YES

Problem:
 f(f(a(),x),a()) -> f(a(),f(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#(f(a(),f(a(),a())),x)
   f#(f(a(),x),a()) -> f#(a(),f(f(a(),f(a(),a())),x))
  TRS:
   f(f(a(),x),a()) -> f(a(),f(f(a(),f(a(),a())),x))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
    [f#](x0, x1) = [0 1 0]x0 + [0 0 1]x1,
    
                  [0 0 0]     [0 0 0]     [0]
    [f](x0, x1) = [1 1 1]x0 + [1 0 1]x1 + [1]
                  [0 0 0]     [0 0 0]     [0],
    
          [1]
    [a] = [1]
          [1]
   orientation:
    f#(f(a(),x),a()) = [1 0 1]x + [5] >= [2] = f#(a(),a())
    
    f#(f(a(),x),a()) = [1 0 1]x + [5] >= [1] = f#(a(),f(a(),a()))
    
    f#(f(a(),x),a()) = [1 0 1]x + [5] >= [0 0 1]x + [4] = f#(f(a(),f(a(),a())),x)
    
    f#(f(a(),x),a()) = [1 0 1]x + [5] >= [1] = f#(a(),f(f(a(),f(a(),a())),x))
    
                      [0 0 0]    [0]    [0]                                
    f(f(a(),x),a()) = [1 0 1]x + [7] >= [4] = f(a(),f(f(a(),f(a(),a())),x))
                      [0 0 0]    [0]    [0]                                
   problem:
    DPs:
     
    TRS:
     f(f(a(),x),a()) -> f(a(),f(f(a(),f(a(),a())),x))
   Qed