proposition: Lemma big_int_cond m n (P : pred int) F : \big[op/idx]_(m <= i < n :> int | P i) F i = \big[op/idx]_(m <= i < n :> int | (m <= i < n) && P i) F i.
proof: Proof. by rewrite big_seq_cond; apply: eq_bigl => i; rewrite mem_index_iotaz. Qed.
imports: mathcomp(all_ssreflect all_algebra) extra_mathcomp