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Why is predicate “all” as in all(SET) true if the SET is empty?

In don't quite understand this quantification over the empty set:

$\forall y \in \emptyset: Q(y)$

The book says that this is always TRUE regardless of the value of the predicate $Q(y)$, and it explain that this is because this quantification adds no predicate at all, and therefore can be considered the weakest predicate possible, which is TRUE.

I know that TRUE is the weakest predicate because $ $P$ \Rightarrow$ TRUE is TRUE for every $P$. I don't see what is the relationship between this weakest predicate and the quantification.

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marked as duplicate by MJD, Henry T. Horton, Marvis, Douglas S. Stones, Ittay Weiss Jan 19 '13 at 4:53

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5 Answers 5

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When you see a quantification like ‘$\forall \phi x : \psi x$’, this is shorthand for ‘$\forall x : \phi x \to \psi x$’. Since ‘$x \in \emptyset$’ is false for all ‘$x$’, the antecedent of ‘$x \in \emptyset \to \phi x$’ will always be false, thus the entire conditional statement is always true.

Edit: In the case of existential quantification, the statement ‘$\exists \phi x : \psi x$’ is shorthand for ‘$\exists x : \phi x \land \psi x$’, so in this case, the conjunction ‘$x \in \emptyset \land \phi x$’ will always be false.

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Where can I find a little more explanation about those shorthands? because if becomes very clear when I know this shorthands. But I can't figure out how can I get those shorthands notations – Victor Jose Arana Rodriguez Jan 19 '13 at 3:29
The concept is discussed here: – c.w.chambers Jan 19 '13 at 4:42

Well if $\forall y \in \emptyset : Q(y)$ were false, we would be able to find some $y \in \emptyset$ such that $Q(y)$ were false. However, there are no $y \in \emptyset$. So $\forall y \in \emptyset : Q(y)$ should be true.

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A peculiar explanation.

Whatever $Q(y)$ may be, it is true that for all $y\in \emptyset$, the sentence $Q(y)$ is true. For the empty set is $\dots$ empty. Every unicorn likes wine.

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Because there are no members to the set, anything you say about a member can be considered trivially true. It's not really to say that there are actually members for which Q, but rather for all y, Q(y)--i.e. there are no circumstances where y and not Q(y).

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I think I got the idea, but it confuses me when I compare this reasoning with the existencial quantifier over the empty set $ \exists x \in \emptyset : P(x) $

I think this is TRUE because I'm quantifying over the empty set too.

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You think it's true that there is an element of $\emptyset$ for which $P$ holds? Which element would that be? – MJD Jan 19 '13 at 3:17
You're right but I'm still very confused sorry =(. – Victor Jose Arana Rodriguez Jan 19 '13 at 3:30

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