Termination w.r.t. Q of the given QTRS could be proven:

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 PisEmptyProof (⇔)
        ↳13 TRUE
    ↳14 QDP
        ↳15 QDPOrderProof (⇔)
        ↳16 QDP
        ↳17 PisEmptyProof (⇔)
        ↳18 TRUE

(0) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
s(X) → n__s(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
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:

2ND(cons(X, XS)) → HEAD(activate(XS))
2ND(cons(X, XS)) → ACTIVATE(XS)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
SEL(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
s(X) → n__s(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
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 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
s(X) → n__s(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
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.


ACTIVATE(n__from(X)) → ACTIVATE(X)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__from(x1)  =  n__from(x1)
n__s(x1)  =  x1
n__take(x1, x2)  =  n__take(x1, x2)
TAKE(x1, x2)  =  TAKE(x1, x2)
activate(x1)  =  x1
s(x1)  =  x1
cons(x1, x2)  =  x2
from(x1)  =  from(x1)
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil

Lexicographic path order with status [LPO].
Quasi-Precedence:
[nfrom1, from1]
0 > [ntake2, TAKE2, take2, nil]

Status:
nfrom1: [1]
ntake2: [1,2]
TAKE2: [1,2]
from1: [1]
take2: [1,2]
0: []
nil: []


The following usable rules [FROCOS05] were oriented:

activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
s(X) → n__s(X)
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
take(0, XS) → nil
take(X1, X2) → n__take(X1, X2)

(7) Obligation:

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

ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
s(X) → n__s(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
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__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
s(X) → n__s(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
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.


ACTIVATE(n__s(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__s(x1)  =  n__s(x1)

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

Status:
ns1: [1]


The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
s(X) → n__s(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

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

(12) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(13) TRUE

(14) Obligation:

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

SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
s(X) → n__s(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  SEL(x1, x2)
s(x1)  =  s(x1)
cons(x1, x2)  =  x2
activate(x1)  =  x1
n__from(x1)  =  n__from
from(x1)  =  from
n__s(x1)  =  n__s(x1)
n__take(x1, x2)  =  n__take(x1, x2)
take(x1, x2)  =  take(x1, x2)
0  =  0
nil  =  nil

Lexicographic path order with status [LPO].
Quasi-Precedence:
[SEL2, s1, ns1] > [ntake2, take2] > nil
[nfrom, from]

Status:
SEL2: [2,1]
s1: [1]
nfrom: []
from: []
ns1: [1]
ntake2: [2,1]
take2: [2,1]
0: []
nil: []


The following usable rules [FROCOS05] were oriented:

activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
s(X) → n__s(X)
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
take(0, XS) → nil
take(X1, X2) → n__take(X1, X2)

(16) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
s(X) → n__s(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

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

(17) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(18) TRUE