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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 PisEmptyProof (⇔)
↳8 TRUE

(0) Obligation:

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

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

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
after(0, XS) → XS
after(s(N), cons(X, XS)) → after(N, activate(XS))
from(X) → n__from(X)
activate(n__from(X)) → from(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 1 SCC with 2 less nodes.

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

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

(5) 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)  =  AFTER(x1, x2)
s(x1)  =  s(x1)
cons(x1, x2)  =  cons(x2)
activate(x1)  =  activate(x1)
from(x1)  =  from(x1)
n__from(x1)  =  n__from(x1)
after(x1, x2)  =  after(x1, x2)
0  =  0

Lexicographic path order with status [LPO].
Precedence:
AFTER2 > activate1 > from1 > cons1 > s1
AFTER2 > activate1 > from1 > nfrom1 > s1
after2 > activate1 > from1 > cons1 > s1
after2 > activate1 > from1 > nfrom1 > s1
0 > s1

Status:
from1: [1]
cons1: [1]
nfrom1: [1]
AFTER2: [1,2]
after2: [1,2]
activate1: [1]
s1: [1]
0: []

The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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

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

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

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE