Termination w.r.t. Q proof of TRS/TRCSREx4_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.

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__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 [13,14,18] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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)))
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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE₁(x₁)) = 3·x₁
POL(n__from₁(x₁)) = 3 + x₁
POL(n__s₁(x₁)) = 3 + 2·x₁

The following usable rules [14] 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:

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

            ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(AFTER₂(x₁, x₂)) = 3·x₁ + 3·x₂
POL(activate₁(x₁)) = x₁
POL(cons₂(x₁, x₂)) = 2·x₂
POL(from₁(x₁)) = 0
POL(n__from₁(x₁)) = 0
POL(n__s₁(x₁)) = 3 + 2·x₁
POL(s₁(x₁)) = 3 + 2·x₁

The following usable rules [14] were oriented:

s₁(X) -> n__s₁(X)
activate₁(n__s₁(X)) -> s₁(activate₁(X))
from₁(X) -> n__from₁(X)
activate₁(n__from₁(X)) -> from₁(activate₁(X))
from₁(X) -> cons₂(X, n__from₁(n__s₁(X)))
activate₁(X) -> X

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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.