proposition: Lemma domD D1 D2 x : [predI (dom D1) & (dom D2)] =1 pred0 -> (x \in dom (D1 + D2)) = (x \in dom D1) || (x \in dom D2).
proof: Proof. by move/domD_perm_eq/perm_mem/(_ x); rewrite mem_cat. Qed.
imports: HB(structures) mathcomp(ssreflect ssrfun ssrbool eqtype ssrnat seq choice) mathcomp(fintype bigop order generic_quotient) mathcomp(ssralg ssrnum ssrint)