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:

active(minus(0, Y)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(geq(X, 0)) → mark(true)
active(geq(0, s(Y))) → mark(false)
active(geq(s(X), s(Y))) → mark(geq(X, Y))
active(div(0, s(Y))) → mark(0)
active(div(s(X), s(Y))) → mark(if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(s(X)) → s(active(X))
active(div(X1, X2)) → div(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
s(mark(X)) → mark(s(X))
div(mark(X1), X2) → mark(div(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(geq(X1, X2)) → geq(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(div(X1, X2)) → div(proper(X1), proper(X2))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
s(ok(X)) → ok(s(X))
geq(ok(X1), ok(X2)) → ok(geq(X1, X2))
div(ok(X1), ok(X2)) → ok(div(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

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

(2) BOUNDS(n^1, INF)