Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/nonterminSLASHCSRSLASHEx14

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)))
length₁(nil) -> 0
length₁(cons₂(X, Y)) -> s₁(length1₁(Y))
length1₁(X) -> length₁(X)

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)))
length₁(nil) -> 0
length₁(cons₂(X, Y)) -> s₁(length1₁(Y))
length1₁(X) -> length₁(X)

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)))
length₁(nil) -> 0
length₁(cons₂(X, Y)) -> s₁(length1₁(Y))
length1₁(X) -> length₁(X)

The set Q consists of the following terms:

from₁(x0)
length₁(nil)
length₁(cons₂(x0, x1))
length1₁(x0)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

LENGTH₁(cons₂(X, Y)) -> LENGTH1₁(Y)
LENGTH1₁(X) -> LENGTH₁(X)
FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
length₁(nil) -> 0
length₁(cons₂(X, Y)) -> s₁(length1₁(Y))
length1₁(X) -> length₁(X)

The set Q consists of the following terms:

from₁(x0)
length₁(nil)
length₁(cons₂(x0, x1))
length1₁(x0)

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:

LENGTH₁(cons₂(X, Y)) -> LENGTH1₁(Y)
LENGTH1₁(X) -> LENGTH₁(X)
FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
length₁(nil) -> 0
length₁(cons₂(X, Y)) -> s₁(length1₁(Y))
length1₁(X) -> length₁(X)

The set Q consists of the following terms:

from₁(x0)
length₁(nil)
length₁(cons₂(x0, x1))
length1₁(x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

LENGTH₁(cons₂(X, Y)) -> LENGTH1₁(Y)
LENGTH1₁(X) -> LENGTH₁(X)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
length₁(nil) -> 0
length₁(cons₂(X, Y)) -> s₁(length1₁(Y))
length1₁(X) -> length₁(X)

The set Q consists of the following terms:

from₁(x0)
length₁(nil)
length₁(cons₂(x0, x1))
length1₁(x0)

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.

LENGTH₁(cons₂(X, Y)) -> LENGTH1₁(Y)

The remaining pairs can at least by weakly be oriented.

LENGTH1₁(X) -> LENGTH₁(X)

Used ordering: Combined order from the following AFS and order.
LENGTH₁(x1) = LENGTH₁(x1)
cons₂(x1, x2) = cons₁(x2)
LENGTH1₁(x1) = LENGTH1₁(x1)

Lexicographic Path Order [19].
Precedence:

[LENGTH₁, LENGTH1_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

              ↳ QDP

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

LENGTH1₁(X) -> LENGTH₁(X)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
length₁(nil) -> 0
length₁(cons₂(X, Y)) -> s₁(length1₁(Y))
length1₁(X) -> length₁(X)

The set Q consists of the following terms:

from₁(x0)
length₁(nil)
length₁(cons₂(x0, x1))
length1₁(x0)

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

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ 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)))
length₁(nil) -> 0
length₁(cons₂(X, Y)) -> s₁(length1₁(Y))
length1₁(X) -> length₁(X)

The set Q consists of the following terms:

from₁(x0)
length₁(nil)
length₁(cons₂(x0, x1))
length1₁(x0)

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