# HG changeset patch
# User blanchet
# Date 1325513746 -3600
# Node ID 86e6e9d42ad7abc1ae823397959cfd2fa441941a
# Parent  a109eb27f54f123171ce66b0463b17299e9a48e7
ported "Sets" example to "set" type constructor

diff -r a109eb27f54f -r 86e6e9d42ad7 src/HOL/Metis_Examples/Sets.thy
--- a/src/HOL/Metis_Examples/Sets.thy	Mon Jan 02 15:08:40 2012 +0100
+++ b/src/HOL/Metis_Examples/Sets.thy	Mon Jan 02 15:15:46 2012 +0100
@@ -24,50 +24,50 @@
 sledgehammer_params [isar_proof, isar_shrink_factor = 1]
 
 (*multiple versions of this example*)
-(* lemma (*equal_union: *)
+lemma (*equal_union: *)
    "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))"
 proof -
-  have F1: "\<forall>(x\<^isub>2\<Colon>'b \<Rightarrow> bool) x\<^isub>1\<Colon>'b \<Rightarrow> bool. x\<^isub>1 \<subseteq> x\<^isub>1 \<union> x\<^isub>2" by (metis Un_commute Un_upper2)
-  have F2a: "\<forall>(x\<^isub>2\<Colon>'b \<Rightarrow> bool) x\<^isub>1\<Colon>'b \<Rightarrow> bool. x\<^isub>1 \<subseteq> x\<^isub>2 \<longrightarrow> x\<^isub>2 = x\<^isub>2 \<union> x\<^isub>1" by (metis Un_commute subset_Un_eq)
-  have F2: "\<forall>(x\<^isub>2\<Colon>'b \<Rightarrow> bool) x\<^isub>1\<Colon>'b \<Rightarrow> bool. x\<^isub>1 \<subseteq> x\<^isub>2 \<and> x\<^isub>2 \<subseteq> x\<^isub>1 \<longrightarrow> x\<^isub>1 = x\<^isub>2" by (metis F2a subset_Un_eq)
+  have F1: "\<forall>(x\<^isub>2\<Colon>'b set) x\<^isub>1\<Colon>'b set. x\<^isub>1 \<subseteq> x\<^isub>1 \<union> x\<^isub>2" by (metis Un_commute Un_upper2)
+  have F2a: "\<forall>(x\<^isub>2\<Colon>'b set) x\<^isub>1\<Colon>'b set. x\<^isub>1 \<subseteq> x\<^isub>2 \<longrightarrow> x\<^isub>2 = x\<^isub>2 \<union> x\<^isub>1" by (metis Un_commute subset_Un_eq)
+  have F2: "\<forall>(x\<^isub>2\<Colon>'b set) x\<^isub>1\<Colon>'b set. x\<^isub>1 \<subseteq> x\<^isub>2 \<and> x\<^isub>2 \<subseteq> x\<^isub>1 \<longrightarrow> x\<^isub>1 = x\<^isub>2" by (metis F2a subset_Un_eq)
   { assume "\<not> Z \<subseteq> X"
-    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
+    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
   moreover
   { assume AA1: "Y \<union> Z \<noteq> X"
     { assume "\<not> Y \<subseteq> X"
-      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis F1) }
+      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis F1) }
     moreover
     { assume AAA1: "Y \<subseteq> X \<and> Y \<union> Z \<noteq> X"
       { assume "\<not> Z \<subseteq> X"
-        hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
+        hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
       moreover
       { assume "(Z \<subseteq> X \<and> Y \<subseteq> X) \<and> Y \<union> Z \<noteq> X"
         hence "Y \<union> Z \<subseteq> X \<and> X \<noteq> Y \<union> Z" by (metis Un_subset_iff)
         hence "Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> Y \<union> Z" by (metis F2)
-        hence "\<exists>x\<^isub>1\<Colon>'a \<Rightarrow> bool. Y \<subseteq> x\<^isub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^isub>1 \<union> Z" by (metis F1)
-        hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
-      ultimately have "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AAA1) }
-    ultimately have "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AA1) }
+        hence "\<exists>x\<^isub>1\<Colon>'a set. Y \<subseteq> x\<^isub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^isub>1 \<union> Z" by (metis F1)
+        hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
+      ultimately have "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AAA1) }
+    ultimately have "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AA1) }
   moreover
-  { assume "\<exists>x\<^isub>1\<Colon>'a \<Rightarrow> bool. (Z \<subseteq> x\<^isub>1 \<and> Y \<subseteq> x\<^isub>1) \<and> \<not> X \<subseteq> x\<^isub>1"
+  { assume "\<exists>x\<^isub>1\<Colon>'a set. (Z \<subseteq> x\<^isub>1 \<and> Y \<subseteq> x\<^isub>1) \<and> \<not> X \<subseteq> x\<^isub>1"
     { assume "\<not> Y \<subseteq> X"
-      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis F1) }
+      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis F1) }
     moreover
     { assume AAA1: "Y \<subseteq> X \<and> Y \<union> Z \<noteq> X"
       { assume "\<not> Z \<subseteq> X"
-        hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
+        hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
       moreover
       { assume "(Z \<subseteq> X \<and> Y \<subseteq> X) \<and> Y \<union> Z \<noteq> X"
         hence "Y \<union> Z \<subseteq> X \<and> X \<noteq> Y \<union> Z" by (metis Un_subset_iff)
         hence "Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> Y \<union> Z" by (metis F2)
-        hence "\<exists>x\<^isub>1\<Colon>'a \<Rightarrow> bool. Y \<subseteq> x\<^isub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^isub>1 \<union> Z" by (metis F1)
-        hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
-      ultimately have "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AAA1) }
-    ultimately have "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by blast }
+        hence "\<exists>x\<^isub>1\<Colon>'a set. Y \<subseteq> x\<^isub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^isub>1 \<union> Z" by (metis F1)
+        hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
+      ultimately have "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AAA1) }
+    ultimately have "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by blast }
   moreover
   { assume "\<not> Y \<subseteq> X"
-    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis F1) }
-  ultimately show "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by metis
+    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis F1) }
+  ultimately show "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by metis
 qed
 
 sledgehammer_params [isar_proof, isar_shrink_factor = 2]
@@ -75,36 +75,36 @@
 lemma (*equal_union: *)
    "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))"
 proof -
-  have F1: "\<forall>(x\<^isub>2\<Colon>'b \<Rightarrow> bool) x\<^isub>1\<Colon>'b \<Rightarrow> bool. x\<^isub>1 \<subseteq> x\<^isub>2 \<and> x\<^isub>2 \<subseteq> x\<^isub>1 \<longrightarrow> x\<^isub>1 = x\<^isub>2" by (metis Un_commute subset_Un_eq)
-  { assume AA1: "\<exists>x\<^isub>1\<Colon>'a \<Rightarrow> bool. (Z \<subseteq> x\<^isub>1 \<and> Y \<subseteq> x\<^isub>1) \<and> \<not> X \<subseteq> x\<^isub>1"
+  have F1: "\<forall>(x\<^isub>2\<Colon>'b set) x\<^isub>1\<Colon>'b set. x\<^isub>1 \<subseteq> x\<^isub>2 \<and> x\<^isub>2 \<subseteq> x\<^isub>1 \<longrightarrow> x\<^isub>1 = x\<^isub>2" by (metis Un_commute subset_Un_eq)
+  { assume AA1: "\<exists>x\<^isub>1\<Colon>'a set. (Z \<subseteq> x\<^isub>1 \<and> Y \<subseteq> x\<^isub>1) \<and> \<not> X \<subseteq> x\<^isub>1"
     { assume AAA1: "Y \<subseteq> X \<and> Y \<union> Z \<noteq> X"
       { assume "\<not> Z \<subseteq> X"
-        hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
+        hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
       moreover
       { assume "Y \<union> Z \<subseteq> X \<and> X \<noteq> Y \<union> Z"
-        hence "\<exists>x\<^isub>1\<Colon>'a \<Rightarrow> bool. Y \<subseteq> x\<^isub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^isub>1 \<union> Z" by (metis F1 Un_commute Un_upper2)
-        hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
-      ultimately have "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AAA1 Un_subset_iff) }
+        hence "\<exists>x\<^isub>1\<Colon>'a set. Y \<subseteq> x\<^isub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^isub>1 \<union> Z" by (metis F1 Un_commute Un_upper2)
+        hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
+      ultimately have "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AAA1 Un_subset_iff) }
     moreover
     { assume "\<not> Y \<subseteq> X"
-      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_commute Un_upper2) }
-    ultimately have "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AA1 Un_subset_iff) }
+      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_commute Un_upper2) }
+    ultimately have "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AA1 Un_subset_iff) }
   moreover
   { assume "\<not> Z \<subseteq> X"
-    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
+    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
   moreover
   { assume "\<not> Y \<subseteq> X"
-    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_commute Un_upper2) }
+    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_commute Un_upper2) }
   moreover
   { assume AA1: "Y \<subseteq> X \<and> Y \<union> Z \<noteq> X"
     { assume "\<not> Z \<subseteq> X"
-      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
+      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
     moreover
     { assume "Y \<union> Z \<subseteq> X \<and> X \<noteq> Y \<union> Z"
-      hence "\<exists>x\<^isub>1\<Colon>'a \<Rightarrow> bool. Y \<subseteq> x\<^isub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^isub>1 \<union> Z" by (metis F1 Un_commute Un_upper2)
-      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
-    ultimately have "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AA1 Un_subset_iff) }
-  ultimately show "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by metis
+      hence "\<exists>x\<^isub>1\<Colon>'a set. Y \<subseteq> x\<^isub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^isub>1 \<union> Z" by (metis F1 Un_commute Un_upper2)
+      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
+    ultimately have "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AA1 Un_subset_iff) }
+  ultimately show "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by metis
 qed
 
 sledgehammer_params [isar_proof, isar_shrink_factor = 3]
@@ -112,16 +112,16 @@
 lemma (*equal_union: *)
    "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))"
 proof -
-  have F1a: "\<forall>(x\<^isub>2\<Colon>'b \<Rightarrow> bool) x\<^isub>1\<Colon>'b \<Rightarrow> bool. x\<^isub>1 \<subseteq> x\<^isub>2 \<longrightarrow> x\<^isub>2 = x\<^isub>2 \<union> x\<^isub>1" by (metis Un_commute subset_Un_eq)
-  have F1: "\<forall>(x\<^isub>2\<Colon>'b \<Rightarrow> bool) x\<^isub>1\<Colon>'b \<Rightarrow> bool. x\<^isub>1 \<subseteq> x\<^isub>2 \<and> x\<^isub>2 \<subseteq> x\<^isub>1 \<longrightarrow> x\<^isub>1 = x\<^isub>2" by (metis F1a subset_Un_eq)
+  have F1a: "\<forall>(x\<^isub>2\<Colon>'b set) x\<^isub>1\<Colon>'b set. x\<^isub>1 \<subseteq> x\<^isub>2 \<longrightarrow> x\<^isub>2 = x\<^isub>2 \<union> x\<^isub>1" by (metis Un_commute subset_Un_eq)
+  have F1: "\<forall>(x\<^isub>2\<Colon>'b set) x\<^isub>1\<Colon>'b set. x\<^isub>1 \<subseteq> x\<^isub>2 \<and> x\<^isub>2 \<subseteq> x\<^isub>1 \<longrightarrow> x\<^isub>1 = x\<^isub>2" by (metis F1a subset_Un_eq)
   { assume "(Z \<subseteq> X \<and> Y \<subseteq> X) \<and> Y \<union> Z \<noteq> X"
-    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis F1 Un_commute Un_subset_iff Un_upper2) }
+    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis F1 Un_commute Un_subset_iff Un_upper2) }
   moreover
-  { assume AA1: "\<exists>x\<^isub>1\<Colon>'a \<Rightarrow> bool. (Z \<subseteq> x\<^isub>1 \<and> Y \<subseteq> x\<^isub>1) \<and> \<not> X \<subseteq> x\<^isub>1"
+  { assume AA1: "\<exists>x\<^isub>1\<Colon>'a set. (Z \<subseteq> x\<^isub>1 \<and> Y \<subseteq> x\<^isub>1) \<and> \<not> X \<subseteq> x\<^isub>1"
     { assume "(Z \<subseteq> X \<and> Y \<subseteq> X) \<and> Y \<union> Z \<noteq> X"
-      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis F1 Un_commute Un_subset_iff Un_upper2) }
-    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AA1 Un_commute Un_subset_iff Un_upper2) }
-  ultimately show "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_commute Un_upper2)
+      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis F1 Un_commute Un_subset_iff Un_upper2) }
+    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AA1 Un_commute Un_subset_iff Un_upper2) }
+  ultimately show "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_commute Un_upper2)
 qed
 
 sledgehammer_params [isar_proof, isar_shrink_factor = 4]
@@ -129,15 +129,15 @@
 lemma (*equal_union: *)
    "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))"
 proof -
-  have F1: "\<forall>(x\<^isub>2\<Colon>'b \<Rightarrow> bool) x\<^isub>1\<Colon>'b \<Rightarrow> bool. x\<^isub>1 \<subseteq> x\<^isub>2 \<and> x\<^isub>2 \<subseteq> x\<^isub>1 \<longrightarrow> x\<^isub>1 = x\<^isub>2" by (metis Un_commute subset_Un_eq)
+  have F1: "\<forall>(x\<^isub>2\<Colon>'b set) x\<^isub>1\<Colon>'b set. x\<^isub>1 \<subseteq> x\<^isub>2 \<and> x\<^isub>2 \<subseteq> x\<^isub>1 \<longrightarrow> x\<^isub>1 = x\<^isub>2" by (metis Un_commute subset_Un_eq)
   { assume "\<not> Y \<subseteq> X"
-    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_commute Un_upper2) }
+    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_commute Un_upper2) }
   moreover
   { assume AA1: "Y \<subseteq> X \<and> Y \<union> Z \<noteq> X"
-    { assume "\<exists>x\<^isub>1\<Colon>'a \<Rightarrow> bool. Y \<subseteq> x\<^isub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^isub>1 \<union> Z"
-      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
-    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AA1 F1 Un_commute Un_subset_iff Un_upper2) }
-  ultimately show "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a \<Rightarrow> bool. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_subset_iff Un_upper2)
+    { assume "\<exists>x\<^isub>1\<Colon>'a set. Y \<subseteq> x\<^isub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^isub>1 \<union> Z"
+      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_upper2) }
+    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AA1 F1 Un_commute Un_subset_iff Un_upper2) }
+  ultimately show "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V\<Colon>'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_subset_iff Un_upper2)
 qed
 
 sledgehammer_params [isar_proof, isar_shrink_factor = 1]
@@ -172,8 +172,8 @@
   assume "\<forall>x \<in> S. \<Union>S \<subseteq> x"
   hence "\<forall>x\<^isub>1. x\<^isub>1 \<subseteq> \<Union>S \<and> x\<^isub>1 \<in> S \<longrightarrow> x\<^isub>1 = \<Union>S" by (metis set_eq_subset)
   hence "\<forall>x\<^isub>1. x\<^isub>1 \<in> S \<longrightarrow> x\<^isub>1 = \<Union>S" by (metis Union_upper)
-  hence "\<forall>x\<^isub>1\<Colon>('a \<Rightarrow> bool) \<Rightarrow> bool. \<Union>S \<in> x\<^isub>1 \<longrightarrow> S \<subseteq> x\<^isub>1" by (metis subsetI)
-  hence "\<forall>x\<^isub>1\<Colon>('a \<Rightarrow> bool) \<Rightarrow> bool. S \<subseteq> insert (\<Union>S) x\<^isub>1" by (metis insert_iff)
+  hence "\<forall>x\<^isub>1\<Colon>('a set) set. \<Union>S \<in> x\<^isub>1 \<longrightarrow> S \<subseteq> x\<^isub>1" by (metis subsetI)
+  hence "\<forall>x\<^isub>1\<Colon>('a set) set. S \<subseteq> insert (\<Union>S) x\<^isub>1" by (metis insert_iff)
   thus "\<exists>z. S \<subseteq> {z}" by metis
 qed
 
@@ -194,12 +194,11 @@
       "(\<forall>C. (0, 0) \<in> C \<and> (\<forall>x y. (x, y) \<in> C \<longrightarrow> (Suc x, Suc y) \<in> C) \<longrightarrow> (n, m) \<in> C) \<and> Q n \<longrightarrow> Q m"
        apply (metis all_not_in_conv)
       apply (metis all_not_in_conv)
-     apply (metis mem_def)
+     apply (metis mem_Collect_eq)
     apply (metis less_int_def singleton_iff)
-   apply (metis mem_def)
-  apply (metis mem_def)
+   apply (metis mem_Collect_eq)
+  apply (metis mem_Collect_eq)
  apply (metis all_not_in_conv)
 by (metis pair_in_Id_conv)
-*)
 
 end