Talk:Drinker paradox

message: The article contains the phrase "either no one is drinking, or everyone is not drinking".  These two mean exactly the same thing.  Perhaps whoever wrote that intended to write "either no one is drinking, or not everyone is drinking, but at least one person is".  —Preceding unsigned comment added by 69.171.176.43 (talk) 21:31, 2 December 2010 (UTC)
thread_title:  Poor Wording 

message: 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)
thread_title:  Infinite undecidable drunken health care 

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
thread_title:  Nonintuitive 

message: 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 (talk • contribs) 04:16, 9 January 2007 (UTC)
replies: text: 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)
thread_title:  has anyone come up with it yet? 

message: 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)
thread_title:  Why is this a paradox? 

message: 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)
# ISBN-10: 0671628321
# ISBN-13: 978-0671628321
or
What is the name of this book?: The riddle of Dracula and other logical puzzles (Unknown Binding)
# ISBN-10: 0139550887
# ISBN-13: 978-0139550881
Root4(one) 20:29, 30 January 2008 (UTC)
replies: text: The 0140135111 isbn currently given in the article works fine in WorldCat and points to the 1990 Penguin reprint . The actual problem is that book I've looked in, having  by the way, says "(c) 1978". Easily confirmed in WordlCat  So I think the year and ISBN should be fixed. Actually, I've done that because I've added page numbers from my copy and they might not be the same in the Penguin reprint. Tijfo098 (talk) 20:47, 27 October 2012 (UTC)
thread_title:  ISBN Wrong 

message: 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)
thread_title:  It's merely a trick of language 

message: jesus christ, can somebody delete this? i'm too lazy to figure how such is done, but this entirely unsourced article is one incomprehensible and futile attempt to prove how something that is obviously not a paradox, is a paradox.
this article is complete and utter nonsense that doesnt meet almost any of the criteria of wikipedia. please delete this crap· Lygophile has spoken 23:51, 7 February 2011 (UTC)
replies: text: You could nominate the article for deletion. I believe it pretty clearly is a paradox: one would not expect it to be the case that "there is such a person that if he is drinking, everybody is drinking", although that is truly the case. It highlights a curiosity of formal logic that contradicts everyday language (the nature of "If P, then Q"), and so something that seems nonsensical is technically correct. So, while I do feel that this is in fact a paradox, the article is also poorly written. The introduction to the paradox and description of it make sense, but everything after that is quite awful. Chris3145 (talk) 00:01, 12 February 2011 (UTC)
thread_title:  delete please 

message: It seems to me that the source of the paradox is the fact that the logical formula is not transcribed properly into English.  It should be "there is a person such that if we assume that 'he is drinking' then everybody is drinking".    Tkuvho (talk) 04:54, 27 March 2011 (UTC)
replies: text: I am not a native speaker, but I doubt the translation is correct. It is not about what we assume, it is about him drinking. -- Zz (talk) 21:30, 27 March 2011 (UTC)
thread_title:  Inaccurate transcription into English 

message: Can anyone explain what is the relevance of this recently added reference?
Coscoy, Yann; Kahn, Gilles; Théry, Laurent (1995), Typed Lambda Calculi and Applications, Lecture Notes in Computer Science, pp. 109-123, doi:10.1007/BFb0014048, http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.36.8114&rep=rep1&type=pdf .
If no-one can suggest any relevance, I think it should be deleted --Jamesrmeyer (talk) 11:03, 28 March 2011 (UTC)
replies: text: They seem to mention the paradox as an example. I agree it's pretty tangential. — Carl (CBM · talk) 16:53, 28 March 2011 (UTC)
replies: text: The reference is relevant because it includes a proof of the paradox, which should dispel the above concern that \exists x.[\neg D(x) \vee \forall y.D(y)] is not a theorem of classical predicate logic. Obviously, better sources are desirable, but I think this one should stay in the mean time. James, do you have any sources you would like to add? Sławomir Biały (talk) 01:41, 29 March 2011 (UTC)
thread_title:  Relevance of reference? 

message: .2 | |__________ ~Ex(Px -> VyPy) . New Subproof
.3 | | |________ Pa. . . . . . . . New Subproof
.4 | | | |_____b . . . . . . . . . new variable for universal introduction
.5 | | | | |____ ~Pb . . . . . . . New Subproof
.6 | | | | | |__ Pb. . . . . . . . New Subproof
.7 | | | | | | . _|_ . . . . . . . 5,6 _|_ Introduction
.8 | | | | | | . VyPy. . . . . . . 7 _|_ Elimination
.. | | | | | <-------------------- end subproof
.9 | | | | | . . Pb -> VyPy. . . . 6-8 -> Introduction
10 | | | | | . . Ex(Px -> VyPy). . 9 Existentional Introduction
11 | | | | | . . _|_ . . . . . . . 2,10 _|_ Introduction
.. | | | | <---------------------- end subproof
12 | | | | . . . ~~Pb. . . . . . . 5-11 ~ Introduction
13 | | | | . . . Pb. . . . . . . . 12 ~~ Elimination
.. | | | <------------------------ end subproof
14 | | | . . . . VyPy. . . . . . . 4-13 Universal Introduction
.. | | <-------------------------- end subproof
15 | | . . . . . Pa -> VyPy. . . . 3-14 -> Introduction
16 | | . . . . . Ex(Px -> VyPy). . 15 Existentional Introduction
17 | | . . . . . _|_ . . . . . . . 2,16 _|_ Introduction
.. | <---------------------------- end subproof
18 | . . . . . . ~~Ex(Px -> VyPy). 2-17 ~ Introduction
19 | . . . . . . Ex(Px -> VyPy). . 18 ~~ Elimination
:What is the point of this?First we would need it in a reliable source. Secondly the proof would have to be notable in itself as Wikipedia does not include proofs, it is not a textbook. Dmcq (talk) 12:15, 24 May 2014 (UTC)
thread_title:  [[Fitch-style calculus]] proof 

message: Somebody left an incomplete proof in game semantics. This is a complete version:
  	  	O        	  	P
1.	  	         	  	\exists x.(Drinks(x) \to \forall y.Drinks(y))
2.	1?	         	2!	Drinks(m) \to \forall y.Drinks(y)
3.	2?	Drinks(m)	3!	\forall y.Drinks(y)
4.	3?	n        	2!	Drinks(n) \to \forall y.Drinks(y)
5.	4?	Drinks(n)	4!	Drinks(n)
At stage 4, the proponent goes back to stage 2 in order to defend against that attack again. For classical logic, that is allowed. In intuitionistic logic, more background is needed. -- Zz (talk) 11:49, 29 March 2011 (UTC)
replies: text: I don't think this is interesting because the winning strategy for the verifier is to pick a non-drinker if it exists or a drinker otherwise, which is really just a restatement of the simple proof. Tijfo098 (talk) 17:24, 28 October 2012 (UTC)
thread_title:  Yet another proof 

message: I looked through the material at the main page and this talkpage and did not find the following explanation:  let's list everybody in the bar in a master list in such a way that all drinkers come before all non-drinkers (other than this condition the order is arbitrary within the set of all drinkers, and similarly for all non-drinkers).  Then if we choose the last person on the list, we can affirm convincingly that if he is drinking then everybody is drinking.  Doesn't seem that paradoxical, either, when you put it that way.  Any comments? Tkuvho (talk) 16:04, 29 March 2011 (UTC)
replies: text: That is a good introduction why it is not so counter-intuitive. But we do not even need a master list connected with a kind of determinism. The last person to drink (and that can change in different scenarios) is the one who, if he drinks, then everyone drinks. If there is not going to be a last person, because too few are drinking anyway, the statement is true in the first place. -- Zz (talk) 09:08, 30 March 2011 (UTC)
replies: text: This only works if there are finitely many people in the pub, but the theorem is true for any (nonempty) domain. Sławomir Biały (talk) 11:43, 30 March 2011 (UTC)
replies: text: Just make sure the ordering has a last element, by pulling someone from the final ω chain if necessary. — Carl (CBM · talk) 11:45, 30 March 2011 (UTC)
replies: text: Ok, well if we're going to use the axiom of choice anyway, may as well do it properly: by well-ordering, the set of all nondrinkers on the has a least element. :-D  Sławomir Biały (talk) 12:11, 30 March 2011 (UTC)

text: It is only paradoxical in that 'person' might be taken to mean a particular person who doesn't change identity. It's like saying there's a woman giving birth every five minutes (or however long it is) in Britain. As to the idea of ordering - any explanations should be based on something that can be cited. Dmcq (talk) 09:26, 30 March 2011 (UTC)
replies: text: BTW I see it is about one and a third babies born every minute in Britain, I should have just remembered the phrase "There's one born every minute". :) Dmcq (talk) 11:58, 30 March 2011 (UTC)
replies: text: Is there a consensus that this is helpful in explaining the paradox? While I have no evidence that this is in the sources, if everyone agrees that this is "common knowledge" perhaps we can include it in the page anyway.  Tkuvho (talk) 12:22, 30 March 2011 (UTC)
replies: text: I think I'd be against it without a source that was somewhat along those lines. Dmcq (talk) 14:21, 30 March 2011 (UTC)
thread_title:  explanation of paradox? 

message: The article is written in such a way that most readers can't understand what is going on.  This is a contrived problem based on two nonintuitive aspects of formal logic:
* The logical statement is considering only one configuration D of the pub!  In other words, the statement assumes that D is fixed, whatever its values may be.  Of course, in "real life" that is completely uninteresting.  We generally don't make statements "if ... then ..." about single instances!   When we read There is someone in the pub such that, if he is drinking, everyone in the pub is drinking., then in "real life" we assume that this statement applies to all possible states of the pub.  (In which case, the statement is false because that person would have to keep changing to ensure that she is a nondrinker.)  However, the formal logical statement presented assumes that D is fixed, and so it is only considering a single state of the pub, "the actual state".  What this means is that the English utterance is being badly modeled.  The logical statement should include "for all D..." for it to model what we mean.  It does not! And this has been done intentionally by the creator of the paradox for the sake of confusing people.  It's a problem of intentionally poor modeling.
* This is compounded by the nonintuitive use of "if X then Y" by which, in formal logic, the statement is always true with one exception: when X is true but Y is false.  This means that any "nondrinker" will establish the truth that "if X is not drinking, then everybody is drinking".  But this is secondary.  And note that we can't be sure that any particular "someone" is not drinking!  So this logic is only true when we can choose our "someone" to be a nondrinker.  This is not true, in general, if we are considering all configurations of the pub, as normally one would assume.
The point is that you can have logical statements that don't apply to real life, and then you can pretend that those logical statements are good models.
The above two issues are what's relevant here.  The other discussion on "excluded middle", "nonempty pubs" is peripheral and greatly adds to the confusion.
I'm doing a study of paradoxes and the article was so confusing that I spent much more time than I expected to.  It should be deleted or explained for what it is, a contrivance, intentionally confusing.  It's miseducation and makes Wikipedia look bad.
AndriusKulikauskas (talk) 19:37, 29 January 2012 (UTC)
replies: text: Non-emptiness of the pub has actually a crucial importance, because in an empty domain any ∃-formula will evaluate to false, a method of inference used in the proof will not be applicable and the theorem therefore will not hold. Incnis Mrsi (talk) 21:25, 29 January 2012 (UTC)
replies: text:  You think that's what makes this paradox interesting? or a paradox? or worthy of an article? or mention in an article?  You surely have references!  In general, is there any other reference to the Drinker paradox beside the creator's book?  — Preceding unsigned comment added by AndriusKulikauskas (talk • contribs) 21:35, 29 January 2012 (UTC)
replies: text:  Surely, it worths a small article or at least a section somewhere in the Wikipedia. What else references do you want: mentioning and proof in the paper of Coscoy, Kahn and Théry is not enough? If you feel that Drinker paradox#Excluded middle is off-topical (and probably even WP:OR), then [ go forth] and delete it. Incnis Mrsi (talk) 22:06, 29 January 2012 (UTC)
replies: text: I've removed most of the excluded middle section as I think it was probably OR, I've put in a cite which actually mentions this though. The business about the empty set though seems to be referenced a number of times. Dmcq (talk) 01:27, 30 January 2012 (UTC)
replies: text:  The "Clausification in Coq" paper says that proof by resolution uses the Excluded Middle, the Axiom of Choice (AC), and non-emptiness of the domain, but that by using an obscure result of Kleene, proof without AC is possible. I can't say I've grokked the last part yet. It refers for further reading on that to a paper I've not read yet: "G. Dowek. Automated theorem proving in type theory. In Course notes for the 2nd International Summer School in Logic for Computer Science, University of Chambery, France, 1994", and which (unusually for a CS paper) is nowhere on the interwebs, although Dowek lists it here. Tijfo098 (talk) 12:32, 27 October 2012 (UTC)
thread_title:  The main point of the "paradox" is never mentioned

message: it is a riddle, for he is the only man in said pub. therefore if he is drinking everyone in the pub is drinking.  — Preceding unsigned comment added by 201.200.37.126 (talk) 03:04, 26 June 2012 (UTC)
thread_title:  obviously a riddle 

message: Escardo and Oliva have an interesting result, showing that the "Drinker paradox" (as they call it) is perhaps more important than a mere amusement.
But we don't seem to have any articles on the related notions: searchable set, Dubuc-Penon compactness, or principle of omniscience although we have Limited principle of omniscience. Tijfo098 (talk) 16:48, 27 October 2012 (UTC)
thread_title:  Do we have any articles on the related notions? 

message: This is a ridiculously long article which, in spite of Wikipedia:Be bold, I do not feel bold enough to edit and yet which, in spite of its great length, and the length of its talk page, does not at any time explain the so-called "paradox" in simple terms.
The "paradox" arises because correlation does not imply causation.  So when the "paradox" claims that "it can't be wrong to say that if that particular person is drinking, then everyone in the pub is drinking" it can be wrong - simply put, it would be more accurate to say "if that particular person is drinking, and everyone in the pub is drinking".  Remove the requirement that any given drinker "causes" the rest of the pub to drink, and the "paradox" disappears.  87.194.209.163 (talk) 04:39, 22 November 2012 (UTC)
replies: text: No that's not it. An easier way to understand it is to rephrase it as: "there exists a person such that either: (1) he is not drinking [period] or (2) he is drinking and everyone else is drinking."  That way it seems trivially true. Tijfo098 (talk) 04:56, 22 November 2012 (UTC)
replies: text: I think we should be open to the possibility that there are multiple ways of understanding this paradox. Personally, I think it makes more sense as a "paradox" when spoken in natural language, and I don't really like the inclusion of logic-type in the lede (defined as the "actual theorem".) I fear that these symbols will turn off the average reader, and I don't think they particularly convey the fun paradox that applies to the statement in question. More broadly, I don't agree with the interpretation currently presented that the theorem is the "actual" theorem behind the statement. groupuscule (talk) 15:05, 23 November 2012 (UTC)
replies: text:  Unfortunately all the sources discussing this topic are discussing it as a theorem in FOL. You're welcome to try find other sources, but I rather doubt you'll turn up any—I've look for such sources myself, unsuccessfully. As for the rest of your proposal, see WP:OR. Tijfo098 (talk) 06:26, 24 November 2012 (UTC)
thread_title:  Beginners - read this first! 

message: Smullyan's Original Drinker Principle:
"[T]here exists someone such that whenever he (or she) drinks, everybody drinks."
--R. Smullyan, "What is the name of this book?" p. 209 
We begin by proving, for arbitrary predicate D and proposition Q:
\exists x: [D(x) \implies Q] 
We make the very reasonable assumption that somewhere there exists even a single non-drinker. Then we would have:
\exists x: \neg D(x).
Applying existential specification, we can infer:
\neg D(a)
For any proposition Q whatsoever, be it true or false, we also then have:
\neg D(a) \lor Q.
Applying the definition of logical implication, we have:
D(a) \implies Q.
Generalizing, we conclude: 
\exists x: [D(x) \implies Q]
Note again that Q need not even be true. It could be false or contradictory. Clearly, such constructs have very limited application in the "real world." Here, anything goes. 
Example 1
As in Smullyan's original "Drinker Principle," we could set:
Q~\equiv ~ \forall y: D(y) 
Here, Q contradicts our initial assumption!
As in Smullyan's original "Drinker Principle," we would have:
\exists x: [D(x) \implies \forall y: D(y)]
Example 2
Alternatively, as in other popular versions of DP which explicitly mention a pub, we could set:
 Q~ \equiv ~ \forall y: [P(y) \implies  D(y)]  
Where " P(y)" means y is in the pub.
Then we would have:
\exists x: [D(x) \implies \forall y: [P(y) \implies  D(y)]]
------------------
A Set-theoretic Variation of DP
If we assume the non-existence of a universal set, then every set must exclude something. Thus, similar to the above argument, for any set D and proposition Q, we must have: 
\exists x:[x\in D \implies Q] 
--Danchristensen (talk) 17:27, 17 April 2024 (UTC)
thread_title:  Trivial resolution if there exists even a single non-drinker 

message: In the current article the 2nd paragraph includes a claim unsupported by any reference: "The formal statement of the theorem is timeless, eliminating the second objection because the person the statement holds true for at one instant is not necessarily the same person it holds true for at any other instant."
 
There are obvious difficulties with a claim that the formal statement applies to the real physical world for an instant of time. One is that according to the general theory of relativity, there is no such thing as a single instant that applies over a region of space. Another is defining when a person is drinking and when they are not. How many molecules of the drink have to pass the lips of a person before they are considered to be drinking? What precise line on the constantly moving lips is to be the determining line? What are the exact locations of the moving drink molecules that exhibit quantum behavior? The notion of the formal statement being true in the real world at an instant in time is ludicrous. --Jamesrmeyer (talk) 20:51, 24 November 2019 (UTC)
thread_title:  Simply because the formal statement is timeless does not mean it can apply to the real world