Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/TRCSRSLASHEx4_7_56_Bor03

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:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
after₂(0, XS) -> XS
after₂(s₁(N), cons₂(X, XS)) -> after₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
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:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
after₂(0, XS) -> XS
after₂(s₁(N), cons₂(X, XS)) -> after₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
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__from₁(X)) -> FROM₁(activate₁(X))
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
AFTER₂(s₁(N), cons₂(X, XS)) -> AFTER₂(N, activate₁(XS))
AFTER₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
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)))
after₂(0, XS) -> XS
after₂(s₁(N), cons₂(X, XS)) -> after₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
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__from₁(X)) -> FROM₁(activate₁(X))
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
AFTER₂(s₁(N), cons₂(X, XS)) -> AFTER₂(N, activate₁(XS))
AFTER₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
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)))
after₂(0, XS) -> XS
after₂(s₁(N), cons₂(X, XS)) -> after₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
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 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ 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)))
after₂(0, XS) -> XS
after₂(s₁(N), cons₂(X, XS)) -> after₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
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: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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)))
after₂(0, XS) -> XS
after₂(s₁(N), cons₂(X, XS)) -> after₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
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: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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)))
after₂(0, XS) -> XS
after₂(s₁(N), cons₂(X, XS)) -> after₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

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

AFTER₂(s₁(N), cons₂(X, XS)) -> AFTER₂(N, activate₁(XS))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
after₂(0, XS) -> XS
after₂(s₁(N), cons₂(X, XS)) -> after₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
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.

AFTER₂(s₁(N), cons₂(X, XS)) -> AFTER₂(N, activate₁(XS))

Used argument filtering: AFTER₂(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)
cons₂(x1, x2) = cons
Used ordering: Precedence:

activate₁ > s₁ > n_{_s₁}
activate₁ > from > n_{_from}
activate₁ > from > 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)))
after₂(0, XS) -> XS
after₂(s₁(N), cons₂(X, XS)) -> after₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
s₁(X) -> n__s₁(X)
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.