带量词公式中连词消除的Z3策略

是否有一种Z3策略可以将目标（例如，一个带有量词和未解释函数的断言）重写为一组无连接的公式？例如，以下公式：

(declare-fun P (Int) Bool)
(declare-fun Q (Int Int) Bool)
(declare-fun R (Int Int) Bool)
(assert (forall ((x Int)) (! 
  (or 
    (P x) 
    (forall ((x Int) (y Int)) (! 
      (and 
        (Q x y) 
        (R x y)) 
     :pattern ((Q x y)) )) ) 
 :pattern ((P x)))))

应该改写为

(declare-fun P (Int) Bool)
(declare-fun Q (Int Int) Bool)
(declare-fun R (Int Int) Bool)
(assert (forall ((x Int)) (! 
  (or 
    (P x) 
    (forall ((x Int) (y Int)) (! 
      (Q x y) 
     :pattern ((Q x y))))) 
 :pattern ((P x)))))
(assert (forall ((x Int)) (! 
  (or 
    (P x) 
    (forall ((x Int) (y Int)) (! 
      (R x y) 
     :pattern ((Q x y))))) 
 :pattern ((P x)))))

---

您可以通过

（帮助战术）

获得战术列表。搜索“连接词”会产生一些点击，包括使用选项

elim_和
的策略
simplify
，这会从示例中删除所有连接词

重写量词是另一回事。我尝试了一些听起来可能有关联的策略，但没有一个有任何有用的效果

; (help-tactic)

(declare-fun P (Int) Bool)
(declare-fun Q (Int Int) Bool)
(declare-fun R (Int Int) Bool)
(assert (forall ((x Int)) (! 
  (or 
    (P x) 
    (forall ((x Int) (y Int)) (! 
      (and 
        (Q x y) 
        (R x y)) 
     :pattern ((Q x y)) )) ) 
 :pattern ((P x)))))

; (apply (try-for qe-light 5000)) ; no effect
; (apply (try-for qe_rec 5000)) ; appears to fail
; (apply (try-for qe2 5000)) ; appears to fail
 
(apply
  (then
    distribute-forall ; no effect
    (using-params simplify :elim_and true)
))
;;; RESULT:
; (forall ((x Int))
  ; (! (let ((a!1 (forall ((x!1 Int) (y Int))
    ; (! (not (or (not (Q x!1 y)) (not (R x!1 y))))
    ; :pattern ((Q x!1 y))))))
 ; (or (P x) a!1))
 ; :pattern ((P x))))

谢谢你的提示！我也尝试了一些策略和选项，但到目前为止，我无法让Z3拆分这个量词。