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
  Subterm Criterion Processor:
   simple projection:
    pi(a#) = 0
   problem:
    DPs:
     
    TRS:
     a(b(x)) -> b(a(x))
     a(c(x)) -> x
   Qed