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)


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 contains 2 SCCs.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LENGTH1(cons2(X, Y)) -> LENGTH11(Y)
Used argument filtering: LENGTH1(x1)  =  x1
cons2(x1, x2)  =  cons1(x2)
LENGTH11(x1)  =  x1
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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 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.