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:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(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__first(n__from(X110_3), X2)) →⁺ first(cons(activate(X110_3), n__from(n__s(activate(X110_3)))), activate(X2))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0].
The pumping substitution is [X110_3 / n__first(n__from(X110_3), X2)].
The result substitution is [ ].

The rewrite sequence
activate(n__first(n__from(X110_3), X2)) →⁺ first(cons(activate(X110_3), n__from(n__s(activate(X110_3)))), activate(X2))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1,0,0].
The pumping substitution is [X110_3 / n__first(n__from(X110_3), X2)].
The result substitution is [ ].

(2) BOUNDS(2^n, INF)