YES

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

Proof:
 DP Processor:
  DPs:
   f#(a(),f(f(a(),x),a())) -> f#(a(),f(a(),x))
   f#(a(),f(f(a(),x),a())) -> f#(f(a(),f(a(),x)),a())
  TRS:
   f(a(),f(f(a(),x),a())) -> f(f(a(),f(a(),x)),a())
  Matrix Interpretation Processor: dim=1
   
   usable rules:
    f(a(),f(f(a(),x),a())) -> f(f(a(),f(a(),x)),a())
   interpretation:
    [f#](x0, x1) = 4x1,
    
    [f](x0, x1) = x0 + 2x1 + 2,
    
    [a] = 1
   orientation:
    f#(a(),f(f(a(),x),a())) = 8x + 28 >= 8x + 12 = f#(a(),f(a(),x))
    
    f#(a(),f(f(a(),x),a())) = 8x + 28 >= 4 = f#(f(a(),f(a(),x)),a())
    
    f(a(),f(f(a(),x),a())) = 4x + 17 >= 4x + 13 = f(f(a(),f(a(),x)),a())
   problem:
    DPs:
     
    TRS:
     f(a(),f(f(a(),x),a())) -> f(f(a(),f(a(),x)),a())
   Qed