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(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
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)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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(first(0, Z)) → MARK(nil)
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
ACTIVE(first(s(X), cons(Y, Z))) → CONS(Y, first(X, Z))
ACTIVE(first(s(X), cons(Y, Z))) → FIRST(X, Z)
ACTIVE(sel(0, cons(X, Z))) → MARK(X)
ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
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(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(first(X1, X2)) → FIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(0) → ACTIVE(0)
MARK(nil) → ACTIVE(nil)
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
MARK(sel(X1, X2)) → SEL(mark(X1), mark(X2))
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X2)
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)
FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(X1, mark(X2)) → FIRST(X1, X2)
FIRST(active(X1), X2) → FIRST(X1, X2)
FIRST(X1, active(X2)) → FIRST(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(active(X1), X2) → SEL(X1, X2)
SEL(X1, active(X2)) → SEL(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
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)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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 6 SCCs with 14 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(active(X1), X2) → SEL(X1, X2)
SEL(X1, active(X2)) → SEL(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
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)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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:

FIRST(X1, mark(X2)) → FIRST(X1, X2)
FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(active(X1), X2) → FIRST(X1, X2)
FIRST(X1, active(X2)) → FIRST(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
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)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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:

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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
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)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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:

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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
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)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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:

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(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
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)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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

(10) 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(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(sel(0, cons(X, Z))) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(first(0, Z)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(sel(0, cons(X, Z))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
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)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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