YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [s](x0) = x0 + 6,
   
   [+](x0, x1) = x0 + x1 + 5,
   
   [0] = 6
  orientation:
   +(x,0()) = x + 11 >= x = x
   
   +(x,s(y)) = x + y + 11 >= x + y + 11 = s(+(x,y))
   
   +(0(),s(y)) = y + 17 >= y + 6 = s(y)
   
   s(+(0(),y)) = y + 17 >= y + 6 = s(y)
  problem:
   +(x,s(y)) -> s(+(x,y))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]     [0]
    [s](x0) = [0 0 0]x0 + [1]
              [0 1 1]     [0],
    
                  [1 0 0]     [1 1 1]  
    [+](x0, x1) = [0 0 0]x0 + [0 1 0]x1
                  [1 0 0]     [0 0 1]  
   orientation:
                [1 0 0]    [1 1 1]    [1]    [1 0 0]    [1 1 1]    [0]            
    +(x,s(y)) = [0 0 0]x + [0 0 0]y + [1] >= [0 0 0]x + [0 0 0]y + [1] = s(+(x,y))
                [1 0 0]    [0 1 1]    [0]    [1 0 0]    [0 1 1]    [0]            
   problem:
    
   Qed