YES

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

Proof:
 DP Processor:
  DPs:
   f#(f(x,a()),a()) -> f#(a(),a())
   f#(f(x,a()),a()) -> f#(a(),f(a(),a()))
   f#(f(x,a()),a()) -> f#(f(a(),f(a(),a())),a())
   f#(f(x,a()),a()) -> f#(f(f(a(),f(a(),a())),a()),x)
  TRS:
   f(f(x,a()),a()) -> f(f(f(a(),f(a(),a())),a()),x)
  Usable Rule Processor:
   DPs:
    f#(f(x,a()),a()) -> f#(a(),a())
    f#(f(x,a()),a()) -> f#(a(),f(a(),a()))
    f#(f(x,a()),a()) -> f#(f(a(),f(a(),a())),a())
    f#(f(x,a()),a()) -> f#(f(f(a(),f(a(),a())),a()),x)
   TRS:
    
   Matrix Interpretation Processor: dim=4
    
    usable rules:
     
    interpretation:
     [f#](x0, x1) = [0 0 1 1]x0 + [0 0 1 0]x1,
     
                   [0 0 0 1]     [0 0 0 1]  
                   [1 0 1 0]     [0 0 1 0]  
     [f](x0, x1) = [0 0 0 0]x0 + [1 0 0 0]x1
                   [0 0 1 0]     [1 0 0 0]  ,
     
           [1]
           [0]
     [a] = [1]
           [0]
    orientation:
     f#(f(x,a()),a()) = [0 0 1 0]x + [3] >= [2] = f#(a(),a())
     
     f#(f(x,a()),a()) = [0 0 1 0]x + [3] >= [2] = f#(a(),f(a(),a()))
     
     f#(f(x,a()),a()) = [0 0 1 0]x + [3] >= [2] = f#(f(a(),f(a(),a())),a())
     
     f#(f(x,a()),a()) = [0 0 1 0]x + [3] >= [0 0 1 0]x + [2] = f#(f(f(a(),f(a(),a())),a()),x)
    problem:
     DPs:
      
     TRS:
      
    Qed