Termination w.r.t. Q proof of TRS/nontermin/CSREx14

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

  ↳ 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)

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:

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)

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:

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)

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.

↳ 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)

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.

LENGTH₁(cons₂(X, Y)) -> LENGTH1₁(Y)

The remaining pairs can at least be oriented weakly.

LENGTH1₁(X) -> LENGTH₁(X)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( LENGTH₁(x₁) ) = x₁

POL( cons₂(x₁, x₂) ) = x₂ + 1

POL( LENGTH1₁(x₁) ) = x₁

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:

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)

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

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)

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