Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/TRCSRSLASHEx6_Luc98

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(X2))
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(X2))
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(X) -> X

Q is empty.

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

ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__from₁(X)) -> FROM₁(activate₁(X))
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(X2))
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
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__first₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__from₁(X)) -> FROM₁(activate₁(X))
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(X2))
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

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

ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(X2))
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__first₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(activate₁(X1), activate₁(X2))

Used argument filtering: ACTIVATE₁(x1) = x1
n__first₂(x1, x2) = n__first₂(x1, x2)
FIRST₂(x1, x2) = x2
activate₁(x1) = x1
n__from₁(x1) = x1
cons₂(x1, x2) = x2
n__s₁(x1) = x1
first₂(x1, x2) = first₂(x1, x2)
from₁(x1) = x1
s₁(x1) = x1
0 = 0
nil = nil
Used ordering: Quasi Precedence: [n__first_2, first_2, nil]

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

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

FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(X2))
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPAfsSolverProof

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:

first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(X2))
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

Used argument filtering: ACTIVATE₁(x1) = x1
n__from₁(x1) = x1
n__s₁(x1) = n__s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPAfsSolverProof

                    ↳ QDP

                      ↳ QDPAfsSolverProof

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:

first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(X2))
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: ACTIVATE₁(x1) = x1
n__from₁(x1) = n__from₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPAfsSolverProof

                    ↳ QDP

                      ↳ QDPAfsSolverProof

                        ↳ QDP

                          ↳ PisEmptyProof

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

first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(X2))
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
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.