(* 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 following conditions are met: 1. Redistributions of source code must retain the above copyright notice, this list of conditions and the following disclaimer. 2. Redistributions in binary form must reproduce the above copyright notice, this list of conditions and the following disclaimer in the documentation and/or other materials provided with the distribution. 3. The name of the author may not be used to endorse or promote products derived from this software without specific prior written permission. THIS SOFTWARE IS PROVIDED BY THE AUTHOR ``AS IS'' AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE. *) section ‹Ordered groups - introduction› theory OrderedGroup_ZF imports Group_ZF_1 AbelianGroup_ZF Order_ZF Finite_ZF_1 begin text{*This theory file defines and shows the basic properties of (partially or linearly) ordered groups. We define the set of nonnegative elements and the absolute value function. We show that in linearly ordered groups finite sets are bounded and provide a sufficient condition for bounded sets to be finite. This allows to show in @{text "Int_ZF_IML.thy"} that subsets of integers are bounded iff they are finite. *} subsection{*Ordered groups*} text{*This section defines ordered groups and various related notions.*} text{*An ordered group is a group equipped with a partial order that is "translation invariant", that is if $a\leq b$ then $a\cdot g \leq b\cdot g$ and $g\cdot a \leq g\cdot b$.*} definition "IsAnOrdGroup(G,P,r) ≡ (IsAgroup(G,P) ∧ r⊆G×G ∧ IsPartOrder(G,r) ∧ (∀g∈G. ∀a b. ⟨ a,b⟩ ∈ r ⟶ ⟨ P`⟨ a,g⟩,P`⟨ b,g⟩ ⟩ ∈ r ∧ ⟨ P`⟨ g,a⟩,P`⟨ g,b⟩ ⟩ ∈ r ) )" text{*We define the set of nonnegative elements in the obvious way as $G^+ =\{x\in G: 1 \leq x\}$.*} definition "Nonnegative(G,P,r) ≡ {x∈G. ⟨ TheNeutralElement(G,P),x⟩ ∈ r}" text{* The @{text "PositiveSet(G,P,r)"} is a set similar to @{text "Nonnegative(G,P,r)"}, but without the unit.*} definition "PositiveSet(G,P,r) ≡ {x∈G. ⟨ TheNeutralElement(G,P),x⟩ ∈ r ∧ TheNeutralElement(G,P)≠ x}" text{*We also define the absolute value as a ZF-function that is the identity on $G^+$ and the group inverse on the rest of the group.*} definition "AbsoluteValue(G,P,r) ≡ id(Nonnegative(G,P,r)) ∪ restrict(GroupInv(G,P),G - Nonnegative(G,P,r))" text{*The odd functions are defined as those having property $f(a^{-1})=(f(a))^{-1}$. This looks a bit strange in the multiplicative notation, I have to admit. For linearly oredered groups a function $f$ defined on the set of positive elements iniquely defines an odd function of the whole group. This function is called an odd extension of $f$*} definition "OddExtension(G,P,r,f) ≡ (f ∪ {⟨a, GroupInv(G,P)`(f`(GroupInv(G,P)`(a)))⟩. a ∈ GroupInv(G,P)``(PositiveSet(G,P,r))} ∪ {⟨TheNeutralElement(G,P),TheNeutralElement(G,P)⟩})" text{*We will use a similar notation for ordered groups as for the generic groups. @{text "G⇧^{+}"} denotes the set of nonnegative elements (that satisfy $1\leq a$) and @{text "G⇩_{+}"} is the set of (strictly) positive elements. @{text "\<sm>A"} is the set inverses of elements from $A$. I hope that using additive notation for this notion is not too shocking here. The symbol @{text "f°"} denotes the odd extension of $f$. For a function defined on $G_+$ this is the unique odd function on $G$ that is equal to $f$ on $G_+$. *} locale group3 = fixes G and P and r assumes ordGroupAssum: "IsAnOrdGroup(G,P,r)" fixes unit ("𝟭") defines unit_def [simp]: "𝟭 ≡ TheNeutralElement(G,P)" fixes groper (infixl "⋅" 70) defines groper_def [simp]: "a ⋅ b ≡ P`⟨ a,b⟩" fixes inv ("_¯ " [90] 91) defines inv_def [simp]: "x¯ ≡ GroupInv(G,P)`(x)" 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" fixes nonnegative ("G⇧^{+}") defines nonnegative_def [simp]: "G⇧^{+}≡ Nonnegative(G,P,r)" fixes positive ("G⇩_{+}") defines positive_def [simp]: "G⇩_{+}≡ PositiveSet(G,P,r)" fixes setinv ("\<sm> _" 72) defines setninv_def [simp]: "\<sm>A ≡ GroupInv(G,P)``(A)" fixes abs ("| _ |") defines abs_def [simp]: "|a| ≡ AbsoluteValue(G,P,r)`(a)" fixes oddext ("_ °") defines oddext_def [simp]: "f° ≡ OddExtension(G,P,r,f)" text{*In @{text "group3"} context we can use the theorems proven in the @{text "group0"} context.*} lemma (in group3) OrderedGroup_ZF_1_L1: shows "group0(G,P)" using ordGroupAssum IsAnOrdGroup_def group0_def by simp text{*Ordered group (carrier) is not empty. This is a property of monoids, but it is good to have it handy in the @{text "group3"} context.*} lemma (in group3) OrderedGroup_ZF_1_L1A: shows "G≠0" using OrderedGroup_ZF_1_L1 group0.group0_2_L1 monoid0.group0_1_L3A by blast text{*The next lemma is just to see the definition of the nonnegative set in our notation.*} lemma (in group3) OrderedGroup_ZF_1_L2: shows "g∈G⇧^{+}⟷ 𝟭\<lsq>g" using ordGroupAssum IsAnOrdGroup_def Nonnegative_def by auto text{*The next lemma is just to see the definition of the positive set in our notation.*} lemma (in group3) OrderedGroup_ZF_1_L2A: shows "g∈G⇩_{+}⟷ (𝟭\<lsq>g ∧ g≠𝟭)" using ordGroupAssum IsAnOrdGroup_def PositiveSet_def by auto text{*For total order if $g$ is not in $G^{+}$, then it has to be less or equal the unit.*} lemma (in group3) OrderedGroup_ZF_1_L2B: assumes A1: "r {is total on} G" and A2: "a∈G-G⇧^{+}" shows "a\<lsq>𝟭" proof - from A2 have "a∈G" "𝟭 ∈ G" "¬(𝟭\<lsq>a)" using OrderedGroup_ZF_1_L1 group0.group0_2_L2 OrderedGroup_ZF_1_L2 by auto with A1 show ?thesis using IsTotal_def by auto qed text{*The group order is reflexive.*} lemma (in group3) OrderedGroup_ZF_1_L3: assumes "g∈G" shows "g\<lsq>g" using ordGroupAssum assms IsAnOrdGroup_def IsPartOrder_def refl_def by simp text{* $1$ is nonnegative.*} lemma (in group3) OrderedGroup_ZF_1_L3A: shows "𝟭∈G⇧^{+}" using OrderedGroup_ZF_1_L2 OrderedGroup_ZF_1_L3 OrderedGroup_ZF_1_L1 group0.group0_2_L2 by simp text{*In this context $a \leq b$ implies that both $a$ and $b$ belong to $G$.*} lemma (in group3) OrderedGroup_ZF_1_L4: assumes "a\<lsq>b" shows "a∈G" "b∈G" using ordGroupAssum assms IsAnOrdGroup_def by auto text{*It is good to have transitivity handy.*} lemma (in group3) Group_order_transitive: assumes A1: "a\<lsq>b" "b\<lsq>c" shows "a\<lsq>c" proof - from ordGroupAssum have "trans(r)" using IsAnOrdGroup_def IsPartOrder_def by simp moreover from A1 have "⟨ a,b⟩ ∈ r ∧ ⟨ b,c⟩ ∈ r" by simp ultimately have "⟨ a,c⟩ ∈ r" by (rule Fol1_L3) thus ?thesis by simp qed text{*The order in an ordered group is antisymmetric.*} lemma (in group3) group_order_antisym: assumes A1: "a\<lsq>b" "b\<lsq>a" shows "a=b" proof - from ordGroupAssum A1 have "antisym(r)" "⟨ a,b⟩ ∈ r" "⟨ b,a⟩ ∈ r" using IsAnOrdGroup_def IsPartOrder_def by auto then show "a=b" by (rule Fol1_L4) qed text{*Transitivity for the strict order: if $a < b$ and $b\leq c$, then $a < c$. *} lemma (in group3) OrderedGroup_ZF_1_L4A: assumes A1: "a\<ls>b" and A2: "b\<lsq>c" shows "a\<ls>c" proof - from A1 A2 have "a\<lsq>b" "b\<lsq>c" by auto then have "a\<lsq>c" by (rule Group_order_transitive) moreover from A1 A2 have "a≠c" using group_order_antisym by auto ultimately show "a\<ls>c" by simp qed text{*Another version of transitivity for the strict order: if $a\leq b$ and $b < c$, then $a < c$. *} lemma (in group3) group_strict_ord_transit: assumes A1: "a\<lsq>b" and A2: "b\<ls>c" shows "a\<ls>c" proof - from A1 A2 have "a\<lsq>b" "b\<lsq>c" by auto then have "a\<lsq>c" by (rule Group_order_transitive) moreover from A1 A2 have "a≠c" using group_order_antisym by auto ultimately show "a\<ls>c" by simp qed text{*Strict order is preserved by translations.*} lemma (in group3) group_strict_ord_transl_inv: assumes "a\<ls>b" and "c∈G" shows "a⋅c \<ls> b⋅c" "c⋅a \<ls> c⋅b" using ordGroupAssum assms IsAnOrdGroup_def OrderedGroup_ZF_1_L4 OrderedGroup_ZF_1_L1 group0.group0_2_L19 by auto text{*If the group order is total, then the group is ordered linearly.*} lemma (in group3) group_ord_total_is_lin: assumes "r {is total on} G" shows "IsLinOrder(G,r)" using assms ordGroupAssum IsAnOrdGroup_def Order_ZF_1_L3 by simp text{*For linearly ordered groups elements in the nonnegative set are greater than those in the complement.*} lemma (in group3) OrderedGroup_ZF_1_L4B: assumes "r {is total on} G" and "a∈G⇧^{+}" and "b ∈ G-G⇧^{+}" shows "b\<lsq>a" proof - from assms have "b\<lsq>𝟭" "𝟭\<lsq>a" using OrderedGroup_ZF_1_L2 OrderedGroup_ZF_1_L2B by auto then show ?thesis by (rule Group_order_transitive) qed text{*If $a\leq 1$ and $a\neq 1$, then $a \in G\setminus G^{+}$. *} lemma (in group3) OrderedGroup_ZF_1_L4C: assumes A1: "a\<lsq>𝟭" and A2: "a≠𝟭" shows "a ∈ G-G⇧^{+}" proof - { assume "a ∉ G-G⇧^{+}" with ordGroupAssum A1 A2 have False using OrderedGroup_ZF_1_L4 OrderedGroup_ZF_1_L2 OrderedGroup_ZF_1_L4 IsAnOrdGroup_def IsPartOrder_def antisym_def by auto } thus ?thesis by auto qed text{*An element smaller than an element in $G\setminus G^+$ is in $G\setminus G^+$.*} lemma (in group3) OrderedGroup_ZF_1_L4D: assumes A1: "a∈G-G⇧^{+}" and A2: "b\<lsq>a" shows "b∈G-G⇧^{+}" proof - { assume "b ∉ G - G⇧^{+}" with A2 have "𝟭\<lsq>b" "b\<lsq>a" using OrderedGroup_ZF_1_L4 OrderedGroup_ZF_1_L2 by auto then have "𝟭\<lsq>a" by (rule Group_order_transitive) with A1 have False using OrderedGroup_ZF_1_L2 by simp } thus ?thesis by auto qed text{*The nonnegative set is contained in the group.*} lemma (in group3) OrderedGroup_ZF_1_L4E: shows "G⇧^{+}⊆ G" using OrderedGroup_ZF_1_L2 OrderedGroup_ZF_1_L4 by auto text{*Taking the inverse on both sides reverses the inequality.*} lemma (in group3) OrderedGroup_ZF_1_L5: assumes A1: "a\<lsq>b" shows "b¯\<lsq>a¯" proof - from A1 have T1: "a∈G" "b∈G" "a¯∈G" "b¯∈G" using OrderedGroup_ZF_1_L4 OrderedGroup_ZF_1_L1 group0.inverse_in_group by auto with A1 ordGroupAssum have "a⋅a¯\<lsq>b⋅a¯" using IsAnOrdGroup_def by simp with T1 ordGroupAssum have "b¯⋅𝟭\<lsq>b¯⋅(b⋅a¯)" using OrderedGroup_ZF_1_L1 group0.group0_2_L6 IsAnOrdGroup_def by simp with T1 show ?thesis using OrderedGroup_ZF_1_L1 group0.group0_2_L2 group0.group_oper_assoc group0.group0_2_L6 by simp qed text{*If an element is smaller that the unit, then its inverse is greater.*} lemma (in group3) OrderedGroup_ZF_1_L5A: assumes A1: "a\<lsq>𝟭" shows "𝟭\<lsq>a¯" proof - from A1 have "𝟭¯\<lsq>a¯" using OrderedGroup_ZF_1_L5 by simp then show ?thesis using OrderedGroup_ZF_1_L1 group0.group_inv_of_one by simp qed text{*If an the inverse of an element is greater that the unit, then the element is smaller.*} lemma (in group3) OrderedGroup_ZF_1_L5AA: assumes A1: "a∈G" and A2: "𝟭\<lsq>a¯" shows "a\<lsq>𝟭" proof - from A2 have "(a¯)¯\<lsq>𝟭¯" using OrderedGroup_ZF_1_L5 by simp with A1 show "a\<lsq>𝟭" using OrderedGroup_ZF_1_L1 group0.group_inv_of_inv group0.group_inv_of_one by simp qed text{*If an element is nonnegative, then the inverse is not greater that the unit. Also shows that nonnegative elements cannot be negative*} lemma (in group3) OrderedGroup_ZF_1_L5AB: assumes A1: "𝟭\<lsq>a" shows "a¯\<lsq>𝟭" and "¬(a\<lsq>𝟭 ∧ a≠𝟭)" proof - from A1 have "a¯\<lsq>𝟭¯" using OrderedGroup_ZF_1_L5 by simp then show "a¯\<lsq>𝟭" using OrderedGroup_ZF_1_L1 group0.group_inv_of_one by simp { assume "a\<lsq>𝟭" and "a≠𝟭" with A1 have False using group_order_antisym by blast } then show "¬(a\<lsq>𝟭 ∧ a≠𝟭)" by auto qed text{*If two elements are greater or equal than the unit, then the inverse of one is not greater than the other.*} lemma (in group3) OrderedGroup_ZF_1_L5AC: assumes A1: "𝟭\<lsq>a" "𝟭\<lsq>b" shows "a¯ \<lsq> b" proof - from A1 have "a¯\<lsq>𝟭" "𝟭\<lsq>b" using OrderedGroup_ZF_1_L5AB by auto then show "a¯ \<lsq> b" by (rule Group_order_transitive) qed subsection{*Inequalities*} text{*This section developes some simple tools to deal with inequalities.*} text{*Taking negative on both sides reverses the inequality, case with an inverse on one side.*} lemma (in group3) OrderedGroup_ZF_1_L5AD: assumes A1: "b ∈ G" and A2: "a\<lsq>b¯" shows "b \<lsq> a¯" proof - from A2 have "(b¯)¯ \<lsq> a¯" using OrderedGroup_ZF_1_L5 by simp with A1 show "b \<lsq> a¯" using OrderedGroup_ZF_1_L1 group0.group_inv_of_inv by simp qed text{*We can cancel the same element on both sides of an inequality.*} lemma (in group3) OrderedGroup_ZF_1_L5AE: assumes A1: "a∈G" "b∈G" "c∈G" and A2: "a⋅b \<lsq> a⋅c" shows "b\<lsq>c" proof - from ordGroupAssum A1 A2 have "a¯⋅(a⋅b) \<lsq> a¯⋅(a⋅c)" using OrderedGroup_ZF_1_L1 group0.inverse_in_group IsAnOrdGroup_def by simp with A1 show "b\<lsq>c" using OrderedGroup_ZF_1_L1 group0.inv_cancel_two by simp qed text{*We can cancel the same element on both sides of an inequality, a version with an inverse on both sides.*} lemma (in group3) OrderedGroup_ZF_1_L5AF: assumes A1: "a∈G" "b∈G" "c∈G" and A2: "a⋅b¯ \<lsq> a⋅c¯" shows "c\<lsq>b" proof - from A1 A2 have "(c¯)¯ \<lsq> (b¯)¯" using OrderedGroup_ZF_1_L1 group0.inverse_in_group OrderedGroup_ZF_1_L5AE OrderedGroup_ZF_1_L5 by simp with A1 show "c\<lsq>b" using OrderedGroup_ZF_1_L1 group0.group_inv_of_inv by simp qed text{*Taking negative on both sides reverses the inequality, another case with an inverse on one side.*} lemma (in group3) OrderedGroup_ZF_1_L5AG: assumes A1: "a ∈ G" and A2: "a¯\<lsq>b" shows "b¯ \<lsq> a" proof - from A2 have "b¯ \<lsq> (a¯)¯" using OrderedGroup_ZF_1_L5 by simp with A1 show "b¯ \<lsq> a" using OrderedGroup_ZF_1_L1 group0.group_inv_of_inv by simp qed text{*We can multiply the sides of two inequalities.*} lemma (in group3) OrderedGroup_ZF_1_L5B: assumes A1: "a\<lsq>b" and A2: "c\<lsq>d" shows "a⋅c \<lsq> b⋅d" proof - from A1 A2 have "c∈G" "b∈G" using OrderedGroup_ZF_1_L4 by auto with A1 A2 ordGroupAssum have "a⋅c\<lsq> b⋅c" "b⋅c\<lsq>b⋅d" using IsAnOrdGroup_def by auto then show "a⋅c \<lsq> b⋅d" by (rule Group_order_transitive) qed text{*We can replace first of the factors on one side of an inequality with a greater one.*} lemma (in group3) OrderedGroup_ZF_1_L5C: assumes A1: "c∈G" and A2: "a\<lsq>b⋅c" and A3: "b\<lsq>b⇩_{1}" shows "a\<lsq>b⇩_{1}⋅c" proof - from A1 A3 have "b⋅c \<lsq> b⇩_{1}⋅c" using OrderedGroup_ZF_1_L3 OrderedGroup_ZF_1_L5B by simp with A2 show "a\<lsq>b⇩_{1}⋅c" by (rule Group_order_transitive) qed text{*We can replace second of the factors on one side of an inequality with a greater one.*} lemma (in group3) OrderedGroup_ZF_1_L5D: assumes A1: "b∈G" and A2: "a \<lsq> b⋅c" and A3: "c\<lsq>b⇩_{1}" shows "a \<lsq> b⋅b⇩_{1}" proof - from A1 A3 have "b⋅c \<lsq> b⋅b⇩_{1}" using OrderedGroup_ZF_1_L3 OrderedGroup_ZF_1_L5B by auto with A2 show "a\<lsq>b⋅b⇩_{1}" by (rule Group_order_transitive) qed text{*We can replace factors on one side of an inequality with greater ones.*} lemma (in group3) OrderedGroup_ZF_1_L5E: assumes A1: "a \<lsq> b⋅c" and A2: "b\<lsq>b⇩_{1}" "c\<lsq>c⇩_{1}" shows "a \<lsq> b⇩_{1}⋅c⇩_{1}" proof - from A2 have "b⋅c \<lsq> b⇩_{1}⋅c⇩_{1}" using OrderedGroup_ZF_1_L5B by simp with A1 show "a\<lsq>b⇩_{1}⋅c⇩_{1}" by (rule Group_order_transitive) qed text{*We don't decrease an element of the group by multiplying by one that is nonnegative.*} lemma (in group3) OrderedGroup_ZF_1_L5F: assumes A1: "𝟭\<lsq>a" and A2: "b∈G" shows "b\<lsq>a⋅b" "b\<lsq>b⋅a" proof - from ordGroupAssum A1 A2 have "𝟭⋅b\<lsq>a⋅b" "b⋅𝟭\<lsq>b⋅a" using IsAnOrdGroup_def by auto with A2 show "b\<lsq>a⋅b" "b\<lsq>b⋅a" using OrderedGroup_ZF_1_L1 group0.group0_2_L2 by auto qed text{*We can multiply the right hand side of an inequality by a nonnegative element.*} lemma (in group3) OrderedGroup_ZF_1_L5G: assumes A1: "a\<lsq>b" and A2: "𝟭\<lsq>c" shows "a\<lsq>b⋅c" "a\<lsq>c⋅b" proof - from A1 A2 have I: "b\<lsq>b⋅c" and II: "b\<lsq>c⋅b" using OrderedGroup_ZF_1_L4 OrderedGroup_ZF_1_L5F by auto from A1 I show "a\<lsq>b⋅c" by (rule Group_order_transitive) from A1 II show "a\<lsq>c⋅b" by (rule Group_order_transitive) qed text{*We can put two elements on the other side of inequality, changing their sign.*} lemma (in group3) OrderedGroup_ZF_1_L5H: assumes A1: "a∈G" "b∈G" and A2: "a⋅b¯ \<lsq> c" shows "a \<lsq> c⋅b" "c¯⋅a \<lsq> b" proof - from A2 have T: "c∈G" "c¯ ∈ G" using OrderedGroup_ZF_1_L4 OrderedGroup_ZF_1_L1 group0.inverse_in_group by auto from ordGroupAssum A1 A2 have "a⋅b¯⋅b \<lsq> c⋅b" using IsAnOrdGroup_def by simp with A1 show "a \<lsq> c⋅b" using OrderedGroup_ZF_1_L1 group0.inv_cancel_two by simp with ordGroupAssum A2 T have "c¯⋅a \<lsq> c¯⋅(c⋅b)" using IsAnOrdGroup_def by simp with A1 T show "c¯⋅a \<lsq> b" using OrderedGroup_ZF_1_L1 group0.inv_cancel_two by simp qed text{*We can multiply the sides of one inequality by inverse of another.*} lemma (in group3) OrderedGroup_ZF_1_L5I: assumes "a\<lsq>b" and "c\<lsq>d" shows "a⋅d¯ \<lsq> b⋅c¯" using assms OrderedGroup_ZF_1_L5 OrderedGroup_ZF_1_L5B by simp text{*We can put an element on the other side of an inequality changing its sign, version with the inverse.*} lemma (in group3) OrderedGroup_ZF_1_L5J: assumes A1: "a∈G" "b∈G" and A2: "c \<lsq> a⋅b¯" shows "c⋅b \<lsq> a" proof - from ordGroupAssum A1 A2 have "c⋅b \<lsq> a⋅b¯⋅b" using IsAnOrdGroup_def by simp with A1 show "c⋅b \<lsq> a" using OrderedGroup_ZF_1_L1 group0.inv_cancel_two by simp qed text{*We can put an element on the other side of an inequality changing its sign, version with the inverse.*} lemma (in group3) OrderedGroup_ZF_1_L5JA: assumes A1: "a∈G" "b∈G" and A2: "c \<lsq> a¯⋅b" shows "a⋅c\<lsq> b" proof - from ordGroupAssum A1 A2 have "a⋅c \<lsq> a⋅(a¯⋅b)" using IsAnOrdGroup_def by simp with A1 show "a⋅c\<lsq> b" using OrderedGroup_ZF_1_L1 group0.inv_cancel_two by simp qed text{*A special case of @{text "OrderedGroup_ZF_1_L5J"} where $c=1$.*} corollary (in group3) OrderedGroup_ZF_1_L5K: assumes A1: "a∈G" "b∈G" and A2: "𝟭 \<lsq> a⋅b¯" shows "b \<lsq> a" proof - from A1 A2 have "𝟭⋅b \<lsq> a" using OrderedGroup_ZF_1_L5J by simp with A1 show "b \<lsq> a" using OrderedGroup_ZF_1_L1 group0.group0_2_L2 by simp qed text{*A special case of @{text "OrderedGroup_ZF_1_L5JA"} where $c=1$.*} corollary (in group3) OrderedGroup_ZF_1_L5KA: assumes A1: "a∈G" "b∈G" and A2: "𝟭 \<lsq> a¯⋅b" shows "a \<lsq> b" proof - from A1 A2 have "a⋅𝟭 \<lsq> b" using OrderedGroup_ZF_1_L5JA by simp with A1 show "a \<lsq> b" using OrderedGroup_ZF_1_L1 group0.group0_2_L2 by simp qed text{*If the order is total, the elements that do not belong to the positive set are negative. We also show here that the group inverse of an element that does not belong to the nonnegative set does belong to the nonnegative set.*} lemma (in group3) OrderedGroup_ZF_1_L6: assumes A1: "r {is total on} G" and A2: "a∈G-G⇧^{+}" shows "a\<lsq>𝟭" "a¯ ∈ G⇧^{+}" "restrict(GroupInv(G,P),G-G⇧^{+})`(a) ∈ G⇧^{+}" proof - from A2 have T1: "a∈G" "a∉G⇧^{+}" "𝟭∈G" using OrderedGroup_ZF_1_L1 group0.group0_2_L2 by auto with A1 show "a\<lsq>𝟭" using OrderedGroup_ZF_1_L2 IsTotal_def by auto then show "a¯ ∈ G⇧^{+}" using OrderedGroup_ZF_1_L5A OrderedGroup_ZF_1_L2 by simp with A2 show "restrict(GroupInv(G,P),G-G⇧^{+})`(a) ∈ G⇧^{+}" using restrict by simp qed text{*If a property is invariant with respect to taking the inverse and it is true on the nonnegative set, than it is true on the whole group.*} lemma (in group3) OrderedGroup_ZF_1_L7: assumes A1: "r {is total on} G" and A2: "∀a∈G⇧^{+}.∀b∈G⇧^{+}. Q(a,b)" and A3: "∀a∈G.∀b∈G. Q(a,b)⟶Q(a¯,b)" and A4: "∀a∈G.∀b∈G. Q(a,b)⟶Q(a,b¯)" and A5: "a∈G" "b∈G" shows "Q(a,b)" proof - { assume A6: "a∈G⇧^{+}" have "Q(a,b)" proof - { assume "b∈G⇧^{+}" with A6 A2 have "Q(a,b)" by simp } moreover { assume "b∉G⇧^{+}" with A1 A2 A4 A5 A6 have "Q(a,(b¯)¯)" using OrderedGroup_ZF_1_L6 OrderedGroup_ZF_1_L1 group0.inverse_in_group by simp with A5 have "Q(a,b)" using OrderedGroup_ZF_1_L1 group0.group_inv_of_inv by simp } ultimately show "Q(a,b)" by auto qed } moreover { assume "a∉G⇧^{+}" with A1 A5 have T1: "a¯ ∈ G⇧^{+}" using OrderedGroup_ZF_1_L6 by simp have "Q(a,b)" proof - { assume "b∈G⇧^{+}" with A2 A3 A5 T1 have "Q((a¯)¯,b)" using OrderedGroup_ZF_1_L1 group0.inverse_in_group by simp with A5 have "Q(a,b)" using OrderedGroup_ZF_1_L1 group0.group_inv_of_inv by simp } moreover { assume "b∉G⇧^{+}" with A1 A2 A3 A4 A5 T1 have "Q((a¯)¯,(b¯)¯)" using OrderedGroup_ZF_1_L6 OrderedGroup_ZF_1_L1 group0.inverse_in_group by simp with A5 have "Q(a,b)" using OrderedGroup_ZF_1_L1 group0.group_inv_of_inv by simp } ultimately show "Q(a,b)" by auto qed } ultimately show "Q(a,b)" by auto qed text{*A lemma about splitting the ordered group "plane" into 6 subsets. Useful for proofs by cases.*} lemma (in group3) OrdGroup_6cases: assumes A1: "r {is total on} G" and A2: "a∈G" "b∈G" shows "𝟭\<lsq>a ∧ 𝟭\<lsq>b ∨ a\<lsq>𝟭 ∧ b\<lsq>𝟭 ∨ a\<lsq>𝟭 ∧ 𝟭\<lsq>b ∧ 𝟭 \<lsq> a⋅b ∨ a\<lsq>𝟭 ∧ 𝟭\<lsq>b ∧ a⋅b \<lsq> 𝟭 ∨ 𝟭\<lsq>a ∧ b\<lsq>𝟭 ∧ 𝟭 \<lsq> a⋅b ∨ 𝟭\<lsq>a ∧ b\<lsq>𝟭 ∧ a⋅b \<lsq> 𝟭" proof - from A1 A2 have "𝟭\<lsq>a ∨ a\<lsq>𝟭" "𝟭\<lsq>b ∨ b\<lsq>𝟭" "𝟭 \<lsq> a⋅b ∨ a⋅b \<lsq> 𝟭" using OrderedGroup_ZF_1_L1 group0.group_op_closed group0.group0_2_L2 IsTotal_def by auto then show ?thesis by auto qed text{*The next lemma shows what happens when one element of a totally ordered group is not greater or equal than another.*} lemma (in group3) OrderedGroup_ZF_1_L8: assumes A1: "r {is total on} G" and A2: "a∈G" "b∈G" and A3: "¬(a\<lsq>b)" shows "b \<lsq> a" "a¯ \<lsq> b¯" "a≠b" "b\<ls>a" proof - from A1 A2 A3 show I: "b \<lsq> a" using IsTotal_def by auto then show "a¯ \<lsq> b¯" using OrderedGroup_ZF_1_L5 by simp from A2 have "a \<lsq> a" using OrderedGroup_ZF_1_L3 by simp with I A3 show "a≠b" "b \<ls> a" by auto qed text{*If one element is greater or equal and not equal to another, then it is not smaller or equal.*} lemma (in group3) OrderedGroup_ZF_1_L8AA: assumes A1: "a\<lsq>b" and A2: "a≠b" shows "¬(b\<lsq>a)" proof - { note A1 moreover assume "b\<lsq>a" ultimately have "a=b" by (rule group_order_antisym) with A2 have False by simp } thus "¬(b\<lsq>a)" by auto qed text{*A special case of @{text "OrderedGroup_ZF_1_L8"} when one of the elements is the unit.*} corollary (in group3) OrderedGroup_ZF_1_L8A: assumes A1: "r {is total on} G" and A2: "a∈G" and A3: "¬(𝟭\<lsq>a)" shows "𝟭 \<lsq> a¯" "𝟭≠a" "a\<lsq>𝟭" proof - from A1 A2 A3 have I: "r {is total on} G" "𝟭∈G" "a∈G" "¬(𝟭\<lsq>a)" using OrderedGroup_ZF_1_L1 group0.group0_2_L2 by auto then have "𝟭¯ \<lsq> a¯" by (rule OrderedGroup_ZF_1_L8) then show "𝟭 \<lsq> a¯" using OrderedGroup_ZF_1_L1 group0.group_inv_of_one by simp from I show "𝟭≠a" by (rule OrderedGroup_ZF_1_L8) from A1 I show "a\<lsq>𝟭" using IsTotal_def by auto qed text{*A negative element can not be nonnegative.*} lemma (in group3) OrderedGroup_ZF_1_L8B: assumes A1: "a\<lsq>𝟭" and A2: "a≠𝟭" shows "¬(𝟭\<lsq>a)" proof - { assume "𝟭\<lsq>a" with A1 have "a=𝟭" using group_order_antisym by auto with A2 have False by simp } thus ?thesis by auto qed text{*An element is greater or equal than another iff the difference is nonpositive.*} lemma (in group3) OrderedGroup_ZF_1_L9: assumes A1: "a∈G" "b∈G" shows "a\<lsq>b ⟷ a⋅b¯ \<lsq> 𝟭" proof assume "a \<lsq> b" with ordGroupAssum A1 have "a⋅b¯ \<lsq> b⋅b¯" using OrderedGroup_ZF_1_L1 group0.inverse_in_group IsAnOrdGroup_def by simp with A1 show "a⋅b¯ \<lsq> 𝟭" using OrderedGroup_ZF_1_L1 group0.group0_2_L6 by simp next assume A2: "a⋅b¯ \<lsq> 𝟭" with ordGroupAssum A1 have "a⋅b¯⋅b \<lsq> 𝟭⋅b" using IsAnOrdGroup_def by simp with A1 show "a \<lsq> b" using OrderedGroup_ZF_1_L1 group0.inv_cancel_two group0.group0_2_L2 by simp qed text{*We can move an element to the other side of an inequality.*} lemma (in group3) OrderedGroup_ZF_1_L9A: assumes A1: "a∈G" "b∈G" "c∈G" shows "a⋅b \<lsq> c ⟷ a \<lsq> c⋅b¯" proof assume "a⋅b \<lsq> c" with ordGroupAssum A1 have "a⋅b⋅b¯ \<lsq> c⋅b¯" using OrderedGroup_ZF_1_L1 group0.inverse_in_group IsAnOrdGroup_def by simp with A1 show "a \<lsq> c⋅b¯" using OrderedGroup_ZF_1_L1 group0.inv_cancel_two by simp next assume "a \<lsq> c⋅b¯" with ordGroupAssum A1 have "a⋅b \<lsq> c⋅b¯⋅b" using OrderedGroup_ZF_1_L1 group0.inverse_in_group IsAnOrdGroup_def by simp with A1 show "a⋅b \<lsq> c" using OrderedGroup_ZF_1_L1 group0.inv_cancel_two by simp qed text{*A one side version of the previous lemma with weaker assuptions.*} lemma (in group3) OrderedGroup_ZF_1_L9B: assumes A1: "a∈G" "b∈G" and A2: "a⋅b¯ \<lsq> c" shows "a \<lsq> c⋅b" proof - from A1 A2 have "a∈G" "b¯∈G" "c∈G" using OrderedGroup_ZF_1_L1 group0.inverse_in_group OrderedGroup_ZF_1_L4 by auto with A1 A2 show "a \<lsq> c⋅b" using OrderedGroup_ZF_1_L9A OrderedGroup_ZF_1_L1 group0.group_inv_of_inv by simp qed text{*We can put en element on the other side of inequality, changing its sign.*} lemma (in group3) OrderedGroup_ZF_1_L9C: assumes A1: "a∈G" "b∈G" and A2: "c\<lsq>a⋅b" shows "c⋅b¯ \<lsq> a" "a¯⋅c \<lsq> b" proof - from ordGroupAssum A1 A2 have "c⋅b¯ \<lsq> a⋅b⋅b¯" "a¯⋅c \<lsq> a¯⋅(a⋅b)" using OrderedGroup_ZF_1_L1 group0.inverse_in_group IsAnOrdGroup_def by auto with A1 show "c⋅b¯ \<lsq> a" "a¯⋅c \<lsq> b" using OrderedGroup_ZF_1_L1 group0.inv_cancel_two by auto qed text{*If an element is greater or equal than another then the difference is nonnegative.*} lemma (in group3) OrderedGroup_ZF_1_L9D: assumes A1: "a\<lsq>b" shows "𝟭 \<lsq> b⋅a¯" proof - from A1 have T: "a∈G" "b∈G" "a¯ ∈ G" using OrderedGroup_ZF_1_L4 OrderedGroup_ZF_1_L1 group0.inverse_in_group by auto with ordGroupAssum A1 have "a⋅a¯ \<lsq> b⋅a¯" using IsAnOrdGroup_def by simp with T show "𝟭 \<lsq> b⋅a¯" using OrderedGroup_ZF_1_L1 group0.group0_2_L6 by simp qed text{*If an element is greater than another then the difference is positive.*} lemma (in group3) OrderedGroup_ZF_1_L9E: assumes A1: "a\<lsq>b" "a≠b" shows "𝟭 \<lsq> b⋅a¯" "𝟭 ≠ b⋅a¯" "b⋅a¯ ∈ G⇩_{+}" proof - from A1 have T: "a∈G" "b∈G" using OrderedGroup_ZF_1_L4 by auto from A1 show I: "𝟭 \<lsq> b⋅a¯" using OrderedGroup_ZF_1_L9D by simp { assume "b⋅a¯ = 𝟭" with T have "a=b" using OrderedGroup_ZF_1_L1 group0.group0_2_L11A by auto with A1 have False by simp } then show "𝟭 ≠ b⋅a¯" by auto then have "b⋅a¯ ≠ 𝟭" by auto with I show "b⋅a¯ ∈ G⇩_{+}" using OrderedGroup_ZF_1_L2A by simp qed text{*If the difference is nonnegative, then $a\leq b$. *} lemma (in group3) OrderedGroup_ZF_1_L9F: assumes A1: "a∈G" "b∈G" and A2: "𝟭 \<lsq> b⋅a¯" shows "a\<lsq>b" proof - from A1 A2 have "𝟭⋅a \<lsq> b" using OrderedGroup_ZF_1_L4 OrderedGroup_ZF_1_L9A by simp with A1 show "a\<lsq>b" using OrderedGroup_ZF_1_L1 group0.group0_2_L2 by simp qed text{*If we increase the middle term in a product, the whole product increases.*} lemma (in group3) OrderedGroup_ZF_1_L10: assumes "a∈G" "b∈G" and "c\<lsq>d" shows "a⋅c⋅b \<lsq> a⋅d⋅b" using ordGroupAssum assms IsAnOrdGroup_def by simp text{*A product of (strictly) positive elements is not the unit.*} lemma (in group3) OrderedGroup_ZF_1_L11: assumes A1: "𝟭\<lsq>a" "𝟭\<lsq>b" and A2: "𝟭 ≠ a" "𝟭 ≠ b" shows "𝟭 ≠ a⋅b" proof - from A1 have T1: "a∈G" "b∈G" using OrderedGroup_ZF_1_L4 by auto { assume "𝟭 = a⋅b" with A1 T1 have "a\<lsq>𝟭" "𝟭\<lsq>a" using OrderedGroup_ZF_1_L1 group0.group0_2_L9 OrderedGroup_ZF_1_L5AA by auto then have "a = 𝟭" by (rule group_order_antisym) with A2 have False by simp } then show "𝟭 ≠ a⋅b" by auto qed text{*A product of nonnegative elements is nonnegative.*} lemma (in group3) OrderedGroup_ZF_1_L12: assumes A1: "𝟭 \<lsq> a" "𝟭 \<lsq> b" shows "𝟭 \<lsq> a⋅b" proof - from A1 have "𝟭⋅𝟭 \<lsq> a⋅b" using OrderedGroup_ZF_1_L5B by simp then show "𝟭 \<lsq> a⋅b" using OrderedGroup_ZF_1_L1 group0.group0_2_L2 by simp qed text{*If $a$ is not greater than $b$, then $1$ is not greater than $b\cdot a^{-1}$.*} lemma (in group3) OrderedGroup_ZF_1_L12A: assumes A1: "a\<lsq>b" shows "𝟭 \<lsq> b⋅a¯" proof - from A1 have T: "𝟭 ∈ G" "a∈G" "b∈G" using OrderedGroup_ZF_1_L4 OrderedGroup_ZF_1_L1 group0.group0_2_L2 by auto with A1 have "𝟭⋅a \<lsq> b" using OrderedGroup_ZF_1_L1 group0.group0_2_L2 by simp with T show "𝟭 \<lsq> b⋅a¯" using OrderedGroup_ZF_1_L9A by simp qed text{*We can move an element to the other side of a strict inequality.*} lemma (in group3) OrderedGroup_ZF_1_L12B: assumes A1: "a∈G" "b∈G" and A2: "a⋅b¯ \<ls> c" shows "a \<ls> c⋅b" proof - from A1 A2 have "a⋅b¯⋅b \<ls> c⋅b" using group_strict_ord_transl_inv by auto moreover from A1 have "a⋅b¯⋅b = a" using OrderedGroup_ZF_1_L1 group0.inv_cancel_two by simp ultimately show "a \<ls> c⋅b" by auto qed (*lemma (in group3) OrderedGroup_ZF_1_L12B: assumes A1: "a∈G" "b∈G" and A2: "a⋅b¯ \<lsq> c" "a⋅b¯ ≠ c" shows "a \<lsq> c⋅b" "a ≠ c⋅b" proof - from A1 A2 have "a⋅b¯⋅b \<lsq> c⋅b" "a⋅b¯⋅b ≠ c⋅b" using group_strict_ord_transl_inv by auto moreover from A1 have "a⋅b¯⋅b = a" using OrderedGroup_ZF_1_L1 group0.inv_cancel_two by simp ultimately show "a \<lsq> c⋅b" "a ≠ c⋅b" by auto qed;*) text{*We can multiply the sides of two inequalities, first of them strict and we get a strict inequality.*} lemma (in group3) OrderedGroup_ZF_1_L12C: assumes A1: "a\<ls>b" and A2: "c\<lsq>d" shows "a⋅c \<ls> b⋅d" proof - from A1 A2 have T: "a∈G" "b∈G" "c∈G" "d∈G" using OrderedGroup_ZF_1_L4 by auto with ordGroupAssum A2 have "a⋅c \<lsq> a⋅d" using IsAnOrdGroup_def by simp moreover from A1 T have "a⋅d \<ls> b⋅d" using group_strict_ord_transl_inv by simp ultimately show "a⋅c \<ls> b⋅d" by (rule group_strict_ord_transit) qed text{*We can multiply the sides of two inequalities, second of them strict and we get a strict inequality.*} lemma (in group3) OrderedGroup_ZF_1_L12D: assumes A1: "a\<lsq>b" and A2: "c\<ls>d" shows "a⋅c \<ls> b⋅d" proof - from A1 A2 have T: "a∈G" "b∈G" "c∈G" "d∈G" using OrderedGroup_ZF_1_L4 by auto with A2 have "a⋅c \<ls> a⋅d" using group_strict_ord_transl_inv by simp moreover from ordGroupAssum A1 T have "a⋅d \<lsq> b⋅d" using IsAnOrdGroup_def by simp ultimately show "a⋅c \<ls> b⋅d" by (rule OrderedGroup_ZF_1_L4A) qed subsection{*The set of positive elements*} text{*In this section we study @{text "G⇩_{+}"} - the set of elements that are (strictly) greater than the unit. The most important result is that every linearly ordered group can decomposed into $\{1\}$, @{text "G⇩_{+}"} and the set of those elements $a\in G$ such that $a^{-1}\in$@{text "G⇩_{+}"}. Another property of linearly ordered groups that we prove here is that if @{text "G⇩_{+}"}$\neq \emptyset$, then it is infinite. This allows to show that nontrivial linearly ordered groups are infinite. *} text{*The positive set is closed under the group operation.*} lemma (in group3) OrderedGroup_ZF_1_L13: shows "G⇩_{+}{is closed under} P" proof - { fix a b assume "a∈G⇩_{+}" "b∈G⇩_{+}" then have T1: "𝟭 \<lsq> a⋅b" and "𝟭 ≠ a⋅b" using PositiveSet_def OrderedGroup_ZF_1_L11 OrderedGroup_ZF_1_L12 by auto moreover from T1 have "a⋅b ∈ G" using OrderedGroup_ZF_1_L4 by simp ultimately have "a⋅b ∈ G⇩_{+}" using PositiveSet_def by simp } then show "G⇩_{+}{is closed under} P" using IsOpClosed_def by simp qed text{*For totally ordered groups every nonunit element is positive or its inverse is positive.*} lemma (in group3) OrderedGroup_ZF_1_L14: assumes A1: "r {is total on} G" and A2: "a∈G" shows "a=𝟭 ∨ a∈G⇩_{+}∨ a¯∈G⇩_{+}" proof - { assume A3: "a≠𝟭" moreover from A1 A2 have "a\<lsq>𝟭 ∨ 𝟭\<lsq>a" using IsTotal_def OrderedGroup_ZF_1_L1 group0.group0_2_L2 by simp moreover from A3 A2 have T1: "a¯ ≠ 𝟭" using OrderedGroup_ZF_1_L1 group0.group0_2_L8B by simp ultimately have "a¯∈G⇩_{+}∨ a∈G⇩_{+}" using OrderedGroup_ZF_1_L5A OrderedGroup_ZF_1_L2A by auto } thus "a=𝟭 ∨ a∈G⇩_{+}∨ a¯∈G⇩_{+}" by auto qed text{*If an element belongs to the positive set, then it is not the unit and its inverse does not belong to the positive set.*} lemma (in group3) OrderedGroup_ZF_1_L15: assumes A1: "a∈G⇩_{+}" shows "a≠𝟭" "a¯∉G⇩_{+}" proof - from A1 show T1: "a≠𝟭" using PositiveSet_def by auto { assume "a¯ ∈ G⇩_{+}" with A1 have "a\<lsq>𝟭" "𝟭\<lsq>a" using OrderedGroup_ZF_1_L5AA PositiveSet_def by auto then have "a=𝟭" by (rule group_order_antisym) with T1 have False by simp } then show "a¯∉G⇩_{+}" by auto qed text{*If $a^{-1}$ is positive, then $a$ can not be positive or the unit.*} lemma (in group3) OrderedGroup_ZF_1_L16: assumes A1: "a∈G" and A2: "a¯∈G⇩_{+}" shows "a≠𝟭" "a∉G⇩_{+}" proof - from A2 have "a¯≠𝟭" "(a¯)¯ ∉ G⇩_{+}" using OrderedGroup_ZF_1_L15 by auto with A1 show "a≠𝟭" "a∉G⇩_{+}" using OrderedGroup_ZF_1_L1 group0.group0_2_L8C group0.group_inv_of_inv by auto qed text{*For linearly ordered groups each element is either the unit, positive or its inverse is positive.*} lemma (in group3) OrdGroup_decomp: assumes A1: "r {is total on} G" and A2: "a∈G" shows "Exactly_1_of_3_holds (a=𝟭,a∈G⇩_{+},a¯∈G⇩_{+})" proof - from A1 A2 have "a=𝟭 ∨ a∈G⇩_{+}∨ a¯∈G⇩_{+}" using OrderedGroup_ZF_1_L14 by simp moreover from A2 have "a=𝟭 ⟶ (a∉G⇩_{+}∧ a¯∉G⇩_{+})" using OrderedGroup_ZF_1_L1 group0.group_inv_of_one PositiveSet_def by simp moreover from A2 have "a∈G⇩_{+}⟶ (a≠𝟭 ∧ a¯∉G⇩_{+})" using OrderedGroup_ZF_1_L15 by simp moreover from A2 have "a¯∈G⇩_{+}⟶ (a≠𝟭 ∧ a∉G⇩_{+})" using OrderedGroup_ZF_1_L16 by simp ultimately show "Exactly_1_of_3_holds (a=𝟭,a∈G⇩_{+},a¯∈G⇩_{+})" by (rule Fol1_L5) qed text{*A if $a$ is a nonunit element that is not positive, then $a^{-1}$ is is positive. This is useful for some proofs by cases.*} lemma (in group3) OrdGroup_cases: assumes A1: "r {is total on} G" and A2: "a∈G" and A3: "a≠𝟭" "a∉G⇩_{+}" shows "a¯ ∈ G⇩_{+}" proof - from A1 A2 have "a=𝟭 ∨ a∈G⇩_{+}∨ a¯∈G⇩_{+}" using OrderedGroup_ZF_1_L14 by simp with A3 show "a¯ ∈ G⇩_{+}" by auto qed text{*Elements from $G\setminus G_+$ are not greater that the unit.*} lemma (in group3) OrderedGroup_ZF_1_L17: assumes A1: "r {is total on} G" and A2: "a ∈ G-G⇩_{+}" shows "a\<lsq>𝟭" proof - { assume "a=𝟭" with A2 have "a\<lsq>𝟭" using OrderedGroup_ZF_1_L3 by simp } moreover { assume "a≠𝟭" with A1 A2 have "a\<lsq>𝟭" using PositiveSet_def OrderedGroup_ZF_1_L8A by auto } ultimately show "a\<lsq>𝟭" by auto qed text{*The next lemma allows to split proofs that something holds for all $a\in G$ into cases $a=1$, $a\in G_+$, $-a\in G_+$.*} lemma (in group3) OrderedGroup_ZF_1_L18: assumes A1: "r {is total on} G" and A2: "b∈G" and A3: "Q(𝟭)" and A4: "∀a∈G⇩_{+}. Q(a)" and A5: "∀a∈G⇩_{+}. Q(a¯)" shows "Q(b)" proof - from A1 A2 A3 A4 A5 have "Q(b) ∨ Q((b¯)¯)" using OrderedGroup_ZF_1_L14 by auto with A2 show "Q(b)" using OrderedGroup_ZF_1_L1 group0.group_inv_of_inv by simp qed text{*All elements greater or equal than an element of @{text "G⇩_{+}"} belong to @{text "G⇩_{+}"}.*} lemma (in group3) OrderedGroup_ZF_1_L19: assumes A1: "a ∈ G⇩_{+}" and A2: "a\<lsq>b" shows "b ∈ G⇩_{+}" proof - from A1 have I: "𝟭\<lsq>a" and II: "a≠𝟭" using OrderedGroup_ZF_1_L2A by auto from I A2 have "𝟭\<lsq>b" by (rule Group_order_transitive) moreover have "b≠𝟭" proof - { assume "b=𝟭" with I A2 have "𝟭\<lsq>a" "a\<lsq>𝟭" by auto then have "𝟭=a" by (rule group_order_antisym) with II have False by simp } then show "b≠𝟭" by auto qed ultimately show "b ∈ G⇩_{+}" using OrderedGroup_ZF_1_L2A by simp qed text{*The inverse of an element of @{text "G⇩_{+}"} cannot be in @{text "G⇩_{+}"}.*} lemma (in group3) OrderedGroup_ZF_1_L20: assumes A1: "r {is total on} G" and A2: "a ∈ G⇩_{+}" shows "a¯ ∉ G⇩_{+}" proof - from A2 have "a∈G" using PositiveSet_def by simp with A1 have "Exactly_1_of_3_holds (a=𝟭,a∈G⇩_{+},a¯∈G⇩_{+})" using OrdGroup_decomp by simp with A2 show "a¯ ∉ G⇩_{+}" by (rule Fol1_L7) qed text{*The set of positive elements of a nontrivial linearly ordered group is not empty.*} lemma (in group3) OrderedGroup_ZF_1_L21: assumes A1: "r {is total on} G" and A2: "G ≠ {𝟭}" shows "G⇩_{+}≠ 0" proof - have "𝟭 ∈ G" using OrderedGroup_ZF_1_L1 group0.group0_2_L2 by simp with A2 obtain a where "a∈G" "a≠𝟭" by auto with A1 have "a∈G⇩_{+}∨ a¯∈G⇩_{+}" using OrderedGroup_ZF_1_L14 by auto then show "G⇩_{+}≠ 0" by auto qed text{*If $b\in$@{text "G⇩_{+}"}, then $a < a\cdot b$. Multiplying $a$ by a positive elemnt increases $a$. *} lemma (in group3) OrderedGroup_ZF_1_L22: assumes A1: "a∈G" "b∈G⇩_{+}" shows "a\<lsq>a⋅b" "a ≠ a⋅b" "a⋅b ∈ G" proof - from ordGroupAssum A1 have "a⋅𝟭 \<lsq> a⋅b" using OrderedGroup_ZF_1_L2A IsAnOrdGroup_def by simp with A1 show "a\<lsq>a⋅b" using OrderedGroup_ZF_1_L1 group0.group0_2_L2 by simp then show "a⋅b ∈ G" using OrderedGroup_ZF_1_L4 by simp { from A1 have "a∈G" "b∈G" using PositiveSet_def by auto moreover assume "a = a⋅b" ultimately have "b = 𝟭" using OrderedGroup_ZF_1_L1 group0.group0_2_L7 by simp with A1 have False using PositiveSet_def by simp } then show "a ≠ a⋅b" by auto qed text{*If $G$ is a nontrivial linearly ordered hroup, then for every element of $G$ we can find one in @{text "G⇩_{+}"} that is greater or equal.*} lemma (in group3) OrderedGroup_ZF_1_L23: assumes A1: "r {is total on} G" and A2: "G ≠ {𝟭}" and A3: "a∈G" shows "∃b∈G⇩_{+}. a\<lsq>b" proof - { assume A4: "a∈G⇩_{+}" then have "a\<lsq>a" using PositiveSet_def OrderedGroup_ZF_1_L3 by simp with A4 have "∃b∈G⇩_{+}. a\<lsq>b" by auto } moreover { assume "a∉G⇩_{+}" with A1 A3 have I: "a\<lsq>𝟭" using OrderedGroup_ZF_1_L17 by simp from A1 A2 obtain b where II: "b∈G⇩_{+}" using OrderedGroup_ZF_1_L21 by auto then have "𝟭\<lsq>b" using PositiveSet_def by simp with I have "a\<lsq>b" by (rule Group_order_transitive) with II have "∃b∈G⇩_{+}. a\<lsq>b" by auto } ultimately show ?thesis by auto qed text{*The @{text "G⇧^{+}"} is @{text "G⇩_{+}"} plus the unit.*} lemma (in group3) OrderedGroup_ZF_1_L24: shows "G⇧^{+}= G⇩_{+}∪{𝟭}" using OrderedGroup_ZF_1_L2 OrderedGroup_ZF_1_L2A OrderedGroup_ZF_1_L3A by auto text{*What is $-G_+$, really?*} lemma (in group3) OrderedGroup_ZF_1_L25: shows "(\<sm>G⇩_{+}) = {a¯. a∈G⇩_{+}}" "(\<sm>G⇩_{+}) ⊆ G" proof - from ordGroupAssum have I: "GroupInv(G,P) : G→G" using IsAnOrdGroup_def group0_2_T2 by simp moreover have "G⇩_{+}⊆ G" using PositiveSet_def by auto ultimately show "(\<sm>G⇩_{+}) = {a¯. a∈G⇩_{+}}" "(\<sm>G⇩_{+}) ⊆ G" using func_imagedef func1_1_L6 by auto qed text{*If the inverse of $a$ is in @{text "G⇩_{+}"}, then $a$ is in the inverse of @{text "G⇩_{+}"}.*} lemma (in group3) OrderedGroup_ZF_1_L26: assumes A1: "a∈G" and A2: "a¯ ∈ G⇩_{+}" shows "a ∈ (\<sm>G⇩_{+})" proof - from A1 have "a¯ ∈ G" "a = (a¯)¯" using OrderedGroup_ZF_1_L1 group0.inverse_in_group group0.group_inv_of_inv by auto with A2 show "a ∈ (\<sm>G⇩_{+})" using OrderedGroup_ZF_1_L25 by auto qed text{*If $a$ is in the inverse of @{text "G⇩_{+}"}, then its inverse is in $G_+$.*} lemma (in group3) OrderedGroup_ZF_1_L27: assumes "a ∈ (\<sm>G⇩_{+})" shows "a¯ ∈ G⇩_{+}" using assms OrderedGroup_ZF_1_L25 PositiveSet_def OrderedGroup_ZF_1_L1 group0.group_inv_of_inv by auto text{*A linearly ordered group can be decomposed into $G_+$, $\{1\}$ and $-G_+$*} lemma (in group3) OrdGroup_decomp2: assumes A1: "r {is total on} G" shows "G = G⇩_{+}∪ (\<sm>G⇩_{+})∪ {𝟭}" "G⇩_{+}∩(\<sm>G⇩_{+}) = 0" "𝟭 ∉ G⇩_{+}∪(\<sm>G⇩_{+})" proof - { fix a assume A2: "a∈G" with A1 have "a∈G⇩_{+}∨ a¯∈G⇩_{+}∨ a=𝟭" using OrderedGroup_ZF_1_L14 by auto with A2 have "a∈G⇩_{+}∨ a∈(\<sm>G⇩_{+}) ∨ a=𝟭" using OrderedGroup_ZF_1_L26 by auto then have "a ∈ (G⇩_{+}∪ (\<sm>G⇩_{+})∪ {𝟭})" by auto } then have "G ⊆ G⇩_{+}∪ (\<sm>G⇩_{+})∪ {𝟭}" by auto moreover have "G⇩_{+}∪ (\<sm>G⇩_{+})∪ {𝟭} ⊆ G" using OrderedGroup_ZF_1_L25 PositiveSet_def OrderedGroup_ZF_1_L1 group0.group0_2_L2 by auto ultimately show "G = G⇩_{+}∪ (\<sm>G⇩_{+})∪ {𝟭}" by auto { let ?A = "G⇩_{+}∩(\<sm>G⇩_{+})" assume "G⇩_{+}∩(\<sm>G⇩_{+}) ≠ 0" then have "?A≠0" by simp then obtain a where "a∈?A" by blast then have False using OrderedGroup_ZF_1_L15 OrderedGroup_ZF_1_L27 by auto } then show "G⇩_{+}∩(\<sm>G⇩_{+}) = 0" by auto show "𝟭 ∉ G⇩_{+}∪(\<sm>G⇩_{+})" using OrderedGroup_ZF_1_L27 OrderedGroup_ZF_1_L1 group0.group_inv_of_one OrderedGroup_ZF_1_L2A by auto qed text{*If $a\cdot b^{-1}$ is nonnegative, then $b \leq a$. This maybe used to recover the order from the set of nonnegative elements and serve as a way to define order by prescibing that set (see the "Alternative definitions" section).*} lemma (in group3) OrderedGroup_ZF_1_L28: assumes A1: "a∈G" "b∈G" and A2: "a⋅b¯ ∈ G⇧^{+}" shows "b\<lsq>a" proof - from A2 have "𝟭 \<lsq> a⋅b¯" using OrderedGroup_ZF_1_L2 by simp with A1 show "b\<lsq>a" using OrderedGroup_ZF_1_L5K by simp qed text{*A special case of @{text "OrderedGroup_ZF_1_L28"} when $a\cdot b^{-1}$ is positive.*} corollary (in group3) OrderedGroup_ZF_1_L29: assumes A1: "a∈G" "b∈G" and A2: "a⋅b¯ ∈ G⇩_{+}" shows "b\<lsq>a" "b≠a" proof - from A2 have "𝟭 \<lsq> a⋅b¯" and I: "a⋅b¯ ≠ 𝟭" using OrderedGroup_ZF_1_L2A by auto with A1 show "b\<lsq>a" using OrderedGroup_ZF_1_L5K by simp from A1 I show "b≠a" using OrderedGroup_ZF_1_L1 group0.group0_2_L6 by auto qed text{*A bit stronger that @{text "OrderedGroup_ZF_1_L29"}, adds case when two elements are equal.*} lemma (in group3) OrderedGroup_ZF_1_L30: assumes "a∈G" "b∈G" and "a=b ∨ b⋅a¯ ∈ G⇩_{+}" shows "a\<lsq>b" using assms OrderedGroup_ZF_1_L3 OrderedGroup_ZF_1_L29 by auto text{*A different take on decomposition: we can have $a=b$ or $a<b$ or $b<a$.*} lemma (in group3) OrderedGroup_ZF_1_L31: assumes A1: "r {is total on} G" and A2: "a∈G" "b∈G" shows "a=b ∨ (a\<lsq>b ∧ a≠b) ∨ (b\<lsq>a ∧ b≠a)" proof - from A2 have "a⋅b¯ ∈ G" using OrderedGroup_ZF_1_L1 group0.inverse_in_group group0.group_op_closed by simp with A1 have "a⋅b¯ = 𝟭 ∨ a⋅b¯ ∈ G⇩_{+}∨ (a⋅b¯)¯ ∈ G⇩_{+}" using OrderedGroup_ZF_1_L14 by simp moreover { assume "a⋅b¯ = 𝟭" then have "a⋅b¯⋅b = 𝟭⋅b" by simp with A2 have "a=b ∨ (a\<lsq>b ∧ a≠b) ∨ (b\<lsq>a ∧ b≠a)" using OrderedGroup_ZF_1_L1 group0.inv_cancel_two group0.group0_2_L2 by auto } moreover { assume "a⋅b¯ ∈ G⇩_{+}" with A2 have "a=b ∨ (a\<lsq>b ∧ a≠b) ∨ (b\<lsq>a ∧ b≠a)" using OrderedGroup_ZF_1_L29 by auto } moreover { assume "(a⋅b¯)¯ ∈ G⇩_{+}" with A2 have "b⋅a¯ ∈ G⇩_{+}" using OrderedGroup_ZF_1_L1 group0.group0_2_L12 by simp with A2 have "a=b ∨ (a\<lsq>b ∧ a≠b) ∨ (b\<lsq>a ∧ b≠a)" using OrderedGroup_ZF_1_L29 by auto } ultimately show "a=b ∨ (a\<lsq>b ∧ a≠b) ∨ (b\<lsq>a ∧ b≠a)" by auto qed subsection{*Intervals and bounded sets*} text{*Intervals here are the closed intervals of the form $\{x\in G. a\leq x \leq b \}$. *} text{*A bounded set can be translated to put it in $G^+$ and then it is still bounded above. *} lemma (in group3) OrderedGroup_ZF_2_L1: assumes A1: "∀g∈A. L\<lsq>g ∧ g\<lsq>M" and A2: "S = RightTranslation(G,P,L¯)" and A3: "a ∈ S``(A)" shows "a \<lsq> M⋅L¯" "𝟭\<lsq>a" proof - from A3 have "A≠0" using func1_1_L13A by fast then obtain g where "g∈A" by auto with A1 have T1: "L∈G" "M∈G" "L¯∈G" using OrderedGroup_ZF_1_L4 OrderedGroup_ZF_1_L1 group0.inverse_in_group by auto with A2 have "S : G→G" using OrderedGroup_ZF_1_L1 group0.group0_5_L1 by simp moreover from A1 have T2: "A⊆G" using OrderedGroup_ZF_1_L4 by auto ultimately have "S``(A) = {S`(b). b∈A}" using func_imagedef by simp with A3 obtain b where T3: "b∈A" "a = S`(b)" by auto with A1 ordGroupAssum T1 have "b⋅L¯\<lsq>M⋅L¯" "L⋅L¯\<lsq>b⋅L¯" using IsAnOrdGroup_def by auto with T3 A2 T1 T2 show "a\<lsq>M⋅L¯" "𝟭\<lsq>a" using OrderedGroup_ZF_1_L1 group0.group0_5_L2 group0.group0_2_L6 by auto qed text{*Every bounded set is an image of a subset of an interval that starts at $1$.*} lemma (in group3) OrderedGroup_ZF_2_L2: assumes A1: "IsBounded(A,r)" shows "∃B.∃g∈G⇧^{+}.∃T∈G→G. A = T``(B) ∧ B ⊆ Interval(r,𝟭,g)" proof - { assume A2: "A=0" let ?B = "0" let ?g = "𝟭" let ?T = "ConstantFunction(G,𝟭)" have "?g∈G⇧^{+}" using OrderedGroup_ZF_1_L3A by simp moreover have "?T : G→G" using func1_3_L1 OrderedGroup_ZF_1_L1 group0.group0_2_L2 by simp moreover from A2 have "A = T``(?B)" by simp moreover have "?B ⊆ Interval(r,𝟭,?g)" by simp ultimately have "∃B.∃g∈G⇧^{+}.∃T∈G→G. A = T``(B) ∧ B ⊆ Interval(r,𝟭,g)" by auto } moreover { assume A3: "A≠0" with A1 have "∃L. ∀x∈A. L\<lsq>x" and "∃U. ∀x∈A. x\<lsq>U" using IsBounded_def IsBoundedBelow_def IsBoundedAbove_def by auto then obtain L U where D1: "∀x∈A. L\<lsq>x ∧ x\<lsq>U " by auto with A3 have T1: "A⊆G" using OrderedGroup_ZF_1_L4 by auto from A3 obtain a where "a∈A" by auto with D1 have T2: "L\<lsq>a" "a\<lsq>U" by auto then have T3: "L∈G" "L¯∈ G" "U∈G" using OrderedGroup_ZF_1_L4 OrderedGroup_ZF_1_L1 group0.inverse_in_group by auto let ?T = "RightTranslation(G,P,L)" let ?B = "RightTranslation(G,P,L¯)``(A)" let ?g = "U⋅L¯" have "?g∈G⇧^{+}" proof - from T2 have "L\<lsq>U" using Group_order_transitive by fast with ordGroupAssum T3 have "L⋅L¯\<lsq>?g" using IsAnOrdGroup_def by simp with T3 show ?thesis using OrderedGroup_ZF_1_L1 group0.group0_2_L6 OrderedGroup_ZF_1_L2 by simp qed moreover from T3 have "?T : G→G" using OrderedGroup_ZF_1_L1 group0.group0_5_L1 by simp moreover have "A = ?T``(?B)" proof - from T3 T1 have "?T``(?B) = {a⋅L¯⋅L. a∈A}" using OrderedGroup_ZF_1_L1 group0.group0_5_L6 by simp moreover from T3 T1 have "∀a∈A. a⋅L¯⋅L = a⋅(L¯⋅L)" using OrderedGroup_ZF_1_L1 group0.group_oper_assoc by auto ultimately have "?T``(?B) = {a⋅(L¯⋅L). a∈A}" by simp with T3 have "?T``(?B) = {a⋅𝟭. a∈A}" using OrderedGroup_ZF_1_L1 group0.group0_2_L6 by simp moreover from T1 have "∀a∈A. a⋅𝟭=a" using OrderedGroup_ZF_1_L1 group0.group0_2_L2 by auto ultimately show ?thesis by simp qed moreover have "?B ⊆ Interval(r,𝟭,?g)" proof fix y assume A4: "y ∈ ?B" let ?S = "RightTranslation(G,P,L¯)" from D1 have T4: "∀x∈A. L\<lsq>x ∧ x\<lsq>U" by simp moreover have T5: "?S = RightTranslation(G,P,L¯)" by simp moreover from A4 have T6: "y ∈ ?S``(A)" by simp ultimately have "y\<lsq>U⋅L¯" using OrderedGroup_ZF_2_L1 by blast moreover from T4 T5 T6 have "𝟭\<lsq>y" by (rule OrderedGroup_ZF_2_L1) ultimately show "y ∈ Interval(r,𝟭,?g)" using Interval_def by auto qed ultimately have "∃B.∃g∈G⇧^{+}.∃T∈G→G. A = T``(B) ∧ B ⊆ Interval(r,𝟭,g)" by auto } ultimately show ?thesis by auto qed text{*If every interval starting at $1$ is finite, then every bounded set is finite. I find it interesting that this does not require the group to be linearly ordered (the order to be total).*} theorem (in group3) OrderedGroup_ZF_2_T1: assumes A1: "∀g∈G⇧^{+}. Interval(r,𝟭,g) ∈ Fin(G)" and A2: "IsBounded(A,r)" shows "A ∈ Fin(G)" proof - from A2 have "∃B.∃g∈G⇧^{+}.∃T∈G→G. A = T``(B) ∧ B ⊆ Interval(r,𝟭,g)" using OrderedGroup_ZF_2_L2 by simp then obtain B g T where D1: "g∈G⇧^{+}" "B ⊆ Interval(r,𝟭,g)" and D2: "T : G→G" "A = T``(B)" by auto from D1 A1 have "B∈Fin(G)" using Fin_subset_lemma by blast with D2 show ?thesis using Finite1_L6A by simp qed text{*In linearly ordered groups finite sets are bounded.*} theorem (in group3) ord_group_fin_bounded: assumes "r {is total on} G" and "B∈Fin(G)" shows "IsBounded(B,r)" using ordGroupAssum assms IsAnOrdGroup_def IsPartOrder_def Finite_ZF_1_T1 by simp text{*For nontrivial linearly ordered groups if for every element $G$ we can find one in $A$ that is greater or equal (not necessarily strictly greater), then $A$ can neither be finite nor bounded above.*} lemma (in group3) OrderedGroup_ZF_2_L2A: assumes A1: "r {is total on} G" and A2: "G ≠ {𝟭}" and A3: "∀a∈G. ∃b∈A. a\<lsq>b" shows "∀a∈G. ∃b∈A. a≠b ∧ a\<lsq>b" "¬IsBoundedAbove(A,r)" "A ∉ Fin(G)" proof - { fix a from A1 A2 obtain c where "c ∈ G⇩_{+}" using OrderedGroup_ZF_1_L21 by auto moreover assume "a∈G" ultimately have "a⋅c ∈ G" and I: "a \<ls> a⋅c" using OrderedGroup_ZF_1_L22 by auto with A3 obtain b where II: "b∈A" and III: "a⋅c \<lsq> b" by auto moreover from I III have "a\<ls>b" by (rule OrderedGroup_ZF_1_L4A) ultimately have "∃b∈A. a≠b ∧ a\<lsq>b" by auto } thus "∀a∈G. ∃b∈A. a≠b ∧ a\<lsq>b" by simp with ordGroupAssum A1 show "¬IsBoundedAbove(A,r)" "A ∉ Fin(G)" using IsAnOrdGroup_def IsPartOrder_def OrderedGroup_ZF_1_L1A Order_ZF_3_L14 Finite_ZF_1_1_L3 by auto qed text{*Nontrivial linearly ordered groups are infinite. Recall that @{text "Fin(A)"} is the collection of finite subsets of $A$. In this lemma we show that $G\notin$ @{text "Fin(G)"}, that is that $G$ is not a finite subset of itself. This is a way of saying that $G$ is infinite. We also show that for nontrivial linearly ordered groups @{text "G⇩_{+}"} is infinite.*} theorem (in group3) Linord_group_infinite: assumes A1: "r {is total on} G" and A2: "G ≠ {𝟭}" shows "G⇩_{+}∉ Fin(G)" "G ∉ Fin(G)" proof - from A1 A2 show I: "G⇩_{+}∉ Fin(G)" using OrderedGroup_ZF_1_L23 OrderedGroup_ZF_2_L2A by simp { assume "G ∈ Fin(G)" moreover have "G⇩_{+}⊆ G" using PositiveSet_def by auto ultimately have "G⇩_{+}∈ Fin(G)" using Fin_subset_lemma by blast with I have False by simp } then show "G ∉ Fin(G)" by auto qed text{*A property of nonempty subsets of linearly ordered groups that don't have a maximum: for any element in such subset we can find one that is strictly greater.*} lemma (in group3) OrderedGroup_ZF_2_L2B: assumes A1: "r {is total on} G" and A2: "A⊆G" and A3: "¬HasAmaximum(r,A)" and A4: "x∈A" shows "∃y∈A. x\<ls>y" proof - from ordGroupAssum assms have "antisym(r)" "r {is total on} G" "A⊆G" "¬HasAmaximum(r,A)" "x∈A" using IsAnOrdGroup_def IsPartOrder_def by auto then have "∃y∈A. ⟨x,y⟩ ∈ r ∧ y≠x" using Order_ZF_4_L16 by simp then show "∃y∈A. x\<ls>y" by auto qed text{*In linearly ordered groups $G\setminus G_+$ is bounded above. *} lemma (in group3) OrderedGroup_ZF_2_L3: assumes A1: "r {is total on} G" shows "IsBoundedAbove(G-G⇩_{+},r)" proof - from A1 have "∀a∈G-G⇩_{+}. a\<lsq>𝟭" using OrderedGroup_ZF_1_L17 by simp then show "IsBoundedAbove(G-G⇩_{+},r)" using IsBoundedAbove_def by auto qed text{*In linearly ordered groups if $A\cap G_+$ is finite, then $A$ is bounded above.*} lemma (in group3) OrderedGroup_ZF_2_L4: assumes A1: "r {is total on} G" and A2: "A⊆G" and A3: "A ∩ G⇩_{+}∈ Fin(G)" shows "IsBoundedAbove(A,r)" proof - have "A ∩ (G-G⇩_{+}) ⊆ G-G⇩_{+}" by auto with A1 have "IsBoundedAbove(A ∩ (G-G⇩_{+}),r)" using OrderedGroup_ZF_2_L3 Order_ZF_3_L13 by blast moreover from A1 A3 have "IsBoundedAbove(A ∩ G⇩_{+},r)" using ord_group_fin_bounded IsBounded_def by simp moreover from A1 ordGroupAssum have "r {is total on} G" "trans(r)" "r⊆G×G" using IsAnOrdGroup_def IsPartOrder_def by auto ultimately have "IsBoundedAbove(A ∩ (G-G⇩_{+}) ∪ A ∩ G⇩_{+},r)" using Order_ZF_3_L3 by simp moreover from A2 have "A = A ∩ (G-G⇩_{+}) ∪ A ∩ G⇩_{+}" by auto ultimately show "IsBoundedAbove(A,r)" by simp qed text{*If a set $-A\subseteq G$ is bounded above, then $A$ is bounded below.*} lemma (in group3) OrderedGroup_ZF_2_L5: assumes A1: "A⊆G" and A2: "IsBoundedAbove(\<sm>A,r)" shows "IsBoundedBelow(A,r)" proof - { assume "A = 0" then have "IsBoundedBelow(A,r)" using IsBoundedBelow_def by auto } moreover { assume A3: "A≠0" from ordGroupAssum have I: "GroupInv(G,P) : G→G" using IsAnOrdGroup_def group0_2_T2 by simp with A1 A2 A3 obtain u where D: "∀a∈(\<sm>A). a\<lsq>u" using func1_1_L15A IsBoundedAbove_def by auto { fix b assume "b∈A" with A1 I D have "b¯ \<lsq> u" and T: "b∈G" using func_imagedef by auto then have "u¯\<lsq>(b¯)¯" using OrderedGroup_ZF_1_L5 by simp with T have "u¯\<lsq>b" using OrderedGroup_ZF_1_L1 group0.group_inv_of_inv by simp } then have "∀b∈A. ⟨u¯,b⟩ ∈ r" by simp then have "IsBoundedBelow(A,r)" using Order_ZF_3_L9 by blast } ultimately show ?thesis by auto qed text{*If $a\leq b$, then the image of the interval $a..b$ by any function is nonempty.*} lemma (in group3) OrderedGroup_ZF_2_L6: assumes "a\<lsq>b" and "f:G→G" shows "f``(Interval(r,a,b)) ≠ 0" using ordGroupAssum assms OrderedGroup_ZF_1_L4 Order_ZF_2_L6 Order_ZF_2_L2A IsAnOrdGroup_def IsPartOrder_def func1_1_L15A by auto end