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