Termination w.r.t. Q proof of TRS/TRCSREx1_Luc02b

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(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)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, 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(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)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
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:

SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))
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:

from₁(X) -> cons₂(X, n__from₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, 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:

SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))
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:

from₁(X) -> cons₂(X, n__from₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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:

from₁(X) -> cons₂(X, n__from₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
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 interpretation [21]:

POL(ACTIVATE₁(x₁)) = 3·x₁
POL(FIRST₂(x₁, x₂)) = 2 + 3·x₁ + 3·x₂
POL(cons₂(x₁, x₂)) = 2 + 3·x₂
POL(n__first₂(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(s₁(x₁)) = 3

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

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

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

POL(0) = 0
POL(SEL₂(x₁, x₂)) = 3·x₁
POL(activate₁(x₁)) = 0
POL(cons₂(x₁, x₂)) = 0
POL(first₂(x₁, x₂)) = 0
POL(from₁(x₁)) = 0
POL(n__first₂(x₁, x₂)) = 2·x₂
POL(n__from₁(x₁)) = 0
POL(nil) = 0
POL(s₁(x₁)) = 3 + x₁

The following usable rules [14] were oriented: none

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