YES

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

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