Z3给出的序列错误模型

我将以下代码放入

结果是

sat
(model 
  (define-fun s () (Seq Int)
    (seq.++ (seq.unit 14) (seq.++ (seq.unit 16) (seq.unit 18))))
  (define-fun t2 () (Seq Int)
    (seq.++ (seq.unit 16) (seq.unit 18)))
  (define-fun n () Int
    3)
  (define-fun t1 () (Seq Int)
    (seq.++ (seq.unit 16) (seq.unit 18)))
)

这种模式是错误的<代码>（seq.extract s0（-n1））应该是

[14；16]

而不是

[16；18]

---

我是否误解了Z3的任何内容，或者这是Z3的错误？

自Z3的rise4fun版本发布以来，Sequence logic在Z3中看到了很多变化，修复了很多错误。通过最近的构建，我得到：

sat
(model
  (define-fun s () (Seq Int)
    (seq.++ (seq.unit 4) (seq.unit 4) (seq.unit 4)))
  (define-fun t2 () (Seq Int)
    (seq.++ (seq.unit 4) (seq.unit 4)))
  (define-fun t1 () (Seq Int)
    (seq.++ (seq.unit 4) (seq.unit 4)))
  (define-fun n () Int
    3)
)

在我看来这是正确的

长话短说：rise4fun的z3版本对于这个问题来说太旧了，而且有bug。看看是否可以从下载更新的版本（甚至是夜间版本）

sat
(model
  (define-fun s () (Seq Int)
    (seq.++ (seq.unit 4) (seq.unit 4) (seq.unit 4)))
  (define-fun t2 () (Seq Int)
    (seq.++ (seq.unit 4) (seq.unit 4)))
  (define-fun t1 () (Seq Int)
    (seq.++ (seq.unit 4) (seq.unit 4)))
  (define-fun n () Int
    3)
)