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:

2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(n__s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
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:

2nd(cons(X, n__cons(Y, Z))) → activate(Y)
from(X) → cons(X, n__from(n__s(X)))
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
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(2ⁿ):
The rewrite sequence
activate(n__cons(n__from(X128_0), X2)) →⁺ cons(cons(activate(X128_0), n__from(n__s(activate(X128_0)))), X2)
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0].
The pumping substitution is [X128_0 / n__cons(n__from(X128_0), X2)].
The result substitution is [ ].

The rewrite sequence
activate(n__cons(n__from(X128_0), X2)) →⁺ cons(cons(activate(X128_0), n__from(n__s(activate(X128_0)))), X2)
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1,0,0].
The pumping substitution is [X128_0 / n__cons(n__from(X128_0), X2)].
The result substitution is [ ].

(4) BOUNDS(2^n, INF)