(*

This file is a part of IsarMathLib -

a library of formalized mathematics for Isabelle/Isar.

Copyright (C) 2005 - 2008 Slawomir Kolodynski

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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