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Can anyone explain why the predicate all is true for an empty set? If the set is empty, there are no elements in it, so there is not really any elements to apply the predicate on? So it feels to me it should be false rather than true.

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Roughly: if $\forall x \in S (R(x))$ is false, then there must be $x \in S$ such that $R(x)$ is false. Since there is no such $x$ for empty $S$, the statement $\forall x \in S (R(x))$ is true. –  Levon Haykazyan Sep 25 '12 at 21:23
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Okay... but you can also argue the other way around: if ∀x∈S(R(x)) is true, then all x∈S must have R(x) true. Since there is no such x for empty set S, the statement ∀x∈S(R(x)) is false. ... What happened? –  bodacydo Sep 25 '12 at 21:27
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The statement $\forall x \in S (R(x))$ does not imply that there is an $x$ for which $R(x)$ holds. It just says that whenever you give me an $x$ from $S$, I can demonstrate that $R(x)$ holds. In the end, it is just a matter of convention. However presuming vacuous universal quantification to be true results in smoother theory. I am just explaining the logic behind it. –  Levon Haykazyan Sep 25 '12 at 21:40
    
I see. I don't think I understand but I'll nod. –  bodacydo Sep 25 '12 at 21:49
    

6 Answers 6

"All of my children are rock stars."

"If we go through the list of my children, one at a time, you will never find one that is not a rock star."

Do you want the above two sentences to mean the same thing?

Also, do you want

"Not all of my children are rock stars."

to mean the same as

"At least one of my children is not a rockstar"?

Because in the situation that I have no children, the last statement is false, so we would want "all of my children are rock stars" to be true to preserve dichotomy.

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Congratulations on your children's success! –  Quinn Culver Sep 25 '12 at 22:11
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Thank you, they have all topped the charts and become pop icons! All of them. –  alex.jordan Sep 26 '12 at 2:23
    
I'd rather have a predicate on "all of my children" to have an undefined result since you have no children. And then I would extend set theory to have undefined-ness result in FALSE value because it only makes sense. You have no children, therefore you have no rockstar/scientist/redheaded/smart/awesome children. –  pwned Aug 2 '13 at 11:06

It kind of makes sense. If I understand correctly, I think you want to prove:

$\forall x (x\in \phi \rightarrow Q)$

where:

$\forall x (x\notin \phi)$

Q is any proposition whatsoever.

Proof: Suppose $y\in \phi$. We want to prove that $Q$ is true for any proposition $Q$ whatsoever. Suppose to the contrary that $Q$ is false.

Applying the definition of $\phi$ to $y$, we obtain the contradiction $y\notin \phi$. Therefore, by contradiction, $Q$ must be true. We have:

$y\in \phi \rightarrow Q$

Generalizing, we obtain, as required:

$\forall x (x\in \phi \rightarrow Q)$

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It hinges on the Law of the Excluded Middle. The claim itself is either TRUE or FALSE, one way or the other, not both, not neither.

Pretend that I am asserting "For every $x\in S$, property $P(x)$ holds." How could you declare me to be a liar? You would have to produce an element of the set ($S=\varnothing$, in this case) that does not have the property $P(x)$. Only then can you declare my assertion FALSE. Since you cannot do that here, my assertion is TRUE. I essentially spoke the truth by NOT speaking a lie.

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Okay that's pretty interesting argument. But why doesn't it work the other way around? -- Pretend that I'm asserting "There exists an x∈S such that P(x) doesn't hold." How could you declare me to be a liar? You would have to produce an element of the set (S=∅, in this case) that does have the property P(x). Since S is the empty set, you can't convince me that there is an x such that P(x) holds, therefore my assertion is true. ... I'm easily getting lost in logic. –  bodacydo Sep 25 '12 at 21:48
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If you're asserting "there exists such-and-such" and I declare you to be a liar, it is your job to show me an actual something and defend the proposition that this particular something is a such-and-such. Someone claiming $\exists$ is a liar by default (just as someone who claims $\lor$ is), whereas someone claiming $\forall$ is right unless his opponent can find a counterexample (just as someone who claims $\land$ can relax until his opponent names one of the conjuncts that he claims is false). –  Henning Makholm Sep 25 '12 at 21:53
    
@HenningMakholm Suppose I claim that there is a well-ordering of the reals, and you declare me be to a liar ... –  Peter Smith Sep 25 '12 at 22:03
    
@PeterSmith: Then you make a call upon your good friend Axiom of Choice, and he goes out into his back room and does something magical whereupon he returns with a something that you then present to me. I have no idea how he does it. –  Henning Makholm Sep 25 '12 at 22:12
    
@HenningMakholm Indeed! :-) The serious point, though, is that issues pro or contra a constructivist reading of disjunction/existentials are one thing, and issues about the treatment of vacuous quantifiers surely something else. –  Peter Smith Sep 25 '12 at 22:23

It could be taken the other way, but it's simpler this way.

Say we believe that all rubies are red, and we consider some some collection of rubies, called $R$; say $R$ is all my rubies.

We would like to conclude that all my rubies are red. This seems very reasonable, since all rubies are red. But with your idea, this conclusion might be false! At best we can say that all my rubies are red, if I have any rubies.

This qualification doesn't add anything to the analysis. It doesn't illuminate any subtle point. It just complicates the discussion with an uninteresting special case.

Since the purpose of formal logic is to model plausible reasoning as closely and as simply as possible, we agree to the convention that "all my rubies are red" is deemed to be true even when I have no rubies, so that we don't have to qualify a lot of claims with "… if there are any such rubies".

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Another approach: the 'vacuous truth' for $\forall$ is roughly the logical equivalent of an empty product being defined as 1 or an empty sum being defined as 0. Just as we want $\sum_{i=1}^{n+1} a_i = a_{n+1} + \sum_{i=1}^{n} a_i$ (and want this to hold in every case, even the 'base case' where $n=0$) and want $\prod_{i=1}^{n+1} a_i = a_{n+1}\cdot \prod_{i=1}^{n} a_i$, so too we want $\forall x\in (S\cup \{z\})\ P(x) \Longleftrightarrow \bigl(\ (\forall x\in S\ P(x))\ \wedge P(z)\bigr)$ to hold even in the 'base case' where $S$ is empty. You should be able to convince yourself (through some relatively straightforward logical manipulation) that this is requires defining $\forall x\in\emptyset \ P(x)$ to be true for all predicates $P()$.

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An extension of the comment to bwsullivan's answer:

Suppose for all elements in a set P(x) holds and P(x) doesn't hold, i.e.

$\forall x \in A: P(x) \land \forall x \in A: \neg P(x)$

Suppose $y \in A$

then $P(y)\land \neg P(y)$

a contradiction.

So $\forall y: \neg y \in A$

I.e. $A = \emptyset$

Now if you made $\forall x \in A: P(x) $ or $ \forall x \in A: \neg P(x)$ not true for the empty set, you couldn't conclude this.

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