YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [+](x0, x1) = x0 + x1 + 2,
   
   [0] = 1,
   
   [f](x0) = x0 + 4
  orientation:
   f(+(x,0())) = x + 7 >= x + 4 = f(x)
   
   +(x,+(y,z)) = x + y + z + 4 >= x + y + z + 4 = +(+(x,y),z)
  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