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__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
mark(sieve(from(X133734_4))) →⁺ a__sieve(cons(mark(mark(X133734_4)), from(s(mark(X133734_4)))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0,0].
The pumping substitution is [X133734_4 / sieve(from(X133734_4))].
The result substitution is [ ].

The rewrite sequence
mark(sieve(from(X133734_4))) →⁺ a__sieve(cons(mark(mark(X133734_4)), from(s(mark(X133734_4)))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1,0,0].
The pumping substitution is [X133734_4 / sieve(from(X133734_4))].
The result substitution is [ ].

(2) BOUNDS(2^n, INF)