YES

Problem:
 a(c(d(x))) -> c(x)
 u(b(d(d(x)))) -> b(x)
 v(a(a(x))) -> u(v(x))
 v(a(c(x))) -> u(b(d(x)))
 v(c(x)) -> b(x)
 w(a(a(x))) -> u(w(x))
 w(a(c(x))) -> u(b(d(x)))
 w(c(x)) -> b(x)

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