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:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(xs) → sum2(xs, 0)
sum2(xs, y) → ifsum(isNil(xs), isZero(head(xs)), xs, y)
ifsum(true, b, xs, y) → y
ifsum(false, b, xs, y) → ifsum2(b, xs, y)
ifsum2(true, xs, y) → sum2(tail(xs), y)
ifsum2(false, xs, y) → sum2(cons(p(head(xs)), tail(xs)), s(y))
isNil(nil) → true
isNil(cons(x, xs)) → false
tail(nil) → nil
tail(cons(x, xs)) → xs
head(cons(x, xs)) → x
head(nil) → error
isZero(0) → true
isZero(s(0)) → false
isZero(s(s(x))) → isZero(s(x))
p(0) → s(s(0))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
a → c
a → d

Rewrite Strategy: INNERMOST

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
sum2(cons(0, xs653324_1), y) →⁺ sum2(xs653324_1, y)
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [xs653324_1 / cons(0, xs653324_1)].
The result substitution is [ ].

(2) BOUNDS(n^1, INF)