NO

Problem:
 f(x) -> g(x)
 f(x) -> h(f(x))
 h(f(x)) -> h(g(x))
 g(x) -> h(g(x))

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