diff -r 71af1fd6a5e4 -r be3e1cc5005c src/HOL/Hahn_Banach/Zorn_Lemma.thy
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/src/HOL/Hahn_Banach/Zorn_Lemma.thy	Wed Jun 24 21:46:54 2009 +0200
@@ -0,0 +1,57 @@
+(*  Title:      HOL/Hahn_Banach/Zorn_Lemma.thy
+    Author:     Gertrud Bauer, TU Munich
+*)
+
+header {* Zorn's Lemma *}
+
+theory Zorn_Lemma
+imports Zorn
+begin
+
+text {*
+  Zorn's Lemmas states: if every linear ordered subset of an ordered
+  set @{text S} has an upper bound in @{text S}, then there exists a
+  maximal element in @{text S}.  In our application, @{text S} is a
+  set of sets ordered by set inclusion. Since the union of a chain of
+  sets is an upper bound for all elements of the chain, the conditions
+  of Zorn's lemma can be modified: if @{text S} is non-empty, it
+  suffices to show that for every non-empty chain @{text c} in @{text
+  S} the union of @{text c} also lies in @{text S}.
+*}
+
+theorem Zorn's_Lemma:
+  assumes r: "\<And>c. c \<in> chain S \<Longrightarrow> \<exists>x. x \<in> c \<Longrightarrow> \<Union>c \<in> S"
+    and aS: "a \<in> S"
+  shows "\<exists>y \<in> S. \<forall>z \<in> S. y \<subseteq> z \<longrightarrow> y = z"
+proof (rule Zorn_Lemma2)
+  show "\<forall>c \<in> chain S. \<exists>y \<in> S. \<forall>z \<in> c. z \<subseteq> y"
+  proof
+    fix c assume "c \<in> chain S"
+    show "\<exists>y \<in> S. \<forall>z \<in> c. z \<subseteq> y"
+    proof cases
+
+      txt {* If @{text c} is an empty chain, then every element in
+	@{text S} is an upper bound of @{text c}. *}
+
+      assume "c = {}"
+      with aS show ?thesis by fast
+
+      txt {* If @{text c} is non-empty, then @{text "\<Union>c"} is an upper
+	bound of @{text c}, lying in @{text S}. *}
+
+    next
+      assume "c \<noteq> {}"
+      show ?thesis
+      proof
+        show "\<forall>z \<in> c. z \<subseteq> \<Union>c" by fast
+        show "\<Union>c \<in> S"
+        proof (rule r)
+          from `c \<noteq> {}` show "\<exists>x. x \<in> c" by fast
+	  show "c \<in> chain S" by fact
+        qed
+      qed
+    qed
+  qed
+qed
+
+end