Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

from1(X) -> cons2(X, n__from1(n__s1(X)))
length1(n__nil) -> 0
length1(n__cons2(X, Y)) -> s1(length11(activate1(Y)))
length11(X) -> length1(activate1(X))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
nil -> n__nil
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__nil) -> nil
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
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)))
length1(n__nil) -> 0
length1(n__cons2(X, Y)) -> s1(length11(activate1(Y)))
length11(X) -> length1(activate1(X))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
nil -> n__nil
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__nil) -> nil
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(X) -> X

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ACTIVATE1(n__cons2(X1, X2)) -> ACTIVATE1(X1)
LENGTH1(n__cons2(X, Y)) -> LENGTH11(activate1(Y))
ACTIVATE1(n__nil) -> NIL
LENGTH11(X) -> LENGTH1(activate1(X))
ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
LENGTH1(n__cons2(X, Y)) -> S1(length11(activate1(Y)))
LENGTH11(X) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
LENGTH1(n__cons2(X, Y)) -> ACTIVATE1(Y)
FROM1(X) -> CONS2(X, n__from1(n__s1(X)))
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__cons2(X1, X2)) -> CONS2(activate1(X1), X2)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
length1(n__nil) -> 0
length1(n__cons2(X, Y)) -> s1(length11(activate1(Y)))
length11(X) -> length1(activate1(X))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
nil -> n__nil
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__nil) -> nil
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
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__cons2(X1, X2)) -> ACTIVATE1(X1)
LENGTH1(n__cons2(X, Y)) -> LENGTH11(activate1(Y))
ACTIVATE1(n__nil) -> NIL
LENGTH11(X) -> LENGTH1(activate1(X))
ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
LENGTH1(n__cons2(X, Y)) -> S1(length11(activate1(Y)))
LENGTH11(X) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
LENGTH1(n__cons2(X, Y)) -> ACTIVATE1(Y)
FROM1(X) -> CONS2(X, n__from1(n__s1(X)))
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__cons2(X1, X2)) -> CONS2(activate1(X1), X2)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
length1(n__nil) -> 0
length1(n__cons2(X, Y)) -> s1(length11(activate1(Y)))
length11(X) -> length1(activate1(X))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
nil -> n__nil
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__nil) -> nil
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 8 less nodes.

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

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

ACTIVATE1(n__cons2(X1, X2)) -> ACTIVATE1(X1)
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)))
length1(n__nil) -> 0
length1(n__cons2(X, Y)) -> s1(length11(activate1(Y)))
length11(X) -> length1(activate1(X))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
nil -> n__nil
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__nil) -> nil
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVATE1(n__cons2(X1, X2)) -> ACTIVATE1(X1)
The remaining pairs can at least be oriented weakly.

ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ACTIVATE1(x1) ) = x1


POL( n__cons2(x1, x2) ) = x1 + 1


POL( n__from1(x1) ) = x1


POL( n__s1(x1) ) = x1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ 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)))
length1(n__nil) -> 0
length1(n__cons2(X, Y)) -> s1(length11(activate1(Y)))
length11(X) -> length1(activate1(X))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
nil -> n__nil
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__nil) -> nil
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
The remaining pairs can at least be oriented weakly.

ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ACTIVATE1(x1) ) = x1


POL( n__from1(x1) ) = x1 + 1


POL( n__s1(x1) ) = x1



The following usable rules [14] were oriented: none



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

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

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

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
length1(n__nil) -> 0
length1(n__cons2(X, Y)) -> s1(length11(activate1(Y)))
length11(X) -> length1(activate1(X))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
nil -> n__nil
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__nil) -> nil
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ACTIVATE1(x1) ) = x1


POL( n__s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
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)))
length1(n__nil) -> 0
length1(n__cons2(X, Y)) -> s1(length11(activate1(Y)))
length11(X) -> length1(activate1(X))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
nil -> n__nil
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__nil) -> nil
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
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

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

LENGTH1(n__cons2(X, Y)) -> LENGTH11(activate1(Y))
LENGTH11(X) -> LENGTH1(activate1(X))

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
length1(n__nil) -> 0
length1(n__cons2(X, Y)) -> s1(length11(activate1(Y)))
length11(X) -> length1(activate1(X))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
nil -> n__nil
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__nil) -> nil
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(X) -> X

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