YES

Problem:
 from(X) -> cons(X,n__from(s(X)))
 sel(0(),cons(X,Y)) -> X
 sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
 from(X) -> n__from(X)
 activate(n__from(X)) -> from(X)
 activate(X) -> X

Proof:
 DP Processor:
  DPs:
   sel#(s(X),cons(Y,Z)) -> activate#(Z)
   sel#(s(X),cons(Y,Z)) -> sel#(X,activate(Z))
   activate#(n__from(X)) -> from#(X)
  TRS:
   from(X) -> cons(X,n__from(s(X)))
   sel(0(),cons(X,Y)) -> X
   sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
   from(X) -> n__from(X)
   activate(n__from(X)) -> from(X)
   activate(X) -> X
  LPO Processor:
   argument filtering:
    pi(from) = [0]
    pi(s) = [0]
    pi(n__from) = [0]
    pi(cons) = [0,1]
    pi(0) = []
    pi(sel) = [0,1]
    pi(activate) = [0]
    pi(from#) = []
    pi(sel#) = 0
    pi(activate#) = []
   precedence:
    sel > activate > from > s > activate# > sel# ~ from# ~ 0 ~ cons ~ n__from
   problem:
    DPs:
     
    TRS:
     from(X) -> cons(X,n__from(s(X)))
     sel(0(),cons(X,Y)) -> X
     sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
     from(X) -> n__from(X)
     activate(n__from(X)) -> from(X)
     activate(X) -> X
   Qed