Theory Set_Theory

theory Set_Theory
imports Main
(*  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, Schröder-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 Schröder-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