YES

Problem:
 f(+(x,0())) -> f(x)
 +(x,+(y,z)) -> +(+(x,y),z)

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 0]  
   [f](x0) = [0 0 0]x0
             [0 0 0]  ,
   
                 [1 0 0]     [1 0 0]     [0]
   [+](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [1]
                 [0 1 0]     [0 1 1]     [0],
   
         [1]
   [0] = [0]
         [0]
  orientation:
                 [1 0 0]    [1]    [1 0 0]        
   f(+(x,0())) = [0 0 0]x + [0] >= [0 0 0]x = f(x)
                 [0 0 0]    [0]    [0 0 0]        
   
                 [1 0 0]    [1 0 0]    [1 0 0]    [0]    [1 0 0]    [1 0 0]    [1 0 0]    [0]              
   +(x,+(y,z)) = [0 0 0]x + [0 0 0]y + [0 0 0]z + [1] >= [0 0 0]x + [0 0 0]y + [0 0 0]z + [1] = +(+(x,y),z)
                 [0 1 0]    [0 1 0]    [0 1 1]    [1]    [0 0 0]    [0 0 0]    [0 1 1]    [1]              
  problem:
   +(x,+(y,z)) -> +(+(x,y),z)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [+](x0, x1) = x0 + 2x1 + 5
   orientation:
    +(x,+(y,z)) = x + 2y + 4z + 15 >= x + 2y + 2z + 10 = +(+(x,y),z)
   problem:
    
   Qed