NO
0.07/0.17	NO
0.07/0.18	
0.07/0.18	Problem:
0.07/0.18	 after(s(N),cons(X)) -> after(N,XS)
0.07/0.18	 from(X) -> cons(X)
0.07/0.18	 after(0(),XS) -> XS
0.07/0.18	
0.07/0.18	Proof:
0.07/0.18	 Fresh Variable Processor:
0.07/0.18	  loop length: 1
0.07/0.18	  terms:
0.07/0.18	   after(s(N),cons(X))
0.07/0.18	  context: after(N,[])
0.07/0.18	  substitution:
0.07/0.18	   XS -> after(s(N),cons(X))
0.07/0.18	  Qed
0.07/0.18	EOF