(* 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 nat_comp : forall(a : nat), exists x : nat, (a = (2 * x) \/ a = (S (2 * x))).