(*
Author: Jeremy Dawson and Gerwin Klein, NICTA
Consequences of type definition theorems, and of extended type definition.
*)
section \<open>Type Definition Theorems\<close>
theory Misc_Typedef
imports Main
begin
section "More lemmas about normal type definitions"
lemma
tdD1: "type_definition Rep Abs A \<Longrightarrow> \<forall>x. Rep x \<in> A" and
tdD2: "type_definition Rep Abs A \<Longrightarrow> \<forall>x. Abs (Rep x) = x" and
tdD3: "type_definition Rep Abs A \<Longrightarrow> \<forall>y. y \<in> A \<longrightarrow> Rep (Abs y) = y"
by (auto simp: type_definition_def)
lemma td_nat_int:
"type_definition int nat (Collect (op <= 0))"
unfolding type_definition_def by auto
context type_definition
begin
declare Rep [iff] Rep_inverse [simp] Rep_inject [simp]
lemma Abs_eqD: "Abs x = Abs y ==> x \<in> A ==> y \<in> A ==> x = y"
by (simp add: Abs_inject)
lemma Abs_inverse':
"r : A ==> Abs r = a ==> Rep a = r"
by (safe elim!: Abs_inverse)
lemma Rep_comp_inverse:
"Rep o f = g ==> Abs o g = f"
using Rep_inverse by auto
lemma Rep_eqD [elim!]: "Rep x = Rep y ==> x = y"
by simp
lemma Rep_inverse': "Rep a = r ==> Abs r = a"
by (safe intro!: Rep_inverse)
lemma comp_Abs_inverse:
"f o Abs = g ==> g o Rep = f"
using Rep_inverse by auto
lemma set_Rep:
"A = range Rep"
proof (rule set_eqI)
fix x
show "(x \<in> A) = (x \<in> range Rep)"
by (auto dest: Abs_inverse [of x, symmetric])
qed
lemma set_Rep_Abs: "A = range (Rep o Abs)"
proof (rule set_eqI)
fix x
show "(x \<in> A) = (x \<in> range (Rep o Abs))"
by (auto dest: Abs_inverse [of x, symmetric])
qed
lemma Abs_inj_on: "inj_on Abs A"
unfolding inj_on_def
by (auto dest: Abs_inject [THEN iffD1])
lemma image: "Abs ` A = UNIV"
by (auto intro!: image_eqI)
lemmas td_thm = type_definition_axioms
lemma fns1:
"Rep o fa = fr o Rep | fa o Abs = Abs o fr ==> Abs o fr o Rep = fa"
by (auto dest: Rep_comp_inverse elim: comp_Abs_inverse simp: o_assoc)
lemmas fns1a = disjI1 [THEN fns1]
lemmas fns1b = disjI2 [THEN fns1]
lemma fns4:
"Rep o fa o Abs = fr ==>
Rep o fa = fr o Rep & fa o Abs = Abs o fr"
by auto
end
interpretation nat_int: type_definition int nat "Collect (op <= 0)"
by (rule td_nat_int)
declare
nat_int.Rep_cases [cases del]
nat_int.Abs_cases [cases del]
nat_int.Rep_induct [induct del]
nat_int.Abs_induct [induct del]
subsection "Extended form of type definition predicate"
lemma td_conds:
"norm o norm = norm ==> (fr o norm = norm o fr) =
(norm o fr o norm = fr o norm & norm o fr o norm = norm o fr)"
apply safe
apply (simp_all add: comp_assoc)
apply (simp_all add: o_assoc)
done
lemma fn_comm_power:
"fa o tr = tr o fr ==> fa ^^ n o tr = tr o fr ^^ n"
apply (rule ext)
apply (induct n)
apply (auto dest: fun_cong)
done
lemmas fn_comm_power' =
ext [THEN fn_comm_power, THEN fun_cong, unfolded o_def]
locale td_ext = type_definition +
fixes norm
assumes eq_norm: "\<And>x. Rep (Abs x) = norm x"
begin
lemma Abs_norm [simp]:
"Abs (norm x) = Abs x"
using eq_norm [of x] by (auto elim: Rep_inverse')
lemma td_th:
"g o Abs = f ==> f (Rep x) = g x"
by (drule comp_Abs_inverse [symmetric]) simp
lemma eq_norm': "Rep o Abs = norm"
by (auto simp: eq_norm)
lemma norm_Rep [simp]: "norm (Rep x) = Rep x"
by (auto simp: eq_norm' intro: td_th)
lemmas td = td_thm
lemma set_iff_norm: "w : A <-> w = norm w"
by (auto simp: set_Rep_Abs eq_norm' eq_norm [symmetric])
lemma inverse_norm:
"(Abs n = w) = (Rep w = norm n)"
apply (rule iffI)
apply (clarsimp simp add: eq_norm)
apply (simp add: eq_norm' [symmetric])
done
lemma norm_eq_iff:
"(norm x = norm y) = (Abs x = Abs y)"
by (simp add: eq_norm' [symmetric])
lemma norm_comps:
"Abs o norm = Abs"
"norm o Rep = Rep"
"norm o norm = norm"
by (auto simp: eq_norm' [symmetric] o_def)
lemmas norm_norm [simp] = norm_comps
lemma fns5:
"Rep o fa o Abs = fr ==>
fr o norm = fr & norm o fr = fr"
by (fold eq_norm') auto
(* following give conditions for converses to td_fns1
the condition (norm o fr o norm = fr o norm) says that
fr takes normalised arguments to normalised results,
(norm o fr o norm = norm o fr) says that fr
takes norm-equivalent arguments to norm-equivalent results,
(fr o norm = fr) says that fr
takes norm-equivalent arguments to the same result, and
(norm o fr = fr) says that fr takes any argument to a normalised result
*)
lemma fns2:
"Abs o fr o Rep = fa ==>
(norm o fr o norm = fr o norm) = (Rep o fa = fr o Rep)"
apply (fold eq_norm')
apply safe
prefer 2
apply (simp add: o_assoc)
apply (rule ext)
apply (drule_tac x="Rep x" in fun_cong)
apply auto
done
lemma fns3:
"Abs o fr o Rep = fa ==>
(norm o fr o norm = norm o fr) = (fa o Abs = Abs o fr)"
apply (fold eq_norm')
apply safe
prefer 2
apply (simp add: comp_assoc)
apply (rule ext)
apply (drule_tac f="a o b" for a b in fun_cong)
apply simp
done
lemma fns:
"fr o norm = norm o fr ==>
(fa o Abs = Abs o fr) = (Rep o fa = fr o Rep)"
apply safe
apply (frule fns1b)
prefer 2
apply (frule fns1a)
apply (rule fns3 [THEN iffD1])
prefer 3
apply (rule fns2 [THEN iffD1])
apply (simp_all add: comp_assoc)
apply (simp_all add: o_assoc)
done
lemma range_norm:
"range (Rep o Abs) = A"
by (simp add: set_Rep_Abs)
end
lemmas td_ext_def' =
td_ext_def [unfolded type_definition_def td_ext_axioms_def]
end