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 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE

(0) Obligation:

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

from(X) → cons(X, n__from(s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__first(X1, X2)) → first(X1, 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:

FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__first(X1, X2)) → first(X1, 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 2 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__first(X1, X2)) → first(X1, 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__first(X1, X2)) → FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__first(x1, x2)  =  n__first(x2)
FIRST(x1, x2)  =  FIRST(x2)
s(x1)  =  s(x1)
cons(x1, x2)  =  cons(x1, x2)
from(x1)  =  from(x1)
n__from(x1)  =  n__from(x1)
first(x1, x2)  =  first(x2)
0  =  0
nil  =  nil
activate(x1)  =  activate(x1)
sel(x1, x2)  =  sel(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:
sel2 > [from1, first1, activate1] > [nfirst1, s1, cons2] > FIRST1 > [0, nil]
sel2 > [from1, first1, activate1] > nfrom1 > [0, nil]

Status:
nfirst1: multiset
FIRST1: [1]
s1: multiset
cons2: multiset
from1: [1]
nfrom1: multiset
first1: [1]
0: multiset
nil: multiset
activate1: [1]
sel2: [1,2]


The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, n__from(s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

(7) Obligation:

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

from(X) → cons(X, n__from(s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
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)  =  cons(x1, x2)
activate(x1)  =  activate(x1)
from(x1)  =  from(x1)
n__from(x1)  =  x1
first(x1, x2)  =  first(x1, x2)
0  =  0
nil  =  nil
n__first(x1, x2)  =  n__first(x1, x2)
sel(x1, x2)  =  sel(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:
SEL2 > activate1 > from1 > s1 > cons2
[first2, nfirst2] > activate1 > from1 > s1 > cons2
[first2, nfirst2] > [0, nil]
sel2 > activate1 > from1 > s1 > cons2

Status:
SEL2: [1,2]
s1: multiset
cons2: multiset
activate1: multiset
from1: [1]
first2: [1,2]
0: multiset
nil: multiset
nfirst2: [1,2]
sel2: [1,2]


The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, n__from(s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

(12) Obligation:

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

from(X) → cons(X, n__from(s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE