YES

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

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