src/HOL/Metis_Examples/Sets.thy
author wenzelm
Sat, 07 Apr 2012 16:41:59 +0200
changeset 47389 e8552cba702d
parent 46077 86e6e9d42ad7
child 48050 72acba14c12b
permissions -rw-r--r--
explicit checks stable_finished_theory/stable_command allow parallel asynchronous command transactions; tuned;
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(*  Title:      HOL/Metis_Examples/Sets.thy
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    Author:     Lawrence C. Paulson, Cambridge University Computer Laboratory
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    Author:     Jasmin Blanchette, TU Muenchen
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Metis example featuring typed set theory.
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*)
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header {* Metis Example Featuring Typed Set Theory *}
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theory Sets
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imports Main
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begin
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declare [[metis_new_skolemizer]]
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lemma "EX x X. ALL y. EX z Z. (~P(y,y) | P(x,x) | ~S(z,x)) &
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               (S(x,y) | ~S(y,z) | Q(Z,Z))  &
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               (Q(X,y) | ~Q(y,Z) | S(X,X))"
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by metis
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lemma "P(n::nat) ==> ~P(0) ==> n ~= 0"
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by metis
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sledgehammer_params [isar_proof, isar_shrink_factor = 1]
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(*multiple versions of this example*)
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lemma (*equal_union: *)
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   "(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))"
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proof -
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  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)
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  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)
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  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)
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  { assume "\<not> Z \<subseteq> X"
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    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) }
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  moreover
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  { assume AA1: "Y \<union> Z \<noteq> X"
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    { assume "\<not> Y \<subseteq> X"
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      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) }
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    moreover
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    { assume AAA1: "Y \<subseteq> X \<and> Y \<union> Z \<noteq> X"
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      { assume "\<not> Z \<subseteq> X"
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        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) }
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      moreover
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      { assume "(Z \<subseteq> X \<and> Y \<subseteq> X) \<and> Y \<union> Z \<noteq> X"
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        hence "Y \<union> Z \<subseteq> X \<and> X \<noteq> Y \<union> Z" by (metis Un_subset_iff)
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        hence "Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> Y \<union> Z" by (metis F2)
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        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)
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        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) }
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      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) }
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    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) }
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  moreover
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  { 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"
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    { assume "\<not> Y \<subseteq> X"
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      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) }
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    moreover
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    { assume AAA1: "Y \<subseteq> X \<and> Y \<union> Z \<noteq> X"
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      { assume "\<not> Z \<subseteq> X"
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        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) }
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      moreover
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      { assume "(Z \<subseteq> X \<and> Y \<subseteq> X) \<and> Y \<union> Z \<noteq> X"
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        hence "Y \<union> Z \<subseteq> X \<and> X \<noteq> Y \<union> Z" by (metis Un_subset_iff)
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        hence "Y \<union> Z \<noteq> X \<and> \<not> X \<subseteq> Y \<union> Z" by (metis F2)
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        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)
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        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) }
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      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) }
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    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 }
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  moreover
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  { assume "\<not> Y \<subseteq> X"
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    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) }
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  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
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qed
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sledgehammer_params [isar_proof, isar_shrink_factor = 2]
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lemma (*equal_union: *)
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   "(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))"
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proof -
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  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)
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  { 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"
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    { assume AAA1: "Y \<subseteq> X \<and> Y \<union> Z \<noteq> X"
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      { assume "\<not> Z \<subseteq> X"
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        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) }
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      moreover
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      { assume "Y \<union> Z \<subseteq> X \<and> X \<noteq> Y \<union> Z"
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        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)
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        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) }
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      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) }
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    moreover
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    { assume "\<not> Y \<subseteq> X"
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      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) }
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    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) }
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  moreover
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  { assume "\<not> Z \<subseteq> X"
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    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) }
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  moreover
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  { assume "\<not> Y \<subseteq> X"
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    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) }
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  moreover
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  { assume AA1: "Y \<subseteq> X \<and> Y \<union> Z \<noteq> X"
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    { assume "\<not> Z \<subseteq> X"
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      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) }
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    moreover
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    { assume "Y \<union> Z \<subseteq> X \<and> X \<noteq> Y \<union> Z"
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      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)
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      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) }
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    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) }
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  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
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qed
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sledgehammer_params [isar_proof, isar_shrink_factor = 3]
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lemma (*equal_union: *)
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   "(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))"
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   114
proof -
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  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)
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   116
  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)
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  { assume "(Z \<subseteq> X \<and> Y \<subseteq> X) \<and> Y \<union> Z \<noteq> X"
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    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) }
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  moreover
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  { 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"
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    { assume "(Z \<subseteq> X \<and> Y \<subseteq> X) \<and> Y \<union> Z \<noteq> X"
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      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) }
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   123
    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) }
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  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)
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qed
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sledgehammer_params [isar_proof, isar_shrink_factor = 4]
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lemma (*equal_union: *)
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   "(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))"
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   131
proof -
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  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)
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  { assume "\<not> Y \<subseteq> X"
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    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) }
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  moreover
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  { assume AA1: "Y \<subseteq> X \<and> Y \<union> Z \<noteq> X"
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   137
    { 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"
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      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) }
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   139
    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) }
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  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)
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qed
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sledgehammer_params [isar_proof, isar_shrink_factor = 1]
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lemma (*equal_union: *)
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   146
   "(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))"
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by (metis Un_least Un_upper1 Un_upper2 set_eq_subset)
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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))"
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by (metis Int_greatest Int_lower1 Int_lower2 subset_antisym)
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lemma fixedpoint: "\<exists>!x. f (g x) = x \<Longrightarrow> \<exists>!y. g (f y) = y"
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by metis
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lemma (* fixedpoint: *) "\<exists>!x. f (g x) = x \<Longrightarrow> \<exists>!y. g (f y) = y"
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   156
proof -
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  assume "\<exists>!x\<Colon>'a. f (g x) = x"
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   158
  thus "\<exists>!y\<Colon>'b. g (f y) = y" by metis
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qed
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lemma (* singleton_example_2: *)
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     "\<forall>x \<in> S. \<Union>S \<subseteq> x \<Longrightarrow> \<exists>z. S \<subseteq> {z}"
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by (metis Set.subsetI Union_upper insertCI set_eq_subset)
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lemma (* singleton_example_2: *)
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     "\<forall>x \<in> S. \<Union>S \<subseteq> x \<Longrightarrow> \<exists>z. S \<subseteq> {z}"
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   167
by (metis Set.subsetI Union_upper insert_iff set_eq_subset)
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lemma singleton_example_2:
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     "\<forall>x \<in> S. \<Union>S \<subseteq> x \<Longrightarrow> \<exists>z. S \<subseteq> {z}"
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   171
proof -
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  assume "\<forall>x \<in> S. \<Union>S \<subseteq> x"
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   173
  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)
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   174
  hence "\<forall>x\<^isub>1. x\<^isub>1 \<in> S \<longrightarrow> x\<^isub>1 = \<Union>S" by (metis Union_upper)
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   175
  hence "\<forall>x\<^isub>1\<Colon>('a set) set. \<Union>S \<in> x\<^isub>1 \<longrightarrow> S \<subseteq> x\<^isub>1" by (metis subsetI)
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   176
  hence "\<forall>x\<^isub>1\<Colon>('a set) set. S \<subseteq> insert (\<Union>S) x\<^isub>1" by (metis insert_iff)
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   177
  thus "\<exists>z. S \<subseteq> {z}" by metis
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qed
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text {*
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  From W. W. Bledsoe and Guohui Feng, SET-VAR. JAR 11 (3), 1993, pages
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  293-314.
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*}
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(* Notes: (1) The numbering doesn't completely agree with the paper.
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   (2) We must rename set variables to avoid type clashes. *)
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lemma "\<exists>B. (\<forall>x \<in> B. x \<le> (0::int))"
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      "D \<in> F \<Longrightarrow> \<exists>G. \<forall>A \<in> G. \<exists>B \<in> F. A \<subseteq> B"
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      "P a \<Longrightarrow> \<exists>A. (\<forall>x \<in> A. P x) \<and> (\<exists>y. y \<in> A)"
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      "a < b \<and> b < (c::int) \<Longrightarrow> \<exists>B. a \<notin> B \<and> b \<in> B \<and> c \<notin> B"
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      "P (f b) \<Longrightarrow> \<exists>s A. (\<forall>x \<in> A. P x) \<and> f s \<in> A"
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      "P (f b) \<Longrightarrow> \<exists>s A. (\<forall>x \<in> A. P x) \<and> f s \<in> A"
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      "\<exists>A. a \<notin> A"
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      "(\<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"
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       apply (metis all_not_in_conv)
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      apply (metis all_not_in_conv)
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     apply (metis mem_Collect_eq)
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    apply (metis less_int_def singleton_iff)
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   apply (metis mem_Collect_eq)
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  apply (metis mem_Collect_eq)
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 apply (metis all_not_in_conv)
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by (metis pair_in_Id_conv)
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end