YES

Problem:
 2nd(cons1(X,cons(Y,Z))) -> Y
 2nd(cons(X,X1)) -> 2nd(cons1(X,activate(X1)))
 from(X) -> cons(X,n__from(s(X)))
 from(X) -> n__from(X)
 activate(n__from(X)) -> from(X)
 activate(X) -> X

Proof:
 DP Processor:
  DPs:
   2nd#(cons(X,X1)) -> activate#(X1)
   2nd#(cons(X,X1)) -> 2nd#(cons1(X,activate(X1)))
   activate#(n__from(X)) -> from#(X)
  TRS:
   2nd(cons1(X,cons(Y,Z))) -> Y
   2nd(cons(X,X1)) -> 2nd(cons1(X,activate(X1)))
   from(X) -> cons(X,n__from(s(X)))
   from(X) -> n__from(X)
   activate(n__from(X)) -> from(X)
   activate(X) -> X
  Usable Rule Processor:
   DPs:
    2nd#(cons(X,X1)) -> activate#(X1)
    2nd#(cons(X,X1)) -> 2nd#(cons1(X,activate(X1)))
    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:
     [from#](x0) = x0,
     
     [activate#](x0) = x0,
     
     [2nd#](x0) = x0,
     
     [n__from](x0) = 1x0,
     
     [s](x0) = 2x0,
     
     [from](x0) = x0 + -12,
     
     [activate](x0) = 1x0 + -3,
     
     [cons1](x0, x1) = x0,
     
     [cons](x0, x1) = 1x0 + 4x1 + 0
    orientation:
     2nd#(cons(X,X1)) = 1X + 4X1 + 0 >= X1 = activate#(X1)
     
     2nd#(cons(X,X1)) = 1X + 4X1 + 0 >= X = 2nd#(cons1(X,activate(X1)))
     
     activate#(n__from(X)) = 1X >= X = from#(X)
     
     activate(n__from(X)) = 2X + -3 >= X + -12 = from(X)
     
     activate(X) = 1X + -3 >= X = X
     
     from(X) = X + -12 >= 7X + 0 = cons(X,n__from(s(X)))
     
     from(X) = X + -12 >= 1X = 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