Prolog:在 Prolog 中定义逻辑运算符作为其他运算符的占位符 [英] Prolog: define logical operator in Prolog as placeholder for other operator

问题描述

我的目标是在 prolog 中编写一个小证明助手.我的第一步是定义逻辑连接词如下:

my aim is to write a little prove assistant in prolog. My first step is to define the logical connectives as follows:

:-op(800, fx, -).

:-op(801, xfy, &).

:-op(802, xfy, v).

:-op(803, xfy, ->).

:-op(804, xfy, <->).

:-op(800, xfy, #).

最后一个操作符 # 只是作为 &、v、-> 或 < 的占位符;->.我的问题是，我不知道如何在序言中定义它.我尝试通过以下方式解决我的问题:

The last operator # just has the meaning to be the placeholder for &, v, -> or <->. My problem is, I don't know how I it is possible to define this in prolog. I tried to solve my problem in the following way:

X # Y :- X v Y; X & Y; X -> Y; X <-> Y.

但是下面的定义:

proposition(X) :- atomicproposition(X).

proposition(X # Y) :- proposition(X), proposition(Y).

proposition(- X) :- proposition(X). 

与

atomicproposition(a).

给予

?- proposition(a v -a).
false

我做错了什么?

推荐答案

您不能以这种方式定义句法同义词.当你定义类似

You cannot define syntactic synonyms this way. When you define something like

X # Y :-
    X & Y.

你定义语义:执行X#Y，执行X&是".但是您还没有定义任何方式来执行 X &Y":

you define semantics: "to execute X # Y, execute X & Y". But you haven't defined any way to "execute X & Y":

?- X # Y.
ERROR: Undefined procedure: (&)/2
ERROR: In:
ERROR:    [9] _2406&_2408
ERROR:    [8] _2432#_2434 at /home/isabelle/op.pl:13
ERROR:    [7] <user>

(即使你为它定义了一些含义，它也可能不是你想要的.)

(And even if you had defined some meaning for it, it might not be what you want.)

您正在寻找的是一种不仅可以定义T 是带有二元运算符的术语"，而且理想情况下还可以定义T 的操作数是 X 和 Y".像这样:

What you are looking for instead is a way to define a notion of not only "T is a term with a binary operator", but ideally also "T's operands are X and Y". Like this:

binary_x_y(X  v  Y, X, Y).
binary_x_y(X  &  Y, X, Y).
binary_x_y(X  -> Y, X, Y).
binary_x_y(X <-> Y, X, Y).

然后，用:

proposition(X) :-
    atomicproposition(X).
proposition(Binary) :-
    binary_x_y(Binary, X, Y),
    proposition(X),
    proposition(Y).
proposition(- X) :-
    proposition(X).

atomicproposition(a).

我们得到:

?- proposition(a v -a).
true ;
false.

?- proposition(P).
P = a ;
P =  (a v a) ;
P =  (a v a v a) ;
P =  (a v a v a v a) ;
P =  (a v a v a v a v a) ;
P =  (a v a v a v a v a v a) .  % unfair enumeration

还有其他方法可以用更少的输入来表达相同的关系，例如:

There are other ways of expressing the same relation with a bit less typing, for example:

binary_x_y(Binary, X, Y) :-
    Binary =.. [Op, X, Y],  % Binary is of the form Op(X, Y)
    member(Op, [v, &, ->, <->]).

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