1.1 --- a/src/HOL/Metis_Examples/Sets.thy Mon Jan 02 15:08:40 2012 +0100
1.2 +++ b/src/HOL/Metis_Examples/Sets.thy Mon Jan 02 15:15:46 2012 +0100
1.3 @@ -24,50 +24,50 @@
1.4 sledgehammer_params [isar_proof, isar_shrink_factor = 1]
1.5
1.6 (*multiple versions of this example*)
1.7 -(* lemma (*equal_union: *)
1.8 +lemma (*equal_union: *)
1.9 "(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))"
1.10 proof -
1.11 - 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)
1.12 - 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)
1.13 - 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)
1.14 + 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)
1.15 + 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)
1.16 + 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)
1.17 { assume "\<not> Z \<subseteq> X"
1.18 - 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) }
1.19 + 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) }
1.20 moreover
1.21 { assume AA1: "Y \<union> Z \<noteq> X"
1.22 { assume "\<not> Y \<subseteq> X"
1.23 - 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) }
1.24 + 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) }
1.25 moreover
1.26 { assume AAA1: "Y \<subseteq> X \<and> Y \<union> Z \<noteq> X"
1.27 { assume "\<not> Z \<subseteq> X"
1.28 - 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) }
1.29 + 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) }
1.30 moreover
1.31 { assume "(Z \<subseteq> X \<and> Y \<subseteq> X) \<and> Y \<union> Z \<noteq> X"
1.32 hence "Y \<union> Z \<subseteq> X \<and> X \<noteq> Y \<union> Z" by (metis Un_subset_iff)
1.33 hence "Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> Y \<union> Z" by (metis F2)
1.34 - 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)
1.35 - 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) }
1.36 - 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) }
1.37 - 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) }
1.38 + 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)
1.39 + 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) }
1.40 + 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) }
1.41 + 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) }
1.42 moreover
1.43 - { 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"
1.44 + { 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"
1.45 { assume "\<not> Y \<subseteq> X"
1.46 - 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) }
1.47 + 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) }
1.48 moreover
1.49 { assume AAA1: "Y \<subseteq> X \<and> Y \<union> Z \<noteq> X"
1.50 { assume "\<not> Z \<subseteq> X"
1.51 - 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) }
1.52 + 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) }
1.53 moreover
1.54 { assume "(Z \<subseteq> X \<and> Y \<subseteq> X) \<and> Y \<union> Z \<noteq> X"
1.55 hence "Y \<union> Z \<subseteq> X \<and> X \<noteq> Y \<union> Z" by (metis Un_subset_iff)
1.56 hence "Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> Y \<union> Z" by (metis F2)
1.57 - 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)
1.58 - 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) }
1.59 - 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) }
1.60 - 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 }
1.61 + 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)
1.62 + 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) }
1.63 + 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) }
1.64 + 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 }
1.65 moreover
1.66 { assume "\<not> Y \<subseteq> X"
1.67 - 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) }
1.68 - 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
1.69 + 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) }
1.70 + 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
1.71 qed
1.72
1.73 sledgehammer_params [isar_proof, isar_shrink_factor = 2]
1.74 @@ -75,36 +75,36 @@
1.75 lemma (*equal_union: *)
1.76 "(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))"
1.77 proof -
1.78 - 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)
1.79 - { 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"
1.80 + 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)
1.81 + { 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"
1.82 { assume AAA1: "Y \<subseteq> X \<and> Y \<union> Z \<noteq> X"
1.83 { assume "\<not> Z \<subseteq> X"
1.84 - 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) }
1.85 + 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) }
1.86 moreover
1.87 { assume "Y \<union> Z \<subseteq> X \<and> X \<noteq> Y \<union> Z"
1.88 - 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)
1.89 - 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) }
1.90 - 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) }
1.91 + 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)
1.92 + 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) }
1.93 + 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) }
1.94 moreover
1.95 { assume "\<not> Y \<subseteq> X"
1.96 - 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) }
1.97 - 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) }
1.98 + 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) }
1.99 + 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) }
1.100 moreover
1.101 { assume "\<not> Z \<subseteq> X"
1.102 - 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) }
1.103 + 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) }
1.104 moreover
1.105 { assume "\<not> Y \<subseteq> X"
1.106 - 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) }
1.107 + 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) }
1.108 moreover
1.109 { assume AA1: "Y \<subseteq> X \<and> Y \<union> Z \<noteq> X"
1.110 { assume "\<not> Z \<subseteq> X"
1.111 - 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) }
1.112 + 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) }
1.113 moreover
1.114 { assume "Y \<union> Z \<subseteq> X \<and> X \<noteq> Y \<union> Z"
1.115 - 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)
1.116 - 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) }
1.117 - 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) }
1.118 - 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
1.119 + 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)
1.120 + 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) }
1.121 + 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) }
1.122 + 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
1.123 qed
1.124
1.125 sledgehammer_params [isar_proof, isar_shrink_factor = 3]
1.126 @@ -112,16 +112,16 @@
1.127 lemma (*equal_union: *)
1.128 "(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))"
1.129 proof -
1.130 - 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)
1.131 - 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)
1.132 + 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)
1.133 + 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)
1.134 { assume "(Z \<subseteq> X \<and> Y \<subseteq> X) \<and> Y \<union> Z \<noteq> X"
1.135 - 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) }
1.136 + 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) }
1.137 moreover
1.138 - { 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"
1.139 + { 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"
1.140 { assume "(Z \<subseteq> X \<and> Y \<subseteq> X) \<and> Y \<union> Z \<noteq> X"
1.141 - 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) }
1.142 - 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) }
1.143 - 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)
1.144 + 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) }
1.145 + 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) }
1.146 + 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)
1.147 qed
1.148
1.149 sledgehammer_params [isar_proof, isar_shrink_factor = 4]
1.150 @@ -129,15 +129,15 @@
1.151 lemma (*equal_union: *)
1.152 "(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))"
1.153 proof -
1.154 - 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)
1.155 + 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)
1.156 { assume "\<not> Y \<subseteq> X"
1.157 - 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) }
1.158 + 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) }
1.159 moreover
1.160 { assume AA1: "Y \<subseteq> X \<and> Y \<union> Z \<noteq> X"
1.161 - { 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"
1.162 - 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) }
1.163 - 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) }
1.164 - 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)
1.165 + { 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"
1.166 + 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) }
1.167 + 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) }
1.168 + 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)
1.169 qed
1.170
1.171 sledgehammer_params [isar_proof, isar_shrink_factor = 1]
1.172 @@ -172,8 +172,8 @@
1.173 assume "\<forall>x \<in> S. \<Union>S \<subseteq> x"
1.174 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)
1.175 hence "\<forall>x\<^isub>1. x\<^isub>1 \<in> S \<longrightarrow> x\<^isub>1 = \<Union>S" by (metis Union_upper)
1.176 - 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)
1.177 - hence "\<forall>x\<^isub>1\<Colon>('a \<Rightarrow> bool) \<Rightarrow> bool. S \<subseteq> insert (\<Union>S) x\<^isub>1" by (metis insert_iff)
1.178 + hence "\<forall>x\<^isub>1\<Colon>('a set) set. \<Union>S \<in> x\<^isub>1 \<longrightarrow> S \<subseteq> x\<^isub>1" by (metis subsetI)
1.179 + hence "\<forall>x\<^isub>1\<Colon>('a set) set. S \<subseteq> insert (\<Union>S) x\<^isub>1" by (metis insert_iff)
1.180 thus "\<exists>z. S \<subseteq> {z}" by metis
1.181 qed
1.182
1.183 @@ -194,12 +194,11 @@
1.184 "(\<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"
1.185 apply (metis all_not_in_conv)
1.186 apply (metis all_not_in_conv)
1.187 - apply (metis mem_def)
1.188 + apply (metis mem_Collect_eq)
1.189 apply (metis less_int_def singleton_iff)
1.190 - apply (metis mem_def)
1.191 - apply (metis mem_def)
1.192 + apply (metis mem_Collect_eq)
1.193 + apply (metis mem_Collect_eq)
1.194 apply (metis all_not_in_conv)
1.195 by (metis pair_in_Id_conv)
1.196 -*)
1.197
1.198 end