[英]Pure Prolog Peano Number Apartness
Lets assume there is pure_2 Prolog with dif/2 and pure_1 Prolog without dif/2.让我们假设 pure_2 Prolog 有 dif/2 和 pure_1 Prolog 没有 dif/2。 Can we realize Peano apartness for values, ie Peano numbers, without using dif/2?
我们可以在不使用 dif/2 的情况下实现值的 Peano 分离性,即 Peano 数吗? Thus lets assume we have Peano apartness like this in pure_2 Prolog:
因此,假设我们在 pure_2 Prolog 中有这样的 Peano 公寓:
/* pure_2 Prolog */
neq(X, Y) :- dif(X, Y).
Can we replace neq(X,Y) by a more pure definition, namely from pure_1 Prolog that doesn't use dif/2?我们能否用更纯粹的定义替换 neq(X,Y),即来自不使用 dif/2 的 pure_1 Prolog? So that we have a terminating neq/2 predicate that can decide inequality for Peano numbers?
这样我们就有了一个终止 neq/2 谓词来决定 Peano 数的不等式? So what would be its definition?
那么它的定义是什么?
/* pure_1 Prolog */
neq(X, Y) :- ??
Using less
from this comment :从此评论中使用
less
:
less(0, s(_)).
less(s(X), s(Y)) :- less(X, Y).
neq(X, Y) :- less(X, Y); less(Y, X).
I had something else in mind, which is derived from two of the Peano Axioms , which is also part of Robinson Arithmetic.我想到了其他一些东西,它来自两个Peano Axioms ,这也是 Robinson Arithmetic 的一部分。 The first axiom is already a Horn clause talking about apartness:
第一个公理已经是一个关于分离性的霍恩子句:
∀x(0 ≠ S(x))
∀x∀y(S(x) = S(y) ⇒ x = y)
Applying contraposition to the second axiom gives.将对立应用于第二个公理给出。
The axiom is now a Horn clause talking about apartness:该公理现在是一个关于分离性的霍恩子句:
∀x∀y(x ≠ y ⇒ S(x) ≠ S(y))
Now we have everything to write some Prolog code.现在我们已经准备好编写一些 Prolog 代码了。
Adding some symmetry we get:添加一些对称性,我们得到:
neq(0, s(_)).
neq(s(_), 0).
neq(s(X), s(Y)) :- neq(X, Y).
Here are some example queries.以下是一些示例查询。 Whether the predicate leaves a choice
谓词是否留下选择
point depends on the Prolog system.点取决于 Prolog 系统。 I get:
我得到:
SWI-Prolog 8.3.15 (some choice point): SWI-Prolog 8.3.15(一些选择点):
?- neq(s(s(0)), s(s(0))).
false.
?- neq(s(s(0)), s(0)).
true ;
false.
Jekejeke Prolog 1.4.6 (no choice point): Jekejeke Prolog 1.4.6(无选择点):
?- neq(s(s(0)), s(s(0))).
No
?- neq(s(s(0)), s(0)).
Yes
Just removing the unwanted choicepoint (in swi-prolog) from user502187's answer:只需从 user502187 的回答中删除不需要的选择点(在 swi-prolog 中):
neq(0, s(_)).
neq(s(N), M) :-
% Switch args, to use first-arg indexing
neq_(M, s(N)).
neq_(0, s(_)).
neq_(s(N), s(M)) :-
% Switch args back, to fix choicepoint
neq(M, N).
Results in swi-prolog: swi-prolog 中的结果:
?- neq(s(s(0)), s(0)).
true.
?- neq(s(0), s(s(0))).
true.
?- neq(N, M).
N = 0,
M = s(_) ;
N = s(_),
M = 0 ;
N = s(s(_)),
M = s(0) ;
N = s(0),
M = s(s(_)) ;
N = s(s(0)),
M = s(s(s(_))) ;
N = s(s(s(_))),
M = s(s(0)) ;
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