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:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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:

A__ZPRIMES → A__NATS(s(s(0)))
MARK(filter(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2, X3)) → MARK(X3)
MARK(filter(X1, X2, X3)) → MARK(X2)
A__NATS(N) → MARK(N)
MARK(nats(X)) → A__NATS(mark(X))
MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(s(X)) → MARK(X)
A__ZPRIMES → A__SIEVE(a__nats(s(s(0))))
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(nats(X)) → MARK(X)
MARK(zprimes) → A__ZPRIMES
A__SIEVE(cons(s(N), Y)) → MARK(N)
A__FILTER(cons(X, Y), s(N), M) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(sieve(X)) → MARK(X)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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:

A__ZPRIMES → A__NATS(s(s(0)))
MARK(filter(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2, X3)) → MARK(X3)
MARK(filter(X1, X2, X3)) → MARK(X2)
A__NATS(N) → MARK(N)
MARK(nats(X)) → A__NATS(mark(X))
MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(s(X)) → MARK(X)
A__ZPRIMES → A__SIEVE(a__nats(s(s(0))))
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(nats(X)) → MARK(X)
MARK(zprimes) → A__ZPRIMES
A__SIEVE(cons(s(N), Y)) → MARK(N)
A__FILTER(cons(X, Y), s(N), M) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(sieve(X)) → MARK(X)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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

          ↳ QDPOrderProof

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

A__ZPRIMES → A__NATS(s(s(0)))
MARK(filter(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2, X3)) → MARK(X3)
MARK(filter(X1, X2, X3)) → MARK(X2)
A__NATS(N) → MARK(N)
MARK(nats(X)) → A__NATS(mark(X))
MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(s(X)) → MARK(X)
A__ZPRIMES → A__SIEVE(a__nats(s(s(0))))
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(nats(X)) → MARK(X)
A__SIEVE(cons(s(N), Y)) → MARK(N)
MARK(zprimes) → A__ZPRIMES
A__FILTER(cons(X, Y), s(N), M) → MARK(X)
MARK(sieve(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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.

MARK(filter(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2, X3)) → MARK(X3)
MARK(filter(X1, X2, X3)) → MARK(X2)
A__FILTER(cons(X, Y), s(N), M) → MARK(X)

The remaining pairs can at least be oriented weakly.

A__ZPRIMES → A__NATS(s(s(0)))
A__NATS(N) → MARK(N)
MARK(nats(X)) → A__NATS(mark(X))
MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(s(X)) → MARK(X)
A__ZPRIMES → A__SIEVE(a__nats(s(s(0))))
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(nats(X)) → MARK(X)
A__SIEVE(cons(s(N), Y)) → MARK(N)
MARK(zprimes) → A__ZPRIMES
MARK(sieve(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

Used ordering: Combined order from the following AFS and order.
A__ZPRIMES = A__ZPRIMES
A__NATS(x1) = x1
s(x1) = x1
0 = 0
MARK(x1) = x1
filter(x1, x2, x3) = filter(x1, x2, x3)
nats(x1) = x1
mark(x1) = x1
A__FILTER(x1, x2, x3) = A__FILTER(x1, x2, x3)
A__SIEVE(x1) = x1
a__nats(x1) = x1
sieve(x1) = x1
cons(x1, x2) = x1
zprimes = zprimes
a__filter(x1, x2, x3) = a__filter(x1, x2, x3)
a__zprimes = a__zprimes
a__sieve(x1) = x1

Lexicographic path order with status [19].
Quasi-Precedence:

[A_{_ZPRIMES}, 0, filter₃, A_{_FILTER₃}, zprimes, a_{_filter₃}, a_{_zprimes]}

Status:

filter₃: [2,3,1]
a_{_filter₃}: [2,3,1]
zprimes: multiset
A_{_FILTER₃}: [2,3,1]
A_{_ZPRIMES}: multiset
0: multiset
a_{_zprimes}: multiset

The following usable rules [14] were oriented:

a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
mark(0) → 0
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(sieve(X)) → a__sieve(mark(X))
mark(s(X)) → s(mark(X))
a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__nats(N) → cons(mark(N), nats(s(N)))
mark(nats(X)) → a__nats(mark(X))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__zprimes → zprimes
mark(zprimes) → a__zprimes
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

A__ZPRIMES → A__NATS(s(s(0)))
MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(s(X)) → MARK(X)
A__ZPRIMES → A__SIEVE(a__nats(s(s(0))))
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(nats(X)) → MARK(X)
MARK(zprimes) → A__ZPRIMES
A__SIEVE(cons(s(N), Y)) → MARK(N)
A__NATS(N) → MARK(N)
MARK(nats(X)) → A__NATS(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
MARK(sieve(X)) → MARK(X)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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

A__ZPRIMES → A__NATS(s(s(0)))
MARK(s(X)) → MARK(X)
A__ZPRIMES → A__SIEVE(a__nats(s(s(0))))
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(nats(X)) → MARK(X)
MARK(zprimes) → A__ZPRIMES
A__SIEVE(cons(s(N), Y)) → MARK(N)
A__NATS(N) → MARK(N)
MARK(nats(X)) → A__NATS(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
MARK(sieve(X)) → MARK(X)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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.

A__ZPRIMES → A__NATS(s(s(0)))
A__ZPRIMES → A__SIEVE(a__nats(s(s(0))))
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(nats(X)) → MARK(X)
MARK(nats(X)) → A__NATS(mark(X))
MARK(sieve(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.

MARK(s(X)) → MARK(X)
MARK(zprimes) → A__ZPRIMES
A__SIEVE(cons(s(N), Y)) → MARK(N)
A__NATS(N) → MARK(N)
MARK(cons(X1, X2)) → MARK(X1)

Used ordering: Combined order from the following AFS and order.
A__ZPRIMES = A__ZPRIMES
A__NATS(x1) = x1
s(x1) = x1
0 = 0
MARK(x1) = x1
A__SIEVE(x1) = x1
a__nats(x1) = a__nats(x1)
sieve(x1) = sieve(x1)
mark(x1) = x1
nats(x1) = nats(x1)
zprimes = zprimes
cons(x1, x2) = x1
a__filter(x1, x2, x3) = a__filter(x1, x3)
filter(x1, x2, x3) = filter(x1, x3)
a__zprimes = a__zprimes
a__sieve(x1) = a__sieve(x1)

Lexicographic path order with status [19].
Quasi-Precedence:

[A_{_ZPRIMES}, zprimes, a_{_zprimes]} > [a_{_nats₁}, nats_1]
[A_{_ZPRIMES}, zprimes, a_{_zprimes]} > [sieve₁, a_{_{sieve_1]}} > 0
[a_{_filter₂}, filter_2] > 0

Status:

sieve₁: [1]
filter₂: [1,2]
zprimes: multiset
A_{_ZPRIMES}: multiset
0: multiset
a_{_filter₂}: [1,2]
nats₁: [1]
a_{_sieve₁}: [1]
a_{_nats₁}: [1]
a_{_zprimes}: multiset

The following usable rules [14] were oriented:

a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
mark(0) → 0
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(sieve(X)) → a__sieve(mark(X))
mark(s(X)) → s(mark(X))
a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__nats(N) → cons(mark(N), nats(s(N)))
mark(nats(X)) → a__nats(mark(X))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__zprimes → zprimes
mark(zprimes) → a__zprimes
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

MARK(s(X)) → MARK(X)
A__NATS(N) → MARK(N)
A__SIEVE(cons(s(N), Y)) → MARK(N)
MARK(zprimes) → A__ZPRIMES
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

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

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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.

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK(x1)
s(x1) = s(x1)
cons(x1, x2) = cons(x1, x2)

Lexicographic path order with status [19].
Quasi-Precedence:

cons₂ > [MARK₁, s_1]

Status:

MARK₁: [1]
s₁: [1]
cons₂: [2,1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ PisEmptyProof

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

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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.