z3量词消除了SMTLIB语法

Question

我有以下使用z3py消除量词的示例。但是我想用SMTLIB语法（python代码下面的代码）重写它。不知何故，我得到的输出与我从python中得到的公式相同。我想知道是否有人能指出我的问题。

from z3 import * a, five = Ints('a five') cmp = Bool('cmp')
j = Goal() j.add(Exists([five, cmp], And(five == a, cmp == (five < 1000), False == cmp)))
t = Tactic('qe') print(t(j)) # output [[1000 <= a]]

(declare-fun five () Int) (declare-fun a () Int) (declare-fun cmp () Bool) (assert (exists ((five Int) (cmp Bool)) (and (= five a) (= cmp (< five 1000)) (= cmp false) ))) (apply (then qe smt))

输出 （进球 （目标   ：精度精确：深度1） ）

Answer 1

我太快问了这个问题。经过更多搜索（Quantifier Elimination - More questions）后，我在下面找到了解决方案。

(declare-fun five () Int) (declare-fun a () Int) (declare-fun cmp () Bool) (assert (exists ((five Int) (cmp Bool)) (and (= five a) (= cmp (< five 1000)) (= cmp false) ))) (apply (using-params qe :qe-nonlinear true))