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(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)
nil -> n__nil
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__nil) -> nil
activate1(n__cons2(X1, X2)) -> cons2(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(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)
nil -> n__nil
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__nil) -> nil
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(X) -> X

Q is empty.

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

ACTIVATE1(n__cons2(X1, X2)) -> CONS2(X1, X2)
LENGTH11(X) -> ACTIVATE1(X)
LENGTH1(n__cons2(X, Y)) -> LENGTH11(activate1(Y))
LENGTH1(n__cons2(X, Y)) -> ACTIVATE1(Y)
ACTIVATE1(n__nil) -> NIL
ACTIVATE1(n__from1(X)) -> FROM1(X)
FROM1(X) -> CONS2(X, n__from1(s1(X)))
LENGTH11(X) -> LENGTH1(activate1(X))

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(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)
nil -> n__nil
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__nil) -> nil
activate1(n__cons2(X1, X2)) -> cons2(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)) -> CONS2(X1, X2)
LENGTH11(X) -> ACTIVATE1(X)
LENGTH1(n__cons2(X, Y)) -> LENGTH11(activate1(Y))
LENGTH1(n__cons2(X, Y)) -> ACTIVATE1(Y)
ACTIVATE1(n__nil) -> NIL
ACTIVATE1(n__from1(X)) -> FROM1(X)
FROM1(X) -> CONS2(X, n__from1(s1(X)))
LENGTH11(X) -> LENGTH1(activate1(X))

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(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)
nil -> n__nil
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__nil) -> nil
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
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(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)
nil -> n__nil
cons2(X1, X2) -> n__cons2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__nil) -> nil
activate1(n__cons2(X1, X2)) -> cons2(X1, X2)
activate1(X) -> X

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