Theory Sets

theory Sets
imports Main
(*  Title:      HOL/Metis_Examples/Sets.thy
Author: Lawrence C. Paulson, Cambridge University Computer Laboratory
Author: Jasmin Blanchette, TU Muenchen

Metis example featuring typed set theory.
*)


header {* Metis Example Featuring Typed Set Theory *}

theory Sets
imports Main
begin

declare [[metis_new_skolem]]

lemma "EX x X. ALL y. EX z Z. (~P(y,y) | P(x,x) | ~S(z,x)) &
(S(x,y) | ~S(y,z) | Q(Z,Z)) &
(Q(X,y) | ~Q(y,Z) | S(X,X))"

by metis

lemma "P(n::nat) ==> ~P(0) ==> n ~= 0"
by metis

sledgehammer_params [isar_proofs, isar_compress = 1]

(*multiple versions of this example*)
lemma (*equal_union: *)
"(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))"
proof -
have F1: "∀(x2::'b set) x1::'b set. x1 ⊆ x1 ∪ x2" by (metis Un_commute Un_upper2)
have F2a: "∀(x2::'b set) x1::'b set. x1 ⊆ x2 --> x2 = x2 ∪ x1" by (metis Un_commute subset_Un_eq)
have F2: "∀(x2::'b set) x1::'b set. x1 ⊆ x2 ∧ x2 ⊆ x1 --> x1 = x2" by (metis F2a subset_Un_eq)
{ assume "¬ Z ⊆ X"
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_upper2) }
moreover
{ assume AA1: "Y ∪ Z ≠ X"
{ assume "¬ Y ⊆ X"
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis F1) }
moreover
{ assume AAA1: "Y ⊆ X ∧ Y ∪ Z ≠ X"
{ assume "¬ Z ⊆ X"
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_upper2) }
moreover
{ assume "(Z ⊆ X ∧ Y ⊆ X) ∧ Y ∪ Z ≠ X"
hence "Y ∪ Z ⊆ X ∧ X ≠ Y ∪ Z" by (metis Un_subset_iff)
hence "Y ∪ Z ≠ X ∧ ¬ X ⊆ Y ∪ Z" by (metis F2)
hence "∃x1::'a set. Y ⊆ x1 ∪ Z ∧ Y ∪ Z ≠ X ∧ ¬ X ⊆ x1 ∪ Z" by (metis F1)
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_upper2) }
ultimately have "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis AAA1) }
ultimately have "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis AA1) }
moreover
{ assume "∃x1::'a set. (Z ⊆ x1 ∧ Y ⊆ x1) ∧ ¬ X ⊆ x1"
{ assume "¬ Y ⊆ X"
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis F1) }
moreover
{ assume AAA1: "Y ⊆ X ∧ Y ∪ Z ≠ X"
{ assume "¬ Z ⊆ X"
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_upper2) }
moreover
{ assume "(Z ⊆ X ∧ Y ⊆ X) ∧ Y ∪ Z ≠ X"
hence "Y ∪ Z ⊆ X ∧ X ≠ Y ∪ Z" by (metis Un_subset_iff)
hence "Y ∪ Z ≠ X ∧ ¬ X ⊆ Y ∪ Z" by (metis F2)
hence "∃x1::'a set. Y ⊆ x1 ∪ Z ∧ Y ∪ Z ≠ X ∧ ¬ X ⊆ x1 ∪ Z" by (metis F1)
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_upper2) }
ultimately have "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis AAA1) }
ultimately have "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by blast }
moreover
{ assume "¬ Y ⊆ X"
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis F1) }
ultimately show "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by metis
qed

sledgehammer_params [isar_proofs, isar_compress = 2]

lemma (*equal_union: *)
"(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))"
proof -
have F1: "∀(x2::'b set) x1::'b set. x1 ⊆ x2 ∧ x2 ⊆ x1 --> x1 = x2" by (metis Un_commute subset_Un_eq)
{ assume AA1: "∃x1::'a set. (Z ⊆ x1 ∧ Y ⊆ x1) ∧ ¬ X ⊆ x1"
{ assume AAA1: "Y ⊆ X ∧ Y ∪ Z ≠ X"
{ assume "¬ Z ⊆ X"
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_upper2) }
moreover
{ assume "Y ∪ Z ⊆ X ∧ X ≠ Y ∪ Z"
hence "∃x1::'a set. Y ⊆ x1 ∪ Z ∧ Y ∪ Z ≠ X ∧ ¬ X ⊆ x1 ∪ Z" by (metis F1 Un_commute Un_upper2)
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_upper2) }
ultimately have "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis AAA1 Un_subset_iff) }
moreover
{ assume "¬ Y ⊆ X"
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_commute Un_upper2) }
ultimately have "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis AA1 Un_subset_iff) }
moreover
{ assume "¬ Z ⊆ X"
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_upper2) }
moreover
{ assume "¬ Y ⊆ X"
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_commute Un_upper2) }
moreover
{ assume AA1: "Y ⊆ X ∧ Y ∪ Z ≠ X"
{ assume "¬ Z ⊆ X"
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_upper2) }
moreover
{ assume "Y ∪ Z ⊆ X ∧ X ≠ Y ∪ Z"
hence "∃x1::'a set. Y ⊆ x1 ∪ Z ∧ Y ∪ Z ≠ X ∧ ¬ X ⊆ x1 ∪ Z" by (metis F1 Un_commute Un_upper2)
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_upper2) }
ultimately have "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis AA1 Un_subset_iff) }
ultimately show "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by metis
qed

sledgehammer_params [isar_proofs, isar_compress = 3]

lemma (*equal_union: *)
"(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))"
proof -
have F1a: "∀(x2::'b set) x1::'b set. x1 ⊆ x2 --> x2 = x2 ∪ x1" by (metis Un_commute subset_Un_eq)
have F1: "∀(x2::'b set) x1::'b set. x1 ⊆ x2 ∧ x2 ⊆ x1 --> x1 = x2" by (metis F1a subset_Un_eq)
{ assume "(Z ⊆ X ∧ Y ⊆ X) ∧ Y ∪ Z ≠ X"
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis F1 Un_commute Un_subset_iff Un_upper2) }
moreover
{ assume AA1: "∃x1::'a set. (Z ⊆ x1 ∧ Y ⊆ x1) ∧ ¬ X ⊆ x1"
{ assume "(Z ⊆ X ∧ Y ⊆ X) ∧ Y ∪ Z ≠ X"
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis F1 Un_commute Un_subset_iff Un_upper2) }
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis AA1 Un_commute Un_subset_iff Un_upper2) }
ultimately show "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_commute Un_upper2)
qed

sledgehammer_params [isar_proofs, isar_compress = 4]

lemma (*equal_union: *)
"(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))"
proof -
have F1: "∀(x2::'b set) x1::'b set. x1 ⊆ x2 ∧ x2 ⊆ x1 --> x1 = x2" by (metis Un_commute subset_Un_eq)
{ assume "¬ Y ⊆ X"
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_commute Un_upper2) }
moreover
{ assume AA1: "Y ⊆ X ∧ Y ∪ Z ≠ X"
{ assume "∃x1::'a set. Y ⊆ x1 ∪ Z ∧ Y ∪ Z ≠ X ∧ ¬ X ⊆ x1 ∪ Z"
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_upper2) }
hence "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis AA1 F1 Un_commute Un_subset_iff Un_upper2) }
ultimately show "(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V::'a set. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))" by (metis Un_subset_iff Un_upper2)
qed

sledgehammer_params [isar_proofs, isar_compress = 1]

lemma (*equal_union: *)
"(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V. Y ⊆ V ∧ Z ⊆ V --> X ⊆ V))"
by (metis Un_least Un_upper1 Un_upper2 set_eq_subset)

lemma "(X = Y ∩ Z) = (X ⊆ Y ∧ X ⊆ Z ∧ (∀V. V ⊆ Y ∧ V ⊆ Z --> V ⊆ X))"
by (metis Int_greatest Int_lower1 Int_lower2 subset_antisym)

lemma fixedpoint: "∃!x. f (g x) = x ==> ∃!y. g (f y) = y"
by metis

lemma (* fixedpoint: *) "∃!x. f (g x) = x ==> ∃!y. g (f y) = y"
proof -
assume "∃!x::'a. f (g x) = x"
thus "∃!y::'b. g (f y) = y" by metis
qed

lemma (* singleton_example_2: *)
"∀x ∈ S. \<Union>S ⊆ x ==> ∃z. S ⊆ {z}"
by (metis Set.subsetI Union_upper insertCI set_eq_subset)

lemma (* singleton_example_2: *)
"∀x ∈ S. \<Union>S ⊆ x ==> ∃z. S ⊆ {z}"
by (metis Set.subsetI Union_upper insert_iff set_eq_subset)

lemma singleton_example_2:
"∀x ∈ S. \<Union>S ⊆ x ==> ∃z. S ⊆ {z}"
proof -
assume "∀x ∈ S. \<Union>S ⊆ x"
hence "∀x1. x1 ⊆ \<Union>S ∧ x1 ∈ S --> x1 = \<Union>S" by (metis set_eq_subset)
hence "∀x1. x1 ∈ S --> x1 = \<Union>S" by (metis Union_upper)
hence "∀x1::('a set) set. \<Union>S ∈ x1 --> S ⊆ x1" by (metis subsetI)
hence "∀x1::('a set) set. S ⊆ insert (\<Union>S) x1" by (metis insert_iff)
thus "∃z. S ⊆ {z}" by metis
qed

text {*
From W. W. Bledsoe and Guohui Feng, SET-VAR. JAR 11 (3), 1993, pages
293-314.
*}


(* Notes: (1) The numbering doesn't completely agree with the paper.
(2) We must rename set variables to avoid type clashes. *)

lemma "∃B. (∀x ∈ B. x ≤ (0::int))"
"D ∈ F ==> ∃G. ∀A ∈ G. ∃B ∈ F. A ⊆ B"
"P a ==> ∃A. (∀x ∈ A. P x) ∧ (∃y. y ∈ A)"
"a < b ∧ b < (c::int) ==> ∃B. a ∉ B ∧ b ∈ B ∧ c ∉ B"
"P (f b) ==> ∃s A. (∀x ∈ A. P x) ∧ f s ∈ A"
"P (f b) ==> ∃s A. (∀x ∈ A. P x) ∧ f s ∈ A"
"∃A. a ∉ A"
"(∀C. (0, 0) ∈ C ∧ (∀x y. (x, y) ∈ C --> (Suc x, Suc y) ∈ C) --> (n, m) ∈ C) ∧ Q n --> Q m"
apply (metis all_not_in_conv)
apply (metis all_not_in_conv)
apply (metis mem_Collect_eq)
apply (metis less_le singleton_iff)
apply (metis mem_Collect_eq)
apply (metis mem_Collect_eq)
apply (metis all_not_in_conv)
by (metis pair_in_Id_conv)

end