![]() | This article is rated Start-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | ||||||||||||||||||||||
|
I'm a little confused as to why Carroll considered this a paradox. In his notes he says
The paradox is a very real difficulty in the Theory of Hypotheticals....Are two Hypotheticals, of the forms If A then B and If A then not-B, compatible?
Of course they are! Why should (P > Q) and (P > ~Q) be contradictory? The contradiction of (P > Q) is ~(P > Q), which resolves to (P . ~Q), which is not the same as (P > ~Q) at all. So my question is: were the rules of logic different in Carroll's day? Was it not considered a standard law that from a falsity you can prove anything (F > R)? And if so, should we make it clear that he made the paradox under different logical rules? —
Asbestos |
Talk
(RFC)
14:39, 19 May 2006 (UTC)
I'm by no means an expert, so I can't be sure this wasn't covered by Radgeek, but I have a very sort of layman solution to the problem (which, most likely, does not go to the heart of the issue, but at least proves that a better wording of the problem is, in my opinion, necessary):
The conditions state that it is necessary that A) Someone is in the shop at any given time B) Allen can not walk to the shop (and be in the shop) without Brown
Based on the above conditions, it's possible to conclude that Allen never goes to the barber shop at all; IF he did, he would go with Brown. However, in this scenario, only Carr and Brown can switch off shifts (or work together) and none of the imposed conditions are false.
Even so, if we add the condition that C) Allen must work sometimes, there is nothing in the conditions that prevents all three barbers from being in the shop at some time, allowing Carr to switch off with Brown and Allen, who work always as a team. (the conditions state that the 3 barbers are not all always in the shop; that does not mean that they can not all 3 be together in the shop for some time to go on/off shifts)
Anyways, just a small contribution by an inquisitive layman. I can't guarantee that I'm right, and I'm fairly certain that this "solution" does not assess the actual problem proposed; however, it seems that for a philosophical paradox to exist, the conditions have to be spelled out precisely to the letter, and in this case, I believe there has been an ommission.
-Wasted —The preceding unsigned comment was added by 70.71.61.212 ( talk) 07:29, 18 February 2007 (UTC).
I think my logic really stinks here. How can we adequately express the idea that the sick guy A has to be in the shop at all times unless B is with him? Since it was stated all 3 are never there all at once (put potentially, 2 could be) it means that when A is out, C is in. It's not really a conflict to say 'if A is out' because why should he be? Of course, if the brothers had seen A out and about they should assume B was out too (and probably should have seen him with him) so it would be right to assume C was in. Man, Carrol is complex, I need to get some foundations and come back to these semantics. Tyciol 20:46, 7 March 2007 (UTC)
So from what I can tell the hypothesis referenced in the above two comments that someone must be in the shop at all times is misleading. The hypothesis listed in the article, "since someone must be in the shop for it to be open" is correct and the subtle distinction clears the air of all the switching off of shifts problems. Another point of confusion is in the actual article: it took a few minutes for me to realize that the fact "that Allen's recently been very ill" was not a hypothesis of the sort 'Allen cannot be the one manning the shop at the time of the story since he is ill.' I think it would be simpler to omit that sentence and just state that Allen never leaves the house without Brown, maybe playing up his nervousness and using the word 'fever' as Carroll did.
That brings me to propose major changes to this article. I'd like to add a section Simplifying Carroll's story, with terminology and a list of propositions like A - Allen is in. I think the two main points (that the contradiction that Joe says clinches his proof is between (~A > B) and (~A > ~B) and that this was Carroll's main dilemma, which has now been cleared up by the law of implication the Russell championed) are not made clear enough. Basically these changes won't affect the spirit of the current article, just incorporate the well written talk posts and make it easier to read for nonexperts (they shouldn't even have to read the original article linked to in the lede). - Callowschoolboy
The problem seems to be a misunderstanding the entire story! Lewis Carroll put forth a symbolic logic argument that is false in reality. That shouldn't be possible--symbolic logic should be perfect logic. Whether the theory of hypotheticals in use in 1894 was thusly flawed or whether Carroll misunderstood the application of the theory is a question better left for historical research, but this paradox doesn't pass muster by proper application of the theory by modern logicians.
Article needs to be scrapped and re-written. CAHeyden ( talk) 19:31, 22 May 2010 (UTC)
I like the shorthand ya'll use here in the Talk, and it has historical support in Principia and all, but now that I've finally tracked down the more formal symbolism and how to implement it I think I'll update the article to use it.
-
Callowschoolboy
17:22, 6 July 2007 (UTC)
Ok so we have the rules set up. Here is where I'm hitting the problem and I think this is what Carroll was trying to prove. So far we have that If A is out, B must be in to have the shop open. It is then reasoned that if A is out than B is out because A won't leave without B. The paradox that I think is suggested is that the rules as given say that if A is out of the shop then B must be there and at the same time not be there. However that's not a paradox at all because B does not have to be out of the store simply because A is not there. There isn't a rule saying that A can't be left at home by himself. If C is not at the shop and neither is A, there is no reason that B can't be at the shop alone. H2P ( Yell at me for what I've done) 07:10, 10 July 2007 (UTC)
Ok I see the problem. "If C is out" does NOT cause a paradox because there is nothing stopping A from being in. The paradox comes from "If C is out and If A is out" causing two solutions. The first rule states that in this instance, B must be in to run the shop. The second rule states that B must be out with A. B can't be both in and out and thus the paradox. However, there aren't any rules stating that when C is out, A must also be out. Actually, a new rule is formed by this paradox: If C is out A is in.
H2P (
Yell at me for
what I've done)
08:17, 10 July 2007 (UTC)
Your first comment was dead on (not saying the second isn't but this whole thing is so confusing that I have to read everything several times ;). And I think in your second comment your basically saying that although the confusion between Allen being out of the shop and Allen being out of his house, "out and about" so to speak, also complicates the discussion about whether the "hypotheticals" that Uncle Jim cooks up are contradictory. I didn't help matters I guess when I stated Uncle Joe's second axiom as "Allen goes nowhere without Brown." I was trying to strengthen the narrative foundation for the logic question Carroll poses. You're right that even taking this strong statement it could be that Allen stays in one place while Brown goes elsewhere (eg A and B go together to the shop, then A stays in the shop while B galavants about town). Although Allen would be stranded, this would allow a narrative solution, rather than addressing the logical conundrum. In that sense it's not helpful to the reader to have this loophole in the story. Any suggestions how to close it? - Callowschoolboy 13:34, 11 July 2007 (UTC)
I would like to go back to something more like what was there before, in order to not skip steps. This is an extremely slippery subject and non-logicians might not easily see that (C ∨ A ∨ B) ∧ (C ∨ A ∨ ¬B) can be a true statement if Allen is in, nor would they seperate the issues we enumerated above from the underlying logic.
-
Callowschoolboy
14:45, 11 July 2007 (UTC)
@ HeirToPendragon and Callowschoolboy: H2P I agree you've hit the nail on the head there ("If C is out A is in"). I'm trying to clear up some of the confusions here, see new section below. Comments welcome. FrankP ( talk) 20:33, 16 December 2019 (UTC)
I think the next step should be to identify the most important points (off the top of my head: LC wrote a story knowing that it was a paradox but intuitively shouldn't be with Jim as a straw man, underlying the story is a pure logic scenario, that scenario is not a paradox under modern logic which reconciles our intuition, etc).
Another list to make and emphasize (at the cost of any unimportant pieces of the article) would be points of confusion, closely related to that initial list. One of the big ones that just came up has to do with commonsense solutions such as "Why couldn't Allen be in, as long as Brown walks him to the shop?"
I started trying to summarize these sorts of objections, when I realized how big a problem my strong statement of Axiom 1 is. Using the result (C ∨ A ∨ B) ∧ (C ∨ A ∨ ¬B), which is really just (Axiom 1) ∧ (C ∨ A ∨ ¬B), I broke down the truth values of some possible commonsense solutions:
In the shop | Not in the shop | (A ∨ ¬B ∨ C) | Notes |
---|---|---|---|
Allen | Brown, Carr | (T ∨ ¬F ∨ F) ≡ T | B picked A up at A's house, they went to the shop together but B left at some point, A not compelled to go with him (argument against strong Axiom 2) |
Brown | Allen, Carr | (F ∨ ¬T ∨ F) ≡ F | Again run into problem with the strong statement of Ax2, but what if Allen stays at his home, Brown goes in, and who knows what Carr does. |
Allen, Carr | Brown | (T ∨ ¬F ∨ T) ≡ T | All this requires is that, as above, Brown escort Allen to the shop but not necessarily man the shop with him (and Carr in this case). |
What if both A and B are in the shop and C is out? I fail to see how this is a contradiction to the initial conditions. -- 86.127.22.71 12:30, 2 August 2007 (UTC)
It would be nice if the article made it clearer why the paradox is not really a paradox. Rather than using paragraphs of symbols, we laymen would prefer an explanation in English. If the shop is open, then at least one of A, B and C must be in. A never goes anywhere without B, so it follows that if A is in then B must be in, and it follows that if A is out then B must be out. So the only possible outcomes from these conditions are (1) The shop is shut; (2) all three barbers are in; (3) C is in; (4) A and B are in. That's as complicated as it gets, as far as I'm concerned. In the story the shop is open, so that rules out (1) as an outcome. According to this, it's never going to happen that C is out and A is out, so "if C is out and A is out..." is a false premise. I have made it simple and not used any symbols, so does that mean I can't contribute to the debate? Brequinda 13:44, 27 August 2007 (UTC)
I agree with Brequinda's analysis. There are only 8 possibilities when you have 3 binary variables: 1 A.B.C -- all in 2 A.B.notC -- A.B are in 3 A.notB.C 4 notA.B.C 5 A.notB.notC 6 notA.B.notC 7 notA.notb.C -- neither A+B is in 8 notA.notB.notC -- none in, shop is closed
I can't understand why this posed a problem. It is either a simple problem in combinatorics or Boolean Algebra. Perhaps Boole's work hadn't percolated very far at that point? Diakron99 20:05, 13 November 2007 (UTC)
Either it is a misstatement of the conditions, or just plain wrong. It seems clear that Allen and Brown can be in the store, without Carr. So the assertion that Carr MUST be there seems false. Am I missing something? —Preceding unsigned comment added by Bigmac31 ( talk • contribs) 20:39, 4 August 2008 (UTC)
From a layman's point of view:
I read Carroll's original text given by the link, and I think, Carroll wanted to arrive at something else: I surmise he wanted to question the validity of the statements ("Are these really valid sets of rules?"), or whether one could arrive at an incoherent system though building it up from individually sensible rules -- but I don't fully grasp what he was up to.
Someone more knowledgeable out there to fill in the gaps...? -- Syzygy ( talk) 10:53, 25 August 2010 (UTC) (edited for sig)
I have take this from the article main page. Firstly, it's OR. (Please, no discussions about "it's so obviously logical, it don't need references".) Secondly, and more importantly, IMHO it is beside the point.
Obviously, the "paradox" at its face value can be resolved with no big problem. But if this was the case, nobody would seriously consider it a philosophical problem anymore. I feel that Carroll was up to something different, namely the inconsistencies created by a consistent set of rules, but I fail to graps what he exactly wanted to say. Someone more knowledgable would have to step in here. -- Syzygy ( talk) 08:51, 27 September 2010 (UTC)
But if this was the case, nobody would seriously consider it a philosophical problem anymore. -- That's like saying that, if Hitler wasn't onto something, there wouldn't be any neo-Nazis. People with "philosophical problems" with material implication end up in the same soup as Carroll ... suggesting that the problems they see are illusory. Much as people might feel uncomfortable with the assertion that "If squares are round then my mother is a tomato" is true, that's the only reasonable and consistent way to view it. -- 71.102.133.72 ( talk) 07:11, 15 September 2014 (UTC)
Condition 1: There must be at least one person home as Joe and Jim clearly see that the shop is open.
Condition 2: Allen is eccentric and cannot leave the house without Brown (He can't leave the house with just Carr, apparently).
Here are all the permutations and whether they are logically sound or not.
In the shop | Not in the shop | Logically sound or unsound | Reasoning |
---|---|---|---|
Allen, Brown, Carr | Unsound | Violates condition 1 | |
Allen | Brown, Carr* | Sound | Meets both conditions |
Brown | Allen, Carr | Unsound | Violates condition 2 |
Carr | Allen, Brown | Sound | Meets both conditions |
Allen, Brown | Carr* | Sound | Meets both conditions |
Allen, Carr | Brown | Sound | Meets both conditions |
Brown, Carr | Allen | Unsound | Violates condition 2 |
Allen, Brown, Carr | Sound | Meets both conditions |
As one can see, there are two ways for the shop to be open without Carr being home [¬C ⇒ (A ∧ ¬B) ∨ (A ∧ B)], therefore, the logic that Uncle Joe used to assume that Carr must be home is erroneous, therefore there is no 'contradiction' and therefore no paradox. -- HakuGaara ( talk) 14:58, 29 June 2011 (UTC)
It's simple. This is NO paradox. You can simply have Allen and Brown together in the shop. — Preceding unsigned comment added by UltimateDragonMaster ( talk • contribs) 20:02, 6 July 2011 (UTC)
You don't even need to have Allen and Brown together in the shop. Allen never leaves the shop without Brown. It doesn't say anything about Brown never leaving the shop without Allen. So either Carr is there, Allen is there, Carr and Allen are there or Allen and Brown are there. Brown is the only one who cannot be by himself, because that would require Allen to be out, which he is not keen on. So there is no paradox, just a flaw in logic. -- Jahkayhla ( talk) 00:43, 31 January 2012 (UTC)
For more than four years now, this talk page has pointed out why the topic of this article, as presented, is no paradox at all; and this can be easily seen by simply considering all seven options. It is, in fact, a rather obvious mistake: the fallacy of False dilemma, known since long before the 1890s. The only paradox I see in all of this is how anyone can rate this article as Mid-Importance: "The article covers a topic that has a strong but not vital role in the history of philosophy". As it stands it is simply a brain fart that fails the test of notability.
It is possible that Carroll was well aware of his mistake when writing the original article and that his point was something else entirely. If so, PLEASE TELL THE READER WHAT THE POINT IS. It's also possible that the Barbershop Paradox is commonly cited as a famous mistake akin to miasmas or the four classical elements, but if so PLEASE TELL THE READER WHY.
I appreceate that work has been put into this article, but to be honest, I think it lacks crucial parts to make it notable. -- Spearman ( talk) 12:57, 14 February 2012 (UTC)
As far as I can see, these are the possible scenarios that do not contradict the rules:
At least one person is in the shop; and, If A is out, B is out
Shop: A, B, C Out:
Shop:A, B Out: C
Shop: A, C Out: B
Shop: A Out: B, C
Shop: C Out: A, B
It is very easy to see that in two of these scenarios, C is out. So where is the paradox? In reference to what someone said before, it says nowhere that they cannot all be in the shop at once. Which also means if there was ever a problem with shifts, C could go to the shop from wherever he was, followed by A and B going to the shop from wherever they are (note that someone is of course already in the shop), followed by any of the above scenarios occurring by having a person or people leave as necessary. I cannot find how this is a paradox even slightly. It seems to be based on the assumption that A and B are NOT together in the shop, which is nowhere in the rules. — Preceding unsigned comment added by 74.15.25.97 ( talk) 06:54, 9 March 2012 (UTC)
It's obvious that ¬C => A ∧ B (¬C => A ∨ B and A <=> B), so based on the hypothesis ¬C, saying ¬A is a contradiction by itself and can't be used in the implication. Or, in simpler terms, if C is out, then both A and B are in (because they can't be separate), so if we suppose that C is out, it is illogical to suppose that A could be out. This article should either be deleted, or it should be clearly stated that his paradox is false. 184.144.193.166 ( talk) 03:09, 11 November 2014 (UTC)
Hello all, I've followed a link to here from Project Mathematics. There's obviously been a bit of frustration over this article, but I do think I can see a way forward. So I'm first going to address some of the important points made above, and explain how I see the problem, before ( boldly) having a go at the article to see if I can improve its clarity and coverage.
FrankP ( talk) 19:20, 16 December 2019 (UTC)
The claim that it is universally true that, if Allen is not in then Brown is not in, is incorrect! No constraint was placed on Brown, only Allen. Allen can't go anywhere without Brown also going. But Brown can go somewhere without Allen going. Brown can go to the shop without Allen. But Allen can't go to the shop without Brown. So the possibilities are:
1. Allen and Brown go to the shop and they are both in.
2. Brown goes to the shop while Allen stays home and only Brown is in the shop.
3. Neither Brown or Allen go to the shop and neither is in.
So, has the fourth possibility, Allen is in while brown is out, been eliminated and is not possible?
No it hasn't because it's more complex and further things can happen.
Allen and Brown can both go to the shop. Since Brown is not constrained he can then leave the shop without Allen. Now only Allen is in the shop.
— Preceding unsigned comment added by 2603:3024:204:B00:E0E1:5AAD:4A60:8B50 ( talk) 17:11, 6 March 2020 (UTC)
![]() | This article is rated Start-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | ||||||||||||||||||||||
|
I'm a little confused as to why Carroll considered this a paradox. In his notes he says
The paradox is a very real difficulty in the Theory of Hypotheticals....Are two Hypotheticals, of the forms If A then B and If A then not-B, compatible?
Of course they are! Why should (P > Q) and (P > ~Q) be contradictory? The contradiction of (P > Q) is ~(P > Q), which resolves to (P . ~Q), which is not the same as (P > ~Q) at all. So my question is: were the rules of logic different in Carroll's day? Was it not considered a standard law that from a falsity you can prove anything (F > R)? And if so, should we make it clear that he made the paradox under different logical rules? —
Asbestos |
Talk
(RFC)
14:39, 19 May 2006 (UTC)
I'm by no means an expert, so I can't be sure this wasn't covered by Radgeek, but I have a very sort of layman solution to the problem (which, most likely, does not go to the heart of the issue, but at least proves that a better wording of the problem is, in my opinion, necessary):
The conditions state that it is necessary that A) Someone is in the shop at any given time B) Allen can not walk to the shop (and be in the shop) without Brown
Based on the above conditions, it's possible to conclude that Allen never goes to the barber shop at all; IF he did, he would go with Brown. However, in this scenario, only Carr and Brown can switch off shifts (or work together) and none of the imposed conditions are false.
Even so, if we add the condition that C) Allen must work sometimes, there is nothing in the conditions that prevents all three barbers from being in the shop at some time, allowing Carr to switch off with Brown and Allen, who work always as a team. (the conditions state that the 3 barbers are not all always in the shop; that does not mean that they can not all 3 be together in the shop for some time to go on/off shifts)
Anyways, just a small contribution by an inquisitive layman. I can't guarantee that I'm right, and I'm fairly certain that this "solution" does not assess the actual problem proposed; however, it seems that for a philosophical paradox to exist, the conditions have to be spelled out precisely to the letter, and in this case, I believe there has been an ommission.
-Wasted —The preceding unsigned comment was added by 70.71.61.212 ( talk) 07:29, 18 February 2007 (UTC).
I think my logic really stinks here. How can we adequately express the idea that the sick guy A has to be in the shop at all times unless B is with him? Since it was stated all 3 are never there all at once (put potentially, 2 could be) it means that when A is out, C is in. It's not really a conflict to say 'if A is out' because why should he be? Of course, if the brothers had seen A out and about they should assume B was out too (and probably should have seen him with him) so it would be right to assume C was in. Man, Carrol is complex, I need to get some foundations and come back to these semantics. Tyciol 20:46, 7 March 2007 (UTC)
So from what I can tell the hypothesis referenced in the above two comments that someone must be in the shop at all times is misleading. The hypothesis listed in the article, "since someone must be in the shop for it to be open" is correct and the subtle distinction clears the air of all the switching off of shifts problems. Another point of confusion is in the actual article: it took a few minutes for me to realize that the fact "that Allen's recently been very ill" was not a hypothesis of the sort 'Allen cannot be the one manning the shop at the time of the story since he is ill.' I think it would be simpler to omit that sentence and just state that Allen never leaves the house without Brown, maybe playing up his nervousness and using the word 'fever' as Carroll did.
That brings me to propose major changes to this article. I'd like to add a section Simplifying Carroll's story, with terminology and a list of propositions like A - Allen is in. I think the two main points (that the contradiction that Joe says clinches his proof is between (~A > B) and (~A > ~B) and that this was Carroll's main dilemma, which has now been cleared up by the law of implication the Russell championed) are not made clear enough. Basically these changes won't affect the spirit of the current article, just incorporate the well written talk posts and make it easier to read for nonexperts (they shouldn't even have to read the original article linked to in the lede). - Callowschoolboy
The problem seems to be a misunderstanding the entire story! Lewis Carroll put forth a symbolic logic argument that is false in reality. That shouldn't be possible--symbolic logic should be perfect logic. Whether the theory of hypotheticals in use in 1894 was thusly flawed or whether Carroll misunderstood the application of the theory is a question better left for historical research, but this paradox doesn't pass muster by proper application of the theory by modern logicians.
Article needs to be scrapped and re-written. CAHeyden ( talk) 19:31, 22 May 2010 (UTC)
I like the shorthand ya'll use here in the Talk, and it has historical support in Principia and all, but now that I've finally tracked down the more formal symbolism and how to implement it I think I'll update the article to use it.
-
Callowschoolboy
17:22, 6 July 2007 (UTC)
Ok so we have the rules set up. Here is where I'm hitting the problem and I think this is what Carroll was trying to prove. So far we have that If A is out, B must be in to have the shop open. It is then reasoned that if A is out than B is out because A won't leave without B. The paradox that I think is suggested is that the rules as given say that if A is out of the shop then B must be there and at the same time not be there. However that's not a paradox at all because B does not have to be out of the store simply because A is not there. There isn't a rule saying that A can't be left at home by himself. If C is not at the shop and neither is A, there is no reason that B can't be at the shop alone. H2P ( Yell at me for what I've done) 07:10, 10 July 2007 (UTC)
Ok I see the problem. "If C is out" does NOT cause a paradox because there is nothing stopping A from being in. The paradox comes from "If C is out and If A is out" causing two solutions. The first rule states that in this instance, B must be in to run the shop. The second rule states that B must be out with A. B can't be both in and out and thus the paradox. However, there aren't any rules stating that when C is out, A must also be out. Actually, a new rule is formed by this paradox: If C is out A is in.
H2P (
Yell at me for
what I've done)
08:17, 10 July 2007 (UTC)
Your first comment was dead on (not saying the second isn't but this whole thing is so confusing that I have to read everything several times ;). And I think in your second comment your basically saying that although the confusion between Allen being out of the shop and Allen being out of his house, "out and about" so to speak, also complicates the discussion about whether the "hypotheticals" that Uncle Jim cooks up are contradictory. I didn't help matters I guess when I stated Uncle Joe's second axiom as "Allen goes nowhere without Brown." I was trying to strengthen the narrative foundation for the logic question Carroll poses. You're right that even taking this strong statement it could be that Allen stays in one place while Brown goes elsewhere (eg A and B go together to the shop, then A stays in the shop while B galavants about town). Although Allen would be stranded, this would allow a narrative solution, rather than addressing the logical conundrum. In that sense it's not helpful to the reader to have this loophole in the story. Any suggestions how to close it? - Callowschoolboy 13:34, 11 July 2007 (UTC)
I would like to go back to something more like what was there before, in order to not skip steps. This is an extremely slippery subject and non-logicians might not easily see that (C ∨ A ∨ B) ∧ (C ∨ A ∨ ¬B) can be a true statement if Allen is in, nor would they seperate the issues we enumerated above from the underlying logic.
-
Callowschoolboy
14:45, 11 July 2007 (UTC)
@ HeirToPendragon and Callowschoolboy: H2P I agree you've hit the nail on the head there ("If C is out A is in"). I'm trying to clear up some of the confusions here, see new section below. Comments welcome. FrankP ( talk) 20:33, 16 December 2019 (UTC)
I think the next step should be to identify the most important points (off the top of my head: LC wrote a story knowing that it was a paradox but intuitively shouldn't be with Jim as a straw man, underlying the story is a pure logic scenario, that scenario is not a paradox under modern logic which reconciles our intuition, etc).
Another list to make and emphasize (at the cost of any unimportant pieces of the article) would be points of confusion, closely related to that initial list. One of the big ones that just came up has to do with commonsense solutions such as "Why couldn't Allen be in, as long as Brown walks him to the shop?"
I started trying to summarize these sorts of objections, when I realized how big a problem my strong statement of Axiom 1 is. Using the result (C ∨ A ∨ B) ∧ (C ∨ A ∨ ¬B), which is really just (Axiom 1) ∧ (C ∨ A ∨ ¬B), I broke down the truth values of some possible commonsense solutions:
In the shop | Not in the shop | (A ∨ ¬B ∨ C) | Notes |
---|---|---|---|
Allen | Brown, Carr | (T ∨ ¬F ∨ F) ≡ T | B picked A up at A's house, they went to the shop together but B left at some point, A not compelled to go with him (argument against strong Axiom 2) |
Brown | Allen, Carr | (F ∨ ¬T ∨ F) ≡ F | Again run into problem with the strong statement of Ax2, but what if Allen stays at his home, Brown goes in, and who knows what Carr does. |
Allen, Carr | Brown | (T ∨ ¬F ∨ T) ≡ T | All this requires is that, as above, Brown escort Allen to the shop but not necessarily man the shop with him (and Carr in this case). |
What if both A and B are in the shop and C is out? I fail to see how this is a contradiction to the initial conditions. -- 86.127.22.71 12:30, 2 August 2007 (UTC)
It would be nice if the article made it clearer why the paradox is not really a paradox. Rather than using paragraphs of symbols, we laymen would prefer an explanation in English. If the shop is open, then at least one of A, B and C must be in. A never goes anywhere without B, so it follows that if A is in then B must be in, and it follows that if A is out then B must be out. So the only possible outcomes from these conditions are (1) The shop is shut; (2) all three barbers are in; (3) C is in; (4) A and B are in. That's as complicated as it gets, as far as I'm concerned. In the story the shop is open, so that rules out (1) as an outcome. According to this, it's never going to happen that C is out and A is out, so "if C is out and A is out..." is a false premise. I have made it simple and not used any symbols, so does that mean I can't contribute to the debate? Brequinda 13:44, 27 August 2007 (UTC)
I agree with Brequinda's analysis. There are only 8 possibilities when you have 3 binary variables: 1 A.B.C -- all in 2 A.B.notC -- A.B are in 3 A.notB.C 4 notA.B.C 5 A.notB.notC 6 notA.B.notC 7 notA.notb.C -- neither A+B is in 8 notA.notB.notC -- none in, shop is closed
I can't understand why this posed a problem. It is either a simple problem in combinatorics or Boolean Algebra. Perhaps Boole's work hadn't percolated very far at that point? Diakron99 20:05, 13 November 2007 (UTC)
Either it is a misstatement of the conditions, or just plain wrong. It seems clear that Allen and Brown can be in the store, without Carr. So the assertion that Carr MUST be there seems false. Am I missing something? —Preceding unsigned comment added by Bigmac31 ( talk • contribs) 20:39, 4 August 2008 (UTC)
From a layman's point of view:
I read Carroll's original text given by the link, and I think, Carroll wanted to arrive at something else: I surmise he wanted to question the validity of the statements ("Are these really valid sets of rules?"), or whether one could arrive at an incoherent system though building it up from individually sensible rules -- but I don't fully grasp what he was up to.
Someone more knowledgeable out there to fill in the gaps...? -- Syzygy ( talk) 10:53, 25 August 2010 (UTC) (edited for sig)
I have take this from the article main page. Firstly, it's OR. (Please, no discussions about "it's so obviously logical, it don't need references".) Secondly, and more importantly, IMHO it is beside the point.
Obviously, the "paradox" at its face value can be resolved with no big problem. But if this was the case, nobody would seriously consider it a philosophical problem anymore. I feel that Carroll was up to something different, namely the inconsistencies created by a consistent set of rules, but I fail to graps what he exactly wanted to say. Someone more knowledgable would have to step in here. -- Syzygy ( talk) 08:51, 27 September 2010 (UTC)
But if this was the case, nobody would seriously consider it a philosophical problem anymore. -- That's like saying that, if Hitler wasn't onto something, there wouldn't be any neo-Nazis. People with "philosophical problems" with material implication end up in the same soup as Carroll ... suggesting that the problems they see are illusory. Much as people might feel uncomfortable with the assertion that "If squares are round then my mother is a tomato" is true, that's the only reasonable and consistent way to view it. -- 71.102.133.72 ( talk) 07:11, 15 September 2014 (UTC)
Condition 1: There must be at least one person home as Joe and Jim clearly see that the shop is open.
Condition 2: Allen is eccentric and cannot leave the house without Brown (He can't leave the house with just Carr, apparently).
Here are all the permutations and whether they are logically sound or not.
In the shop | Not in the shop | Logically sound or unsound | Reasoning |
---|---|---|---|
Allen, Brown, Carr | Unsound | Violates condition 1 | |
Allen | Brown, Carr* | Sound | Meets both conditions |
Brown | Allen, Carr | Unsound | Violates condition 2 |
Carr | Allen, Brown | Sound | Meets both conditions |
Allen, Brown | Carr* | Sound | Meets both conditions |
Allen, Carr | Brown | Sound | Meets both conditions |
Brown, Carr | Allen | Unsound | Violates condition 2 |
Allen, Brown, Carr | Sound | Meets both conditions |
As one can see, there are two ways for the shop to be open without Carr being home [¬C ⇒ (A ∧ ¬B) ∨ (A ∧ B)], therefore, the logic that Uncle Joe used to assume that Carr must be home is erroneous, therefore there is no 'contradiction' and therefore no paradox. -- HakuGaara ( talk) 14:58, 29 June 2011 (UTC)
It's simple. This is NO paradox. You can simply have Allen and Brown together in the shop. — Preceding unsigned comment added by UltimateDragonMaster ( talk • contribs) 20:02, 6 July 2011 (UTC)
You don't even need to have Allen and Brown together in the shop. Allen never leaves the shop without Brown. It doesn't say anything about Brown never leaving the shop without Allen. So either Carr is there, Allen is there, Carr and Allen are there or Allen and Brown are there. Brown is the only one who cannot be by himself, because that would require Allen to be out, which he is not keen on. So there is no paradox, just a flaw in logic. -- Jahkayhla ( talk) 00:43, 31 January 2012 (UTC)
For more than four years now, this talk page has pointed out why the topic of this article, as presented, is no paradox at all; and this can be easily seen by simply considering all seven options. It is, in fact, a rather obvious mistake: the fallacy of False dilemma, known since long before the 1890s. The only paradox I see in all of this is how anyone can rate this article as Mid-Importance: "The article covers a topic that has a strong but not vital role in the history of philosophy". As it stands it is simply a brain fart that fails the test of notability.
It is possible that Carroll was well aware of his mistake when writing the original article and that his point was something else entirely. If so, PLEASE TELL THE READER WHAT THE POINT IS. It's also possible that the Barbershop Paradox is commonly cited as a famous mistake akin to miasmas or the four classical elements, but if so PLEASE TELL THE READER WHY.
I appreceate that work has been put into this article, but to be honest, I think it lacks crucial parts to make it notable. -- Spearman ( talk) 12:57, 14 February 2012 (UTC)
As far as I can see, these are the possible scenarios that do not contradict the rules:
At least one person is in the shop; and, If A is out, B is out
Shop: A, B, C Out:
Shop:A, B Out: C
Shop: A, C Out: B
Shop: A Out: B, C
Shop: C Out: A, B
It is very easy to see that in two of these scenarios, C is out. So where is the paradox? In reference to what someone said before, it says nowhere that they cannot all be in the shop at once. Which also means if there was ever a problem with shifts, C could go to the shop from wherever he was, followed by A and B going to the shop from wherever they are (note that someone is of course already in the shop), followed by any of the above scenarios occurring by having a person or people leave as necessary. I cannot find how this is a paradox even slightly. It seems to be based on the assumption that A and B are NOT together in the shop, which is nowhere in the rules. — Preceding unsigned comment added by 74.15.25.97 ( talk) 06:54, 9 March 2012 (UTC)
It's obvious that ¬C => A ∧ B (¬C => A ∨ B and A <=> B), so based on the hypothesis ¬C, saying ¬A is a contradiction by itself and can't be used in the implication. Or, in simpler terms, if C is out, then both A and B are in (because they can't be separate), so if we suppose that C is out, it is illogical to suppose that A could be out. This article should either be deleted, or it should be clearly stated that his paradox is false. 184.144.193.166 ( talk) 03:09, 11 November 2014 (UTC)
Hello all, I've followed a link to here from Project Mathematics. There's obviously been a bit of frustration over this article, but I do think I can see a way forward. So I'm first going to address some of the important points made above, and explain how I see the problem, before ( boldly) having a go at the article to see if I can improve its clarity and coverage.
FrankP ( talk) 19:20, 16 December 2019 (UTC)
The claim that it is universally true that, if Allen is not in then Brown is not in, is incorrect! No constraint was placed on Brown, only Allen. Allen can't go anywhere without Brown also going. But Brown can go somewhere without Allen going. Brown can go to the shop without Allen. But Allen can't go to the shop without Brown. So the possibilities are:
1. Allen and Brown go to the shop and they are both in.
2. Brown goes to the shop while Allen stays home and only Brown is in the shop.
3. Neither Brown or Allen go to the shop and neither is in.
So, has the fourth possibility, Allen is in while brown is out, been eliminated and is not possible?
No it hasn't because it's more complex and further things can happen.
Allen and Brown can both go to the shop. Since Brown is not constrained he can then leave the shop without Allen. Now only Allen is in the shop.
— Preceding unsigned comment added by 2603:3024:204:B00:E0E1:5AAD:4A60:8B50 ( talk) 17:11, 6 March 2020 (UTC)