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_Theoryimports Mainbegintext{*  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 blastlemma "(X = Y ∩ Z) =    (X ⊆ Y ∧ X ⊆ Z ∧ (∀V. V ⊆ Y ∧ V ⊆ Z --> V ⊆ X))"  by blasttext {*  Trivial example of term synthesis: apparently hard for some provers!*}schematic_lemma "a ≠ b ==> a ∈ ?X ∧ b ∉ ?X"  by blastsubsection {* 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 blastlemma "(\<Inter>x ∈ C. f x ∩ g x) = \<Inter>(f  C) ∩ \<Inter>(g  C)"  -- {* Inter-image, called @{text Int_Inter_image} in Main HOL *}  by blastlemma singleton_example_1:     "!!S::'a set set. ∀x ∈ S. ∀y ∈ S. x ⊆ y ==> ∃z. S ⊆ {z}"  by blastlemma singleton_example_2:     "∀x ∈ S. \<Union>S ⊆ x ==> ∃z. S ⊆ {z}"  -- {*Variant of the problem above. *}  by blastlemma "∃!x. f (g x) = x ==> ∃!y. g (f y) = y"  -- {* A unique fixpoint theorem --- @{text fast}/@{text best}/@{text meson} all fail. *}  by metissubsection {* 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 bestschematic_lemma "∀f:: 'a => 'a set. ∀x. f x ≠ ?S f"  -- {*This form displays the diagonal term. *}  by bestschematic_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 bestsubsection {* The Schröder-Berstein Theorem *}lemma disj_lemma: "- (f  X) = g  (-X) ==> f a = g b ==> a ∈ X ==> b ∈ X"  by blastlemma surj_if_then_else:  "-(f  X) = g  (-X) ==> surj (λz. if z ∈ X then f z else g z)"  by (simp add: surj_def) blastlemma 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)  donelemma decomposition: "∃X. X = - (g  (- (f  X)))"  apply (rule exI)  apply (rule lfp_unfold)  apply (rule monoI, blast)  donetheorem 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])  donesubsection {* A simple party theorem *}text{* \emph{At any party there are two people who know the samenumber of people}. Provided the party consists of at least two peopleand the knows relation is symmetric. Knowing yourself does not count--- otherwise knows needs to be reflexive. (From Freek Wiedijk's talkat 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  qedqedtext {*  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 forcelemma "D ∈ F ==> ∃G. ∀A ∈ G. ∃B ∈ F. A ⊆ B"  -- {* Example 2. *}  by forcelemma "P a ==> ∃A. (∀x ∈ A. P x) ∧ (∃y. y ∈ A)"  -- {* Example 3. *}  by forcelemma "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 forcelemma "P (f b) ==> ∃s A. (∀x ∈ A. P x) ∧ f s ∈ A"  -- {* Example 6. *}  by forcelemma "∃A. a ∉ A"  -- {* Example 7. *}  by forcelemma "(∀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 autolemma  "(∀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+  doneend`