Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → 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(n__cons(X, Y)) → ACTIVATE(Y)
LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
FROM(X) → CONS(X, n__from(s(X)))
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)
ACTIVATE(n__nil) → NIL
ACTIVATE(n__from(X)) → FROM(X)
LENGTH1(X) → LENGTH(activate(X))
LENGTH1(X) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → 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(n__cons(X, Y)) → ACTIVATE(Y)
LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
FROM(X) → CONS(X, n__from(s(X)))
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)
ACTIVATE(n__nil) → NIL
ACTIVATE(n__from(X)) → FROM(X)
LENGTH1(X) → LENGTH(activate(X))
LENGTH1(X) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(X) → LENGTH(activate(X))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule LENGTH1(X) → LENGTH(activate(X)) at position [0] we obtained the following new rules:

LENGTH1(n__cons(x0, x1)) → LENGTH(cons(x0, x1))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__from(x0)) → LENGTH(from(x0))
LENGTH1(n__nil) → LENGTH(nil)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing

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

LENGTH1(n__cons(x0, x1)) → LENGTH(cons(x0, x1))
LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__nil) → LENGTH(nil)
LENGTH1(n__from(x0)) → LENGTH(from(x0))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule LENGTH1(n__cons(x0, x1)) → LENGTH(cons(x0, x1)) at position [0] we obtained the following new rules:

LENGTH1(n__cons(x0, x1)) → LENGTH(n__cons(x0, x1))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ Narrowing

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__cons(x0, x1)) → LENGTH(n__cons(x0, x1))
LENGTH1(n__nil) → LENGTH(nil)
LENGTH1(n__from(x0)) → LENGTH(from(x0))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule LENGTH1(n__from(x0)) → LENGTH(from(x0)) at position [0] we obtained the following new rules:

LENGTH1(n__from(x0)) → LENGTH(n__from(x0))
LENGTH1(n__from(x0)) → LENGTH(cons(x0, n__from(s(x0))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ DependencyGraphProof

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

LENGTH1(n__from(x0)) → LENGTH(n__from(x0))
LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__cons(x0, x1)) → LENGTH(n__cons(x0, x1))
LENGTH1(n__nil) → LENGTH(nil)
LENGTH1(n__from(x0)) → LENGTH(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__cons(x0, x1)) → LENGTH(n__cons(x0, x1))
LENGTH1(n__nil) → LENGTH(nil)
LENGTH1(n__from(x0)) → LENGTH(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule LENGTH1(n__nil) → LENGTH(nil) at position [0] we obtained the following new rules:

LENGTH1(n__nil) → LENGTH(n__nil)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__cons(x0, x1)) → LENGTH(n__cons(x0, x1))
LENGTH1(n__from(x0)) → LENGTH(cons(x0, n__from(s(x0))))
LENGTH1(n__nil) → LENGTH(n__nil)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ NonTerminationProof

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__cons(x0, x1)) → LENGTH(n__cons(x0, x1))
LENGTH1(n__from(x0)) → LENGTH(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__cons(x0, x1)) → LENGTH(n__cons(x0, x1))
LENGTH1(n__from(x0)) → LENGTH(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X


s = LENGTH1(activate(n__from(x0))) evaluates to t =LENGTH1(activate(n__from(s(x0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

LENGTH1(activate(n__from(x0))) → LENGTH1(n__from(x0))
with rule activate(X) → X at position [0] and matcher [X / n__from(x0)]

LENGTH1(n__from(x0)) → LENGTH(cons(x0, n__from(s(x0))))
with rule LENGTH1(n__from(x0')) → LENGTH(cons(x0', n__from(s(x0')))) at position [] and matcher [x0' / x0]

LENGTH(cons(x0, n__from(s(x0)))) → LENGTH(n__cons(x0, n__from(s(x0))))
with rule cons(X1, X2) → n__cons(X1, X2) at position [0] and matcher [X2 / n__from(s(x0)), X1 / x0]

LENGTH(n__cons(x0, n__from(s(x0)))) → LENGTH1(activate(n__from(s(x0))))
with rule LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.