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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
  ↳6 QDP
  ↳7 QDP
  ↳8 QDP
  ↳9 QDP
  ↳10 QDP
  ↳11 QDP
    ↳12 QDPOrderProof (⇔)
    ↳13 QDP
    ↳14 DependencyGraphProof (⇔)
    ↳15 QDP

(0) Obligation:

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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(nats) → MARK(adx(zeros))
ACTIVE(nats) → ADX(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
ACTIVE(zeros) → CONS(0, zeros)
ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
ACTIVE(incr(cons(X, Y))) → CONS(s(X), incr(Y))
ACTIVE(incr(cons(X, Y))) → S(X)
ACTIVE(incr(cons(X, Y))) → INCR(Y)
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
ACTIVE(adx(cons(X, Y))) → INCR(cons(X, adx(Y)))
ACTIVE(adx(cons(X, Y))) → CONS(X, adx(Y))
ACTIVE(adx(cons(X, Y))) → ADX(Y)
ACTIVE(hd(cons(X, Y))) → MARK(X)
ACTIVE(tl(cons(X, Y))) → MARK(Y)
MARK(nats) → ACTIVE(nats)
MARK(adx(X)) → ACTIVE(adx(mark(X)))
MARK(adx(X)) → ADX(mark(X))
MARK(adx(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
MARK(0) → ACTIVE(0)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(X))
MARK(hd(X)) → ACTIVE(hd(mark(X)))
MARK(hd(X)) → HD(mark(X))
MARK(hd(X)) → MARK(X)
MARK(tl(X)) → ACTIVE(tl(mark(X)))
MARK(tl(X)) → TL(mark(X))
MARK(tl(X)) → MARK(X)
ADX(mark(X)) → ADX(X)
ADX(active(X)) → ADX(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)
INCR(mark(X)) → INCR(X)
INCR(active(X)) → INCR(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
HD(mark(X)) → HD(X)
HD(active(X)) → HD(X)
TL(mark(X)) → TL(X)
TL(active(X)) → TL(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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 7 SCCs with 13 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

TL(active(X)) → TL(X)
TL(mark(X)) → TL(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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:

HD(active(X)) → HD(X)
HD(mark(X)) → HD(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(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:

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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:

INCR(active(X)) → INCR(X)
INCR(mark(X)) → INCR(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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:

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(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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:

ADX(active(X)) → ADX(X)
ADX(mark(X)) → ADX(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(11) Obligation:

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(adx(X)) → MARK(X)
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(adx(zeros))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(hd(cons(X, Y))) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(X))
ACTIVE(tl(cons(X, Y))) → MARK(Y)
MARK(hd(X)) → ACTIVE(hd(mark(X)))
MARK(hd(X)) → MARK(X)
MARK(tl(X)) → ACTIVE(tl(mark(X)))
MARK(tl(X)) → MARK(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(12) 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(X1, X2))
MARK(s(X)) → ACTIVE(s(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK
adx(x1)  =  adx
ACTIVE(x1)  =  x1
mark(x1)  =  mark
incr(x1)  =  incr
cons(x1, x2)  =  cons
s(x1)  =  s
nats  =  nats
zeros  =  zeros
0  =  0
hd(x1)  =  hd
tl(x1)  =  tl
active(x1)  =  active

Lexicographic Path Order [LPO].
Precedence:
[mark, active] > [MARK, adx, incr, nats, zeros, hd, tl] > 0 > [cons, s]


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

(13) Obligation:

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

MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(adx(X)) → MARK(X)
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(adx(zeros))
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(hd(cons(X, Y))) → MARK(X)
MARK(incr(X)) → MARK(X)
ACTIVE(tl(cons(X, Y))) → MARK(Y)
MARK(hd(X)) → ACTIVE(hd(mark(X)))
MARK(hd(X)) → MARK(X)
MARK(tl(X)) → ACTIVE(tl(mark(X)))
MARK(tl(X)) → MARK(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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

(14) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(15) Obligation:

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

ACTIVE(incr(cons(X, Y))) → MARK(cons(s(X), incr(Y)))
MARK(adx(X)) → ACTIVE(adx(mark(X)))
ACTIVE(adx(cons(X, Y))) → MARK(incr(cons(X, adx(Y))))
MARK(adx(X)) → MARK(X)
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(adx(zeros))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(hd(cons(X, Y))) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(hd(X)) → ACTIVE(hd(mark(X)))
ACTIVE(tl(cons(X, Y))) → MARK(Y)
MARK(hd(X)) → MARK(X)
MARK(tl(X)) → ACTIVE(tl(mark(X)))
MARK(tl(X)) → MARK(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
mark(nats) → active(nats)
mark(adx(X)) → active(adx(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(X))
mark(hd(X)) → active(hd(mark(X)))
mark(tl(X)) → active(tl(mark(X)))
adx(mark(X)) → adx(X)
adx(active(X)) → adx(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
hd(mark(X)) → hd(X)
hd(active(X)) → hd(X)
tl(mark(X)) → tl(X)
tl(active(X)) → tl(X)

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