YES

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

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