Termination w.r.t. Q proof of /tmp/tmpVyHTcD/Ex14_AEGL02_FR.trs

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:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
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(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
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:

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH(n__cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__nil) → NIL
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)
LENGTH(n__cons(X, Y)) → S(length1(activate(Y)))
LENGTH1(X) → LENGTH(activate(X))
LENGTH1(X) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
FROM(X) → CONS(X, n__from(n__s(X)))
ACTIVATE(n__from(X)) → FROM(activate(X))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
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:

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH(n__cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__nil) → NIL
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)
LENGTH(n__cons(X, Y)) → S(length1(activate(Y)))
LENGTH1(X) → LENGTH(activate(X))
LENGTH1(X) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
FROM(X) → CONS(X, n__from(n__s(X)))
ACTIVATE(n__from(X)) → FROM(activate(X))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
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 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:

ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(n__from(x₁)) = 3/2 + (11/4)x_1
POL(n__cons(x₁, x₂)) = 1/4 + (7/4)x_1
POL(n__s(x₁)) = 3/2 + (4)x_1
POL(ACTIVATE(x₁)) = (4)x_1

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(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:

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(X) → LENGTH(activate(X))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

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