Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/nonterminSLASHCSRSLASHEx7

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

  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

from₁(x0)
head₁(cons₂(x0, x1))
2nd₁(cons₂(x0, x1))
take₂(0, x0)
take₂(s₁(x0), cons₂(x1, x2))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))

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)

The set Q consists of the following terms:

from₁(x0)
head₁(cons₂(x0, x1))
2nd₁(cons₂(x0, x1))
take₂(0, x0)
take₂(s₁(x0), cons₂(x1, x2))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

from₁(x0)
head₁(cons₂(x0, x1))
2nd₁(cons₂(x0, x1))
take₂(0, x0)
take₂(s₁(x0), cons₂(x1, x2))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))

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

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

from₁(x0)
head₁(cons₂(x0, x1))
2nd₁(cons₂(x0, x1))
take₂(0, x0)
take₂(s₁(x0), cons₂(x1, x2))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SEL₂(x1, x2) = x1
s₁(x1) = s₁(x1)
cons₂(x1, x2) = x1

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

from₁(x0)
head₁(cons₂(x0, x1))
2nd₁(cons₂(x0, x1))
take₂(0, x0)
take₂(s₁(x0), cons₂(x1, x2))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

from₁(x0)
head₁(cons₂(x0, x1))
2nd₁(cons₂(x0, x1))
take₂(0, x0)
take₂(s₁(x0), cons₂(x1, x2))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
TAKE₂(x1, x2) = x1
s₁(x1) = s₁(x1)
cons₂(x1, x2) = x1

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

from₁(x0)
head₁(cons₂(x0, x1))
2nd₁(cons₂(x0, x1))
take₂(0, x0)
take₂(s₁(x0), cons₂(x1, x2))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

from₁(x0)
head₁(cons₂(x0, x1))
2nd₁(cons₂(x0, x1))
take₂(0, x0)
take₂(s₁(x0), cons₂(x1, x2))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))

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