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.