The Runtime Complexity (full) 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:

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

Rewrite Strategy: FULL

(1) CpxTrsToCpxRelTrsProof (BOTH BOUNDS(ID, ID) transformation)

Transformed TRS to relative TRS where S is empty.

(2) Obligation:

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

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
activate(n__take(X1, X2)) →⁺ take(activate(X1), activate(X2))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [X1 / n__take(X1, X2)].
The result substitution is [ ].

(4) BOUNDS(n^1, INF)