The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(n__s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(n__s(n__s(X)), activate(Z)), n__cons(Y, n__filter(X, n__sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, n__sieve(activate(Y))))
from(X) → n__from(X)
s(X) → n__s(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
sieve(X) → n__sieve(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__filter(X1, X2)) → filter(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__sieve(X)) → sieve(activate(X))
activate(X) → X

Rewrite Strategy: INNERMOST

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
activate(n__from(X)) →⁺ from(activate(X))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [X / n__from(X)].
The result substitution is [ ].

(2) BOUNDS(n^1, INF)