YES

Problem:
 w(r(x)) -> r(w(x))
 b(r(x)) -> r(b(x))
 b(w(x)) -> w(b(x))

Proof:
 KBO Processor:
  weight function:
   w0 = 1
   w(b) = w(w) = w(r) = 1
  precedence:
   b > w > r
  problem:
   
  Qed