section‹Set Theory examples: Cantor's Theorem, Schröder-Bernstein Theorem, etc.›
theory Set_Theory imports Main begin
text‹
These two are cited in Benzmueller and Kohlhase's system description
of LEO, CADE-15, 1998 (pages 139-143) as theorems LEO could not
prove. ›
lemma"(X = Y ∪ Z) = (Y ⊆ X ∧ Z ⊆ X ∧ (∀V. Y ⊆ V ∧ Z ⊆ V ⟶ X ⊆ V))" by blast
lemma"(X = Y ∩ Z) = (X ⊆ Y ∧ X ⊆ Z ∧ (∀V. V ⊆ Y ∧ V ⊆ Z ⟶ V ⊆ X))" by blast
text‹
Trivial example of term synthesis: apparently hard for some provers! ›
schematic_goal "a ≠ b ==> a ∈ ?X ∧ b ∉ ?X" by blast
subsection‹Examples for the ‹blast› paper›
lemma"(∪x ∈ C. f x ∪ g x) = ∪(f ` C) ∪∪(g ` C)"
― ‹Union-image, called ‹Un_Union_image› in Main HOL› by blast
lemma"(∩x ∈ C. f x ∩ g x) = ∩(f ` C) ∩∩(g ` C)"
― ‹Inter-image, called ‹Int_Inter_image› in Main HOL› by blast
lemma singleton_example_1: "∧S::'a set set. ∀x ∈ S. ∀y ∈ S. x ⊆ y ==>∃z. S ⊆ {z}" by blast
lemma singleton_example_2: "∀x ∈ S. ∪S ⊆ x ==>∃z. S ⊆ {z}"
― ‹Variant of the problem above.› by blast
lemma"∃!x. f (g x) = x ==>∃!y. g (f y) = y"
― ‹A unique fixpoint theorem --- ‹fast›/‹best›/‹meson› all fail.› by metis
subsection‹Cantor's Theorem: There is no surjection from a set to its powerset›
lemma cantor1: "¬ (∃f:: 'a ==> 'a set. ∀S. ∃x. f x = S)"
― ‹Requires best-first search because it is undirectional.› by best
schematic_goal "∀f:: 'a ==> 'a set. ∀x. f x ≠ ?S f"
― ‹This form displays the diagonal term.› by best
schematic_goal "?S ∉ range (f :: 'a ==> 'a set)"
― ‹This form exploits the set constructs.› by (rule notI, erule rangeE, best)
schematic_goal "?S ∉ range (f :: 'a ==> 'a set)"
― ‹Or just this!› by best
subsection‹The Schröder-Bernstein Theorem›
lemma decomposition: obtains X where"X = - (g ` (- (f ` X)))" using lfp_unfold [OF monoI, of "λX. - g ` (- f ` X)"] by blast
theorem Schroeder_Bernstein: fixes f :: "'a ==> 'b"and g :: "'b ==> 'a" assumes"inj f""inj g" obtains h:: "'a ==> 'b"where"inj h""surj h" proof (rule decomposition) fix X assume X: "X = - (g ` (- (f ` X)))" let ?h = "λz. if z ∈ X then f z else inv g z" show thesis proof have"inj_on (inv g) (-X)" by (metis X ‹inj g› bij_betw_def double_complement inj_imp_bij_betw_inv) with‹inj f›show"inj ?h" unfolding inj_on_def by (metis Compl_iff X ‹inj g› imageI image_inv_f_f) show"surj ?h" using‹inj g› X image_iff surj_def by fastforce qed qed
subsection‹A simple party theorem›
text‹\emph{At any party there are two people who know the same
of people}. Provided the party consists of at least two people
the knows relation is symmetric. Knowing yourself does not count
-- otherwise knows needs to be reflexive. (From Freek Wiedijk's talk
TPHOLs 2007.)›
lemma equal_number_of_acquaintances: assumes"Domain R <= A"and"sym R"and"card A ≥ 2" shows"¬ inj_on (%a. card(R `` {a} - {a})) A" proof - let ?N = "%a. card(R `` {a} - {a})" let ?n = "card A" have"finite A"using‹card A ≥ 2›by(auto intro:ccontr) have0: "R `` A <= A"using‹sym R›‹Domain R <= A› unfolding Domain_unfold sym_def by blast have h: "∀a∈A. R `` {a} <= A"using0by blast hence1: "∀a∈A. finite(R `` {a})"using‹finite A› by(blast intro: finite_subset) have sub: "?N ` A <= {0..<?n}" proof - have"∀a∈A. R `` {a} - {a} < A"using h by blast thus ?thesis using psubset_card_mono[OF ‹finite A›] by auto qed show"~ inj_on ?N A" (is"~ ?I") proof assume ?I hence"?n = card(?N ` A)"by(rule card_image[symmetric]) with sub ‹finite A›have2[simp]: "?N ` A = {0..<?n}" using subset_card_intvl_is_intvl[of _ 0] by(auto) have"0 ∈ ?N ` A"and"?n - 1 ∈ ?N ` A"using‹card A ≥ 2›by simp+ thenobtain a b where ab: "a∈A""b∈A"and Na: "?N a = 0"and Nb: "?N b = ?n - 1" by (auto simp del: 2) have"a ≠ b"using Na Nb ‹card A ≥ 2›by auto have"R `` {a} - {a} = {}"by (metis 1 Na ab card_eq_0_iff finite_Diff) hence"b ∉ R `` {a}"using‹a≠b›by blast hence"a ∉ R `` {b}"by (metis Image_singleton_iff assms(2) sym_def) hence3: "R `` {b} - {b} <= A - {a,b}"using0 ab by blast have4: "finite (A - {a,b})"using‹finite A›by simp have"?N b <= ?n - 2"using ab ‹a≠b›‹finite A› card_mono[OF 43] by simp thenshow False using Nb ‹card A ≥ 2›by arith qed qed
text‹
From W. W. Bledsoe and Guohui Feng, SET-VAR. JAR 11 (3), 1993, pages
293-314.
Isabelle can prove the easy examples without any special mechanisms,
but it can't prove the hard ones. ›
lemma"∃A. (∀x ∈ A. x ≤ (0::int))"
― ‹Example 1, page 295.› by force
lemma"D ∈ F ==>∃G. ∀A ∈ G. ∃B ∈ F. A ⊆ B"
― ‹Example 2.› by force
lemma"P a ==>∃A. (∀x ∈ A. P x) ∧ (∃y. y ∈ A)"
― ‹Example 3.› by force
lemma"a < b ∧ b < (c::int) ==>∃A. a ∉ A ∧ b ∈ A ∧ c ∉ A"
― ‹Example 4.› by auto ― ‹slow›
lemma"P (f b) ==>∃s A. (∀x ∈ A. P x) ∧ f s ∈ A"
― ‹Example 5, page 298.› by force
lemma"P (f b) ==>∃s A. (∀x ∈ A. P x) ∧ f s ∈ A"
― ‹Example 6.› by force
lemma"∃A. a ∉ A"
― ‹Example 7.› by force
lemma"(∀u v. u < (0::int) ⟶ u ≠∣v∣) ⟶ (∃A::int set. -2 ∈ A & (∀y. ∣y∣∉ A))"
― ‹Example 8 needs a small hint.› by force
― ‹not ‹blast›, which can't simplify ‹-2 < 0››
text‹Example 9 omitted (requires the reals).›
text‹The paper has no Example 10!›
lemma"(∀A. 0 ∈ A ∧ (∀x ∈ A. Suc x ∈ A) ⟶ n ∈ A) ∧ P 0 ∧ (∀x. P x ⟶ P (Suc x)) ⟶ P n"
― ‹Example 11: needs a hint.› by(metis nat.induct)
lemma "(∀A. (0, 0) ∈ A ∧ (∀x y. (x, y) ∈ A ⟶ (Suc x, Suc y) ∈ A) ⟶ (n, m) ∈ A) ∧ P n ⟶ P m"
― ‹Example 12.› by auto
lemma "(∀x. (∃u. x = 2 * u) = (¬ (∃v. Suc x = 2 * v))) ⟶ (∃A. ∀x. (x ∈ A) = (Suc x ∉ A))"
― ‹Example EO1: typo in article, and with the obvious fix it seems
to require arithmetic reasoning. 2024-06-19: now trivial for sledgehammer (LCP)› by (metis even_Suc mem_Collect_eq)
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