(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(after(0, XS)) → MARK(XS)
ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))
ACTIVE(after(s(N), cons(X, XS))) → AFTER(N, XS)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))
MARK(after(X1, X2)) → AFTER(mark(X1), mark(X2))
MARK(after(X1, X2)) → MARK(X1)
MARK(after(X1, X2)) → MARK(X2)
MARK(0) → ACTIVE(0)
FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(X)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(X)
AFTER(mark(X1), X2) → AFTER(X1, X2)
AFTER(X1, mark(X2)) → AFTER(X1, X2)
AFTER(active(X1), X2) → AFTER(X1, X2)
AFTER(X1, active(X2)) → AFTER(X1, X2)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 9 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
AFTER(X1, mark(X2)) → AFTER(X1, X2)
AFTER(mark(X1), X2) → AFTER(X1, X2)
AFTER(active(X1), X2) → AFTER(X1, X2)
AFTER(X1, active(X2)) → AFTER(X1, X2)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(active(X)) → S(X)
S(mark(X)) → S(X)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FROM(active(X)) → FROM(X)
FROM(mark(X)) → FROM(X)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(after(0, XS)) → MARK(XS)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))
MARK(after(X1, X2)) → MARK(X1)
MARK(after(X1, X2)) → MARK(X2)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.