YES

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

Proof:
 DP Processor:
  DPs:
   w#(r(x)) -> w#(x)
   b#(r(x)) -> b#(x)
   b#(w(x)) -> b#(x)
   b#(w(x)) -> w#(b(x))
  TRS:
   w(r(x)) -> r(w(x))
   b(r(x)) -> r(b(x))
   b(w(x)) -> w(b(x))
  KBO Processor:
   weight function:
    w0 = 1
    w(b#) = w(b) = w(w) = w(r) = 1
    w(w#) = 0
   precedence:
    w# > b > w > b# ~ r
   problem:
    DPs:
     
    TRS:
     
   Qed