proposition: Lemma char0_natf_div : char F =i pred0 -> forall m d, d %| m -> (m %/ d)%:R = m%:R / d%:R :> F.
proof: Proof. move/charf0P=> char0F m [|d] d_dv_m; first by rewrite divn0 invr0 mulr0. by rewrite natr_div // unitfE char0F. Qed.
imports: HB(structures) mathcomp(ssreflect ssrfun ssrbool eqtype ssrnat div seq) mathcomp(choice fintype finfun bigop prime binomial)