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:

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

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
mark(filter(nats(X72712_4), X2, X3)) →⁺ a__filter(cons(mark(mark(X72712_4)), nats(s(mark(X72712_4)))), mark(X2), mark(X3))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0,0].
The pumping substitution is [X72712_4 / filter(nats(X72712_4), X2, X3)].
The result substitution is [ ].

The rewrite sequence
mark(filter(nats(X72712_4), X2, X3)) →⁺ a__filter(cons(mark(mark(X72712_4)), nats(s(mark(X72712_4)))), mark(X2), mark(X3))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1,0,0].
The pumping substitution is [X72712_4 / filter(nats(X72712_4), X2, X3)].
The result substitution is [ ].

(2) BOUNDS(2^n, INF)