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)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
s(X) → n__s(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__zWquot(X1, X2)) → zWquot(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)