NO

Problem:
 f(n__a(),n__b(),X) -> f(X,X,X)
 c() -> a()
 c() -> b()
 a() -> n__a()
 b() -> n__b()
 activate(n__a()) -> a()
 activate(n__b()) -> b()
 activate(X) -> X

Proof:
 Unfolding Processor:
  loop length: 5
  terms:
   f(c(),c(),c())
   f(c(),b(),c())
   f(c(),n__b(),c())
   f(a(),n__b(),c())
   f(n__a(),n__b(),c())
  context: []
  substitution:
   
  Qed