YES

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

Proof:
 DP Processor:
  DPs:
   a#(d(x)) -> a#(x)
   a#(d(x)) -> b#(a(x))
   b#(c(x)) -> b#(x)
   b#(c(x)) -> a#(b(x))
  TRS:
   a(d(x)) -> d(c(b(a(x))))
   b(c(x)) -> c(d(a(b(x))))
   a(c(x)) -> x
   b(d(x)) -> x
  CDG Processor:
   DPs:
    a#(d(x)) -> a#(x)
    a#(d(x)) -> b#(a(x))
    b#(c(x)) -> b#(x)
    b#(c(x)) -> a#(b(x))
   TRS:
    a(d(x)) -> d(c(b(a(x))))
    b(c(x)) -> c(d(a(b(x))))
    a(c(x)) -> x
    b(d(x)) -> x
   graph:
    
   Qed