Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/TRCSRSLASHEx14_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(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 strictly oriented and are deleted.

ACTIVATE₁(n__cons₂(X1, X2)) -> ACTIVATE₁(X1)

The remaining pairs can at least by weakly be oriented.

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

Used ordering: Combined order from the following AFS and order.
ACTIVATE₁(x1) = ACTIVATE₁(x1)
n__cons₂(x1, x2) = n__cons₁(x1)
n__from₁(x1) = x1
n__s₁(x1) = x1

Lexicographic Path Order [19].
Precedence:

[ACTIVATE₁, n_{_{cons_1]}}

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 strictly oriented and are deleted.

ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The remaining pairs can at least by weakly be oriented.

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

Used ordering: Combined order from the following AFS and order.
ACTIVATE₁(x1) = ACTIVATE₁(x1)
n__from₁(x1) = x1
n__s₁(x1) = n__s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

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__from₁(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 strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACTIVATE₁(x1) = ACTIVATE₁(x1)
n__from₁(x1) = n__from₁(x1)

Lexicographic Path Order [19].
Precedence:

n_{_from₁} > ACTIVATE₁

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.