简单的布尔前pression测试 [英] Simple boolean expression testing

问题描述

| ?- [user].
compiling user for byte code...
formula_0(P, Q):- (P; Q), \+ P.

user compiled, 2 lines read - 768 bytes written, 37208 ms

yes
| ?- formula_0(P, Q).
uncaught exception: error(instantiation_error,formula_0/2)

所有我基本上想要做的就是要问的是一组前pressions的{P或Q，〜p}​​满足的？的

不过，该错误信息不会在这里帮助...

But the error message is not helping here...

PS。答案应该是是，这个公式是满足当p =假，Q =真。

PS. The answer should be "yes", this formula is satisfied when P = false and Q = true.

推荐答案

究其原因，你得到一个实例化的错误就是太知之甚少 P 和<$ C $ ç>问：说在它们被用作目标的时候他们什么。当你问比如你是正确的：

The reason you get an instantiation error is that too little is known about P and Q to say anything about them at the time they are used as goals. You are right that when you ask for example:

?- G.

然后 G =真是使查询成功的解决方案。但这样是例如 G =（A = A），因为 A = A 也是如此。因为它是不可能枚举成功的所有目标，你会得到一个实例的错误。但是，例如通知，当你明确给出有约束力的，你得到你所期望的结果：

then G = true is a solution that makes the query succeed. But so is for example G = (a=a) because a=a is also true. Since it is not possible to enumerate all goals G that succeed, you get an instantiation error. But notice for example that when you explicitly give a binding, you get the result you expect:

?- G = true, G.
G = true.

所以，你应该提供一套价值你感兴趣的是：

So, you should provide a set of values you are interested in:

?- maplist(boolean, [P,Q]), formula(P, Q).

和定义，例如：

boolean(true).
boolean(false).

要获得具体的解决方案。或者使用的限制的，这让你在使用之前约束变量的域。

to obtain concrete solutions. Or use constraints, which let you constrain the domain of variables before using them.

修改：既然有这个一些讨论，我去到更多的细节。所出现的主要问题是：为什么查询

EDIT: Since there was some discussion about this, I go into more detail. The main question that arose is: Why does the query

?- Q.

与鞋底的解决方案不会成功。

not succeed with the sole solution

Q = true.

因为很明显，真正成功，因此是一个有效的解决方案？答：还有其他可能的答案为好，因为真正不是成功的唯一目标。例如，结合（真的，真）成功也是如此。假设现在 Q =真是上述查询的唯一解决方案，那么这将是的情况是：

since obviously, true succeeds and is thus a valid solution? Answer: There are other possible answers as well, since true is not the only goal that succeeds. For example, the conjunction (true,true) succeeds as well. Suppose now that Q = true were the only solution to the above query, then it would be the case that:

?- Q, Q = (true, true).

的失败的（因为 DIF（TRUE（真，真））本身是真实的），只是简单的交流，比如

fails (since dif(true, (true,true)) is itself true), but simply exchanging the goals like

?- Q = (true,true), Q.

的成功的，否则 - ？真的，真的将也有失败，并不能如此。这将违反配合可交换已经是最基本的predicate（统一）。

succeeds because otherwise ?- true, true would also have to fail and that cannot be the case. This would violate commutativity of conjunction already for the most basic predicate (unification).

请注意，虽然真正和（真的，真）的两个的执行时成功作为目标，他们显然的不同的术语的，不能在一般的相互取代。同样，术语追加（[]，[]，[]）从长期不同真正，虽然两者当作为目标执行成功。因此，你得到一个实例化错误，因为太知之甚少问：在查询像 - ？：Q 给有用的结果。

Note that while true and (true,true) both succeed when executed as goals, they are obviously different terms and cannot replace each other in general. Similarly, the term append([], [], []) is different from the term true, although both succeed when executed as goals. You therefore get an instantiation error, since too little is known about Q in a query like ?- Q. to give useful results.

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