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:

from1(X) -> cons2(X, n__from1(n__s1(X)))
after2(0, XS) -> XS
after2(s1(N), cons2(X, XS)) -> after2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(X) -> X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

from1(X) -> cons2(X, n__from1(n__s1(X)))
after2(0, XS) -> XS
after2(s1(N), cons2(X, XS)) -> after2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(X) -> X

Q is empty.

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

ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
AFTER2(s1(N), cons2(X, XS)) -> AFTER2(N, activate1(XS))
AFTER2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
after2(0, XS) -> XS
after2(s1(N), cons2(X, XS)) -> after2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(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:

ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
AFTER2(s1(N), cons2(X, XS)) -> AFTER2(N, activate1(XS))
AFTER2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
after2(0, XS) -> XS
after2(s1(N), cons2(X, XS)) -> after2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ QDP

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

ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
after2(0, XS) -> XS
after2(s1(N), cons2(X, XS)) -> after2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
Used argument filtering: ACTIVATE1(x1)  =  x1
n__from1(x1)  =  x1
n__s1(x1)  =  n__s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
          ↳ QDP

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

ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
after2(0, XS) -> XS
after2(s1(N), cons2(X, XS)) -> after2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
Used argument filtering: ACTIVATE1(x1)  =  x1
n__from1(x1)  =  n__from1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP

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

from1(X) -> cons2(X, n__from1(n__s1(X)))
after2(0, XS) -> XS
after2(s1(N), cons2(X, XS)) -> after2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof

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

AFTER2(s1(N), cons2(X, XS)) -> AFTER2(N, activate1(XS))

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
after2(0, XS) -> XS
after2(s1(N), cons2(X, XS)) -> after2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

AFTER2(s1(N), cons2(X, XS)) -> AFTER2(N, activate1(XS))
Used argument filtering: AFTER2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
activate1(x1)  =  x1
n__from1(x1)  =  n__from
from1(x1)  =  from
n__s1(x1)  =  n__s1(x1)
cons2(x1, x2)  =  cons
Used ordering: Quasi Precedence: [s_1, n__s_1] [n__from, from, cons] 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof

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

from1(X) -> cons2(X, n__from1(n__s1(X)))
after2(0, XS) -> XS
after2(s1(N), cons2(X, XS)) -> after2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(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.