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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
le(s(X), s(Y)) →⁺ le(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)