YES

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

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