(* 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: 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 ‹Binary operations› 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 $X$.› subsection‹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 subsection‹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 subsection‹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 subsection‹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 ‹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 subsection‹Identity function› text‹In this section we show some additional facts about the identity function defined in the standard Isabelle's ‹Perm› theory. Note there is also ‹image_id_same› lemma in ‹func1› 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 surprisingly 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 subsection‹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 ‹fst(p)› and ‹snd(p)›, resp. Recall that in Isabelle/ZF ‹fst(p)› and ‹snd(p)› denote the first and second components of an ordered pair $p$. See the lemma ‹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 ‹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 ‹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 {is associative on} X" and A2: "F = f {lifted to subsets of} X" shows "F {is associative on} Pow(X)" proof - from A1 have "f : X×X → X" unfolding IsAssociative_def by simp with A2 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 ‹f:X×X → X› A2 ‹A ∈ Pow(X)› ‹B ∈ Pow(X)› have "F`⟨A,B⟩ ∈ Pow(X)" using lift_subsets_binop apply_funtype by blast with ‹f:X×X → X› A2 ‹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 ‹f:X×X → X› A2 ‹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 A1 ‹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 ‹f:X×X → X› A2 ‹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 ‹f:X×X → X› A2 ‹B ∈ Pow(X)› ‹C ∈ Pow(X)› have "F`⟨B,C⟩ ⊆ X" using lift_subsets_binop apply_funtype by blast moreover note ‹f:X×X → X› A2 ‹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 ‹f:X×X → X› A2 ‹B ∈ Pow(X)› ‹C ∈ Pow(X)› have "F`⟨B,C⟩ ∈ Pow(X)" using lift_subsets_binop apply_funtype by blast with ‹f:X×X → X› A2 ‹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 ‹f:X×X → X› A2 ‹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 A1 ‹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 ‹f:X×X → X› A2 ‹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 ‹f:X×X → X› A2 ‹A ∈ Pow(X)› ‹B ∈ Pow(X)› have "F`⟨A,B⟩ ⊆ X" using lift_subsets_binop apply_funtype by blast moreover note ‹f:X×X → X› A2 ‹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 subsection‹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 ‹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