Termination w.r.t. Q proof of TRS/TRCSREx7_BLR02

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)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(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)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(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:

ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(X1, X2)
TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
SEL₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))
2ND₁(cons₂(X, XS)) -> HEAD₁(activate₁(XS))
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
2ND₁(cons₂(X, XS)) -> ACTIVATE₁(XS)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(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:

ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(X1, X2)
TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
SEL₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)
SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))
2ND₁(cons₂(X, XS)) -> HEAD₁(activate₁(XS))
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
2ND₁(cons₂(X, XS)) -> ACTIVATE₁(XS)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(X1, X2)
TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(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.

TAKE₂(s₁(N), cons₂(X, XS)) -> ACTIVATE₁(XS)

The remaining pairs can at least be oriented weakly.

ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(X1, X2)

Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE₁(x₁)) = x₁
POL(TAKE₂(x₁, x₂)) = x₁ + x₂
POL(cons₂(x₁, x₂)) = x₂
POL(n__take₂(x₁, x₂)) = x₁ + x₂
POL(s₁(x₁)) = 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(X1, X2)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(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 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

SEL₂(s₁(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(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(N), cons₂(X, XS)) -> SEL₂(N, activate₁(XS))

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

POL(0) = 0
POL(SEL₂(x₁, x₂)) = x₁
POL(activate₁(x₁)) = 0
POL(cons₂(x₁, x₂)) = 0
POL(from₁(x₁)) = 0
POL(n__from₁(x₁)) = 0
POL(n__take₂(x₁, x₂)) = 0
POL(nil) = 0
POL(s₁(x₁)) = 1 + x₁
POL(take₂(x₁, x₂)) = 0

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)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(activate₁(XS))
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, n__take₂(N, activate₁(XS)))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, activate₁(XS))
from₁(X) -> n__from₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(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.