NO

Problem:
 I(x) -> I(J(x))
 J(x) -> J(K(J(x)))
 H(I(x)) -> K(J(x))
 J(x) -> K(J(x))

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