YES

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

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