NO

Problem:
 from(X) -> cons(X,from(s(X)))
 after(0(),XS) -> XS
 after(s(N),cons(X,XS)) -> after(N,XS)

Proof:
 Containment Processor: loop length: 1
                        terms:
                         from(X)
                        context: cons(X,[])
                        substitution:
                         X -> s(X)
  Qed