proposition: Lemma BRabs_cQ x: ratp x -> BRabs (BR_of_Q x) = BR_of_Q (BQabs x).
proof: Proof. move => h; case /BQ_i1P: (h) => ha. + by rewrite ha BQabs_0 BRabs_0. + by rewrite (BQabs_pos (BQps_sBQp ha)) BRabs_poss //; apply:RpsS_of_Q. + by rewrite (BQabs_negs ha) (BRabs_negs (RmsS_of_Q ha)) BRopp_Q. Qed.
imports: mathcomp(ssreflect ssrfun ssrbool eqtype ssrnat)