(* 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.

Theorem lognat_ge_monotonic_b : forall(a : nat)(pf1 : (a <> 0))(pf2: (S a) <> 0), (lognat pf2) >= (lognat pf1).