YES

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

Proof:
 DP Processor:
  DPs:
   f#(x,f(f(a(),a()),a())) -> f#(a(),f(a(),a()))
   f#(x,f(f(a(),a()),a())) -> f#(f(a(),f(a(),a())),x)
  TRS:
   f(x,f(f(a(),a()),a())) -> f(f(a(),f(a(),a())),x)
  Usable Rule Processor:
   DPs:
    f#(x,f(f(a(),a()),a())) -> f#(a(),f(a(),a()))
    f#(x,f(f(a(),a()),a())) -> f#(f(a(),f(a(),a())),x)
   TRS:
    
   CDG Processor:
    DPs:
     f#(x,f(f(a(),a()),a())) -> f#(a(),f(a(),a()))
     f#(x,f(f(a(),a()),a())) -> f#(f(a(),f(a(),a())),x)
    TRS:
     
    graph:
     
    Qed