proposition: Lemma mdeg_max m1 m2 : mdeg (m1 `|` m2)%O = maxn (mdeg m1) (mdeg m2).
proof: Proof. have [/lemc_mdeg|Hgt] := leP; first by case: ltngtP. by apply/esym/maxn_idPl; apply/contra_lt_leq: Hgt => /lt_mdeg_ltmc /ltW. Qed.
imports: HB(structures) mathcomp(ssreflect ssrfun ssrbool eqtype ssrnat seq path) mathcomp(choice fintype tuple finfun bigop finset binomial) mathcomp(order fingroup perm ssralg zmodp poly ssrint) mathcomp(matrix vector) mathcomp(bigenough) ssrcomplements freeg