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
  Usable Rule 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:
    activate(n__from(X)) -> from(X)
    activate(X) -> X
    from(X) -> cons(X,n__from(s(X)))
    from(X) -> n__from(X)
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [activate#](x0) = 1,
     
     [sel#](x0, x1) = 4x0,
     
     [from#](x0) = 0,
     
     [activate](x0) = x0 + 0,
     
     [cons](x0, x1) = x0 + x1 + 0,
     
     [n__from](x0) = x0 + 0,
     
     [s](x0) = 1x0 + 0,
     
     [from](x0) = x0 + 0
    orientation:
     sel#(s(X),cons(Y,Z)) = 5X + 4 >= 1 = activate#(Z)
     
     sel#(s(X),cons(Y,Z)) = 5X + 4 >= 4X = sel#(X,activate(Z))
     
     activate#(n__from(X)) = 1 >= 0 = from#(X)
     
     activate(n__from(X)) = X + 0 >= X + 0 = from(X)
     
     activate(X) = X + 0 >= X = X
     
     from(X) = X + 0 >= 1X + 0 = cons(X,n__from(s(X)))
     
     from(X) = X + 0 >= X + 0 = n__from(X)
    problem:
     DPs:
      
     TRS:
      activate(n__from(X)) -> from(X)
      activate(X) -> X
      from(X) -> cons(X,n__from(s(X)))
      from(X) -> n__from(X)
    Qed