NO

Problem:
 f(x,y) -> f(g(x),g(x))
 g(x) -> h(x)
 F(g(x),x) -> F(x,g(x))
 F(h(x),x) -> F(x,h(x))

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