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:

minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))

Rewrite Strategy: INNERMOST

(1) DecreasingLoopProof (EQUIVALENT transformation)

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