(* Title: HOL/ex/Set_Theory.thy

Author: Tobias Nipkow and Lawrence C Paulson

Copyright 1991 University of Cambridge

*)

header {* Set Theory examples: Cantor's Theorem, SchrÃ¶der-Bernstein Theorem, etc. *}

theory Set_Theory

imports Main

begin

text{*

These two are cited in Benzmueller and Kohlhase's system description

of LEO, CADE-15, 1998 (pages 139-143) as theorems LEO could not

prove.

*}

lemma "(X = Y ∪ Z) =

(Y ⊆ X ∧ Z ⊆ X ∧ (∀V. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))"

by blast

lemma "(X = Y ∩ Z) =

(X ⊆ Y ∧ X ⊆ Z ∧ (∀V. V ⊆ Y ∧ V ⊆ Z --> V ⊆ X))"

by blast

text {*

Trivial example of term synthesis: apparently hard for some provers!

*}

schematic_lemma "a ≠ b ==> a ∈ ?X ∧ b ∉ ?X"

by blast

subsection {* Examples for the @{text blast} paper *}

lemma "(\<Union>x ∈ C. f x ∪ g x) = \<Union>(f ` C) ∪ \<Union>(g ` C)"

-- {* Union-image, called @{text Un_Union_image} in Main HOL *}

by blast

lemma "(\<Inter>x ∈ C. f x ∩ g x) = \<Inter>(f ` C) ∩ \<Inter>(g ` C)"

-- {* Inter-image, called @{text Int_Inter_image} in Main HOL *}

by blast

lemma singleton_example_1:

"!!S::'a set set. ∀x ∈ S. ∀y ∈ S. x ⊆ y ==> ∃z. S ⊆ {z}"

by blast

lemma singleton_example_2:

"∀x ∈ S. \<Union>S ⊆ x ==> ∃z. S ⊆ {z}"

-- {*Variant of the problem above. *}

by blast

lemma "∃!x. f (g x) = x ==> ∃!y. g (f y) = y"

-- {* A unique fixpoint theorem --- @{text fast}/@{text best}/@{text meson} all fail. *}

by metis

subsection {* Cantor's Theorem: There is no surjection from a set to its powerset *}

lemma cantor1: "¬ (∃f:: 'a => 'a set. ∀S. ∃x. f x = S)"

-- {* Requires best-first search because it is undirectional. *}

by best

schematic_lemma "∀f:: 'a => 'a set. ∀x. f x ≠ ?S f"

-- {*This form displays the diagonal term. *}

by best

schematic_lemma "?S ∉ range (f :: 'a => 'a set)"

-- {* This form exploits the set constructs. *}

by (rule notI, erule rangeE, best)

schematic_lemma "?S ∉ range (f :: 'a => 'a set)"

-- {* Or just this! *}

by best

subsection {* The SchrÃ¶der-Berstein Theorem *}

lemma disj_lemma: "- (f ` X) = g ` (-X) ==> f a = g b ==> a ∈ X ==> b ∈ X"

by blast

lemma surj_if_then_else:

"-(f ` X) = g ` (-X) ==> surj (λz. if z ∈ X then f z else g z)"

by (simp add: surj_def) blast

lemma bij_if_then_else:

"inj_on f X ==> inj_on g (-X) ==> -(f ` X) = g ` (-X) ==>

h = (λz. if z ∈ X then f z else g z) ==> inj h ∧ surj h"

apply (unfold inj_on_def)

apply (simp add: surj_if_then_else)

apply (blast dest: disj_lemma sym)

done

lemma decomposition: "∃X. X = - (g ` (- (f ` X)))"

apply (rule exI)

apply (rule lfp_unfold)

apply (rule monoI, blast)

done

theorem Schroeder_Bernstein:

"inj (f :: 'a => 'b) ==> inj (g :: 'b => 'a)

==> ∃h:: 'a => 'b. inj h ∧ surj h"

apply (rule decomposition [where f=f and g=g, THEN exE])

apply (rule_tac x = "(λz. if z ∈ x then f z else inv g z)" in exI)

--{*The term above can be synthesized by a sufficiently detailed proof.*}

apply (rule bij_if_then_else)

apply (rule_tac [4] refl)

apply (rule_tac [2] inj_on_inv_into)

apply (erule subset_inj_on [OF _ subset_UNIV])

apply blast

apply (erule ssubst, subst double_complement, erule inv_image_comp [symmetric])

done

subsection {* A simple party theorem *}

text{* \emph{At any party there are two people who know the same

number of people}. Provided the party consists of at least two people

and the knows relation is symmetric. Knowing yourself does not count

--- otherwise knows needs to be reflexive. (From Freek Wiedijk's talk

at TPHOLs 2007.) *}

lemma equal_number_of_acquaintances:

assumes "Domain R <= A" and "sym R" and "card A ≥ 2"

shows "¬ inj_on (%a. card(R `` {a} - {a})) A"

proof -

let ?N = "%a. card(R `` {a} - {a})"

let ?n = "card A"

have "finite A" using `card A ≥ 2` by(auto intro:ccontr)

have 0: "R `` A <= A" using `sym R` `Domain R <= A`

unfolding Domain_unfold sym_def by blast

have h: "ALL a:A. R `` {a} <= A" using 0 by blast

hence 1: "ALL a:A. finite(R `` {a})" using `finite A`

by(blast intro: finite_subset)

have sub: "?N ` A <= {0..<?n}"

proof -

have "ALL a:A. R `` {a} - {a} < A" using h by blast

thus ?thesis using psubset_card_mono[OF `finite A`] by auto

qed

show "~ inj_on ?N A" (is "~ ?I")

proof

assume ?I

hence "?n = card(?N ` A)" by(rule card_image[symmetric])

with sub `finite A` have 2[simp]: "?N ` A = {0..<?n}"

using subset_card_intvl_is_intvl[of _ 0] by(auto)

have "0 : ?N ` A" and "?n - 1 : ?N ` A" using `card A ≥ 2` by simp+

then obtain a b where ab: "a:A" "b:A" and Na: "?N a = 0" and Nb: "?N b = ?n - 1"

by (auto simp del: 2)

have "a ≠ b" using Na Nb `card A ≥ 2` by auto

have "R `` {a} - {a} = {}" by (metis 1 Na ab card_eq_0_iff finite_Diff)

hence "b ∉ R `` {a}" using `a≠b` by blast

hence "a ∉ R `` {b}" by (metis Image_singleton_iff assms(2) sym_def)

hence 3: "R `` {b} - {b} <= A - {a,b}" using 0 ab by blast

have 4: "finite (A - {a,b})" using `finite A` by simp

have "?N b <= ?n - 2" using ab `a≠b` `finite A` card_mono[OF 4 3] by simp

then show False using Nb `card A ≥ 2` by arith

qed

qed

text {*

From W. W. Bledsoe and Guohui Feng, SET-VAR. JAR 11 (3), 1993, pages

293-314.

Isabelle can prove the easy examples without any special mechanisms,

but it can't prove the hard ones.

*}

lemma "∃A. (∀x ∈ A. x ≤ (0::int))"

-- {* Example 1, page 295. *}

by force

lemma "D ∈ F ==> ∃G. ∀A ∈ G. ∃B ∈ F. A ⊆ B"

-- {* Example 2. *}

by force

lemma "P a ==> ∃A. (∀x ∈ A. P x) ∧ (∃y. y ∈ A)"

-- {* Example 3. *}

by force

lemma "a < b ∧ b < (c::int) ==> ∃A. a ∉ A ∧ b ∈ A ∧ c ∉ A"

-- {* Example 4. *}

by auto --{*slow*}

lemma "P (f b) ==> ∃s A. (∀x ∈ A. P x) ∧ f s ∈ A"

-- {*Example 5, page 298. *}

by force

lemma "P (f b) ==> ∃s A. (∀x ∈ A. P x) ∧ f s ∈ A"

-- {* Example 6. *}

by force

lemma "∃A. a ∉ A"

-- {* Example 7. *}

by force

lemma "(∀u v. u < (0::int) --> u ≠ abs v)

--> (∃A::int set. -2 ∈ A & (∀y. abs y ∉ A))"

-- {* Example 8 needs a small hint. *}

by force

-- {* not @{text blast}, which can't simplify @{text "-2 < 0"} *}

text {* Example 9 omitted (requires the reals). *}

text {* The paper has no Example 10! *}

lemma "(∀A. 0 ∈ A ∧ (∀x ∈ A. Suc x ∈ A) --> n ∈ A) ∧

P 0 ∧ (∀x. P x --> P (Suc x)) --> P n"

-- {* Example 11: needs a hint. *}

by(metis nat.induct)

lemma

"(∀A. (0, 0) ∈ A ∧ (∀x y. (x, y) ∈ A --> (Suc x, Suc y) ∈ A) --> (n, m) ∈ A)

∧ P n --> P m"

-- {* Example 12. *}

by auto

lemma

"(∀x. (∃u. x = 2 * u) = (¬ (∃v. Suc x = 2 * v))) -->

(∃A. ∀x. (x ∈ A) = (Suc x ∉ A))"

-- {* Example EO1: typo in article, and with the obvious fix it seems

to require arithmetic reasoning. *}

apply clarify

apply (rule_tac x = "{x. ∃u. x = 2 * u}" in exI, auto)

apply metis+

done

end