Termination w.r.t. Q proof of TRS/TRCSREx6_Luc98_Z.trs

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

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

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(s₁(X)))
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(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 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

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

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( FIRST₂(x₁, x₂) ) = 2x₂ + 3

POL( s₁(x₁) ) = max{0, x₁ - 3}

POL( cons₂(x₁, x₂) ) = 2x₁ + x₂

POL( ACTIVATE₁(x₁) ) = 2x₁ + 1

POL( n__first₂(x₁, x₂) ) = x₁ + 2x₂ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ 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(s₁(X)))
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(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.