YES

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

Proof:
 WPO Processor:
  algebra: Pol
  weight function:
   w0 = 0
   w(a) = 4
   w(b) = 1
   w(c) = w(d) = 0
  status function:
   st(c) = st(b) = st(a) = st(d) = [0]
  subterm coefficient function:
   sc(c, 0) = sc(d, 0) = 4
   sc(b, 0) = sc(a, 0) = 1
  precedence:
   c ~ a > b ~ d
  problem:
   
  Qed