Termination w.r.t. Q proof of TRS/TRCSREx14_AEGL02

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.

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)) -> 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 [13,14,18] 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.