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()))
  KBO Processor:
   argument filtering:
    pi(a) = []
    pi(f) = [0,1]
    pi(f#) = 0
   usable rules:
    f(f(a(),f(x,a())),a()) -> f(a(),f(f(x,a()),a()))
   weight function:
    w0 = 1
    w(f#) = w(a) = 1
    w(f) = 0
   precedence:
    f# ~ f ~ a
   problem:
    DPs:
     
    TRS:
     f(f(a(),f(x,a())),a()) -> f(a(),f(f(x,a()),a()))
   Qed