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:

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(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

Rewrite Strategy: FULL

(1) CpxTrsToCpxRelTrsProof (BOTH BOUNDS(ID, ID) transformation)

Transformed TRS to relative TRS where S is empty.

(2) Obligation:

Runtime Complexity Relative 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(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

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

(4) BOUNDS(n^1, INF)