YES

Problem:
 b(c(a(x1))) -> a(b(x1))
 b(b(b(x1))) -> c(a(c(x1)))
 c(d(x1)) -> d(c(x1))
 c(d(b(x1))) -> d(c(c(x1)))
 d(c(x1)) -> b(b(b(x1)))
 c(b(x1)) -> d(a(x1))
 d(b(c(x1))) -> a(a(x1))
 d(a(x1)) -> b(x1)

Proof:
 WPO Processor:
  algebra: Sum
  weight function:
   w0 = 0
   w(d) = 2
   w(b) = w(c) = 1
   w(a) = 0
  status function:
   st(d) = st(b) = st(c) = st(a) = [0]
  precedence:
   c > d > b ~ a
  problem:
   
  Qed