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
  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
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)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)


Using Dependency Pairs [1,13] 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)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ EdgeDeletionProof

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)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
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)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ 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)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

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.


LENGTH(cons(X, Y)) → LENGTH1(Y)
The remaining pairs can at least be oriented weakly.

LENGTH1(X) → LENGTH(X)
Used ordering: Combined order from the following AFS and order.
LENGTH(x1)  =  x1
cons(x1, x2)  =  cons(x2)
LENGTH1(x1)  =  x1

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ DependencyGraphProof
                  ↳ QDP

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

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)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ 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)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

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