proposition: Lemma fmorph_eq0 x : (f x == 0) = (x == 0).
proof: Proof. have [-> | nz_x] := eqVneq x; first by rewrite rmorph0 eqxx. apply/eqP; move/(congr1 ( *%R (f x^-1)))/eqP. by rewrite -rmorphM mulVf // mulr0 rmorph1 ?oner_eq0. Qed.
imports: HB(structures) mathcomp(ssreflect ssrfun ssrbool eqtype ssrnat div seq) mathcomp(choice fintype finfun bigop prime binomial)