# Theory FiniteSeq_ZF

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theory FiniteSeq_ZF
imports Nat_ZF_IML func1
(*    This file is a part of IsarMathLib -     a library of formalized mathematics for Isabelle/Isar.    Copyright (C) 2007  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.*)header{*\isaheader{FiniteSeq\_ZF.thy}*}theory FiniteSeq_ZF imports Nat_ZF_IML func1begintext{* This theory treats finite sequences (i.e. maps $n\rightarrow X$, where  $n=\{0,1,..,n-1\}$ is a natural number) as lists. It defines and proves  the properties of basic operations on lists: concatenation, appending  and element etc.*}section{*Lists as finite sequences*}text{*A natural way of representing (finite) lists in set theory is through   (finite) sequences.  In such view a list of elements of a set $X$ is a   function that maps the set $\{0,1,..n-1\}$ into $X$. Since natural numbers   in set theory are defined so that $n =\{0,1,..n-1\}$, a list of length $n$  can be understood as an element of the function space $n\rightarrow X$.  *}text{*We define the set of lists with values in set $X$ as @{text "Lists(X)"}.*}definition  "Lists(X) ≡ \<Union>n∈nat.(n->X)"text{*The set of nonempty $X$-value listst will be called @{text "NELists(X)"}.*}definition  "NELists(X) ≡ \<Union>n∈nat.(succ(n)->X)"text{*We first define the shift that moves the second sequence  to the domain $\{n,..,n+k-1\}$, where $n,k$ are the lengths of the first   and the second sequence, resp.    To understand the notation in the definitions below recall that in Isabelle/ZF   @{text "pred(n)"} is the previous natural number and     denotes the difference between natural numbers $n$ and $k$.*}definition  "ShiftedSeq(b,n) ≡ {⟨j, b(j #- n)⟩. j ∈ NatInterval(n,domain(b))}"text{*We define concatenation of two sequences as the union of the first sequence   with the shifted second sequence. The result of concatenating lists   $a$ and $b$ is called @{text "Concat(a,b)"}. *}definition  "Concat(a,b) ≡ a ∪ ShiftedSeq(b,domain(a))"text{* For a finite sequence we define the sequence of all elements   except the first one. This corresponds to the "tail" function in Haskell.  We call it @{text "Tail"} here as well.*}definition    "Tail(a) ≡ {⟨k, a(succ(k))⟩. k ∈ pred(domain(a))}"text{*A dual notion to @{text "Tail"} is the list  of all elements of a list except the last one. Borrowing  the terminology from Haskell again, we will call this @{text "Init"}.*}definition  "Init(a) ≡ restrict(a,pred(domain(a)))"text{* Another obvious operation we can talk about is appending an element  at the end of a sequence. This is called @{text "Append"}.*}definition  "Append(a,x) ≡ a ∪ {⟨domain(a),x⟩}"text{*If lists are modeled as finite sequences (i.e. functions on natural   intervals $\{0,1,..,n-1\} = n$) it is easy to get the first element  of a list as the value of the sequence at $0$. The last element is the  value at $n-1$. To hide this behind a familiar name we define the @{text "Last"}  element of a list. *} definition  "Last(a) ≡ a(pred(domain(a)))"text{*Shifted sequence is a function on a the interval of natural numbers.*}lemma shifted_seq_props:   assumes A1: "n ∈ nat"  "k ∈ nat" and A2: "b:k->X"  shows   "ShiftedSeq(b,n): NatInterval(n,k) -> X"  "∀i ∈ NatInterval(n,k). ShiftedSeq(b,n)(i) = b(i #- n)"  "∀j∈k. ShiftedSeq(b,n)(n #+ j) = b(j)"   proof -  let ?I = "NatInterval(n,domain(b))"  from A2 have Fact: "?I = NatInterval(n,k)" using func1_1_L1 by simp  with A1 A2 have "∀j∈ ?I. b(j #- n) ∈ X"     using inter_diff_in_len apply_funtype by simp  then have     "{⟨j, b(j #- n)⟩. j ∈ ?I} : ?I -> X" by (rule ZF_fun_from_total)  with Fact show thesis_1: "ShiftedSeq(b,n): NatInterval(n,k) -> X"    using ShiftedSeq_def by simp  { fix i     from Fact thesis_1 have  "ShiftedSeq(b,n): ?I -> X" by simp    moreover     assume "i ∈ NatInterval(n,k)"    with Fact have "i ∈ ?I" by simp    moreover from Fact have       "ShiftedSeq(b,n) = {⟨i, b(i #- n)⟩. i ∈ ?I}"      using ShiftedSeq_def by simp    ultimately have "ShiftedSeq(b,n)(i) =  b(i #- n)"      by (rule ZF_fun_from_tot_val)  } then show thesis1:       "∀i ∈ NatInterval(n,k). ShiftedSeq(b,n)(i) = b(i #- n)"    by simp  { fix j     let ?i = "n #+ j"    assume A3: "j∈k"    with A1 have "j ∈ nat" using elem_nat_is_nat by blast    then have "?i #- n = j" using diff_add_inverse by simp    with A3 thesis1 have "ShiftedSeq(b,n)(?i) = b(j)"      using NatInterval_def by auto  } then show "∀j∈k. ShiftedSeq(b,n)(n #+ j) = b(j)"    by simpqedtext{*Basis properties of the contatenation of two finite sequences.*}theorem concat_props:  assumes A1: "n ∈ nat"  "k ∈ nat" and A2: "a:n->X"   "b:k->X"  shows  "Concat(a,b): n #+ k -> X"  "∀i∈n. Concat(a,b)(i) = a(i)"  "∀i ∈ NatInterval(n,k). Concat(a,b)(i) =  b(i #- n)"  "∀j ∈ k. Concat(a,b)(n #+ j) = b(j)"proof -  from A1 A2 have    "a:n->X"  and I: "ShiftedSeq(b,n): NatInterval(n,k) -> X"    and "n ∩ NatInterval(n,k) = 0"    using shifted_seq_props length_start_decomp by auto  then have     "a ∪ ShiftedSeq(b,n): n ∪ NatInterval(n,k) -> X ∪ X"    by (rule fun_disjoint_Un)  with A1 A2 show "Concat(a,b): n #+ k -> X"    using func1_1_L1 Concat_def length_start_decomp by auto  { fix i assume "i ∈ n"    with A1 I have "i ∉ domain(ShiftedSeq(b,n))"      using length_start_decomp func1_1_L1 by auto    with A2 have "Concat(a,b)(i) = a(i)"      using func1_1_L1 fun_disjoint_apply1 Concat_def by simp  } thus "∀i∈n. Concat(a,b)(i) = a(i)" by simp  { fix i assume A3: "i ∈ NatInterval(n,k)"    with A1 A2 have "i ∉ domain(a)"       using length_start_decomp func1_1_L1 by auto    with A1 A2 A3 have "Concat(a,b)(i) =  b(i #- n)"      using func1_1_L1 fun_disjoint_apply2 Concat_def shifted_seq_props      by simp  } thus II: "∀i ∈ NatInterval(n,k). Concat(a,b)(i) =  b(i #- n)"    by simp  { fix j    let ?i = "n #+ j"    assume A3: "j∈k"    with A1 have "j ∈ nat" using elem_nat_is_nat by blast    then have "?i #- n = j" using diff_add_inverse by simp     with A3 II have "Concat(a,b)(?i) = b(j)"      using NatInterval_def by auto  } thus "∀j ∈ k. Concat(a,b)(n #+ j) = b(j)"    by simpqedtext{*Properties of concatenating three lists.*}lemma concat_concat_list:   assumes A1: "n ∈ nat"  "k ∈ nat"  "m ∈ nat" and  A2: "a:n->X"   "b:k->X"  "c:m->X" and  A3: "d = Concat(Concat(a,b),c)"  shows  "d : n #+k #+ m -> X"  "∀j ∈ n. d(j) = a(j)"  "∀j ∈ k. d(n #+ j) = b(j)"  "∀j ∈ m. d(n #+ k #+ j) = c(j)"proof -  from A1 A2 have I:    "n #+ k ∈ nat"   "m ∈ nat"    "Concat(a,b): n #+ k -> X"   "c:m->X"    using concat_props by auto  with A3 show "d: n #+k #+ m -> X"    using concat_props by simp  from I have II: "∀i ∈ n #+ k.     Concat(Concat(a,b),c)(i) = Concat(a,b)(i)"    by (rule concat_props)  { fix j assume A4: "j ∈ n"    moreover from A1 have "n ⊆ n #+ k" using add_nat_le by simp    ultimately have "j ∈ n #+ k" by auto    with A3 II have "d(j) =  Concat(a,b)(j)" by simp    with A1 A2 A4 have "d(j) = a(j)"      using concat_props by simp  } thus "∀j ∈ n. d(j) = a(j)" by simp  { fix j assume A5: "j ∈ k"    with A1 A3 II have "d(n #+ j) = Concat(a,b)(n #+ j)"      using add_lt_mono by simp    also from A1 A2 A5 have "… = b(j)"      using concat_props by simp    finally have "d(n #+ j) = b(j)" by simp  } thus "∀j ∈ k. d(n #+ j) = b(j)" by simp  from I have "∀j ∈ m. Concat(Concat(a,b),c)(n #+ k #+ j) = c(j)"    by (rule concat_props)  with A3 show "∀j ∈ m. d(n #+ k #+ j) = c(j)"    by simpqedtext{*Properties of concatenating a list with a concatenation  of two other lists.*}lemma concat_list_concat:   assumes A1: "n ∈ nat"  "k ∈ nat"  "m ∈ nat" and  A2: "a:n->X"   "b:k->X"  "c:m->X" and  A3: "e = Concat(a, Concat(b,c))"  shows   "e : n #+k #+ m -> X"  "∀j ∈ n. e(j) = a(j)"  "∀j ∈ k. e(n #+ j) = b(j)"  "∀j ∈ m. e(n #+ k #+ j) = c(j)"proof -  from A1 A2 have I:     "n ∈ nat"  "k #+ m ∈ nat"    "a:n->X"  "Concat(b,c): k #+ m -> X"    using concat_props by auto  with A3 show  "e : n #+k #+ m -> X"    using concat_props add_assoc by simp  from I have "∀j ∈ n. Concat(a, Concat(b,c))(j) = a(j)"    by (rule concat_props)  with A3 show "∀j ∈ n. e(j) = a(j)" by simp  from I have II:    "∀j ∈ k #+ m. Concat(a, Concat(b,c))(n #+ j) = Concat(b,c)(j)"    by (rule concat_props)  { fix j assume A4: "j ∈ k"    moreover from A1 have "k ⊆ k #+ m" using add_nat_le by simp    ultimately have "j ∈ k #+ m" by auto    with A3 II have "e(n #+ j) =  Concat(b,c)(j)" by simp    also from A1 A2 A4 have "… = b(j)"      using concat_props by simp    finally have "e(n #+ j) = b(j)" by simp  } thus "∀j ∈ k. e(n #+ j) = b(j)" by simp  { fix j assume A5: "j ∈ m"    with A1 II A3 have "e(n #+ k #+ j) = Concat(b,c)(k #+ j)"      using add_lt_mono add_assoc by simp    also from A1 A2 A5 have "… = c(j)"      using concat_props by simp    finally have "e(n #+ k #+ j) = c(j)" by simp  } then show "∀j ∈ m. e(n #+ k #+ j) = c(j)"    by simpqedtext{*Concatenation is associative.*}theorem concat_assoc:   assumes A1: "n ∈ nat"  "k ∈ nat"  "m ∈ nat" and  A2: "a:n->X"   "b:k->X"   "c:m->X"  shows "Concat(Concat(a,b),c) =  Concat(a, Concat(b,c))"proof -  let ?d = "Concat(Concat(a,b),c)"  let ?e = "Concat(a, Concat(b,c))"  from A1 A2 have    "?d : n #+k #+ m -> X" and "?e : n #+k #+ m -> X"    using concat_concat_list concat_list_concat by auto  moreover have "∀i ∈  n #+k #+ m. ?d(i) = ?e(i)"  proof -    { fix i assume "i ∈ n #+k #+ m"      moreover from A1 have 	"n #+k #+ m = n ∪ NatInterval(n,k) ∪ NatInterval(n #+ k,m)"	using adjacent_intervals3 by simp      ultimately have 	"i ∈ n ∨ i ∈ NatInterval(n,k) ∨ i ∈ NatInterval(n #+ k,m)"	by simp      moreover      { assume "i ∈ n"	with A1 A2 have "?d(i) = ?e(i)"	using concat_concat_list concat_list_concat by simp }      moreover      { assume "i ∈ NatInterval(n,k)"	then obtain j where "j∈k" and "i = n #+ j"	  using NatInterval_def by auto	with A1 A2 have "?d(i) = ?e(i)"	  using concat_concat_list concat_list_concat by simp }      moreover      { assume "i ∈ NatInterval(n #+ k,m)"	then obtain j where "j ∈ m" and "i = n #+ k #+ j"	  using NatInterval_def by auto	with A1 A2 have "?d(i) = ?e(i)"	  using concat_concat_list concat_list_concat by simp }      ultimately have "?d(i) = ?e(i)" by auto    } thus ?thesis by simp  qed  ultimately show "?d = ?e" by (rule func_eq)qed    text{*Properties of @{text "Tail"}.*}theorem tail_props:   assumes A1: "n ∈ nat" and A2: "a: succ(n) -> X"  shows  "Tail(a) : n -> X"  "∀k ∈ n. Tail(a)(k) = a(succ(k))"proof -  from A1 A2 have "∀k ∈ n. a(succ(k)) ∈ X"    using succ_ineq apply_funtype by simp  then have "{⟨k, a(succ(k))⟩. k ∈ n} : n -> X"    by (rule ZF_fun_from_total)  with A2 show I: "Tail(a) : n -> X"    using func1_1_L1 pred_succ_eq Tail_def by simp  moreover from A2 have "Tail(a) = {⟨k, a(succ(k))⟩. k ∈ n}"    using func1_1_L1 pred_succ_eq Tail_def by simp  ultimately show "∀k ∈ n. Tail(a)(k) = a(succ(k))"    by (rule ZF_fun_from_tot_val0)qed  text{*Properties of @{text "Append"}. It is a bit surprising that  the we don't need to assume that $n$ is a natural number.*}theorem append_props:  assumes A1: "a: n -> X" and A2: "x∈X" and A3: "b = Append(a,x)"  shows   "b : succ(n) -> X"  "∀k∈n. b(k) = a(k)"  "b(n) = x"proof -  note A1  moreover have I: "n ∉ n" using mem_not_refl by simp  moreover from A1 A3 have II: "b = a ∪ {⟨n,x⟩}"    using func1_1_L1 Append_def by simp  ultimately have "b : n ∪ {n} -> X ∪ {x}"    by (rule func1_1_L11D)  with A2 show "b : succ(n) -> X"    using succ_explained set_elem_add by simp  from A1 I II show "∀k∈n. b(k) = a(k)" and "b(n) = x"    using func1_1_L11D by autoqedtext{*A special case of @{text "append_props"}: appending to a nonempty  list does not change the head (first element) of the list.*}corollary head_of_append:   assumes "n∈ nat" and "a: succ(n) -> X" and "x∈X"  shows "Append(a,x)(0) = a(0)"  using assms append_props empty_in_every_succ by auto(*text{*A bit technical special case of @{text "append_props"} that tells us  what is the value of the appended list at the sucessor of some argument.*}corollary append_val_succ:  assumes "n ∈ nat" and "a: succ(n) -> X" and "x∈X" and "k ∈ n"  shows "Append(a,x)(succ(k)) = a(succ(k))"  using assms succ_ineq append_props by simp*)text{* @{text "Tail"} commutes with @{text "Append"}.*}theorem tail_append_commute:   assumes A1: "n ∈ nat" and A2: "a: succ(n) -> X" and A3: "x∈X"  shows "Append(Tail(a),x) = Tail(Append(a,x))"proof -  let ?b = "Append(Tail(a),x)"  let ?c = "Tail(Append(a,x))"  from A1 A2 have I: "Tail(a) : n -> X" using tail_props    by simp  from A1 A2 A3 have     "succ(n) ∈ nat" and "Append(a,x) : succ(succ(n)) -> X"    using append_props by auto  then have II: "∀k ∈ succ(n). ?c(k) = Append(a,x)(succ(k))"    by (rule tail_props)  from assms have     "?b : succ(n) -> X" and "?c : succ(n) -> X"    using tail_props append_props by auto  moreover have "∀k ∈ succ(n). ?b(k) = ?c(k)"  proof -    { fix k assume "k ∈ succ(n)"      hence "k ∈ n ∨ k = n" by auto      moreover      { assume A4: "k ∈ n"	with assms II have "?c(k) = a(succ(k))"	  using succ_ineq append_props by simp	moreover	from A3 I have "∀k∈n. ?b(k) = Tail(a)(k)"	  using append_props by simp	with A1 A2 A4 have "?b(k) =  a(succ(k))"	  using tail_props by simp	ultimately have "?b(k) = ?c(k)" by simp }      moreover      { assume A5: "k = n"	with A2 A3 I II have "?b(k) = ?c(k)"	  using append_props by auto }      ultimately have "?b(k) = ?c(k)" by auto    } thus ?thesis by simp  qed  ultimately show "?b = ?c" by (rule func_eq)qed  text{*Properties of @{text "Init"}.*}theorem init_props:   assumes A1: "n ∈ nat" and A2: "a: succ(n) -> X"  shows   "Init(a) : n -> X"  "∀k∈n. Init(a)(k) = a(k)"  "a = Append(Init(a), a(n))"proof -  have "n ⊆ succ(n)" by auto  with A2 have "restrict(a,n): n -> X"    using restrict_type2 by simp  moreover from A1 A2 have I: "restrict(a,n) = Init(a)"    using func1_1_L1 pred_succ_eq Init_def by simp  ultimately show thesis1: "Init(a) : n -> X" by simp  { fix k assume "k∈n"    then have "restrict(a,n)(k) = a(k)"      using restrict by simp    with I have "Init(a)(k) = a(k)" by simp  } then show thesis2: "∀k∈n. Init(a)(k) = a(k)" by simp  let ?b = "Append(Init(a), a(n))"  from A2 thesis1 have II:    "Init(a) : n -> X"   "a(n) ∈ X"    "?b = Append(Init(a), a(n))"    using apply_funtype by auto  note A2  moreover from II have "?b : succ(n) -> X"    by (rule append_props)  moreover have "∀k ∈ succ(n). a(k) = ?b(k)"  proof -    { fix k assume A3: "k ∈ n"      from II have "∀j∈n. ?b(j) = Init(a)(j)"	by (rule append_props)      with thesis2 A3 have "a(k) = ?b(k)" by simp }    moreover     from II have "?b(n) = a(n)"      by (rule append_props)    hence " a(n) = ?b(n)" by simp    ultimately show "∀k ∈ succ(n). a(k) = ?b(k)"      by simp  qed  ultimately show "a = ?b" by (rule func_eq)qedtext{*If we take init of the result of append, we get back the same list.*} lemma init_append: assumes A1: "n ∈ nat" and A2: "a:n->X" and A3: "x ∈ X"  shows "Init(Append(a,x)) = a"proof -  from A2 A3 have "Append(a,x): succ(n)->X" using append_props by simp  with A1 have "Init(Append(a,x)):n->X" and "∀k∈n. Init(Append(a,x))(k) = Append(a,x)(k)"      using init_props by auto  with A2 A3 have "∀k∈n. Init(Append(a,x))(k) = a(k)" using append_props by simp  with Init(Append(a,x)):n->X A2 show ?thesis by (rule func_eq)qedtext{*A reformulation of definition of  @{text "Init"}.*}lemma init_def: assumes "n ∈ nat" and "x:succ(n)->X"  shows "Init(x) = restrict(x,n)"  using assms func1_1_L1 Init_def by simp    text{*A lemma about extending a finite sequence by one more value. This is   just a more explicit version of @{text "append_props"}.*}lemma finseq_extend:   assumes  "a:n->X"   "y∈X"   "b = a ∪ {⟨n,y⟩}"  shows  "b: succ(n) -> X"  "∀k∈n. b(k) = a(k)"  "b(n) = y"  using assms Append_def func1_1_L1 append_props by autotext{*The next lemma is a bit displaced as it is mainly   about finite sets. It is proven here because it uses  the notion of @{text "Append"}.  Suppose we have a list of element of $A$ is a bijection.  Then for every element that does not belong to $A$   we can we can construct   a bijection for the set $A \cup \{ x\}$ by appending $x$.  This is just a specialised version of lemma @{text "bij_extend_point"}  from @{text "func1.thy"}.  *}lemma bij_append_point:   assumes A1: "n ∈ nat" and A2: "b ∈ bij(n,X)" and A3: "x ∉ X"  shows "Append(b,x) ∈ bij(succ(n), X ∪ {x})"proof -  from A2 A3 have "b ∪ {⟨n,x⟩} ∈ bij(n ∪ {n},X ∪ {x})"    using mem_not_refl bij_extend_point by simp  moreover have "Append(b,x) = b ∪ {⟨n,x⟩}"  proof -    from A2 have "b:n->X"      using bij_def surj_def by simp    then have "b : n -> X ∪ {x}" using func1_1_L1B      by blast    then show "Append(b,x) = b ∪ {⟨n,x⟩}"      using Append_def func1_1_L1 by simp  qed  ultimately show ?thesis using succ_explained by autoqedtext{*The next lemma rephrases the definition of @{text "Last"}.  Recall that in ZF we have $\{0,1,2,..,n\} = n+1=$@{text "succ"}$(n)$.*}lemma last_seq_elem: assumes "a: succ(n) -> X" shows "Last(a) = a(n)"  using assms func1_1_L1 pred_succ_eq Last_def by simptext{*If two finite sequences are the same when restricted to domain one   shorter than the original and have the same value on the last element,   then they are equal.*}lemma finseq_restr_eq: assumes A1: "n ∈ nat" and   A2: "a: succ(n) -> X"  "b: succ(n) -> X" and  A3: "restrict(a,n) = restrict(b,n)" and  A4: "a(n) = b(n)"  shows "a = b"proof -  { fix k assume "k ∈ succ(n)"    then have "k ∈ n ∨ k = n" by auto    moreover    { assume "k ∈ n"        then have 	"restrict(a,n)(k) = a(k)" and "restrict(b,n)(k) = b(k)"	using restrict by auto      with A3 have "a(k) = b(k)" by simp }    moreover    { assume "k = n"      with A4 have "a(k) = b(k)" by simp }    ultimately have "a(k) = b(k)" by auto  } then have "∀ k ∈ succ(n). a(k) = b(k)" by simp  with A2 show "a = b" by (rule func_eq)qedtext{*Concatenating a list of length $1$ is the same as appending its  first (and only) element. Recall that in ZF set theory   $1 = \{ 0 \}$.*}lemma append_1elem: assumes A1: "n ∈ nat" and   A2: "a: n -> X"  and A3: "b : 1 -> X"  shows "Concat(a,b) = Append(a,b(0))"proof -  let ?C = "Concat(a,b)"  let ?A = "Append(a,b(0))"  from A1 A2 A3 have I:    "n ∈ nat"  "1 ∈ nat"    "a:n->X"   "b:1->X" by auto  have "?C : succ(n) -> X"  proof -    from I have "?C : n #+ 1 -> X"      by (rule concat_props)    with A1 show "?C : succ(n) -> X" by simp  qed  moreover from A2 A3 have "?A : succ(n) -> X"    using apply_funtype append_props by simp  moreover have "∀k ∈ succ(n). ?C(k) = ?A(k)"  proof    fix k assume "k ∈ succ(n)"    moreover    { assume "k ∈ n"      moreover from I have "∀i ∈ n. ?C(i) = a(i)"	by (rule concat_props)      moreover from A2 A3 have "∀i∈n. ?A(i) = a(i)"	using apply_funtype append_props by simp      ultimately have "?C(k) =  ?A(k)" by simp }    moreover have "?C(n) = ?A(n)"    proof -      from I have "∀j ∈ 1. ?C(n #+ j) = b(j)"	by (rule concat_props)      with A1 A2 A3 show "?C(n) = ?A(n)"	using apply_funtype append_props by simp    qed    ultimately show "?C(k) = ?A(k)" by auto  qed  ultimately show "?C = ?A" by (rule func_eq)qedtext{*A simple lemma about lists of length $1$.*}lemma list_len1_singleton: assumes A1: "x∈X"   shows "{⟨0,x⟩} : 1 -> X"proof -  from A1 have "{⟨0,x⟩} : {0} -> X" using pair_func_singleton    by simp  moreover have "{0} = 1" by auto  ultimately show ?thesis by simpqedtext{*A singleton list is in fact a singleton set with a pair as the only element.*}lemma list_singleton_pair: assumes A1: "x:1->X" shows "x = {⟨0,x(0)⟩}"proof -  from A1 have "x = {⟨t,x(t)⟩. t∈1}" by (rule fun_is_set_of_pairs)  hence "x = {⟨t,x(t)⟩. t∈{0} }" by simp  thus ?thesis by simpqed   text{*When we append an element to the empty list we get  a list with length $1$.*}lemma empty_append1: assumes A1: "x∈X"  shows "Append(0,x): 1 -> X" and "Append(0,x)(0) = x"proof -  let ?a = "Append(0,x)"  have "?a = {⟨0,x⟩}" using Append_def by auto  with A1 show "?a : 1 -> X" and "?a(0) = x"    using list_len1_singleton pair_func_singleton    by autoqed  (*text{*Tail of a list of length 1 is a list of length 0.*}lemma list_len1_tail: assumes "a:1->X"  shows "Tail(a) : 0 -> X"  using assms tail_props by blast *)text{*Appending an element is the same as concatenating  with certain pair.*}lemma append_concat_pair:   assumes "n ∈ nat" and "a: n -> X" and "x∈X"  shows "Append(a,x) = Concat(a,{⟨0,x⟩})"  using assms list_len1_singleton append_1elem pair_val  by simptext{*An associativity property involving concatenation   and appending. For proof we just convert appending to  concatenation and use @{text "concat_assoc"}.*}lemma concat_append_assoc: assumes A1: "n ∈ nat"  "k ∈ nat" and   A2: "a:n->X"   "b:k->X" and A3: "x ∈ X"  shows "Append(Concat(a,b),x) = Concat(a, Append(b,x))"proof -  from A1 A2 A3 have     "n #+ k ∈ nat"   "Concat(a,b) : n #+ k -> X"   "x ∈ X"    using concat_props by auto  then have     "Append(Concat(a,b),x) =  Concat(Concat(a,b),{⟨0,x⟩})"    by (rule append_concat_pair)  moreover  from A1 A2 A3 have    "n ∈ nat"  "k ∈ nat"  "1 ∈ nat"     "a:n->X"   "b:k->X"  "{⟨0,x⟩} :  1 -> X"    using list_len1_singleton by auto  then have    "Concat(Concat(a,b),{⟨0,x⟩}) = Concat(a, Concat(b,{⟨0,x⟩}))"    by (rule concat_assoc)  moreover from A1 A2 A3 have "Concat(b,{⟨0,x⟩}) =  Append(b,x)"    using list_len1_singleton append_1elem pair_val by simp  ultimately show "Append(Concat(a,b),x) = Concat(a, Append(b,x))"    by simpqedtext{*An identity involving concatenating with init  and appending the last element.*}lemma concat_init_last_elem:   assumes "n ∈ nat"  "k ∈ nat" and   "a: n -> X"  and "b : succ(k) -> X"  shows "Append(Concat(a,Init(b)),b(k)) = Concat(a,b)"  using assms init_props apply_funtype concat_append_assoc  by simptext{*A lemma about creating lists by composition and how  @{text "Append"} behaves in such case.*}lemma list_compose_append:   assumes A1: "n ∈ nat" and A2: "a : n -> X" and   A3: "x ∈ X" and A4: "c : X -> Y"  shows  "c O Append(a,x) : succ(n) -> Y"  "c O Append(a,x) = Append(c O a, c(x))"proof -  let ?b = "Append(a,x)"  let ?d = "Append(c O a, c(x))"  from A2 A4 have "c O a : n -> Y"    using comp_fun by simp  from A2 A3 have "?b : succ(n) -> X"    using append_props by simp  with A4 show "c O ?b : succ(n) -> Y"    using comp_fun by simp  moreover from A3 A4 c O a : n -> Y have     "?d: succ(n) -> Y"    using apply_funtype append_props by simp  moreover have "∀k ∈ succ(n). (c O ?b) (k) = ?d(k)"  proof -    { fix k assume "k ∈ succ(n)"      with ?b : succ(n) -> X have 	"(c O ?b) (k) = c(?b(k))"	using comp_fun_apply by simp      with A2 A3 A4 c O a : n -> Y c O a : n -> Y k ∈ succ(n)      have "(c O ?b) (k) = ?d(k)"	using append_props comp_fun_apply apply_funtype	by auto    } thus ?thesis by simp  qed  ultimately show "c O ?b = ?d" by (rule func_eq)qedtext{*A lemma about appending an element to a list defined by set  comprehension.*}lemma set_list_append:  assumes   A1: "∀i ∈ succ(k). b(i) ∈ X" and  A2: "a = {⟨i,b(i)⟩. i ∈ succ(k)}"  shows   "a: succ(k) -> X"  "{⟨i,b(i)⟩. i ∈ k}: k -> X"   "a = Append({⟨i,b(i)⟩. i ∈ k},b(k))"proof -  from A1 have "{⟨i,b(i)⟩. i ∈ succ(k)} : succ(k) -> X"     by (rule ZF_fun_from_total)  with A2 show "a: succ(k) -> X" by simp  from A1 have "∀i ∈ k. b(i) ∈ X"    by simp  then show "{⟨i,b(i)⟩. i ∈ k}: k -> X"    by (rule ZF_fun_from_total)  with A2 show "a = Append({⟨i,b(i)⟩. i ∈ k},b(k))"    using func1_1_L1 Append_def by autoqedtext{* An induction theorem for lists.*}lemma list_induct: assumes A1: "∀b∈1->X. P(b)" and   A2: "∀b∈NELists(X). P(b) --> (∀x∈X. P(Append(b,x)))" and  A3: "d ∈ NELists(X)"  shows "P(d)"proof -  { fix n     assume "n∈nat"    moreover from A1 have "∀b∈succ(0)->X. P(b)" by simp     moreover have "∀k∈nat. ((∀b∈succ(k)->X. P(b)) --> (∀c∈succ(succ(k))->X. P(c)))"    proof -      { fix k assume "k ∈ nat" assume "∀b∈succ(k)->X. P(b)"        have "∀c∈succ(succ(k))->X. P(c)"        proof          fix c assume "c: succ(succ(k))->X"          let ?b = "Init(c)"          let ?x = "c(succ(k))"          from k ∈ nat c: succ(succ(k))->X have "?b:succ(k)->X"            using init_props by simp          with A2 k ∈ nat ∀b∈succ(k)->X. P(b) have "∀x∈X. P(Append(?b,x))"            using NELists_def by auto           with c: succ(succ(k))->X have "P(Append(?b,?x))" using apply_funtype by simp           with k ∈ nat c: succ(succ(k))->X show "P(c)"            using init_props by simp         qed      } thus ?thesis by simp     qed    ultimately have "∀b∈succ(n)->X. P(b)" by (rule ind_on_nat)  } with A3 show ?thesis using NELists_def by auto qedsection{*Lists and cartesian products*}text{*Lists of length $n$ of elements of some set $X$ can be thought of as a model of the cartesian product $X^n$ which is more convenient in many applications.*}text{*There is a natural bijection between the space $(n+1)\rightarrow X$ of lists of length $n+1$ of elements of $X$ and the cartesian product $(n\rightarrow X)\times X$.*}lemma lists_cart_prod: assumes "n ∈ nat"  shows "{⟨x,⟨Init(x),x(n)⟩⟩. x ∈ succ(n)->X} ∈ bij(succ(n)->X,(n->X)×X)"proof -  let ?f = "{⟨x,⟨Init(x),x(n)⟩⟩. x ∈ succ(n)->X}"  from assms have "∀x ∈ succ(n)->X. ⟨Init(x),x(n)⟩ ∈ (n->X)×X"    using init_props succ_iff apply_funtype by simp  then have I: "?f: (succ(n)->X)->((n->X)×X)" by (rule ZF_fun_from_total)  moreover from assms I have "∀x∈succ(n)->X.∀y∈succ(n)->X. ?f(x)=?f(y) --> x=y"    using ZF_fun_from_tot_val init_def finseq_restr_eq by auto  moreover have "∀p∈(n->X)×X.∃x∈succ(n)->X. ?f(x) = p"  proof    fix p assume "p ∈ (n->X)×X"    let ?x = "Append(fst(p),snd(p))"    from assms p ∈ (n->X)×X have "?x:succ(n)->X" using append_props by simp    with I have "?f(?x) = ⟨Init(?x),?x(n)⟩" using succ_iff ZF_fun_from_tot_val by simp    moreover from assms p ∈ (n->X)×X have "Init(?x) = fst(p)" and "?x(n) = snd(p)"      using init_append append_props by auto    ultimately have "?f(?x) = ⟨fst(p),snd(p)⟩" by auto    with p ∈ (n->X)×X ?x:succ(n)->X show "∃x∈succ(n)->X. ?f(x) = p" by auto  qed  ultimately show ?thesis using inj_def surj_def bij_def by autoqedtext{*We can identify a set $X$ with lists of length one of elements of $X$.*}lemma singleton_list_bij: shows "{⟨x,x(0)⟩. x∈1->X} ∈ bij(1->X,X)"proof -  let ?f = "{⟨x,x(0)⟩. x∈1->X}"  have "∀x∈1->X. x(0) ∈ X" using apply_funtype by simp  then have I: "?f:(1->X)->X" by  (rule ZF_fun_from_total)  moreover have "∀x∈1->X.∀y∈1->X. ?f(x) = ?f(y) --> x=y"  proof -    { fix x y      assume "x:1->X" "y:1->X" and "?f(x) = ?f(y)"        with I have "x(0) = y(0)" using ZF_fun_from_tot_val by auto      moreover from x:1->X y:1->X have "x = {⟨0,x(0)⟩}" and "y = {⟨0,y(0)⟩}"         using list_singleton_pair by auto      ultimately have "x=y" by simp     } thus ?thesis by auto   qed  moreover have "∀y∈X. ∃x∈1->X. ?f(x)=y"  proof    fix y assume "y∈X"    let ?x = "{⟨0,y⟩}"    from I y∈X have "?x:1->X" and "?f(?x) = y"       using list_len1_singleton ZF_fun_from_tot_val pair_val by auto     thus "∃x∈1->X. ?f(x)=y" by auto  qed  ultimately show ?thesis using inj_def surj_def bij_def by simp qedtext{*We can identify a set of $X$-valued lists of length with $X$.*}lemma list_singleton_bij: shows   "{⟨x,{⟨0,x⟩}⟩.x∈X} ∈ bij(X,1->X)" and   "{⟨y,y(0)⟩. y∈1->X} = converse({⟨x,{⟨0,x⟩}⟩.x∈X})" and  "{⟨x,{⟨0,x⟩}⟩.x∈X} = converse({⟨y,y(0)⟩. y∈1->X})"proof -  let ?f = "{⟨y,y(0)⟩. y∈1->X}"  let ?g = "{⟨x,{⟨0,x⟩}⟩.x∈X}"  have "1 = {0}" by auto  then have "?f ∈ bij(1->X,X)" and "?g:X->(1->X)"     using singleton_list_bij pair_func_singleton ZF_fun_from_total      by auto  moreover have "∀y∈1->X.?g(?f(y)) = y"  proof    fix y assume "y:1->X"    have "?f:(1->X)->X" using singleton_list_bij bij_def inj_def by simp    with 1 = {0} y:1->X ?g:X->(1->X) show "?g(?f(y)) = y"       using ZF_fun_from_tot_val apply_funtype func_singleton_pair      by simp   qed  ultimately show "?g ∈ bij(X,1->X)" and "?f = converse(?g)" and "?g = converse(?f)"    using comp_conv_id by autoqed text{*What is the inverse image of a set by the natural bijection between $X$-valued   singleton lists and $X$? *}lemma singleton_vimage: assumes "U⊆X" shows "{x∈1->X. x(0) ∈ U} = { {⟨0,y⟩}. y∈U}"proof  have "1 = {0}" by auto   { fix x assume "x ∈ {x∈1->X. x(0) ∈ U}"    with 1 = {0} have "x = {⟨0, x(0)⟩}" using func_singleton_pair by auto     } thus "{x∈1->X. x(0) ∈ U} ⊆ { {⟨0,y⟩}. y∈U}" by auto  { fix x assume "x ∈ { {⟨0,y⟩}. y∈U}"    then obtain y where "x = {⟨0,y⟩}" and "y∈U" by auto    with 1 = {0} assms have "x:1->X" using pair_func_singleton by auto  } thus "{ {⟨0,y⟩}. y∈U} ⊆ {x∈1->X. x(0) ∈ U}" by autoqedtext{*A technical lemma about extending a list by values from a set.*} lemma list_append_from: assumes A1: "n ∈ nat" and A2: "U ⊆ n->X" and A3: "V ⊆ X"  shows   "{x ∈ succ(n)->X. Init(x) ∈ U ∧ x(n) ∈ V} = (\<Union>y∈V.{Append(x,y).x∈U})"proof -  { fix x assume "x ∈ {x ∈ succ(n)->X. Init(x) ∈ U ∧ x(n) ∈ V}"    then have "x ∈ succ(n)->X" and "Init(x) ∈ U" and I: "x(n) ∈ V"      by auto    let ?y = "x(n)"    from A1 and x ∈ succ(n)->X  have "x = Append(Init(x),?y)"      using init_props by simp    with I and Init(x) ∈ U have "x ∈ (\<Union>y∈V.{Append(a,y).a∈U})" by auto  }  moreover  { fix x assume "x ∈ (\<Union>y∈V.{Append(a,y).a∈U})"    then obtain a y where "y∈V" and "a∈U" and "x = Append(a,y)" by auto    with A2 A3 have "x: succ(n)->X" using append_props by blast     from A2 A3 y∈V a∈U have "a:n->X" and "y∈X" by auto    with A1 a∈U  y∈V x = Append(a,y) have "Init(x) ∈ U" and  "x(n) ∈ V"      using append_props init_append by auto        with x: succ(n)->X have "x ∈ {x ∈ succ(n)->X. Init(x) ∈ U ∧ x(n) ∈ V}"      by auto  }  ultimately show ?thesis by blastqedend`