proposition: Lemma Cnat_mul_eq1 x y : x \in Cnat -> y \in Cnat -> (x * y == 1) = (x == 1) && (y == 1).
proof: Proof. exact: natr_mul_eq1. Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrfun ssrnat eqtype seq choice) mathcomp(div fintype path bigop finset prime order ssralg) mathcomp(poly polydiv mxpoly generic_quotient countalg) mathcomp(ssrnum closed_field ssrint archimedean rat intdiv) mathcomp(algebraics_fundamentals)