Termination w.r.t. Q proof of /tmp/tmpBi58Cu/Ex9

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, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

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, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

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:

SIEVE(cons(s(N), Y)) → FILTER(Y, N, N)
FILTER(cons(X, Y), 0, M) → FILTER(Y, M, M)
NATS(N) → NATS(s(N))
ZPRIMES → SIEVE(nats(s(s(0))))
ZPRIMES → NATS(s(s(0)))
SIEVE(cons(0, Y)) → SIEVE(Y)
FILTER(cons(X, Y), s(N), M) → FILTER(Y, N, M)
SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, N))

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

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:

SIEVE(cons(s(N), Y)) → FILTER(Y, N, N)
FILTER(cons(X, Y), 0, M) → FILTER(Y, M, M)
NATS(N) → NATS(s(N))
ZPRIMES → SIEVE(nats(s(s(0))))
ZPRIMES → NATS(s(s(0)))
SIEVE(cons(0, Y)) → SIEVE(Y)
FILTER(cons(X, Y), s(N), M) → FILTER(Y, N, M)
SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, N))

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

NATS(N) → NATS(s(N))

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

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.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(FILTER(x₁, x₂, x₃)) = (4)x_1
POL(cons(x₁, x₂)) = 4 + x_2
POL(s(x₁)) = 0
POL(0) = 0

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

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

          ↳ QDP

            ↳ QDPOrderProof

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

SIEVE(cons(0, Y)) → SIEVE(Y)
SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, N))

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

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.

SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, N))

The remaining pairs can at least be oriented weakly.

SIEVE(cons(0, Y)) → SIEVE(Y)

Used ordering: Polynomial interpretation [25,35]:

POL(SIEVE(x₁)) = (2)x_1
POL(cons(x₁, x₂)) = x_1 + (2)x_2
POL(filter(x₁, x₂, x₃)) = (2)x_1
POL(s(x₁)) = 3 + (3)x_1
POL(0) = 0

The value of delta used in the strict ordering is 6.
The following usable rules [17] were oriented:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

SIEVE(cons(0, Y)) → SIEVE(Y)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

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.

SIEVE(cons(0, Y)) → SIEVE(Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(SIEVE(x₁)) = (2)x_1
POL(cons(x₁, x₂)) = 1 + (4)x_2
POL(0) = 3

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

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.