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
          ↳ QDPSizeChangeProof

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.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: