YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [s](x0) = x0 + 2,
   
   [+](x0, x1) = x0 + 2x1 + 1,
   
   [0] = 0
  orientation:
   +(0(),y) = 2y + 1 >= y = y
   
   +(s(x),0()) = x + 3 >= x + 2 = s(x)
   
   +(s(x),s(y)) = x + 2y + 7 >= x + 2y + 7 = s(+(s(x),+(y,0())))
  problem:
   +(s(x),s(y)) -> s(+(s(x),+(y,0())))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]     [0]
    [s](x0) = [0 0 0]x0 + [0]
              [0 0 0]     [1],
    
                  [1 0 0]     [1 0 1]  
    [+](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                  [0 0 0]     [0 0 1]  ,
    
          [0]
    [0] = [0]
          [0]
   orientation:
                   [1 0 0]    [1 0 0]    [1]    [1 0 0]    [1 0 0]    [0]                      
    +(s(x),s(y)) = [0 0 0]x + [0 0 0]y + [0] >= [0 0 0]x + [0 0 0]y + [0] = s(+(s(x),+(y,0())))
                   [0 0 0]    [0 0 0]    [1]    [0 0 0]    [0 0 0]    [1]                      
   problem:
    
   Qed