(* This file is a part of IsarMathLib - a library of formalized mathematics written for Isabelle/Isar. Copyright (C) 2005 - 2022 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 ‹Functions - introduction› theory func1 imports ZF.func Fol1 ZF1 begin text‹This theory covers basic properties of function spaces. A set of functions with domain $X$ and values in the set $Y$ is denoted in Isabelle as $X\rightarrow Y$. It just happens that the colon ":" is a synonym of the set membership symbol $\in$ in Isabelle/ZF so we can write $f:X\rightarrow Y$ instead of $f \in X\rightarrow Y$. This is the only case that we use the colon instead of the regular set membership symbol.› subsection‹Properties of functions, function spaces and (inverse) images.› text‹Functions in ZF are sets of pairs. This means that if $f: X\rightarrow Y $ then $f\subseteq X\times Y$. This section is mostly about consequences of this understanding of the notion of function. › text‹We define the notion of function that preserves a collection here. Given two collection of sets a function preserves the collections if the inverse image of sets in one collection belongs to the second one. This notion does not have a name in romantic math. It is used to define continuous functions in ‹Topology_ZF_2› theory. We define it here so that we can use it for other purposes, like defining measurable functions. Recall that ‹f-``(A)› means the inverse image of the set $A$.› definition "PresColl(f,S,T) ≡ ∀ A∈T. f-``(A)∈S" text‹A definition that allows to get the first factor of the domain of a binary function $f: X\times Y \rightarrow Z$.› definition "fstdom(f) ≡ domain(domain(f))" text‹If a function maps $A$ into another set, then $A$ is the domain of the function.› lemma func1_1_L1: assumes "f:A→C" shows "domain(f) = A" using assms domain_of_fun by simp text‹Standard Isabelle defines a ‹function(f)› predicate. The next lemma shows that our functions satisfy that predicate. It is a special version of Isabelle's ‹fun_is_function›.› lemma fun_is_fun: assumes "f:X→Y" shows "function(f)" using assms fun_is_function by simp text‹A lemma explains what ‹fstdom› is for.› lemma fstdomdef: assumes A1: "f: X×Y → Z" and A2: "Y≠0" shows "fstdom(f) = X" proof - from A1 have "domain(f) = X×Y" using func1_1_L1 by simp with A2 show "fstdom(f) = X" unfolding fstdom_def by auto qed text‹A version of the ‹Pi_type› lemma from the standard Isabelle/ZF library.› lemma func1_1_L1A: assumes A1: "f:X→Y" and A2: "∀x∈X. f`(x) ∈ Z" shows "f:X→Z" proof - { fix x assume "x∈X" with A2 have "f`(x) ∈ Z" by simp } with A1 show "f:X→Z" by (rule Pi_type) qed text‹A variant of ‹func1_1_L1A›.› lemma func1_1_L1B: assumes A1: "f:X→Y" and A2: "Y⊆Z" shows "f:X→Z" proof - from A1 A2 have "∀x∈X. f`(x) ∈ Z" using apply_funtype by auto with A1 show "f:X→Z" using func1_1_L1A by blast qed text‹There is a value for each argument.› lemma func1_1_L2: assumes A1: "f:X→Y" "x∈X" shows "∃y∈Y. ⟨x,y⟩ ∈ f" proof- from A1 have "f`(x) ∈ Y" using apply_type by simp moreover from A1 have "⟨ x,f`(x)⟩∈ f" using apply_Pair by simp ultimately show ?thesis by auto qed text‹The inverse image is the image of converse. True for relations as well.› lemma vimage_converse: shows "r-``(A) = converse(r)``(A)" using vimage_iff image_iff converse_iff by auto text‹The image is the inverse image of converse.› lemma image_converse: shows "converse(r)-``(A) = r``(A)" using vimage_iff image_iff converse_iff by auto text‹The inverse image by a composition is the composition of inverse images.› lemma vimage_comp: shows "(r O s)-``(A) = s-``(r-``(A))" using vimage_converse converse_comp image_comp image_converse by simp text‹A version of ‹vimage_comp› for three functions.› lemma vimage_comp3: shows "(r O s O t)-``(A) = t-``(s-``(r-``(A)))" using vimage_comp by simp text‹Inverse image of any set is contained in the domain.› lemma func1_1_L3: assumes A1: "f:X→Y" shows "f-``(D) ⊆ X" proof- have "∀x. x∈f-``(D) ⟶ x ∈ domain(f)" using vimage_iff domain_iff by auto with A1 have "∀x. (x ∈ f-``(D)) ⟶ (x∈X)" using func1_1_L1 by simp then show ?thesis by auto qed text‹The inverse image of the range is the domain.› lemma func1_1_L4: assumes "f:X→Y" shows "f-``(Y) = X" using assms func1_1_L3 func1_1_L2 vimage_iff by blast text‹The arguments belongs to the domain and values to the range.› lemma func1_1_L5: assumes A1: "⟨ x,y⟩ ∈ f" and A2: "f:X→Y" shows "x∈X ∧ y∈Y" proof from A1 A2 show "x∈X" using apply_iff by simp with A2 have "f`(x)∈ Y" using apply_type by simp with A1 A2 show "y∈Y" using apply_iff by simp qed text‹Function is a subset of cartesian product.› lemma fun_subset_prod: assumes A1: "f:X→Y" shows "f ⊆ X×Y" proof fix p assume "p ∈ f" with A1 have "∃x∈X. p = ⟨x, f`(x)⟩" using Pi_memberD by simp then obtain x where I: "p = ⟨x, f`(x)⟩" by auto with A1 ‹p ∈ f› have "x∈X ∧ f`(x) ∈ Y" using func1_1_L5 by blast with I show "p ∈ X×Y" by auto qed text‹The (argument, value) pair belongs to the graph of the function.› lemma func1_1_L5A: assumes A1: "f:X→Y" "x∈X" "y = f`(x)" shows "⟨x,y⟩ ∈ f" "y ∈ range(f)" proof - from A1 show "⟨x,y⟩ ∈ f" using apply_Pair by simp then show "y ∈ range(f)" using rangeI by simp qed text‹The next theorem illustrates the meaning of the concept of function in ZF.› theorem fun_is_set_of_pairs: assumes A1: "f:X→Y" shows "f = {⟨x, f`(x)⟩. x ∈ X}" proof from A1 show "{⟨x, f`(x)⟩. x ∈ X} ⊆ f" using func1_1_L5A by auto next { fix p assume "p ∈ f" with A1 have "p ∈ X×Y" using fun_subset_prod by auto with A1 ‹p ∈ f› have "p ∈ {⟨x, f`(x)⟩. x ∈ X}" using apply_equality by auto } thus "f ⊆ {⟨x, f`(x)⟩. x ∈ X}" by auto qed text‹The range of function that maps $X$ into $Y$ is contained in $Y$.› lemma func1_1_L5B: assumes A1: "f:X→Y" shows "range(f) ⊆ Y" proof fix y assume "y ∈ range(f)" then obtain x where "⟨ x,y⟩ ∈ f" using range_def converse_def domain_def by auto with A1 show "y∈Y" using func1_1_L5 by blast qed text‹The image of any set is contained in the range.› lemma func1_1_L6: assumes A1: "f:X→Y" shows "f``(B) ⊆ range(f)" and "f``(B) ⊆ Y" proof - show "f``(B) ⊆ range(f)" using image_iff rangeI by auto with A1 show "f``(B) ⊆ Y" using func1_1_L5B by blast qed text‹The inverse image of any set is contained in the domain.› lemma func1_1_L6A: assumes A1: "f:X→Y" shows "f-``(A)⊆X" proof fix x assume A2: "x∈f-``(A)" then obtain y where "⟨ x,y⟩ ∈ f" using vimage_iff by auto with A1 show "x∈X" using func1_1_L5 by fast qed text‹Image of a greater set is greater.› lemma func1_1_L8: assumes A1: "A⊆B" shows "f``(A)⊆ f``(B)" using assms image_Un by auto text‹A set is contained in the the inverse image of its image. There is similar theorem in ‹equalities.thy› (‹function_image_vimage›) which shows that the image of inverse image of a set is contained in the set.› lemma func1_1_L9: assumes A1: "f:X→Y" and A2: "A⊆X" shows "A ⊆ f-``(f``(A))" proof - from A1 A2 have "∀x∈A. ⟨ x,f`(x)⟩ ∈ f" using apply_Pair by auto then show ?thesis using image_iff by auto qed text‹The inverse image of the image of the domain is the domain.› lemma inv_im_dom: assumes A1: "f:X→Y" shows "f-``(f``(X)) = X" proof from A1 show "f-``(f``(X)) ⊆ X" using func1_1_L3 by simp from A1 show "X ⊆ f-``(f``(X))" using func1_1_L9 by simp qed text‹A technical lemma needed to make the ‹func1_1_L11› proof more clear.› lemma func1_1_L10: assumes A1: "f ⊆ X×Y" and A2: "∃!y. (y∈Y ∧ ⟨x,y⟩ ∈ f)" shows "∃!y. ⟨x,y⟩ ∈ f" proof from A2 show "∃y. ⟨x, y⟩ ∈ f" by auto fix y n assume "⟨x,y⟩ ∈ f" and "⟨x,n⟩ ∈ f" with A1 A2 show "y=n" by auto qed text‹If $f\subseteq X\times Y$ and for every $x\in X$ there is exactly one $y\in Y$ such that $(x,y)\in f$ then $f$ maps $X$ to $Y$.› lemma func1_1_L11: assumes "f ⊆ X×Y" and "∀x∈X. ∃!y. y∈Y ∧ ⟨x,y⟩ ∈ f" shows "f: X→Y" using assms func1_1_L10 Pi_iff_old by simp text‹A set defined by a lambda-type expression is a fuction. There is a similar lemma in func.thy, but I had problems with lambda expressions syntax so I could not apply it. This lemma is a workaround for this. Besides, lambda expressions are not readable. › lemma func1_1_L11A: assumes A1: "∀x∈X. b(x) ∈ Y" shows "{⟨x,y⟩ ∈ X×Y. b(x) = y} : X→Y" proof - let ?f = "{⟨ x,y⟩ ∈ X×Y. b(x) = y}" have "?f ⊆ X×Y" by auto moreover have "∀x∈X. ∃!y. y∈Y ∧ ⟨ x,y⟩ ∈ ?f" proof fix x assume A2: "x∈X" show "∃!y. y∈Y ∧ ⟨x, y⟩ ∈ {⟨x,y⟩ ∈ X×Y . b(x) = y}" proof from A2 A1 show "∃y. y∈Y ∧ ⟨x, y⟩ ∈ {⟨x,y⟩ ∈ X×Y . b(x) = y}" by simp next fix y y1 assume "y∈Y ∧ ⟨x, y⟩ ∈ {⟨x,y⟩ ∈ X×Y . b(x) = y}" and "y1∈Y ∧ ⟨x, y1⟩ ∈ {⟨x,y⟩ ∈ X×Y . b(x) = y}" then show "y = y1" by simp qed qed ultimately show "{⟨ x,y⟩ ∈ X×Y. b(x) = y} : X→Y" using func1_1_L11 by simp qed text‹The next lemma will replace ‹func1_1_L11A› one day.› lemma ZF_fun_from_total: assumes A1: "∀x∈X. b(x) ∈ Y" shows "{⟨x,b(x)⟩. x∈X} : X→Y" proof - let ?f = "{⟨x,b(x)⟩. x∈X}" { fix x assume A2: "x∈X" have "∃!y. y∈Y ∧ ⟨x, y⟩ ∈ ?f" proof from A1 A2 show "∃y. y∈Y ∧ ⟨x, y⟩ ∈ ?f" by simp next fix y y1 assume "y∈Y ∧ ⟨x, y⟩ ∈ ?f" and "y1∈Y ∧ ⟨x, y1⟩ ∈ ?f" then show "y = y1" by simp qed } then have "∀x∈X. ∃!y. y∈Y ∧ ⟨ x,y⟩ ∈ ?f" by simp moreover from A1 have "?f ⊆ X×Y" by auto ultimately show ?thesis using func1_1_L11 by simp qed text‹The value of a function defined by a meta-function is this meta-function (deprecated, use ‹ZF_fun_from_tot_val(1)› instead).› lemma func1_1_L11B: assumes A1: "f:X→Y" "x∈X" and A2: "f = {⟨x,y⟩ ∈ X×Y. b(x) = y}" shows "f`(x) = b(x)" proof - from A1 have "⟨ x,f`(x)⟩ ∈ f" using apply_iff by simp with A2 show ?thesis by simp qed text‹The next lemma will replace ‹func1_1_L11B› one day.› lemma ZF_fun_from_tot_val: assumes "f:X→Y" "x∈X" and "f = {⟨x,b(x)⟩. x∈X}" shows "f`(x) = b(x)" and "b(x)∈Y" proof - from assms(1,2) have "⟨x,f`(x)⟩ ∈ f" using apply_iff by simp with assms(3) show "f`(x) = b(x)" by simp from assms(1,2) have "f`(x)∈Y" by (rule apply_funtype) with ‹f`(x) = b(x)› show "b(x)∈Y" by simp qed text‹Identical meaning as ‹ZF_fun_from_tot_val›, but phrased a bit differently.› lemma ZF_fun_from_tot_val0: assumes "f:X→Y" and "f = {⟨x,b(x)⟩. x∈X}" shows "∀x∈X. f`(x) = b(x)" using assms ZF_fun_from_tot_val by simp text‹Another way of expressing that lambda expression is a function.› lemma lam_is_fun_range: assumes "f={⟨x,g(x)⟩. x∈X}" shows "f:X→range(f)" proof - have "∀x∈X. g(x) ∈ range({⟨x,g(x)⟩. x∈X})" unfolding range_def by auto then have "{⟨x,g(x)⟩. x∈X} : X→range({⟨x,g(x)⟩. x∈X})" by (rule ZF_fun_from_total) with assms show ?thesis by auto qed text‹Yet another way of expressing value of a function.› lemma ZF_fun_from_tot_val1: assumes "x∈X" shows "{⟨x,b(x)⟩. x∈X}`(x)=b(x)" proof - let ?f = "{⟨x,b(x)⟩. x∈X}" have "?f:X→range(?f)" using lam_is_fun_range by simp with assms show ?thesis using ZF_fun_from_tot_val0 by simp qed text‹An hypotheses-free form of ‹ZF_fun_from_tot_val1›: the value of a function $X\ni x \mapsto p(x)$ is $p(x)$ for all $x\in X$. › lemma ZF_fun_from_tot_val2: shows "∀x∈X. {⟨x,b(x)⟩. x∈X}`(x) = b(x)" using ZF_fun_from_tot_val1 by simp text‹The range of a function defined by set comprehension is the set of its values."› lemma range_fun: shows "range({⟨x,b(x)⟩. x∈X}) = {b(x). x∈X}" by blast text‹In Isabelle/ZF and Metamath if $x$ is not in the domain of a function $f$ then $f(x)$ is the empty set. This allows us to conclude that if $y\in f(x)$, then $x$ must be en element of the domain of $f$. › lemma arg_in_domain: assumes "f:X→Y" "y∈f`(x)" shows "x∈X" proof - { assume "x∉X" with assms have False using func1_1_L1 apply_0 by simp } thus ?thesis by auto qed text‹We can extend a function by specifying its values on a set disjoint with the domain.› lemma func1_1_L11C: assumes A1: "f:X→Y" and A2: "∀x∈A. b(x)∈B" and A3: "X∩A = 0" and Dg: "g = f ∪ {⟨x,b(x)⟩. x∈A}" shows "g : X∪A → Y∪B" "∀x∈X. g`(x) = f`(x)" "∀x∈A. g`(x) = b(x)" proof - let ?h = "{⟨x,b(x)⟩. x∈A}" from A1 A2 A3 have I: "f:X→Y" "?h : A→B" "X∩A = 0" using ZF_fun_from_total by auto then have "f∪?h : X∪A → Y∪B" by (rule fun_disjoint_Un) with Dg show "g : X∪A → Y∪B" by simp { fix x assume A4: "x∈A" with A1 A3 have "(f∪?h)`(x) = ?h`(x)" using func1_1_L1 fun_disjoint_apply2 by blast moreover from I A4 have "?h`(x) = b(x)" using ZF_fun_from_tot_val by simp ultimately have "(f∪?h)`(x) = b(x)" by simp } with Dg show "∀x∈A. g`(x) = b(x)" by simp { fix x assume A5: "x∈X" with A3 I have "x ∉ domain(?h)" using func1_1_L1 by auto then have "(f∪?h)`(x) = f`(x)" using fun_disjoint_apply1 by simp } with Dg show "∀x∈X. g`(x) = f`(x)" by simp qed text‹We can extend a function by specifying its value at a point that does not belong to the domain.› lemma func1_1_L11D: assumes A1: "f:X→Y" and A2: "a∉X" and Dg: "g = f ∪ {⟨a,b⟩}" shows "g : X∪{a} → Y∪{b}" "∀x∈X. g`(x) = f`(x)" "g`(a) = b" proof - let ?h = "{⟨a,b⟩}" from A1 A2 Dg have I: "f:X→Y" "∀x∈{a}. b∈{b}" "X∩{a} = 0" "g = f ∪ {⟨x,b⟩. x∈{a}}" by auto then show "g : X∪{a} → Y∪{b}" by (rule func1_1_L11C) from I show "∀x∈X. g`(x) = f`(x)" by (rule func1_1_L11C) from I have "∀x∈{a}. g`(x) = b" by (rule func1_1_L11C) then show "g`(a) = b" by auto qed text‹A technical lemma about extending a function both by defining on a set disjoint with the domain and on a point that does not belong to any of those sets.› lemma func1_1_L11E: assumes A1: "f:X→Y" and A2: "∀x∈A. b(x)∈B" and A3: "X∩A = 0" and A4: "a∉ X∪A" and Dg: "g = f ∪ {⟨x,b(x)⟩. x∈A} ∪ {⟨a,c⟩}" shows "g : X∪A∪{a} → Y∪B∪{c}" "∀x∈X. g`(x) = f`(x)" "∀x∈A. g`(x) = b(x)" "g`(a) = c" proof - let ?h = "f ∪ {⟨x,b(x)⟩. x∈A}" from assms show "g : X∪A∪{a} → Y∪B∪{c}" using func1_1_L11C func1_1_L11D by simp from A1 A2 A3 have I: "f:X→Y" "∀x∈A. b(x)∈B" "X∩A = 0" "?h = f ∪ {⟨x,b(x)⟩. x∈A}" by auto from assms have II: "?h : X∪A → Y∪B" "a∉ X∪A" "g = ?h ∪ {⟨a,c⟩}" using func1_1_L11C by auto then have III: "∀x∈X∪A. g`(x) = ?h`(x)" by (rule func1_1_L11D) moreover from I have "∀x∈X. ?h`(x) = f`(x)" by (rule func1_1_L11C) ultimately show "∀x∈X. g`(x) = f`(x)" by simp from I have "∀x∈A. ?h`(x) = b(x)" by (rule func1_1_L11C) with III show "∀x∈A. g`(x) = b(x)" by simp from II show "g`(a) = c" by (rule func1_1_L11D) qed text‹A way of defining a function on a union of two possibly overlapping sets. We decompose the union into two differences and the intersection and define a function separately on each part.› lemma fun_union_overlap: assumes "∀x∈A∩B. h(x) ∈ Y" "∀x∈A-B. f(x) ∈ Y" "∀x∈B-A. g(x) ∈ Y" shows "{⟨x,if x∈A-B then f(x) else if x∈B-A then g(x) else h(x)⟩. x ∈ A∪B}: A∪B → Y" proof - let ?F = "{⟨x,if x∈A-B then f(x) else if x∈B-A then g(x) else h(x)⟩. x ∈ A∩B}" from assms have "∀x∈A∪B. (if x∈A-B then f(x) else if x∈B-A then g(x) else h(x)) ∈ Y" by auto then show ?thesis by (rule ZF_fun_from_total) qed text‹Inverse image of intersection is the intersection of inverse images.› lemma invim_inter_inter_invim: assumes "f:X→Y" shows "f-``(A∩B) = f-``(A) ∩ f-``(B)" using assms fun_is_fun function_vimage_Int by simp text‹The inverse image of an intersection of a nonempty collection of sets is the intersection of the inverse images. This generalizes ‹invim_inter_inter_invim› which is proven for the case of two sets.› lemma func1_1_L12: assumes A1: "B ⊆ Pow(Y)" and A2: "B≠0" and A3: "f:X→Y" shows "f-``(⋂B) = (⋂U∈B. f-``(U))" proof from A2 show "f-``(⋂B) ⊆ (⋂U∈B. f-``(U))" by blast show "(⋂U∈B. f-``(U)) ⊆ f-``(⋂B)" proof fix x assume A4: "x ∈ (⋂U∈B. f-``(U))" from A3 have "∀U∈B. f-``(U) ⊆ X" using func1_1_L6A by simp with A4 have "∀U∈B. x∈X" by auto with A2 have "x∈X" by auto with A3 have "∃!y. ⟨ x,y⟩ ∈ f" using Pi_iff_old by simp with A2 A4 show "x ∈ f-``(⋂B)" using vimage_iff by blast qed qed text‹The inverse image of a set does not change when we intersect the set with the image of the domain.› lemma inv_im_inter_im: assumes "f:X→Y" shows "f-``(A ∩ f``(X)) = f-``(A)" using assms invim_inter_inter_invim inv_im_dom func1_1_L6A by blast text‹If the inverse image of a set is not empty, then the set is not empty. Proof by contradiction.› lemma func1_1_L13: assumes A1:"f-``(A) ≠ 0" shows "A≠0" using assms by auto text‹If the image of a set is not empty, then the set is not empty. Proof by contradiction.› lemma func1_1_L13A: assumes A1: "f``(A)≠0" shows "A≠0" using assms by auto text‹What is the inverse image of a singleton?› lemma func1_1_L14: assumes "f∈X→Y" shows "f-``({y}) = {x∈X. f`(x) = y}" using assms func1_1_L6A vimage_singleton_iff apply_iff by auto text‹A lemma that can be used instead ‹fun_extension_iff› to show that two functions are equal› lemma func_eq: assumes "f: X→Y" "g: X→Z"and "∀x∈X. f`(x) = g`(x)" shows "f = g" using assms fun_extension_iff by simp text‹An alternative syntax for defining a function: instead of writing $\{\langle x,p(x)\rangle. x\in X\}$ we can write $\lambda x\in X. p(x)$. › lemma lambda_fun_alt: shows "{⟨x,p(x)⟩. x∈X} = (λx∈X. p(x))" proof - let ?L = "{⟨x,p(x)⟩. x∈X}" let ?R = "λx∈X. p(x)" have "?L:X→range(?L)" and "?R:X→range(?L)" using lam_is_fun_range range_fun lam_funtype by simp_all moreover have "∀x∈X. ?L`(x) = ?R`(x)" using ZF_fun_from_tot_val1 beta by simp ultimately show "?L = ?R" using func_eq by blast qed text‹If a function is equal to an expression $b(x)$ on $X$, then it has to be of the form $\{ \langle x, b(x)\rangle | x\in X\}$. › lemma func_eq_set_of_pairs: assumes "f:X→Y" "∀x∈X. f`(x) = b(x)" shows "f = {⟨x, b(x)⟩. x ∈ X}" proof - from assms(1) have "f = {⟨x, f`(x)⟩. x ∈ X}" using fun_is_set_of_pairs by simp with assms(2) show ?thesis by simp qed text‹Function defined on a singleton is a single pair.› lemma func_singleton_pair: assumes A1: "f : {a}→X" shows "f = {⟨a, f`(a)⟩}" proof - let ?g = "{⟨a, f`(a)⟩}" note A1 moreover have "?g : {a} → {f`(a)}" using singleton_fun by simp moreover have "∀x ∈ {a}. f`(x) = ?g`(x)" using singleton_apply by simp ultimately show "f = ?g" by (rule func_eq) qed text‹A single pair is a function on a singleton. This is similar to ‹singleton_fun› from standard Isabelle/ZF.› lemma pair_func_singleton: assumes A1: "y ∈ Y" shows "{⟨x,y⟩} : {x} → Y" proof - have "{⟨x,y⟩} : {x} → {y}" using singleton_fun by simp moreover from A1 have "{y} ⊆ Y" by simp ultimately show "{⟨x,y⟩} : {x} → Y" by (rule func1_1_L1B) qed text‹The value of a pair on the first element is the second one.› lemma pair_val: shows "{⟨x,y⟩}`(x) = y" using singleton_fun apply_equality by simp text‹A more familiar definition of inverse image.› lemma func1_1_L15: assumes A1: "f:X→Y" shows "f-``(A) = {x∈X. f`(x) ∈ A}" proof - have "f-``(A) = (⋃y∈A . f-``{y})" by (rule vimage_eq_UN) with A1 show ?thesis using func1_1_L14 by auto qed text‹A more familiar definition of image.› lemma func_imagedef: assumes A1: "f:X→Y" and A2: "A⊆X" shows "f``(A) = {f`(x). x ∈ A}" proof from A1 show "f``(A) ⊆ {f`(x). x ∈ A}" using image_iff apply_iff by auto show "{f`(x). x ∈ A} ⊆ f``(A)" proof fix y assume "y ∈ {f`(x). x ∈ A}" then obtain x where "x∈A" and "y = f`(x)" by auto with A1 A2 have "⟨x,y⟩ ∈ f" using apply_iff by force with A1 A2 ‹x∈A› show "y ∈ f``(A)" using image_iff by auto qed qed text‹A technical lemma about graphs of functions: if we have two disjoint sets $A$ and $B$ then the cartesian product of the inverse image of $A$ and $B$ is disjoint with (the graph of) $f$.› lemma vimage_prod_dis_graph: assumes "f:X→Y" "A∩B = 0" shows "f-``(A)×B ∩ f = 0" proof - { assume "f-``(A)×B ∩ f ≠ 0" then obtain p where "p ∈ f-``(A)×B" and "p∈f" by blast from assms(1) ‹p∈f› have "p ∈ {⟨x, f`(x)⟩. x ∈ X}" using fun_is_set_of_pairs by simp then obtain x where "p = ⟨x, f`(x)⟩" by blast with assms ‹p ∈ f-``(A)×B› have False using func1_1_L15 by auto } thus ?thesis by auto qed text‹ For two functions with the same domain $X$ and the codomain $Y,Z$ resp., we can define a third one that maps $X$ to the cartesian product of $Y$ and $Z$. › lemma prod_fun_val: assumes "{⟨x,p(x)⟩. x∈X}: X→Y" "{⟨x,q(x)⟩. x∈X}: X→Z" defines "h ≡ {⟨x,⟨p(x),q(x)⟩⟩. x∈X}" shows "h:X→Y×Z" and "∀x∈X. h`(x) = ⟨p(x),q(x)⟩" proof - from assms(1,2) have "∀x∈X. ⟨p(x),q(x)⟩ ∈ Y×Z" using ZF_fun_from_tot_val(2) by auto with assms(3) show "h:X→Y×Z" using ZF_fun_from_total by simp with assms(3) show "∀x∈X. h`(x) = ⟨p(x),q(x)⟩" using ZF_fun_from_tot_val0 by simp qed text‹Suppose we have two functions $f:X\rightarrow Y$ and $g:X\rightarrow Z$ and the third one is defined as $h:X\rightarrow Y\times Z$, $x\mapsto \langle f(x),g(x)\rangle$. Given two sets $U$, $V$ we have $h^{-1}(U\times V) = (f^{-1}(U)) \cap (g^{-1}(V))$. We also show that the set where the function $f,g$ are equal is the same as $h^{-1}(\{ \langle y,y\rangle : y\in X\}$. It is a bit surprising that we get the last identity without the assumption that $Y=Z$. › lemma vimage_prod: assumes "f:X→Y" "g:X→Z" defines "h ≡ {⟨x,⟨f`(x),g`(x)⟩⟩. x∈X}" shows "h:X→Y×Z" "∀x∈X. h`(x) = ⟨f`(x),g`(x)⟩" "h-``(U×V) = f-``(U) ∩ g-``(V)" "{x∈X. f`(x) = g`(x)} = h-``({⟨y,y⟩. y∈Y})" proof - from assms show "h:X→Y×Z" using apply_funtype ZF_fun_from_total by simp with assms(3) show I: "∀x∈X. h`(x) = ⟨f`(x),g`(x)⟩" using ZF_fun_from_tot_val by simp with assms(1,2) ‹h:X→Y×Z› show "h-``(U×V) = f-``(U) ∩ g-``(V)" using func1_1_L15 by auto from assms(1) I ‹h:X→Y×Z› show "{x∈X. f`(x) = g`(x)} = h-``({⟨y,y⟩. y∈Y})" using apply_funtype func1_1_L15 by auto qed text‹The image of a set contained in domain under identity is the same set.› lemma image_id_same: assumes "A⊆X" shows "id(X)``(A) = A" using assms id_type id_conv by auto text‹The inverse image of a set contained in domain under identity is the same set.› lemma vimage_id_same: assumes "A⊆X" shows "id(X)-``(A) = A" using assms id_type id_conv by auto text‹What is the image of a singleton?› lemma singleton_image: assumes "f∈X→Y" and "x∈X" shows "f``{x} = {f`(x)}" using assms func_imagedef by auto text‹If an element of the domain of a function belongs to a set, then its value belongs to the image of that set.› lemma func1_1_L15D: assumes "f:X→Y" "x∈A" "A⊆X" shows "f`(x) ∈ f``(A)" using assms func_imagedef by auto text‹Range is the image of the domain. Isabelle/ZF defines ‹range(f)› as ‹domain(converse(f))›, and that's why we have something to prove here.› lemma range_image_domain: assumes A1: "f:X→Y" shows "f``(X) = range(f)" proof show "f``(X) ⊆ range(f)" using image_def by auto { fix y assume "y ∈ range(f)" then obtain x where "⟨y,x⟩ ∈ converse(f)" by auto with A1 have "x∈X" using func1_1_L5 by blast with A1 have "f`(x) ∈ f``(X)" using func_imagedef by auto with A1 ‹⟨y,x⟩ ∈ converse(f)› have "y ∈ f``(X)" using apply_equality by auto } then show "range(f) ⊆ f``(X)" by auto qed text‹The difference of images is contained in the image of difference.› lemma diff_image_diff: assumes A1: "f: X→Y" and A2: "A⊆X" shows "f``(X) - f``(A) ⊆ f``(X-A)" proof fix y assume "y ∈ f``(X) - f``(A)" hence "y ∈ f``(X)" and I: "y ∉ f``(A)" by auto with A1 obtain x where "x∈X" and II: "y = f`(x)" using func_imagedef by auto with A1 A2 I have "x∉A" using func1_1_L15D by auto with ‹x∈X› have "x ∈ X-A" "X-A ⊆ X" by auto with A1 II show "y ∈ f``(X-A)" using func1_1_L15D by simp qed text‹The image of an intersection is contained in the intersection of the images.› lemma image_of_Inter: assumes A1: "f:X→Y" and A2: "I≠0" and A3: "∀i∈I. P(i) ⊆ X" shows "f``(⋂i∈I. P(i)) ⊆ ( ⋂i∈I. f``(P(i)) )" proof fix y assume A4: "y ∈ f``(⋂i∈I. P(i))" from A1 A2 A3 have "f``(⋂i∈I. P(i)) = {f`(x). x ∈ ( ⋂i∈I. P(i) )}" using ZF1_1_L7 func_imagedef by simp with A4 obtain x where "x ∈ ( ⋂i∈I. P(i) )" and "y = f`(x)" by auto with A1 A2 A3 show "y ∈ ( ⋂i∈I. f``(P(i)) )" using func_imagedef by auto qed text‹The image of union is the union of images.› lemma image_of_Union: assumes A1: "f:X→Y" and A2: "∀A∈M. A⊆X" shows "f``(⋃M) = ⋃{f``(A). A∈M}" proof from A2 have "⋃M ⊆ X" by auto { fix y assume "y ∈ f``(⋃M)" with A1 ‹⋃M ⊆ X› obtain x where "x∈⋃M" and I: "y = f`(x)" using func_imagedef by auto then obtain A where "A∈M" and "x∈A" by auto with assms I have "y ∈ ⋃{f``(A). A∈M}" using func_imagedef by auto } thus "f``(⋃M) ⊆ ⋃{f``(A). A∈M}" by auto { fix y assume "y ∈ ⋃{f``(A). A∈M}" then obtain A where "A∈M" and "y ∈ f``(A)" by auto with assms ‹⋃M ⊆ X› have "y ∈ f``(⋃M)" using func_imagedef by auto } thus "⋃{f``(A). A∈M} ⊆ f``(⋃M)" by auto qed text‹If the domain of a function is nonempty, then the codomain is as well.› lemma codomain_nonempty: assumes "f:X→Y" "X≠0" shows "Y≠0" using assms apply_funtype by blast text‹The image of a nonempty subset of domain is nonempty.› lemma func1_1_L15A: assumes A1: "f: X→Y" and A2: "A⊆X" and A3: "A≠0" shows "f``(A) ≠ 0" proof - from A3 obtain x where "x∈A" by auto with A1 A2 have "f`(x) ∈ f``(A)" using func_imagedef by auto then show "f``(A) ≠ 0" by auto qed text‹The next lemma allows to prove statements about the values in the domain of a function given a statement about values in the range.› lemma func1_1_L15B: assumes "f:X→Y" and "A⊆X" and "∀y∈f``(A). P(y)" shows "∀x∈A. P(f`(x))" using assms func_imagedef by simp text‹An image of an image is the image of a composition.› lemma func1_1_L15C: assumes A1: "f:X→Y" and A2: "g:Y→Z" and A3: "A⊆X" shows "g``(f``(A)) = {g`(f`(x)). x∈A}" "g``(f``(A)) = (g O f)``(A)" proof - from A1 A3 have "{f`(x). x∈A} ⊆ Y" using apply_funtype by auto with A2 have "g``{f`(x). x∈A} = {g`(f`(x)). x∈A}" using func_imagedef by auto with A1 A3 show I: "g``(f``(A)) = {g`(f`(x)). x∈A}" using func_imagedef by simp from A1 A3 have "∀x∈A. (g O f)`(x) = g`(f`(x))" using comp_fun_apply by auto with I have "g``(f``(A)) = {(g O f)`(x). x∈A}" by simp moreover from A1 A2 A3 have "(g O f)``(A) = {(g O f)`(x). x∈A}" using comp_fun func_imagedef by blast ultimately show "g``(f``(A)) = (g O f)``(A)" by simp qed text‹What is the image of a set defined by a meta-fuction?› lemma func1_1_L17: assumes A1: "f ∈ X→Y" and A2: "∀x∈A. b(x) ∈ X" shows "f``({b(x). x∈A}) = {f`(b(x)). x∈A}" proof - from A2 have "{b(x). x∈A} ⊆ X" by auto with A1 show ?thesis using func_imagedef by auto qed text‹What are the values of composition of three functions?› lemma func1_1_L18: assumes A1: "f:A→B" "g:B→C" "h:C→D" and A2: "x∈A" shows "(h O g O f)`(x) ∈ D" "(h O g O f)`(x) = h`(g`(f`(x)))" proof - from A1 have "(h O g O f) : A→D" using comp_fun by blast with A2 show "(h O g O f)`(x) ∈ D" using apply_funtype by simp from A1 A2 have "(h O g O f)`(x) = h`( (g O f)`(x))" using comp_fun comp_fun_apply by blast with A1 A2 show "(h O g O f)`(x) = h`(g`(f`(x)))" using comp_fun_apply by simp qed text‹A composition of functions is a function. This is a slight generalization of standard Isabelle's ‹comp_fun› › lemma comp_fun_subset: assumes A1: "g:A→B" and A2: "f:C→D" and A3: "B ⊆ C" shows "f O g : A → D" proof - from A1 A3 have "g:A→C" by (rule func1_1_L1B) with A2 show "f O g : A → D" using comp_fun by simp qed text‹This lemma supersedes the lemma ‹comp_eq_id_iff› in Isabelle/ZF. Contributed by Victor Porton.› lemma comp_eq_id_iff1: assumes A1: "g: B→A" and A2: "f: A→C" shows "(∀y∈B. f`(g`(y)) = y) ⟷ f O g = id(B)" proof - from assms have "f O g: B→C" and "id(B): B→B" using comp_fun id_type by auto then have "(∀y∈B. (f O g)`y = id(B)`(y)) ⟷ f O g = id(B)" by (rule fun_extension_iff) moreover from A1 have "∀y∈B. (f O g)`y = f`(g`y)" and "∀y∈B. id(B)`(y) = y" by auto ultimately show "(∀y∈B. f`(g`y) = y) ⟷ f O g = id(B)" by simp qed text‹A lemma about a value of a function that is a union of some collection of functions.› lemma fun_Union_apply: assumes A1: "⋃F : X→Y" and A2: "f∈F" and A3: "f:A→B" and A4: "x∈A" shows "(⋃F)`(x) = f`(x)" proof - from A3 A4 have "⟨x, f`(x)⟩ ∈ f" using apply_Pair by simp with A2 have "⟨x, f`(x)⟩ ∈ ⋃F" by auto with A1 show "(⋃F)`(x) = f`(x)" using apply_equality by simp qed subsection‹Functions restricted to a set› text‹Standard Isabelle/ZF defines the notion ‹restrict(f,A)› of to mean a function (or relation) $f$ restricted to a set. This means that if $f$ is a function defined on $X$ and $A$ is a subset of $X$ then ‹restrict(f,A)› is a function whith the same values as $f$, but whose domain is $A$.› text‹What is the inverse image of a set under a restricted fuction?› lemma func1_2_L1: assumes A1: "f:X→Y" and A2: "B⊆X" shows "restrict(f,B)-``(A) = f-``(A) ∩ B" proof - let ?g = "restrict(f,B)" from A1 A2 have "?g:B→Y" using restrict_type2 by simp with A2 A1 show "?g-``(A) = f-``(A) ∩ B" using func1_1_L15 restrict_if by auto qed text‹A criterion for when one function is a restriction of another. The lemma below provides a result useful in the actual proof of the criterion and applications.› lemma func1_2_L2: assumes A1: "f:X→Y" and A2: "g ∈ A→Z" and A3: "A⊆X" and A4: "f ∩ A×Z = g" shows "∀x∈A. g`(x) = f`(x)" proof fix x assume "x∈A" with A2 have "⟨x,g`(x)⟩ ∈ g" using apply_Pair by simp with A4 A1 show "g`(x) = f`(x)" using apply_iff by auto qed text‹Here is the actual criterion.› lemma func1_2_L3: assumes A1: "f:X→Y" and A2: "g:A→Z" and A3: "A⊆X" and A4: "f ∩ A×Z = g" shows "g = restrict(f,A)" proof from A4 show "g ⊆ restrict(f, A)" using restrict_iff by auto show "restrict(f, A) ⊆ g" proof fix z assume A5:"z ∈ restrict(f,A)" then obtain x y where D1:"z∈f ∧ x∈A ∧ z = ⟨x,y⟩" using restrict_iff by auto with A1 have "y = f`(x)" using apply_iff by auto with A1 A2 A3 A4 D1 have "y = g`(x)" using func1_2_L2 by simp with A2 D1 show "z∈g" using apply_Pair by simp qed qed text‹Which function space a restricted function belongs to?› lemma func1_2_L4: assumes A1: "f:X→Y" and A2: "A⊆X" and A3: "∀x∈A. f`(x) ∈ Z" shows "restrict(f,A) : A→Z" proof - let ?g = "restrict(f,A)" from A1 A2 have "?g : A→Y" using restrict_type2 by simp moreover { fix x assume "x∈A" with A1 A3 have "?g`(x) ∈ Z" using restrict by simp} ultimately show ?thesis by (rule Pi_type) qed text‹A simpler case of ‹func1_2_L4›, where the range of the original and restricted function are the same.› corollary restrict_fun: assumes A1: "f:X→Y" and A2: "A⊆X" shows "restrict(f,A) : A → Y" proof - from assms have "∀x∈A. f`(x) ∈ Y" using apply_funtype by auto with assms show ?thesis using func1_2_L4 by simp qed text‹A function restricted to its domain is itself.› lemma restrict_domain: assumes "f:X→Y" shows "restrict(f,X) = f" proof - have "∀x∈X. restrict(f,X)`(x) = f`(x)" using restrict by simp with assms show ?thesis using func_eq restrict_fun by blast qed text‹Suppose a function $f:X\rightarrow Y$ is defined by an expression $q$, i.e. $f = \{\langle x,y\rangle : x\in X\}$. Then a function that is defined by the same expression, but on a smaller set is the same as the restriction of $f$ to that smaller set.› lemma restrict_def_alt: assumes "A⊆X" shows "restrict({⟨x,q(x)⟩. x∈X},A) = {⟨x,q(x)⟩. x∈A}" proof - let ?Y = "{q(x). x∈X}" let ?f = "{⟨x,q(x)⟩. x∈X}" have "∀x∈X. q(x)∈?Y" by blast with assms have "?f:X→?Y" using ZF_fun_from_total by simp with assms have "restrict(?f,A):A→?Y" using restrict_fun by simp moreover from assms have "∀x∈A. q(x)∈?Y" by blast then have "{⟨x,q(x)⟩. x∈A}:A→?Y" using ZF_fun_from_total by simp moreover from assms have "∀x∈A. restrict(?f,A)`(x) = {⟨x,q(x)⟩. x∈A}`(x)" using restrict ZF_fun_from_tot_val1 by auto ultimately show ?thesis by (rule func_eq) qed text‹A composition of two functions is the same as composition with a restriction.› lemma comp_restrict: assumes A1: "f : A→B" and A2: "g : X → C" and A3: "B⊆X" shows "g O f = restrict(g,B) O f" proof - from assms have "g O f : A → C" using comp_fun_subset by simp moreover from assms have "restrict(g,B) O f : A → C" using restrict_fun comp_fun by simp moreover from A1 have "∀x∈A. (g O f)`(x) = (restrict(g,B) O f)`(x)" using comp_fun_apply apply_funtype restrict by simp ultimately show "g O f = restrict(g,B) O f" by (rule func_eq) qed text‹A way to look at restriction. Contributed by Victor Porton.› lemma right_comp_id_any: shows "r O id(C) = restrict(r,C)" unfolding restrict_def by auto subsection‹Constant functions› text‹Constant functions are trivial, but still we need to prove some properties to shorten proofs.› text‹We define constant($=c$) functions on a set $X$ in a natural way as ConstantFunction$(X,c)$.› definition "ConstantFunction(X,c) ≡ X×{c}" text‹Constant function is a function (i.e. belongs to a function space).› lemma func1_3_L1: assumes A1: "c∈Y" shows "ConstantFunction(X,c) : X→Y" proof - from A1 have "X×{c} = {⟨ x,y⟩ ∈ X×Y. c = y}" by auto with A1 show ?thesis using func1_1_L11A ConstantFunction_def by simp qed text‹Constant function is equal to the constant on its domain.› lemma func1_3_L2: assumes A1: "x∈X" shows "ConstantFunction(X,c)`(x) = c" proof - have "ConstantFunction(X,c) ∈ X→{c}" using func1_3_L1 by simp moreover from A1 have "⟨x,c⟩ ∈ ConstantFunction(X,c)" using ConstantFunction_def by simp ultimately show ?thesis using apply_iff by simp qed text‹Another way of looking at the constant function - it's a set of pairs $\langle x,c\rangle$ as $x$ ranges over $X$. › lemma const_fun_def_alt: shows "ConstantFunction(X,c) = {⟨x,c⟩. x∈X}" unfolding ConstantFunction_def by auto text‹If $c\in A$ then the inverse image of $A$ by the constant function $x\mapsto c$ is the whole domain. › lemma const_vimage_domain: assumes "c∈A" shows "ConstantFunction(X,c)-``(A) = X" proof - let ?C = "ConstantFunction(X,c)" have "?C-``(A) = {x∈X. ?C`(x) ∈ A}" using func1_3_L1 func1_1_L15 by blast with assms show ?thesis using func1_3_L2 by simp qed text‹If $c$ is not an element of $A$ then the inverse image of $A$ by the constant function $x\mapsto c$ is empty. › lemma const_vimage_empty: assumes "c∉A" shows "ConstantFunction(X,c)-``(A) = 0" proof - let ?C = "ConstantFunction(X,c)" have "?C-``(A) = {x∈X. ?C`(x) ∈ A}" using func1_3_L1 func1_1_L15 by blast with assms show ?thesis using func1_3_L2 by simp qed subsection‹Injections, surjections, bijections etc.› text‹In this section we prove the properties of the spaces of injections, surjections and bijections that we can't find in the standard Isabelle's ‹Perm.thy›.› text‹For injections the image a difference of two sets is the difference of images› lemma inj_image_dif: assumes A1: "f ∈ inj(A,B)" and A2: "C ⊆ A" shows "f``(A-C) = f``(A) - f``(C)" proof show "f``(A - C) ⊆ f``(A) - f``(C)" proof fix y assume A3: "y ∈ f``(A - C)" from A1 have "f:A→B" using inj_def by simp moreover have "A-C ⊆ A" by auto ultimately have "f``(A-C) = {f`(x). x ∈ A-C}" using func_imagedef by simp with A3 obtain x where I: "f`(x) = y" and "x ∈ A-C" by auto hence "x∈A" by auto with ‹f:A→B› I have "y ∈ f``(A)" using func_imagedef by auto moreover have "y ∉ f``(C)" proof - { assume "y ∈ f``(C)" with A2 ‹f:A→B› obtain x⇩_{0}where II: "f`(x⇩_{0}) = y" and "x⇩_{0}∈ C" using func_imagedef by auto with A1 A2 I ‹x∈A› have "f ∈ inj(A,B)" "f`(x) = f`(x⇩_{0})" "x∈A" "x⇩_{0}∈ A" by auto then have "x = x⇩_{0}" by (rule inj_apply_equality) with ‹x ∈ A-C› ‹x⇩_{0}∈ C› have False by simp } thus ?thesis by auto qed ultimately show "y ∈ f``(A) - f``(C)" by simp qed from A1 A2 show "f``(A) - f``(C) ⊆ f``(A-C)" using inj_def diff_image_diff by auto qed text‹For injections the image of intersection is the intersection of images.› lemma inj_image_inter: assumes A1: "f ∈ inj(X,Y)" and A2: "A⊆X" "B⊆X" shows "f``(A∩B) = f``(A) ∩ f``(B)" proof show "f``(A∩B) ⊆ f``(A) ∩ f``(B)" using image_Int_subset by simp { from A1 have "f:X→Y" using inj_def by simp fix y assume "y ∈ f``(A) ∩ f``(B)" then have "y ∈ f``(A)" and "y ∈ f``(B)" by auto with A2 ‹f:X→Y› obtain x⇩_{A}x⇩_{B}where "x⇩_{A}∈ A" "x⇩_{B}∈ B" and I: "y = f`(x⇩_{A})" "y = f`(x⇩_{B})" using func_imagedef by auto with A2 have "x⇩_{A}∈ X" "x⇩_{B}∈ X" and " f`(x⇩_{A}) = f`(x⇩_{B})" by auto with A1 have "x⇩_{A}= x⇩_{B}" using inj_def by auto with ‹x⇩_{A}∈ A› ‹x⇩_{B}∈ B› have "f`(x⇩_{A}) ∈ {f`(x). x ∈ A∩B}" by auto moreover from A2 ‹f:X→Y› have "f``(A∩B) = {f`(x). x ∈ A∩B}" using func_imagedef by blast ultimately have "f`(x⇩_{A}) ∈ f``(A∩B)" by simp with I have "y ∈ f``(A∩B)" by simp } thus "f``(A) ∩ f``(B) ⊆ f``(A ∩ B)" by auto qed text‹For surjection from $A$ to $B$ the image of the domain is $B$.› lemma surj_range_image_domain: assumes A1: "f ∈ surj(A,B)" shows "f``(A) = B" proof - from A1 have "f``(A) = range(f)" using surj_def range_image_domain by auto with A1 show "f``(A) = B" using surj_range by simp qed text‹For injections the inverse image of an image is the same set.› lemma inj_vimage_image: assumes "f ∈ inj(X,Y)" and "A⊆X" shows "f-``(f``(A)) = A" proof - have "f-``(f``(A)) = (converse(f) O f)``(A)" using vimage_converse image_comp by simp with assms show ?thesis using left_comp_inverse image_id_same by simp qed text‹For surjections the image of an inverse image is the same set.› lemma surj_image_vimage: assumes A1: "f ∈ surj(X,Y)" and A2: "A⊆Y" shows "f``(f-``(A)) = A" proof - have "f``(f-``(A)) = (f O converse(f))``(A)" using vimage_converse image_comp by simp with assms show ?thesis using right_comp_inverse image_id_same by simp qed text‹A lemma about how a surjection maps collections of subsets in domain and range.› lemma surj_subsets: assumes A1: "f ∈ surj(X,Y)" and A2: "B ⊆ Pow(Y)" shows "{ f``(U). U ∈ {f-``(V). V∈B} } = B" proof { fix W assume "W ∈ { f``(U). U ∈ {f-``(V). V∈B} }" then obtain U where I: "U ∈ {f-``(V). V∈B}" and II: "W = f``(U)" by auto then obtain V where "V∈B" and "U = f-``(V)" by auto with II have "W = f``(f-``(V))" by simp moreover from assms ‹V∈B› have "f ∈ surj(X,Y)" and "V⊆Y" by auto ultimately have "W=V" using surj_image_vimage by simp with ‹V∈B› have "W ∈ B" by simp } thus "{ f``(U). U ∈ {f-``(V). V∈B} } ⊆ B" by auto { fix W assume "W∈B" let ?U = "f-``(W)" from ‹W∈B› have "?U ∈ {f-``(V). V∈B}" by auto moreover from A1 A2 ‹W∈B› have "W = f``(?U)" using surj_image_vimage by auto ultimately have "W ∈ { f``(U). U ∈ {f-``(V). V∈B} }" by auto } thus "B ⊆ { f``(U). U ∈ {f-``(V). V∈B} }" by auto qed text‹Restriction of an bijection to a set without a point is a a bijection.› lemma bij_restrict_rem: assumes A1: "f ∈ bij(A,B)" and A2: "a∈A" shows "restrict(f, A-{a}) ∈ bij(A-{a}, B-{f`(a)})" proof - let ?C = "A-{a}" from A1 have "f ∈ inj(A,B)" "?C ⊆ A" using bij_def by auto then have "restrict(f,?C) ∈ bij(?C, f``(?C))" using restrict_bij by simp moreover have "f``(?C) = B-{f`(a)}" proof - from A2 ‹f ∈ inj(A,B)› have "f``(?C) = f``(A) - f``{a}" using inj_image_dif by simp moreover from A1 have "f``(A) = B" using bij_def surj_range_image_domain by auto moreover from A1 A2 have "f``{a} = {f`(a)}" using bij_is_fun singleton_image by blast ultimately show "f``(?C) = B-{f`(a)}" by simp qed ultimately show ?thesis by simp qed text‹The domain of a bijection between $X$ and $Y$ is $X$.› lemma domain_of_bij: assumes A1: "f ∈ bij(X,Y)" shows "domain(f) = X" proof - from A1 have "f:X→Y" using bij_is_fun by simp then show "domain(f) = X" using func1_1_L1 by simp qed text‹The value of the inverse of an injection on a point of the image of a set belongs to that set.› lemma inj_inv_back_in_set: assumes A1: "f ∈ inj(A,B)" and A2: "C⊆A" and A3: "y ∈ f``(C)" shows "converse(f)`(y) ∈ C" "f`(converse(f)`(y)) = y" proof - from A1 have I: "f:A→B" using inj_is_fun by simp with A2 A3 obtain x where II: "x∈C" "y = f`(x)" using func_imagedef by auto with A1 A2 show "converse(f)`(y) ∈ C" using left_inverse by auto from A1 A2 I II show "f`(converse(f)`(y)) = y" using func1_1_L5A right_inverse by auto qed text‹For injections if a value at a point belongs to the image of a set, then the point belongs to the set.› lemma inj_point_of_image: assumes A1: "f ∈ inj(A,B)" and A2: "C⊆A" and A3: "x∈A" and A4: "f`(x) ∈ f``(C)" shows "x ∈ C" proof - from A1 A2 A4 have "converse(f)`(f`(x)) ∈ C" using inj_inv_back_in_set by simp moreover from A1 A3 have "converse(f)`(f`(x)) = x" using left_inverse_eq by simp ultimately show "x ∈ C" by simp qed text‹For injections the image of intersection is the intersection of images.› lemma inj_image_of_Inter: assumes A1: "f ∈ inj(A,B)" and A2: "I≠0" and A3: "∀i∈I. P(i) ⊆ A" shows "f``(⋂i∈I. P(i)) = ( ⋂i∈I. f``(P(i)) )" proof from A1 A2 A3 show "f``(⋂i∈I. P(i)) ⊆ ( ⋂i∈I. f``(P(i)) )" using inj_is_fun image_of_Inter by auto from A1 A2 A3 have "f:A→B" and "( ⋂i∈I. P(i) ) ⊆ A" using inj_is_fun ZF1_1_L7 by auto then have I: "f``(⋂i∈I. P(i)) = { f`(x). x ∈ ( ⋂i∈I. P(i) ) }" using func_imagedef by simp { fix y assume A4: "y ∈ ( ⋂i∈I. f``(P(i)) )" let ?x = "converse(f)`(y)" from A2 obtain i⇩_{0}where "i⇩_{0}∈ I" by auto with A1 A4 have II: "y ∈ range(f)" using inj_is_fun func1_1_L6 by auto with A1 have III: "f`(?x) = y" using right_inverse by simp from A1 II have IV: "?x ∈ A" using inj_converse_fun apply_funtype by blast { fix i assume "i∈I" with A3 A4 III have "P(i) ⊆ A" and "f`(?x) ∈ f``(P(i))" by auto with A1 IV have "?x ∈ P(i)" using inj_point_of_image by blast } then have "∀i∈I. ?x ∈ P(i)" by simp with A2 I have "f`(?x) ∈ f``( ⋂i∈I. P(i) )" by auto with III have "y ∈ f``( ⋂i∈I. P(i) )" by simp } then show "( ⋂i∈I. f``(P(i)) ) ⊆ f``( ⋂i∈I. P(i) )" by auto qed text‹An injection is injective onto its range. Suggested by Victor Porton.› lemma inj_inj_range: assumes "f ∈ inj(A,B)" shows "f ∈ inj(A,range(f))" using assms inj_def range_of_fun by auto text‹An injection is a bijection on its range. Suggested by Victor Porton.› lemma inj_bij_range: assumes "f ∈ inj(A,B)" shows "f ∈ bij(A,range(f))" proof - from assms have "f ∈ surj(A,range(f))" using inj_def fun_is_surj by auto with assms show ?thesis using inj_inj_range bij_def by simp qed text‹A lemma about extending a surjection by one point.› lemma surj_extend_point: assumes A1: "f ∈ surj(X,Y)" and A2: "a∉X" and A3: "g = f ∪ {⟨a,b⟩}" shows "g ∈ surj(X∪{a},Y∪{b})" proof - from A1 A2 A3 have "g : X∪{a} → Y∪{b}" using surj_def func1_1_L11D by simp moreover have "∀y ∈ Y∪{b}. ∃x ∈ X∪{a}. y = g`(x)" proof fix y assume "y ∈ Y ∪ {b}" then have "y ∈ Y ∨ y = b" by auto moreover { assume "y ∈ Y" with A1 obtain x where "x∈X" and "y = f`(x)" using surj_def by auto with A1 A2 A3 have "x ∈ X∪{a}" and "y = g`(x)" using surj_def func1_1_L11D by auto then have "∃x ∈ X∪{a}. y = g`(x)" by auto } moreover { assume "y = b" with A1 A2 A3 have "y = g`(a)" using surj_def func1_1_L11D by auto then have "∃x ∈ X∪{a}. y = g`(x)" by auto } ultimately show "∃x ∈ X∪{a}. y = g`(x)" by auto qed ultimately show "g ∈ surj(X∪{a},Y∪{b})" using surj_def by auto qed text‹A lemma about extending an injection by one point. Essentially the same as standard Isabelle's ‹inj_extend›. › lemma inj_extend_point: assumes "f ∈ inj(X,Y)" "a∉X" "b∉Y" shows "(f ∪ {⟨a,b⟩}) ∈ inj(X∪{a},Y∪{b})" proof - from assms have "cons(⟨a,b⟩,f) ∈ inj(cons(a, X), cons(b, Y))" using assms inj_extend by simp moreover have "cons(⟨a,b⟩,f) = f ∪ {⟨a,b⟩}" and "cons(a, X) = X∪{a}" and "cons(b, Y) = Y∪{b}" by auto ultimately show ?thesis by simp qed text‹A lemma about extending a bijection by one point.› lemma bij_extend_point: assumes "f ∈ bij(X,Y)" "a∉X" "b∉Y" shows "(f ∪ {⟨a,b⟩}) ∈ bij(X∪{a},Y∪{b})" using assms surj_extend_point inj_extend_point bij_def by simp text‹A quite general form of the $a^{-1}b = 1$ implies $a=b$ law.› lemma comp_inv_id_eq: assumes A1: "converse(b) O a = id(A)" and A2: "a ⊆ A×B" "b ∈ surj(A,B)" shows "a = b" proof - from A1 have "(b O converse(b)) O a = b O id(A)" using comp_assoc by simp with A2 have "id(B) O a = b O id(A)" using right_comp_inverse by simp moreover from A2 have "a ⊆ A×B" and "b ⊆ A×B" using surj_def fun_subset_prod by auto then have "id(B) O a = a" and "b O id(A) = b" using left_comp_id right_comp_id by auto ultimately show "a = b" by simp qed text‹A special case of ‹comp_inv_id_eq› - the $a^{-1}b = 1$ implies $a=b$ law for bijections.› lemma comp_inv_id_eq_bij: assumes A1: "a ∈ bij(A,B)" "b ∈ bij(A,B)" and A2: "converse(b) O a = id(A)" shows "a = b" proof - from A1 have "a ⊆ A×B" and "b ∈ surj(A,B)" using bij_def surj_def fun_subset_prod by auto with A2 show "a = b" by (rule comp_inv_id_eq) qed text‹Converse of a converse of a bijection is the same bijection. This is a special case of ‹converse_converse› from standard Isabelle's ‹equalities› theory where it is proved for relations.› lemma bij_converse_converse: assumes "a ∈ bij(A,B)" shows "converse(converse(a)) = a" proof - from assms have "a ⊆ A×B" using bij_def surj_def fun_subset_prod by simp then show ?thesis using converse_converse by simp qed text‹If a composition of bijections is identity, then one is the inverse of the other.› lemma comp_id_conv: assumes A1: "a ∈ bij(A,B)" "b ∈ bij(B,A)" and A2: "b O a = id(A)" shows "a = converse(b)" and "b = converse(a)" proof - from A1 have "a ∈ bij(A,B)" and "converse(b) ∈ bij(A,B)" using bij_converse_bij by auto moreover from assms have "converse(converse(b)) O a = id(A)" using bij_converse_converse by simp ultimately show "a = converse(b)" by (rule comp_inv_id_eq_bij) with assms show "b = converse(a)" using bij_converse_converse by simp qed text‹A version of ‹comp_id_conv› with weaker assumptions.› lemma comp_conv_id: assumes A1: "a ∈ bij(A,B)" and A2: "b:B→A" and A3: "∀x∈A. b`(a`(x)) = x" shows "b ∈ bij(B,A)" and "a = converse(b)" and "b = converse(a)" proof - have "b ∈ surj(B,A)" proof - have "∀x∈A. ∃y∈B. b`(y) = x" proof - { fix x assume "x∈A" let ?y = "a`(x)" from A1 A3 ‹x∈A› have "?y∈B" and "b`(?y) = x" using bij_def inj_def apply_funtype by auto hence "∃y∈B. b`(y) = x" by auto } thus ?thesis by simp qed with A2 show "b ∈ surj(B,A)" using surj_def by simp qed moreover have "b ∈ inj(B,A)" proof - have "∀w∈B.∀y∈B. b`(w) = b`(y) ⟶ w=y" proof - { fix w y assume "w∈B" "y∈B" and I: "b`(w) = b`(y)" from A1 have "a ∈ surj(A,B)" unfolding bij_def by simp with ‹w∈B› obtain x⇩_{w}where "x⇩_{w}∈ A" and II: "a`(x⇩_{w}) = w" using surj_def by auto with I have "b`(a`(x⇩_{w})) = b`(y)" by simp moreover from ‹a ∈ surj(A,B)› ‹y∈B› obtain x⇩_{y}where "x⇩_{y}∈ A" and III: "a`(x⇩_{y}) = y" using surj_def by auto moreover from A3 ‹x⇩_{w}∈ A› ‹x⇩_{y}∈ A› have "b`(a`(x⇩_{w})) = x⇩_{w}" and "b`(a`(x⇩_{y})) = x⇩_{y}" by auto ultimately have "x⇩_{w}= x⇩_{y}" by simp with II III have "w=y" by simp } thus ?thesis by auto qed with A2 show "b ∈ inj(B,A)" using inj_def by auto qed ultimately show "b ∈ bij(B,A)" using bij_def by simp from assms have "b O a = id(A)" using bij_def inj_def comp_eq_id_iff1 by auto with A1 ‹b ∈ bij(B,A)› show "a = converse(b)" and "b = converse(a)" using comp_id_conv by auto qed text‹For a surjection the union if images of singletons is the whole range.› lemma surj_singleton_image: assumes A1: "f ∈ surj(X,Y)" shows "(⋃x∈X. {f`(x)}) = Y" proof from A1 show "(⋃x∈X. {f`(x)}) ⊆ Y" using surj_def apply_funtype by auto next { fix y assume "y ∈ Y" with A1 have "y ∈ (⋃x∈X. {f`(x)})" using surj_def by auto } then show "Y ⊆ (⋃x∈X. {f`(x)})" by auto qed subsection‹Functions of two variables› text‹In this section we consider functions whose domain is a cartesian product of two sets. Such functions are called functions of two variables (although really in ZF all functions admit only one argument). For every function of two variables we can define families of functions of one variable by fixing the other variable. This section establishes basic definitions and results for this concept.› text‹We can create functions of two variables by combining functions of one variable.› lemma cart_prod_fun: assumes "f⇩_{1}:X⇩_{1}→Y⇩_{1}" "f⇩_{2}:X⇩_{2}→Y⇩_{2}" and "g = {⟨p,⟨f⇩_{1}`(fst(p)),f⇩_{2}`(snd(p))⟩⟩. p ∈ X⇩_{1}×X⇩_{2}}" shows "g: X⇩_{1}×X⇩_{2}→ Y⇩_{1}×Y⇩_{2}" using assms apply_funtype ZF_fun_from_total by simp text‹A reformulation of ‹cart_prod_fun› above in a sligtly different notation.› lemma prod_fun: assumes "f:X⇩_{1}→X⇩_{2}" "g:X⇩_{3}→X⇩_{4}" shows "{⟨⟨x,y⟩,⟨f`x,g`y⟩⟩. ⟨x,y⟩∈X⇩_{1}×X⇩_{3}}:X⇩_{1}×X⇩_{3}→X⇩_{2}×X⇩_{4}" proof - have "{⟨⟨x,y⟩,⟨f`x,g`y⟩⟩. ⟨x,y⟩∈X⇩_{1}×X⇩_{3}} = {⟨p,⟨f`(fst(p)),g`(snd(p))⟩⟩. p ∈ X⇩_{1}×X⇩_{3}}" by auto with assms show ?thesis using cart_prod_fun by simp qed text‹Product of two surjections is a surjection.› theorem prod_functions_surj: assumes "f∈surj(A,B)" "g∈surj(C,D)" shows "{⟨⟨a1,a2⟩,⟨f`a1,g`a2⟩⟩.⟨a1,a2⟩∈A×C} ∈ surj(A×C,B×D)" proof - let ?h = "{⟨⟨x, y⟩, f`(x), g`(y)⟩ . ⟨x,y⟩ ∈ A × C}" from assms have fun: "f:A→B""g:C→D" unfolding surj_def by auto then have pfun: "?h : A × C → B × D" using prod_fun by auto { fix b assume "b∈B×D" then obtain b1 b2 where "b=⟨b1,b2⟩" "b1∈B" "b2∈D" by auto with assms obtain a1 a2 where "f`(a1)=b1" "g`(a2)=b2" "a1∈A" "a2∈C" unfolding surj_def by blast hence "⟨⟨a1,a2⟩,⟨b1,b2⟩⟩ ∈ ?h" by auto with pfun have "?h`⟨a1,a2⟩=⟨b1,b2⟩" using apply_equality by auto with ‹b=⟨b1,b2⟩› ‹a1∈A› ‹a2∈C› have "∃a∈A×C. ?h`(a)=b" by auto } hence "∀b∈B×D. ∃a∈A×C. ?h`(a) = b" by auto with pfun show ?thesis unfolding surj_def by auto qed text‹For a function of two variables created from functions of one variable as in ‹cart_prod_fun› above, the inverse image of a cartesian product of sets is the cartesian product of inverse images.› lemma cart_prod_fun_vimage: assumes "f⇩_{1}:X⇩_{1}→Y⇩_{1}" "f⇩_{2}:X⇩_{2}→Y⇩_{2}" and "g = {⟨p,⟨f⇩_{1}`(fst(p)),f⇩_{2}`(snd(p))⟩⟩. p ∈ X⇩_{1}×X⇩_{2}}" shows "g-``(A⇩_{1}×A⇩_{2}) = f⇩_{1}-``(A⇩_{1}) × f⇩_{2}-``(A⇩_{2})" proof - from assms have "g: X⇩_{1}×X⇩_{2}→ Y⇩_{1}×Y⇩_{2}" using cart_prod_fun by simp then have "g-``(A⇩_{1}×A⇩_{2}) = {p ∈ X⇩_{1}×X⇩_{2}. g`(p) ∈ A⇩_{1}×A⇩_{2}}" using func1_1_L15 by simp with assms ‹g: X⇩_{1}×X⇩_{2}→ Y⇩_{1}×Y⇩_{2}› show "g-``(A⇩_{1}×A⇩_{2}) = f⇩_{1}-``(A⇩_{1}) × f⇩_{2}-``(A⇩_{2})" using ZF_fun_from_tot_val func1_1_L15 by auto qed text‹For a function of two variables defined on $X\times Y$, if we fix an $x\in X$ we obtain a function on $Y$. Note that if ‹domain(f)› is $X\times Y$, ‹range(domain(f))› extracts $Y$ from $X\times Y$.› definition "Fix1stVar(f,x) ≡ {⟨y,f`⟨x,y⟩⟩. y ∈ range(domain(f))}" text‹For every $y\in Y$ we can fix the second variable in a binary function $f: X\times Y \rightarrow Z$ to get a function on $X$.› definition "Fix2ndVar(f,y) ≡ {⟨x,f`⟨x,y⟩⟩. x ∈ domain(domain(f))}" text‹We defined ‹Fix1stVar› and ‹Fix2ndVar› so that the domain of the function is not listed in the arguments, but is recovered from the function. The next lemma is a technical fact that makes it easier to use this definition.› lemma fix_var_fun_domain: assumes A1: "f : X×Y → Z" shows "x∈X ⟶ Fix1stVar(f,x) = {⟨y,f`⟨x,y⟩⟩. y ∈ Y}" "y∈Y ⟶ Fix2ndVar(f,y) = {⟨x,f`⟨x,y⟩⟩. x ∈ X}" proof - from A1 have I: "domain(f) = X×Y" using func1_1_L1 by simp { assume "x∈X" with I have "range(domain(f)) = Y" by auto then have "Fix1stVar(f,x) = {⟨y,f`⟨x,y⟩⟩. y ∈ Y}" using Fix1stVar_def by simp } then show "x∈X ⟶ Fix1stVar(f,x) = {⟨y,f`⟨x,y⟩⟩. y ∈ Y}" by simp { assume "y∈Y" with I have "domain(domain(f)) = X" by auto then have "Fix2ndVar(f,y) = {⟨x,f`⟨x,y⟩⟩. x ∈ X}" using Fix2ndVar_def by simp } then show "y∈Y ⟶ Fix2ndVar(f,y) = {⟨x,f`⟨x,y⟩⟩. x ∈ X}" by simp qed text‹If we fix the first variable, we get a function of the second variable.› lemma fix_1st_var_fun: assumes A1: "f : X×Y → Z" and A2: "x∈X" shows "Fix1stVar(f,x) : Y → Z" proof - from A1 A2 have "∀y∈Y. f`⟨x,y⟩ ∈ Z" using apply_funtype by simp then have "{⟨y,f`⟨x,y⟩⟩. y ∈ Y} : Y → Z" using ZF_fun_from_total by simp with A1 A2 show "Fix1stVar(f,x) : Y → Z" using fix_var_fun_domain by simp qed text‹If we fix the second variable, we get a function of the first variable.› lemma fix_2nd_var_fun: assumes A1: "f : X×Y → Z" and A2: "y∈Y" shows "Fix2ndVar(f,y) : X → Z" proof - from A1 A2 have "∀x∈X. f`⟨x,y⟩ ∈ Z" using apply_funtype by simp then have "{⟨x,f`⟨x,y⟩⟩. x ∈ X} : X → Z" using ZF_fun_from_total by simp with A1 A2 show "Fix2ndVar(f,y) : X → Z" using fix_var_fun_domain by simp qed text‹What is the value of ‹Fix1stVar(f,x)› at $y\in Y$ and the value of ‹Fix2ndVar(f,y)› at $x\in X$"?› lemma fix_var_val: assumes A1: "f : X×Y → Z" and A2: "x∈X" "y∈Y" shows "Fix1stVar(f,x