Termination w.r.t. Q proof of TRS/nontermin/CSREx7

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(XS)
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, take₂(N, XS))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, XS)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(XS)
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, take₂(N, XS))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, XS)

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(N), cons₂(X, XS)) -> SEL₂(N, XS)
TAKE₂(s₁(N), cons₂(X, XS)) -> TAKE₂(N, XS)
2ND₁(cons₂(X, XS)) -> HEAD₁(XS)
FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(XS)
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, take₂(N, XS))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, XS)

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(N), cons₂(X, XS)) -> SEL₂(N, XS)
TAKE₂(s₁(N), cons₂(X, XS)) -> TAKE₂(N, XS)
2ND₁(cons₂(X, XS)) -> HEAD₁(XS)
FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(XS)
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, take₂(N, XS))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, XS)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(XS)
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, take₂(N, XS))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, XS)

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, XS)

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

POL(SEL₂(x₁, x₂)) = x₁ + 2·x₁·x₂
POL(cons₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(XS)
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, take₂(N, XS))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, XS)

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

          ↳ QDP

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

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

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(XS)
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, take₂(N, XS))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, XS)

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)) -> TAKE₂(N, XS)

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

POL(TAKE₂(x₁, x₂)) = x₁ + 2·x₁·x₂
POL(cons₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(XS)
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, take₂(N, XS))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, XS)

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

          ↳ QDP

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

FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, XS)) -> X
2nd₁(cons₂(X, XS)) -> head₁(XS)
take₂(0, XS) -> nil
take₂(s₁(N), cons₂(X, XS)) -> cons₂(X, take₂(N, XS))
sel₂(0, cons₂(X, XS)) -> X
sel₂(s₁(N), cons₂(X, XS)) -> sel₂(N, XS)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.