YES

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

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