z3用量词解决约束集是否昂贵？

我正在使用z3（v4.3.1）解决一个约束集，该约束集包含带有量词的未解释函数。问题是z3的内存使用率非常高。以下是我的限制：

(solver
(v0 0)
(forall ((i Int))
(let ((a!1 (=> (v0 i)
               (and (not (v1 i))
                    (not (v2 i))
                    (not (v3 i))
                    (not (v4 i))
                    (not (v5 i))))))
  (=> (>= i 0) a!1)))
(forall ((i Int)) (=> (>= i 0) (=> (v0 i) (v1 (+ i 1)))))
(forall ((i Int))
(let ((a!1 (=> (v1 i)
               (and (not (v0 i))
                    (not (v2 i))
                    (not (v3 i))
                    (not (v4 i))
                    (not (v5 i))))))
  (=> (>= i 0) a!1)))
forall ((i Int))
(let ((a!1 (=> (v1 i) (or (v2 (+ i 1)) (v5 (+ i 1))))))
  (=> (>= i 0) a!1)))
(forall ((i Int))
(let ((a!1 (=> (v2 i)
               (and (not (v0 i))
                    (not (v1 i))
                    (not (v3 i))
                    (not (v4 i))
                    (not (v5 i))))))
  (=> (>= i 0) a!1)))
(forall ((i Int)) (=> (>= i 0) (=> (v2 i) (v3 (+ i 1)))))
(forall ((i Int))
(let ((a!1 (=> (v3 i)
               (and (not (v0 i))
                    (not (v1 i))
                    (not (v2 i))
                    (not (v4 i))
                    (not (v5 i))))))
  (=> (>= i 0) a!1)))
(forall ((i Int)) (=> (>= i 0) (=> (v3 i) (v4 (+ i 1)))))
(forall ((i Int))
(let ((a!1 (=> (v4 i)
               (and (not (v0 i))
                    (not (v1 i))
                    (not (v2 i))
                    (not (v3 i))
                    (not (v5 i))))))
  (=> (>= i 0) a!1)))
(forall ((i Int)) (=> (>= i 0) (=> (v4 i) (v1 (+ i 1)))))
(forall ((i Int))
(let ((a!1 (=> (v5 i)
               (and (not (v0 i))
                    (not (v1 i))
                    (not (v2 i))
                    (not (v3 i))
                    (not (v4 i))))))
  (=> (>= i 0) a!1)))
(forall ((i Int)) (=> (>= i 0) (=> (v5 i) (v5 (+ i 1)))))
(exists ((i Int)) (=> (>= i 0) (v5 i)))
(forall ((i Int))
(let ((a!1 (not (v5 (- (+ i 3) 1)))))
(let ((a!2 (=> (and (v0 i) (v1 (+ i 1))) a!1)))
  (=> (>= i 0) a!2))))
(forall ((i Int))
(let ((a!1 (not (v5 (- (+ i 4) 1)))))
(let ((a!2 (=> (and (v3 i) (v4 (+ i 1)) (v1 (+ i 2))) a!1)))

当我将约束集写入smt2文件并使用z3解决它时，我很快得到了“未知”的结果。下面是smt2文件的内容

(declare-fun v0 (Int) Bool)
(declare-fun v1 (Int) Bool)
(declare-fun v2 (Int) Bool)
(declare-fun v3 (Int) Bool)
(declare-fun v4 (Int) Bool)
(declare-fun v5 (Int) Bool)
(assert (v0 0))
(assert  (forall ((i Int)) (=> (v0 i)     (and (not (v1 i)) (not (v2 i)) (not (v3 i))    (not (v4 i)) (not (v5 i))))))
(assert  (forall ((i Int)) (=> (v1 i)     (and (not (v0 i)) (not (v2 i)) (not (v3 i))   (not (v4 i)) (not (v5 i))))))
assert  (forall ((i Int)) (=> (v2 i)     (and (not (v1 i)) (not (v0 i)) (not (v3 i)) (not (v4 i)) (not (v5 i))))))
(assert  (forall ((i Int)) (=> (v3 i)     (and (not (v1 i)) (not (v2 i)) (not (v0 i)) (not (v4 i)) (not (v5 i))))))
(assert  (forall ((i Int)) (=> (v4 i)     (and (not (v1 i)) (not (v2 i)) (not (v3 i)) (not (v0 i)) (not (v5 i))))))
(assert  (forall ((i Int)) (=> (v5 i)     (and (not (v1 i)) (not (v2 i)) (not (v3 i)) (not (v4 i)) (not (v0 i))))))
(assert  (forall ((i Int)) (=> (>= i 0) (=> (v0 i) (v1 (+ i 1))))))
(assert  (forall ((i Int)) (=> (>= i 0) (=> (v1 i) (or (v2 (+ i 1)) (v5 (+ i 1)))))))
(assert  (forall ((i Int)) (=> (>= i 0) (=> (v2 i) (v3 (+ i 1))))))
(assert  (forall ((i Int)) (=> (>= i 0) (=> (v3 i) (v4 (+ i 1))))))
(assert  (forall ((i Int)) (=> (>= i 0) (=> (v4 i) (v1 (+ i 1))))))
(assert  (forall ((i Int)) (=> (>= i 0) (=> (v5 i) (v5 (+ i 1))))))
(assert  (exists ((i Int)) (=> (>= i 0) (v5 i))))
(assert (forall ((i Int)) (=> (>= i 0) (=> (and (v0 i) (v1 (+ i 1))) (not (v5 (+ i 2)))))))
(assert (forall ((i Int)) (=> (>= i 0) (=> (and (v3 i) (v4 (+ i 1)) (v1 (+ i 2))) (not (v5 (+ i 3)))))))
(check-sat)

---

我只是想知道是我写错了什么，还是z3很难解决这个问题。

只有最后一个断言导致了这个问题，如果您从代码中删除它，那么您将获得“sat”

但是，必须添加最后一个断言，而且该约束集实际上是未设置的。如果仅删除第一个断言，并保留最后一个断言，则再次获得“sat”。