NO

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

Proof:
 Fresh Variable Processor:
  loop length: 1
  terms:
   after(s(N),cons(X))
  context: after(N,[])
  substitution:
   XS -> after(s(N),cons(X))
  Qed