NO

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

Proof:
 Unfolding Processor:
  loop length: 3
  terms:
   f(c(),c(),c())
   f(c(),b(),c())
   f(a(),b(),c())
  context: []
  substitution:
   
  Qed