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:

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

Rewrite Strategy: INNERMOST

(1) DecreasingLoopProof (EQUIVALENT transformation)

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

(2) BOUNDS(n^1, INF)