proposition: Lemma Forall_and : forall {A: Type} [P Q: A -> Prop] [l], Forall P l -> Forall Q l -> Forall (fun x => P x /\ Q x) l.
proof: Proof. intros. rewrite Forall_forall in *. intuition. Qed.
imports: Cdcl.PatriciaR Cdcl.KeyInt Cdcl.ReifClasses Cdcl.Lib Bool Setoid ZifyBool ZArith Uint63 Lia List Cdcl.Syntax Cdcl.Clause