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

(0) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(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:

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

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(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 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(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.


ACTIVATE(n__s(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACTIVATE(x1)  =  ACTIVATE(x1)

Tags:
ACTIVATE has tags [0]

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
n__s(x1)  =  n__s(x1)
n__from(x1)  =  x1

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

Status:
ns1: [1]


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

ACTIVATE(n__from(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

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

(8) 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)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ACTIVATE(x1)  =  ACTIVATE(x1)

Tags:
ACTIVATE has tags [0]

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Lexicographic path order with status [LPO].
Quasi-Precedence:
trivial

Status:
nfrom1: [1]


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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(n__s(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

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

(13) 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
SEL(x1, x2)  =  SEL(x1)

Tags:
SEL has tags [1,1]

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
s(x1)  =  s(x1)
cons(x1, x2)  =  cons(x1, x2)
activate(x1)  =  activate(x1)
n__from(x1)  =  n__from(x1)
from(x1)  =  from
n__s(x1)  =  n__s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[cons2, activate1] > s1 > ns1
[cons2, activate1] > [nfrom1, from] > ns1

Status:
s1: [1]
cons2: [1,2]
activate1: [1]
nfrom1: [1]
from: []
ns1: [1]


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE