"There are two different people who have visited exactly the same websites"

Question

Let $W(x,y)$ mean that student x has visited website y, where the domain for $x$ consists of all students in your school and the domain for $y$ consists of all websites. Express the statement $∃x∃y∀z((x≠y)∧(W(x,z)↔W(y,z)))$ using a simple English sentence.

Answer: There are two different people who have visited exactly the same websites.

I think the given answer is wrong. Let's suppose that the domain of x and y (students) is {Eric, John, Albin} and the domain of z (websites) is {A, B, C}.

According to the answer, if, for example, Eric and Albin both visited only website A (which means that both of them did not visit website B and C), the statement

∃x∃y∀z((x≠y)∧(W(x,z)↔W(y,z)))

should be true. Let's see if it's true...

(Eric≠Albin)∧(W(Eric,A)↔W(Albin,A))

(Eric≠Albin)∧((W(Eric,A)→W(Albin,A))∧(W(Albin,A)→W(Eric,A)))

Eric is not equal to Albin, so Eric≠Albin is true. Moving to the next,

Since Eric visited website A, we have to test whether Albin also visited website A. Because I already assumed that Albin visited website A, the compound proposition ((W(Eric,A)→W(Albin,A)) is true.

Next compound proposition (W(Albin,A)→W(Eric,A)) should also be true, so the statement will return True.

However, what if Eric, Albin, and John did not visit any websites?

If we express the compound proposition, if becomes: (I only wrote the case where x and y are not equal since the statment will definitely return false if x and y are equal due to "x≠y")

Reminder: domain of x,y = {Eric, John, Albin}, domain of z = {A, B, C}.

{(W(Eric,A)→W(John,A))∧(W(John,A)→W(Eric,A)) ∧

(W(Eric,B)→W(John,B))∧(W(John,B)→W(Eric,B)) ∧

(W(Eric,C)→W(John,C))∧(W(John,C)→W(Eric,C))}

∨

{(W(Eric,A)→W(Albin,A))∧(W(Albin,A)→W(Eric,A)) ∧

(W(Eric,B)→W(Albin,B))∧(W(Albin,B)→W(Eric,B)) ∧

(W(Eric,C)→W(Albin,C))∧(W(Albin,C)→W(Eric,C))}.

Now, let's suppose that Eric, Albin, and John did not visit any websites. If the proposition means "There are two different people who have visited exactly the same websites.", then the proposition should return FALSE since no one has visited any websites.

Checking from the first row,

(W(Eric,A)→W(John,A))∧(W(John,A)→W(Eric,A))

returns TRUE since Eric did not visit website A. Therefore, W(Eric,A)→W(John,A) is vacuously true. Applying the same logic to (W(John,A)→W(Eric,A), it is also vacuously true. Therefore, the proposition returns TRUE.

Applying the same logic to every single compound propositions, they all returns TRUE. Therefore, the compound propositon ∃x∃y∀z((x≠y)∧(W(x,z)↔W(y,z))) returns TRUE when nobody visited websites, which is a false.

Is there anything that I could have interpreted the question in a wrong way? If so, I would be very appreciated to nofify my mistake and understand it in the right way.

Answer 1

$∃x\,∃y\,∀z\:((x≠y)∧(W(x,z)↔W(y,z)))$

Answer: There are two different people who have visited exactly the same websites.

I don't think you have a strong case:

• "Eric and Albin have visited the same website" (singular 'website'; $\exists w\,(Ew\land Aw)$) does indicate that they have visited some website;

• "Eric and Albin have visited the same websites" is reasonably understood as "the websites that Eric and Albin have visited are the same", i.e., "for every website, Eric has visited it iff Albin has visited it"; if Eric and Albin have not visited any website, then the statement is vacuously true.

  "Eric and Albin have visited the same websites" can also be read as "Eric and Albin have visited the same set of websites"; since the empty set is indeed a set, Eric and Albin are again not precluded from not having visited any website.

Or just write a clearer answer than the given one, something like

• There are two different students in your school such that the set of all websites that each has visited is the same.

On a related note: while translating Boris hasn't tried anything better than chocolate recently, I prefaced my efforts by introducing the axiom "Boris has tried some chocolate"; in that answer, unlike here, that axiom did genuinely help to sidestep complication.

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Reply to the OP's comments below

"Eric and Albin have visited the same websites" - for me, if you break down this sentence, it states that Eric has visited some websites

No.

and Albin has visited some websites

No.

and the websites they have visited is the same.

Yes.

Also, why would they even state “Eric and Albin have visited the same websites” if both of them have not visited any website?

Because then that statement is vacuously true, as is the statement "Every unicorn is invisible". Sure, these sentences are misleading, but they are nonetheless correct.

It is pointless and for me, it just seems like the question is tricking the students.

Isn't it illogical to blame the question when you disagree merely with the officially-suggested answer? Anyhow, see whether my alternative phrasings above are more acceptable.