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:

min(0, y) → 0
min(s(x), 0) → 0
min(s(x), s(y)) → min(x, y)
len(nil) → 0
len(cons(x, xs)) → s(len(xs))
sum(x, 0) → x
sum(x, s(y)) → s(sum(x, y))
le(0, x) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
take(0, cons(y, ys)) → y
take(s(x), cons(y, ys)) → take(x, ys)
addList(x, y) → if(le(0, min(len(x), len(y))), 0, x, y, nil)
if(false, c, x, y, z) → z
if(true, c, xs, ys, z) → if(le(s(c), min(len(xs), len(ys))), s(c), xs, ys, cons(sum(take(c, xs), take(c, ys)), z))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
min(s(x), s(y)) →⁺ min(x, y)
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [x / s(x), y / s(y)].
The result substitution is [ ].

(2) BOUNDS(n^1, INF)