(* 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_prod_sum_gen : forall (a b: nat)(pf1 : (a <> 0))(pf2 : (b <> 0))(pf3 : a * b <> 0), lognat pf1 + lognat pf2 <= lognat pf3.