proposition: Lemma sum_cond1 (I : finType) (r : seq I) (P Q : I -> bool) : \sum_(x <- r | Q x ) P x = \sum_(x <- r | Q x && P x) 1.
proof: Proof. by rewrite [RHS]big_mkcondr; apply: eq_bigr => x _; case: (P x). Qed.
imports: Coq(Setoid Morphisms) mathcomp(all_ssreflect) GraphTheory(edone)