我想用 Prolog 证明一些定理，但是，它总是 returns "Out of global stack"

Question

我正在做证明代数群论的AI作业。

定理可以表示为：

A1. i(e,X) = X                   (identity)
A2. i(X, e) = X                  (identity)
A3. i(comp(X),X) = e             (complement)
A4. i(X, comp(X)) = e            (complement)
A5. i(X, i(Y,Z)) = i(i(X,Y),Z)   (associativity)

THEOREM: If G is a group such that for every X,
A6. i(X,X) = e,
then G is commutative, i.e., for every X; Y ,
i(X,Y) = i(Y,X):

and the commutative part can be represented as 
A7. i(a, b, c)                          clause derived from negated conclusion
A8. -i(b, a, c)                         clause derived from negated conclusion

并且我将它们转换成Prolog格式如下：

% A7
i(a, b, c).
% A1
i(e, X, X) .
%A2
i(X, e, X).
% A3
i(comp(X), X, e).
% A4
i(X, comp(X), e).
% A51
i(U, Z, W) :- i(X, Y, U), i(Y, Z, V), i(X, V, W).
% A52
i(X, V, W) :- i(X, Y, U), i(Y, Z, V), i(U, Z, W).
% A6
i(X, X, e).

然后我想证明这个定理，所以我在 Prolog 控制台输入 "i(b,a,c)"，然后我得到以下错误信息：

?- i(b,a,c).
ERROR: Out of global-stack.
ERROR: No room for exception term.  Aborting.
ERROR: Out of global-stack.
ERROR: No room for exception term.  Aborting.
ERROR: Out of global-stack.
ERROR: No room for exception term.  Aborting.
% Execution Aborted

请帮帮我，非常感谢！

Answer 1

A51 和 A52 子句是左递归，这会导致堆栈外错误。在 Prolog 中处理左递归的规范解决方案是使用支持 tabling 的系统（例如 XSB、YAP、SWI-Prolog、B-Prolog 或 Ciao）。但是您的代码中还有另一个问题。 A3 和 A4 子句会导致创建循环项。例如，仅加载子句 A3:

?- i(X, X, Y), cyclic_term(X).
X = comp(X),
Y = e.

如果注释掉 A3 和 A4 子句并在源文件顶部添加指令：

:- table(i/3).

您将获得：

?- i(b,a,c).
true.

我想用 Prolog 证明一些定理，但是，它总是 returns "Out of global stack"

I want to prove some theorem using Prolog, however, it always returns "Out of global stack"

prolog

theorem-proving

swi-prolog