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:

and(tt, T) → T
isNatIList(IL) → isNatList(activate(IL))
isNat(n__0) → tt
isNat(n__s(N)) → isNat(activate(N))
isNat(n__length(L)) → isNatList(activate(L))
isNatIList(n__zeros) → tt
isNatIList(n__cons(N, IL)) → and(isNat(activate(N)), isNatIList(activate(IL)))
isNatList(n__nil) → tt
isNatList(n__cons(N, L)) → and(isNat(activate(N)), isNatList(activate(L)))
isNatList(n__take(N, IL)) → and(isNat(activate(N)), isNatIList(activate(IL)))
zeros → cons(0, n__zeros)
take(0, IL) → uTake1(isNatIList(IL))
uTake1(tt) → nil
take(s(M), cons(N, IL)) → uTake2(and(isNat(M), and(isNat(N), isNatIList(activate(IL)))), M, N, activate(IL))
uTake2(tt, M, N, IL) → cons(activate(N), n__take(activate(M), activate(IL)))
length(cons(N, L)) → uLength(and(isNat(N), isNatList(activate(L))), activate(L))
uLength(tt, L) → s(length(activate(L)))
0 → n__0
s(X) → n__s(X)
length(X) → n__length(X)
zeros → n__zeros
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
take(X1, X2) → n__take(X1, X2)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__length(X)) → length(activate(X))
activate(n__zeros) → zeros
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
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__s(n__length(n__cons(X148473_3, X248474_3)))) →⁺ s(uLength(and(isNat(activate(X148473_3)), isNatList(activate(X248474_3))), activate(X248474_3)))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0,1,0].
The pumping substitution is [X248474_3 / n__s(n__length(n__cons(X148473_3, X248474_3)))].
The result substitution is [ ].

The rewrite sequence
activate(n__s(n__length(n__cons(X148473_3, X248474_3)))) →⁺ s(uLength(and(isNat(activate(X148473_3)), isNatList(activate(X248474_3))), activate(X248474_3)))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1].
The pumping substitution is [X248474_3 / n__s(n__length(n__cons(X148473_3, X248474_3)))].
The result substitution is [ ].

(2) BOUNDS(2^n, INF)