44

Suppose we have a line of people that starts with person #1 and goes for a (finite or infinite) number of people behind him/her, and this property holds for every person in the line:

If everyone in front of you is bald, then you are bald.

Without further assumptions, does this mean that the first person is necessarily bald? Does it say anything about the first person at all?

In my opinion, it means:

If there exist anyone in front of you and they're all bald, then you're bald.

Generally, for a statement that consists of a subject and a predicate, if the subject doesn't exist, then does the statement have a truth value?

I think there's a convention in math that if the subject doesn't exist, then the statement is right.

I don't have a problem with this convention (in the same way that I don't have a problem with the meaning of 'or' in math). My question is whether it's a clear logical implication of the facts, or we have to define the truth value for these subject-less statements.


Addendum:

You can read up on this matter here (too).

Færd
  • 579
  • 18
    https://en.wikipedia.org/wiki/Vacuous_truth – The Chaz 2.0 Feb 23 '16 at 18:17
  • 2
    This is called strong induction. – DanielV Feb 23 '16 at 18:27
  • 3
    Why don't they just touch their heads and find out... – Nikunj Feb 23 '16 at 18:29
  • 2
    If the sentence were a mathematical statement, the answer would be clear, yes, bald. However, it is not a mathematical statement. – André Nicolas Feb 23 '16 at 18:41
  • @AndréNicolas So, what now? Do you mean that in math, a statement whose subject doesn't exist is defined to be true? – Færd Feb 23 '16 at 18:58
  • 6
    @MJF In that case, a universal statement ('for all') is always true, and an existential ('there exists') statement is always false. By convention. https://en.wikipedia.org/wiki/Universal_quantification#The_empty_set and https://en.wikipedia.org/wiki/Existential_quantification#The_empty_set – Řídící Feb 23 '16 at 19:02
  • @MJF: For formal mathematical English, Kinnisal Mountain Chicken has summarized things very well. – André Nicolas Feb 23 '16 at 19:05
  • @KinnisalMountainChicken Would you mind posting that as an answer so I can accept it? :) – Færd Feb 23 '16 at 19:08
  • @KinnisalMountainChicken, I would encourage you to post an answer. – Barry Cipra Feb 23 '16 at 20:42
  • @Nikunj: Why find stuff out on your own when all the answers are in the Book? – mathreadler Feb 23 '16 at 20:44
  • I like The Chaz 2.0 comment, why don't you write an answer based on it? – mathreadler Feb 23 '16 at 21:00
  • 2
    Let's not rule out that there is no one in line. – TTT Feb 23 '16 at 22:22
  • 1
    @TTT: The problem statement does say the line "starts with person #1", so we're given the existence of at least one person in line. That might not be the case in the original problem the questioner was working on, but it's definitely the case in the problem as posted. – user2357112 Feb 23 '16 at 22:46
  • @user2357112 - you're right. Missed that. So then isn't this the same thing: "There is a line, and everyone in line is bald." – TTT Feb 23 '16 at 22:50
  • Some of these answers are writing as if you asked about the situation where no one is standing in front of the first person, but I don't see that anywhere in here. I don't see a question at all, actually. – DCShannon Feb 24 '16 at 02:07
  • 1
    @DanielV: even better, it's transfinite induction. – Steve Jessop Feb 24 '16 at 02:59
  • 2
    @DCShannon: There can't be anyone standing in front of the first person, by the definition of "first". If there was anyone in front of the first person, one of them would be first. Also, the question is in the title, as well as restated in the body. Look at the parts with the question marks. – user2357112 Feb 24 '16 at 03:26
  • 1
    @user2357112 Ah, I read the description of the line wrong. I imagined all the people in front of the first person, not behind them. That makes way more sense. – DCShannon Feb 24 '16 at 03:57
  • I believe this is automatically true if you're the only person in the bathroom and you're staring at yourself in the mirror. –  Feb 24 '16 at 08:24
  • 1
    If you're looking for more discussion about the issue, you may find the search term "existential import" helpful. – 2012rcampion Feb 24 '16 at 18:35
  • @DanielV it's strong induction without the base case. – asmeurer Feb 24 '16 at 19:31
  • 1
    @asmeurer Strong induction doesn't have a "base case" like weak induction does. Read the answers below to see why. – DanielV Feb 24 '16 at 21:31
  • @2012rcampion That was helpful indeed. Glad to know that this matter has been under serious study. – Færd Feb 25 '16 at 02:33
  • @2012rcampion I added your link in the body. – Færd Feb 25 '16 at 02:38
  • "In my opinion, it means" -- If that is your opinion of what it means, then it doesn't imply that the first person is bald, because the antecedent is false. And there is no objective matter of fact as to what it does mean, because it is an ambiguous sentence of the English language, not a formal statement in logic. – Jim Balter Feb 25 '16 at 05:55
  • "if the subject doesn't exist, then does the statement have a truth value?" -- it depends on the statement. "I think there's a convention in math that if the subject doesn't exist, then the statement is right." -- you are quite confused. If the antecedent of a material implication is false, then the implication is true. Your more general statement is nonsensical. – Jim Balter Feb 25 '16 at 05:58
  • "But what does a logical statement suggest beyond itself? " -- suggestion is a psychological, not logical, property. There are no intrinsic suggestions. – Jim Balter Feb 25 '16 at 06:04
  • @DanielV ah yes, you are right. – asmeurer Feb 26 '16 at 19:49
  • What so much interesting in it ?? – Archis Welankar Feb 28 '16 at 14:45

6 Answers6

44

You can see what's going on by reformulating the assumption in its equivalent contrapositive form:

If I'm not bald, then there is someone in front of me who is not bald.

Now the first person in line finds himself thinking, "There is no one in front of me. So it's not true that there is someone in front of me who is not bald. So it's not true that I'm not bald. So I must be bald!"

Barry Cipra
  • 81,321
  • 5
    Also "if there is nobody in front of you with hair, then you are bald." – Dan Brumleve Feb 23 '16 at 18:29
  • 1
    @DanBrumleve, nice alternative. – Barry Cipra Feb 23 '16 at 18:35
  • 9
    That's a good trick. But the correct way to reformulate it is: If I'm not bald, then it's not true that everyone in front of me is bald. So is it false for the first person? No, not false either, because there is no one in front of him. it simply doesn't have a truth value, I'd say. We have to define the value here. – Færd Feb 23 '16 at 18:36
  • 11
    @MJF, the logical negation of a "for all" statement is a "there exists" statement, i.e., $\lnot\forall x(B(x))\iff \exists x(\lnot B(x))$. – Barry Cipra Feb 23 '16 at 18:40
  • 1
    Barry, that is logically equivalent to what I'm questioning. It's not an answer. If there's no x, then we have to define what the negation of "for all x" means. Maybe I should read up on this matter through the links below my question and then continue the discussion... – Færd Feb 23 '16 at 18:49
  • 19
    There is a form of logic (at least in some philosophical logic system, or so I've heard) in which "every $X$ is $Y$" implies that there is at least one $X$. In that interpretation, if you're the first person in line, it's not true that everyone in front of you is bald. But in the mathematical interpretation as I learned it, if there is nobody in front of you, then everyone in front of you is bald. In fact, everyone in front of you is a pink eight-legged unicorn with purple polka dots. – David K Feb 23 '16 at 18:55
  • 1
    @DavidK In some logic systems, it is assumed that the universe isn't empty. From that, it follows that $\forall x ~:~ P(x)$ implies $\exists x ~:~ P(x)$. But there is no coherent logic system that assumes $\forall x \in S ~:~ P(x) \implies \exists x \in S ~:~ P(x)$ just for the reason that $S$ could be empty. – DanielV Feb 23 '16 at 18:57
  • 1
    @DanielV The "form of logic" I referred to is not any kind of modern mathematics I know of. I usually see it attributed to Aristotle, for example here: http://onemorebrown.com/2010/09/27/1310/ As far as I am concerned, $\forall x\in \emptyset : P(x)$ is true. – David K Feb 23 '16 at 19:10
  • @DanielV A non-empty domain of discourse is part of the standard semantics for classical first-order logic. My answer to another question, along with the comments on it, discuss the standard interpretation (non-empty domain of discourse, vacuous truth, etc.), as well as some of the alternatives, such as free logic. – Joshua Taylor Feb 23 '16 at 21:21
  • @DanielV In fact, a variant of the formula you present (depending on where exactly you put the brackets) is a theorem of classical first-order logic with equality. Suppose $\forall x P(x)$. Then by universal specialization ($\forall$-elim), infer that, for some constant $c$, $P(c)$. From $P(c)$, infer $\exists x P(x)$. Then $\forall x P(x) \to \exists x P(x)$. But you're right that once you add a guard, this doesn't hold: $(\forall x S(x) \to P(x)) \not\rightarrow (\exists x S(x) \wedge P(x))$. – Joshua Taylor Feb 23 '16 at 21:27
  • 1
    The truth of $\forall x \in \emptyset: P(x)$ is explained in the classes I'm aware of as being a result of $(\forall x \in \emptyset: P(x)) := (\forall x: x\in \emptyset\Rightarrow P(x))$. If someone accepts the principle of explosion, they should accept vacuous truths, assuming the language they are talking in is first-order logic. @MJF The semantics of logic used by most math people is well-defined, and takes care of cases like quantifiers over empty sets. – G. Bach Feb 24 '16 at 13:07
  • 1
    "it simply doesn't have a truth value, I'd say" -- you would say wrong. It's bizarre, or tragic, that this got 6 upvotes. – Jim Balter Feb 25 '16 at 06:12
  • 1
    @JoshuaTaylor: your proof assumes that there is always some constant or variable available. This holds in some presentations of predicate logic, but not all. For instance, in the categorical logic tradition, it is more common to always work in a “context”, i.e. the finite set of variables that are in scope. Then in context of at least one variable, universal instantiation is valid (by your argument, using a variable). But in the empty context (i.e. considered as a closed formula), universal instantiation is not valid. Being used to this approach, I find it very clean and more natural. – Peter LeFanu Lumsdaine Feb 25 '16 at 14:26
  • @PeterLeFanuLumsdaine Yes, the principal of existential import is something that's much easier to include or exclude in categorical logic. In syntactic/"traditional" approaches, there are always some constant symbols or variables around. (I've mentioned constant symbols, but variables would work as well; because an interpretation function maps every constant symbol and variable to a value in the universe of discourse. Even if there are no constants, there are still variables (or else we're essentially in propositional logic).) – Joshua Taylor Feb 25 '16 at 14:32
  • 1
    @JoshuaTaylor: it’s not at all difficult to incorporate this into standard syntax; you just have to add contexts. In the sequent calculus form of natural deduction, for instance, you just change the basic form of sequents from $\varphi_1, \ldots, \varphi_n \vdash \psi\ $ to $\ \varphi_1, \ldots, \varphi_n \vdash_{x_1,\ldots,x_m} \psi$; essentially nothing else changes, and the quantifier rules now involve a change in the context. This has the side benefit of explaining very clearly the variable-freeness conditions in the quantifier rules, which otherwise are a bit tricky to remember. – Peter LeFanu Lumsdaine Feb 25 '16 at 14:41
  • @PeterLeFanuLumsdaine In the "typical" semantics, the truth of a sentence is defined with respect to an interpretation function. The interpretation function constants and variables, relation symbols, function symbols, etc., to domain elements, relations, functions, etc. Then, $(\forall x) P(x)$ is true in an interpretation $\mathcal I$ if $P(x)$ is true in every interpretation $\mathcal I'$ that differs from $\mathcal I$ in at most the mapping of $x$. $\mathcal I$ and $\mathcal I'$ have to map $x$ to something, though, which requires a non-empty domain. – Joshua Taylor Feb 25 '16 at 15:08
  • 1
    @JoshuaTaylor: yes, agreed. The semantics too must be adapted to incorporate contexts, but again, the change is not large. Instead of including values of variables as part of the interpretation from the start, one defines an interpretation as just giving interpretation of constants, functions, and predicate symbols. Then one interprets a formula $\varphi$ in context $X = {x_1,\ldots,x_m}$ as a function from values for these variables to truth values; equivalently, as a subset of $A^X$ (where $A$ is the domain of the interpretation). [cont’d] – Peter LeFanu Lumsdaine Feb 25 '16 at 16:08
  • 1
    Again, this seems quite natural: when we speak of “the standard interpretation of PA in $\mathbb{N}$”, for instance, we have specific interpretations of the function and predicate symbols in mind, but not any particular values for variables. // I can go into more detail if you’d like, but we should probably not continue here; you can find my email address fairly easily with google. – Peter LeFanu Lumsdaine Feb 25 '16 at 16:08
42

Mathematical logic defines a statement about all elements of an empty set to be true. This is called vacuous truth. It may be somewhat confusing since it doesn't agree with common everyday usage, where making a statement tends to suggest that there is some object for which the statement actually holds (like the person in front of you in your example).

But it is exactly the right thing to do in a formal setup, for several reasons. One reason is that logical statements don't suggest anything: you must not assume any meaning in excess of what's stated explicitly. Another reason is that it makes several conversions possible without special cases. For example,

$$\forall x\in(A\cup B):P(x)\;\Leftrightarrow \forall x\in A:P(x)\;\wedge\;\forall x\in B:P(x)$$

holds even if $A$ (or $B$) happens to be the empty set. Another example is the conversion between universal and existential quantification Barry Cipra used:

$$\forall x\in A:\neg P(x)\;\Leftrightarrow \neg\exists x\in A:P(x)$$

If you are into programming, then the following pseudocode snippet may also help explaining this:

bool universal(set, property) {
  for (element in set)
    if (not property(element))
      return false
  return true
}

As you can see, the universally quantified statement is only false if there exists an element of the set for which it does not hold. Conversely, you could define

bool existential(set, property) {
  for (element in set)
    if (property(element))
      return true
  return false
}

This is also similar to other empty-set definitions like

$$\sum_{x\in\emptyset}f(x)=0\qquad\prod_{x\in\emptyset}f(x)=1$$

If everyone in front of you is bald, then you are bald.

Applying the above to the statement from your question: from

$$\bigl(\forall y\in\text{People in front of }x: \operatorname{bald}(y) \bigr)\implies\operatorname{bald}(x)$$

one can derive

$$\emptyset=\text{People in front of }x\implies\operatorname{bald}(x)$$

so yes, the first person must be bald because there is noone in front of him.

Some formalisms prefer to write the “People in front of” as a pair of predicates instead of a set. In such a setup, you'd see fewer sets and more implications:

$$\Bigl(\forall y: \bigl(\operatorname{person}(y)\wedge(y\operatorname{infrontof}x)\bigr)\implies\operatorname{bald}(y) \Bigr)\implies\operatorname{bald}(x)$$

If there is no $y$ satisfying both predicates, then the left hand side of the first implication is always false, rendering the implication as a whole always true, thus allowing us to conclude the baldness of the first person. The fact that an implication with a false antecedent is always true is another form of vacuous truth.

Note to self: this comment indicates that Alice in Wonderland was dealing with vacuous truth at some point. I should re-read that book and quote any interesting examples when I find the time.

MvG
  • 44,006
  • I accepted the whole of your answer except for the first reason you mentioned (that "logical statements don't suggest anything"), and the last note (because I haven't read Alice yet!). But they are minor issues from the mathematical standpoint, so this is a good enough answer. Thanks. :) – Færd Feb 24 '16 at 05:11
  • A function from an empty set to an empty set is of the same flavor. – A.S. Feb 24 '16 at 05:47
  • I read the Wikipedia article, but I cannot understand the reasoning. The statement is true if the precondition is known to be false. However, in this case the precondition is neither true nor false. If your logic follows, that the first person is bold, then the second person must also be bold, and so is everyone else in the line, and the statement essentially means everyone is bold. For your logic to hold true, there has to be a mathematic statement where if the precondition is undetermined, the the statement is true. That's not vacuous truth, is it? – daniel Feb 24 '16 at 11:11
  • I just realized that I just answered my own question. The logic is only true if you checked every person in the line and arrived with the statement. And you can only arrive with that conclusion if everyone is bold or if every one is hairy. So it's true that the first person is bold, therefore everyone is bold. It's also true that the first person is hairy and that everyone is hairy. – daniel Feb 24 '16 at 11:40
  • 6
    -1 for not answering a yes/no question with a yes/no response anywhere in the answer. The answer is hairy but would benefit from a bald leader. – crokusek Feb 24 '16 at 20:27
  • 2
    I agree with crokusek. It took me a while to conclude that the answer was yes. Even a "TLDR: Yes" at the top would improve the answer. – Ivo Feb 24 '16 at 20:38
  • 2
    Agree with MvG not posting yes/no. A purpose of the website is to provide "next step" type assistance with math problems, not to commission answers. A "yes" or "no" wouldn't help the original poster understand if he can't figure out whether it is yes or no on his own, so it is better not to post it. I ag – DanielV Feb 24 '16 at 21:34
  • 1
    I would downvote, but I don't have the rep. I understand what you're saying about vacuous truths being useful, but they pose a problem in this situation. So we should interpret the conditional statement as having an understood/unstated condition that prevents the conditional from applying to the first person in line. – jpmc26 Feb 25 '16 at 00:07
  • 2
    Also agree with @crokusek. It's a yes/no question. That doesn't mean there can't (or shouldn't) be an explanation as part of the answer, but somewhere there should also be a "yes, the first person can only be bald", or a "no, the first person may or may not be bald and there's not enough information to determine which". – aroth Feb 25 '16 at 06:07
  • So if the original statement is reduced to "If the person in front of you is bald, then you are bald." (instead of everyone in front), the conclusion that the first person is bald (and therefore that everyone is bald) still holds (when using standard logic conventions). Is that correct? – KIAaze Feb 25 '16 at 15:17
  • 1
    @KIAaze: not neccessarily. “The person in front of you” sounds like a function to me: input a person, output the person in front. But in case of a single row, it's only a partial function, making the outcome undefined. If you model that partial function as a relation, you'd get no statement for the person, since you need a pair of people to satisfy the relation, and can't say anything about individual persons without forming such pairs. – MvG Feb 25 '16 at 15:27
  • Just to be more exact, I meant: "If the person just in front of you is bald, then you are bald.". If I understood your reply correctly, the baldness of the first person is also undefined if the statement had been: "If any person in front of you is bald, then you are bald.", i.e. unless there is a way to re-formulate the statement with a \forall, the baldness of the first person is undefined. Correct? (although initially my thinking was "forall y in {x-1}" which would have been empty for the first person, ergo first person bald) – KIAaze Feb 25 '16 at 15:48
  • @KIAaze: “If any person …” translates to an existential quantifier, which is defined to be false for the empty set. “Any”, “some”, “all”, “every” and “no” indicate quantifiers, while “the” (at least as used in your first alternative) indicates a function. The statement “any person in front of you is bald” is false for the first person. (Which means the statement “If any person in front of you is bald, then you are bald.” is vacuously true for the first person, no matter whether they is hairy or bald.) – MvG Feb 25 '16 at 16:48
  • @KIAaze "If the person in front of you is bald, then so are you" is undefined for the first person. That is why in weak induction, it is usually phrased as "if you are bald, then so is the person behind you" -- that is, $P(n) \implies P(n+1)$, because $P(n - 1)$ isn't defined in the base case. – DanielV Feb 25 '16 at 17:02
  • My question was "Without further assumptions, does this mean that the first person is necessarily bald?", and the answer was (as the previous version of your answer implied) : "No, unless we add a new definition (= make a further assumption).". I unmarked your answer because this courageous No matters to me, not because I think your answer is faulty. – Færd Feb 25 '16 at 21:17
  • +1 Good answer. Unfortunately, a lot of people responding, including the OP and especially jpmc26, are inept and are unable to appreciate or understand it. – Jim Balter Feb 26 '16 at 05:30
  • 1
    @MJF: Funny, I always meant my answer as “Yes, the first person is bold, at least if it's mathematical logic we're talking about.” Because in mathematical logic, the definition of $\forall x\in\emptyset\ldots$ is standard. In a way, the “further assumption” in this case was derived from the fact that you posted this on a math forum, where I'd expect mathematical rules to hold more weight than ambiguous everyday language interpretations. – MvG Feb 26 '16 at 11:47
  • You're right. Maybe it's my fault not to confine my views of logical aspects of human language to just mathematical logic. I'm being off-topic and expect people to keep up with me. It's not an special vote or anything, but your informative answer still has my +1. – Færd Feb 26 '16 at 20:25
  • @MJF: I'm not sure what other areas your question would stray into if it's about human language. Anglistic, linguistics, philosophy, psychology come to my mind. I think there is a Stack Exchange for each of these. If you decide to cross-post your question to any of them, be sure to provide a link somewhere on this page. (I don't mean to imply that you should cross-post, though.) – MvG Feb 26 '16 at 23:48
21

It's worth looking at the reputations of the users who've given the contradicting answers, because different groups of people use language differently and on this occasion rep seems to neatly illustrate that [Edit: update, since I wrote that there's now at least one low-rep user answering that the front person is bald, so the neat division has broken down]

So far as mathematicians and mathematical writing are concerned, universal quantifiers and vacuous truth do say that the front person must be bald (although beware clever alternative logics).

Supposing for a moment that there are no unicorns, then the statement "every unicorn has a horn" is true, and for that matter the statement "every unicorn does not have a horn" is also true. So if you're at the front, then "everyone in front of you is bald" is true. "Everyone in front of you is hairy" is also true.

The reason is that we want "for all things, X is true" to be equivalent to "there does not exist a thing for which X is false", and "there exists a thing for which X is true" to be equivalent to "it's false that for all things, X is false". We don't want a funny special case where there doesn't exist a hairy person in front of you, but it still fails to be true that "everyone who is in front of you is bald".

However, this is not always the meaning in general-purpose plain English. Real people might consider the statement "every unicorn has a horn" to be false if there are no unicorns, or they might consider it undefined whether or not it's true. That's fine, we just have to be careful interpreting what civilians say into formal logic.

Anyway when reading mathematics, it's not possible for "If everyone in front of you is bald, then you are bald" and "If everyone in front of you is hairy, then you are hairy" to both be true, since they contradict as to the baldness/hairiness of the front person. If you want to make mathematical statements along these lines, with the meanings you prefer, then you should explicitly exclude the front person.

However, there's that little word "too" at the end of your sentence, which throws a spanner in the works. The word "too" implies "as well as something else". But it doesn't really belong if this were a mathematical proposition. It's one thing to say someone is bald when there's nobody in front of them, and it's another thing entirely to say they're bald "as well as nobody". That doesn't make sense, and might cause us to reject the whole thing as unclear.

Steve Jessop
  • 4,156
  • Nice catch! I'll edit the too out. Thanks. – Færd Feb 24 '16 at 04:54
  • "Anyway when reading mathematics, it's not possible for "If everyone in front of you is bald, then you are bald" and "If everyone in front of you is hairy, then you are hairy" to both be true" - If you are at the head of the line, both are vacuously true. They do not conflict if the line has 0 or 1 people. – emory Feb 24 '16 at 23:13
  • 3
    @emory "If you are at the head of the line, both are vacuously true" -- No, the antecedents are vacuously true, making the statements contradictory. – Jim Balter Feb 25 '16 at 06:16
  • @JimBalter If the line is empty or if there is one person in line, then I do not see the contradiction. If there are 2 or more people in line, then yes the statements are contradictory. – emory Feb 25 '16 at 10:24
  • 1
    @emory: the problem states that there is at least one person in the line. That person cannot be both bald and hairy. So it cannot be the case that both "if everyone in front of you is X then you are bald" and "if everyone in front of you is Y then you are hairy" for any values of X and Y, because the first statement implies that the front person is bald and the second that the front person is hairy. If the line were empty, then sure, we could admit both statements, but I took from the question without explicitly restating, that there is a front person. – Steve Jessop Feb 25 '16 at 11:05
  • "Anyway when reading mathematics, it's not possible for "If everyone in front of you is bald, then you are bald" and "If everyone in front of you is hairy, then you are hairy" to both be true" This is technically not correct, as they are "if" statements instead of "iff" (if and only if). In this case, it means that if you are third in line, and you see one bald and one hairy ahead, the consequence is that you don't have a clue about yourself. – polettix Feb 25 '16 at 11:49
  • @polettix: Question says, "this property holds for every person in the line", and when I say "true" I mean "true in general of people in the line". Agreed, they could both hold for the third person in line (indeed they could both hold for the third person in line as iffs). It's the front person where there's a problem – Steve Jessop Feb 25 '16 at 11:51
  • I see, thanks. So the whole thing about "impossibility" actually stems from the ordering I guess? – polettix Feb 25 '16 at 11:56
  • 1
    @polettix: well, the front person is in the unique position that both "everyone in front you is bald" and "everyone in front of you is hairy" are true. For anyone else, at most one and possibly neither of those is true. So yes, you could say it stems from the line being a well-ordering. Certainly it's a property of the order whether or not a set has a minimal element. – Steve Jessop Feb 25 '16 at 12:09
  • @emory " I do not see the contradiction" -- how odd. "If [vacuously true] then X" and "If [vacuously true] then not X" are clearly contradictory. – Jim Balter Feb 25 '16 at 20:39
  • @JimBalter I see it now. It could only be true for a zero person line - which is disallowed by the question. – emory Feb 25 '16 at 20:45
5

So we have starting point

If everyone in front of you is bald, then you are bald too.

The first person in line sees that two properties hold:

  1. Everyone in front of the first person is bald.

and

  1. Everyone in front of the first person is not bald.

However, the starting point only applies to the case where everyone in front of a person is bald, so the first property, and nothing about the second property where every person is not bald.

Therefore, the first person must be bald by a combination of the starting point and property 1.

If there would be a rule depicting that

If everyone in front of you is not bald, then you are not bald too.

then this would give a contradiction. However, there is no rule about property 2, so it plays no role.

1

Suppose we have a line of people that starts with person #1 and goes for a (finite or infinite) number of people behind him/her, and this property holds for every person in the line:

If everyone in front of you is bald, then you are bald.

Without further assumptions, does this mean that the first person is necessarily bald?

Yes.

The above property can be formalized as follows:

$\forall a\in \mathbb{N}:[\forall b\in \mathbb{N}:[b<a \implies B(b)] \implies B(a)]$

where $B(n)$ means person #$n$ is bald.

Specifying $a=1$, we have:

$\forall b\in \mathbb{N}:[b<1 \implies B(b)] \implies B(1)$

The antecedent here is vacuously true, therefore $B(1)$ is true, i.e. person #$1$ must be bald.

EDIT 1: This will also work for finite lines. Just substitute $\{1, 2, \cdots n\}$ for $\mathbb{N}$ where $n\geq 1$.

EDIT 2: Re: Vacuously true. Since $b<1$ will always be false, the implication $b<1 \implies B(b)$ will always be true. See my previous answer on this topic at Understanding Vacuously True (Truth Table)

  • 1
    You may want to be more specific for the newcomers here, there are a lot of them. Rather than "The antecedent here is vacuously true" you can say "b < 1 is always false, so b<1 implies B(b) is always true". That is the entire point of the question after all. – DanielV Feb 27 '16 at 07:59
  • @DanielV Good point. See my EDIT #2. – Dan Christensen Feb 28 '16 at 14:32
-2

Let's try to isolate the problematic part of the question. Imagine you are the first person in line. Your cell phone rings; the man on the other end says to you, "My psychic powers tell me that everyone in front of you is bald. Am I wrong?" How do you answer?

From the standpoint of mathematical logic, the answer is unequivocally No, he is not wrong. This is not a "convention", but rather an inevitable consequence of how logic works: The person on the other end of the phone would only be wrong if there were a non-bald person in front of you; since there is no non-bald person in front of you, he is not wrong.

Now here is where it gets tricky: If he is not wrong, is he right?

If we assume a logical system in which every statement has a truth value, and there are only two possible truth values, then if he is not wrong, he must be right. So in such a logical system it must be true that everybody in front of the first person in line is bald.

For similar reasons, the largest natural number is odd, and the final digit of the decimal expansion of $\pi$ is a 4. These statements are true because they are not false; they are not false because there is no largest natural number and $\pi$ has no final digit.

But human language does not operate according to the principles of formal logic. In the real world, we would not tell the person on the cell phone "You are wrong" or "You are right"; what we would say is "There is nobody in front of me." More precisely, we would infer the tacit additional premise "There is at least one person in front of you", and reject the entire construction because that premise is false.

mweiss
  • 24,547
  • 3
    Vacuous truth is certainly a standard part of mathematics, but phrases such as "the largest natural number" are not necessarily represented by vacuous quantifiers. The vacuous truth in the original question here is about vacuous quantification over an empty set, rather than about terms that fail to denote any object. So it is coherent for mathematicians to accept that "every negative natural number is also irrational" is true while not accepting that "the largest natural number is off" has any truth value. Not every sentence has a truth value, of course, e.g. "this sentence is false". – Carl Mummert Feb 25 '16 at 20:09
  • I parse "The largest natural number is odd" as "If $x$ is the largest natural number, then $x$ is odd." I am not sure how else to interpret it. – mweiss Feb 25 '16 at 22:27
  • 1
    One common interpretation is as a definite description, as in https://en.wikipedia.org/wiki/Definite_description . When such a description refers, it can be safely expressed by a universal or existential quantifier: "for every $n$, if $n$ is the smallest natural number then $n$ is even" or "there is an $n$ such that $n$ is the smallest natural number and $n$ is even". These have the same truth value because the description refers. When the description does not refer, they give opposite truth values, and it is not obvious from plain English which is intended. – Carl Mummert Feb 26 '16 at 01:45
  • 1
    It isn't actually correct to say "if it isn't false, then it must be true". It could be undefined, which is what most of those new to this problem are having trouble with. For example, both statements from the phone "the person in front of you is bald" and "all people in front of you are bald" are not false. But only the second one is true. – DanielV Feb 27 '16 at 08:02