(* 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_succ_h : forall a n (pf1 : n <> 0)(pf2: (S n) <> 0), a = lognat pf1 -> lognat pf2 = a \/ lognat pf2 = S (a).