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:

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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Precedence:

filter₃ > cons₁
filter₃ > 0
zprimes > 0
zprimes > s₁
zprimes > sieve₁
zprimes > nats₁ > cons₁

Status:

filter₃: multiset
cons₁: [1]
0: multiset
s₁: multiset
sieve₁: multiset
nats₁: multiset
zprimes: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE