YES

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

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