(* Title: HOL/Metis_Examples/Abstraction.thy
Author: Lawrence C. Paulson, Cambridge University Computer Laboratory
Author: Jasmin Blanchette, TU Muenchen
Example featuring Metis's support for lambda-abstractions.
*)
section \<open>Example Featuring Metis's Support for Lambda-Abstractions\<close>
theory Abstraction
imports "HOL-Library.FuncSet"
begin
(* For Christoph Benzmüller *)
lemma "x < 1 \<and> ((=) = (=)) \<Longrightarrow> ((=) = (=)) \<and> x < (2::nat)"
by (metis nat_1_add_1 trans_less_add2)
lemma "((=) ) = (\<lambda>x y. y = x)"
by metis
consts
monotone :: "['a => 'a, 'a set, ('a *'a)set] => bool"
pset :: "'a set => 'a set"
order :: "'a set => ('a * 'a) set"
lemma "a \<in> {x. P x} \<Longrightarrow> P a"
proof -
assume "a \<in> {x. P x}"
thus "P a" by (metis mem_Collect_eq)
qed
lemma Collect_triv: "a \<in> {x. P x} \<Longrightarrow> P a"
by (metis mem_Collect_eq)
lemma "a \<in> {x. P x --> Q x} \<Longrightarrow> a \<in> {x. P x} \<Longrightarrow> a \<in> {x. Q x}"
by (metis Collect_imp_eq ComplD UnE)
lemma "(a, b) \<in> Sigma A B \<Longrightarrow> a \<in> A \<and> b \<in> B a"
proof -
assume A1: "(a, b) \<in> Sigma A B"
hence F1: "b \<in> B a" by (metis mem_Sigma_iff)
have F2: "a \<in> A" by (metis A1 mem_Sigma_iff)
have "b \<in> B a" by (metis F1)
thus "a \<in> A \<and> b \<in> B a" by (metis F2)
qed
lemma Sigma_triv: "(a, b) \<in> Sigma A B \<Longrightarrow> a \<in> A & b \<in> B a"
by (metis SigmaD1 SigmaD2)
lemma "(a, b) \<in> (SIGMA x:A. {y. x = f y}) \<Longrightarrow> a \<in> A \<and> a = f b"
by (metis (full_types, lifting) CollectD SigmaD1 SigmaD2)
lemma "(a, b) \<in> (SIGMA x:A. {y. x = f y}) \<Longrightarrow> a \<in> A \<and> a = f b"
proof -
assume A1: "(a, b) \<in> (SIGMA x:A. {y. x = f y})"
hence F1: "a \<in> A" by (metis mem_Sigma_iff)
have "b \<in> {R. a = f R}" by (metis A1 mem_Sigma_iff)
hence "a = f b" by (metis (full_types) mem_Collect_eq)
thus "a \<in> A \<and> a = f b" by (metis F1)
qed
lemma "(cl, f) \<in> CLF \<Longrightarrow> CLF = (SIGMA cl: CL.{f. f \<in> pset cl}) \<Longrightarrow> f \<in> pset cl"
by (metis Collect_mem_eq SigmaD2)
lemma "(cl, f) \<in> CLF \<Longrightarrow> CLF = (SIGMA cl: CL.{f. f \<in> pset cl}) \<Longrightarrow> f \<in> pset cl"
proof -
assume A1: "(cl, f) \<in> CLF"
assume A2: "CLF = (SIGMA cl:CL. {f. f \<in> pset cl})"
have "\<forall>v u. (u, v) \<in> CLF \<longrightarrow> v \<in> {R. R \<in> pset u}" by (metis A2 mem_Sigma_iff)
hence "\<forall>v u. (u, v) \<in> CLF \<longrightarrow> v \<in> pset u" by (metis mem_Collect_eq)
thus "f \<in> pset cl" by (metis A1)
qed
lemma
"(cl, f) \<in> (SIGMA cl: CL. {f. f \<in> pset cl \<rightarrow> pset cl}) \<Longrightarrow>
f \<in> pset cl \<rightarrow> pset cl"
by (metis (no_types) Collect_mem_eq Sigma_triv)
lemma
"(cl, f) \<in> (SIGMA cl: CL. {f. f \<in> pset cl \<rightarrow> pset cl}) \<Longrightarrow>
f \<in> pset cl \<rightarrow> pset cl"
proof -
assume A1: "(cl, f) \<in> (SIGMA cl:CL. {f. f \<in> pset cl \<rightarrow> pset cl})"
have "f \<in> {R. R \<in> pset cl \<rightarrow> pset cl}" using A1 by simp
thus "f \<in> pset cl \<rightarrow> pset cl" by (metis mem_Collect_eq)
qed
lemma
"(cl, f) \<in> (SIGMA cl: CL. {f. f \<in> pset cl \<inter> cl}) \<Longrightarrow>
f \<in> pset cl \<inter> cl"
by (metis (no_types) Collect_conj_eq Int_def Sigma_triv inf_idem)
lemma
"(cl, f) \<in> (SIGMA cl: CL. {f. f \<in> pset cl \<inter> cl}) \<Longrightarrow>
f \<in> pset cl \<inter> cl"
proof -
assume A1: "(cl, f) \<in> (SIGMA cl:CL. {f. f \<in> pset cl \<inter> cl})"
have "f \<in> {R. R \<in> pset cl \<inter> cl}" using A1 by simp
hence "f \<in> Id_on cl `` pset cl" by (metis Int_commute Image_Id_on mem_Collect_eq)
hence "f \<in> cl \<inter> pset cl" by (metis Image_Id_on)
thus "f \<in> pset cl \<inter> cl" by (metis Int_commute)
qed
lemma
"(cl, f) \<in> (SIGMA cl: CL. {f. f \<in> pset cl \<rightarrow> pset cl & monotone f (pset cl) (order cl)}) \<Longrightarrow>
(f \<in> pset cl \<rightarrow> pset cl) & (monotone f (pset cl) (order cl))"
by auto
lemma
"(cl, f) \<in> CLF \<Longrightarrow>
CLF \<subseteq> (SIGMA cl: CL. {f. f \<in> pset cl \<inter> cl}) \<Longrightarrow>
f \<in> pset cl \<inter> cl"
by (metis (lifting) CollectD Sigma_triv subsetD)
lemma
"(cl, f) \<in> CLF \<Longrightarrow>
CLF = (SIGMA cl: CL. {f. f \<in> pset cl \<inter> cl}) \<Longrightarrow>
f \<in> pset cl \<inter> cl"
by (metis (lifting) CollectD Sigma_triv)
lemma
"(cl, f) \<in> CLF \<Longrightarrow>
CLF \<subseteq> (SIGMA cl': CL. {f. f \<in> pset cl' \<rightarrow> pset cl'}) \<Longrightarrow>
f \<in> pset cl \<rightarrow> pset cl"
by (metis (lifting) CollectD Sigma_triv subsetD)
lemma
"(cl, f) \<in> CLF \<Longrightarrow>
CLF = (SIGMA cl: CL. {f. f \<in> pset cl \<rightarrow> pset cl}) \<Longrightarrow>
f \<in> pset cl \<rightarrow> pset cl"
by (metis (lifting) CollectD Sigma_triv)
lemma
"(cl, f) \<in> CLF \<Longrightarrow>
CLF = (SIGMA cl: CL. {f. f \<in> pset cl \<rightarrow> pset cl & monotone f (pset cl) (order cl)}) \<Longrightarrow>
(f \<in> pset cl \<rightarrow> pset cl) & (monotone f (pset cl) (order cl))"
by auto
lemma "map (\<lambda>x. (f x, g x)) xs = zip (map f xs) (map g xs)"
apply (induct xs)
apply (metis list.map(1) zip_Nil)
by auto
lemma
"map (\<lambda>w. (w \<rightarrow> w, w \<times> w)) xs =
zip (map (\<lambda>w. w \<rightarrow> w) xs) (map (\<lambda>w. w \<times> w) xs)"
apply (induct xs)
apply (metis list.map(1) zip_Nil)
by auto
lemma "(\<lambda>x. Suc (f x)) ` {x. even x} \<subseteq> A \<Longrightarrow> \<forall>x. even x --> Suc (f x) \<in> A"
by (metis mem_Collect_eq image_eqI subsetD)
lemma
"(\<lambda>x. f (f x)) ` ((\<lambda>x. Suc(f x)) ` {x. even x}) \<subseteq> A \<Longrightarrow>
(\<forall>x. even x --> f (f (Suc(f x))) \<in> A)"
by (metis mem_Collect_eq imageI rev_subsetD)
lemma "f \<in> (\<lambda>u v. b \<times> u \<times> v) ` A \<Longrightarrow> \<forall>u v. P (b \<times> u \<times> v) \<Longrightarrow> P(f y)"
by (metis (lifting) imageE)
lemma image_TimesA: "(\<lambda>(x, y). (f x, g y)) ` (A \<times> B) = (f ` A) \<times> (g ` B)"
by (metis map_prod_def map_prod_surj_on)
lemma image_TimesB:
"(\<lambda>(x, y, z). (f x, g y, h z)) ` (A \<times> B \<times> C) = (f ` A) \<times> (g ` B) \<times> (h ` C)"
by force
lemma image_TimesC:
"(\<lambda>(x, y). (x \<rightarrow> x, y \<times> y)) ` (A \<times> B) =
((\<lambda>x. x \<rightarrow> x) ` A) \<times> ((\<lambda>y. y \<times> y) ` B)"
by (metis image_TimesA)
end