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量词，嵌套，作用域-自由变量和绑定变量

[英]Quantifiers, nesting, scope - free and bound variables

原文 2016-09-02 12:27:44 5 1 logic/ predicate/ quantifiers/ first-order-logic

I couldn't find a proper explanation to several questions i have about the scope of quantifiers, free and bound variables. 我找不到关于量词，自由变量和绑定变量的范围的几个问题的正确解释。

1) ∃X(p(X,Y) & ∃Y(X,Y)) - I am wondering about the second X. Is it still in the scope of the ∃X quantifier,because it's actually nested under ∃Y? 1）∃X（p（X，Y）＆∃Y（X，Y））-我想知道第二个X。它是否仍在∃X量词的范围内，因为它实际上嵌套在∃Y下？ If it is not in the scope of ∃X then is it a free variable? 如果它不在∃X的范围内，那么它是一个自由变量吗？ I am thinking it's should be bound to ∃X. 我认为应该绑定到X。
2)∃X(p(Y,Z) & ∃X(q(Y,Z)) - is the second ∃X needed at all? Does it change anything for the free variables Y and Z? 2）∃X（p（Y，Z）＆∃X（q（Y，Z））-根本需要第二个∃X吗？它对自由变量Y和Z有什么改变吗？

I have more questions of this type but those two bother me the most currently. 我有更多这类问题，但是这两个问题目前最困扰我。 And answering them might make me get the idea in general and answer myself the next questions. 回答这些问题可能会让我大致了解这个主意，并回答下一个问题。

Also if you have a resource where stuff of this type are explained in detail I will be thankful! 另外，如果您有详细说明此类资料的资源，我将不胜感激！

Thank you in advance! 先感谢您！

1 个解决方案

Yes, the second X is bound by ∃X . 是的，第二个X由∃X约束。 (If you had ∃X p(X,Y) & ∃Y(X,Y) instead, it depends on the convention used; in some books it's understood as ∃X(p(X,Y) & ∃Y(X,Y)) , in others as (∃X p(X,Y)) & ∃Y(X,Y) ). （如果∃X p(X,Y) & ∃Y(X,Y) ，则取决于所使用的约定；在某些书中，它被理解为∃X(p(X,Y) & ∃Y(X,Y)) ，其他形式为(∃X p(X,Y)) & ∃Y(X,Y) ）。
No, the second ∃X doesn't change anything (the first one doesn't either). 不，第二个∃X不会更改任何内容（第一个∃X也不更改）。
Free extra example: in ∃X(p(Y,Z) & ∃X(q(X,Z)) , the outer ∃X doesn't bind anything (and thus removing it gives an equivalent formula) because the only X is already bound by the inner one. 免费的额外示例：在∃X(p(Y,Z) & ∃X(q(X,Z)) ，外部∃X不绑定任何内容（因此将其删除即可得到等效公式），因为唯一的X是已经被内在的束缚了。