NO

Problem:
 f(a(),a()) -> b()
 a() -> a'()
 f(a'(),x) -> f(x,x)
 f(x,a'()) -> f(x,x)
 f(a'(),a'()) -> b()
 b() -> f(a'(),a'())

Proof:
 Unfolding Processor:
  loop length: 1
  terms:
   f(a'(),a'())
  context: []
  substitution:
   
  Qed