YES

Problem:
 f(s(x),y) -> f(x,s(x))
 f(x,s(y)) -> f(y,x)

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
                 [1 1 1]     [1 0 1]  
   [f](x0, x1) = [0 1 1]x0 + [0 1 0]x1
                 [0 0 0]     [0 0 0]  ,
   
             [1 0 1]     [1]
   [s](x0) = [1 1 1]x0 + [1]
             [0 1 1]     [0]
  orientation:
               [2 2 3]    [1 0 1]    [2]    [2 2 3]    [1]            
   f(s(x),y) = [1 2 2]x + [0 1 0]y + [1] >= [1 2 2]x + [1] = f(x,s(x))
               [0 0 0]    [0 0 0]    [0]    [0 0 0]    [0]            
   
               [1 1 1]    [1 1 2]    [1]    [1 0 1]    [1 1 1]          
   f(x,s(y)) = [0 1 1]x + [1 1 1]y + [1] >= [0 1 0]x + [0 1 1]y = f(y,x)
               [0 0 0]    [0 0 0]    [0]    [0 0 0]    [0 0 0]          
  problem:
   
  Qed