src/HOL/NSA/Filter.thy
author huffman
Fri Mar 30 12:32:35 2012 +0200 (2012-03-30)
changeset 47220 52426c62b5d0
parent 46008 c296c75f4cf4
child 47486 4d49f3ffe97e
permissions -rw-r--r--
replace lemmas eval_nat_numeral with a simpler reformulation
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(*  Title:      HOL/NSA/Filter.thy
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    Author:     Jacques D. Fleuriot, University of Cambridge
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    Author:     Lawrence C Paulson
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    Author:     Brian Huffman
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*) 
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header {* Filters and Ultrafilters *}
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theory Filter
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imports "~~/src/HOL/Library/Zorn" "~~/src/HOL/Library/Infinite_Set"
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begin
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subsection {* Definitions and basic properties *}
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subsubsection {* Filters *}
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locale filter =
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  fixes F :: "'a set set"
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  assumes UNIV [iff]:  "UNIV \<in> F"
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  assumes empty [iff]: "{} \<notin> F"
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  assumes Int:         "\<lbrakk>u \<in> F; v \<in> F\<rbrakk> \<Longrightarrow> u \<inter> v \<in> F"
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  assumes subset:      "\<lbrakk>u \<in> F; u \<subseteq> v\<rbrakk> \<Longrightarrow> v \<in> F"
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lemma (in filter) memD: "A \<in> F \<Longrightarrow> - A \<notin> F"
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proof
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  assume "A \<in> F" and "- A \<in> F"
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  hence "A \<inter> (- A) \<in> F" by (rule Int)
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  thus "False" by simp
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qed
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lemma (in filter) not_memI: "- A \<in> F \<Longrightarrow> A \<notin> F"
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by (drule memD, simp)
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lemma (in filter) Int_iff: "(x \<inter> y \<in> F) = (x \<in> F \<and> y \<in> F)"
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by (auto elim: subset intro: Int)
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subsubsection {* Ultrafilters *}
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locale ultrafilter = filter +
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  assumes ultra: "A \<in> F \<or> - A \<in> F"
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lemma (in ultrafilter) memI: "- A \<notin> F \<Longrightarrow> A \<in> F"
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by (cut_tac ultra [of A], simp)
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lemma (in ultrafilter) not_memD: "A \<notin> F \<Longrightarrow> - A \<in> F"
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by (rule memI, simp)
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lemma (in ultrafilter) not_mem_iff: "(A \<notin> F) = (- A \<in> F)"
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by (rule iffI [OF not_memD not_memI])
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lemma (in ultrafilter) Compl_iff: "(- A \<in> F) = (A \<notin> F)"
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by (rule iffI [OF not_memI not_memD])
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lemma (in ultrafilter) Un_iff: "(x \<union> y \<in> F) = (x \<in> F \<or> y \<in> F)"
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 apply (rule iffI)
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  apply (erule contrapos_pp)
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  apply (simp add: Int_iff not_mem_iff)
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 apply (auto elim: subset)
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done
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subsubsection {* Free Ultrafilters *}
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locale freeultrafilter = ultrafilter +
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  assumes infinite: "A \<in> F \<Longrightarrow> infinite A"
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lemma (in freeultrafilter) finite: "finite A \<Longrightarrow> A \<notin> F"
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by (erule contrapos_pn, erule infinite)
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lemma (in freeultrafilter) singleton: "{x} \<notin> F"
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by (rule finite, simp)
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lemma (in freeultrafilter) insert_iff [simp]: "(insert x A \<in> F) = (A \<in> F)"
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apply (subst insert_is_Un)
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apply (subst Un_iff)
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apply (simp add: singleton)
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done
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lemma (in freeultrafilter) filter: "filter F" ..
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lemma (in freeultrafilter) ultrafilter: "ultrafilter F" ..
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subsection {* Collect properties *}
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lemma (in filter) Collect_ex:
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  "({n. \<exists>x. P n x} \<in> F) = (\<exists>X. {n. P n (X n)} \<in> F)"
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proof
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  assume "{n. \<exists>x. P n x} \<in> F"
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  hence "{n. P n (SOME x. P n x)} \<in> F"
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    by (auto elim: someI subset)
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  thus "\<exists>X. {n. P n (X n)} \<in> F" by fast
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next
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  show "\<exists>X. {n. P n (X n)} \<in> F \<Longrightarrow> {n. \<exists>x. P n x} \<in> F"
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    by (auto elim: subset)
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qed
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lemma (in filter) Collect_conj:
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  "({n. P n \<and> Q n} \<in> F) = ({n. P n} \<in> F \<and> {n. Q n} \<in> F)"
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by (subst Collect_conj_eq, rule Int_iff)
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lemma (in ultrafilter) Collect_not:
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  "({n. \<not> P n} \<in> F) = ({n. P n} \<notin> F)"
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by (subst Collect_neg_eq, rule Compl_iff)
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lemma (in ultrafilter) Collect_disj:
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  "({n. P n \<or> Q n} \<in> F) = ({n. P n} \<in> F \<or> {n. Q n} \<in> F)"
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by (subst Collect_disj_eq, rule Un_iff)
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lemma (in ultrafilter) Collect_all:
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  "({n. \<forall>x. P n x} \<in> F) = (\<forall>X. {n. P n (X n)} \<in> F)"
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apply (rule Not_eq_iff [THEN iffD1])
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apply (simp add: Collect_not [symmetric])
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apply (rule Collect_ex)
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done
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subsection {* Maximal filter = Ultrafilter *}
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text {*
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   A filter F is an ultrafilter iff it is a maximal filter,
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   i.e. whenever G is a filter and @{term "F \<subseteq> G"} then @{term "F = G"}
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*}
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text {*
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  Lemmas that shows existence of an extension to what was assumed to
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  be a maximal filter. Will be used to derive contradiction in proof of
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  property of ultrafilter.
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*}
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lemma extend_lemma1: "UNIV \<in> F \<Longrightarrow> A \<in> {X. \<exists>f\<in>F. A \<inter> f \<subseteq> X}"
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by blast
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lemma extend_lemma2: "F \<subseteq> {X. \<exists>f\<in>F. A \<inter> f \<subseteq> X}"
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by blast
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lemma (in filter) extend_filter:
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assumes A: "- A \<notin> F"
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shows "filter {X. \<exists>f\<in>F. A \<inter> f \<subseteq> X}" (is "filter ?X")
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proof (rule filter.intro)
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  show "UNIV \<in> ?X" by blast
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next
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  show "{} \<notin> ?X"
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  proof (clarify)
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    fix f assume f: "f \<in> F" and Af: "A \<inter> f \<subseteq> {}"
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    from Af have fA: "f \<subseteq> - A" by blast
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    from f fA have "- A \<in> F" by (rule subset)
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    with A show "False" by simp
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  qed
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next
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  fix u and v
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  assume u: "u \<in> ?X" and v: "v \<in> ?X"
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  from u obtain f where f: "f \<in> F" and Af: "A \<inter> f \<subseteq> u" by blast
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  from v obtain g where g: "g \<in> F" and Ag: "A \<inter> g \<subseteq> v" by blast
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  from f g have fg: "f \<inter> g \<in> F" by (rule Int)
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  from Af Ag have Afg: "A \<inter> (f \<inter> g) \<subseteq> u \<inter> v" by blast
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  from fg Afg show "u \<inter> v \<in> ?X" by blast
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next
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  fix u and v
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  assume uv: "u \<subseteq> v" and u: "u \<in> ?X"
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  from u obtain f where f: "f \<in> F" and Afu: "A \<inter> f \<subseteq> u" by blast
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  from Afu uv have Afv: "A \<inter> f \<subseteq> v" by blast
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  from f Afv have "\<exists>f\<in>F. A \<inter> f \<subseteq> v" by blast
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  thus "v \<in> ?X" by simp
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qed
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lemma (in filter) max_filter_ultrafilter:
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assumes max: "\<And>G. \<lbrakk>filter G; F \<subseteq> G\<rbrakk> \<Longrightarrow> F = G"
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shows "ultrafilter_axioms F"
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proof (rule ultrafilter_axioms.intro)
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  fix A show "A \<in> F \<or> - A \<in> F"
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  proof (rule disjCI)
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    let ?X = "{X. \<exists>f\<in>F. A \<inter> f \<subseteq> X}"
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    assume AF: "- A \<notin> F"
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    from AF have X: "filter ?X" by (rule extend_filter)
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    from UNIV have AX: "A \<in> ?X" by (rule extend_lemma1)
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    have FX: "F \<subseteq> ?X" by (rule extend_lemma2)
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    from X FX have "F = ?X" by (rule max)
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    with AX show "A \<in> F" by simp
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  qed
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qed
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lemma (in ultrafilter) max_filter:
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assumes G: "filter G" and sub: "F \<subseteq> G" shows "F = G"
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proof
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  show "F \<subseteq> G" using sub .
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  show "G \<subseteq> F"
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  proof
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    fix A assume A: "A \<in> G"
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    from G A have "- A \<notin> G" by (rule filter.memD)
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    with sub have B: "- A \<notin> F" by blast
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    thus "A \<in> F" by (rule memI)
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  qed
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qed
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subsection {* Ultrafilter Theorem *}
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text "A locale makes proof of ultrafilter Theorem more modular"
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locale UFT =
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  fixes   frechet :: "'a set set"
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  and     superfrechet :: "'a set set set"
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  assumes infinite_UNIV: "infinite (UNIV :: 'a set)"
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  defines frechet_def: "frechet \<equiv> {A. finite (- A)}"
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  and     superfrechet_def: "superfrechet \<equiv> {G. filter G \<and> frechet \<subseteq> G}"
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lemma (in UFT) superfrechetI:
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  "\<lbrakk>filter G; frechet \<subseteq> G\<rbrakk> \<Longrightarrow> G \<in> superfrechet"
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by (simp add: superfrechet_def)
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lemma (in UFT) superfrechetD1:
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  "G \<in> superfrechet \<Longrightarrow> filter G"
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by (simp add: superfrechet_def)
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lemma (in UFT) superfrechetD2:
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  "G \<in> superfrechet \<Longrightarrow> frechet \<subseteq> G"
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by (simp add: superfrechet_def)
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text {* A few properties of free filters *}
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lemma filter_cofinite:
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assumes inf: "infinite (UNIV :: 'a set)"
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shows "filter {A:: 'a set. finite (- A)}" (is "filter ?F")
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proof (rule filter.intro)
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  show "UNIV \<in> ?F" by simp
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next
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  show "{} \<notin> ?F" using inf by simp
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next
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  fix u v assume "u \<in> ?F" and "v \<in> ?F"
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  thus "u \<inter> v \<in> ?F" by simp
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next
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  fix u v assume uv: "u \<subseteq> v" and u: "u \<in> ?F"
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  from uv have vu: "- v \<subseteq> - u" by simp
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  from u show "v \<in> ?F"
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    by (simp add: finite_subset [OF vu])
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qed
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text {*
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   We prove: 1. Existence of maximal filter i.e. ultrafilter;
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             2. Freeness property i.e ultrafilter is free.
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             Use a locale to prove various lemmas and then 
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             export main result: The ultrafilter Theorem
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*}
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lemma (in UFT) filter_frechet: "filter frechet"
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by (unfold frechet_def, rule filter_cofinite [OF infinite_UNIV])
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lemma (in UFT) frechet_in_superfrechet: "frechet \<in> superfrechet"
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by (rule superfrechetI [OF filter_frechet subset_refl])
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lemma (in UFT) lemma_mem_chain_filter:
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  "\<lbrakk>c \<in> chain superfrechet; x \<in> c\<rbrakk> \<Longrightarrow> filter x"
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by (unfold chain_def superfrechet_def, blast)
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subsubsection {* Unions of chains of superfrechets *}
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text "In this section we prove that superfrechet is closed
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with respect to unions of non-empty chains. We must show
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  1) Union of a chain is a filter,
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  2) Union of a chain contains frechet.
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Number 2 is trivial, but 1 requires us to prove all the filter rules."
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lemma (in UFT) Union_chain_UNIV:
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"\<lbrakk>c \<in> chain superfrechet; c \<noteq> {}\<rbrakk> \<Longrightarrow> UNIV \<in> \<Union>c"
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proof -
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  assume 1: "c \<in> chain superfrechet" and 2: "c \<noteq> {}"
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  from 2 obtain x where 3: "x \<in> c" by blast
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  from 1 3 have "filter x" by (rule lemma_mem_chain_filter)
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  hence "UNIV \<in> x" by (rule filter.UNIV)
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  with 3 show "UNIV \<in> \<Union>c" by blast
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qed
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lemma (in UFT) Union_chain_empty:
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"c \<in> chain superfrechet \<Longrightarrow> {} \<notin> \<Union>c"
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proof
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  assume 1: "c \<in> chain superfrechet" and 2: "{} \<in> \<Union>c"
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  from 2 obtain x where 3: "x \<in> c" and 4: "{} \<in> x" ..
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  from 1 3 have "filter x" by (rule lemma_mem_chain_filter)
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  hence "{} \<notin> x" by (rule filter.empty)
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  with 4 show "False" by simp
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qed
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lemma (in UFT) Union_chain_Int:
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"\<lbrakk>c \<in> chain superfrechet; u \<in> \<Union>c; v \<in> \<Union>c\<rbrakk> \<Longrightarrow> u \<inter> v \<in> \<Union>c"
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proof -
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  assume c: "c \<in> chain superfrechet"
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  assume "u \<in> \<Union>c"
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    then obtain x where ux: "u \<in> x" and xc: "x \<in> c" ..
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  assume "v \<in> \<Union>c"
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    then obtain y where vy: "v \<in> y" and yc: "y \<in> c" ..
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  from c xc yc have "x \<subseteq> y \<or> y \<subseteq> x" by (rule chainD)
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  with xc yc have xyc: "x \<union> y \<in> c"
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    by (auto simp add: Un_absorb1 Un_absorb2)
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  with c have fxy: "filter (x \<union> y)" by (rule lemma_mem_chain_filter)
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  from ux have uxy: "u \<in> x \<union> y" by simp
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  from vy have vxy: "v \<in> x \<union> y" by simp
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  from fxy uxy vxy have "u \<inter> v \<in> x \<union> y" by (rule filter.Int)
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  with xyc show "u \<inter> v \<in> \<Union>c" ..
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qed
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lemma (in UFT) Union_chain_subset:
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"\<lbrakk>c \<in> chain superfrechet; u \<in> \<Union>c; u \<subseteq> v\<rbrakk> \<Longrightarrow> v \<in> \<Union>c"
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proof -
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  assume c: "c \<in> chain superfrechet"
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     and u: "u \<in> \<Union>c" and uv: "u \<subseteq> v"
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  from u obtain x where ux: "u \<in> x" and xc: "x \<in> c" ..
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  from c xc have fx: "filter x" by (rule lemma_mem_chain_filter)
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  from fx ux uv have vx: "v \<in> x" by (rule filter.subset)
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  with xc show "v \<in> \<Union>c" ..
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qed
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lemma (in UFT) Union_chain_filter:
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assumes chain: "c \<in> chain superfrechet" and nonempty: "c \<noteq> {}"
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shows "filter (\<Union>c)"
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proof (rule filter.intro)
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  show "UNIV \<in> \<Union>c" using chain nonempty by (rule Union_chain_UNIV)
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next
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  show "{} \<notin> \<Union>c" using chain by (rule Union_chain_empty)
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next
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  fix u v assume "u \<in> \<Union>c" and "v \<in> \<Union>c"
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  with chain show "u \<inter> v \<in> \<Union>c" by (rule Union_chain_Int)
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next
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  fix u v assume "u \<in> \<Union>c" and "u \<subseteq> v"
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  with chain show "v \<in> \<Union>c" by (rule Union_chain_subset)
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qed
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lemma (in UFT) lemma_mem_chain_frechet_subset:
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  "\<lbrakk>c \<in> chain superfrechet; x \<in> c\<rbrakk> \<Longrightarrow> frechet \<subseteq> x"
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by (unfold superfrechet_def chain_def, blast)
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lemma (in UFT) Union_chain_superfrechet:
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  "\<lbrakk>c \<noteq> {}; c \<in> chain superfrechet\<rbrakk> \<Longrightarrow> \<Union>c \<in> superfrechet"
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proof (rule superfrechetI)
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  assume 1: "c \<in> chain superfrechet" and 2: "c \<noteq> {}"
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  thus "filter (\<Union>c)" by (rule Union_chain_filter)
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  from 2 obtain x where 3: "x \<in> c" by blast
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  from 1 3 have "frechet \<subseteq> x" by (rule lemma_mem_chain_frechet_subset)
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  also from 3 have "x \<subseteq> \<Union>c" by blast
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  finally show "frechet \<subseteq> \<Union>c" .
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qed
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subsubsection {* Existence of free ultrafilter *}
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lemma (in UFT) max_cofinite_filter_Ex:
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  "\<exists>U\<in>superfrechet. \<forall>G\<in>superfrechet. U \<subseteq> G \<longrightarrow> U = G"
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proof (rule Zorn_Lemma2 [rule_format])
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  fix c assume c: "c \<in> chain superfrechet"
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  show "\<exists>U\<in>superfrechet. \<forall>G\<in>c. G \<subseteq> U" (is "?U")
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  proof (cases)
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    assume "c = {}"
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    with frechet_in_superfrechet show "?U" by blast
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  next
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    assume A: "c \<noteq> {}"
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    from A c have "\<Union>c \<in> superfrechet"
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      by (rule Union_chain_superfrechet)
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    thus "?U" by blast
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  qed
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qed
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lemma (in UFT) mem_superfrechet_all_infinite:
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  "\<lbrakk>U \<in> superfrechet; A \<in> U\<rbrakk> \<Longrightarrow> infinite A"
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proof
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  assume U: "U \<in> superfrechet" and A: "A \<in> U" and fin: "finite A"
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  from U have fil: "filter U" and fre: "frechet \<subseteq> U"
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    by (simp_all add: superfrechet_def)
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  from fin have "- A \<in> frechet" by (simp add: frechet_def)
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  with fre have cA: "- A \<in> U" by (rule subsetD)
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  from fil A cA have "A \<inter> - A \<in> U" by (rule filter.Int)
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  with fil show "False" by (simp add: filter.empty)
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qed
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text {* There exists a free ultrafilter on any infinite set *}
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lemma (in UFT) freeultrafilter_ex:
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  "\<exists>U::'a set set. freeultrafilter U"
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proof -
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  from max_cofinite_filter_Ex obtain U
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    where U: "U \<in> superfrechet"
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      and max [rule_format]: "\<forall>G\<in>superfrechet. U \<subseteq> G \<longrightarrow> U = G" ..
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  from U have fil: "filter U" by (rule superfrechetD1)
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  from U have fre: "frechet \<subseteq> U" by (rule superfrechetD2)
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  have ultra: "ultrafilter_axioms U"
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  proof (rule filter.max_filter_ultrafilter [OF fil])
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    fix G assume G: "filter G" and UG: "U \<subseteq> G"
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    from fre UG have "frechet \<subseteq> G" by simp
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    with G have "G \<in> superfrechet" by (rule superfrechetI)
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    from this UG show "U = G" by (rule max)
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  qed
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  have free: "freeultrafilter_axioms U"
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  proof (rule freeultrafilter_axioms.intro)
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    fix A assume "A \<in> U"
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    with U show "infinite A" by (rule mem_superfrechet_all_infinite)
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  qed
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  from fil ultra free have "freeultrafilter U"
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    by (rule freeultrafilter.intro [OF ultrafilter.intro])
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    (* FIXME: unfold_locales should use chained facts *)
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  then show ?thesis ..
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qed
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lemmas freeultrafilter_Ex = UFT.freeultrafilter_ex [OF UFT.intro]
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hide_const (open) filter
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end