proposition: Lemma in_fset_cons x (xs : seq K) a : (a \in [fset[key] x in x :: xs]) = ((a == x) || (a \in [fset[key] x in xs])).
proof: Proof. by rewrite !mem_imfset. Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrnat eqtype ssrfun seq) mathcomp(choice path finset finfun fintype bigop)