Talk:Drinker paradox

From Wikipedia, the free encyclopedia

Socrates This article is within the scope of the WikiProject Philosophy, which collaborates on articles related to philosophy. To participate, you can edit this article or visit the project page for more details.
??? This article has not yet received a rating on the quality scale.
Low This article has been rated as low-importance on the importance scale.
This article may be too technical for a general audience.
Please help improve this article by providing more context and better explanations of technical details to make it more accessible, without removing technical details.

Contents

[edit] This is not a paradox!

This is not a paradox by any stretch of the imagination; it's a riddle where we're merely being confused due to the wording. The emphasized text shows where the problem exists:

Suppose, on the other hand, at least one person isn't drinking. For that particular person, it still can't be wrong to say that if that particular person is drinking, then everyone in the pub is drinking — because that person is, in fact, not drinking.

As you can see, there is an inconsistency in the logic. For us to make inquiry or assumptions about "that particular person" is fallacious when in fact there may be more than one person not drinking.

As simply as I can put it:

If in our pub there exists more than one non-drinker, we can not refer to that demographic(non-drinkers) as "that particular person". Thus the entire arguement that follows is inherently wrong. If no one objects, I suggest the article be removed or at least be put under a more appropriate category; perhaps riddles. --131.191.106.81 02:22, 16 September 2007 (UTC)


The first step of the proof, stating that either everybody drinks

or someone doesn't drink, is an application of the principle of  
excluded middle, so the paradox is true only in classical logic 
but false in intuitionistic logic; 

I don't agree. This is valid in intuitionistic logic. The set of people in a bar is finite (!) and it should be decidable by direct observation whether someone is drinking or not*. No appeal to the law of the excluded middle is needed.

The last step of the proof uses the notion of material 
implication, according to which a hypothetical statement in the  
form if A then B is always true if A is false. 

This seems to be the essence of the paradox. It's about fallacies of relevance, not constructivism.

Paradoxes of this kind are used by the intuitionist school in the
philosophy of mathematics to argue against the the laws of 
classical logic.

I'm curious about that. Does one of the references cited argue this way?

-Dan 13:49, 17 May 2006 (UTC)

* Even if not, it's not the essence of the paradox -- just change it to something directly decidable, like "in every non-empty hospital, there is a patient who, if he is declared dead, every patient is declared dead", or if you still object that a declaration of death might conceivably be ambiguous, you can't argue with "in every finite non-empty set of natural numbers there is a number which, if it is prime, all numbers in the set are prime."

It is true that excluded middle is not needed because the set of drinkers is finite, but the logical formula doesn't assume finiteness. Since finiteness is not formalizable in first order logic, the only way to prove the paradox is to use excluded middle.
(Anyway, I like your hospital example even better than the drinker. We could add it to the page, but the paradox is commonly known as the drinker paradox.)
About its relevance to intuitionism, maybe that's a silly part of the article. Some intuitionists argued in that way to me, but it is not to be found in any published work. So it probably must be deleted.
Eubulide 14:04, 17 May 2006 (UTC)
No, it's not the only way to prove it. We can still do without LEM.
  • For a bar/hospital: every customer/patient is a human. The human population is less than one trillion (true now, almost certainly true in my lifetime, and can be increased as desired). "Less than one trillion" is formalizable in first-order logic.
  • For prime numbers: we can talk about finite sequences of natural numbers in Heyting arithmetic just fine. I do believe the sentence "in every finite non-empty sequence of natural numbers there is a number which, if it is prime, all numbers in the sequence are prime" is expressible in HA and is a theorem!
I would never make this argument against LEM, and I consider myself a constructivist. Now I don't consider myself an intuitionist, still, Brouwer himself considered LEM to be something from our experience of finite space and time (and used improperly in discussing the infinite), so presumably he would not like this argument either. Please hit your intuitionist friends over the head! -Dan 16:09, 17 May 2006 (UTC)
It just occurred to me how your hospital interpretation actually shows the paradox to be essentially classic. Just change your interpretation a little: "In every hospital there is a patient such that, if he recovers, all the patients recover". So the doctors only need to cure that one patient! Of course, it is undecidable which one that patient is. Eubulide 14:21, 17 May 2006 (UTC)
A medical breakthrough! -Dan 16:09, 17 May 2006 (UTC)

[edit] Infinite undecidable drunken health care

Speaking of Ralph Klein, is there a version of this which a) is classically (but not constructively) equivalent to the original statement that every non-empty hospital has a patient who, if cured, everyone is cured, and b) is constructively provable, even without assuming that the hospital is finite, or that death is decidable?

Answer: Yes. If, in a hospital, there isn't a patient who dies if any patient dies, then that hospital has no patients at all.

[\not \exist p \in H : (\exist p' \in H: D(p')) \to D(p)] \to \not \exist p \in H

Proof: exercise for the reader :-) -Dan 19:44, 18 May 2006 (UTC)

[edit] LME Redux

The paradox can be stated with less parentheses, i.e. (Ex) Drink(x) -> (x) Drink(x). Further, the argument seems to rely on the closed world assumption, and not on the excuded middle.

The Coq prover page uses sets and that goes beyond the scope of logic. So, in order to be certain, can somebody give a derivation of the paradox in classical logic, and if possible, in intuitionistic logic? Actually, I wonder if either is possible. If it is not possible, please correct the claims on the page. -- 129.247.247.238 16:37, 22 May 2006 (UTC)

I'm not sure that's a good way to write that, as it might be read (\exist x Drink(x)) \to (\forall x Drink(x)) rather than \exist x (Drink(x) \to (\forall x Drink(x))). As for a proof, alright:


SOLUTION TO QUESTION POSED ABOVE











In constructive free logic:

\begin{matrix}
\mbox{1.} & \exist x. \top & \mbox{assumption} \\
\mbox{2.} & \neg \exist x.((\exist y.Dies(y)) \to Dies(x)) & \mbox{assumption} \\
\mbox{3.} & \forall x. \neg((\exist y.Dies(y)) \to Dies(x)) & \mbox{2, constructive equivalent} \\
\mbox{4.} & \forall x. (\neg \neg \exist y.Dies(y)) \wedge (\neg Dies(x)) & \mbox{3, constructive equivalent} \\
\mbox{5.} & (\forall x. \neg \neg \exist y.Dies(y)) \wedge (\forall x. \neg Dies(x)) & \mbox{4, free logic equivalent} \\
\mbox{6.} & (\forall x. \neg \neg \exist y.Dies(y)) \wedge (\neg \exist x. Dies(x)) & \mbox{5, constructive equivalent} \\
\mbox{7.} & (\neg \neg \exist y.Dies(y)) \wedge (\neg \exist x. Dies(x)) & \mbox{1, 6, free logic inference} \\
\mbox{8.} & \bot & \mbox{7,law of non-contradiction} \\
\mbox{9.} & \neg \exist x. \top & \mbox{8, discharge assumption 1} \\
\mbox{10.} & (\neg \exist x.((\exist y.Dies(y)) \to Dies(x))) \to \neg \exist x. \top & \mbox{9, discharge assumption 2} \\
\end{matrix}

Which was my statement above. I trust you can take it from here to classical logic. Bonus points: where is the paradox of material implication (aka fallacy of relevance)? -Dan 20:17, 22 May 2006 (UTC)
I admit I am confused now. I have a straightforward proof using game semantics showing that (Ex) (Drink(x) -> (y) Drink(y)) is neither intuitionistically nor classically derivable:
  0                       (Ex) (Drink(x) -> (y) Drink(y))
  1  0?                   Drink(n) -> (y) Drink(y)
  2  1? Drink(n)          (y) Drink(y)
  3  2? m                 Drink(m)
  4  3?
Am I missing something badly? I think we have a case of hidden axioms which are needed to prove the proposition. -- 129.247.247.238 10:02, 23 May 2006 (UTC)

Yes, I was missing something -- there is a condition, in any non-empty-bar... This makes the domain non-empty and closes the domain of discourse, too. What is wrong, is that the formal version does not list these conditions formally.

The article should be modified accordingly. And the claim, that the proof relies on LME can be deleted, as shown by Dan. -- 129.247.247.238 14:15, 23 May 2006 (UTC)

I do not understand this. If \forall x.Px \or \exists x.\neg Px , then  \exists x. \forall y.(Px \to Py) holds. Why must the bar be non-empty?

[edit] Nonintuitive

Shouldn't it be mentioned that while the original definition seems counter-intuitive, the contrapositive makes perfect sense (that is, if not (everybody drinks) then (someone doesn't drink))? Or am I wrong about that? MagiMaster

Contraposition is not an equivalence in intuitionistic logic. Precisely, (p \rightarrow q) \rightarrow (\neg q \rightarrow \neg p) holds, but the reverse, (\neg q \rightarrow \neg p) \rightarrow (p \rightarrow q), doesn't. Therefore, the drinker's principle only implies your preferred contrapositive version in all logics. I have no opinion on which version of the classical statement is more "intuitive" because I tend to find classical logic non-intuitive. Kaustuv Chaudhuri 03:24, 16 July 2006 (UTC)
Actually, I don't think that's even the correct contrapositive. The implication occurs inside the existential quantifier. The contrapositive of \exist x.(D(x) \to \forall y. D(y)) should be \exist x.(\neg(\forall y. D(y)) \to \neg D(x)), which is (non-constructively) equivalent to \exist x.((\exist y. \neg D(y)) \to \neg D(x)). That is, "there exists someone who, if anyone doesn't drink, he also doesn't drink." 192.75.48.150 12:58, 17 July 2006 (UTC)
Hmm, I think the original comment was essentially correct. I interpreted the statement as talking about the body of the outer \exists. Kaustuv Chaudhuri 13:47, 17 July 2006 (UTC)
Ah, never mind, I see what you're getting at. Yeah, \neg (\forall y.\ D(y)) \rightarrow \exists x.\ \neg D(x) is only implied by the contrapositive of the drinker principle. The original statement was imprecise because "someone doesn't drink" can be interpreted as an \exists. Kaustuv Chaudhuri 14:20, 17 July 2006 (UTC)
Sorry, I forgot to sign my first comment. This is what I was thinking about when I wrote that though. I wasn't sure if it was right, because I don't remember exactly how negating qualifiers works. (Also, I don't know much about intuistionistic logic. I'm a computer scientist, so I studied classical logic.) MagiMaster 19:58, 17 July 2006 (UTC)

[edit] laymen

Could someone write out the reasoning in plain speech as opposed to latex? I'm not familiar with all the logic symbols, and furthermore, I prefer, as many probably do, the construction of natural language.

I hope this helps. 192.75.48.150 15:37, 14 August 2006 (UTC)
I guess not. We got a {{technical}} tag again. Please feel free to ask questions here. We don't bite. 192.75.48.150 21:06, 17 August 2006 (UTC)

(transcribed from WP:RfF --192.75.48.150 19:17, 28 August 2006 (UTC))

  • It would be unwise for any newcomers to logic to leap straight in at this point, so the application of the technical tag is probably not appropriate. It is certainly a lot more accessible than certain other articles in this subject area. That said, there are a couple of points that could be cleared up:
    • In the proof section, I don't think it would be redundant to re-explain the proof (although I'm struggling to come up with alternate phrasing).
    • The proof diagram is scary - perhaps move it to its own subsection in the proof section, so readers can see that they don't need to be able to decipher that to understand the concept.
    • The reference to Goldbach's conjecture is awkward - you should make clear that it is unsolved, can not be solved like this because this reasoning is invalid, and introduce the section more clearly, something like "For instance, it would allow us to solve Goldbach's conjecture which has taxed mathematicians for hundreds of years...".
    • I'd always explain in English before using notations: the section "Material versus indicative conditional" in particular, would probably read better if the order of the sentences was changed and the notation explained.
    • In some cases you might like to point readers to appropriate connected themes explicitly rather than just a wikilink.
    • I also have a feeling that the title may draw an unintended audience, so some stress in the lead that it has nothing to do with pub culture or drinking games (or rather, overstressing the connection to first order logic) won't hurt.
Hope this helps. Yomanganitalk 11:03, 25 August 2006 (UTC)

I have simply removed the proof diagrams as in the present form they add no clarity to the article. I disagree with nearly all of the mumbo jumbo that was added as the logical issue seems fairly shallow to me, but I'll leave it be. Kaustuv Chaudhuri 12:56, 3 September 2006 (UTC)

The paradox is less paradoxical and easier to understand when it is explained the following way: assume there is a person who would be the last to start drinking. If and when this person drinks, then everybody drinks.
While reminiscent of Zorn's Lemma, one need not delve into set theory at this point. Although the connection might be used in order to explain why the "paradox" need not arise in intuitionistic logic. -- Zz 13:13, 5 September 2006 (UTC)
Eh. I can see where you're headed with the last person to drink, but I'm not seeing the connection to Zorn's lemma. 192.75.48.150 17:27, 5 September 2006 (UTC)
The connection lies in the maximal element existing for every non-empty partially ordered set (e.g. the last one to drink). For finite sets the existence of such an element is obvious. For infinite sets, however, the argument presented might fail, unless the lemma guarantees the existence of this element. This echoes the objection listed for intuitionistic logic - for finite sets, the drinker paradox is provable (see above), for potentially infinite sets, this need not be so. -- Zz 02:09, 6 September 2006 (UTC)
Sorry Kaustuv, did you mean your only objection to the mumbo-jumbo is that it is unneeded, because the issue is shallow? Or was there something else? 192.75.48.150 17:27, 5 September 2006 (UTC)

[edit] has anyone come up with it yet?

if you apply these assumptions to a pub, where at least two people are NOT drinking, then the statement that everyone is drinking is false, because if he or she is drinking, there is still one more which might be not drinking :) !!! —The preceding unsigned comment was added by Katoa (talkcontribs) 04:16, 9 January 2007 (UTC).


== confusing or error?

"Suppose, on the other hand, at least one person isn't drinking. "

Right up to here.

"For any particular person,"

I think this "For any" is at least confusing if not wrong. _There exists_ a person for which "if that particular person is drinking, then everony else" holds true. But it doesnt hold true "For any" person.

"... it still can't be wrong to say that if that particular person is drinking, then everyone in the pub is drinking -- because that person is, in fact, not drinking."


The first comment above makes sense to me. If all but two people are drinking, either one could drink without the either drinking, correct? If this is the case then either one of them can drink without everyone drinking. Since everyone else was already drinking, obviously any of them can drink without everyone drinking. If any person in the room can drink at times when others aren't drinking, how can at be that there must be someone who only drinks when everyone is drinking?--Eloil 20:44, 17 April 2007 (UTC)

[edit] so much fuss for nothing

I am not a logician, but as an interested reader, this page appears to me as mind-bogglingly technical over a simple thing. That's fine with me, just as long as there is also a good initial paragraph with common-sense language that would explain that:

1. If everyone is drinking, then all the premises are true. The special drinker is drinking and the entire pub is also drinking;

2. If one person is not drinking, then none of the premises are true. We can deduce that such person must be the special drinker (though the paradox doesn't ask us that), because if it was anyone else, there shouldn't be anyone not drinking;

3. If there is more than one person not drinking, then none of the premises are true, and we can deduce that one of those people who are not drinking is the "special" drinker.

We can also say:

1. If the set has zero people, then the premises are irrelevant, but not necessarily untrue; 2. If the set has one person, then the premises are true: if he is drinking, everyone is drinking; 3. If the set has two people or more, we can infer who's the special drinker by going through the previous 3 points.

What's more than this about it? Sorry people, can't understand well. What's so special about it? I think that that is what this text lacks.

Luis Dias, 14-02-2007 —The preceding unsigned comment was added by 195.23.224.70 (talk) 13:35, 14 February 2007 (UTC).

I like the way you explain it. The technical details are obvious to the logically trained mind, which in turn proves that a training in formal logic can make things difficult. However, there is a reason why the paradox gets the attention of logicians... If no objections are raised, I will shorten the discussion in the article substantially and use your explanation instead. -- Zz 16:16, 24 April 2007 (UTC)
P.S.: Oh my, the explanation is already in the article. Nonetheless, I will shorten the technical details if no objections are raised.
How is the following true? "If there is more than one person not drinking, then none of the premises are true, and we can deduce that one of those people who are not drinking is the "special" drinker." I can understand that you may not be able to disprove this to be the case for each of the non-drinkers, but you wouldn't normally deduce something like that. Is there one person one earth who, if they're inside their mothers womb, all other people will be inside their mother's womb? Since there is more than one person not in their mother's womb can we deduce that this statement is correct? Constan69 (talk) 08:22, 12 February 2008 (UTC)

[edit] Why is this a paradox?

All it says is that there's always at least one person we're waiting on to take a drink in a given round, and when they decide whether they're taking a drink we'll know whether everyone's drinking that round or not. It doesn't seem paradoxical at all. 80.192.29.107 14:15, 25 August 2007 (UTC)

[edit] ISBN Wrong

The sole reference for this paradox (bookwise) seems to have the wrong ISBN, unless this book is contained within some other book. The ISBNs I get for the book "WHAT IS THE NAME OF THIS BOOK?" by Raymond Smullyan (Author) are (from Amazon):

(Paperback)

  1. ISBN-10: 0671628321
  2. ISBN-13: 978-0671628321

or What is the name of this book?: The riddle of Dracula and other logical puzzles (Unknown Binding)

  1. ISBN-10: 0139550887
  2. ISBN-13: 978-0139550881

Root4(one) 20:29, 30 January 2008 (UTC)

[edit] It's merely a trick of language

One of the premises is "there is at least one person who is not drinking", but this is then sneakily changed (by implication) into "there is exactly one person who is not drinking", which is invalid since there could be five people, or thirty, who are not drinking. Therefore the entire problem is completely imaginary.

All it's saying is "if every person who is not drinking were to start, then everyone would be drinking", which is obvious.

86.63.16.62 (talk) 09:02, 1 May 2008 (UTC)