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:

eq(0, 0) → true
eq(0, s(Y)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
min(cons(0, nil)) → 0
min(cons(s(N), nil)) → s(N)
min(cons(N, cons(M, L))) → ifmin(le(N, M), cons(N, cons(M, L)))
ifmin(true, cons(N, cons(M, L))) → min(cons(N, L))
ifmin(false, cons(N, cons(M, L))) → min(cons(M, L))
replace(N, M, nil) → nil
replace(N, M, cons(K, L)) → ifrepl(eq(N, K), N, M, cons(K, L))
ifrepl(true, N, M, cons(K, L)) → cons(M, L)
ifrepl(false, N, M, cons(K, L)) → cons(K, replace(N, M, L))
selsort(nil) → nil
selsort(cons(N, L)) → ifselsort(eq(N, min(cons(N, L))), cons(N, L))
ifselsort(true, cons(N, L)) → cons(N, selsort(L))
ifselsort(false, cons(N, L)) → cons(min(cons(N, L)), selsort(replace(min(cons(N, L)), 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
eq(s(X), s(Y)) →⁺ eq(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)