1
$\begingroup$

$L(x,y)$ be "x eats lunch with y." $C(x,y)$ be "x has class with y." $R(x,y)$ be "x is roommates with y." The domain for x and y is the students at a university. Translate the following into English:

b) $\exists x \forall y(((x \ne y) \land C(x,y))\rightarrow\lnot L(x,y))$

My Answer: For each student, if they are who is not themself, and has class with every student, then this student doesn't live with every student.

c) $\forall x \exists y ((x \ne y) \land ((C(x,y) \lor R(x,y)) \land \lnot L(x,y)) $

My Answer: Every student, who is not themself, and every student has class with this one student or lives with this one student, and every student doesn't live with this student.

My problem with this translation is I don't know if I am supposed to say for $x \ne y$ that these students are not themselves or they're unique?

Also, how to use English for $\exists$ and $\forall$. Some helpful tips and double checking to see if my English is right would be greatly apprecaited!

$\endgroup$
2
  • $\begingroup$ I have no idea what "Every student, who is not themself" could possibly mean! $\endgroup$
    – user247327
    Commented Sep 11, 2020 at 20:32
  • $\begingroup$ Wrong phrasing with English with this. Struggling with the idea of translating $x \ne y$. $\endgroup$
    – user750949
    Commented Sep 11, 2020 at 20:34

2 Answers 2

Reset to default
2
$\begingroup$

Use these:

  • $\exists x$ = "there is a student"
  • $\forall y$ = "any student"
  • $x\ne y$ = "other"
  • Combine the first two statements with "such that"

So then the first sentence becomes "There is a student such that any other student who has a class with the first will not eat lunch with the first student". Or, in a simplified version, "There is a student who does not eat lunch with anyone else in his class"

$\endgroup$
6
  • $\begingroup$ Hmm, I thought that it would be more like "There is a student with whom nobody else in his class has lunch". $\endgroup$
    – Stefan Octavian
    Commented Sep 11, 2020 at 20:24
  • $\begingroup$ I think it's the same. But I'm not a native English speaker. $\endgroup$
    – Andrei
    Commented Sep 11, 2020 at 20:30
  • 1
    $\begingroup$ Neither am I, but I guess it depends on whether $L$ is symmetric. $\endgroup$
    – Stefan Octavian
    Commented Sep 11, 2020 at 20:33
  • $\begingroup$ So my first interpretation was more precise than my simplified version. Thanks for pointing that out. $\endgroup$
    – Andrei
    Commented Sep 11, 2020 at 20:35
  • $\begingroup$ Actually, I think that you simplified version is more precise than my comment, since $L$ is defined as "$x$ has lunch with $y$" and not "$y$ has lunch with $x$". But if $L$ is considered symmetric, there is no difference between the two. $\endgroup$
    – Stefan Octavian
    Commented Sep 11, 2020 at 20:40
0
$\begingroup$

First, remember what $L$ stands for. It's "Lunch" not "Lives with". As for your translations, they are kind of confusing. $\forall$ reads as "for all" or "every", $\exists$ read as "exists" or "there is (at least) one". Implication reads as an if statement, and at times, conjunctions, disjunctions and other relations (such as $\ne$) can be rephrased to words more common. How I would translate them

b) There exists a student for which, for every student, if they are not the same and they have a class together, then they don't have lunch together.

Which can be rephrased as

b) There exists a student for which every other student that has a class with them, doesn't have lunch with them.

c) For every student, there exists a student which is not themselves and they have a class with them or they are roommates and they don't have lunch together.

Again, the $\ne$ can be rephrased, the repetition of "and" can be avoided and the word "even" can be added to increase the naturality of the language (assuming that the condition of being roommates is, in some sense, stronger than just having a class with them).

c) For every student, there exists another student that they have a class with or are even roommates with and they don't have lunch together.

$\endgroup$

You must log in to answer this question.