How to Prove It - Ch2 Sec 2 Exercise 2a Question: I am uncertain if my initial formulation of the logical form of the below statement is correct. 
How to Prove It (Velleman) Chapter 2, Section 2, Exercise 2a


Negate these statements and then reexpress the results as equivalent positive statements. (it is implied that you put into logical form first based on worked examples in the chapter)
2(a) "There is someone in the freshman class who doesn't have a roommate."


I have written this (prior to negation)
$$\begin{equation}\begin{aligned}
\exists x[F(x) \rightarrow \forall y \neg R(x,y)]
\\ \\
F(x): \text{x is in the freshman class}\\
R(x, y): \text{x is roommates with y}\\
\end{aligned}\end{equation}\tag{1}$$
I thought this was the correct formulation however, I found two blogs online that formulate the phrase as follows (again, before negation)
$$\begin{equation}\begin{aligned}
\exists x[F(x) \wedge \neg\exists yR(x,y)]
\end{aligned}\end{equation}\tag{2}$$
I checked to make sure equations (1) and (2) are not equivalent
$$\begin{equation}\begin{aligned}
\exists x[F(x) \rightarrow \forall y \neg R(x,y)] \qquad &\text{(1)}\\
\exists x[F(x) \rightarrow \neg \exists yR(x,y)] \qquad &\text{quantifier negation}\\
\exists x[\neg F(x) \vee \neg \exists yR(x,y)] \qquad &\text{conditional law}\\
\exists x\neg[F(x) \wedge \exists yR(x,y)] \qquad &\text{DeMorgans}\\
\neg\forall x[F(x) \wedge \exists yR(x,y)] \qquad &\text{quantifier negation}\\
\end{aligned}\end{equation}$$
I am trying to convince myself that (2) is correct, but can't see why (1) is incorrect. Is it because it's somewhat speculative ("There exists someone x, where if that person x is a freshman then for all people y, person x and person y are not roommates") vs declarative ("There is someone who is a freshman and for all people y, that person and person y are not roommates")?
 A: You are quite correct in your final paragraph, but perhaps could do with a more concrete argument? Consider a sophomore, say. They're not a freshman. So in fact they satisfy equation (1) - the implication holds true trivially because they are not a freshman at all! That is why (2) is the correct formulation - (1) allows the sophomore. (How many roommates they have doesn't matter at all!)
EDIT: Perhaps I could be clearer. Your equation (1) says 'there exists a student who, if they are a freshman, must not have any roommates'. My point is that a sophomore satisfies this condition, and so this does not correspond to the logical statement you were aiming for.
A: This year every freshman has a roommate.
Dan, it so happens, is not a freshman.
If he were a freshman, he wouldn't have a roommate.
Thusly Dan satisfies your statement even though
there is no freshman without a roommate.
A: Universal quantifiers are restricted by a conditional; existential quantifiers are restricted by conjunction".    Thus you use form (2), as follows.

"There is someone in the freshman class who doesn't have a roommate."

That is to say: "There exists a person, who, is in the freshman class and does not have a roommate."
$$\exists x~(F(x)\land \lnot\exists y~R(x,y))$$
