YES

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

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