Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/TRCSRSLASHEx1_Luc02b_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:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
sel₂(0, cons₂(X, Z)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(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)))
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
sel₂(0, cons₂(X, Z)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(X2))
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))
SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))
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:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
sel₂(0, cons₂(X, Z)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(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__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))
SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))
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:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
sel₂(0, cons₂(X, Z)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(X2))
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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))
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:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
sel₂(0, cons₂(X, Z)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(X2))
activate₁(X) -> X

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

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

SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
sel₂(0, cons₂(X, Z)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(X2))
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.

SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))

Used argument filtering: SEL₂(x1, x2) = x1
s₁(x1) = s₁(x1)
activate₁(x1) = activate₁(x1)
n__from₁(x1) = n__from
from₁(x1) = from
n__s₁(x1) = n__s₁(x1)
n__first₂(x1, x2) = n__first
first₂(x1, x2) = first
0 = 0
nil = nil
cons₂(x1, x2) = cons
Used ordering: Precedence:

activate₁ > s₁ > n_{_s₁}
activate₁ > from > n_{_from}
activate₁ > from > cons
activate₁ > first > n_{_first}
activate₁ > first > nil
activate₁ > first > cons

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

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

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
sel₂(0, cons₂(X, Z)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__first₂(X1, X2)) -> first₂(activate₁(X1), activate₁(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.