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