$\forall x ~\exists y ~(R(x,y) \to \forall z(\neg L(y,z)))$ says "For everyone ($x$) there exists someone ($y$) who, if they ($y$) are the first person's ($x$) roommate, then they ($y$) will dislike everyone."   Thus it is not what you want.
In section 2.2, it asked us to negate the statement: Everyone has a roommate who dislike everyone.
Recall, universal statements are restricted by conditional and existential statements are restricted by a conjunction. $$\forall x\in A: P(x)\iff \forall x~(x\in A\to P(x))\\\exists y\in B: Q(y)\iff \exists y~(y\in B\land Q(y))$$
You want to say "For every person there exists a person, who is their roommate and dislikes everyone". Here the implcit domain is people and the restriction is on the existential (to roommates of the first entity). It is, basically: $$\forall x{\in}\text{People}~\exists y{\in}\text{RoommatesOf}(x)~\forall z{\in}\text{People}:\lnot L(y,z)$$
Hence:$$\forall x~\exists y~(R(x,y)\land\forall z~(\lnot L(y,z)))$$
I am confused because in another question: Everyone who is majoring in math has a friend who needs help with his homework. I wrote the logical form as
$$\forall x (M(x) \to \exists y(F(x,y) \land H(y)))$$.
where $M(x)$ means $x$ is a math major, $F(x,y)$ means $x$ is a friend of $y$ and $H(y)$ means $y$ needs help on homework.
Now, here you are restricting both the universal entities and the existential entities:
$$\forall x{\in}\text{MathMajors}~\exists y{\in}\text{FriendsOf}(x): H(y)$$
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"
Best Answer
Not quite. Your statement reads something like
You might fix your statement by replacing $\neg L(y,x)$, which reads “$y$ dislikes $x$,” with $\forall z \neg L(y,z)$, which reads “$y$ dislikes everyone.”
Not to get nitpicky, but something about this problem bothers me: it assumes that $\neg L(x,y)$ reads “$x$ dislikes $y$,” when, in reality, “not liking someone” is not necessarily the same as “disliking” them. However, this is just a semantic point and not wholly relevant.