IsarMathLib

Proofs by humans, for humans, formally verified by Isabelle/ZF proof assistant

theory Finite_ZF imports ZF1 Nat_ZF_IML ZF.Cardinal func1

begin

Standard Isabelle Finite.thy contains a very useful notion of finite powerset: the set of finite subsets of a given set. The definition, however, is specific to Isabelle and based on the notion of "datatype", obviously not something that belongs to ZF set theory. This theory file devolops the notion of finite powerset similarly as in Finite.thy, but based on standard library's Cardinal.thy. This theory file is intended to replace IsarMathLib's Finite1 and Finite_ZF_1 theories that are currently derived from the "datatype" approach.

Definition and basic properties of finite powerset

The goal of this section is to prove an induction theorem about finite powersets: if the empty set has some property and this property is preserved by adding a single element of a set, then this property is true for all finite subsets of this set.

We defined the finite powerset $FinPow (X)$ as those elements of the powerset that are finite.

definition

$FinPow (X) \equiv {A \in P o w (X) . F i n i t e (A)}$

The cardinality of an element of finite powerset is a natural number.

lemma card_fin_is_nat:

assumes $A \in FinPow (X)$

shows

| A | \in n a t

and

A \approx | A |

using assms, FinPow_def, Finite_def, cardinal_cong, nat_into_Card, Card_cardinal_eq

The cardinality of a finite set is a natural number.

lemma card_fin_is_nat1:

assumes $F i n i t e (A)$

shows

| A | \in n a t

using assms, card_fin_is_nat(1) unfolding FinPow_def

A reformulation of card_fin_is_nat: for a finit set $A$ there is a bijection between $| A |$ and $A$ .

lemma fin_bij_card:

assumes A1: $A \in FinPow (X)$

shows

\exists b . b \in bij (| A |, A)

proof

from A1 have

| A | \approx A

using card_fin_is_nat, eqpoll_sym

then show

t h e s i s

using eqpoll_def

qed

If a set has the same number of elements as $n \in N$ , then its cardinality is $n$ . Recall that in set theory a natural number $n$ is a set that has $n$ elements.

lemma card_card:

assumes $A \approx n$ and $n \in n a t$

shows

| A | = n

using assms, cardinal_cong, nat_into_Card, Card_cardinal_eq

If we add a point to a finite set, the cardinality increases by one. To understand the second assertion $| A \cup {a} | = | A | \cup {| A |}$ recall that the cardinality $| A |$ of $A$ is a natural number and for natural numbers we have $n + 1 = n \cup {n}$ .

lemma card_fin_add_one:

assumes A1: $A \in FinPow (X)$ and A2: $a \in X - A$

shows

| A \cup {a} | = s u c c (| A |)

| A \cup {a} | = | A | \cup {| A |}

proof

from A1, A2 have

c o n s (a, A) \approx c o n s (| A |, | A |)

using card_fin_is_nat, mem_not_refl, cons_eqpoll_cong

moreover

have

c o n s (a, A) = A \cup {a}

by (rule consdef )

moreover

have

c o n s (| A |, | A |) = | A | \cup {| A |}

by (rule consdef )

ultimately have

A \cup {a} \approx s u c c (| A |)

using succ_explained

with A1 show

| A \cup {a} | = s u c c (| A |)

and

| A \cup {a} | = | A | \cup {| A |}

using card_fin_is_nat, card_card

qed

We can decompose the finite powerset into collection of sets of the same natural cardinalities.

lemma finpow_decomp:

shows

FinPow (X) = (⋃ n \in n a t . {A \in P o w (X) . A \approx n})

using Finite_def, FinPow_def

Finite powerset is the union of sets of cardinality bounded by natural numbers.

lemma finpow_union_card_nat:

shows

FinPow (X) = (⋃ n \in n a t . {A \in P o w (X) . A ⪯ n})

proof

have

FinPow (X) \subseteq (⋃ n \in n a t . {A \in P o w (X) . A ⪯ n})

using finpow_decomp, FinPow_def, eqpoll_imp_lepoll

moreover

have

(⋃ n \in n a t . {A \in P o w (X) . A ⪯ n}) \subseteq FinPow (X)

using lepoll_nat_imp_Finite, FinPow_def

ultimately show

t h e s i s

qed

A different form of finpow_union_card_nat (see above) - a subset that has not more elements than a given natural number is in the finite powerset.

lemma lepoll_nat_in_finpow:

assumes $n \in n a t$ , $A \subseteq X$ , $A ⪯ n$

shows

A \in FinPow (X)

using assms, finpow_union_card_nat

Natural numbers are finite subsets of the set of natural numbers.

lemma nat_finpow_nat:

assumes $n \in n a t$

shows

n \in FinPow (n a t)

using assms, nat_into_Finite, nat_subset_nat, FinPow_def

A finite subset is a finite subset of itself.

lemma fin_finpow_self:

assumes $A \in FinPow (X)$

shows

A \in FinPow (A)

using assms, FinPow_def

A set is finite iff it is in its finite powerset.

lemma fin_finpow_iff:

shows

F i n i t e (A) ⟷ A \in FinPow (A)

unfolding FinPow_def

If we remove an element and put it back we get the set back.

lemma rem_add_eq:

assumes $a \in A$

shows

(A - {a}) \cup {a} = A

using assms

Induction for finite powerset. This is smilar to the standard Isabelle's Fin_induct.

theorem FinPow_induct:

assumes A1: $P (0)$ and A2: $\forall A \in FinPow (X) . P (A) ⟶ (\forall a \in X . P (A \cup {a}))$ and A3: $B \in FinPow (X)$

shows

P (B)

proof

{

fix

n

assume

n \in n a t

moreover

from A1 have I:

\forall B \in P o w (X) . B ⪯ 0 ⟶ P (B)

using lepoll_0_is_0

moreover

have

\forall k \in n a t .

(\forall B \in P o w (X) . (B ⪯ k ⟶ P (B))) ⟶

(\forall B \in P o w (X) . (B ⪯ s u c c (k) ⟶ P (B)))

proof

{

fix

k

assume A4:

k \in n a t

assume A5:

\forall B \in P o w (X) . (B ⪯ k ⟶ P (B))

fix

B

assume A6:

B \in P o w (X)

B ⪯ s u c c (k)

have

P (B)

proof

have

B = 0 ⟶ P (B)

proof

{

assume

B = 0

then have

B ⪯ 0

using lepoll_0_iff

with I, A6 have

P (B)

}

thus

B = 0 ⟶ P (B)

qed

moreover

have

B \neq 0 ⟶ P (B)

proof

{

assume

B \neq 0

then obtain

a

where II:

a \in B

let

A = B - {a}

from A6, II have

A \subseteq X

and

A ⪯ k

using Diff_sing_lepoll

with A4, A5 have

A \in FinPow (X)

and

P (A)

using lepoll_nat_in_finpow, finpow_decomp

with A2, A6, II have

P (A \cup {a})

moreover

from II have

A \cup {a} = B

ultimately have

P (B)

}

thus

B \neq 0 ⟶ P (B)

qed

ultimately show

P (B)

qed

}

thus

t h e s i s

qed

ultimately have

\forall B \in P o w (X) . (B ⪯ n ⟶ P (B))

by (rule ind_on_nat )

}

then have

\forall n \in n a t . \forall B \in P o w (X) . (B ⪯ n ⟶ P (B))

with A3 show

P (B)

using finpow_union_card_nat

qed

A subset of a finite subset is a finite subset.

lemma subset_finpow:

assumes $A \in FinPow (X)$ and $B \subseteq A$

shows

B \in FinPow (X)

using assms, FinPow_def, subset_Finite

If we subtract anything from a finite set, the resulting set is finite.

lemma diff_finpow:

assumes $A \in FinPow (X)$

shows

A - B \in FinPow (X)

using assms, subset_finpow

If we remove a point from a finites subset, we get a finite subset.

corollary fin_rem_point_fin:

assumes $A \in FinPow (X)$

shows

A - {a} \in FinPow (X)

using assms, diff_finpow

Cardinality of a nonempty finite set is a successsor of some natural number.

lemma card_non_empty_succ:

assumes A1: $A \in FinPow (X)$ and A2: $A \neq 0$

shows

\exists n \in n a t . | A | = s u c c (n)

proof

from A2 obtain

a

where

a \in A

let

B = A - {a}

from A1,

a \in A

have

B \in FinPow (X)

and

a \in X - B

using FinPow_def, fin_rem_point_fin

then have

| B \cup {a} | = s u c c (| B |)

using card_fin_add_one

moreover

from

a \in A

B \in FinPow (X)

have

A = B \cup {a}

and

| B | \in n a t

using card_fin_is_nat

ultimately show

\exists n \in n a t . | A | = s u c c (n)

qed

Nonempty set has non-zero cardinality. This is probably true without the assumption that the set is finite, but I couldn't derive it from standard Isabelle theorems.

lemma card_non_empty_non_zero:

assumes $A \in FinPow (X)$ and $A \neq 0$

shows

| A | \neq 0

proof

from assms obtain

n

where

| A | = s u c c (n)

using card_non_empty_succ

then show

| A | \neq 0

using succ_not_0

qed

Another variation on the induction theme: If we can show something holds for the empty set and if it holds for all finite sets with at most $k$ elements then it holds for all finite sets with at most $k + 1$ elements, the it holds for all finite sets.

theorem FinPow_card_ind:

assumes A1: $P (0)$ and A2: $\forall k \in n a t .$ $(\forall A \in FinPow (X) . A ⪯ k ⟶ P (A)) ⟶$ $(\forall A \in FinPow (X) . A ⪯ s u c c (k) ⟶ P (A))$ and A3: $A \in FinPow (X)$

shows

P (A)

proof

from A3 have

| A | \in n a t

and

A \in FinPow (X)

and

A ⪯ | A |

using card_fin_is_nat, eqpoll_imp_lepoll

moreover

have

\forall n \in n a t . (\forall A \in FinPow (X) .

A ⪯ n ⟶ P (A))

proof

fix

n

assume

n \in n a t

moreover

from A1 have

\forall A \in FinPow (X) . A ⪯ 0 ⟶ P (A)

using lepoll_0_is_0

moreover

note A2

ultimately show

\forall A \in FinPow (X) . A ⪯ n ⟶ P (A)

by (rule ind_on_nat )

qed

ultimately show

P (A)

qed

Another type of induction (or, maybe recursion). In the induction step we try to find a point in the set that if we remove it, the fact that the property holds for the smaller set implies that the property holds for the whole set.

lemma FinPow_ind_rem_one:

assumes A1: $P (0)$ and A2: $\forall A \in FinPow (X) . A \neq 0 ⟶ (\exists a \in A . P (A - {a}) ⟶ P (A))$ and A3: $B \in FinPow (X)$

shows

P (B)

proof

note A1

moreover

have

\forall k \in n a t .

(\forall B \in FinPow (X) . B ⪯ k ⟶ P (B)) ⟶

(\forall C \in FinPow (X) . C ⪯ s u c c (k) ⟶ P (C))

proof

{

fix

k

assume

k \in n a t

assume A4:

\forall B \in FinPow (X) . B ⪯ k ⟶ P (B)

have

\forall C \in FinPow (X) . C ⪯ s u c c (k) ⟶ P (C)

proof

{

fix

C

assume

C \in FinPow (X)

assume

C ⪯ s u c c (k)

note A1

moreover {

assume

C \neq 0

with A2,

C \in FinPow (X)

obtain

a

where

a \in C

and

P (C - {a}) ⟶ P (C)

with A4,

C \in FinPow (X)

C ⪯ s u c c (k)

have

P (C)

using Diff_sing_lepoll, fin_rem_point_fin

} ultimately have

P (C)

}

thus

t h e s i s

qed

}

thus

t h e s i s

qed

moreover

note A3

ultimately show

P (B)

by (rule FinPow_card_ind )

qed

Yet another induction theorem. This is similar, but slightly more complicated than FinPow_ind_rem_one. The difference is in the treatment of the empty set to allow to show properties that are not true for empty set.

lemma FinPow_rem_ind:

assumes A1: $\forall A \in FinPow (X) .$ $A = 0 \lor (\exists a \in A . A = {a} \lor P (A - {a}) ⟶ P (A))$ and A2: $A \in FinPow (X)$ and A3: $A \neq 0$

shows

P (A)

proof

have

0 = 0 \lor P (0)

moreover

have

\forall k \in n a t .

(\forall B \in FinPow (X) . B ⪯ k ⟶ (B = 0 \lor P (B))) ⟶

(\forall A \in FinPow (X) . A ⪯ s u c c (k) ⟶ (A = 0 \lor P (A)))

proof

{

fix

k

assume

k \in n a t

assume A4:

\forall B \in FinPow (X) . B ⪯ k ⟶ (B = 0 \lor P (B))

have

\forall A \in FinPow (X) . A ⪯ s u c c (k) ⟶ (A = 0 \lor P (A))

proof

{

fix

A

assume

A \in FinPow (X)

assume

A ⪯ s u c c (k)

A \neq 0

from A1,

A \in FinPow (X)

A \neq 0

obtain

a

where

a \in A

and

A = {a} \lor P (A - {a}) ⟶ P (A)

let

B = A - {a}

from A4,

A \in FinPow (X)

A ⪯ s u c c (k)

a \in A

have

B = 0 \lor P (B)

using Diff_sing_lepoll, fin_rem_point_fin

with

a \in A

A = {a} \lor P (A - {a}) ⟶ P (A)

have

P (A)

}

thus

t h e s i s

qed

}

thus

t h e s i s

qed

moreover

note A2

ultimately have

A = 0 \lor P (A)

by (rule FinPow_card_ind )

with A3 show

P (A)

qed

If a family of sets is closed with respect to taking intersections of two sets then it is closed with respect to taking intersections of any nonempty finite collection.

lemma inter_two_inter_fin:

assumes A1: $\forall V \in T . \forall W \in T . V \cap W \in T$ and A2: $N \neq 0$ and A3: $N \in FinPow (T)$

shows

(⋂ N \in T)

proof

have

0 = 0 \lor (⋂ 0 \in T)

moreover

have

\forall M \in FinPow (T) . (M = 0 \lor ⋂ M \in T) ⟶

(\forall W \in T . M \cup {W} = 0 \lor ⋂ (M \cup {W}) \in T)

proof

{

fix

M

assume

M \in FinPow (T)

assume A4:

M = 0 \lor ⋂ M \in T

{

assume

M = 0

hence

\forall W \in T . M \cup {W} = 0 \lor ⋂ (M \cup {W}) \in T

}

moreover {

assume

M \neq 0

with A4 have

⋂ M \in T

{

fix

W

assume

W \in T

from

M \neq 0

have

⋂ (M \cup {W}) = (⋂ M) \cap W

with A1,

⋂ M \in T

W \in T

have

⋂ (M \cup {W}) \in T

}

hence

\forall W \in T . M \cup {W} = 0 \lor ⋂ (M \cup {W}) \in T

} ultimately have

\forall W \in T . M \cup {W} = 0 \lor ⋂ (M \cup {W}) \in T

}

thus

t h e s i s

qed

moreover

note

N \in FinPow (T)

ultimately have

N = 0 \lor (⋂ N \in T)

by (rule FinPow_induct )

with A2 show

(⋂ N \in T)

qed

If a family of sets contains the empty set and is closed with respect to taking unions of two sets then it is closed with respect to taking unions of any finite collection.

lemma union_two_union_fin:

assumes A1: $0 \in C$ and A2: $\forall A \in C . \forall B \in C . A \cup B \in C$ and A3: $N \in FinPow (C)$

shows

⋃ N \in C

proof

from

0 \in C

have

⋃ 0 \in C

moreover

have

\forall M \in FinPow (C) . ⋃ M \in C ⟶ (\forall A \in C . ⋃ (M \cup {A}) \in C)

proof

{

fix

M

assume

M \in FinPow (C)

assume

⋃ M \in C

fix

A

assume

A \in C

have

⋃ (M \cup {A}) = (⋃ M) \cup A

with A2,

⋃ M \in C

A \in C

have

⋃ (M \cup {A}) \in C

}

thus

t h e s i s

qed

moreover

note

N \in FinPow (C)

ultimately show

⋃ N \in C

by (rule FinPow_induct )

qed

Empty set is in finite power set, hence finite power set is never empty.

lemma empty_in_finpow:

shows

\emptyset \in FinPow (X)

and

FinPow (X) \neq \emptyset

using FinPow_def

Singleton is in the finite powerset.

lemma singleton_in_finpow:

assumes $x \in X$

shows

{x} \in FinPow (X)

using assms, FinPow_def

If a set is nonempty then its finite power set contains a nonempty set.

lemma finpow_nempty_nempty:

assumes $X \neq \emptyset$

shows

FinPow (X) ∖ {\emptyset} \neq \emptyset

using assms, singleton_in_finpow

Union of two finite subsets is a finite subset.

lemma union_finpow:

assumes $A \in FinPow (X)$ and $B \in FinPow (X)$

shows

A \cup B \in FinPow (X)

using assms, FinPow_def

Union of finite number of finite sets is finite.

lemma fin_union_finpow:

assumes $M \in FinPow (FinPow (X))$

shows

⋃ M \in FinPow (X)

using assms, empty_in_finpow, union_finpow, union_two_union_fin

If a set is finite after removing one element, then it is finite.

lemma rem_point_fin_fin:

assumes A1: $x \in X$ and A2: $A - {x} \in FinPow (X)$

shows

A \in FinPow (X)

proof

from assms have

(A - {x}) \cup {x} \in FinPow (X)

using singleton_in_finpow, union_finpow

moreover

have

A \subseteq (A - {x}) \cup {x}

ultimately show

A \in FinPow (X)

using FinPow_def, subset_Finite

qed

An image of a finite set is finite.

lemma fin_image_fin:

assumes $\forall V \in B . K (V) \in C$ and $N \in FinPow (B)$

shows

{K (V) . V \in N} \in FinPow (C)

proof

have

{K (V) . V \in 0} \in FinPow (C)

using FinPow_def

moreover

have

\forall A \in FinPow (B) .

{K (V) . V \in A} \in FinPow (C) ⟶ (\forall a \in B . {K (V) . V \in (A \cup {a})} \in FinPow (C))

proof

{

fix

A

assume

A \in FinPow (B)

assume

{K (V) . V \in A} \in FinPow (C)

fix

a

assume

a \in B

have

{K (V) . V \in (A \cup {a})} \in FinPow (C)

proof

have

{K (V) . V \in (A \cup {a})} = {K (V) . V \in A} \cup {K (a)}

moreover

note

{K (V) . V \in A} \in FinPow (C)

moreover

from

\forall V \in B . K (V) \in C

a \in B

have

{K (a)} \in FinPow (C)

using singleton_in_finpow

ultimately show

t h e s i s

using union_finpow

qed

}

thus

t h e s i s

qed

moreover

note

N \in FinPow (B)

ultimately show

{K (V) . V \in N} \in FinPow (C)

by (rule FinPow_induct )

qed

If a set $X$ is finite then the set ${K (x) . x \in X}$ is also finite. Its basically standard Isalelle/ZF Finite_RepFun in nicer notation.

lemma fin_rep_fin:

assumes $F i n i t e (X)$

shows

F i n i t e ({K (x) . x \in X})

using assms, Finite_RepFun

The image of a singleton by any function is finite. It's of course either empty or has exactly one element, but showing that it's a finite subset of the codomain is good enough for us.

lemma image_singleton_fin:

assumes $f : X \to Y$

shows

f {x} \in FinPow (Y)

proof

{

assume

x \in X

with assms have

f {x} \in FinPow (Y)

using singleton_image, singleton_in_finpow

}

moreover {

assume

x \notin X

with assms have

f {x} \in FinPow (Y)

using arg_not_in_domain(1), empty_in_finpow

} ultimately show

t h e s i s

qed

Union of a finite indexed family of finite sets is finite.

lemma union_fin_list_fin:

assumes A1: $n \in n a t$ and A2: $\forall k \in n . N (k) \in FinPow (X)$

shows

{N (k) . k \in n} \in FinPow (FinPow (X))

and

(⋃ k \in n . N (k)) \in FinPow (X)

proof

from A1 have

n \in FinPow (n)

using nat_finpow_nat, fin_finpow_self

with A2 show

{N (k) . k \in n} \in FinPow (FinPow (X))

by (rule fin_image_fin )

then show

(⋃ k \in n . N (k)) \in FinPow (X)

using fin_union_finpow

qed

end

Definition of FinPow:

FinPow (X) \equiv {A \in P o w (X) . F i n i t e (A)}

lemma card_fin_is_nat:

assumes $A \in FinPow (X)$

shows

| A | \in n a t

and

A \approx | A |

lemma card_fin_is_nat:

assumes $A \in FinPow (X)$

shows

| A | \in n a t

and

A \approx | A |

lemma consdef: shows

c o n s (a, A) = A \cup {a}

lemma succ_explained: shows

s u c c (n) = n \cup {n}

lemma card_card:

assumes $A \approx n$ and $n \in n a t$

shows

| A | = n

lemma finpow_decomp: shows

FinPow (X) = (⋃ n \in n a t . {A \in P o w (X) . A \approx n})

lemma finpow_union_card_nat: shows

FinPow (X) = (⋃ n \in n a t . {A \in P o w (X) . A ⪯ n})

lemma nat_subset_nat:

assumes $n \in n a t$

shows

n \subseteq n a t

lemma lepoll_nat_in_finpow:

assumes $n \in n a t$ , $A \subseteq X$ , $A ⪯ n$

shows

A \in FinPow (X)

theorem ind_on_nat:

assumes $n \in n a t$ and $P (0)$ and $\forall k \in n a t . P (k) ⟶ P (s u c c (k))$

shows

P (n)

lemma subset_finpow:

assumes $A \in FinPow (X)$ and $B \subseteq A$

shows

B \in FinPow (X)

lemma diff_finpow:

assumes $A \in FinPow (X)$

shows

A - B \in FinPow (X)

corollary fin_rem_point_fin:

assumes $A \in FinPow (X)$

shows

A - {a} \in FinPow (X)

lemma card_fin_add_one:

assumes $A \in FinPow (X)$ and $a \in X - A$

shows

| A \cup {a} | = s u c c (| A |)

| A \cup {a} | = | A | \cup {| A |}

lemma card_non_empty_succ:

assumes $A \in FinPow (X)$ and $A \neq 0$

shows

\exists n \in n a t . | A | = s u c c (n)

theorem FinPow_card_ind:

assumes $P (0)$ and $\forall k \in n a t .$ $(\forall A \in FinPow (X) . A ⪯ k ⟶ P (A)) ⟶$ $(\forall A \in FinPow (X) . A ⪯ s u c c (k) ⟶ P (A))$ and $A \in FinPow (X)$