Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(tail(X)) → active(tail(mark(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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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(zeros) → MARK(cons(0, zeros))
ACTIVE(zeros) → CONS(0, zeros)
ACTIVE(tail(cons(X, XS))) → MARK(XS)
MARK(zeros) → ACTIVE(zeros)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(0) → ACTIVE(0)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → TAIL(mark(X))
MARK(tail(X)) → MARK(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)
TAIL(mark(X)) → TAIL(X)
TAIL(active(X)) → TAIL(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(tail(X)) → active(tail(mark(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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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 3 SCCs with 4 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TAIL(active(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(tail(X)) → active(tail(mark(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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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:

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(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(tail(X)) → active(tail(mark(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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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:

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(tail(cons(X, XS))) → MARK(XS)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → MARK(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(tail(X)) → active(tail(mark(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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK
cons(x1, x2)  =  cons
ACTIVE(x1)  =  x1
mark(x1)  =  mark
tail(x1)  =  tail
zeros  =  zeros
0  =  0
active(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:

[MARK, mark, tail, zeros] > cons
[MARK, mark, tail, zeros] > 0

Status:

MARK: []
cons: []
mark: []
tail: []
zeros: []
0: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(tail(X)) → active(tail(mark(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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVE(tail(cons(X, XS))) → MARK(XS)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → MARK(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(tail(X)) → active(tail(mark(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)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.