NO

Problem:
 F(H(x),y) -> F(H(x),I(I(y)))
 F(x,G(y)) -> F(I(x),G(y))
 I(x) -> x

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