The Runtime Complexity (innermost) 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:

plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
sum(xs) → sumIter(xs, 0)
sumIter(xs, x) → ifSum(isempty(xs), xs, x, plus(x, head(xs)))
ifSum(true, xs, x, y) → x
ifSum(false, xs, x, y) → sumIter(tail(xs), y)
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
a → b
a → c

Rewrite Strategy: INNERMOST

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
le(s(x), s(y)) →⁺ le(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)