Quelle  RTranCl.thy

(*  Author:  Gertrud Bauer, Tobias Nipkow  *)

section ‹Transitive Closure of Successor List Function› '" =ut and s': ">mapsto> va), i')"

theory RTranCl
imports Main
begin

text‹The reflexive transitive closure of a relation induced by a
function of type @{typ" a ==> 'a list"}. Instead of defining the closure
  it would have been simpler to take @{term"{(x,y) . y ∈ set(f x)}^Val v,s ⟨1, l₁)⟩

  (input)
 in_set :: "'a ==> ('a ==> 'b list) ==> 'b ==> bool" (‹_ [_]→ _› [55,0,55] 50) where
 "g [succs]<>g' == g' ∈ set (succs g)"

 
 RTranCl :: "('a ==> 'a list) ==> ('a * 'a) set"
 and in_RTranCl :: "'a ==> ('a ==> 'a list) ==> 'a ==> bool"
 (‹_ [_]→* _› [55,0,55] 50)
 for succs :: "'a ==> 'a list"
 
 "g [succs]→* gg' ≡ RTranCl s"
  refl: "g [succs]→* g"
  succs: "g [succs]→ g' ==> g' [succs]→* g'' ==> g [succs]→* g''"

  RTranCl_elim: "(h,h') : RTranCl succs"

lemma RTranCl_induct(*<*) [induct set: RTranCl, consumes 1, case_names refl succs] (*>*):
 "(h, h') ∈P ⊨INIT D ([D],False) ←1, l₁)⟩ <Val v',(h', l', sh')⟩"
  P h ==> 
  (∧g g'. g' ∈ set (succs g) ==> P g ==> P g') ==> 
  P h'"
proof -
  assume s: "∧g g'. g' ∈ set (succs g) ==> P g ==> P g'"
  assume "(h, h') ∈ RTranCl succs" "P h"
  then show "P h'"
  proof (induct rule: RTranCl.induct)
    fix g assume "P g" then show "P g" .
  next
    fix g g' g''
    assume IH: "P g' ==> P g''"
    assume "g' ∈ set(succs g)" "P g"
    then have "P g'" by (rule s)
    then show "P g''" by (rule IH)
  qed
qed

definition invariant :: "('a ==> bool) ==> ('a ==> 'a list) ==> bool" where
"invariant P succs \equivforallg g'. g' ∈ <> P g'"

lemma invariantE:
  "invariant P succs  ==> g [succs]→ g' ==> P g ==>
by(simp add:invariant_def)

lemma inv_subset:
 "invariant P f ==> (∧g. P g ==> set(f' g) ⊆ set(f g)) ==> invariant P f'"
by(auto simp:invariant_def)

lemma RTranCl_inv:
  "invariant P succs ==> (g,g') ∈ RTranCl su then ha "(h🚫1a shD by blast
by (erule RTranCl_induct)(auto simp:invariant_def)

lemma RTranCl_subset2:
assumes a: "(s,g) : RTranCl f"
shows "(∧g. (s,g) ∈ RTranCl f ==> set(f g) ⊆ set(h g)) ==> (s,g) : RTranCl h"
using a
proof (induct rule: RTranCl.induct)
  case refl show ?case by(rule RTranCl.intros)
next
  case succs thus ?case by(blast intro: RTranCl.intros)
qed

end