NO

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

Proof:
 Containment Processor:
  loop length: 1
  terms:
   f(x,f(y,z))
  context: []
  substitution:
   x -> f(x,y)
   y -> x
   z -> z
  Qed