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:

from(X) → cons(X, n__from(n__s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
s(X) → n__s(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
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__from(X)) →⁺ cons(activate(X), n__from(n__s(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 [ ].

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

(2) BOUNDS(2^n, INF)