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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).

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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!"

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    $\begingroup$ Also "if there is nobody in front of you with hair, then you are bald." $\endgroup$ – Dan Brumleve Feb 23 '16 at 18:29
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    $\begingroup$ @DanBrumleve, nice alternative. $\endgroup$ – Barry Cipra Feb 23 '16 at 18:35
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    $\begingroup$ 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. $\endgroup$ – Færd Feb 23 '16 at 18:36
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    $\begingroup$ @MJF, the logical negation of a "for all" statement is a "there exists" statement, i.e., $\lnot\forall x(B(x))\iff \exists x(\lnot B(x))$. $\endgroup$ – Barry Cipra Feb 23 '16 at 18:40
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    $\begingroup$ 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. $\endgroup$ – David K Feb 23 '16 at 18:55
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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.

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    $\begingroup$ -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. $\endgroup$ – crokusek Feb 24 '16 at 20:27
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    $\begingroup$ 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. $\endgroup$ – Ivo Beckers Feb 24 '16 at 20:38
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    $\begingroup$ 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 $\endgroup$ – DanielV Feb 24 '16 at 21:34
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    $\begingroup$ 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. $\endgroup$ – jpmc26 Feb 25 '16 at 0:07
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    $\begingroup$ 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". $\endgroup$ – aroth Feb 25 '16 at 6:07
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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.

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  • $\begingroup$ Nice catch! I'll edit the too out. Thanks. $\endgroup$ – Færd Feb 24 '16 at 4:54
  • $\begingroup$ "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. $\endgroup$ – emory Feb 24 '16 at 23:13
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    $\begingroup$ @emory "If you are at the head of the line, both are vacuously true" -- No, the antecedents are vacuously true, making the statements contradictory. $\endgroup$ – Jim Balter Feb 25 '16 at 6:16
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    $\begingroup$ @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. $\endgroup$ – Steve Jessop Feb 25 '16 at 11:05
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    $\begingroup$ @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. $\endgroup$ – Steve Jessop Feb 25 '16 at 12:09
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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.

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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)

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    $\begingroup$ 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. $\endgroup$ – DanielV Feb 27 '16 at 7:59
  • $\begingroup$ @DanielV Good point. See my EDIT #2. $\endgroup$ – Dan Christensen Feb 28 '16 at 14:32
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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.

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    $\begingroup$ 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". $\endgroup$ – Carl Mummert Feb 25 '16 at 20:09
  • $\begingroup$ 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. $\endgroup$ – mweiss Feb 25 '16 at 22:27
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    $\begingroup$ One common interpretation is as a definite description, as in 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. $\endgroup$ – Carl Mummert Feb 26 '16 at 1:45
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    $\begingroup$ 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. $\endgroup$ – DanielV Feb 27 '16 at 8:02

protected by J. M. is a poor mathematician Jul 26 '16 at 12:14

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