YES

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

Proof:
 DP Processor:
  DPs:
   f#(a(),f(f(a(),x),a())) -> f#(a(),f(a(),x))
   f#(a(),f(f(a(),x),a())) -> f#(f(a(),f(a(),x)),a())
  TRS:
   f(a(),f(f(a(),x),a())) -> f(f(a(),f(a(),x)),a())
  Subterm Criterion Processor:
   simple projection:
    pi(f#) = 1
   problem:
    DPs:
     
    TRS:
     f(a(),f(f(a(),x),a())) -> f(f(a(),f(a(),x)),a())
   Qed