src/HOL/Word/Misc_Typedef.thy
author wenzelm
Sun, 20 Nov 2011 20:59:30 +0100
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(* 
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  Author:     Jeremy Dawson and Gerwin Klein, NICTA
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  Consequences of type definition theorems, and of extended type definition. theorems
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*)
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header {* Type Definition Theorems *}
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theory Misc_Typedef
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imports Main
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begin
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section "More lemmas about normal type definitions"
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lemma
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  tdD1: "type_definition Rep Abs A \<Longrightarrow> \<forall>x. Rep x \<in> A" and
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  tdD2: "type_definition Rep Abs A \<Longrightarrow> \<forall>x. Abs (Rep x) = x" and
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  tdD3: "type_definition Rep Abs A \<Longrightarrow> \<forall>y. y \<in> A \<longrightarrow> Rep (Abs y) = y"
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  by (auto simp: type_definition_def)
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lemma td_nat_int: 
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  "type_definition int nat (Collect (op <= 0))"
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  unfolding type_definition_def by auto
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context type_definition
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begin
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declare Rep [iff] Rep_inverse [simp] Rep_inject [simp]
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lemma Abs_eqD: "Abs x = Abs y ==> x \<in> A ==> y \<in> A ==> x = y"
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  by (simp add: Abs_inject)
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lemma Abs_inverse': 
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  "r : A ==> Abs r = a ==> Rep a = r"
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  by (safe elim!: Abs_inverse)
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lemma Rep_comp_inverse: 
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  "Rep o f = g ==> Abs o g = f"
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  using Rep_inverse by auto
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lemma Rep_eqD [elim!]: "Rep x = Rep y ==> x = y"
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  by simp
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lemma Rep_inverse': "Rep a = r ==> Abs r = a"
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  by (safe intro!: Rep_inverse)
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lemma comp_Abs_inverse: 
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  "f o Abs = g ==> g o Rep = f"
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  using Rep_inverse by auto
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lemma set_Rep: 
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  "A = range Rep"
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proof (rule set_eqI)
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  fix x
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  show "(x \<in> A) = (x \<in> range Rep)"
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    by (auto dest: Abs_inverse [of x, symmetric])
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qed  
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lemma set_Rep_Abs: "A = range (Rep o Abs)"
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proof (rule set_eqI)
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  fix x
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  show "(x \<in> A) = (x \<in> range (Rep o Abs))"
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    by (auto dest: Abs_inverse [of x, symmetric])
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qed  
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lemma Abs_inj_on: "inj_on Abs A"
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  unfolding inj_on_def 
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  by (auto dest: Abs_inject [THEN iffD1])
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lemma image: "Abs ` A = UNIV"
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  by (auto intro!: image_eqI)
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lemmas td_thm = type_definition_axioms
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lemma fns1: 
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  "Rep o fa = fr o Rep | fa o Abs = Abs o fr ==> Abs o fr o Rep = fa"
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  by (auto dest: Rep_comp_inverse elim: comp_Abs_inverse simp: o_assoc)
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lemmas fns1a = disjI1 [THEN fns1]
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lemmas fns1b = disjI2 [THEN fns1]
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lemma fns4:
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  "Rep o fa o Abs = fr ==> 
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   Rep o fa = fr o Rep & fa o Abs = Abs o fr"
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  by auto
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end
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interpretation nat_int: type_definition int nat "Collect (op <= 0)"
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  by (rule td_nat_int)
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declare
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  nat_int.Rep_cases [cases del]
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  nat_int.Abs_cases [cases del]
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  nat_int.Rep_induct [induct del]
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  nat_int.Abs_induct [induct del]
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subsection "Extended form of type definition predicate"
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lemma td_conds:
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  "norm o norm = norm ==> (fr o norm = norm o fr) = 
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    (norm o fr o norm = fr o norm & norm o fr o norm = norm o fr)"
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  apply safe
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    apply (simp_all add: o_assoc [symmetric])
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   apply (simp_all add: o_assoc)
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  done
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lemma fn_comm_power:
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  "fa o tr = tr o fr ==> fa ^^ n o tr = tr o fr ^^ n" 
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  apply (rule ext) 
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  apply (induct n)
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   apply (auto dest: fun_cong)
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  done
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lemmas fn_comm_power' =
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  ext [THEN fn_comm_power, THEN fun_cong, unfolded o_def]
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locale td_ext = type_definition +
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  fixes norm
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  assumes eq_norm: "\<And>x. Rep (Abs x) = norm x"
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begin
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lemma Abs_norm [simp]: 
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  "Abs (norm x) = Abs x"
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  using eq_norm [of x] by (auto elim: Rep_inverse')
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lemma td_th:
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  "g o Abs = f ==> f (Rep x) = g x"
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  by (drule comp_Abs_inverse [symmetric]) simp
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lemma eq_norm': "Rep o Abs = norm"
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  by (auto simp: eq_norm)
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lemma norm_Rep [simp]: "norm (Rep x) = Rep x"
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  by (auto simp: eq_norm' intro: td_th)
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lemmas td = td_thm
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lemma set_iff_norm: "w : A <-> w = norm w"
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  by (auto simp: set_Rep_Abs eq_norm' eq_norm [symmetric])
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lemma inverse_norm: 
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  "(Abs n = w) = (Rep w = norm n)"
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  apply (rule iffI)
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   apply (clarsimp simp add: eq_norm)
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  apply (simp add: eq_norm' [symmetric])
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  done
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lemma norm_eq_iff: 
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  "(norm x = norm y) = (Abs x = Abs y)"
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  by (simp add: eq_norm' [symmetric])
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lemma norm_comps: 
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  "Abs o norm = Abs" 
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  "norm o Rep = Rep" 
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  "norm o norm = norm"
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  by (auto simp: eq_norm' [symmetric] o_def)
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lemmas norm_norm [simp] = norm_comps
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lemma fns5: 
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  "Rep o fa o Abs = fr ==> 
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  fr o norm = fr & norm o fr = fr"
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  by (fold eq_norm') auto
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(* following give conditions for converses to td_fns1
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  the condition (norm o fr o norm = fr o norm) says that 
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  fr takes normalised arguments to normalised results,
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  (norm o fr o norm = norm o fr) says that fr 
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  takes norm-equivalent arguments to norm-equivalent results,
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  (fr o norm = fr) says that fr 
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  takes norm-equivalent arguments to the same result, and 
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  (norm o fr = fr) says that fr takes any argument to a normalised result 
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  *)
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lemma fns2: 
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  "Abs o fr o Rep = fa ==> 
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   (norm o fr o norm = fr o norm) = (Rep o fa = fr o Rep)"
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  apply (fold eq_norm')
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  apply safe
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   prefer 2
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   apply (simp add: o_assoc)
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  apply (rule ext)
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  apply (drule_tac x="Rep x" in fun_cong)
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  apply auto
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  done
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lemma fns3: 
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  "Abs o fr o Rep = fa ==> 
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   (norm o fr o norm = norm o fr) = (fa o Abs = Abs o fr)"
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  apply (fold eq_norm')
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  apply safe
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   prefer 2
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   apply (simp add: o_assoc [symmetric])
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  apply (rule ext)
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  apply (drule fun_cong)
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  apply simp
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  done
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lemma fns: 
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  "fr o norm = norm o fr ==> 
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    (fa o Abs = Abs o fr) = (Rep o fa = fr o Rep)"
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  apply safe
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   apply (frule fns1b)
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   prefer 2 
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   apply (frule fns1a) 
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   apply (rule fns3 [THEN iffD1])
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     prefer 3
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     apply (rule fns2 [THEN iffD1])
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       apply (simp_all add: o_assoc [symmetric])
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   apply (simp_all add: o_assoc)
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  done
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lemma range_norm:
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  "range (Rep o Abs) = A"
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  by (simp add: set_Rep_Abs)
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end
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lemmas td_ext_def' =
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  td_ext_def [unfolded type_definition_def td_ext_axioms_def]
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end
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