one_is_nonneg: shows "\<zero> \<lsq> \<one>"

proof -

have "\<zero> \<lsq> (\<one>⇧^{2})" using Ring_ZF_1_L2 OrdRing_ZF_1_L10

by simp;

then show "\<zero> \<lsq> \<one>" using Ring_ZF_1_L2 Ring_ZF_1_L3

by simp;

qed;

text{*In nontrivial rings one is positive.*}

lemma (in ring1) ordring_one_is_pos: assumes "\<zero>≠\<one>"

shows "\<one> ∈ R⇩_{+}"

using assms Ring_ZF_1_L2 ordring_one_is_nonneg PositiveSet_def

by auto;

text{*Nonnegative is not negative. Property of ordered groups.*}

lemma (in ring1) OrdRing_ZF_1_L11: assumes "\<zero>\<lsq>a"

shows "¬(a\<lsq>\<zero> ∧ a≠\<zero>)"

using assms OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L5AB

by simp;

text{*A negative element cannot be a square.*}

lemma (in ring1) OrdRing_ZF_1_L12:

assumes A1: "a\<lsq>\<zero>" "a≠\<zero>"

shows "¬(∃b∈R. a = (b⇧^{2}))"

proof -

{ assume "∃b∈R. a = (b⇧^{2})"

with A1 have False using OrdRing_ZF_1_L10 OrdRing_ZF_1_L11

by auto;

} then show ?thesis by auto;

qed;

text{*If $a\leq b$, then $0\leq b-a$.*}

lemma (in ring1) OrdRing_ZF_1_L13: assumes "a\<lsq>b"

shows "\<zero> \<lsq> b\<rs>a"

using assms OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L9D

by simp;

text{*If $a<b$, then $0 < b-a$.*}

lemma (in ring1) OrdRing_ZF_1_L14: assumes "a\<lsq>b" "a≠b"

shows

"\<zero> \<lsq> b\<rs>a" "\<zero> ≠ b\<rs>a"

"b\<rs>a ∈ R⇩_{+}"

using assms OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L9E

by auto;

text{*If the difference is nonnegative, then $a\leq b$. *}

lemma (in ring1) OrdRing_ZF_1_L15:

assumes "a∈R" "b∈R" and "\<zero> \<lsq> b\<rs>a"

shows "a\<lsq>b"

using assms OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L9F

by simp;

text{*A nonnegative number is does not decrease when multiplied by

a number greater or equal $1$.*}

lemma (in ring1) OrdRing_ZF_1_L16:

assumes A1: "\<zero>\<lsq>a" and A2: "\<one>\<lsq>b"

shows "a\<lsq>a·b"

proof -

from A1 A2 have T: "a∈R" "b∈R" "a·b ∈ R"

using OrdRing_ZF_1_L3 Ring_ZF_1_L4 by auto

from A1 A2 have "\<zero> \<lsq> a·(b\<rs>\<one>)"

using OrdRing_ZF_1_L13 OrdRing_ZF_1_L5 by simp;

with T show "a\<lsq>a·b"

using Ring_ZF_1_L8 Ring_ZF_1_L2 Ring_ZF_1_L3 OrdRing_ZF_1_L15

by simp;

qed;

text{*We can multiply the right hand side of an inequality between

nonnegative ring elements by an element greater or equal $1$.*}

lemma (in ring1) OrdRing_ZF_1_L17:

assumes A1: "\<zero>\<lsq>a" and A2: "a\<lsq>b" and A3: "\<one>\<lsq>c"

shows "a\<lsq>b·c"

proof -

from A1 A2 have "\<zero>\<lsq>b" by (rule ring_ord_transitive);

with A3 have "b\<lsq>b·c" using OrdRing_ZF_1_L16

by simp;

with A2 show "a\<lsq>b·c" by (rule ring_ord_transitive);

qed;

text{*Strict order is preserved by translations.*}

lemma (in ring1) ring_strict_ord_trans_inv:

assumes "a\<ls>b" and "c∈R"

shows

"a\<ra>c \<ls> b\<ra>c"

"c\<ra>a \<ls> c\<ra>b"

using assms OrdRing_ZF_1_L4 group3.group_strict_ord_transl_inv

by auto;

text{*We can put an element on the other side of a strict inequality,

changing its sign.*}

lemma (in ring1) OrdRing_ZF_1_L18:

assumes "a∈R" "b∈R" and "a\<rs>b \<ls> c"

shows "a \<ls> c\<ra>b"

using assms OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L12B

by simp;

text{*We can add the sides of two inequalities,

the first of them strict, and we get a strict inequality.

Property of ordered groups.*}

lemma (in ring1) OrdRing_ZF_1_L19:

assumes "a\<ls>b" and "c\<lsq>d"

shows "a\<ra>c \<ls> b\<ra>d"

using assms OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L12C

by simp;

text{*We can add the sides of two inequalities,

the second of them strict and we get a strict inequality.

Property of ordered groups.*}

lemma (in ring1) OrdRing_ZF_1_L20:

assumes "a\<lsq>b" and "c\<ls>d"

shows "a\<ra>c \<ls> b\<ra>d"

using assms OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L12D

by simp;

section{*Absolute value for ordered rings*}

text{*Absolute value is defined for ordered groups as a function

that is the identity on the nonnegative set and the negative of the element

(the inverse in the multiplicative notation) on the rest. In this section

we consider properties of absolute value related to multiplication in

ordered rings.*}

text{*Absolute value of a product is the product of absolute values:

the case when both elements of the ring are nonnegative.*}

lemma (in ring1) OrdRing_ZF_2_L1:

assumes "\<zero>\<lsq>a" "\<zero>\<lsq>b"

shows "|a·b| = |a|·|b|"

using assms OrdRing_ZF_1_L5 OrdRing_ZF_1_L4

group3.OrderedGroup_ZF_1_L2 group3.OrderedGroup_ZF_3_L2

by simp;

text{*The absolue value of an element and its negative are the same.*}

lemma (in ring1) OrdRing_ZF_2_L2: assumes "a∈R"

shows "|\<rm>a| = |a|"

using assms OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_3_L7A by simp;

text{*The next lemma states that

$|a\cdot (-b)| = |(-a)\cdot b| = |(-a)\cdot (-b)| = |a\cdot b|$.*}

lemma (in ring1) OrdRing_ZF_2_L3:

assumes "a∈R" "b∈R"

shows

"|(\<rm>a)·b| = |a·b|"

"|a·(\<rm>b)| = |a·b|"

"|(\<rm>a)·(\<rm>b)| = |a·b|"

using assms Ring_ZF_1_L4 Ring_ZF_1_L7 Ring_ZF_1_L7A

OrdRing_ZF_2_L2 by auto;

text{*This lemma allows to prove theorems for the case of positive and

negative elements of the ring separately.*}

lemma (in ring1) OrdRing_ZF_2_L4: assumes "a∈R" and "¬(\<zero>\<lsq>a)"

shows "\<zero> \<lsq> (\<rm>a)" "\<zero>≠a"

using assms OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L8A

by auto;

text{*Absolute value of a product is the product of absolute values.*}

lemma (in ring1) OrdRing_ZF_2_L5:

assumes A1: "a∈R" "b∈R"

shows "|a·b| = |a|·|b|"

proof -

{ assume A2: "\<zero>\<lsq>a" have "|a·b| = |a|·|b|"

proof -

{ assume "\<zero>\<lsq>b"

with A2 have "|a·b| = |a|·|b|"

using OrdRing_ZF_2_L1 by simp }

moreover

{ assume "¬(\<zero>\<lsq>b)"

with A1 A2 have "|a·(\<rm>b)| = |a|·|\<rm>b|"

using OrdRing_ZF_2_L4 OrdRing_ZF_2_L1 by simp

with A1 have "|a·b| = |a|·|b|"

using OrdRing_ZF_2_L2 OrdRing_ZF_2_L3 by simp }

ultimately show ?thesis by blast

qed; }

moreover

{ assume "¬(\<zero>\<lsq>a)"

with A1 have A3: "\<zero> \<lsq> (\<rm>a)"

using OrdRing_ZF_2_L4 by simp;

have "|a·b| = |a|·|b|"

proof -

{ assume "\<zero>\<lsq>b"

with A3 have "|(\<rm>a)·b| = |\<rm>a|·|b|"

using OrdRing_ZF_2_L1 by simp;

with A1 have "|a·b| = |a|·|b|"

using OrdRing_ZF_2_L2 OrdRing_ZF_2_L3 by simp }

moreover

{ assume "¬(\<zero>\<lsq>b)"

with A1 A3 have "|(\<rm>a)·(\<rm>b)| = |\<rm>a|·|\<rm>b|"

using OrdRing_ZF_2_L4 OrdRing_ZF_2_L1 by simp;

with A1 have "|a·b| = |a|·|b|"

using OrdRing_ZF_2_L2 OrdRing_ZF_2_L3 by simp }

ultimately show ?thesis by blast

qed }

ultimately show ?thesis by blast

qed;

text{*Triangle inequality. Property of linearly ordered abelian groups.*}

lemma (in ring1) ord_ring_triangle_ineq: assumes "a∈R" "b∈R"

shows "|a\<ra>b| \<lsq> |a|\<ra>|b|"

using assms OrdRing_ZF_1_L4 group3.OrdGroup_triangle_ineq

by simp;

text{*If $a\leq c$ and $b\leq c$, then $a+b\leq 2\cdot c$.*}

lemma (in ring1) OrdRing_ZF_2_L6:

assumes "a\<lsq>c" "b\<lsq>c" shows "a\<ra>b \<lsq> \<two>·c"

using assms OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L5B

OrdRing_ZF_1_L3 Ring_ZF_1_L3 by simp;

section{*Positivity in ordered rings*}

text{*This section is about properties of the set of positive

elements @{text "R⇩_{+}"}. *}

text{*The set of positive elements is closed under ring addition.

This is a property of ordered groups, we just reference a theorem

from @{text "OrderedGroup_ZF"} theory in the proof.*}

lemma (in ring1) OrdRing_ZF_3_L1: shows "R⇩_{+}{is closed under} A"

using OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L13

by simp;

text{*Every element of a ring can be either in the postitive set, equal to

zero or its opposite (the additive inverse) is in the positive set.

This is a property of ordered groups, we just reference a theorem

from @{text "OrderedGroup_ZF"} theory.*}

lemma (in ring1) OrdRing_ZF_3_L2: assumes "a∈R"

shows "Exactly_1_of_3_holds (a=\<zero>, a∈R⇩_{+}, (\<rm>a) ∈ R⇩_{+})"

using assms OrdRing_ZF_1_L4 group3.OrdGroup_decomp

by simp;

text{*If a ring element $a\neq 0$, and it is not positive, then

$-a$ is positive.*}

lemma (in ring1) OrdRing_ZF_3_L2A: assumes "a∈R" "a≠\<zero>" "a ∉ R⇩_{+}"

shows "(\<rm>a) ∈ R⇩_{+}"

using assms OrdRing_ZF_1_L4 group3.OrdGroup_cases

by simp;

text{*@{text "R⇩_{+}"} is closed under

multiplication iff the ring has no zero divisors. *}

lemma (in ring1) OrdRing_ZF_3_L3:

shows "(R⇩_{+}{is closed under} M)<-> HasNoZeroDivs(R,A,M)"

proof;

assume A1: "HasNoZeroDivs(R,A,M)"

{ fix a b assume "a∈R⇩_{+}" "b∈R⇩_{+}"

then have "\<zero>\<lsq>a" "a≠\<zero>" "\<zero>\<lsq>b" "b≠\<zero>"

using PositiveSet_def by auto;

with A1 have "a·b ∈ R⇩_{+}"

using OrdRing_ZF_1_L5 Ring_ZF_1_L2 OrdRing_ZF_1_L3 Ring_ZF_1_L12

OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L2A

by simp;

} then show "R⇩_{+}{is closed under} M" using IsOpClosed_def

by simp;

next assume A2: "R⇩_{+}{is closed under} M"

{ fix a b assume A3: "a∈R" "b∈R" and "a≠\<zero>" "b≠\<zero>"

with A2 have "|a·b| ∈ R⇩_{+}"

using OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_3_L12 IsOpClosed_def

OrdRing_ZF_2_L5 by simp;

with A3 have "a·b ≠ \<zero>"

using PositiveSet_def Ring_ZF_1_L4

OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_3_L2A

by auto;

} then show "HasNoZeroDivs(R,A,M)" using HasNoZeroDivs_def

by auto;

qed;

text{*Another (in addition to @{text "OrdRing_ZF_1_L6"} sufficient condition

that defines order in an ordered ring starting from the positive set.*}

theorem (in ring0) ring_ord_by_positive_set:

assumes

A1: "M {is commutative on} R" and

A2: "P⊆R" "P {is closed under} A" "\<zero> ∉ P" and

A3: "∀a∈R. a≠\<zero> --> (a∈P) Xor ((\<rm>a) ∈ P)" and

A4: "P {is closed under} M" and

A5: "r = OrderFromPosSet(R,A,P)"

shows

"IsAnOrdGroup(R,A,r)"

"IsAnOrdRing(R,A,M,r)"

"r {is total on} R"

"PositiveSet(R,A,r) = P"

"Nonnegative(R,A,r) = P ∪ {\<zero>}"

"HasNoZeroDivs(R,A,M)"

proof -

from A2 A3 A5 show

I: "IsAnOrdGroup(R,A,r)" "r {is total on} R" and

II: "PositiveSet(R,A,r) = P" and

III: "Nonnegative(R,A,r) = P ∪ {\<zero>}"

using Ring_ZF_1_L1 group0.Group_ord_by_positive_set

by auto;

from A2 A4 III have "Nonnegative(R,A,r) {is closed under} M"

using Ring_ZF_1_L16 by simp;

with ringAssum A1 I show "IsAnOrdRing(R,A,M,r)"

using OrdRing_ZF_1_L6 by simp;

with A4 II show "HasNoZeroDivs(R,A,M)"

using OrdRing_ZF_1_L2 ring1.OrdRing_ZF_3_L3

by auto;

qed;

text{*Nontrivial ordered rings are infinite. More precisely we assume

that the neutral

element of the additive operation is not equal to the multiplicative neutral

element and show that the the set of positive elements of the ring is not a

finite subset of the ring and the ring is not a finite subset of itself.*}

theorem (in ring1) ord_ring_infinite: assumes "\<zero>≠\<one>"

shows

"R⇩_{+}∉ Fin(R)"

"R ∉ Fin(R)"

using assms Ring_ZF_1_L17 OrdRing_ZF_1_L4 group3.Linord_group_infinite

by auto

text{*If every element of a nontrivial ordered ring can be dominated

by an element from $B$, then we $B$ is not bounded and not finite.*}

lemma (in ring1) OrdRing_ZF_3_L4:

assumes "\<zero>≠\<one>" and "∀a∈R. ∃b∈B. a\<lsq>b"

shows

"¬IsBoundedAbove(B,r)"

"B ∉ Fin(R)"

using assms Ring_ZF_1_L17 OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_2_L2A

by auto;

text{*If $m$ is greater or equal the multiplicative unit, then the set

$\{m\cdot n: n\in R\}$ is infinite (unless the ring is trivial).*}

lemma (in ring1) OrdRing_ZF_3_L5: assumes A1: "\<zero>≠\<one>" and A2: "\<one>\<lsq>m"

shows

"{m·x. x∈R⇩_{+}} ∉ Fin(R)"

"{m·x. x∈R} ∉ Fin(R)"

"{(\<rm>m)·x. x∈R} ∉ Fin(R)"

proof -

from A2 have T: "m∈R" using OrdRing_ZF_1_L3 by simp

from A2 have "\<zero>\<lsq>\<one>" "\<one>\<lsq>m"

using ordring_one_is_nonneg by auto;

then have I: "\<zero>\<lsq>m" by (rule ring_ord_transitive);

let ?B = "{m·x. x∈R⇩_{+}}"

{ fix a assume A3: "a∈R"

then have "a\<lsq>\<zero> ∨ (\<zero>\<lsq>a ∧ a≠\<zero>)"

using ord_ring_split2 by simp;

moreover

{ assume A4: "a\<lsq>\<zero>"

from A1 have "m·\<one> ∈ ?B" using ordring_one_is_pos

by auto;

with T have "m∈?B" using Ring_ZF_1_L3 by simp;

moreover from A4 I have "a\<lsq>m" by (rule ring_ord_transitive);

ultimately have "∃b∈?B. a\<lsq>b" by blast }

moreover

{ assume A4: "\<zero>\<lsq>a ∧ a≠\<zero>"

with A3 have "m·a ∈ ?B" using PositiveSet_def

by auto;

moreover

from A2 A4 have "\<one>·a \<lsq> m·a" using OrdRing_ZF_1_L9

by simp;

with A3 have "a \<lsq> m·a" using Ring_ZF_1_L3

by simp;

ultimately have "∃b∈?B. a\<lsq>b" by auto }

ultimately have "∃b∈?B. a\<lsq>b" by auto;

} then have "∀a∈R. ∃b∈?B. a\<lsq>b"

by simp;

with A1 show "?B ∉ Fin(R)" using OrdRing_ZF_3_L4

by simp;

moreover have "?B ⊆ {m·x. x∈R}"

using PositiveSet_def by auto;

ultimately show "{m·x. x∈R} ∉ Fin(R)" using Fin_subset

by auto;

with T show "{(\<rm>m)·x. x∈R} ∉ Fin(R)" using Ring_ZF_1_L18

by simp

qed;

text{*If $m$ is less or equal than the negative of

multiplicative unit, then the set

$\{m\cdot n: n\in R\}$ is infinite (unless the ring is trivial).*}

lemma (in ring1) OrdRing_ZF_3_L6: assumes A1: "\<zero>≠\<one>" and A2: "m \<lsq> \<rm>\<one>"

shows "{m·x. x∈R} ∉ Fin(R)"

proof -

from A2 have "(\<rm>(\<rm>\<one>)) \<lsq> \<rm>m"

using OrdRing_ZF_1_L4B by simp;

with A1 have "{(\<rm>m)·x. x∈R} ∉ Fin(R)"

using Ring_ZF_1_L2 Ring_ZF_1_L3 OrdRing_ZF_3_L5

by simp;

with A2 show "{m·x. x∈R} ∉ Fin(R)"

using OrdRing_ZF_1_L3 Ring_ZF_1_L18 by simp;

qed;

text{*All elements greater or equal than an element of @{text "R⇩_{+}"}

belong to @{text "R⇩_{+}"}. Property of ordered groups.*}

lemma (in ring1) OrdRing_ZF_3_L7: assumes A1: "a ∈ R⇩_{+}" and A2: "a\<lsq>b"

shows "b ∈ R⇩_{+}"

proof -

from A1 A2 have

"group3(R,A,r)"

"a ∈ PositiveSet(R,A,r)"

"⟨a,b⟩ ∈ r"

using OrdRing_ZF_1_L4 by auto;

then have "b ∈ PositiveSet(R,A,r)"

by (rule group3.OrderedGroup_ZF_1_L19);

then show "b ∈ R⇩_{+}" by simp;

qed;

text{*A special case of @{text "OrdRing_ZF_3_L7"}: a ring element greater

or equal than $1$ is positive.*}

corollary (in ring1) OrdRing_ZF_3_L8: assumes A1: "\<zero>≠\<one>" and A2: "\<one>\<lsq>a"

shows "a ∈ R⇩_{+}"

proof -

from A1 A2 have "\<one> ∈ R⇩_{+}" "\<one>\<lsq>a"

using ordring_one_is_pos by auto

then show "a ∈ R⇩_{+}" by (rule OrdRing_ZF_3_L7);

qed;

text{*Adding a positive element to $a$ strictly increases $a$.

Property of ordered groups.*}

lemma (in ring1) OrdRing_ZF_3_L9: assumes A1: "a∈R" "b∈R⇩_{+}"

shows "a \<lsq> a\<ra>b" "a ≠ a\<ra>b"

using assms OrdRing_ZF_1_L4 group3.OrderedGroup_ZF_1_L22

by auto;

text{*A special case of @{text " OrdRing_ZF_3_L9"}: in nontrivial

rings adding one to $a$ increases $a$. *}

corollary (in ring1) OrdRing_ZF_3_L10: assumes A1: "\<zero>≠\<one>" and A2: "a∈R"

shows "a \<lsq> a\<ra>\<one>" "a ≠ a\<ra>\<one>"

using assms ordring_one_is_pos OrdRing_ZF_3_L9

by auto;

text{*If $a$ is not greater than $b$, then it is strictly less than

$b+1$.*}

lemma (in ring1) OrdRing_ZF_3_L11: assumes A1: "\<zero>≠\<one>" and A2: "a\<lsq>b"

shows "a\<ls> b\<ra>\<one>"

proof -

from A1 A2 have I: "b \<ls> b\<ra>\<one>"

using OrdRing_ZF_1_L3 OrdRing_ZF_3_L10 by auto;

with A2 show "a\<ls> b\<ra>\<one>" by (rule ring_strict_ord_transit);

qed;

text{*For any ring element $a$ the greater of $a$ and $1$ is a positive

element that is greater or equal than $m$. If we add $1$ to it we

get a positive element that is strictly greater than $m$. This holds

in nontrivial rings.*}

lemma (in ring1) OrdRing_ZF_3_L12: assumes A1: "\<zero>≠\<one>" and A2: "a∈R"

shows

"a \<lsq> GreaterOf(r,\<one>,a)"

"GreaterOf(r,\<one>,a) ∈ R⇩_{+}"

"GreaterOf(r,\<one>,a) \<ra> \<one> ∈ R⇩_{+}"

"a \<lsq> GreaterOf(r,\<one>,a) \<ra> \<one>" "a ≠ GreaterOf(r,\<one>,a) \<ra> \<one>"

proof -

from linord have "r {is total on} R" using IsLinOrder_def

by simp

moreover from A2 have "\<one> ∈ R" "a∈R"

using Ring_ZF_1_L2 by auto

ultimately have

"\<one> \<lsq> GreaterOf(r,\<one>,a)" and

I: "a \<lsq> GreaterOf(r,\<one>,a)"

using Order_ZF_3_L2 by auto;

with A1 show

"a \<lsq> GreaterOf(r,\<one>,a)" and

"GreaterOf(r,\<one>,a) ∈ R⇩_{+}"

using OrdRing_ZF_3_L8 by auto;

with A1 show "GreaterOf(r,\<one>,a) \<ra> \<one> ∈ R⇩_{+}"

using ordring_one_is_pos OrdRing_ZF_3_L1 IsOpClosed_def

by simp;

from A1 I show

"a \<lsq> GreaterOf(r,\<one>,a) \<ra> \<one>" "a ≠ GreaterOf(r,\<one>,a) \<ra> \<one>"

using OrdRing_ZF_3_L11 by auto;

qed;

text{*We can multiply strict inequality by a positive element.*}

lemma (in ring1) OrdRing_ZF_3_L13:

assumes A1: "HasNoZeroDivs(R,A,M)" and

A2: "a\<ls>b" and A3: "c∈R⇩_{+}"

shows

"a·c \<ls> b·c"

"c·a \<ls> c·b"

proof -

from A2 A3 have T: "a∈R" "b∈R" "c∈R" "c≠\<zero>"

using OrdRing_ZF_1_L3 PositiveSet_def by auto;

from A2 A3 have "a·c \<lsq> b·c" using OrdRing_ZF_1_L9A

by simp;

moreover from A1 A2 T have "a·c ≠ b·c"

using Ring_ZF_1_L12A by auto;

ultimately show "a·c \<ls> b·c" by simp

moreover from mult_commut T have "a·c = c·a" and "b·c = c·b"

using IsCommutative_def by auto;

ultimately show "c·a \<ls> c·b" by simp;

qed;

text{*A sufficient condition for an element to be in the set

of positive ring elements. *}

lemma (in ring1) OrdRing_ZF_3_L14: assumes "\<zero>\<lsq>a" and "a≠\<zero>"

shows "a ∈ R⇩_{+}"

using assms OrdRing_ZF_1_L3 PositiveSet_def

by auto;

text{*If a ring has no zero divisors, the square of a nonzero

element is positive.*}

lemma (in ring1) OrdRing_ZF_3_L15:

assumes "HasNoZeroDivs(R,A,M)" and "a∈R" "a≠\<zero>"

shows "\<zero> \<lsq> a⇧^{2}" "a⇧^{2}≠ \<zero>" "a⇧^{2}∈ R⇩_{+}"

using assms OrdRing_ZF_1_L10 Ring_ZF_1_L12 OrdRing_ZF_3_L14

by auto;

text{*In rings with no zero divisors we can (strictly) increase a

positive element by multiplying it by an element that is greater than $1$.*}

lemma (in ring1) OrdRing_ZF_3_L16:

assumes "HasNoZeroDivs(R,A,M)" and "a ∈ R⇩_{+}" and "\<one>\<lsq>b" "\<one>≠b"

shows "a\<lsq>a·b" "a ≠ a·b"

using assms PositiveSet_def OrdRing_ZF_1_L16 OrdRing_ZF_1_L3

Ring_ZF_1_L12C by auto;

text{*If the right hand side of an inequality is positive we can multiply it

by a number that is greater than one.*}

lemma (in ring1) OrdRing_ZF_3_L17:

assumes A1: "HasNoZeroDivs(R,A,M)" and A2: "b∈R⇩_{+}" and

A3: "a\<lsq>b" and A4: "\<one>\<ls>c"

shows "a\<ls>b·c"

proof -

from A1 A2 A4 have "b \<ls> b·c"

using OrdRing_ZF_3_L16 by auto;

with A3 show "a\<ls>b·c" by (rule ring_strict_ord_transit);

qed;

text{*We can multiply a right hand side of an inequality between

positive numbers by a number that is greater than one.*}

lemma (in ring1) OrdRing_ZF_3_L18:

assumes A1: "HasNoZeroDivs(R,A,M)" and A2: "a ∈ R⇩_{+}" and

A3: "a\<lsq>b" and A4: "\<one>\<ls>c"

shows "a\<ls>b·c"

proof -

from A2 A3 have "b ∈ R⇩_{+}" using OrdRing_ZF_3_L7

by blast;

with A1 A3 A4 show "a\<ls>b·c"

using OrdRing_ZF_3_L17 by simp

qed;

text{*In ordered rings with no zero divisors if at

least one of $a,b$ is not zero, then

$0 < a^2+b^2$, in particular $a^2+b^2\neq 0$.*}

lemma (in ring1) OrdRing_ZF_3_L19:

assumes A1: "HasNoZeroDivs(R,A,M)" and A2: "a∈R" "b∈R" and

A3: "a ≠ \<zero> ∨ b ≠ \<zero>"

shows "\<zero> \<ls> a⇧^{2}\<ra> b⇧^{2}"

proof -

{ assume "a ≠ \<zero>"

with A1 A2 have "\<zero> \<lsq> a⇧^{2}" "a⇧^{2}≠ \<zero>"

using OrdRing_ZF_3_L15 by auto;

then have "\<zero> \<ls> a⇧^{2}" by auto;

moreover from A2 have "\<zero> \<lsq> b⇧^{2}"

using OrdRing_ZF_1_L10 by simp

ultimately have "\<zero> \<ra> \<zero> \<ls> a⇧^{2}\<ra> b⇧^{2}"

using OrdRing_ZF_1_L19 by simp;

then have "\<zero> \<ls> a⇧^{2}\<ra> b⇧^{2}"

using Ring_ZF_1_L2 Ring_ZF_1_L3 by simp }

moreover

{ assume A4: "a = \<zero>"

then have "a⇧^{2}\<ra> b⇧^{2}= \<zero> \<ra> b⇧^{2}"

using Ring_ZF_1_L2 Ring_ZF_1_L6 by simp;

also from A2 have "… = b⇧^{2}"

using Ring_ZF_1_L4 Ring_ZF_1_L3 by simp

finally have "a⇧^{2}\<ra> b⇧^{2}= b⇧^{2}" by simp;

moreover

from A3 A4 have "b ≠ \<zero>" by simp;

with A1 A2 have "\<zero> \<lsq> b⇧^{2}" and "b⇧^{2}≠ \<zero>"

using OrdRing_ZF_3_L15 by auto;

hence "\<zero> \<ls> b⇧^{2}" by auto;

ultimately have "\<zero> \<ls> a⇧^{2}\<ra> b⇧^{2}" by simp }

ultimately show "\<zero> \<ls> a⇧^{2}\<ra> b⇧^{2}"

by auto;

qed;

end