YES

Problem:
 f(f(a(),f(x,a())),a()) -> f(a(),f(f(x,a()),a()))

Proof:
 DP Processor:
  DPs:
   f#(f(a(),f(x,a())),a()) -> f#(f(x,a()),a())
   f#(f(a(),f(x,a())),a()) -> f#(a(),f(f(x,a()),a()))
  TRS:
   f(f(a(),f(x,a())),a()) -> f(a(),f(f(x,a()),a()))
  Subterm Criterion Processor:
   simple projection:
    pi(f#) = 0
   problem:
    DPs:
     
    TRS:
     f(f(a(),f(x,a())),a()) -> f(a(),f(f(x,a()),a()))
   Qed