Theory Nat_ZF_IML

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

theory Nat_ZF_IML imports Arith

begin

text{*The  ZF set theory constructs natural numbers from the empty set
  and the notion of a one-element set. Namely, zero of natural numbers
  is defined as the empty set. For each natural number $n$ the next natural
  number is defined as $n\cup \{n\}$. With this definition for every
  non-zero natural number we get the identity $n = \{0,1,2,..,n-1\}$.
  It is good to remember that when we see an expression like
  $f: n \rightarrow X$. Also, with this definition 
  the relation "less or equal than" becomes "$\subseteq$" and 
  the relation "less than" becomes "$\in$".
  *}

section{*Induction*}

text{*The induction lemmas in the standard Isabelle's Nat.thy file like 
  for example @{text "nat_induct"} require the induction step to 
  be a higher order 
  statement (the one that uses the $\Longrightarrow$ sign). I found it 
  difficult to apply from Isar, which is perhaps more of an indication of 
  my Isar skills than anything else. Anyway, here we provide a first order
  version that is easier to reference in Isar declarative style proofs.*}

text{*The next theorem is a version of induction on natural numbers
  that I was thought in school.*}

theorem ind_on_nat: 
  assumes A1: "n∈nat" and A2: "P(0)" and A3: "∀k∈nat. P(k)-->P(succ(k))"
  shows "P(n)"
proof -
  note A1 A2
  moreover
  { fix x
    assume "x∈nat"  "P(x)"
    with A3 have "P(succ(x))" by simp }
  ultimately show  "P(n)" by (rule nat_induct)
qed

text{*A nonzero natural number has a predecessor.*}

lemma Nat_ZF_1_L3: assumes A1: "n ∈ nat" and A2: "n≠0"
  shows "∃k∈nat. n = succ(k)"
proof -
  from A1 have "n ∈ {0} ∪ {succ(k). k∈nat}"
    using nat_unfold by simp
  with A2 show ?thesis by simp
qed

text{*What is @{text "succ"}, anyway?*}

lemma succ_explained: shows "succ(n) = n ∪ {n}"
  using succ_iff by auto

text{*Empty set is an element of every natural number which is not zero.*}

lemma empty_in_every_succ: assumes A1: "n ∈ nat"
  shows "0 ∈ succ(n)"
proof -
  note A1
  moreover have "0 ∈ succ(0)" by simp
  moreover
  { fix k assume "k ∈ nat" and A2: "0 ∈ succ(k)"
    then have "succ(k) ⊆ succ(succ(k))" by auto
    with A2 have "0 ∈ succ(succ(k))" by auto
  } then have "∀k ∈ nat. 0 ∈ succ(k) --> 0 ∈ succ(succ(k))"
    by simp
  ultimately show "0 ∈ succ(n)" by (rule ind_on_nat)
qed  

text{*If one natural number is less than another then their successors
  are in the same relation.*}

lemma succ_ineq: assumes A1: "n ∈ nat"
  shows "∀i ∈ n. succ(i) ∈ succ(n)"
proof -
  note A1
  moreover have "∀k ∈ 0. succ(k) ∈ succ(0)" by simp 
  moreover
  { fix k assume A2: "∀i∈k. succ(i) ∈ succ(k)"
    { fix i assume "i ∈ succ(k)"
      then have "i ∈ k ∨ i = k" by auto
      moreover
      { assume "i∈k"
	with A2 have "succ(i) ∈ succ(k)" by simp
	hence "succ(i) ∈ succ(succ(k))" by auto }
      moreover
      { assume "i = k"
	then have "succ(i) ∈ succ(succ(k))" by auto }
      ultimately have "succ(i) ∈ succ(succ(k))" by auto
    } then have "∀i ∈ succ(k). succ(i) ∈ succ(succ(k))"
      by simp
  } then have "∀k ∈ nat. 
      ( (∀i∈k. succ(i) ∈ succ(k)) --> (∀i ∈ succ(k). succ(i) ∈ succ(succ(k))) )"
    by simp
  ultimately show "∀i ∈ n. succ(i) ∈ succ(n)" by (rule ind_on_nat)
qed

text{*For natural numbers if $k\subseteq n$ the similar holds for 
  their successors. *}

lemma succ_subset: assumes A1: "k ∈ nat"  "n ∈ nat" and A2: "k⊆n" 
  shows "succ(k) ⊆ succ(n)"
proof -
  from A1 have T: "Ord(k)" and "Ord(n)"
    using nat_into_Ord by auto
  with A2 have "succ(k) ≤ succ(n)"
    using subset_imp_le by simp
  then show "succ(k) ⊆ succ(n)" using le_imp_subset
    by simp
qed

text{*For any two natural numbers one of them is contained in the other.*}

lemma nat_incl_total: assumes A1: "i ∈ nat"  "j ∈ nat"
  shows "i ⊆ j ∨ j ⊆ i"
proof -
  from A1 have T: "Ord(i)"   "Ord(j)" 
    using nat_into_Ord by auto
  then have "i∈j ∨ i=j ∨ j∈i" using Ord_linear
    by simp
  moreover
  { assume "i∈j"
    with T have "i⊆j ∨ j⊆i"
      using lt_def leI le_imp_subset by simp }
  moreover
  { assume "i=j"
    then have "i⊆j ∨ j⊆i" by simp }
  moreover
  { assume "j∈i"
    with T have "i⊆j ∨ j⊆i" 
      using lt_def leI  le_imp_subset by simp }
  ultimately show "i ⊆ j ∨ j ⊆ i" by auto
qed

text{*The set of natural numbers is the union of all successors of natural
  numbers.*}

lemma nat_union_succ: shows "nat = (\<Union>n ∈ nat. succ(n))"
proof
  show "nat ⊆ (\<Union>n ∈ nat. succ(n))" by auto
next
  { fix k assume A2: "k ∈ (\<Union>n ∈ nat. succ(n))"
    then obtain n where T: "n ∈ nat" and I: "k ∈ succ(n)"
      by auto
    then have "k ≤ n" using nat_into_Ord lt_def
      by simp
    with T have "k ∈ nat" using le_in_nat by simp
  } then show  "(\<Union>n ∈ nat. succ(n)) ⊆ nat" by auto
qed

text{*Successors of natural numbers are subsets of the 
  set of natural numbers.*}

lemma succnat_subset_nat: assumes A1: "n ∈ nat" shows "succ(n) ⊆ nat"
proof -
  from A1 have "succ(n) ⊆ (\<Union>n ∈ nat. succ(n))" by auto
  then show "succ(n) ⊆ nat" using nat_union_succ by simp
qed

text{*Element of a natural number is a natural number.*}

lemma elem_nat_is_nat: assumes A1: "n ∈ nat"  and A2: "k∈n"
  shows "k < n"  "k ∈ nat"  "k ≤ n"  "⟨k,n⟩ ∈ Le"
proof -
  from A1 A2 show "k < n" using nat_into_Ord lt_def by simp
  with A1 show "k ∈ nat" using lt_nat_in_nat by simp
  from `k < n` show "k ≤ n" using leI by simp
  with A1 `k ∈ nat` show "⟨k,n⟩ ∈ Le" using Le_def
    by simp
qed

(*text{*Element of a natural number is less or equal than 
  the number. *}

lemma elem_nat_is_Le: assumes A1: "n ∈ nat"  and A2: "k∈n"
  shows "k ≤ n" and "⟨k,n⟩ ∈ Le"
proof -
  from A1 A2 have "k < n" 
    using elem_nat_is_nat by simp
    and "k ∈ nat"
    using elem_nat_is_nat by auto*)
  


text{*The set of natural numbers is the union of its elements.*}

lemma nat_union_nat: shows "nat = \<Union> nat"
  using elem_nat_is_nat by blast

text{*A natural number is a subset of the set of natural numbers.*}

lemma nat_subset_nat: assumes A1: "n ∈ nat" shows "n ⊆ nat"
proof -
  from A1 have "n ⊆ \<Union> nat" by auto
  then show "n ⊆ nat" using nat_union_nat by simp
qed

text{*Adding a natural numbers does not decrease what we add to.*}

lemma add_nat_le: assumes A1: "n ∈ nat"  and A2: "k ∈ nat"
  shows 
  "n ≤ n #+ k"
  "n ⊆ n #+ k"
  "n ⊆ k #+ n"
proof -
  from A1 A2 have "n ≤ n"  "0 ≤ k"  "n ∈ nat"  "k ∈ nat"
    using nat_le_refl nat_0_le by auto
  then have "n #+ 0 ≤ n #+ k" by (rule add_le_mono)
  with A1 show "n ≤ n #+ k" using add_0_right by simp
  then show "n ⊆ n #+ k" using le_imp_subset by simp
  then show "n ⊆ k #+ n" using add_commute by simp
qed

text{*Result of adding an element of $k$ is smaller than of adding $k$. *}

lemma add_lt_mono: 
  assumes "k ∈ nat" and "j∈k"
  shows 
  "(n #+ j) < (n #+ k)"
  "(n #+ j) ∈ (n #+ k)"
proof -
  from assms have "j < k" using elem_nat_is_nat by blast
  moreover note `k ∈ nat`
  ultimately show "(n #+ j) < (n #+ k)"   "(n #+ j) ∈ (n #+ k)"
    using add_lt_mono2 ltD by auto
qed

text{*A technical lemma about a decomposition of a sum of two natural
  numbers: if a number $i$ is from $m + n$ then it is either from $m$
  or can be written as a sum of $m$ and a number from $n$. 
  The proof by induction w.r.t. to $m$ seems to be a bit heavy-handed, but I could
  not figure out how to do this directly from results from standard Isabelle/ZF.*}

lemma nat_sum_decomp: assumes A1: "n ∈ nat"  and A2: "m ∈ nat"
  shows "∀i ∈ m #+ n. i ∈ m ∨ (∃j ∈ n. i = m #+ j)" 
proof -
  note A1
  moreover from A2 have "∀i ∈ m #+ 0. i ∈ m ∨ (∃j ∈ 0. i = m #+ j)"
    using add_0_right by simp
  moreover have "∀k∈nat.
    (∀i ∈ m #+ k. i ∈ m ∨ (∃j ∈ k. i = m #+ j)) -->
    (∀i ∈ m #+ succ(k). i ∈ m ∨ (∃j ∈ succ(k). i = m #+ j))"
  proof -
    { fix k assume A3: "k ∈ nat"
      { assume A4: "∀i ∈ m #+ k. i ∈ m ∨ (∃j ∈ k. i = m #+ j)"
	  { fix i assume "i ∈  m #+ succ(k)"
	    then have "i ∈ m #+ k ∨ i = m #+ k" using add_succ_right
	      by auto
	    moreover from A4 A3 have
	      "i ∈ m #+ k --> i ∈ m ∨ (∃j ∈ succ(k). i = m #+ j)"
	      by auto
	    ultimately have "i ∈ m ∨ (∃j ∈ succ(k). i = m #+ j)"
	      by auto
	  } then have "∀i ∈ m #+ succ(k). i ∈ m ∨ (∃j ∈ succ(k). i = m #+ j)"
	    by simp
      } then have "(∀i ∈ m #+ k. i ∈ m ∨ (∃j ∈ k. i = m #+ j)) -->
	  (∀i ∈ m #+ succ(k). i ∈ m ∨ (∃j ∈ succ(k). i = m #+ j))"
	by simp
    } then show ?thesis by simp
  qed
  ultimately show "∀i ∈ m #+ n. i ∈ m ∨ (∃j ∈ n. i = m #+ j)"
    by (rule ind_on_nat)
qed

text{*A variant of induction useful for finite sequences.*}

lemma fin_nat_ind: assumes A1: "n ∈ nat" and A2: "k ∈ succ(n)"
  and A3: "P(0)" and A4: "∀j∈n. P(j)  --> P(succ(j))"
  shows "P(k)"
proof -
  from A2 have "k ∈ n ∨ k=n" by auto
  with A1 have "k ∈ nat" using elem_nat_is_nat by blast
  moreover from A3 have "0 ∈ succ(n) --> P(0)" by simp
  moreover from A1 A4 have
    "∀k ∈ nat. (k ∈ succ(n) --> P(k)) --> (succ(k) ∈ succ(n) --> P(succ(k)))"
    using nat_into_Ord Ord_succ_mem_iff by auto
  ultimately have "k ∈ succ(n) --> P(k)"
    by (rule ind_on_nat)
  with A2 show "P(k)" by simp
qed

text{*Some properties of positive natural numbers.*}

lemma succ_plus: assumes "n ∈ nat"  "k ∈ nat"
  shows 
  "succ(n #+ j) ∈ nat" 
  "succ(n) #+ succ(j) = succ(succ(n #+ j))"
  using assms by auto

section{*Intervals*}

text{*In this section we consider intervals of natural numbers i.e. sets
  of the form $\{n+j : j \in 0..k-1\}$. *}

text{*The interval is determined
  by two parameters: starting point and length. Recall that in standard 
  Isabelle's @{text "Arith.thy"} the symbol @{text "#+"} is defined
  as the sum of natural numbers.*}

definition

  "NatInterval(n,k) ≡ {n #+ j. j∈k}"

text{*Subtracting the beginning af the interval results in a number from
  the length of the interval.It may sound weird, but note that the length of
  such interval is a natural number, hence a set. *}

lemma inter_diff_in_len: 
  assumes A1: "k ∈ nat" and A2: "i ∈ NatInterval(n,k)"
  shows "i #- n ∈ k"
proof -
  from A2 obtain j where I: "i = n #+ j" and II: "j ∈ k"
    using NatInterval_def by auto
  from A1 II have "j ∈ nat" using elem_nat_is_nat by blast
  moreover from I have "i #- n = natify(j)" using diff_add_inverse
    by simp
  ultimately have "i #- n = j" by simp
  with II show ?thesis by simp
qed

text{*Intervals don't overlap with their starting point and 
  the union of an interval with its starting point is the sum of the starting
  point and the length of the interval.*}

lemma length_start_decomp: assumes  A1: "n ∈ nat"  "k ∈ nat"
  shows 
  "n ∩ NatInterval(n,k) = 0"
  "n ∪ NatInterval(n,k) = n #+ k"
proof -
  { fix i assume A2: "i ∈ n" and "i ∈ NatInterval(n,k)"
    then obtain j where I: "i = n #+ j" and II: "j ∈ k"
      using NatInterval_def by auto
    from A1 have "k ∈ nat" using elem_nat_is_nat by blast
    with II have "j ∈ nat" using elem_nat_is_nat by blast
    with A1 I have "n ≤ i" using add_nat_le by simp
    moreover from A1 A2 have "i < n" using elem_nat_is_nat by blast
    ultimately have False using le_imp_not_lt by blast
  } thus "n ∩ NatInterval(n,k) = 0" by auto
  from A1 have "n ⊆ n #+ k" using add_nat_le by simp
  moreover
  { fix i assume "i ∈ NatInterval(n,k)"
    then obtain j where III: "i = n #+ j" and IV: "j ∈ k"
      using NatInterval_def by auto
    with A1 have "j < k" using elem_nat_is_nat by blast
    with A1 III have "i ∈ n #+ k" using add_lt_mono2 ltD
      by simp }
  ultimately have "n ∪ NatInterval(n,k) ⊆ n #+ k" by auto
  moreover from A1 have "n #+ k ⊆ n ∪ NatInterval(n,k)" 
    using nat_sum_decomp NatInterval_def by auto
  ultimately show "n ∪ NatInterval(n,k) = n #+ k" by auto
qed

text{*Sme properties of three adjacent intervals.*}

lemma adjacent_intervals3: assumes "n ∈ nat"  "k ∈ nat"  "m ∈ nat"
  shows 
  "n #+ k #+ m = (n #+ k) ∪ NatInterval(n #+ k,m)"
  "n #+ k #+ m = n ∪ NatInterval(n,k #+  m)"
  "n #+ k #+ m = n ∪ NatInterval(n,k) ∪ NatInterval(n #+ k,m)"
  using assms add_assoc length_start_decomp by auto

end