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