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( LENGTH11(x1) ) = x1 + 1


POL( cons2(x1, x2) ) = x2 + 1


POL( LENGTH1(x1) ) = x1 + 1



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.