header {* Warshall's algorithm *}
theory Warshall
imports Main
begin
text {*
Derivation of Warshall's algorithm using program extraction,
based on Berger, Schwichtenberg and Seisenberger \cite{Berger-JAR-2001}.
*}
datatype b = T | F
primrec
is_path' :: "('a => 'a => b) => 'a => 'a list => 'a => bool"
where
"is_path' r x [] z = (r x z = T)"
| "is_path' r x (y # ys) z = (r x y = T ∧ is_path' r y ys z)"
definition
is_path :: "(nat => nat => b) => (nat * nat list * nat) =>
nat => nat => nat => bool"
where
"is_path r p i j k <-> fst p = j ∧ snd (snd p) = k ∧
list_all (λx. x < i) (fst (snd p)) ∧
is_path' r (fst p) (fst (snd p)) (snd (snd p))"
definition
conc :: "('a * 'a list * 'a) => ('a * 'a list * 'a) => ('a * 'a list * 'a)"
where
"conc p q = (fst p, fst (snd p) @ fst q # fst (snd q), snd (snd q))"
theorem is_path'_snoc [simp]:
"!!x. is_path' r x (ys @ [y]) z = (is_path' r x ys y ∧ r y z = T)"
by (induct ys) simp+
theorem list_all_scoc [simp]: "list_all P (xs @ [x]) = (P x ∧ list_all P xs)"
by (induct xs, simp+, iprover)
theorem list_all_lemma:
"list_all P xs ==> (!!x. P x ==> Q x) ==> list_all Q xs"
proof -
assume PQ: "!!x. P x ==> Q x"
show "list_all P xs ==> list_all Q xs"
proof (induct xs)
case Nil
show ?case by simp
next
case (Cons y ys)
hence Py: "P y" by simp
from Cons have Pys: "list_all P ys" by simp
show ?case
by simp (rule conjI PQ Py Cons Pys)+
qed
qed
theorem lemma1: "!!p. is_path r p i j k ==> is_path r p (Suc i) j k"
apply (unfold is_path_def)
apply (simp cong add: conj_cong add: split_paired_all)
apply (erule conjE)+
apply (erule list_all_lemma)
apply simp
done
theorem lemma2: "!!p. is_path r p 0 j k ==> r j k = T"
apply (unfold is_path_def)
apply (simp cong add: conj_cong add: split_paired_all)
apply (case_tac "aa")
apply simp+
done
theorem is_path'_conc: "is_path' r j xs i ==> is_path' r i ys k ==>
is_path' r j (xs @ i # ys) k"
proof -
assume pys: "is_path' r i ys k"
show "!!j. is_path' r j xs i ==> is_path' r j (xs @ i # ys) k"
proof (induct xs)
case (Nil j)
hence "r j i = T" by simp
with pys show ?case by simp
next
case (Cons z zs j)
hence jzr: "r j z = T" by simp
from Cons have pzs: "is_path' r z zs i" by simp
show ?case
by simp (rule conjI jzr Cons pzs)+
qed
qed
theorem lemma3:
"!!p q. is_path r p i j i ==> is_path r q i i k ==>
is_path r (conc p q) (Suc i) j k"
apply (unfold is_path_def conc_def)
apply (simp cong add: conj_cong add: split_paired_all)
apply (erule conjE)+
apply (rule conjI)
apply (erule list_all_lemma)
apply simp
apply (rule conjI)
apply (erule list_all_lemma)
apply simp
apply (rule is_path'_conc)
apply assumption+
done
theorem lemma5:
"!!p. is_path r p (Suc i) j k ==> ~ is_path r p i j k ==>
(∃q. is_path r q i j i) ∧ (∃q'. is_path r q' i i k)"
proof (simp cong add: conj_cong add: split_paired_all is_path_def, (erule conjE)+)
fix xs
assume asms:
"list_all (λx. x < Suc i) xs"
"is_path' r j xs k"
"¬ list_all (λx. x < i) xs"
show "(∃ys. list_all (λx. x < i) ys ∧ is_path' r j ys i) ∧
(∃ys. list_all (λx. x < i) ys ∧ is_path' r i ys k)"
proof
show "!!j. list_all (λx. x < Suc i) xs ==> is_path' r j xs k ==>
¬ list_all (λx. x < i) xs ==>
∃ys. list_all (λx. x < i) ys ∧ is_path' r j ys i" (is "PROP ?ih xs")
proof (induct xs)
case Nil
thus ?case by simp
next
case (Cons a as j)
show ?case
proof (cases "a=i")
case True
show ?thesis
proof
from True and Cons have "r j i = T" by simp
thus "list_all (λx. x < i) [] ∧ is_path' r j [] i" by simp
qed
next
case False
have "PROP ?ih as" by (rule Cons)
then obtain ys where ys: "list_all (λx. x < i) ys ∧ is_path' r a ys i"
proof
from Cons show "list_all (λx. x < Suc i) as" by simp
from Cons show "is_path' r a as k" by simp
from Cons and False show "¬ list_all (λx. x < i) as" by (simp)
qed
show ?thesis
proof
from Cons False ys
show "list_all (λx. x<i) (a#ys) ∧ is_path' r j (a#ys) i" by simp
qed
qed
qed
show "!!k. list_all (λx. x < Suc i) xs ==> is_path' r j xs k ==>
¬ list_all (λx. x < i) xs ==>
∃ys. list_all (λx. x < i) ys ∧ is_path' r i ys k" (is "PROP ?ih xs")
proof (induct xs rule: rev_induct)
case Nil
thus ?case by simp
next
case (snoc a as k)
show ?case
proof (cases "a=i")
case True
show ?thesis
proof
from True and snoc have "r i k = T" by simp
thus "list_all (λx. x < i) [] ∧ is_path' r i [] k" by simp
qed
next
case False
have "PROP ?ih as" by (rule snoc)
then obtain ys where ys: "list_all (λx. x < i) ys ∧ is_path' r i ys a"
proof
from snoc show "list_all (λx. x < Suc i) as" by simp
from snoc show "is_path' r j as a" by simp
from snoc and False show "¬ list_all (λx. x < i) as" by simp
qed
show ?thesis
proof
from snoc False ys
show "list_all (λx. x < i) (ys @ [a]) ∧ is_path' r i (ys @ [a]) k"
by simp
qed
qed
qed
qed (rule asms)+
qed
theorem lemma5':
"!!p. is_path r p (Suc i) j k ==> ¬ is_path r p i j k ==>
¬ (∀q. ¬ is_path r q i j i) ∧ ¬ (∀q'. ¬ is_path r q' i i k)"
by (iprover dest: lemma5)
theorem warshall:
"!!j k. ¬ (∃p. is_path r p i j k) ∨ (∃p. is_path r p i j k)"
proof (induct i)
case (0 j k)
show ?case
proof (cases "r j k")
assume "r j k = T"
hence "is_path r (j, [], k) 0 j k"
by (simp add: is_path_def)
hence "∃p. is_path r p 0 j k" ..
thus ?thesis ..
next
assume "r j k = F"
hence "r j k ~= T" by simp
hence "¬ (∃p. is_path r p 0 j k)"
by (iprover dest: lemma2)
thus ?thesis ..
qed
next
case (Suc i j k)
thus ?case
proof
assume h1: "¬ (∃p. is_path r p i j k)"
from Suc show ?case
proof
assume "¬ (∃p. is_path r p i j i)"
with h1 have "¬ (∃p. is_path r p (Suc i) j k)"
by (iprover dest: lemma5')
thus ?case ..
next
assume "∃p. is_path r p i j i"
then obtain p where h2: "is_path r p i j i" ..
from Suc show ?case
proof
assume "¬ (∃p. is_path r p i i k)"
with h1 have "¬ (∃p. is_path r p (Suc i) j k)"
by (iprover dest: lemma5')
thus ?case ..
next
assume "∃q. is_path r q i i k"
then obtain q where "is_path r q i i k" ..
with h2 have "is_path r (conc p q) (Suc i) j k"
by (rule lemma3)
hence "∃pq. is_path r pq (Suc i) j k" ..
thus ?case ..
qed
qed
next
assume "∃p. is_path r p i j k"
hence "∃p. is_path r p (Suc i) j k"
by (iprover intro: lemma1)
thus ?case ..
qed
qed
extract warshall
text {*
The program extracted from the above proof looks as follows
@{thm [display, eta_contract=false] warshall_def [no_vars]}
The corresponding correctness theorem is
@{thm [display] warshall_correctness [no_vars]}
*}
ML "@{code warshall}"
end