Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.


↳ QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

AFTER(s(N), cons(X, XS)) → ACTIVATE(XS)
AFTER(s(N), cons(X, XS)) → AFTER(N, 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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

AFTER(s(N), cons(X, XS)) → ACTIVATE(XS)
AFTER(s(N), cons(X, XS)) → AFTER(N, 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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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.
We use the reduction pair processor [15].


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.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x1, x2)) = 0   
POL(from(x1)) = 0   
POL(AFTER(x1, x2)) = (1/2)x_1   
POL(n__from(x1)) = 0   
POL(s(x1)) = 1 + (2)x_1   
POL(activate(x1)) = 0   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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