YES

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

Proof:
 DP Processor:
  DPs:
   f#(f(a(),f(a(),a())),x) -> f#(f(a(),a()),a())
   f#(f(a(),f(a(),a())),x) -> f#(x,f(f(a(),a()),a()))
  TRS:
   f(f(a(),f(a(),a())),x) -> f(x,f(f(a(),a()),a()))
  Usable Rule Processor:
   DPs:
    f#(f(a(),f(a(),a())),x) -> f#(f(a(),a()),a())
    f#(f(a(),f(a(),a())),x) -> f#(x,f(f(a(),a()),a()))
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [f#](x0, x1) = 1x0 + 2x1 + 4,
     
     [f](x0, x1) = 2x1,
     
     [a] = 0
    orientation:
     f#(f(a(),f(a(),a())),x) = 2x + 5 >= 4 = f#(f(a(),a()),a())
     
     f#(f(a(),f(a(),a())),x) = 2x + 5 >= 1x + 4 = f#(x,f(f(a(),a()),a()))
    problem:
     DPs:
      
     TRS:
      
    Qed