The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(2^n, INF).

(0) Obligation:

Runtime Complexity TRS:
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(n__filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(n__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)
s(X) → n__s(X)
activate(n__filter(X1, X2, X3)) → filter(activate(X1), activate(X2), activate(X3))
activate(n__sieve(X)) → sieve(activate(X))
activate(n__nats(X)) → nats(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
activate(n__filter(n__nats(X44182_4), X2, X3)) →⁺ filter(cons(activate(X44182_4), n__nats(n__s(activate(X44182_4)))), activate(X2), activate(X3))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0].
The pumping substitution is [X44182_4 / n__filter(n__nats(X44182_4), X2, X3)].
The result substitution is [ ].

The rewrite sequence
activate(n__filter(n__nats(X44182_4), X2, X3)) →⁺ filter(cons(activate(X44182_4), n__nats(n__s(activate(X44182_4)))), activate(X2), activate(X3))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1,0,0].
The pumping substitution is [X44182_4 / n__filter(n__nats(X44182_4), X2, X3)].
The result substitution is [ ].

(2) BOUNDS(2^n, INF)