src/HOL/ex/Set_Theory.thy
author haftmann
Sat Dec 24 15:53:09 2011 +0100 (2011-12-24)
changeset 45966 03ce2b2a29a2
parent 44276 fe769a0fcc96
child 46752 e9e7209eb375
permissions -rw-r--r--
tuned proofs
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(*  Title:      HOL/ex/Set_Theory.thy
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    Author:     Tobias Nipkow and Lawrence C Paulson
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    Copyright   1991  University of Cambridge
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*)
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header {* Set Theory examples: Cantor's Theorem, Schröder-Bernstein Theorem, etc. *}
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theory Set_Theory
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imports Main
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begin
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text{*
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  These two are cited in Benzmueller and Kohlhase's system description
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  of LEO, CADE-15, 1998 (pages 139-143) as theorems LEO could not
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  prove.
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*}
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lemma "(X = Y \<union> Z) =
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    (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))"
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  by blast
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lemma "(X = Y \<inter> Z) =
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    (X \<subseteq> Y \<and> X \<subseteq> Z \<and> (\<forall>V. V \<subseteq> Y \<and> V \<subseteq> Z \<longrightarrow> V \<subseteq> X))"
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  by blast
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text {*
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  Trivial example of term synthesis: apparently hard for some provers!
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*}
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schematic_lemma "a \<noteq> b \<Longrightarrow> a \<in> ?X \<and> b \<notin> ?X"
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  by blast
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subsection {* Examples for the @{text blast} paper *}
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lemma "(\<Union>x \<in> C. f x \<union> g x) = \<Union>(f ` C)  \<union>  \<Union>(g ` C)"
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  -- {* Union-image, called @{text Un_Union_image} in Main HOL *}
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  by blast
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lemma "(\<Inter>x \<in> C. f x \<inter> g x) = \<Inter>(f ` C) \<inter> \<Inter>(g ` C)"
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  -- {* Inter-image, called @{text Int_Inter_image} in Main HOL *}
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  by blast
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lemma singleton_example_1:
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     "\<And>S::'a set set. \<forall>x \<in> S. \<forall>y \<in> S. x \<subseteq> y \<Longrightarrow> \<exists>z. S \<subseteq> {z}"
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  by blast
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lemma singleton_example_2:
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     "\<forall>x \<in> S. \<Union>S \<subseteq> x \<Longrightarrow> \<exists>z. S \<subseteq> {z}"
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  -- {*Variant of the problem above. *}
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  by blast
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lemma "\<exists>!x. f (g x) = x \<Longrightarrow> \<exists>!y. g (f y) = y"
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  -- {* A unique fixpoint theorem --- @{text fast}/@{text best}/@{text meson} all fail. *}
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  by metis
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subsection {* Cantor's Theorem: There is no surjection from a set to its powerset *}
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lemma cantor1: "\<not> (\<exists>f:: 'a \<Rightarrow> 'a set. \<forall>S. \<exists>x. f x = S)"
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  -- {* Requires best-first search because it is undirectional. *}
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  by best
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schematic_lemma "\<forall>f:: 'a \<Rightarrow> 'a set. \<forall>x. f x \<noteq> ?S f"
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  -- {*This form displays the diagonal term. *}
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  by best
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schematic_lemma "?S \<notin> range (f :: 'a \<Rightarrow> 'a set)"
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  -- {* This form exploits the set constructs. *}
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  by (rule notI, erule rangeE, best)
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schematic_lemma "?S \<notin> range (f :: 'a \<Rightarrow> 'a set)"
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  -- {* Or just this! *}
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  by best
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subsection {* The Schröder-Berstein Theorem *}
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lemma disj_lemma: "- (f ` X) = g ` (-X) \<Longrightarrow> f a = g b \<Longrightarrow> a \<in> X \<Longrightarrow> b \<in> X"
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  by blast
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lemma surj_if_then_else:
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  "-(f ` X) = g ` (-X) \<Longrightarrow> surj (\<lambda>z. if z \<in> X then f z else g z)"
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  by (simp add: surj_def) blast
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lemma bij_if_then_else:
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  "inj_on f X \<Longrightarrow> inj_on g (-X) \<Longrightarrow> -(f ` X) = g ` (-X) \<Longrightarrow>
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    h = (\<lambda>z. if z \<in> X then f z else g z) \<Longrightarrow> inj h \<and> surj h"
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  apply (unfold inj_on_def)
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  apply (simp add: surj_if_then_else)
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  apply (blast dest: disj_lemma sym)
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  done
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lemma decomposition: "\<exists>X. X = - (g ` (- (f ` X)))"
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  apply (rule exI)
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  apply (rule lfp_unfold)
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  apply (rule monoI, blast)
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  done
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theorem Schroeder_Bernstein:
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  "inj (f :: 'a \<Rightarrow> 'b) \<Longrightarrow> inj (g :: 'b \<Rightarrow> 'a)
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    \<Longrightarrow> \<exists>h:: 'a \<Rightarrow> 'b. inj h \<and> surj h"
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  apply (rule decomposition [where f=f and g=g, THEN exE])
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  apply (rule_tac x = "(\<lambda>z. if z \<in> x then f z else inv g z)" in exI) 
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    --{*The term above can be synthesized by a sufficiently detailed proof.*}
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  apply (rule bij_if_then_else)
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     apply (rule_tac [4] refl)
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    apply (rule_tac [2] inj_on_inv_into)
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    apply (erule subset_inj_on [OF _ subset_UNIV])
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   apply blast
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  apply (erule ssubst, subst double_complement, erule inv_image_comp [symmetric])
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  done
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subsection {* A simple party theorem *}
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text{* \emph{At any party there are two people who know the same
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number of people}. Provided the party consists of at least two people
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and the knows relation is symmetric. Knowing yourself does not count
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--- otherwise knows needs to be reflexive. (From Freek Wiedijk's talk
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at TPHOLs 2007.) *}
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lemma equal_number_of_acquaintances:
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assumes "Domain R <= A" and "sym R" and "card A \<ge> 2"
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shows "\<not> inj_on (%a. card(R `` {a} - {a})) A"
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proof -
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  let ?N = "%a. card(R `` {a} - {a})"
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  let ?n = "card A"
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  have "finite A" using `card A \<ge> 2` by(auto intro:ccontr)
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  have 0: "R `` A <= A" using `sym R` `Domain R <= A`
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    unfolding Domain_def sym_def by blast
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  have h: "ALL a:A. R `` {a} <= A" using 0 by blast
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  hence 1: "ALL a:A. finite(R `` {a})" using `finite A`
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    by(blast intro: finite_subset)
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  have sub: "?N ` A <= {0..<?n}"
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  proof -
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    have "ALL a:A. R `` {a} - {a} < A" using h by blast
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    thus ?thesis using psubset_card_mono[OF `finite A`] by auto
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  qed
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  show "~ inj_on ?N A" (is "~ ?I")
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  proof
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    assume ?I
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    hence "?n = card(?N ` A)" by(rule card_image[symmetric])
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    with sub `finite A` have 2[simp]: "?N ` A = {0..<?n}"
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      using subset_card_intvl_is_intvl[of _ 0] by(auto)
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    have "0 : ?N ` A" and "?n - 1 : ?N ` A"  using `card A \<ge> 2` by simp+
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    then obtain a b where ab: "a:A" "b:A" and Na: "?N a = 0" and Nb: "?N b = ?n - 1"
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      by (auto simp del: 2)
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    have "a \<noteq> b" using Na Nb `card A \<ge> 2` by auto
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    have "R `` {a} - {a} = {}" by (metis 1 Na ab card_eq_0_iff finite_Diff)
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    hence "b \<notin> R `` {a}" using `a\<noteq>b` by blast
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    hence "a \<notin> R `` {b}" by (metis Image_singleton_iff assms(2) sym_def)
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    hence 3: "R `` {b} - {b} <= A - {a,b}" using 0 ab by blast
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    have 4: "finite (A - {a,b})" using `finite A` by simp
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    have "?N b <= ?n - 2" using ab `a\<noteq>b` `finite A` card_mono[OF 4 3] by simp
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    then show False using Nb `card A \<ge>  2` by arith
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  qed
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qed
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text {*
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  From W. W. Bledsoe and Guohui Feng, SET-VAR. JAR 11 (3), 1993, pages
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  293-314.
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  Isabelle can prove the easy examples without any special mechanisms,
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  but it can't prove the hard ones.
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*}
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lemma "\<exists>A. (\<forall>x \<in> A. x \<le> (0::int))"
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  -- {* Example 1, page 295. *}
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  by force
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lemma "D \<in> F \<Longrightarrow> \<exists>G. \<forall>A \<in> G. \<exists>B \<in> F. A \<subseteq> B"
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  -- {* Example 2. *}
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  by force
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lemma "P a \<Longrightarrow> \<exists>A. (\<forall>x \<in> A. P x) \<and> (\<exists>y. y \<in> A)"
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  -- {* Example 3. *}
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  by force
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lemma "a < b \<and> b < (c::int) \<Longrightarrow> \<exists>A. a \<notin> A \<and> b \<in> A \<and> c \<notin> A"
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  -- {* Example 4. *}
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  by auto --{*slow*}
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lemma "P (f b) \<Longrightarrow> \<exists>s A. (\<forall>x \<in> A. P x) \<and> f s \<in> A"
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  -- {*Example 5, page 298. *}
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  by force
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lemma "P (f b) \<Longrightarrow> \<exists>s A. (\<forall>x \<in> A. P x) \<and> f s \<in> A"
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  -- {* Example 6. *}
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  by force
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lemma "\<exists>A. a \<notin> A"
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  -- {* Example 7. *}
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  by force
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lemma "(\<forall>u v. u < (0::int) \<longrightarrow> u \<noteq> abs v)
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    \<longrightarrow> (\<exists>A::int set. -2 \<in> A & (\<forall>y. abs y \<notin> A))"
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  -- {* Example 8 needs a small hint. *}
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  by force
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    -- {* not @{text blast}, which can't simplify @{text "-2 < 0"} *}
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text {* Example 9 omitted (requires the reals). *}
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text {* The paper has no Example 10! *}
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lemma "(\<forall>A. 0 \<in> A \<and> (\<forall>x \<in> A. Suc x \<in> A) \<longrightarrow> n \<in> A) \<and>
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  P 0 \<and> (\<forall>x. P x \<longrightarrow> P (Suc x)) \<longrightarrow> P n"
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  -- {* Example 11: needs a hint. *}
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by(metis nat.induct)
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lemma
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  "(\<forall>A. (0, 0) \<in> A \<and> (\<forall>x y. (x, y) \<in> A \<longrightarrow> (Suc x, Suc y) \<in> A) \<longrightarrow> (n, m) \<in> A)
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    \<and> P n \<longrightarrow> P m"
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  -- {* Example 12. *}
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  by auto
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lemma
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  "(\<forall>x. (\<exists>u. x = 2 * u) = (\<not> (\<exists>v. Suc x = 2 * v))) \<longrightarrow>
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    (\<exists>A. \<forall>x. (x \<in> A) = (Suc x \<notin> A))"
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  -- {* Example EO1: typo in article, and with the obvious fix it seems
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      to require arithmetic reasoning. *}
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  apply clarify
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  apply (rule_tac x = "{x. \<exists>u. x = 2 * u}" in exI, auto)
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   apply metis+
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  done
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end