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:

dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(n__dbl(activate(X)), n__dbls(activate(Y)))
sel(0, cons(X, Y)) → activate(X)
sel(s(X), cons(Y, Z)) → sel(activate(X), activate(Z))
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
s(X) → n__s(X)
dbl(X) → n__dbl(X)
dbls(X) → n__dbls(X)
sel(X1, X2) → n__sel(X1, X2)
indx(X1, X2) → n__indx(X1, X2)
from(X) → n__from(X)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(activate(X))
activate(n__dbls(X)) → dbls(activate(X))
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(n__indx(X1, X2)) → indx(activate(X1), X2)
activate(n__from(X)) → from(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:

dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(n__dbl(activate(X)), n__dbls(activate(Y)))
sel(0, cons(X, Y)) → activate(X)
sel(s(X), cons(Y, Z)) → sel(activate(X), activate(Z))
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
s(X) → n__s(X)
dbl(X) → n__dbl(X)
dbls(X) → n__dbls(X)
sel(X1, X2) → n__sel(X1, X2)
indx(X1, X2) → n__indx(X1, X2)
from(X) → n__from(X)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(activate(X))
activate(n__dbls(X)) → dbls(activate(X))
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(n__indx(X1, X2)) → indx(activate(X1), X2)
activate(n__from(X)) → from(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
from(n__from(X13499_3)) →⁺ cons(from(X13499_3), n__from(n__s(from(X13499_3))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [X13499_3 / n__from(X13499_3)].
The result substitution is [ ].

The rewrite sequence
from(n__from(X13499_3)) →⁺ cons(from(X13499_3), n__from(n__s(from(X13499_3))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0,0].
The pumping substitution is [X13499_3 / n__from(X13499_3)].
The result substitution is [ ].

(4) BOUNDS(2^n, INF)