proposition: Lemma setIdE A pB : [set x in A | pB x] = A :&: [set x | pB x].
proof: Proof. by apply/setP=> x; rewrite !inE. Qed.
imports: HB(structures) mathcomp(ssreflect ssrbool ssrfun eqtype ssrnat div seq) mathcomp(choice fintype finfun bigop)