Translating "There are two different students in your class who between them have sent an e-mail message to or telephoned everyone else in the class"

Question

Let $M(x,y)$ be “$x$ has sent $y$ an e-mail message” and $T(x,y)$ be “$x$ has telephoned $y$”, where the domain consists of all students in your class. Assume that all e-mail messages that were sent are received, which is not the way things often work. Symbolise this statement in first-order logic:

There are two different students in your class who between them have sent an e-mail message to or telephoned everyone else in the class.

I'm not sure whether my answer is correct; if not, please tell me the reasons: $$∃x∃y(x≠y ∧ ∀z(∀s(M(z,s)∨T(z,s)) ↔ (z = x ∨ z = y))).$$

Answer 1

There are two different students in your class who between them have sent an e-mail message to or telephoned everyone else in the class.

This means: $$∃x∃y\;\Big(x≠y \;∧\; ∀z\;\big(z\ne x \land z\ne y \to Mxz ∨ Txz ∨ Myz ∨ Tyz\big)\Big).\tag1$$

$$∃x∃y(x≠y ∧ ∀z(∀s(Mzs∨Tzs) ↔ (z = x ∨ z = y))).\tag2$$

Sentence $(2)$ is stronger than sentence $(1):$ it additionally claims that each of those two students—and nobody else—has mailed/telephoned every student.

Answer 2

Your Implication is not matching the given Statement. It seems to state that who-ever did the mailing & the telephoning must be $x,y$ , which may not be true when there are many such Students or Pairs of Students.

Consider these instead :

$$\exists x (\forall z((M(x,z) \lor T(x,z))))\tag{1}$$

This says that there is at least 1 student $x$ ( though there may be more ) who did the mailing to very student $z$ or did the telephoning to $z$.

We want at least 2 such students , hence this will work :

$$\exists x \exists y ((x≠y) \land \forall z((M(x,z) \lor T(x,z)) \land (M(y,z) \lor T(y,z))))\tag{2}$$

This makes the same claim about $x$ & $y$ & then it adds the Criteria that these two are not the SAME , hence we have at least 2 such students ( though there may be more ) matching the given Statement.

$$\exists x \exists y ((x≠y) \land \forall z((M(x,z) \lor T(x,z)) \lor (M(y,z) \lor T(y,z))))\tag{3}$$

This makes the same claim every Student has got Communication from either $x$ or $y$ , either by mail or by telephone & then it adds the Criteria that these two are not the SAME , hence we have at least 2 such students ( though there may be more ) matching the given Statement.

[[ I am adding (3) because I see "...who between them..." , which means both of them "Combined together" have Communicated with the other Students. ]]

Do not use Implication here , It is unnecessary here , It is enough to use $M()$ & $T()$ in this Case.

DIAGRAMMATIC EXPLANTION :

Consider 6 Students , $1\cdots6$ , & let the Black lines indicate Communication (Either Email or telephone) where the Arrow Directions indicate who contacted whom.

1 contacted : 2,4,6
2 contacted : 6
3 contacted : 2,4
4 contacted : 1,3
5 contacted : 4,6
6 contacted : 5

Which-ever 2 we take , we do not get all other Students. Let the Blue Dashed line be a new email contact from 3 to 5.

Then (1,3) contacted 2,4,6 + 5 which is all other Students.
Is this what we want ? Question in English is ambiguous : It may be that 1 & 3 must also have been contacted. Then let the Purple Dashed lines be the new contacts between 1 & 3.
We can then state that there are 2 Students (1,3) who between them have contacted every other Student.

OK , let the Green Dashed lines be the new contact between 1 & 5.

Then 1 contacted : 2,3,4,5,6
5 contacted : 1,4,6
We can then state that there are 2 Students (1,5) who between them have contacted every other Student.

It is not unique , then we can not make the implication that which-ever Pair has this Property must be (1,3) , that Pair could be (1,5) too.
When more contacts occur , there may be more such Pair Choices.
In the Extreme Case where Every Student has contacted all Students , Every Pair is valid Choice !

Without Implication , we can simply state that there are 2 Students who have the Property , leaving it open whether that Pair is unique or not.