YES

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

Proof:
 String Reversal Processor:
  r(w(x)) -> w(r(x))
  r(b(x)) -> b(r(x))
  w(b(x)) -> b(w(x))
  KBO Processor:
   weight function:
    w0 = 1
    w(b) = w(w) = 1
    w(r) = 0
   precedence:
    r > w > b
   problem:
    
   Qed