# 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.
*)

section ‹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, 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: "∀(x⇩2::'b set) x⇩1::'b set. x⇩1 ⊆ x⇩1 ∪ x⇩2" by (metis Un_commute Un_upper2)
have F2a: "∀(x⇩2::'b set) x⇩1::'b set. x⇩1 ⊆ x⇩2 ⟶ x⇩2 = x⇩2 ∪ x⇩1" by (metis Un_commute subset_Un_eq)
have F2: "∀(x⇩2::'b set) x⇩1::'b set. x⇩1 ⊆ x⇩2 ∧ x⇩2 ⊆ x⇩1 ⟶ x⇩1 = x⇩2" 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 "∃x⇩1::'a set. Y ⊆ x⇩1 ∪ Z ∧ Y ∪ Z ≠ X ∧ ¬ X ⊆ x⇩1 ∪ 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 "∃x⇩1::'a set. (Z ⊆ x⇩1 ∧ Y ⊆ x⇩1) ∧ ¬ X ⊆ x⇩1"
{ 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 "∃x⇩1::'a set. Y ⊆ x⇩1 ∪ Z ∧ Y ∪ Z ≠ X ∧ ¬ X ⊆ x⇩1 ∪ 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, compress = 2]

lemma (*equal_union: *)
"(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V. Y ⊆ V ∧ Z ⊆ V ⟶ X ⊆ V))"
proof -
have F1: "∀(x⇩2::'b set) x⇩1::'b set. x⇩1 ⊆ x⇩2 ∧ x⇩2 ⊆ x⇩1 ⟶ x⇩1 = x⇩2" by (metis Un_commute subset_Un_eq)
{ assume AA1: "∃x⇩1::'a set. (Z ⊆ x⇩1 ∧ Y ⊆ x⇩1) ∧ ¬ X ⊆ x⇩1"
{ 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 "∃x⇩1::'a set. Y ⊆ x⇩1 ∪ Z ∧ Y ∪ Z ≠ X ∧ ¬ X ⊆ x⇩1 ∪ 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 "∃x⇩1::'a set. Y ⊆ x⇩1 ∪ Z ∧ Y ∪ Z ≠ X ∧ ¬ X ⊆ x⇩1 ∪ 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, compress = 3]

lemma (*equal_union: *)
"(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V. Y ⊆ V ∧ Z ⊆ V ⟶ X ⊆ V))"
proof -
have F1a: "∀(x⇩2::'b set) x⇩1::'b set. x⇩1 ⊆ x⇩2 ⟶ x⇩2 = x⇩2 ∪ x⇩1" by (metis Un_commute subset_Un_eq)
have F1: "∀(x⇩2::'b set) x⇩1::'b set. x⇩1 ⊆ x⇩2 ∧ x⇩2 ⊆ x⇩1 ⟶ x⇩1 = x⇩2" 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: "∃x⇩1::'a set. (Z ⊆ x⇩1 ∧ Y ⊆ x⇩1) ∧ ¬ X ⊆ x⇩1"
{ 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, compress = 4]

lemma (*equal_union: *)
"(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V. Y ⊆ V ∧ Z ⊆ V ⟶ X ⊆ V))"
proof -
have F1: "∀(x⇩2::'b set) x⇩1::'b set. x⇩1 ⊆ x⇩2 ∧ x⇩2 ⊆ x⇩1 ⟶ x⇩1 = x⇩2" 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 "∃x⇩1::'a set. Y ⊆ x⇩1 ∪ Z ∧ Y ∪ Z ≠ X ∧ ¬ X ⊆ x⇩1 ∪ 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, 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. ⋃S ⊆ x ⟹ ∃z. S ⊆ {z}"
by (metis Set.subsetI Union_upper insertCI set_eq_subset)

lemma (* singleton_example_2: *)
"∀x ∈ S. ⋃S ⊆ x ⟹ ∃z. S ⊆ {z}"
by (metis Set.subsetI Union_upper insert_iff set_eq_subset)

lemma singleton_example_2:
"∀x ∈ S. ⋃S ⊆ x ⟹ ∃z. S ⊆ {z}"
proof -
assume "∀x ∈ S. ⋃S ⊆ x"
hence "∀x⇩1. x⇩1 ⊆ ⋃S ∧ x⇩1 ∈ S ⟶ x⇩1 = ⋃S" by (metis set_eq_subset)
hence "∀x⇩1. x⇩1 ∈ S ⟶ x⇩1 = ⋃S" by (metis Union_upper)
hence "∀x⇩1::('a set) set. ⋃S ∈ x⇩1 ⟶ S ⊆ x⇩1" by (metis subsetI)
hence "∀x⇩1::('a set) set. S ⊆ insert (⋃S) x⇩1" 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
```