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:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
f → g
f → h

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

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

(2) BOUNDS(n^1, INF)