YES

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

Proof:
 DP Processor:
  DPs:
   f#(f(a(),f(x,a())),a()) -> f#(f(x,a()),a())
   f#(f(a(),f(x,a())),a()) -> f#(a(),f(f(x,a()),a()))
  TRS:
   f(f(a(),f(x,a())),a()) -> f(a(),f(f(x,a()),a()))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [f#](x0, x1) = 4x0 + 1,
    
    [f](x0, x1) = 2x0 + 2x1 + 2,
    
    [a] = 0
   orientation:
    f#(f(a(),f(x,a())),a()) = 16x + 25 >= 8x + 9 = f#(f(x,a()),a())
    
    f#(f(a(),f(x,a())),a()) = 16x + 25 >= 1 = f#(a(),f(f(x,a()),a()))
    
    f(f(a(),f(x,a())),a()) = 8x + 14 >= 8x + 14 = f(a(),f(f(x,a()),a()))
   problem:
    DPs:
     
    TRS:
     f(f(a(),f(x,a())),a()) -> f(a(),f(f(x,a()),a()))
   Qed