NO

Problem:
 2nd(cons1(X,cons(Y,Z))) -> Y
 2nd(cons(X,X1)) -> 2nd(cons1(X,X1))
 from(X) -> cons(X,from(s(X)))

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