(* Title: HOL/ex/Set_Theory.thy Author: Tobias Nipkow and Lawrence C Paulson Copyright 1991 University of Cambridge *) section {* 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