Quantification over the empty set

Question

Possible Duplicate:
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.

score 7 · Accepted Answer · edited Aug 23 '22 at 08:17

7

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 ‘$x \in \emptyset \to \psi x$’ 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 ‘$x \in \emptyset \land \psi x$’ will always be false.

edited Aug 23 '22 at 08:17

Inertial Ignorance

267

answered Jan 19 '13 at 03:09

c.w.chambers

311

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 03:29
The concept is discussed here: http://en.m.wikipedia.org/wiki/Bounded_quantifier – c.w.chambers Jan 19 '13 at 04:42

André Nicolas · Answer 2 · 2013-01-19T02:51:40.857

2

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.

edited Jan 19 '13 at 02:51

answered Jan 19 '13 at 02:43

André Nicolas

507,029

score 2 · Answer 3 · answered Jan 19 '13 at 02:50

2

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.

answered Jan 19 '13 at 02:50

zrbecker

4,048

score 0 · Answer 4 · answered Jan 19 '13 at 02:50

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).

score 0 · Answer 5 · answered Jan 19 '13 at 03:11

0

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.

answered Jan 19 '13 at 03:11

Victor Jose Arana Rodriguez

525

2

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 03:17
You're right but I'm still very confused sorry =(. – Victor Jose Arana Rodriguez Jan 19 '13 at 03:30

Quantification over the empty set

5 Answers5

Linked