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

(0) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, activate(XS))
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:

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

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, activate(XS))
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)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, activate(XS))
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)
ACTIVATE(n__from(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)
n__from(x1)  =  n__from(x1)
from(x1)  =  from(x1)
cons(x1, x2)  =  x2
after(x1, x2)  =  after(x1, x2)
0  =  0
s(x1)  =  s(x1)
activate(x1)  =  activate(x1)

Recursive path order with status [RPO].
Precedence:

after₂ > activate₁ > from₁ > n_s1
after₂ > activate₁ > from₁ > n_from1
after₂ > activate₁ > s₁ > n_s1

Status:

after₂: [1,2]

The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, n__from(n__s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, activate(XS))
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

(7) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, activate(XS))
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) 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:

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

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, activate(XS))
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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AFTER(x1, x2)  =  x1
s(x1)  =  s(x1)
from(x1)  =  from
cons(x1, x2)  =  x2
n__from(x1)  =  n__from
n__s(x1)  =  n__s(x1)
after(x1, x2)  =  after(x1, x2)
0  =  0
activate(x1)  =  activate(x1)

Recursive path order with status [RPO].
Precedence:

after₂ > activate₁ > s₁ > n_s1
after₂ > activate₁ > from > n_from

Status:

after₂: [1,2]

The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, n__from(n__s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, activate(XS))
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

(12) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, activate(XS))
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) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE