Theory Sets

(*  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 "∃x X. ∀y. ∃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: "∀(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, 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, 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, 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, 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 "∀x1. x1 ⊆ ⋃S ∧ x1 ∈ S ⟶ x1 = ⋃S" by (metis set_eq_subset)
  hence "∀x1. x1 ∈ S ⟶ x1 = ⋃S" by (metis Union_upper)
  hence "∀x1::('a set) set. ⋃S ∈ x1 ⟶ S ⊆ x1" by (metis subsetI)
  hence "∀x1::('a set) set. S ⊆ insert (⋃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