无法推导出可以调用哪个谓词

Question

我正在备考，但在其中一个准备题上卡住了：

问题：

The following Prolog-program is a meta-program. Explain why this program is a meta-program and give the output to the three questions to the program:

?- prove(moving_method(bird,M),N). 
?- prove(moving_method(ross,M),N).
?- prove(moving_method(kim,M),N).

我正在尝试运行代码（在 swish.swi-prolog.org 上），但它只给我这个错误消息：

Sandbox restriction! 
Could not derive which predicate may be called from
      call(C)
      prove(A,B)
      prove(moving_method(bird,A),B)

我们给出的代码：

:- dynamic moving_method/2, is_a/2.

is_a(bird,animal).
is_a(ross,albatross).
is_a(kim,kiwi).
is_a(albatross,bird).

moving_method(bird,fly).
moving_method(kiwi,walk).

prove(Fact,l):-
    Fact,!.
prove(Fact,X):-
    Fact=..[Rel,A1,A2],
    is_a(A1,SA),
    NewFact=..[Rel,SA,A2],
    prove(NewFact,X1),
    X is X1 + 1.

错误消息可能相当简单，但我该如何解决？为什么这是一个元程序？

谢谢！

Answer 1

why is this a meta-program?

参见：SWI-Prolog Meta-Call Predicates

Meta-call predicates are used to call terms constructed at run time.

在这种情况下，传递要调用的谓词，Fact，然后运行它作为目标。

无法推导出可以调用哪个谓词

Could not derive which predicate may be called

metaprogramming

prolog