Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/TRCSRSLASHEx14_AEGL02_GM.trs

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.

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 contains 2 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: A__LENGTH₁(x1) = x1
cons₂(x1, x2) = cons₁(x2)
A__LENGTH1₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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 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.