(* for good measure, we will define exponentiation for nat and prove that lognat is correct *)
Fixpoint expnat(a e : nat) : nat :=
  match e with
    | 0 => 1
    | (S e') => a * (expnat a e')
  end.

Lemma lognat_correct1 : forall(b a : nat)(pf : a <> 0), (lognat pf) = b -> (expnat 2 b) <= a.