Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/nonterminSLASHCSRSLASHEx4

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:

f₁(X) -> cons₂(X, f₁(g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

f₁(X) -> cons₂(X, f₁(g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)

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:

f₁(X) -> cons₂(X, f₁(g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)

The set Q consists of the following terms:

f₁(x0)
g₁(0)
g₁(s₁(x0))
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:

F₁(X) -> F₁(g₁(X))
F₁(X) -> G₁(X)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, Z)
G₁(s₁(X)) -> G₁(X)

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, f₁(g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)

The set Q consists of the following terms:

f₁(x0)
g₁(0)
g₁(s₁(x0))
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:

F₁(X) -> F₁(g₁(X))
F₁(X) -> G₁(X)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, Z)
G₁(s₁(X)) -> G₁(X)

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, f₁(g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)

The set Q consists of the following terms:

f₁(x0)
g₁(0)
g₁(s₁(x0))
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(X), cons₂(Y, Z)) -> SEL₂(X, Z)

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, f₁(g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)

The set Q consists of the following terms:

f₁(x0)
g₁(0)
g₁(s₁(x0))
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(X), cons₂(Y, Z)) -> SEL₂(X, Z)

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:

f₁(X) -> cons₂(X, f₁(g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)

The set Q consists of the following terms:

f₁(x0)
g₁(0)
g₁(s₁(x0))
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:

G₁(s₁(X)) -> G₁(X)

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, f₁(g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)

The set Q consists of the following terms:

f₁(x0)
g₁(0)
g₁(s₁(x0))
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.

G₁(s₁(X)) -> G₁(X)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
G₁(x1) = G₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > G₁

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:

f₁(X) -> cons₂(X, f₁(g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)

The set Q consists of the following terms:

f₁(x0)
g₁(0)
g₁(s₁(x0))
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:

F₁(X) -> F₁(g₁(X))

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, f₁(g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)

The set Q consists of the following terms:

f₁(x0)
g₁(0)
g₁(s₁(x0))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))

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