YES

Problem:
 add(0(),x) -> x
 add(s(x),y) -> s(add(x,y))

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [s](x0) = x0,
   
   [add](x0, x1) = x0 + x1 + 6,
   
   [0] = 1
  orientation:
   add(0(),x) = x + 7 >= x = x
   
   add(s(x),y) = x + y + 6 >= x + y + 6 = s(add(x,y))
  problem:
   add(s(x),y) -> s(add(x,y))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                   [0]
    [s](x0) = x0 + [1]
                   [1],
    
                    [1 0 1]     [1 0 1]  
    [add](x0, x1) = [0 1 0]x0 + [0 1 0]x1
                    [0 1 0]     [0 0 0]  
   orientation:
                  [1 0 1]    [1 0 1]    [1]    [1 0 1]    [1 0 1]    [0]              
    add(s(x),y) = [0 1 0]x + [0 1 0]y + [1] >= [0 1 0]x + [0 1 0]y + [1] = s(add(x,y))
                  [0 1 0]    [0 0 0]    [1]    [0 1 0]    [0 0 0]    [1]              
   problem:
    
   Qed