Z3不能证明群论中的权利取消性质吗？

我试图证明群论中的一些一般性质

例如，左取消属性：（xy=xz）=>（y=z）使用以下代码证明

(declare-sort S)
(declare-fun e () S)
(declare-fun mult (S S) S)
(declare-fun inv (S) S)
(assert (forall ((x S) (y S) (z S)) (= (mult (mult x y) z) (mult x (mult y z)))))
(assert (forall ((x S)) (= (mult e x)  x)))
(assert (forall ((x S)) (= (mult (inv x) x) e)))
(assert (forall ((x S)) (= (mult x e)  x)))
(assert (forall ((x S)) (= (mult x (inv x)) e)))
(check-sat)
(assert (not (forall ((x S) (y S) (z S)) (=> (= (mult x y) (mult x z)) (= y z)))))
(check-sat)

(declare-sort S)
(declare-fun e () S)
(declare-fun mult (S S) S)
(declare-fun inv (S) S)
(assert (forall ((x S) (y S) (z S)) (= (mult (mult x y) z) (mult x (mult y z)))))
(assert (forall ((x S)) (= (mult e x)  x)))
(assert (forall ((x S)) (= (mult (inv x) x) e)))
(assert (forall ((x S)) (= (mult x e)  x)))
(assert (forall ((x S)) (= (mult x (inv x)) e)))

(check-sat)
(assert (not (forall ((x S) (y S) (z S)) (=> (= (mult y x) (mult z x)) (= y z)))))
(check-sat)

相应的输出为：

sat
unsat

sat
unsat

现在，当我试图证明正确的取消属性时：（yx=zx）=>（y=z）使用以下代码

(declare-sort S)
(declare-fun e () S)
(declare-fun mult (S S) S)
(declare-fun inv (S) S)
(assert (forall ((x S) (y S) (z S)) (= (mult (mult x y) z) (mult x (mult y z)))))
(assert (forall ((x S)) (= (mult e x)  x)))
(assert (forall ((x S)) (= (mult (inv x) x) e)))
(assert (forall ((x S)) (= (mult x e)  x)))
(assert (forall ((x S)) (= (mult x (inv x)) e)))
(check-sat)
(assert (not (forall ((x S) (y S) (z S)) (=> (= (mult x y) (mult x z)) (= y z)))))
(check-sat)

(declare-sort S)
(declare-fun e () S)
(declare-fun mult (S S) S)
(declare-fun inv (S) S)
(assert (forall ((x S) (y S) (z S)) (= (mult (mult x y) z) (mult x (mult y z)))))
(assert (forall ((x S)) (= (mult e x)  x)))
(assert (forall ((x S)) (= (mult (inv x) x) e)))
(assert (forall ((x S)) (= (mult x e)  x)))
(assert (forall ((x S)) (= (mult x (inv x)) e)))

(check-sat)
(assert (not (forall ((x S) (y S) (z S)) (=> (= (mult y x) (mult z x)) (= y z)))))
(check-sat)

我正在获得

timeout

---

请让我知道如何证明正确的取消属性，或者不可能使用Z3？

据我所知，这种类型的定理最好由超级位置推理引擎处理。 Z3的旧版本包含一个叠加引擎，但它已从新版本中删除

---

正如我们所看到的，在解决普遍代数中的问题时，几乎没有什么用途。有几个定理证明者专门研究超位置推断，如Vampire、E、SPASS，您可以使用www.tptptp.org提供的工具来尝试这些引擎。

使用Z3的正确取消属性的可能证明如下

(declare-sort S)
(declare-fun e () S)
(declare-fun mult (S S) S)
(declare-fun inv (S) S)
(assert (forall ((x S) (y S) (z S)) (= (mult (mult x y) z) (mult x (mult y z)))))
(assert (forall ((x S) (y S) (z S)) (= (mult x (mult y z)) (mult (mult x y) z))))
(assert (forall ((x S)) (= (mult e x)  x)))
(assert (forall ((x S)) (= (mult (inv x) x) e)))
(assert (forall ((x S)) (= (mult x e)  x)))
(assert (forall ((x S)) (= (mult x (inv x)) e)))
(check-sat)

(push)
(assert (not (forall ((x S) (y S) (z S)) (=> (= (mult (mult y x) (inv x)) 
                                                (mult (mult z x) (inv x) )) (= y z)))))
(check-sat)
(pop)

并且相应的输出是