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:

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

LENGTH(cons(X, Y)) → LENGTH1(Y)
LENGTH1(X) → LENGTH(X)
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(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:

LENGTH(cons(X, Y)) → LENGTH1(Y)
LENGTH1(X) → LENGTH(X)
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

LENGTH(cons(X, Y)) → LENGTH1(Y)
LENGTH1(X) → LENGTH(X)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


LENGTH(cons(X, Y)) → LENGTH1(Y)
LENGTH1(X) → LENGTH(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(LENGTH(x1)) = (3)x_1   
POL(cons(x1, x2)) = 2 + (3)x_2   
POL(LENGTH1(x1)) = 3 + (4)x_1   
The value of delta used in the strict ordering is 3.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

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