p(x)⇒∀x.p(x)是偶然的吗? [英] p(x)⇒∀x.p(x) is contingent?

问题描述

我遇到了一个问题，询问以下句子是否有效/偶然/无法满足:

I've encountered a question asking whether the flowing sentence is valid/contingent/unsatisfiable:

p(x)⇒∀x.p(x) 

我认为答案是正确的句子.在此处的教科书第6.10节中 http://logic.stanford.edu/intrologic/secondary/notes/chapter_06.html说明

I think the answer is the sentence is valid. under section 6.10 of the textbook here http://logic.stanford.edu/intrologic/secondary/notes/chapter_06.htmlsays

带有自由变量的句子等同于所有自由变量都被普遍量化的句子.

a sentence with free variables is equivalent to the sentence in which all of the free variables are universally quantified.

因此，我认为第一个关系语句p(x)等于∀x.p(x)，因此该语句有效，即.总是如此.

Therefore I think the first relational sentence p(x) is equal to ∀x.p(x) and therefore the sentence is valid, ie. it is always true.

但是，正确的答案是句子是偶然的，即.在某些真相分配下，它为真，而在其他一些真相分配下，它为假.

However,the correct answer is that the sentence is contingent viz. under some truth assignment it is true and other some other truth assignment it is false.

那为什么句子是和句的呢?答案是错误的吗?

So why is the sentence contingent?Is the answer wrong?

推荐答案

您有一条声明:

p(x)⇒∀x.p(x)

如果您普遍关闭free变量，则会得到:

If you universally close the free variable, you get:

∀x.(p(x)⇒∀x.p(x))

换句话说:

∀x.(p(x)⇒∀y.p(y))

不是重言式，而是偶然的.用非技术术语来说，它是:

which is not tautology, but is contingent. In non-technical terms, this reads:

对于任何x，

，如果p(x)为true，则对于所有y

for any x, if p(x) is true, then p(y) is true for all y

或将其转换为等效形式:

or, to transform it into an equivalent form:

(∃x.p(x))⇒(∀y.p(y))

它显示为:

如果p(x)对于某些x为true，则p(y)对于所有y

if p(x) is true for some x, then p(y) is true for all y

换句话说，

p(x)始终为true或始终为false

p(x) is either always true or always false

这篇关于p(x)⇒∀x.p(x)是偶然的吗?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！