# HG changeset patch
# User blanchet
# Date 1275656930 -7200
# Node ID c2a44bc874f9d106e2772b87ee242f1f64ccd72f
# Parent  d77250dd24162116a16e4b21fc7fa4c7b9123e4b
redid the Isar proofs using the latest Sledgehammer, eliminating the last occurrences of "neg_clausify" in proofs

diff -r d77250dd2416 -r c2a44bc874f9 src/HOL/Metis_Examples/set.thy
--- a/src/HOL/Metis_Examples/set.thy	Fri Jun 04 14:08:23 2010 +0200
+++ b/src/HOL/Metis_Examples/set.thy	Fri Jun 04 15:08:50 2010 +0200
@@ -21,242 +21,164 @@
 (*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 (neg_clausify)
-fix x
-assume 0: "Y \<subseteq> X \<or> X = Y \<union> Z"
-assume 1: "Z \<subseteq> X \<or> X = Y \<union> Z"
-assume 2: "(\<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X \<or> Y \<subseteq> x) \<or> X \<noteq> Y \<union> Z"
-assume 3: "(\<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X \<or> Z \<subseteq> x) \<or> X \<noteq> Y \<union> Z"
-assume 4: "(\<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X \<or> \<not> X \<subseteq> x) \<or> X \<noteq> Y \<union> Z"
-assume 5: "\<And>V. ((\<not> Y \<subseteq> V \<or> \<not> Z \<subseteq> V) \<or> X \<subseteq> V) \<or> X = Y \<union> Z"
-have 6: "sup Y Z = X \<or> Y \<subseteq> X"
-  by (metis 0)
-have 7: "sup Y Z = X \<or> Z \<subseteq> X"
-  by (metis 1)
-have 8: "\<And>X3. sup Y Z = X \<or> X \<subseteq> X3 \<or> \<not> Y \<subseteq> X3 \<or> \<not> Z \<subseteq> X3"
-  by (metis 5)
-have 9: "Y \<subseteq> x \<or> sup Y Z \<noteq> X \<or> \<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X"
-  by (metis 2)
-have 10: "Z \<subseteq> x \<or> sup Y Z \<noteq> X \<or> \<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X"
-  by (metis 3)
-have 11: "sup Y Z \<noteq> X \<or> \<not> X \<subseteq> x \<or> \<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X"
-  by (metis 4)
-have 12: "Z \<subseteq> X"
-  by (metis Un_upper2 7)
-have 13: "\<And>X3. sup Y Z = X \<or> X \<subseteq> sup X3 Z \<or> \<not> Y \<subseteq> sup X3 Z"
-  by (metis 8 Un_upper2)
-have 14: "Y \<subseteq> X"
-  by (metis Un_upper1 6)
-have 15: "Z \<subseteq> x \<or> sup Y Z \<noteq> X \<or> \<not> Y \<subseteq> X"
-  by (metis 10 12)
-have 16: "Y \<subseteq> x \<or> sup Y Z \<noteq> X \<or> \<not> Y \<subseteq> X"
-  by (metis 9 12)
-have 17: "sup Y Z \<noteq> X \<or> \<not> X \<subseteq> x \<or> \<not> Y \<subseteq> X"
-  by (metis 11 12)
-have 18: "sup Y Z \<noteq> X \<or> \<not> X \<subseteq> x"
-  by (metis 17 14)
-have 19: "Z \<subseteq> x \<or> sup Y Z \<noteq> X"
-  by (metis 15 14)
-have 20: "Y \<subseteq> x \<or> sup Y Z \<noteq> X"
-  by (metis 16 14)
-have 21: "sup Y Z = X \<or> X \<subseteq> sup Y Z"
-  by (metis 13 Un_upper1)
-have 22: "sup Y Z = X \<or> \<not> sup Y Z \<subseteq> X"
-  by (metis equalityI 21)
-have 23: "sup Y Z = X \<or> \<not> Z \<subseteq> X \<or> \<not> Y \<subseteq> X"
-  by (metis 22 Un_least)
-have 24: "sup Y Z = X \<or> \<not> Y \<subseteq> X"
-  by (metis 23 12)
-have 25: "sup Y Z = X"
-  by (metis 24 14)
-have 26: "\<And>X3. X \<subseteq> X3 \<or> \<not> Z \<subseteq> X3 \<or> \<not> Y \<subseteq> X3"
-  by (metis Un_least 25)
-have 27: "Y \<subseteq> x"
-  by (metis 20 25)
-have 28: "Z \<subseteq> x"
-  by (metis 19 25)
-have 29: "\<not> X \<subseteq> x"
-  by (metis 18 25)
-have 30: "X \<subseteq> x \<or> \<not> Y \<subseteq> x"
-  by (metis 26 28)
-have 31: "X \<subseteq> x"
-  by (metis 30 27)
-show "False"
-  by (metis 31 29)
+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)
+  { 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) }
+  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) }
+    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) }
+      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) }
+  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 "\<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) }
+    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) }
+      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 }
+  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
 qed
 
 sledgehammer_params [isar_proof, isar_shrink_factor = 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 (neg_clausify)
-fix x
-assume 0: "Y \<subseteq> X \<or> X = Y \<union> Z"
-assume 1: "Z \<subseteq> X \<or> X = Y \<union> Z"
-assume 2: "(\<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X \<or> Y \<subseteq> x) \<or> X \<noteq> Y \<union> Z"
-assume 3: "(\<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X \<or> Z \<subseteq> x) \<or> X \<noteq> Y \<union> Z"
-assume 4: "(\<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X \<or> \<not> X \<subseteq> x) \<or> X \<noteq> Y \<union> Z"
-assume 5: "\<And>V. ((\<not> Y \<subseteq> V \<or> \<not> Z \<subseteq> V) \<or> X \<subseteq> V) \<or> X = Y \<union> Z"
-have 6: "sup Y Z = X \<or> Y \<subseteq> X"
-  by (metis 0)
-have 7: "Y \<subseteq> x \<or> sup Y Z \<noteq> X \<or> \<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X"
-  by (metis 2)
-have 8: "sup Y Z \<noteq> X \<or> \<not> X \<subseteq> x \<or> \<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X"
-  by (metis 4)
-have 9: "\<And>X3. sup Y Z = X \<or> X \<subseteq> sup X3 Z \<or> \<not> Y \<subseteq> sup X3 Z"
-  by (metis 5 Un_upper2)
-have 10: "Z \<subseteq> x \<or> sup Y Z \<noteq> X \<or> \<not> Y \<subseteq> X"
-  by (metis 3 Un_upper2)
-have 11: "sup Y Z \<noteq> X \<or> \<not> X \<subseteq> x \<or> \<not> Y \<subseteq> X"
-  by (metis 8 Un_upper2)
-have 12: "Z \<subseteq> x \<or> sup Y Z \<noteq> X"
-  by (metis 10 Un_upper1)
-have 13: "sup Y Z = X \<or> X \<subseteq> sup Y Z"
-  by (metis 9 Un_upper1)
-have 14: "sup Y Z = X \<or> \<not> Z \<subseteq> X \<or> \<not> Y \<subseteq> X"
-  by (metis equalityI 13 Un_least)
-have 15: "sup Y Z = X"
-  by (metis 14 1 6)
-have 16: "Y \<subseteq> x"
-  by (metis 7 Un_upper2 Un_upper1 15)
-have 17: "\<not> X \<subseteq> x"
-  by (metis 11 Un_upper1 15)
-have 18: "X \<subseteq> x"
-  by (metis Un_least 15 12 15 16)
-show "False"
-  by (metis 18 17)
+   "(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"
+    { 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) }
+      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) }
+    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) }
+  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) }
+  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) }
+  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) }
+    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
 qed
 
 sledgehammer_params [isar_proof, isar_shrink_factor = 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 (neg_clausify)
-fix x
-assume 0: "Y \<subseteq> X \<or> X = Y \<union> Z"
-assume 1: "Z \<subseteq> X \<or> X = Y \<union> Z"
-assume 2: "(\<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X \<or> Y \<subseteq> x) \<or> X \<noteq> Y \<union> Z"
-assume 3: "(\<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X \<or> Z \<subseteq> x) \<or> X \<noteq> Y \<union> Z"
-assume 4: "(\<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X \<or> \<not> X \<subseteq> x) \<or> X \<noteq> Y \<union> Z"
-assume 5: "\<And>V. ((\<not> Y \<subseteq> V \<or> \<not> Z \<subseteq> V) \<or> X \<subseteq> V) \<or> X = Y \<union> Z"
-have 6: "Z \<subseteq> x \<or> sup Y Z \<noteq> X \<or> \<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X"
-  by (metis 3)
-have 7: "\<And>X3. sup Y Z = X \<or> X \<subseteq> sup X3 Z \<or> \<not> Y \<subseteq> sup X3 Z"
-  by (metis 5 Un_upper2)
-have 8: "Y \<subseteq> x \<or> sup Y Z \<noteq> X \<or> \<not> Y \<subseteq> X"
-  by (metis 2 Un_upper2)
-have 9: "Z \<subseteq> x \<or> sup Y Z \<noteq> X"
-  by (metis 6 Un_upper2 Un_upper1)
-have 10: "sup Y Z = X \<or> \<not> sup Y Z \<subseteq> X"
-  by (metis equalityI 7 Un_upper1)
-have 11: "sup Y Z = X"
-  by (metis 10 Un_least 1 0)
-have 12: "Z \<subseteq> x"
-  by (metis 9 11)
-have 13: "X \<subseteq> x"
-  by (metis Un_least 11 12 8 Un_upper1 11)
-show "False"
-  by (metis 13 4 Un_upper2 Un_upper1 11)
+   "(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)
+  { 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) }
+  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 "(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)
 qed
 
-(*Example included in TPHOLs paper*)
-
 sledgehammer_params [isar_proof, isar_shrink_factor = 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 (neg_clausify)
-fix x
-assume 0: "Y \<subseteq> X \<or> X = Y \<union> Z"
-assume 1: "Z \<subseteq> X \<or> X = Y \<union> Z"
-assume 2: "(\<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X \<or> Y \<subseteq> x) \<or> X \<noteq> Y \<union> Z"
-assume 3: "(\<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X \<or> Z \<subseteq> x) \<or> X \<noteq> Y \<union> Z"
-assume 4: "(\<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X \<or> \<not> X \<subseteq> x) \<or> X \<noteq> Y \<union> Z"
-assume 5: "\<And>V. ((\<not> Y \<subseteq> V \<or> \<not> Z \<subseteq> V) \<or> X \<subseteq> V) \<or> X = Y \<union> Z"
-have 6: "sup Y Z \<noteq> X \<or> \<not> X \<subseteq> x \<or> \<not> Y \<subseteq> X \<or> \<not> Z \<subseteq> X"
-  by (metis 4)
-have 7: "Z \<subseteq> x \<or> sup Y Z \<noteq> X \<or> \<not> Y \<subseteq> X"
-  by (metis 3 Un_upper2)
-have 8: "Z \<subseteq> x \<or> sup Y Z \<noteq> X"
-  by (metis 7 Un_upper1)
-have 9: "sup Y Z = X \<or> \<not> Z \<subseteq> X \<or> \<not> Y \<subseteq> X"
-  by (metis equalityI 5 Un_upper2 Un_upper1 Un_least)
-have 10: "Y \<subseteq> x"
-  by (metis 2 Un_upper2 1 Un_upper1 0 9 Un_upper2 1 Un_upper1 0)
-have 11: "X \<subseteq> x"
-  by (metis Un_least 9 Un_upper2 1 Un_upper1 0 8 9 Un_upper2 1 Un_upper1 0 10)
-show "False"
-  by (metis 11 6 Un_upper2 1 Un_upper1 0 9 Un_upper2 1 Un_upper1 0)
-qed 
+   "(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 "\<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) }
+  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)
+qed
 
-declare [[ atp_problem_prefix = "set__equal_union" ]]
+sledgehammer_params [isar_proof, isar_shrink_factor = 1]
+
 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))" 
-(*One shot proof: hand-reduced. Metis can't do the full proof any more.*)
+   "(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))"
 by (metis Un_least Un_upper1 Un_upper2 set_eq_subset)
 
+lemma "(X = Y \<inter> Z) = (X \<subseteq> Y \<and> X \<subseteq> Z \<and> (\<forall>V. V \<subseteq> Y \<and> V \<subseteq> Z \<longrightarrow> V \<subseteq> X))"
+by (metis Int_greatest Int_lower1 Int_lower2 subset_antisym)
 
-declare [[ atp_problem_prefix = "set__equal_inter" ]]
-lemma "(X = Y \<inter> Z) =
-    (X \<subseteq> Y \<and> X \<subseteq> Z \<and> (\<forall>V. V \<subseteq> Y \<and> V \<subseteq> Z \<longrightarrow> V \<subseteq> X))"
-by (metis Int_greatest Int_lower1 Int_lower2 set_eq_subset)
-
-declare [[ atp_problem_prefix = "set__fixedpoint" ]]
-lemma fixedpoint:
-    "\<exists>!x. f (g x) = x \<Longrightarrow> \<exists>!y. g (f y) = y"
+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 (neg_clausify)
-fix x xa
-assume 0: "f (g x) = x"
-assume 1: "\<And>y. y = x \<or> f (g y) \<noteq> y"
-assume 2: "\<And>x. g (f (xa x)) = xa x \<or> g (f x) \<noteq> x"
-assume 3: "\<And>x. g (f x) \<noteq> x \<or> xa x \<noteq> x"
-have 4: "\<And>X1. g (f X1) \<noteq> X1 \<or> g x \<noteq> X1"
-  by (metis 3 1 2)
-show "False"
-  by (metis 4 0)
+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
 qed
 
-declare [[ atp_problem_prefix = "set__singleton_example" ]]
-lemma (*singleton_example_2:*)
+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)
-  --{*found by SPASS*}
 
-lemma (*singleton_example_2:*)
+lemma (* singleton_example_2: *)
      "\<forall>x \<in> S. \<Union>S \<subseteq> x \<Longrightarrow> \<exists>z. S \<subseteq> {z}"
 by (metis Set.subsetI Union_upper insert_iff set_eq_subset)
 
 lemma singleton_example_2:
      "\<forall>x \<in> S. \<Union>S \<subseteq> x \<Longrightarrow> \<exists>z. S \<subseteq> {z}"
-proof (neg_clausify)
-assume 0: "\<And>x. \<not> S \<subseteq> {x}"
-assume 1: "\<And>x. x \<notin> S \<or> \<Union>S \<subseteq> x"
-have 2: "\<And>X3. X3 = \<Union>S \<or> \<not> X3 \<subseteq> \<Union>S \<or> X3 \<notin> S"
-  by (metis set_eq_subset 1)
-have 3: "\<And>X3. S \<subseteq> insert (\<Union>S) X3"
-  by (metis insert_iff Set.subsetI Union_upper 2 Set.subsetI)
-show "False"
-  by (metis 3 0)
+proof -
+  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)
+  thus "\<exists>z. S \<subseteq> {z}" by metis
 qed
 
-
-
 text {*
   From W. W. Bledsoe and Guohui Feng, SET-VAR. JAR 11 (3), 1993, pages
   293-314.
 *}
 
-declare [[ atp_problem_prefix = "set__Bledsoe_Fung" ]]
-(*Notes: 1, the numbering doesn't completely agree with the paper. 
-2, we must rename set variables to avoid type clashes.*)
+(* Notes: (1) The numbering doesn't completely agree with the paper.
+   (2) We must rename set variables to avoid type clashes. *)
 lemma "\<exists>B. (\<forall>x \<in> B. x \<le> (0::int))"
       "D \<in> F \<Longrightarrow> \<exists>G. \<forall>A \<in> G. \<exists>B \<in> F. A \<subseteq> B"
       "P a \<Longrightarrow> \<exists>A. (\<forall>x \<in> A. P x) \<and> (\<exists>y. y \<in> A)"
@@ -265,13 +187,13 @@
       "P (f b) \<Longrightarrow> \<exists>s A. (\<forall>x \<in> A. P x) \<and> f s \<in> A"
       "\<exists>A. a \<notin> A"
       "(\<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 mem_def)
-apply (metis insertCI singletonE zless_le)
-apply (metis Collect_def Collect_mem_eq)
-apply (metis Collect_def Collect_mem_eq)
-apply (metis DiffE)
-apply (metis pair_in_Id_conv)
-done
+       apply (metis all_not_in_conv)
+      apply (metis all_not_in_conv)
+     apply (metis mem_def)
+    apply (metis less_int_def singleton_iff)
+   apply (metis mem_def)
+  apply (metis mem_def)
+ apply (metis all_not_in_conv)
+by (metis pair_in_Id_conv)
 
 end