YES

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

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