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:
 String Reversal Processor:
  d(c(a(x))) -> c(x)
  d(d(b(u(x)))) -> b(x)
  a(a(v(x))) -> v(u(x))
  c(a(v(x))) -> d(b(u(x)))
  c(v(x)) -> b(x)
  a(a(w(x))) -> w(u(x))
  c(a(w(x))) -> d(b(u(x)))
  c(w(x)) -> b(x)
  KBO Processor:
   weight function:
    w0 = 1
    w(w) = w(v) = w(u) = w(b) = w(a) = w(c) = w(d) = 1
   precedence:
    u > v > a > b > w > c > d
   problem:
    
   Qed