YES

Problem:
 f(f(x,y),z) -> f(x,f(y,z))
 f(i(x1),x1) -> e()

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