src/HOL/Metis_Examples/Sets.thy

 
 (*multiple versions of this example*)
 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\<^sub>2\<Colon>'b set) x\<^sub>1\<Colon>'b set. x\<^sub>1 \<subseteq> x\<^sub>1 \<union> x\<^sub>2" by (metis Un_commute Un_upper2)
-  have F2a: "\<forall>(x\<^sub>2\<Colon>'b set) x\<^sub>1\<Colon>'b set. x\<^sub>1 \<subseteq> x\<^sub>2 \<longrightarrow> x\<^sub>2 = x\<^sub>2 \<union> x\<^sub>1" by (metis Un_commute subset_Un_eq)
-  have F2: "\<forall>(x\<^sub>2\<Colon>'b set) x\<^sub>1\<Colon>'b set. x\<^sub>1 \<subseteq> x\<^sub>2 \<and> x\<^sub>2 \<subseteq> x\<^sub>1 \<longrightarrow> x\<^sub>1 = x\<^sub>2" by (metis F2a subset_Un_eq)
+  have F1: "\<forall>(x\<^sub>2::'b set) x\<^sub>1::'b set. x\<^sub>1 \<subseteq> x\<^sub>1 \<union> x\<^sub>2" by (metis Un_commute Un_upper2)
+  have F2a: "\<forall>(x\<^sub>2::'b set) x\<^sub>1::'b set. x\<^sub>1 \<subseteq> x\<^sub>2 \<longrightarrow> x\<^sub>2 = x\<^sub>2 \<union> x\<^sub>1" by (metis Un_commute subset_Un_eq)
+  have F2: "\<forall>(x\<^sub>2::'b set) x\<^sub>1::'b set. x\<^sub>1 \<subseteq> x\<^sub>2 \<and> x\<^sub>2 \<subseteq> x\<^sub>1 \<longrightarrow> x\<^sub>1 = x\<^sub>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 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::'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 set. 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::'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 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::'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\<^sub>1\<Colon>'a set. Y \<subseteq> x\<^sub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^sub>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\<^sub>1\<Colon>'a set. (Z \<subseteq> x\<^sub>1 \<and> Y \<subseteq> x\<^sub>1) \<and> \<not> X \<subseteq> x\<^sub>1"
+        hence "\<exists>x\<^sub>1::'a set. Y \<subseteq> x\<^sub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^sub>1 \<union> Z" by (metis F1)
+        hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V::'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::'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::'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis AA1) }
+  moreover
+  { assume "\<exists>x\<^sub>1::'a set. (Z \<subseteq> x\<^sub>1 \<and> Y \<subseteq> x\<^sub>1) \<and> \<not> X \<subseteq> x\<^sub>1"
     { assume "\<not> Y \<subseteq> X"
-      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) }
+      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V::'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 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::'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\<^sub>1\<Colon>'a set. Y \<subseteq> x\<^sub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^sub>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 }
+        hence "\<exists>x\<^sub>1::'a set. Y \<subseteq> x\<^sub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^sub>1 \<union> Z" by (metis F1)
+        hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V::'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::'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::'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 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
+    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V::'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::'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by metis
 qed
 
 sledgehammer_params [isar_proofs, compress = 2]
 
 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\<^sub>2\<Colon>'b set) x\<^sub>1\<Colon>'b set. x\<^sub>1 \<subseteq> x\<^sub>2 \<and> x\<^sub>2 \<subseteq> x\<^sub>1 \<longrightarrow> x\<^sub>1 = x\<^sub>2" by (metis Un_commute subset_Un_eq)
-  { assume AA1: "\<exists>x\<^sub>1\<Colon>'a set. (Z \<subseteq> x\<^sub>1 \<and> Y \<subseteq> x\<^sub>1) \<and> \<not> X \<subseteq> x\<^sub>1"
+  have F1: "\<forall>(x\<^sub>2::'b set) x\<^sub>1::'b set. x\<^sub>1 \<subseteq> x\<^sub>2 \<and> x\<^sub>2 \<subseteq> x\<^sub>1 \<longrightarrow> x\<^sub>1 = x\<^sub>2" by (metis Un_commute subset_Un_eq)
+  { assume AA1: "\<exists>x\<^sub>1::'a set. (Z \<subseteq> x\<^sub>1 \<and> Y \<subseteq> x\<^sub>1) \<and> \<not> X \<subseteq> x\<^sub>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 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::'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\<^sub>1\<Colon>'a set. Y \<subseteq> x\<^sub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^sub>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) }
+        hence "\<exists>x\<^sub>1::'a set. Y \<subseteq> x\<^sub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^sub>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::'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::'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 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) }
+      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V::'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::'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 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::'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 set. 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::'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 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::'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\<^sub>1\<Colon>'a set. Y \<subseteq> x\<^sub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^sub>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
+      hence "\<exists>x\<^sub>1::'a set. Y \<subseteq> x\<^sub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^sub>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::'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::'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::'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by metis
 qed
 
 sledgehammer_params [isar_proofs, compress = 3]
 
 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\<^sub>2\<Colon>'b set) x\<^sub>1\<Colon>'b set. x\<^sub>1 \<subseteq> x\<^sub>2 \<longrightarrow> x\<^sub>2 = x\<^sub>2 \<union> x\<^sub>1" by (metis Un_commute subset_Un_eq)
-  have F1: "\<forall>(x\<^sub>2\<Colon>'b set) x\<^sub>1\<Colon>'b set. x\<^sub>1 \<subseteq> x\<^sub>2 \<and> x\<^sub>2 \<subseteq> x\<^sub>1 \<longrightarrow> x\<^sub>1 = x\<^sub>2" by (metis F1a subset_Un_eq)
+  have F1a: "\<forall>(x\<^sub>2::'b set) x\<^sub>1::'b set. x\<^sub>1 \<subseteq> x\<^sub>2 \<longrightarrow> x\<^sub>2 = x\<^sub>2 \<union> x\<^sub>1" by (metis Un_commute subset_Un_eq)
+  have F1: "\<forall>(x\<^sub>2::'b set) x\<^sub>1::'b set. x\<^sub>1 \<subseteq> x\<^sub>2 \<and> x\<^sub>2 \<subseteq> x\<^sub>1 \<longrightarrow> x\<^sub>1 = x\<^sub>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 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\<^sub>1\<Colon>'a set. (Z \<subseteq> x\<^sub>1 \<and> Y \<subseteq> x\<^sub>1) \<and> \<not> X \<subseteq> x\<^sub>1"
+    hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V::'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\<^sub>1::'a set. (Z \<subseteq> x\<^sub>1 \<and> Y \<subseteq> x\<^sub>1) \<and> \<not> X \<subseteq> x\<^sub>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 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)
+      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V::'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::'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::'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_commute Un_upper2)
 qed
 
 sledgehammer_params [isar_proofs, compress = 4]
 
 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\<^sub>2\<Colon>'b set) x\<^sub>1\<Colon>'b set. x\<^sub>1 \<subseteq> x\<^sub>2 \<and> x\<^sub>2 \<subseteq> x\<^sub>1 \<longrightarrow> x\<^sub>1 = x\<^sub>2" by (metis Un_commute subset_Un_eq)
+  have F1: "\<forall>(x\<^sub>2::'b set) x\<^sub>1::'b set. x\<^sub>1 \<subseteq> x\<^sub>2 \<and> x\<^sub>2 \<subseteq> x\<^sub>1 \<longrightarrow> x\<^sub>1 = x\<^sub>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 set. 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::'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\<^sub>1\<Colon>'a set. Y \<subseteq> x\<^sub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^sub>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)
+    { assume "\<exists>x\<^sub>1::'a set. Y \<subseteq> x\<^sub>1 \<union> Z \<and> Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> x\<^sub>1 \<union> Z"
+      hence "(X = Y \<union> Z) = (Y \<subseteq> X \<and> Z \<subseteq> X \<and> (\<forall>V::'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::'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::'a set. Y \<subseteq> V \<and> Z \<subseteq> V \<longrightarrow> X \<subseteq> V))" by (metis Un_subset_iff Un_upper2)
 qed
 
 sledgehammer_params [isar_proofs, compress = 1]
 
 lemma (*equal_union: *)

 lemma fixedpoint: "\<exists>!x. f (g x) = x \<Longrightarrow> \<exists>!y. g (f y) = y"
 by metis
 
 lemma (* fixedpoint: *) "\<exists>!x. f (g x) = x \<Longrightarrow> \<exists>!y. g (f y) = y"
 proof -
-  assume "\<exists>!x\<Colon>'a. f (g x) = x"
-  thus "\<exists>!y\<Colon>'b. g (f y) = y" by metis
+  assume "\<exists>!x::'a. f (g x) = x"
+  thus "\<exists>!y::'b. g (f y) = y" by metis
 qed
 
 lemma (* singleton_example_2: *)
      "\<forall>x \<in> S. \<Union>S \<subseteq> x \<Longrightarrow> \<exists>z. S \<subseteq> {z}"
 by (metis Set.subsetI Union_upper insertCI set_eq_subset)

      "\<forall>x \<in> S. \<Union>S \<subseteq> x \<Longrightarrow> \<exists>z. S \<subseteq> {z}"
 proof -
   assume "\<forall>x \<in> S. \<Union>S \<subseteq> x"
   hence "\<forall>x\<^sub>1. x\<^sub>1 \<subseteq> \<Union>S \<and> x\<^sub>1 \<in> S \<longrightarrow> x\<^sub>1 = \<Union>S" by (metis set_eq_subset)
   hence "\<forall>x\<^sub>1. x\<^sub>1 \<in> S \<longrightarrow> x\<^sub>1 = \<Union>S" by (metis Union_upper)
-  hence "\<forall>x\<^sub>1\<Colon>('a set) set. \<Union>S \<in> x\<^sub>1 \<longrightarrow> S \<subseteq> x\<^sub>1" by (metis subsetI)
-  hence "\<forall>x\<^sub>1\<Colon>('a set) set. S \<subseteq> insert (\<Union>S) x\<^sub>1" by (metis insert_iff)
+  hence "\<forall>x\<^sub>1::('a set) set. \<Union>S \<in> x\<^sub>1 \<longrightarrow> S \<subseteq> x\<^sub>1" by (metis subsetI)
+  hence "\<forall>x\<^sub>1::('a set) set. S \<subseteq> insert (\<Union>S) x\<^sub>1" by (metis insert_iff)
   thus "\<exists>z. S \<subseteq> {z}" by metis
 qed
 
 text {*
   From W. W. Bledsoe and Guohui Feng, SET-VAR. JAR 11 (3), 1993, pages