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 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 DependencyGraphProof (⇔)
    ↳9 QDP
    ↳10 QDPOrderProof (⇔)
    ↳11 QDP
    ↳12 DependencyGraphProof (⇔)
    ↳13 QDP
    ↳14 QDPOrderProof (⇔)
    ↳15 QDP
    ↳16 DependencyGraphProof (⇔)
    ↳17 QDP
  ↳18 QDP

(0) Obligation:

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

pairNs → cons(0, n__incr(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → 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:

PAIRNS → CONS(0, n__incr(oddNs))
PAIRNS → ODDNS
ODDNS → INCR(pairNs)
ODDNS → PAIRNS
INCR(cons(X, XS)) → CONS(s(X), n__incr(activate(XS)))
INCR(cons(X, XS)) → ACTIVATE(XS)
TAKE(s(N), cons(X, XS)) → CONS(X, n__take(N, activate(XS)))
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
ZIP(cons(X, XS), cons(Y, YS)) → CONS(pair(X, Y), n__zip(activate(XS), activate(YS)))
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
TAIL(cons(X, XS)) → ACTIVATE(XS)
REPITEMS(cons(X, XS)) → CONS(X, n__cons(X, n__repItems(activate(XS))))
REPITEMS(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(X)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
ACTIVATE(n__zip(X1, X2)) → ZIP(X1, X2)
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)
ACTIVATE(n__repItems(X)) → REPITEMS(X)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → 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 2 SCCs with 8 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(X)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__zip(X1, X2)) → ZIP(X1, X2)
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ACTIVATE(n__repItems(X)) → REPITEMS(X)
REPITEMS(cons(X, XS)) → ACTIVATE(XS)
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → X

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ZIP(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
INCR(x1)  =  x1
cons(x1, x2)  =  x2
ACTIVATE(x1)  =  x1
n__incr(x1)  =  x1
n__take(x1, x2)  =  x2
TAKE(x1, x2)  =  x2
s(x1)  =  s(x1)
n__zip(x1, x2)  =  n__zip(x1, x2)
ZIP(x1, x2)  =  ZIP(x1, x2)
n__repItems(x1)  =  x1
REPITEMS(x1)  =  x1
pairNs  =  pairNs
0  =  0
oddNs  =  oddNs
incr(x1)  =  x1
activate(x1)  =  x1
take(x1, x2)  =  x2
nil  =  nil
zip(x1, x2)  =  zip(x1, x2)
pair(x1, x2)  =  pair(x1)
tail(x1)  =  tail(x1)
repItems(x1)  =  x1
n__cons(x1, x2)  =  x2

Recursive Path Order [RPO].
Precedence:
s1 > [0, nil, pair1]
[nzip2, ZIP2, zip2] > [0, nil, pair1]
[pairNs, oddNs] > [0, nil, pair1]
tail1 > [0, nil, pair1]


The following usable rules [FROCOS05] were oriented:

pairNs → cons(0, n__incr(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → X

(7) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(X)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__zip(X1, X2)) → ZIP(X1, X2)
ACTIVATE(n__repItems(X)) → REPITEMS(X)
REPITEMS(cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → X

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

(8) DependencyGraphProof (EQUIVALENT transformation)

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

(9) Obligation:

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

ACTIVATE(n__incr(X)) → INCR(X)
INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__repItems(X)) → REPITEMS(X)
REPITEMS(cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → X

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

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__incr(x1)  =  x1
INCR(x1)  =  x1
cons(x1, x2)  =  x2
n__take(x1, x2)  =  n__take(x2)
TAKE(x1, x2)  =  TAKE(x2)
s(x1)  =  s(x1)
n__repItems(x1)  =  x1
REPITEMS(x1)  =  x1
pairNs  =  pairNs
0  =  0
oddNs  =  oddNs
incr(x1)  =  x1
activate(x1)  =  x1
take(x1, x2)  =  take(x2)
nil  =  nil
zip(x1, x2)  =  zip(x1)
pair(x1, x2)  =  pair(x1)
n__zip(x1, x2)  =  n__zip(x1)
tail(x1)  =  tail(x1)
repItems(x1)  =  x1
n__cons(x1, x2)  =  x2

Recursive Path Order [RPO].
Precedence:
s1 > [ntake1, TAKE1, take1] > nil
[pairNs, oddNs] > nil
0 > nil
[zip1, nzip1] > pair1 > nil
tail1 > nil


The following usable rules [FROCOS05] were oriented:

pairNs → cons(0, n__incr(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → X

(11) Obligation:

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

ACTIVATE(n__incr(X)) → INCR(X)
INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
ACTIVATE(n__repItems(X)) → REPITEMS(X)
REPITEMS(cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → X

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

(12) DependencyGraphProof (EQUIVALENT transformation)

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

(13) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(X)
ACTIVATE(n__repItems(X)) → REPITEMS(X)
REPITEMS(cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → X

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

(14) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__repItems(X)) → REPITEMS(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
INCR(x1)  =  INCR(x1)
cons(x1, x2)  =  x2
ACTIVATE(x1)  =  ACTIVATE(x1)
n__incr(x1)  =  x1
n__repItems(x1)  =  n__repItems(x1)
REPITEMS(x1)  =  REPITEMS(x1)
pairNs  =  pairNs
0  =  0
oddNs  =  oddNs
incr(x1)  =  x1
s(x1)  =  s(x1)
activate(x1)  =  x1
take(x1, x2)  =  x1
nil  =  nil
n__take(x1, x2)  =  x1
zip(x1, x2)  =  x1
pair(x1, x2)  =  pair
n__zip(x1, x2)  =  x1
tail(x1)  =  tail(x1)
repItems(x1)  =  repItems(x1)
n__cons(x1, x2)  =  x2

Recursive Path Order [RPO].
Precedence:
[nrepItems1, repItems1] > [INCR1, ACTIVATE1, REPITEMS1] > nil
[pairNs, oddNs] > 0 > nil
s1 > nil
pair > nil
tail1 > nil


The following usable rules [FROCOS05] were oriented:

pairNs → cons(0, n__incr(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → X

(15) Obligation:

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

INCR(cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__incr(X)) → INCR(X)
REPITEMS(cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → X

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

(16) DependencyGraphProof (EQUIVALENT transformation)

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

(17) Obligation:

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

ACTIVATE(n__incr(X)) → INCR(X)
INCR(cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → X

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

(18) Obligation:

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

PAIRNS → ODDNS
ODDNS → PAIRNS

The TRS R consists of the following rules:

pairNs → cons(0, n__incr(oddNs))
oddNs → incr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__zip(X1, X2)) → zip(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__repItems(X)) → repItems(X)
activate(X) → X

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