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:

f(node(s(n), xs)) → f(addchild(select(xs), node(n, xs)))
select(cons(ap, xs)) → ap
select(cons(ap, xs)) → select(xs)
addchild(node(y, ys), node(n, xs)) → node(y, cons(node(n, xs), ys))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
f(node(s(n), cons(node(s(n181_0), ys68_0), xs4_0))) →⁺ f(node(n, cons(node(n181_0, cons(node(n, cons(node(s(n181_0), ys68_0), xs4_0)), ys68_0)), cons(node(s(n181_0), ys68_0), xs4_0))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [n / s(n), n181_0 / s(n181_0)].
The result substitution is [ys68_0 / cons(node(n, cons(node(s(n181_0), ys68_0), xs4_0)), ys68_0), xs4_0 / cons(node(s(n181_0), ys68_0), xs4_0)].

(2) BOUNDS(n^1, INF)