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=3
  
  interpretation:
               
   [s](x0) = x0
               ,
   
                 [1 1 1]     [1 1 1]     [0]
   [+](x0, x1) = [0 1 0]x0 + [0 1 0]x1 + [1]
                 [0 0 1]     [0 0 1]     [0],
   
         [0]
   [0] = [0]
         [1]
  orientation:
              [1 1 1]    [1]         
   +(x,0()) = [0 1 0]x + [1] >= x = x
              [0 0 1]    [1]         
   
               [1 1 1]    [1 1 1]    [0]    [1 1 1]    [1 1 1]    [0]            
   +(x,s(y)) = [0 1 0]x + [0 1 0]y + [1] >= [0 1 0]x + [0 1 0]y + [1] = s(+(x,y))
               [0 0 1]    [0 0 1]    [0]    [0 0 1]    [0 0 1]    [0]            
   
                 [1 1 1]    [1]            
   +(0(),s(y)) = [0 1 0]y + [1] >= y = s(y)
                 [0 0 1]    [1]            
   
                 [1 1 1]    [1]            
   s(+(0(),y)) = [0 1 0]y + [1] >= y = s(y)
                 [0 0 1]    [1]            
  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