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:

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimessieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimessieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → 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:

ACTIVATE(n__nats(X)) → NATS(X)
SIEVE(cons(s(N), Y)) → ACTIVATE(Y)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
ZPRIMESSIEVE(nats(s(s(0))))
SIEVE(cons(0, Y)) → ACTIVATE(Y)
SIEVE(cons(s(N), Y)) → FILTER(activate(Y), N, N)
ZPRIMESNATS(s(s(0)))
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
ACTIVATE(n__filter(X1, X2, X3)) → FILTER(X1, X2, X3)
ACTIVATE(n__sieve(X)) → SIEVE(X)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimessieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ EdgeDeletionProof

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

ACTIVATE(n__nats(X)) → NATS(X)
SIEVE(cons(s(N), Y)) → ACTIVATE(Y)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
ZPRIMESSIEVE(nats(s(s(0))))
SIEVE(cons(0, Y)) → ACTIVATE(Y)
SIEVE(cons(s(N), Y)) → FILTER(activate(Y), N, N)
ZPRIMESNATS(s(s(0)))
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
ACTIVATE(n__filter(X1, X2, X3)) → FILTER(X1, X2, X3)
ACTIVATE(n__sieve(X)) → SIEVE(X)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimessieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

ACTIVATE(n__nats(X)) → NATS(X)
SIEVE(cons(s(N), Y)) → ACTIVATE(Y)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
SIEVE(cons(0, Y)) → ACTIVATE(Y)
ZPRIMESSIEVE(nats(s(s(0))))
SIEVE(cons(s(N), Y)) → FILTER(activate(Y), N, N)
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
ZPRIMESNATS(s(s(0)))
ACTIVATE(n__filter(X1, X2, X3)) → FILTER(X1, X2, X3)
ACTIVATE(n__sieve(X)) → SIEVE(X)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimessieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
QDP

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

SIEVE(cons(s(N), Y)) → ACTIVATE(Y)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
SIEVE(cons(0, Y)) → ACTIVATE(Y)
SIEVE(cons(s(N), Y)) → FILTER(activate(Y), N, N)
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
ACTIVATE(n__filter(X1, X2, X3)) → FILTER(X1, X2, X3)
ACTIVATE(n__sieve(X)) → SIEVE(X)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimessieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X

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