YES

Problem:
 2nd(cons(X,n__cons(Y,Z))) -> activate(Y)
 from(X) -> cons(X,n__from(n__s(X)))
 cons(X1,X2) -> n__cons(X1,X2)
 from(X) -> n__from(X)
 s(X) -> n__s(X)
 activate(n__cons(X1,X2)) -> cons(activate(X1),X2)
 activate(n__from(X)) -> from(activate(X))
 activate(n__s(X)) -> s(activate(X))
 activate(X) -> X

Proof:
 DP Processor:
  DPs:
   2nd#(cons(X,n__cons(Y,Z))) -> activate#(Y)
   from#(X) -> cons#(X,n__from(n__s(X)))
   activate#(n__cons(X1,X2)) -> activate#(X1)
   activate#(n__cons(X1,X2)) -> cons#(activate(X1),X2)
   activate#(n__from(X)) -> activate#(X)
   activate#(n__from(X)) -> from#(activate(X))
   activate#(n__s(X)) -> activate#(X)
   activate#(n__s(X)) -> s#(activate(X))
  TRS:
   2nd(cons(X,n__cons(Y,Z))) -> activate(Y)
   from(X) -> cons(X,n__from(n__s(X)))
   cons(X1,X2) -> n__cons(X1,X2)
   from(X) -> n__from(X)
   s(X) -> n__s(X)
   activate(n__cons(X1,X2)) -> cons(activate(X1),X2)
   activate(n__from(X)) -> from(activate(X))
   activate(n__s(X)) -> s(activate(X))
   activate(X) -> X
  Usable Rule Processor:
   DPs:
    2nd#(cons(X,n__cons(Y,Z))) -> activate#(Y)
    from#(X) -> cons#(X,n__from(n__s(X)))
    activate#(n__cons(X1,X2)) -> activate#(X1)
    activate#(n__cons(X1,X2)) -> cons#(activate(X1),X2)
    activate#(n__from(X)) -> activate#(X)
    activate#(n__from(X)) -> from#(activate(X))
    activate#(n__s(X)) -> activate#(X)
    activate#(n__s(X)) -> s#(activate(X))
   TRS:
    activate(n__cons(X1,X2)) -> cons(activate(X1),X2)
    activate(n__from(X)) -> from(activate(X))
    activate(n__s(X)) -> s(activate(X))
    activate(X) -> X
    cons(X1,X2) -> n__cons(X1,X2)
    from(X) -> cons(X,n__from(n__s(X)))
    from(X) -> n__from(X)
    s(X) -> n__s(X)
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [s#](x0) = 0,
     
     [cons#](x0, x1) = 0,
     
     [from#](x0) = 1,
     
     [activate#](x0) = 1x0,
     
     [2nd#](x0) = 4x0 + 2,
     
     [s](x0) = x0 + 0,
     
     [n__from](x0) = 2x0 + 3,
     
     [n__s](x0) = 2x0 + 3,
     
     [from](x0) = x0 + 0,
     
     [activate](x0) = x0 + 0,
     
     [cons](x0, x1) = x1 + 0,
     
     [n__cons](x0, x1) = 1x0 + 1x1 + 0
    orientation:
     2nd#(cons(X,n__cons(Y,Z))) = 5Y + 5Z + 4 >= 1Y = activate#(Y)
     
     from#(X) = 1 >= 0 = cons#(X,n__from(n__s(X)))
     
     activate#(n__cons(X1,X2)) = 2X1 + 2X2 + 1 >= 1X1 = activate#(X1)
     
     activate#(n__cons(X1,X2)) = 2X1 + 2X2 + 1 >= 0 = cons#(activate(X1),X2)
     
     activate#(n__from(X)) = 3X + 4 >= 1X = activate#(X)
     
     activate#(n__from(X)) = 3X + 4 >= 1 = from#(activate(X))
     
     activate#(n__s(X)) = 3X + 4 >= 1X = activate#(X)
     
     activate#(n__s(X)) = 3X + 4 >= 0 = s#(activate(X))
     
     activate(n__cons(X1,X2)) = 1X1 + 1X2 + 0 >= X2 + 0 = cons(activate(X1),X2)
     
     activate(n__from(X)) = 2X + 3 >= X + 0 = from(activate(X))
     
     activate(n__s(X)) = 2X + 3 >= X + 0 = s(activate(X))
     
     activate(X) = X + 0 >= X = X
     
     cons(X1,X2) = X2 + 0 >= 1X1 + 1X2 + 0 = n__cons(X1,X2)
     
     from(X) = X + 0 >= 4X + 5 = cons(X,n__from(n__s(X)))
     
     from(X) = X + 0 >= 2X + 3 = n__from(X)
     
     s(X) = X + 0 >= 2X + 3 = n__s(X)
    problem:
     DPs:
      
     TRS:
      activate(n__cons(X1,X2)) -> cons(activate(X1),X2)
      activate(n__from(X)) -> from(activate(X))
      activate(n__s(X)) -> s(activate(X))
      activate(X) -> X
      cons(X1,X2) -> n__cons(X1,X2)
      from(X) -> cons(X,n__from(n__s(X)))
      from(X) -> n__from(X)
      s(X) -> n__s(X)
    Qed