(* 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_ge_monotonic_h : forall (c a b : nat)(pfa : a <> 0)(pfb : b <> 0), a >= b -> c = (a - b) -> (lognat pfa) >= (lognat pfb).