YES

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

Proof:
 DP Processor:
  DPs:
   f#(a(),f(f(a(),a()),x)) -> f#(a(),x)
   f#(a(),f(f(a(),a()),x)) -> f#(a(),f(a(),x))
   f#(a(),f(f(a(),a()),x)) -> f#(f(a(),a()),f(a(),f(a(),x)))
  TRS:
   f(a(),f(f(a(),a()),x)) -> f(f(a(),a()),f(a(),f(a(),x)))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
    [f#](x0, x1) = [0 1]x1,
    
                  [3 0]     [0 2]     [1]
    [f](x0, x1) = [2 0]x0 + [0 1]x1 + [0],
    
          [0]
    [a] = [0]
   orientation:
    f#(a(),f(f(a(),a()),x)) = [0 1]x + [2] >= [0 1]x = f#(a(),x)
    
    f#(a(),f(f(a(),a()),x)) = [0 1]x + [2] >= [0 1]x = f#(a(),f(a(),x))
    
    f#(a(),f(f(a(),a()),x)) = [0 1]x + [2] >= [0 1]x = f#(f(a(),a()),f(a(),f(a(),x)))
    
                             [0 2]    [5]    [0 2]    [4]                                
    f(a(),f(f(a(),a()),x)) = [0 1]x + [2] >= [0 1]x + [2] = f(f(a(),a()),f(a(),f(a(),x)))
   problem:
    DPs:
     
    TRS:
     f(a(),f(f(a(),a()),x)) -> f(f(a(),a()),f(a(),f(a(),x)))
   Qed