YES

Problem:
 from(X) -> cons(X,n__from(n__s(X)))
 first(0(),Z) -> nil()
 first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
 sel(0(),cons(X,Z)) -> X
 sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
 from(X) -> n__from(X)
 s(X) -> n__s(X)
 first(X1,X2) -> n__first(X1,X2)
 activate(n__from(X)) -> from(activate(X))
 activate(n__s(X)) -> s(activate(X))
 activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
 activate(X) -> X

Proof:
 WPO Processor:
  algebra: Max
  weight function:
   w0 = 0
   w(sel) = 6
   w(n__first) = w(first) = w(0) = 5
   w(activate) = 4
   w(cons) = w(n__from) = w(from) = 1
   w(s) = w(nil) = w(n__s) = 0
  status function:
   st(sel) = st(n__first) = [0, 1]
   st(activate) = st(s) = [0]
   st(nil) = []
   st(first) = [0, 1]
   st(0) = []
   st(cons) = [0, 1]
   st(n__from) = st(n__s) = st(from) = [0]
  subterm penalty function:
   sp(sel, 1) = 2
   sp(sel, 0) = sp(n__first, 1) = sp(first, 1) = 1
   sp(n__first, 0) = sp(activate, 0) = sp(s, 0) = sp(first, 0) = sp(cons, 1) = sp(
   cons, 0) = sp(n__from, 0) = sp(n__s, 0) = sp(from, 0) = 0
  precedence:
   activate > first ~ from > s > sel ~ n__first ~ nil ~ 0 ~ cons ~ n__from ~ n__s
  problem:
   
  Qed