The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(2^n, INF).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

qsort(nil) → nil
qsort(.(x, y)) → ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) → nil
lowers(x, .(y, z)) → if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) → nil
greaters(x, .(y, z)) → if(<=(y, x), greaters(x, z), .(y, greaters(x, z)))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
lowers(x, .(y, z)) →⁺ if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1,1].
The pumping substitution is [z / .(y, z)].
The result substitution is [ ].

The rewrite sequence
lowers(x, .(y, z)) →⁺ if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
gives rise to a decreasing loop by considering the right hand sides subterm at position [2].
The pumping substitution is [z / .(y, z)].
The result substitution is [ ].

(2) BOUNDS(2^n, INF)