The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

nats → adx(zeros)
zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
hd(cons(X, Y)) → activate(X)
tl(cons(X, Y)) → activate(Y)
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(X) → X

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
incr(cons(n__incr(X42433_1), Y)) →⁺ cons(n__s(incr(X42433_1)), n__incr(activate(Y)))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0].
The pumping substitution is [X42433_1 / cons(n__incr(X42433_1), Y)].
The result substitution is [ ].

(2) BOUNDS(n^1, INF)