YES

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

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