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There exists an $X\in A$ such that $P(X)$. When $A$ is the empty set then this statement is false because there is nothing in $A$ that when plugged in for $X$, makes $P(X)$ come out True.

However, when the quantifier is the universal one, meaning "For all $X\in A$" (which is the empty set), then the statement is true !!

How is that ? .. Because by the reasoning in the first statement, there are not any values of $X$ that makes $P(X)$ come out True.

I'm missing something !!

(I'm studying Electrical Engineering and self studying velleman's how to prove it and the quantifiers are already kicking my butt so simplify the answers for me if possible :))

Zev Chonoles
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Mustafa Adam
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6 Answers6

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The negation of the statement "for all $x\in X$, $P(x)$ holds" is "there exists $x\in X$ such that $P(x)$ does not hold". Well, then if $X$ is empty, then there is not a single $x\in X$ such that $P(X)$ is not true, and thus it is indeed true that for all $x\in X$, $P(X)$ is true.

The confusing part is that in general the validity of a statement of the form $\forall x\in X: P(X)$ does not imply the existence of a single $x\in X$ for which $P(X)$ holds. It only states that if $x\in X$ then $P(x)$ holds.

You can also think of it this way: Suppose I claim that everything in a box in front of you is pink. Can you sue me if you open it up and see a banana in there? Sure, the banana is yellow, so the judge will rule in your favor. But, can you sue me if you open up the box and see nothing at all in there? Well, no, you have nothing to complain about: everything in the box is indeed pink since nothing in it is not pink.

Ittay Weiss
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You are not alone who is puzzled with that. Lewis Carroll (who was a logician by trade) also objected to "vacuous truth". The choice that the modern logic makes is due more to the convenience of drawing implications without worrying if some sets are empty or not at each step, than to the intuition. You want to be able to combine the sentences "Everybody who is not too short can pick an apple on this tree" and "Everybody who is not too tall will be able to enter the vault" into "Everybody who is neither too short, nor too tall will be able to pick the apple and go to the vault with it" without heavy thinking of how the threshold heights in those cases are related to each other. This forces you to assign the "true" value to the last statement even if you need to be at least 6 feet tall to reach the apple and no more than 5 feet tall to enter the vault.

In general, such vacuous usage of "for every" is not uncommon in the real life. When the sheriff in a small town proclaims publicly that "Every criminal will be caught and punished!", the streets are patrolled day and night, and there are no crimes in the town, you don't call him a liar and the police department a worthless waste of taxpayer money, quite the opposite: you suspect that he is so right that even the to-be-criminals believe this statement of his and abstain from misdeeds.

fedja
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  • I don't understand how the last paragraph is related to vacuous truth. The statements made there are exaggerations of truths to the point that they become false, but everybody understand their true true (no typo) meanings. It's more of an example of the vagueness of natural languages than it is of formal logic. – Ittay Weiss May 26 '13 at 19:03
  • It's quite a relief to know I'm not the only one who is puzzled with this :). – Mustafa Adam May 26 '13 at 19:09
  • @Ittay. Well, if we ignore politicians and other professional cheaters, many such statements are made with the hope and the intent that they end up being vacuously true, not just true. This was the point I was trying to make (perhaps, in a rather lame way :)). – fedja May 26 '13 at 20:22
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In classical logic, given a particular structure for some language, if something is not false in the structure, then it's true there. Of course we can easily make things complicated by considering a structure whose universe is made of structures of a particular language, and so on and so forth. So let's stay in the context of one particular structure.

Something is vacuously true if it just cannot be made false. That is to say, there's no counterexample. If it's not false, it has to be true.

So saying "for every $x\in\varnothing$ ..." does not have a counterexample, and therefore it is true; but on the other hand "there exists $x\in\varnothing$ ..." is false because we can prove that there isn't such $x$, and therefore conclude a contradiction.

Asaf Karagila
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    I know you don't mean this, but a naive person reading your response might jump to the (false) conclusion that something is true if it can't be proven to be false. – Robert Israel May 26 '13 at 18:30
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    @RobertIsrael In a way, that's how it is. If for some theory $T$ there's no model of $T \cup {\lnot\phi}$ i.e. no counter-example to $\phi$, then $T \cup {\lnot\phi}$ is inconsistent, and thus $T\vdash \phi$. – fgp May 26 '13 at 19:12
  • @fgp notice the boldface 'proven'. You can't prove AC is false from ZF, but that does not mean that AC is provable in ZF. Having no counterexample is much stronger than not being able to prove. – Ittay Weiss May 27 '13 at 06:36
  • @Ittay: This doesn't contradict what fgp said, though. It's like everyone is saying something true, but pointing that out to the previous guy... :-) – Asaf Karagila May 27 '13 at 06:39
  • @AsafKaragila that's logic for you :) – Ittay Weiss May 27 '13 at 06:41
  • @Ittay: Notice that I'm TA'ing Logic 2 this semester, so this is not logic for me. :-) – Asaf Karagila May 27 '13 at 06:43
  • @IttayWeiss My point is, it depends on who's doing the proving. ZF can't prove $AC$ precisely because there is a model of $ZF+\lnot AC$ or in other words a counter-example - it just takes something stronger than ZF to construct it. – fgp May 27 '13 at 09:06
  • @fgp: That been said, one doesn't have to assume the consistency of $\sf ZF$ in order to prove the consistency and use models of set theory to prove there is a model of $\sf ZF+\lnot AC$. There are purely proof theoretical methods for these things. Recall that $\text{Con}(T)$ is just a number theoretical statement, and $\rm Con(\sf ZF)\to\rm Con(\sf ZF+\lnot AC)$ can be expressed, and I believe proved, within $\sf PA$ without mentioning models or anything. – Asaf Karagila May 27 '13 at 09:26
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Let $A$ be the set of monkeys in the room, let $P(x)$ denote the sentence "x is playing chess".

$\forall x\in A, P(x)$ means "every monkey in the room is playing chess", which is true almost all the time, since most rooms are monkey-free.

$\exists x\in A, P(x)$ means "there is a monkey in the room playing chess", which is almost never true.

vadim123
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I think I'm getting it. How about the conditional form of this though. For all X(If x belongs to the empty set, then P(X)). This is only false when the hypothesis is True and the conclusion is False. But since the hypothesis is always False since there are not any members of the empty set, therefore the conditional is always True. Great . Now when we change the quantifier into the existential one. Why is it wrong ?! The hypothesis is still wrong because since there aren't any elements in the empty set, then there are not even some elements , Therefore shouldn't it be True as well

Mustafa Adam
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    $(\forall X\in A)(P(X))$ is just shorthand notation for $(\forall X)(X\in A\implies P(X))$ and as you've noted, it's true if $A=\varnothing$. About $(\exists X\in A)(P(X))$, this is shorthand notation for $(\exists X)(X\in A \land P(X))$, plugging in $A=\varnothing$ you get a false assertion. – Git Gud May 26 '13 at 18:58
  • I did not realize that at all. The existential quantifier is a conjunction. Is that because there could be values of X in A such that P(X) does not hold ? .. Therefore making the hypothesis either True of False at the same time ?! – Mustafa Adam May 26 '13 at 19:06
  • It's by definition, actually. As I said $(\exists X\in A)(P(X))$ is just shorthand notation for $(\exists X)(X\in A \land P(X))$. The formula $(\exists X\in A)(P(X))$ has no independent meaning on its own, it is equivalent to $(\exists X)(X\in A\land P(X))$ by definition. The last paragraph of this answer might be helpful. – Git Gud May 26 '13 at 19:09
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    You helped me greatly, thanks Git Gud. – Mustafa Adam May 26 '13 at 19:20
  • You're welcome. – Git Gud May 26 '13 at 19:22
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Wittgenstein also had problems with it. I'm not sure how well I can explain this, but I'll give it a go.

Crudely, his claim was that you can't talk about things that don't exist.

He says, in typically abstruse form:

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I dissociate the concept of all from truth-functions.

Frege and Russell introduced generality in association with logical product or sum. This made it difficult to understand the propositions '$(\exists x).fx$' [there exists $x$ and the proposition $f$ holds for it] and $'(x).fx'$ [all $x$ and the proposition $f$ holds for it].

His difficulties are associated with treating "$\exists x$" and "$\forall x$" as valid statements on their own (which may or not be held true these days). The [brackets] are my attempt to interpret the statements in the way he thought was silly. He goes on to say:

5.524

If we are given objects then we are given all objects.

By which he basically means that $\forall$ is redundant and adds nothing to a statement, though he puts it in a very theory laden way. You can always write a statement without it.

But, it seems he would also have a problem with doing this:

$$\forall x : P(x) \rightarrow \nexists x: ¬P(x) $$

or in words, replace "for all $x$, $P$ is true" with "there is no $x$ for which $P$ does not hold". The two statements are mathematically the same (in first order logic). He would, I think, favor

$$P(x)$$

which one might read as "$P$ is true", with the subtext of "for all $x$ that exist, obviously! How could we know if it is true for those things that don't?"

Lucas
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  • Please do let me know if you think I've got any of this wrong. I find him quite hard to understand. – Lucas May 26 '13 at 21:42
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    How can one get Wittgenstein wrong when Wittgenstein himself got himself wrong, negated himself, negated the negation, and did all of that in a perfectly unreadable fashion? You just can't go wrong ... – Ittay Weiss May 27 '13 at 06:38
  • @IttayWeiss A very good point ;) But there is always a "canonical reading". – Lucas May 27 '13 at 14:42