Termination w.r.t. Q proof of TRS/TRCSREx14_AEGL02

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:

a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__length₁(nil) -> 0
a__length₁(cons₂(X, Y)) -> s₁(a__length1₁(Y))
a__length1₁(X) -> a__length₁(X)
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(length₁(X)) -> a__length₁(X)
mark₁(length1₁(X)) -> a__length1₁(X)
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(nil) -> nil
mark₁(0) -> 0
a__from₁(X) -> from₁(X)
a__length₁(X) -> length₁(X)
a__length1₁(X) -> length1₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__length₁(nil) -> 0
a__length₁(cons₂(X, Y)) -> s₁(a__length1₁(Y))
a__length1₁(X) -> a__length₁(X)
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(length₁(X)) -> a__length₁(X)
mark₁(length1₁(X)) -> a__length1₁(X)
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(nil) -> nil
mark₁(0) -> 0
a__from₁(X) -> from₁(X)
a__length₁(X) -> length₁(X)
a__length1₁(X) -> length1₁(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:

MARK₁(s₁(X)) -> MARK₁(X)
A__LENGTH₁(cons₂(X, Y)) -> A__LENGTH1₁(Y)
MARK₁(length₁(X)) -> A__LENGTH₁(X)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
A__LENGTH1₁(X) -> A__LENGTH₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(length1₁(X)) -> A__LENGTH1₁(X)
A__FROM₁(X) -> MARK₁(X)

The TRS R consists of the following rules:

a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__length₁(nil) -> 0
a__length₁(cons₂(X, Y)) -> s₁(a__length1₁(Y))
a__length1₁(X) -> a__length₁(X)
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(length₁(X)) -> a__length₁(X)
mark₁(length1₁(X)) -> a__length1₁(X)
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(nil) -> nil
mark₁(0) -> 0
a__from₁(X) -> from₁(X)
a__length₁(X) -> length₁(X)
a__length1₁(X) -> length1₁(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:

MARK₁(s₁(X)) -> MARK₁(X)
A__LENGTH₁(cons₂(X, Y)) -> A__LENGTH1₁(Y)
MARK₁(length₁(X)) -> A__LENGTH₁(X)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
A__LENGTH1₁(X) -> A__LENGTH₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(length1₁(X)) -> A__LENGTH1₁(X)
A__FROM₁(X) -> MARK₁(X)

The TRS R consists of the following rules:

a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__length₁(nil) -> 0
a__length₁(cons₂(X, Y)) -> s₁(a__length1₁(Y))
a__length1₁(X) -> a__length₁(X)
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(length₁(X)) -> a__length₁(X)
mark₁(length1₁(X)) -> a__length1₁(X)
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(nil) -> nil
mark₁(0) -> 0
a__from₁(X) -> from₁(X)
a__length₁(X) -> length₁(X)
a__length1₁(X) -> length1₁(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 with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

A__LENGTH₁(cons₂(X, Y)) -> A__LENGTH1₁(Y)
A__LENGTH1₁(X) -> A__LENGTH₁(X)

The TRS R consists of the following rules:

a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__length₁(nil) -> 0
a__length₁(cons₂(X, Y)) -> s₁(a__length1₁(Y))
a__length1₁(X) -> a__length₁(X)
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(length₁(X)) -> a__length₁(X)
mark₁(length1₁(X)) -> a__length1₁(X)
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(nil) -> nil
mark₁(0) -> 0
a__from₁(X) -> from₁(X)
a__length₁(X) -> length₁(X)
a__length1₁(X) -> length1₁(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.

A__LENGTH₁(cons₂(X, Y)) -> A__LENGTH1₁(Y)

The remaining pairs can at least be oriented weakly.

A__LENGTH1₁(X) -> A__LENGTH₁(X)

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

POL( A__LENGTH₁(x₁) ) = max{0, x₁ - 2}

POL( cons₂(x₁, x₂) ) = x₂ + 3

POL( A__LENGTH1₁(x₁) ) = max{0, x₁ - 2}

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:

A__LENGTH1₁(X) -> A__LENGTH₁(X)

The TRS R consists of the following rules:

a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__length₁(nil) -> 0
a__length₁(cons₂(X, Y)) -> s₁(a__length1₁(Y))
a__length1₁(X) -> a__length₁(X)
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(length₁(X)) -> a__length₁(X)
mark₁(length1₁(X)) -> a__length1₁(X)
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(nil) -> nil
mark₁(0) -> 0
a__from₁(X) -> from₁(X)
a__length₁(X) -> length₁(X)
a__length1₁(X) -> length1₁(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:

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)

The TRS R consists of the following rules:

a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__length₁(nil) -> 0
a__length₁(cons₂(X, Y)) -> s₁(a__length1₁(Y))
a__length1₁(X) -> a__length₁(X)
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(length₁(X)) -> a__length₁(X)
mark₁(length1₁(X)) -> a__length1₁(X)
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(nil) -> nil
mark₁(0) -> 0
a__from₁(X) -> from₁(X)
a__length₁(X) -> length₁(X)
a__length1₁(X) -> length1₁(X)

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