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:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, n__app(activate(XS), YS))
from(X) → cons(X, n__from(n__s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS)))
prefix(L) → cons(nil, n__zWadr(L, n__prefix(L)))
app(X1, X2) → n__app(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
zWadr(X1, X2) → n__zWadr(X1, X2)
prefix(X) → n__prefix(X)
activate(n__app(X1, X2)) → app(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__zWadr(X1, X2)) → zWadr(activate(X1), activate(X2))
activate(n__prefix(X)) → prefix(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__app(n__from(X13306_0), X2)) →⁺ app(cons(activate(X13306_0), n__from(n__s(activate(X13306_0)))), activate(X2))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0].
The pumping substitution is [X13306_0 / n__app(n__from(X13306_0), X2)].
The result substitution is [ ].

The rewrite sequence
activate(n__app(n__from(X13306_0), X2)) →⁺ app(cons(activate(X13306_0), n__from(n__s(activate(X13306_0)))), 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 [X13306_0 / n__app(n__from(X13306_0), X2)].
The result substitution is [ ].

(2) BOUNDS(2^n, INF)