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:

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(nil) → nil
sort(cons(x, xs)) → cons(max(cons(x, xs)), sort(h(del(max(cons(x, xs)), cons(x, xs)))))
ge(0, 0) → true
ge(s(x), 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
h(nil) → nil
h(cons(x, xs)) → cons(x, h(xs))

Rewrite Strategy: FULL

(1) CpxTrsToCpxRelTrsProof (BOTH BOUNDS(ID, ID) transformation)

Transformed TRS to relative TRS where S is empty.

(2) Obligation:

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

max(nil) → 0
max(cons(x, nil)) → x
max(cons(x, cons(y, xs))) → if1(ge(x, y), x, y, xs)
if1(true, x, y, xs) → max(cons(x, xs))
if1(false, x, y, xs) → max(cons(y, xs))
del(x, nil) → nil
del(x, cons(y, xs)) → if2(eq(x, y), x, y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
sort(nil) → nil
sort(cons(x, xs)) → cons(max(cons(x, xs)), sort(h(del(max(cons(x, xs)), cons(x, xs)))))
ge(0, 0) → true
ge(s(x), 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
h(nil) → nil
h(cons(x, xs)) → cons(x, h(xs))

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

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

(4) BOUNDS(n^1, INF)