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:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [f#](x0, x1) = 2x0 + x1 + 0,
     
     [f](x0, x1) = 3x0 + x1 + 4,
     
     [a] = 0
    orientation:
     f#(x,f(f(a(),a()),a())) = 2x + 7 >= 4 = f#(a(),f(a(),a()))
     
     f#(x,f(f(a(),a()),a())) = 2x + 7 >= x + 6 = f#(f(a(),f(a(),a())),x)
    problem:
     DPs:
      
     TRS:
      
    Qed