# “If everyone in front of you is bald, then you're bald.” Does this logically mean that the first person is bald?

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.

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

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This is called strong induction. – DanielV Feb 23 at 18:27
Why don't they just touch their heads and find out... – Nikunj Feb 23 at 18:29
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 at 18:41
@MJF In that case, a universal statement ('for all') is always true, and an existential ('there exists') statement is always false. By convention. en.wikipedia.org/wiki/Universal_quantification#The_empty_set and en.wikipedia.org/wiki/Existential_quantification#The_empty_set – Keep these mind Feb 23 at 19:02

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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Also "if there is nobody in front of you with hair, then you are bald." – Dan Brumleve Feb 23 at 18:29
@DanBrumleve, nice alternative. – Barry Cipra Feb 23 at 18:35
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 at 18:36
@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 at 18:40
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 at 18:55

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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-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 at 20:27
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 Beckers Feb 24 at 20:38
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 at 21:34
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 at 0:07
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 at 6:07

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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Nice catch! I'll edit the too out. Thanks. – Færd Feb 24 at 4:54
"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 at 23:13
@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 at 6:16
@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 at 11:05
@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 at 12:09

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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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 at 7:59
@DanielV Good point. See my EDIT #2. – Dan Christensen Feb 28 at 14:32

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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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 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 at 22:27
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. – Carl Mummert Feb 26 at 1:45
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 at 8:02

No. The statement does not imply that the first person is bald. A hairy first person does not violate the rule. Having "nobody in front of the first person" is not equivalent to "everybody in front of the first person is bald."

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In mathematics, if there is nobody in front of the first person, then it is actually true that everyone in front of the first person is bald. Also, everyone in front of the first person is an alien from Mars. This is one of the linguistic conventions whose understanding is key to working in math. – Carl Mummert Feb 24 at 11:29
You are of course free as an individual to leave "Every member of an empty set has property P" undefined. However, that would be not only useless, but actually quite frustrating to work with. No one with any experience or competence in logic would join you in that convention. – DanielV Feb 24 at 14:19
@aroth Ironically, you did not answer my question. Anyway, a) 31 people upvoted the "non-answer" and the OP accepted it, and b) this answer is flat out wrong (no one in front of the first person is equivalent to everyone in front of them being bald), so I don't think your O is worth much. – Jim Balter Feb 25 at 7:02
"no one in front of the first person is equivalent to everyone in front of them being bald" - Only through the vacuous truth. If I kidnap someone, blindfold them, numb their scalp, and either shave their head or don't and then stick them in a line and say "if everyone in front of you is bald, then you are too", they don't have enough information to say who's bald or not. The line is a tangible construct, and the vacuous truth only holds if the line-master chose for it to, which isn't something a third-party observer knows (unless they ask the line-master, first). [1/2] – aroth Feb 25 at 7:50
...in formal logic/math, however, there are of course agreed-upon conventions stating that the vacuous truth holds. But the original statement was framed in English, not formal logic. So we don't know if it can be translated into formal logic, using standard conventions, unless we can track down the person who said it and query their intent. Care should be taken to avoid passing convention off as fundamental truth. [2/2] – aroth Feb 25 at 7:50

No. It says nothing about the first person since

"If everyone in front of you is X then Y"

is always false for the first person except for maybe some values of X like "Non-Existent".

It implies that the whole line is bald if the first person is bald.

It could (arguably) be made more applicable to the first person as:

"If no one in front of you has hair (on their head) then you are bald."


Then any person in any line must be bald. Even lines of 1!

This question seems more English related than Math.

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### TL;DR

There's an understood but unstated "for the second person on" in your conditional statement. Interpreting the statement as holding true for the first person can lead to contradictions.

### More complete analysis

Let's define some terms:

• $p_n$ is the person in the $n$th position in the line.
• $P_n$ is the set of all $p_i$ where $i <= n$ and $i$ is a non-negative integer. (E.g., $P_3 = \lbrace p_1, p_2, p_3 \rbrace$.) We further define $P_0 = \lbrace \rbrace$.
• $P_{line}$ is the set of all people in the line.
• $b(p)$ is the statement, "$p$ is bald."

Here's what we get translating the English to formal logic notation:

$$\forall p \in P_{n-1}, b(p) \rightarrow b(p_n)$$

Now let's consider a somewhat tangential situation. Let's assume a different statement is true of the line:

$$\forall p \in P_{line}, b(p) \lor \forall p \in P_{line}, \neg b(p)$$

That is, the line is either all bald or all nonbald. For this particular line of people, if your original conditional statement holds, another one holds as well:

$$\forall p \in P_{n-1}, \neg b(p) \rightarrow \neg b(p_n)$$

So we have

$$\forall p \in P_{n-1}, b(p) \rightarrow b(p_n) \land \forall p \in P_{n-1}, \neg b(p) \rightarrow \neg b(p_n)$$

Let's go ahead and make this specific to the first person:

$$\forall p \in P_0, b(p) \rightarrow b(p_1) \land \forall p \in P_0, \neg b(p) \rightarrow \neg b(p_1)$$

Expanding $P_0$,

$$\forall p \in \lbrace \rbrace, b(p) \rightarrow b(p_1) \land \forall p \in \lbrace \rbrace, \neg b(p) \rightarrow \neg b(p_1)$$

Now substituting $p \rightarrow q$ with $\neg p \lor q$,

$$[ \neg (\forall p \in \lbrace \rbrace, b(p)) \lor b(p_1) ] \land [ \neg (\forall p \in \lbrace \rbrace, \neg b(p)) \lor \neg b(p_1)]$$

Both antecedents are vacuous truths. Therefore,

$$[ \neg (T) \lor b(p_1) ] \land [ \neg (T) \lor \neg b(p_1)]$$

$$[ F \lor b(p_1) ] \land [ F \lor \neg b(p_1)]$$

$$b(p_1) \land \neg b(p_1)$$

So we see that if these conditionals apply to the first person in the line ($p_1$), we could end up with logical contradictions! This suggests we either have an ill formed problem statement or vacuous truths pose a problem.

This makes intuitive sense as well. In a standard induction problem, you first need to show the initial condition and then prove that everything follows from that. Proving the inductive step isn't enough to say anything meaningful about the sequence; you have to prove the initial condition.

MvG's answer shows us why vacuous truths are useful, so we should go back to the original set up. There is something understood but unstated in the original statement: the conditional statement only applies to the second person on. So we should actually write this as:

$$n > 1 \land \forall p \in P_{n-1}, b(p) \rightarrow b(p_n)$$

This avoids attempting to apply the conditional statement to the first person and properly describes the situation. It allows for the first person to be not bald (which would mean that all following people in the line could be any mix of bald and not bald), and then the conditional statement can still hold for instances when the first person is bald.

Obvious information is often left unstated in both English and mathematics. For example, did you take note of the fact that I didn't specify $n$ is an integer? There was no need to; it was clearly implied by the context. This is fine in both English and math. It lets us focus on the important information. Although more caution should be exercised in mathematics. It's better to err on the side of being explicit in that realm, but if you are confident your audience will interpret it correctly, this isn't a problem.

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There is no contradiction in the sentence; if it is true, it implies that everyone in the line, including the first person, is bald – Carl Mummert Feb 25 at 1:53
" if your original conditional statement holds, another one holds as well: $(\forall p \in P_{n-1} ~:~ \lnot b(p)) \implies \lnot b(p_n)$ " No it most certainly does not hold. It fails exactly for the first person, where the statement is equivalent to $\text{True} \implies \lnot b(p_1)$, which is clearly unsound. – DanielV Feb 25 at 10:27
If everyone in the line is bald, then the sentence certainly holds. So, the statement is not contradictory. In fact, the statement implies that everyone in the line, including the first person, is bald. The notation in your argument is sufficiently inscrutable that it is difficult to say exactly where it goes wrong. One error in the negations near the end - the $\lor$ is inside the scope of the $\forall$ there, so the negation is incorrect – Carl Mummert Feb 25 at 12:32
'what you mean by "incoherent."' -- If simply asserting the statement in the title "can lead to contradictions", then logic is broken. "This suggests we either have an ill formed problem statement or vacuous truths pose a problem." -- but neither of those is true; thus, it's your analysis that is wrong. The premise, given in the title, is ∀p∈Pn−1,b(p)→b(pn), from which it follows that ∀p∈{},b(p) →b(p1). It's also (vacuously) true that ∀p∈{},¬b(p) but nothing follows from that; certainly not that ¬b(p1). – Jim Balter Feb 25 at 21:00
" if your original conditional statement holds, another one holds as well: ∀p∈Pn−1,¬b(p)→¬b(pn)" -- No, this is completely, utterly wrong, reminiscent of the Bellman in The Hunting of the Snark. Not only doesn't your inference follow from the title statement, it is contradicted by it ... that is the source of the contradiction you find -- you invented it. See Simon Klaver's answer, where he correctly states that "there is no rule about property 2, so it plays no role". – Jim Balter Feb 26 at 5:37

## edit

• Yes, mathematically the first person is bold
• Which leads to the invariable truth that everyone in the line is bold
• And in real life, the only way you can prove the statement is that you checked everyone in the line and everyone is bold.
• So, there is no contradiction

Now, there could be other arguments, but those would probably lead to contradiction. And I have not read any comment/answer that assert "vacuous truth" is "the truth".

## second edit

There is the problem of context that is not addressed.

If the statement is formulated by a person not well versed in theoretical mathematics, he would tell you that the statement only applies to the second person on. If you were to tell him about vacuous truth and he didn't look at you like you are a total jerk, he would then said the statement needs to be changed.

If the statement is formulated by a mathematician, then she or he would tell you that it means everyone is bold.

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Regardless of whether or not the first person MUST be bald, If the first person IS bald, then everyone in the line IS bald. Which makes it fairly important to determine how this applies to the first person and how the rule is applied to them. Practically speaking this means you need to clearly define how "everyone in front of you is bald" is determined to be true or false. – jmoreno Feb 25 at 1:50
"You are too caught up by the notion of "the first person" when the position is not important at all." -- This is nonsense. The question is specifically about the first person. "says nothing about" -- of course it does. (Everyone is X) implies (not (someone is not X)). – Jim Balter Feb 25 at 6:25
Don't post erroneous answers hastily if you don't like being corrected? – Benjamin Lindqvist Feb 25 at 22:27
@BenjaminLindqvist Yes, it's bizarre that someone would whine about comments pointing out errors in their answer -- that's a large part of why there are comments. But I suppose it is to be expected of someone who thinks that people who are competent at logic are "total jerks". And this drivel about "context" is funny; the context is a question in a math/logic forum about the formal implications of a statement. The context is also about being bald, not "bold". – Jim Balter Feb 26 at 6:00

Some are saying that "everyone in front of you is bald" is a binary 'yes/no' question. This is no more valid than police asking someone "when did you stop beating your wife", in that it is built around an assumption that may or may not be valid.

The word 'everyone' means 'x out of x, for any integer x greater than 1'.

nb: This depends on your dictionary. I am using http://dictionary.reference.com/browse/everyone which redirects to 'each' and then sticks that definition in front of the definition for 'person'

The example would be a lot easier to see if the statement replaced 'everyone' with 'both people'.

If it said both people, then the third person would see the statement as either true or false, and be able to work out their hair color. All that the other people would be able to determine is that there is a flaw in the statement and that either the person making the statement is wrong, or we live in a world where logic is meaningless. They would be unable to draw any conclusion about their hair color.

So the answer depends entirely on your definition of 'everybody', and this question likely belongs on a different part of stack exchange.

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In terms of ordinary mathematical English usage, the question is clear. – Carl Mummert Feb 24 at 11:31
@CarlMummert , I'm sorry, but what's the definition of every in mathematical English? – daniel Feb 25 at 12:16
@daniel: it's the mathematical English that I encounter as I work in mathematics on a daily basis, as I read mathematics papers and books, and as I talk to colleagues and students about mathematics. The essence of the original question, actually, seems to be that the OP was not familiar with the conventions of mathematical English, such as the use of vacuous truth, which is absolutely standard among mathematicians. – Carl Mummert Feb 25 at 12:35
@CarlMummert The comments sections on YouTube videos about .(9) = 1 or the Collatz conjecture are full of mathematical illiterates who are absolutely certain that they know better than all the world's mathematicians, but it's sad to see that such folk are now posting at math.stackexchange – Jim Balter Feb 26 at 6:05