(*

This file is a part of IsarMathLib -

a library of formalized mathematics for Isabelle/Isar.

Copyright (C) 2005 - 2008 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:

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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,

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WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR

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EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.*)

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

theory func_ZF imports func1

begin

text{*In this theory we consider properties of functions that are binary

operations, that is they map $X\times X$ into $X$.*}

section{*Lifting operations to a function space*}

text{*It happens quite often that we have a binary operation on some set and

we need a similar operation that is defined for functions on that set.

For example once we know how to add real numbers we also know how to add

real-valued functions: for $f,g:X \rightarrow \mathbf{R}$ we define

$(f+g)(x) = f(x) + g(x)$. Note that formally the $+$ means something

different on the left hand side of this equality than on the

right hand side.

This section aims at formalizing this process.

We will call it "lifting to a function space", if you have a

suggestion for a better name, please let me know. *}

text{*Since we are writing in generic set notation,

the definition below is a bit complicated. Here it what it says:

Given a set $X$ and another set $f$ (that represents a binary function on $X$)

we are defining $f$ lifted to function space over $X$

as the binary function (a set of pairs) on the space

$F = X \rightarrow \textrm{range}(f)$ such that the value of this function

on pair $\langle a,b \rangle$ of functions on $X$ is another function $c$ on $X$

with values defined by $c(x) = f\langle a(x), b(x)\rangle$.

*}

definition

Lift2FcnSpce (infix "{lifted to function space over}" 65) where

"f {lifted to function space over} X ≡

{⟨ p,{⟨x,f`⟨fst(p)`(x),snd(p)`(x)⟩⟩. x ∈ X}⟩.

p ∈ (X->range(f))×(X->range(f))}";

text{*The result of the lift belongs to the function space.*}

lemma func_ZF_1_L1:

assumes A1: "f : Y×Y->Y"

and A2: "p ∈(X->range(f))×(X->range(f))"

shows

"{⟨x,f`⟨fst(p)`(x),snd(p)`(x)⟩⟩. x ∈ X} : X->range(f)"

proof -;

have "∀x∈X. f`⟨fst(p)`(x),snd(p)`(x)⟩ ∈ range(f)"

proof;

fix x assume "x∈X"

let ?p = "⟨fst(p)`(x),snd(p)`(x)⟩"

from A2 `x∈X` have

"fst(p)`(x) ∈ range(f)" "snd(p)`(x) ∈ range(f)"

using apply_type by auto;

with A1 have "?p ∈ Y×Y"

using func1_1_L5B by blast;

with A1 have "⟨?p, f`(?p)⟩ ∈ f"

using apply_Pair by simp;

with A1 show

"f`(?p) ∈ range(f)"

using rangeI by simp;

qed

then show ?thesis using ZF_fun_from_total by simp;

qed;

text{*The values of the lift are defined by the value of the liftee in a

natural way.*}

lemma func_ZF_1_L2:

assumes A1: "f : Y×Y->Y"

and A2: "p ∈ (X->range(f))×(X->range(f))" and A3: "x∈X"

and A4: "P = {⟨x,f`⟨fst(p)`(x),snd(p)`(x)⟩⟩. x ∈ X}"

shows "P`(x) = f`⟨fst(p)`(x),snd(p)`(x)⟩"

proof -

from A1 A2 have

"{⟨x,f`⟨fst(p)`(x),snd(p)`(x)⟩⟩. x ∈ X} : X -> range(f)"

using func_ZF_1_L1 by simp;

with A4 have "P : X -> range(f)" by simp;

with A3 A4 show "P`(x) = f`⟨fst(p)`(x),snd(p)`(x)⟩"

using ZF_fun_from_tot_val by simp;

qed;

text{*Function lifted to a function space results in function space

operator.*}

theorem func_ZF_1_L3:

assumes "f : Y×Y->Y"

and "F = f {lifted to function space over} X"

shows "F : (X->range(f))×(X->range(f))->(X->range(f))"

using assms Lift2FcnSpce_def func_ZF_1_L1 ZF_fun_from_total

by simp;

text{*The values of the lift are defined by the values of the liftee in

the natural way.*}

theorem func_ZF_1_L4:

assumes A1: "f : Y×Y->Y"

and A2: "F = f {lifted to function space over} X"

and A3: "s:X->range(f)" "r:X->range(f)"

and A4: "x∈X"

shows "(F`⟨s,r⟩)`(x) = f`⟨s`(x),r`(x)⟩"

proof -;

let ?p = "⟨s,r⟩"

let ?P = "{⟨x,f`⟨fst(?p)`(x),snd(?p)`(x)⟩⟩. x ∈ X}";

from A1 A3 A4 have

"f : Y×Y->Y" "?p ∈ (X->range(f))×(X->range(f))"

"x∈X" "?P = {⟨x,f`⟨fst(?p)`(x),snd(?p)`(x)⟩⟩. x ∈ X}"

by auto;

then have "?P`(x) = f`⟨fst(?p)`(x),snd(?p)`(x)⟩"

by (rule func_ZF_1_L2);

hence "?P`(x) = f`⟨s`(x),r`(x)⟩" by auto;

moreover have "?P = F`⟨s,r⟩"

proof -

from A1 A2 have "F : (X->range(f))×(X->range(f))->(X->range(f))"

using func_ZF_1_L3 by simp;

moreover from A3 have "?p ∈ (X->range(f))×(X->range(f))"

by auto;

moreover from A2 have

"F = {⟨p,{⟨x,f`⟨fst(p)`(x),snd(p)`(x)⟩⟩. x ∈ X}⟩.

p ∈ (X->range(f))×(X->range(f))}"

using Lift2FcnSpce_def by simp;

ultimately show ?thesis using ZF_fun_from_tot_val

by simp;

qed;

ultimately show "(F`⟨s,r⟩)`(x) = f`⟨s`(x),r`(x)⟩" by auto;

qed

section{*Associative and commutative operations*}

text{*In this section we define associative and commutative operations

and prove that they remain such when we lift them

to a function space.*}

text{*Typically we say that a binary operation "$\cdot $"

on a set $G$ is ''associative''

if $(x\cdot y)\cdot z = x\cdot (y\cdot z)$ for all $x,y,z \in G$.

Our actual definition below does not use the multiplicative notation

so that we can apply it equally to the additive notation $+$

or whatever infix symbol we may want to use.

Instead, we use the generic set theory notation

and write $P\langle x,y \rangle$ to denote the value of the operation

$P$ on a pair $\langle x,y \rangle \in G\times G$.*}

definition

IsAssociative (infix "{is associative on}" 65) where

"P {is associative on} G ≡ P : G×G->G ∧

(∀ x ∈ G. ∀ y ∈ G. ∀ z ∈ G.

( P`(⟨P`(⟨x,y⟩),z⟩) = P`( ⟨x,P`(⟨y,z⟩)⟩ )))"

text{*A binary function $f: X\times X \rightarrow Y$ is commutative

if $f\langle x,y \rangle = f\langle y,x \rangle$. Note that

in the definition of associativity above we talk about binary

''operation'' and here we say use the term binary ''function''.

This is not set in stone, but usually the word "operation" is used

when the range is a factor of the domain, while the word "function"

allows the range to be a completely unrelated set. *}

definition

IsCommutative (infix "{is commutative on}" 65) where

"f {is commutative on} G ≡ ∀x∈G. ∀y∈G. f`⟨x,y⟩ = f`⟨y,x⟩"

text{*The lift of a commutative function is commutative.*}

lemma func_ZF_2_L1:

assumes A1: "f : G×G->G"

and A2: "F = f {lifted to function space over} X"

and A3: "s : X->range(f)" "r : X->range(f)"

and A4: "f {is commutative on} G"

shows "F`⟨s,r⟩ = F`⟨r,s⟩"

proof -;

from A1 A2 have

"F : (X->range(f))×(X->range(f))->(X->range(f))"

using func_ZF_1_L3 by simp;

with A3 have

"F`⟨s,r⟩ : X->range(f)" and "F`⟨r,s⟩ : X->range(f)"

using apply_type by auto;

moreover have

"∀x∈X. (F`⟨s,r⟩)`(x) = (F`⟨r,s⟩)`(x)"

proof;

fix x assume "x∈X"

from A1 have "range(f)⊆G"

using func1_1_L5B by simp;

with A3 `x∈X` have "s`(x) ∈ G" and "r`(x) ∈ G"

using apply_type by auto;

with A1 A2 A3 A4 `x∈X` show

"(F`⟨s,r⟩)`(x) = (F`⟨r,s⟩)`(x)"

using func_ZF_1_L4 IsCommutative_def by simp;

qed;

ultimately show ?thesis using fun_extension_iff

by simp;

qed;

text{*The lift of a commutative function is commutative

on the function space.*}

lemma func_ZF_2_L2:

assumes "f : G×G->G"

and "f {is commutative on} G"

and "F = f {lifted to function space over} X"

shows "F {is commutative on} (X->range(f))"

using assms IsCommutative_def func_ZF_2_L1 by simp;

text{*The lift of an associative function is associative.*}

lemma func_ZF_2_L3:

assumes A2: "F = f {lifted to function space over} X"

and A3: "s : X->range(f)" "r : X->range(f)" "q : X->range(f)"

and A4: "f {is associative on} G"

shows "F`⟨F`⟨s,r⟩,q⟩ = F`⟨s,F`⟨r,q⟩⟩"

proof -;

from A4 A2 have

"F : (X->range(f))×(X->range(f))->(X->range(f))"

using IsAssociative_def func_ZF_1_L3 by auto;

with A3 have I:

"F`⟨s,r⟩ : X->range(f)"

"F`⟨r,q⟩ : X->range(f)"

"F`⟨F`⟨s,r⟩,q⟩ : X->range(f)"

"F`⟨s,F`⟨r,q⟩⟩: X->range(f)"

using apply_type by auto;

moreover have

"∀x∈X. (F`⟨F`⟨s,r⟩,q⟩)`(x) = (F`⟨s,F`⟨r,q⟩⟩)`(x)"

proof;

fix x assume "x∈X"

from A4 have "f:G×G->G"

using IsAssociative_def by simp;

then have "range(f)⊆G"

using func1_1_L5B by simp;

with A3 `x∈X` have

"s`(x) ∈ G" "r`(x) ∈ G" "q`(x) ∈ G"

using apply_type by auto;

with A2 I A3 A4 `x∈X` `f:G×G->G` show

"(F`⟨F`⟨s,r⟩,q⟩)`(x) = (F`⟨s,F`⟨r,q⟩⟩)`(x)"

using func_ZF_1_L4 IsAssociative_def by simp;

qed;

ultimately show ?thesis using fun_extension_iff

by simp;

qed;

text{*The lift of an associative function is associative

on the function space.*}

lemma func_ZF_2_L4:

assumes A1: "f {is associative on} G"

and A2: "F = f {lifted to function space over} X"

shows "F {is associative on} (X->range(f))"

proof -;

from A1 A2 have

"F : (X->range(f))×(X->range(f))->(X->range(f))"

using IsAssociative_def func_ZF_1_L3 by auto;

moreover from A1 A2 have

"∀s ∈ X->range(f). ∀ r ∈ X->range(f). ∀q ∈ X->range(f).

F`⟨F`⟨s,r⟩,q⟩ = F`⟨s,F`⟨r,q⟩⟩"

using func_ZF_2_L3 by simp;

ultimately show ?thesis using IsAssociative_def

by simp;

qed;

section{*Restricting operations*}

text{*In this section we consider conditions under which

restriction of the operation to a set

inherits properties like commutativity and associativity.*}

text{*The commutativity is inherited when restricting a function to a set.*}

lemma func_ZF_4_L1:

assumes A1: "f:X×X->Y" and A2: "A⊆X"

and A3: "f {is commutative on} X"

shows "restrict(f,A×A) {is commutative on} A"

proof -;

{ fix x y assume "x∈A" and "y∈A"

with A2 have "x∈X" and "y∈X" by auto;

with A3 `x∈A` `y∈A` have

"restrict(f,A×A)`⟨x,y⟩ = restrict(f,A×A)`⟨y,x⟩"

using IsCommutative_def restrict_if by simp }

then show ?thesis using IsCommutative_def by simp;

qed;

text{*Next we define what it means that a set is closed with

respect to an operation.*}

definition

IsOpClosed (infix "{is closed under}" 65) where

"A {is closed under} f ≡ ∀x∈A. ∀y∈A. f`⟨x,y⟩ ∈ A";

text{*Associative operation restricted to a set that is closed with

resp. to this operation is associative.*}

lemma func_ZF_4_L2:assumes A1: "f {is associative on} X"

and A2: "A⊆X" and A3: "A {is closed under} f"

and A4: "x∈A" "y∈A" "z∈A"

and A5: "g = restrict(f,A×A)"

shows "g`⟨g`⟨x,y⟩,z⟩ = g`⟨x,g`⟨y,z⟩⟩"

proof -;

from A4 A2 have I: "x∈X" "y∈X" "z∈X"

by auto;

from A3 A4 A5 have

"g`⟨g`⟨x,y⟩,z⟩ = f`⟨f`⟨x,y⟩,z⟩"

"g`⟨x,g`⟨y,z⟩⟩ = f`⟨x,f`⟨y,z⟩⟩"

using IsOpClosed_def restrict_if by auto;

moreover from A1 I have

"f`⟨f`⟨x,y⟩,z⟩ = f`⟨x,f`⟨y,z⟩⟩"

using IsAssociative_def by simp;

ultimately show ?thesis by simp;

qed;

text{*An associative operation restricted to a set that is closed with

resp. to this operation is associative on the set.*}

lemma func_ZF_4_L3: assumes A1: "f {is associative on} X"

and A2: "A⊆X" and A3: "A {is closed under} f"

shows "restrict(f,A×A) {is associative on} A"

proof -;

let ?g = "restrict(f,A×A)"

from A1 have "f:X×X->X"

using IsAssociative_def by simp;

moreover from A2 have "A×A ⊆ X×X" by auto;

moreover from A3 have "∀p ∈ A×A. ?g`(p) ∈ A"

using IsOpClosed_def restrict_if by auto;

ultimately have "?g : A×A->A"

using func1_2_L4 by simp;

moreover from A1 A2 A3 have

"∀ x ∈ A. ∀ y ∈ A. ∀ z ∈ A.

?g`⟨?g`⟨x,y⟩,z⟩ = ?g`⟨ x,?g`⟨y,z⟩⟩"

using func_ZF_4_L2 by simp;

ultimately show ?thesis

using IsAssociative_def by simp;

qed;

text{*The essential condition to show that if a set $A$ is closed

with respect to an operation,

then it is closed under this operation restricted

to any superset of $A$.*}

lemma func_ZF_4_L4: assumes "A {is closed under} f"

and "A⊆B" and "x∈A" "y∈A" and "g = restrict(f,B×B)"

shows "g`⟨x,y⟩ ∈ A"

using assms IsOpClosed_def restrict by auto;

text{*If a set $A$ is closed under an operation,

then it is closed under this operation restricted

to any superset of $A$. *}

lemma func_ZF_4_L5:

assumes A1: "A {is closed under} f"

and A2: "A⊆B"

shows "A {is closed under} restrict(f,B×B)"

proof -

let ?g = "restrict(f,B×B)"

from A1 A2 have "∀x∈A. ∀y∈A. ?g`⟨x,y⟩ ∈ A"

using func_ZF_4_L4 by simp

then show ?thesis using IsOpClosed_def by simp;

qed;

text{*The essential condition to show that intersection of sets that are

closed with respect to an operation is closed with respect

to the operation.*}

lemma func_ZF_4_L6:

assumes "A {is closed under} f"

and "B {is closed under} f"

and "x ∈ A∩B" "y∈ A∩B"

shows "f`⟨x,y⟩ ∈ A∩B" using assms IsOpClosed_def by auto;

text{*Intersection of sets that are

closed with respect to an operation is closed under

the operation.*}

lemma func_ZF_4_L7:

assumes "A {is closed under} f"

"B {is closed under} f"

shows "A∩B {is closed under} f"

using assms IsOpClosed_def by simp;

section{*Compositions*}

text{*For any set $X$ we can consider a binary operation

on the set of functions

$f:X\rightarrow X$ defined by $C(f,g) = f\circ g$. Composition of functions

(or relations) is defined in the standard Isabelle distribution as a higher

order function and denoted with the letter @{text "O"}.

In this section we consider the corresponding two-argument

ZF-function (binary operation), that is a subset of

$((X\rightarrow X)\times (X\rightarrow X))\times (X\rightarrow X)$.*}

text{*We define the notion of composition on the set $X$ as the

binary operation on the function space $X\rightarrow X$

that takes two functions and creates the their composition.*}

definition

"Composition(X) ≡

{⟨p,fst(p) O snd(p)⟩. p ∈ (X->X)×(X->X)}"

text{*Composition operation is a function that maps

$(X\rightarrow X)\times (X\rightarrow X)$ into $X\rightarrow X$.*}

lemma func_ZF_5_L1: shows "Composition(X) : (X->X)×(X->X)->(X->X)"

using comp_fun Composition_def ZF_fun_from_total by simp;

text{*The value of the composition operation is the composition of arguments.*}

lemma func_ZF_5_L2: assumes "f:X->X" and "g:X->X"

shows "Composition(X)`⟨f,g⟩ = f O g"

proof -

from assms have

"Composition(X) : (X->X)×(X->X)->(X->X)"

"⟨f,g⟩ ∈ (X->X)×(X->X)"

"Composition(X) = {⟨p,fst(p) O snd(p)⟩. p ∈ (X->X)×(X->X)}"

using func_ZF_5_L1 Composition_def by auto;

then show "Composition(X)`⟨f,g⟩ = f O g"

using ZF_fun_from_tot_val by auto;

qed;

text{*What is the value of a composition on an argument?*}

lemma func_ZF_5_L3: assumes "f:X->X" and "g:X->X" and "x∈X"

shows "(Composition(X)`⟨f,g⟩)`(x) = f`(g`(x))"

using assms func_ZF_5_L2 comp_fun_apply by simp;

text{*The essential condition to show that composition is associative.*}

lemma func_ZF_5_L4: assumes A1: "f:X->X" "g:X->X" "h:X->X"

and A2: "C = Composition(X)"

shows "C`⟨C`⟨f,g⟩,h⟩ = C`⟨ f,C`⟨g,h⟩⟩"

proof -;

from A2 have "C : ((X->X)×(X->X))->(X->X)"

using func_ZF_5_L1 by simp;

with A1 have I:

"C`⟨f,g⟩ : X->X"

"C`⟨g,h⟩ : X->X"

"C`⟨C`⟨f,g⟩,h⟩ : X->X"

"C`⟨ f,C`⟨g,h⟩⟩ : X->X"

using apply_funtype by auto;

moreover have

"∀ x ∈ X. C`⟨C`⟨f,g⟩,h⟩`(x) = C`⟨f,C`⟨g,h⟩⟩`(x)"

proof;

fix x assume "x∈X"

with A1 A2 I have

"C`⟨C`⟨f,g⟩,h⟩ ` (x) = f`(g`(h`(x)))"

"C`⟨ f,C`⟨g,h⟩⟩`(x) = f`(g`(h`(x)))"

using func_ZF_5_L3 apply_funtype by auto;

then show "C`⟨C`⟨f,g⟩,h⟩`(x) = C`⟨ f,C`⟨g,h⟩⟩`(x)"

by simp;

qed;

ultimately show ?thesis using fun_extension_iff by simp;

qed;

text{*Composition is an associative operation on $X\rightarrow X$ (the space

of functions that map $X$ into itself).*}

lemma func_ZF_5_L5: shows "Composition(X) {is associative on} (X->X)"

proof -;

let ?C = "Composition(X)"

have "∀f∈X->X. ∀g∈X->X. ∀h∈X->X.

?C`⟨?C`⟨f,g⟩,h⟩ = ?C`⟨f,?C`⟨g,h⟩⟩"

using func_ZF_5_L4 by simp;

then show ?thesis using func_ZF_5_L1 IsAssociative_def

by simp;

qed;

section{*Identity function *}

text{*In this section we show some additional facts about the identity

function defined in the standard Isabelle's @{text "Perm"} theory.*}

text{*A function that maps every point to itself is the identity on its domain.*}

lemma indentity_fun: assumes A1: "f:X->Y" and A2:"∀x∈X. f`(x)=x"

shows "f = id(X)"

proof -

from assms have "f:X->Y" and "id(X):X->X" and "∀x∈X. f`(x) = id(X)`(x)"

using id_type id_conv by auto

then show ?thesis by (rule func_eq)

qed

text{*Composing a function with identity does not change the function.*}

lemma func_ZF_6_L1A: assumes A1: "f : X->X"

shows "Composition(X)`⟨f,id(X)⟩ = f"

"Composition(X)`⟨id(X),f⟩ = f"

proof -;

have "Composition(X) : (X->X)×(X->X)->(X->X)"

using func_ZF_5_L1 by simp;

with A1 have "Composition(X)`⟨id(X),f⟩ : X->X"

"Composition(X)`⟨f,id(X)⟩ : X->X"

using id_type apply_funtype by auto;

moreover note A1

moreover from A1 have

"∀x∈X. (Composition(X)`⟨id(X),f⟩)`(x) = f`(x)"

"∀x∈X. (Composition(X)`⟨f,id(X)⟩)`(x) = f`(x)"

using id_type func_ZF_5_L3 apply_funtype id_conv

by auto;

ultimately show "Composition(X)`⟨id(X),f⟩ = f"

"Composition(X)`⟨f,id(X)⟩ = f"

using fun_extension_iff by auto;

qed;

text{*An intuitively clear, but surprsingly nontrivial fact:identity is the only function from

a singleton to itself.*}

lemma singleton_fun_id: shows "({x} -> {x}) = {id({x})}"

proof;

show "{id({x})} ⊆ ({x} -> {x})"

using id_def by simp;

{ let ?g = "id({x})"

fix f assume "f : {x} -> {x}"

then have "f : {x} -> {x}" and "?g : {x} -> {x}"

using id_def by auto;

moreover from `f : {x} -> {x}` have "∀x ∈ {x}. f`(x) = ?g`(x)"

using apply_funtype id_def by auto;

ultimately have "f = ?g" by (rule func_eq)

} then show "({x} -> {x}) ⊆ {id({x})}" by auto;

qed;

text{*Another trivial fact: identity is the only bijection of a singleton

with itself.*}

lemma single_bij_id: shows "bij({x},{x}) = {id({x})}"

proof;

show "{id({x})} ⊆ bij({x},{x})" using id_bij

by simp;

{ fix f assume "f ∈ bij({x},{x})"

then have "f : {x} -> {x}" using bij_is_fun

by simp;

then have "f ∈ {id({x})}" using singleton_fun_id

by simp;

} then show "bij({x},{x}) ⊆ {id({x})}" by auto;

qed;

text{*A kind of induction for the identity: if a function

$f$ is the identity on a set with a fixpoint of $f$

removed, then it is the indentity on the whole set.*}

lemma id_fixpoint_rem: assumes A1: "f:X->X" and

A2: "p∈X" and A3: "f`(p) = p" and

A4: "restrict(f, X-{p}) = id(X-{p})"

shows "f = id(X)"

proof -

from A1 have "f: X->X" and "id(X) : X->X"

using id_def by auto;

moreover

{ fix x assume "x∈X"

{ assume "x ∈ X-{p}"

then have "f`(x) = restrict(f, X-{p})`(x)"

using restrict by simp

with A4 `x ∈ X-{p}` have "f`(x) = x"

using id_def by simp }

with A2 A3 `x∈X` have "f`(x) = x" by auto;

} then have "∀x∈X. f`(x) = id(X)`(x)"

using id_def by simp;

ultimately show "f = id(X)" by (rule func_eq);

qed

section{*Lifting to subsets*}

text{*Suppose we have a binary operation $f : X \times X \rightarrow X$

written additively as $f\langle x,y\rangle = x + y$. Such operation

naturally defines another binary operation on the subsets of $X$

that satisfies $A+B = \{ x+y : x \in A, y\in B\}$. This new operation

which we will call "$f$ lifted to subsets" inherits many properties

of $f$, such as associativity, commutativity and existence of the

neutral element. This notion is useful for considering interval arithmetics.

*}

text{*The next definition describes the notion of a binary operation

lifted to subsets. It is written in a way that might be a bit unexpected,

but really it is the same as the intuitive definition, but shorter.

In the definition we take a pair $p \in Pow(X)\times Pow(X)$, say

$p = \langle A, B\rangle $, where $A,B \subseteq X$.

Then we assign this pair of sets the set

$\{f\langle x,y \rangle : x\in A, y\in B \} = \{ f(x'): x' \in A\times B\}$

The set on the right hand side is the same as the image

of $A\times B$ under $f$. In the definition we don't use $A$ and $B$ symbols,

but write @{text "fst(p)"} and @{text "snd(p)"}, resp. Recall that in Isabelle/ZF

@{text "fst(p)"} and @{text "snd(p)"} denote the first and second components

of an ordered pair $p$.

See the lemma @{text "lift_subsets_explained"} for a more intuitive

notation.*}

definition

Lift2Subsets (infix "{lifted to subsets of}" 65) where

"f {lifted to subsets of} X ≡

{⟨p, f``(fst(p)×snd(p))⟩. p ∈ Pow(X)×Pow(X)}";

text{*The lift to subsets defines a binary operation on the subsets.*}

lemma lift_subsets_binop: assumes A1: "f : X × X -> Y"

shows "(f {lifted to subsets of} X) : Pow(X) × Pow(X) -> Pow(Y)"

proof -

let ?F = "{⟨p, f``(fst(p)×snd(p))⟩. p ∈ Pow(X)×Pow(X)}"

from A1 have "∀p ∈ Pow(X) × Pow(X). f``(fst(p)×snd(p)) ∈ Pow(Y)"

using func1_1_L6 by simp;

then have "?F : Pow(X) × Pow(X) -> Pow(Y)"

by (rule ZF_fun_from_total);

then show ?thesis unfolding Lift2Subsets_def by simp;

qed

text{*The definition of the lift to subsets rewritten in a more intuitive

notation. We would like to write the last assertion as

@{text "F`⟨A,B⟩ = {f`⟨x,y⟩. x ∈ A, y ∈ B}"}, but Isabelle/ZF does not allow

such syntax.*}

lemma lift_subsets_explained: assumes A1: "f : X×X -> Y"

and A2: "A ⊆ X" "B ⊆ X" and A3: "F = f {lifted to subsets of} X"

shows

"F`⟨A,B⟩ ⊆ Y" and

"F`⟨A,B⟩ = f``(A×B)"

"F`⟨A,B⟩ = {f`(p). p ∈ A×B}"

"F`⟨A,B⟩ = {f`⟨x,y⟩ . ⟨x,y⟩ ∈ A×B}"

proof -

let ?p = "⟨A,B⟩"

from assms have

I: "F : Pow(X) × Pow(X) -> Pow(Y)" and "?p ∈ Pow(X) × Pow(X)"

using lift_subsets_binop by auto;

moreover from A3 have "F = {⟨p, f``(fst(p)×snd(p))⟩. p ∈ Pow(X)×Pow(X)}"

unfolding Lift2Subsets_def by simp

ultimately show "F`⟨A,B⟩ = f``(A×B)"

using ZF_fun_from_tot_val by auto;

also

from A1 A2 have "A×B ⊆ X×X" by auto;

with A1 have "f``(A×B) = {f`(p). p ∈ A×B}"

by (rule func_imagedef)

finally show "F`⟨A,B⟩ = {f`(p) . p ∈ A×B}" by simp;

also

have "∀x∈A. ∀y ∈ B. f`⟨x,y⟩ = f`⟨x,y⟩" by simp;

then have "{f`(p). p ∈ A×B} = {f`⟨x,y⟩. ⟨x,y⟩ ∈ A×B}"

by (rule ZF1_1_L4A);

finally show "F`⟨A,B⟩ = {f`⟨x,y⟩ . ⟨x,y⟩ ∈ A×B}"

by simp

from A2 I show "F`⟨A,B⟩ ⊆ Y" using apply_funtype by blast

qed

text{*A sufficient condition for a point to belong to a result of

lifting to subsets.*}

lemma lift_subset_suff: assumes A1: "f : X × X -> Y" and

A2: "A ⊆ X" "B ⊆ X" and A3: "x∈A" "y∈B" and

A4: "F = f {lifted to subsets of} X"

shows "f`⟨x,y⟩ ∈ F`⟨A,B⟩"

proof -

from A3 have "f`⟨x,y⟩ ∈ {f`(p) . p ∈ A×B}" by auto;

moreover from A1 A2 A4 have "{f`(p). p ∈ A×B} = F`⟨A,B⟩ "

using lift_subsets_explained by simp;

ultimately show "f`⟨x,y⟩ ∈ F`⟨A,B⟩" by simp;

qed;

text{*A kind of converse of @{text "lift_subset_apply"}, providing

a necessary condition for a point to be in the result of lifting to

subsets.*}

lemma lift_subset_nec: assumes A1: "f : X × X -> Y" and

A2: "A ⊆ X" "B ⊆ X" and

A3: "F = f {lifted to subsets of} X" and

A4: "z ∈ F`⟨A,B⟩"

shows "∃x y. x∈A ∧ y∈B ∧ z = f`⟨x,y⟩"

proof -

from A1 A2 A3 have "F`⟨A,B⟩ = {f`(p). p ∈ A×B}"

using lift_subsets_explained by simp;

with A4 show ?thesis by auto;

qed

text{*Lifting to subsets inherits commutativity.*}

lemma lift_subset_comm: assumes A1: "f : X × X -> Y" and

A2: "f {is commutative on} X" and

A3: "F = f {lifted to subsets of} X"

shows "F {is commutative on} Pow(X)"

proof -

have "∀A ∈ Pow(X). ∀B ∈ Pow(X). F`⟨A,B⟩ = F`⟨B,A⟩"

proof -

{ fix A assume "A ∈ Pow(X)"

fix B assume "B ∈ Pow(X)"

have "F`⟨A,B⟩ = F`⟨B,A⟩"

proof -

have "∀z ∈ F`⟨A,B⟩. z ∈ F`⟨B,A⟩"

proof;

fix z assume I: "z ∈ F`⟨A,B⟩"

with A1 A3 `A ∈ Pow(X)` `B ∈ Pow(X)` have

"∃x y. x∈A ∧ y∈B ∧ z = f`⟨x,y⟩"

using lift_subset_nec by simp;

then obtain x y where "x∈A" and "y∈B" and "z = f`⟨x,y⟩"

by auto;

with A2 `A ∈ Pow(X)` `B ∈ Pow(X)` have "z = f`⟨y,x⟩"

using IsCommutative_def by auto;

with A1 A3 I `A ∈ Pow(X)` `B ∈ Pow(X)` `x∈A` `y∈B`

show "z ∈ F`⟨B,A⟩" using lift_subset_suff by simp;

qed

moreover have "∀z ∈ F`⟨B,A⟩. z ∈ F`⟨A,B⟩"

proof;

fix z assume I: "z ∈ F`⟨B,A⟩"

with A1 A3 `A ∈ Pow(X)` `B ∈ Pow(X)` have

"∃x y. x∈B ∧ y∈A ∧ z = f`⟨x,y⟩"

using lift_subset_nec by simp;

then obtain x y where "x∈B" and "y∈A" and "z = f`⟨x,y⟩"

by auto;

with A2 `A ∈ Pow(X)` `B ∈ Pow(X)` have "z = f`⟨y,x⟩"

using IsCommutative_def by auto;

with A1 A3 I `A ∈ Pow(X)` `B ∈ Pow(X)` `x∈B` `y∈A`

show "z ∈ F`⟨A,B⟩" using lift_subset_suff by simp;

qed;

ultimately show "F`⟨A,B⟩ = F`⟨B,A⟩" by auto

qed;

} thus ?thesis by auto;

qed;

then show "F {is commutative on} Pow(X)"

unfolding IsCommutative_def by auto;

qed;

text{*Lifting to subsets inherits associativity.

To show that

$F\langle \langle A,B\rangle C\rangle = F\langle A,F\langle B,C\rangle\rangle$

we prove two inclusions and the proof of the second inclusion is very similar

to the proof of the first one.*}

lemma lift_subset_assoc: assumes A1: "f : X × X -> X" and

A2: "f {is associative on} X" and

A3: "F = f {lifted to subsets of} X"

shows "F {is associative on} Pow(X)"

proof -

from A1 A3 have "F : Pow(X)×Pow(X) -> Pow(X)"

using lift_subsets_binop by simp;

moreover have "∀A ∈ Pow(X).∀B ∈ Pow(X). ∀C ∈ Pow(X).

F`⟨F`⟨A,B⟩,C⟩ = F`⟨A,F`⟨B,C⟩⟩"

proof -

{ fix A B C

assume "A ∈ Pow(X)" "B ∈ Pow(X)" "C ∈ Pow(X)"

have "F`⟨F`⟨A,B⟩,C⟩ ⊆ F`⟨A,F`⟨B,C⟩⟩"

proof;

fix z assume I: "z ∈ F`⟨F`⟨A,B⟩,C⟩"

from A1 A3 `A ∈ Pow(X)` `B ∈ Pow(X)`

have "F`⟨A,B⟩ ∈ Pow(X)"

using lift_subsets_binop apply_funtype by blast;

with A1 A3 `C ∈ Pow(X)` I have

"∃x y. x ∈ F`⟨A,B⟩ ∧ y ∈ C ∧ z = f`⟨x,y⟩"

using lift_subset_nec by simp;

then obtain x y where

II: "x ∈ F`⟨A,B⟩" and "y ∈ C" and III: "z = f`⟨x,y⟩"

by auto;

from A1 A3 `A ∈ Pow(X)` `B ∈ Pow(X)` II have

"∃ s t. s ∈ A ∧ t ∈ B ∧ x = f`⟨s,t⟩"

using lift_subset_nec by auto;

then obtain s t where "s∈A" and "t∈B" and "x = f`⟨s,t⟩"

by auto;

with A2 `A ∈ Pow(X)` `B ∈ Pow(X)` `C ∈ Pow(X)` III

`s∈A` `t∈B` `y∈C` have IV: "z = f`⟨s, f`⟨t,y⟩⟩"

using IsAssociative_def by blast;

from A1 A3 `B ∈ Pow(X)` `C ∈ Pow(X)` `t∈B` `y∈C`

have "f`⟨t,y⟩ ∈ F`⟨B,C⟩" using lift_subset_suff by simp;

moreover from A1 A3 `B ∈ Pow(X)` `C ∈ Pow(X)`

have "F`⟨B,C⟩ ⊆ X" using lift_subsets_binop apply_funtype

by blast;

moreover note A1 A3 `A ∈ Pow(X)` `s∈A` IV

ultimately show "z ∈ F`⟨A,F`⟨B,C⟩⟩"

using lift_subset_suff by simp;

qed;

moreover have "F`⟨A,F`⟨B,C⟩⟩ ⊆ F`⟨F`⟨A,B⟩,C⟩"

proof;

fix z assume I: "z ∈ F`⟨A,F`⟨B,C⟩⟩"

from A1 A3 `B ∈ Pow(X)` `C ∈ Pow(X)`

have "F`⟨B,C⟩ ∈ Pow(X)"

using lift_subsets_binop apply_funtype by blast;

with A1 A3 `A ∈ Pow(X)` I have

"∃x y. x ∈ A ∧ y ∈ F`⟨B,C⟩ ∧ z = f`⟨x,y⟩"

using lift_subset_nec by simp;

then obtain x y where

"x ∈ A" and II: "y ∈ F`⟨B,C⟩" and III: "z = f`⟨x,y⟩"

by auto;

from A1 A3 `B ∈ Pow(X)` `C ∈ Pow(X)` II have

"∃ s t. s ∈ B ∧ t ∈ C ∧ y = f`⟨s,t⟩"

using lift_subset_nec by auto;

then obtain s t where "s∈B" and "t∈C" and "y = f`⟨s,t⟩"

by auto;

with III have "z = f`⟨x,f`⟨s,t⟩⟩" by simp;

moreover from A2 `A ∈ Pow(X)` `B ∈ Pow(X)` `C ∈ Pow(X)`

`x∈A` `s∈B` `t∈C` have "f`⟨f`⟨x,s⟩,t⟩ = f`⟨x,f`⟨s,t⟩⟩"

using IsAssociative_def by blast;

ultimately have IV: "z = f`⟨f`⟨x,s⟩,t⟩" by simp;

from A1 A3 `A ∈ Pow(X)` `B ∈ Pow(X)` `x∈A` `s∈B`

have "f`⟨x,s⟩ ∈ F`⟨A,B⟩" using lift_subset_suff by simp;

moreover from A1 A3 `A ∈ Pow(X)` `B ∈ Pow(X)`

have "F`⟨A,B⟩ ⊆ X" using lift_subsets_binop apply_funtype

by blast;

moreover note A1 A3 `C ∈ Pow(X)` `t∈C` IV;

ultimately show "z ∈ F`⟨F`⟨A,B⟩,C⟩"

using lift_subset_suff by simp;

qed;

ultimately have "F`⟨F`⟨A,B⟩,C⟩ = F`⟨A,F`⟨B,C⟩⟩" by auto;

} thus ?thesis by auto;

qed

ultimately show ?thesis unfolding IsAssociative_def

by auto;

qed;

section{*Distributive operations*}

text{*In this section we deal with pairs of operations such that one is

distributive with respect to the other, that is

$a\cdot (b+c) = a\cdot b + a\cdot c$ and

$(b+c)\cdot a = b\cdot a + c\cdot a$. We show that this property is

preserved under restriction to a set closed with respect to both

operations. In @{text "EquivClass1"} theory we show that this property is

preserved by projections to the quotient space if both operations are

congruent with respect to the equivalence relation.*}

text{*We define distributivity as a statement about three sets. The first

set is the set on which the operations act. The second set is the

additive operation (a ZF function) and the third is the multiplicative

operation.*}

definition

"IsDistributive(X,A,M) ≡ (∀a∈X.∀b∈X.∀c∈X.

M`⟨a,A`⟨b,c⟩⟩ = A`⟨M`⟨a,b⟩,M`⟨a,c⟩⟩ ∧

M`⟨A`⟨b,c⟩,a⟩ = A`⟨M`⟨b,a⟩,M`⟨c,a⟩ ⟩)"

text{*The essential condition to show that distributivity is preserved by

restrictions to sets that are closed with

respect to both operations.*}

lemma func_ZF_7_L1:

assumes A1: "IsDistributive(X,A,M)"

and A2: "Y⊆X"

and A3: "Y {is closed under} A" "Y {is closed under} M"

and A4: "A⇩_{r}= restrict(A,Y×Y)" "M⇩_{r}= restrict(M,Y×Y)"

and A5: "a∈Y" "b∈Y" "c∈Y"

shows "M⇩_{r}`⟨ a,A⇩_{r}`⟨b,c⟩ ⟩ = A⇩_{r}`⟨ M⇩_{r}`⟨a,b⟩,M⇩_{r}`⟨a,c⟩ ⟩ ∧

M⇩_{r}`⟨ A⇩_{r}`⟨b,c⟩,a ⟩ = A⇩_{r}`⟨ M⇩_{r}`⟨b,a⟩, M⇩_{r}`⟨c,a⟩ ⟩"

proof -

from A3 A5 have "A`⟨b,c⟩ ∈ Y" "M`⟨a,b⟩ ∈ Y" "M`⟨a,c⟩ ∈ Y"

"M`⟨b,a⟩ ∈ Y" "M`⟨c,a⟩ ∈ Y" using IsOpClosed_def by auto;

with A5 A4 have

"A⇩_{r}`⟨b,c⟩ ∈ Y" "M⇩_{r}`⟨a,b⟩ ∈ Y" "M⇩_{r}`⟨a,c⟩ ∈ Y"

"M⇩_{r}`⟨b,a⟩ ∈ Y" "M⇩_{r}`⟨c,a⟩ ∈ Y"

using restrict by auto;

with A1 A2 A4 A5 show ?thesis

using restrict IsDistributive_def by auto;

qed;

text{*Distributivity is preserved by restrictions to sets that are closed with

respect to both operations.*}

lemma func_ZF_7_L2:

assumes "IsDistributive(X,A,M)"

and "Y⊆X"

and "Y {is closed under} A"

"Y {is closed under} M"

and "A⇩_{r}= restrict(A,Y×Y)" "M⇩_{r}= restrict(M,Y×Y)"

shows "IsDistributive(Y,A⇩_{r},M⇩_{r})"

proof -

from assms have "∀a∈Y.∀b∈Y.∀c∈Y.

M⇩_{r}`⟨ a,A⇩_{r}`⟨b,c⟩ ⟩ = A⇩_{r}`⟨ M⇩_{r}`⟨a,b⟩,M⇩_{r}`⟨a,c⟩ ⟩ ∧

M⇩_{r}`⟨ A⇩_{r}`⟨b,c⟩,a ⟩ = A⇩_{r}`⟨ M⇩_{r}`⟨b,a⟩,M⇩_{r}`⟨c,a⟩⟩"

using func_ZF_7_L1 by simp;

then show ?thesis using IsDistributive_def by simp;

qed

end;