YES

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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
                 [1 0 0]     [1 1 0]  
   [+](x0, x1) = [1 0 0]x0 + [1 1 0]x1
                 [0 0 0]     [0 0 1]  ,
   
         [0]
   [0] = [0]
         [0],
   
             [1 0 0]     [1]
   [s](x0) = [1 0 0]x0 + [1]
             [0 0 0]     [0]
  orientation:
              [1 1 0]          
   +(0(),y) = [1 1 0]y >= y = y
              [0 0 1]          
   
                 [1 0 0]    [1]    [1 0 0]    [1]       
   +(s(x),0()) = [1 0 0]x + [1] >= [1 0 0]x + [1] = s(x)
                 [0 0 0]    [0]    [0 0 0]    [0]       
   
                  [1 0 0]    [2 0 0]    [3]    [1 0 0]    [2 0 0]    [2]                      
   +(s(x),s(y)) = [1 0 0]x + [2 0 0]y + [3] >= [1 0 0]x + [2 0 0]y + [2] = s(+(s(x),+(y,0())))
                  [0 0 0]    [0 0 0]    [0]    [0 0 0]    [0 0 0]    [0]                      
  problem:
   +(0(),y) -> y
   +(s(x),0()) -> s(x)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                  [1 0 0]          [1]
    [+](x0, x1) = [0 0 0]x0 + x1 + [0]
                  [0 0 1]          [0],
    
          [0]
    [0] = [0]
          [0],
    
              [1 0 0]  
    [s](x0) = [0 0 0]x0
              [0 0 1]  
   orientation:
                   [1]         
    +(0(),y) = y + [0] >= y = y
                   [0]         
    
                  [1 0 0]    [1]    [1 0 0]        
    +(s(x),0()) = [0 0 0]x + [0] >= [0 0 0]x = s(x)
                  [0 0 1]    [0]    [0 0 1]        
   problem:
    
   Qed