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)
eq(0, 0) → true
eq(0, s(y)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
if1(true, x, y, xs) → min(x, xs)
if1(false, x, y, xs) → min(y, xs)
if2(true, x, y, xs) → xs
if2(false, x, y, xs) → cons(y, del(x, xs))
minsort(nil) → nil
minsort(cons(x, y)) → cons(min(x, y), minsort(del(min(x, y), cons(x, y))))
min(x, nil) → x
min(x, cons(y, z)) → if1(le(x, y), x, y, z)
del(x, nil) → nil
del(x, cons(y, z)) → if2(eq(x, y), x, y, z)

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)