YES

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

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