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

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)