Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__FILTER(x1, x2, x3)  =  A__FILTER(x1, x2, x3)
cons(x1, x2)  =  cons(x1)
s(x1)  =  s(x1)
MARK(x1)  =  x1
A__SIEVE(x1)  =  A__SIEVE(x1)
A__NATS(x1)  =  A__NATS(x1)
A__ZPRIMES  =  A__ZPRIMES
a__nats(x1)  =  a__nats(x1)
0  =  0
filter(x1, x2, x3)  =  filter(x1, x2, x3)
mark(x1)  =  x1
sieve(x1)  =  sieve(x1)
nats(x1)  =  nats(x1)
zprimes  =  zprimes
a__filter(x1, x2, x3)  =  a__filter(x1, x2, x3)
a__sieve(x1)  =  a__sieve(x1)
a__zprimes  =  a__zprimes

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

[A_FILTER3, filter₃, a_filter3] > [cons₁, s₁, A_SIEVE1, A_NATS1, a_nats1, 0, sieve₁, nats₁, a_sieve1]
[A_ZPRIMES, zprimes, a_zprimes] > [cons₁, s₁, A_SIEVE1, A_NATS1, a_nats1, 0, sieve₁, nats₁, a_sieve1]

Status:

A_FILTER3: [3,2,1]
cons₁: [1]
s₁: [1]
A_SIEVE1: [1]
A_NATS1: [1]
A_ZPRIMES: []
a_nats1: [1]
0: []
filter₃: [3,2,1]
sieve₁: [1]
nats₁: [1]
zprimes: []
a_filter3: [3,2,1]
a_sieve1: [1]
a_zprimes: []

The following usable rules [FROCOS05] were oriented:

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

(4) Obligation:

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

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

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.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 4 less nodes.

(6) TRUE