YES

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

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