How to formalise “There are two students who between them have chatted with everyone else”?

Question

From Discrete Mathematics and Its Applications by Kenneth H. Rosen, 8th edition :

Let $C(x, y)$ be the predicate “$x$ has chatted with $y$”, where the domain for $x$ and $y$ consists of all the students in the class. Translate the following statement into a logical expression involving predicates and quantifiers:

“There are two students in the class who between them have chatted with everyone else in the class.”

In my opinion, the statement can be parsed and translated as follows:

There are two people $x$ and $y$ in the class such that, for any student $z$ in the class, if $z$ is not $x$ and $z$ is not $y,$ then either $x$ or $y$ has chatted with $z.$ $$\exists x \exists y \big( (x \neq y) \land \forall z \big( (z \neq x \land z \neq y ) \rightarrow ( C(x, z) \lor C(y, z) ) \big) \big) \tag{M} $$

However, the book's answer is $$\exists x \exists y \big( x\neq y \land \forall z \big( C(x, z) \lor C(y, z) \big) \big) \tag{R}$$

Which is the correct answer: M or R?

Answer 1

Let $Cxy$ be the predicate “$x$ has chatted with $y$.”

“There are two persons who between them have chatted with everyone else.”

$$\exists x \exists y \;\big( x \neq y \land \forall z\, \big( (z \neq x \land z \neq y ) \rightarrow ( Cxz \lor Cyz ) \big) \big) \tag{M}$$

$$\exists x \exists y \;\big( x\neq y \land \forall z \,\big( Cxz \lor Cyz \big) \big) \tag{R}$$

Only Statement M accurately formalises the given verbal statement. As a matter of fact, the following statements are pairwise non-equivalent:

• There are two persons who between them have chatted with everyone else. $$\exists x \exists y \;\big( x \neq y \land \forall z\, \big( (z \neq x \land z \neq y ) \rightarrow ( Cxz \lor Cyz ) \big) \big) \tag{M}$$
• There are two persons who between them have chatted with everyone. $$\exists x \exists y \;\big( x\neq y \land \forall z \,\big( Cxz \lor Cyz \big) \big) \tag{R}$$
• There are two persons each of whom has chatted with everyone who is neither of them. $$\exists x \exists y \;\big( x \neq y \land \forall z\, \big( (z \neq x \land z \neq y ) \rightarrow ( Cxz \land Cyz ) \big) \big) \tag{C}$$
• There are two persons each of whom has chatted with everyone who isn't that person. $$\exists x \exists y\; \big( x \neq y \land \forall z\, \big( z \neq x \rightarrow Cxz \big) \land \forall z\, \big(z \neq y \rightarrow Cyz \big) \big) \tag{B}$$
• There are two persons each of whom has chatted with everyone. $$\exists x \exists y \;\big( x \neq y \land \forall z\, \big( Cxz \land Cyz \big) \big) \tag{A}$$

Specifically (the diagram's arrows denote all the entailments between the five statements):

1. Statement A is stronger than Statement B, which is stronger than Statement C, which is stronger than Statement M.
2. Statement B is stronger than Statement R, which is stronger than Statement M.
3. Neither Statement C nor Statement R implies the other.

For instance:

1. the domain $\{a,b,c\}$ with $(a,a),(b,b),(a,c),(c,a)$ as the only chatted pairs satisfies Statements M and R but not Statement C;
2. the domain $\{a,b,c\}$ with $(a,c),(c,a)$ as the only chatted pairs satisfies Statement M but not Statement R.

Answer 2

SUMMARY :
K H Rosen is right.
You are not wrong.

It is more about "Implicit Assumptions" , rather than "Alternate Interpretations" ....

Consider : We make a claim that "X got more marks than everybody in the class" , it would mean "X got more marks than himself" , which is Contradictory.
Hence we usually make the claim that "X got more marks than everybody else in the class" , to exclude "X got more marks than himself" , which is very good.

Like-wise , there are many relations which are inherently not reflexive , that is , X is not related to itself : Eg Rosen is assuming no number is less than itself , no country is trading with itself , nobody will be chatting with himself , nobody has friendship with himself , ETC.
Hence the claims will automatically include "everybody else" , to eliminate the reflexive case ....

Now , when we have 2 elements $x,y$ , then , it will work out to "everybody" , even with "everybody else" , because $y$ is included in "everybody else" for $x$ , while $x$ is included in "everybody else" for $y$ ....

Hence the Rosen version is correct with that "Implicit Assumption" , while yours is correct too without that.
It just that Natural Languages [ like English ] are inherently ambiguous.

ADDENDUM :

Stating what I am trying to explain , in more theoretical way :

Let $C=\{1,2,3,4,5\}$ be the Universal Set.
Here let us consider 2 Distinct Elements $x=1,y=2$ :
What is "everybody else" for $x$ : It is $C-\{x\}=\{2,3,4,5\}$
What is "everybody else" for $y$ : It is $C-\{y\}=\{1,3,4,5\}$

What is "everybody else" for $x,y$ : It is $[C-\{x\}] \cup [C-\{y\}]$
That will eventually boil down to $=\{1,3,4,5\} \cup \{2,3,4,5\} \equiv C$ , that is , "everybody else" becomes "everybody" here. It is easy to visualize that with Venn Diagram.

When we want "everybody else" to mean $[C-\{x\}-\{y\}]=\{3,4,5\}$ , we might say something like "everybody except the both of them" Explicitly , thereby avoiding the Issues with Natural Languages.

Answer 3

You don't care if $x$ chatted with $y$, though this seems to be required.