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:

div(x, y) → div2(x, y, 0)
div2(x, y, i) → if1(le(y, 0), le(y, x), x, y, plus(i, 0), inc(i))
if1(true, b, x, y, i, j) → divZeroError
if1(false, b, x, y, i, j) → if2(b, x, y, i, j)
if2(true, x, y, i, j) → div2(minus(x, y), y, j)
if2(false, x, y, i, j) → i
inc(0) → 0
inc(s(i)) → s(inc(i))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
a → c
a → d

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(i)) →⁺ s(inc(i))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [i / s(i)].
The result substitution is [ ].

(2) BOUNDS(n^1, INF)