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:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(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__fst(n__from(X7745_3), X2)) →⁺ fst(cons(activate(X7745_3), n__from(n__s(activate(X7745_3)))), activate(X2))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0].
The pumping substitution is [X7745_3 / n__fst(n__from(X7745_3), X2)].
The result substitution is [ ].

The rewrite sequence
activate(n__fst(n__from(X7745_3), X2)) →⁺ fst(cons(activate(X7745_3), n__from(n__s(activate(X7745_3)))), 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 [X7745_3 / n__fst(n__from(X7745_3), X2)].
The result substitution is [ ].

(2) BOUNDS(2^n, INF)