Termination w.r.t. Q proof of TRS/TRCSREx9_BLR02

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

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)))
zprimes → sieve(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)))
zprimes → sieve(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)
ZPRIMES → SIEVE(nats(s(s(0))))
SIEVE(cons(0, Y)) → ACTIVATE(Y)
SIEVE(cons(s(N), Y)) → FILTER(activate(Y), N, N)
ZPRIMES → NATS(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)))
zprimes → sieve(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)
ZPRIMES → SIEVE(nats(s(s(0))))
SIEVE(cons(0, Y)) → ACTIVATE(Y)
SIEVE(cons(s(N), Y)) → FILTER(activate(Y), N, N)
ZPRIMES → NATS(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)))
zprimes → sieve(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)
ZPRIMES → SIEVE(nats(s(s(0))))
SIEVE(cons(s(N), Y)) → FILTER(activate(Y), N, N)
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
ZPRIMES → NATS(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)))
zprimes → sieve(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

              ↳ QDPOrderProof

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)))
zprimes → sieve(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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__sieve(X)) → SIEVE(X)

The remaining pairs can at least be oriented weakly.

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)

Used ordering: Combined order from the following AFS and order.
SIEVE(x1) = x1
cons(x1, x2) = x2
ACTIVATE(x1) = x1
FILTER(x1, x2, x3) = x1
activate(x1) = x1
n__filter(x1, x2, x3) = x1
n__sieve(x1) = n__sieve(x1)
filter(x1, x2, x3) = x1
sieve(x1) = sieve(x1)
n__nats(x1) = n__nats
nats(x1) = nats

Lexicographic Path Order [19].
Precedence:

[n_{_sieve₁}, sieve_1]
[n_{_nats}, nats]

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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)

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)))
zprimes → sieve(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

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

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

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

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)))
zprimes → sieve(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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__filter(X1, X2, X3)) → FILTER(X1, X2, X3)

The remaining pairs can at least be oriented weakly.

FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)

Used ordering: Combined order from the following AFS and order.
FILTER(x1, x2, x3) = x1
cons(x1, x2) = x2
ACTIVATE(x1) = x1
n__filter(x1, x2, x3) = n__filter(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ DependencyGraphProof

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

FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)

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)))
zprimes → sieve(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 0 SCCs with 2 less nodes.