proposition: Lemma shiftl_eq_0_iff : forall a n, a << n == 0 <-> a == 0.
proof: Proof. intros a n. rewrite shiftl_mul_pow2. rewrite eq_mul_0. split. - intros [H | H]; trivial. contradict H; order_nz. - intros H. now left. Qed.
imports: Bool NAxioms NSub NPow NDiv NParity NLog