What is the purpose of implication in this scenario? Consider the case:
Let:


*

*S(x) = “x is a student” 

*F(x) = “x is a faculty member” 

*A(x, y) = “x has asked y a question” 

*Dx and Dy = Consists of all people associated with your school. 


Use quantifiers to express this statement:

Some student has never been asked a question by a faculty member.

Attempted Solution:
Exist x, All y ( S(x) AND F(y) AND (NOT A(y, x)) )  
Book Solution:
Exist x ( S(x) AND All y ( F(y) -> NOT A(y, x) ) )
What is the point of using implication here? Would my answer also be correct?
 A: Let's make it slightly simpler.  Fix an $x$ and assume that $x$ is a student.  We want to say that $x$ has never been asked a question by a faculty member.  The relevant part of what you wrote is $\forall y\, (F(y) \mathbin{\And} \neg A(y,x))$.  However, this does not express the desired property of $x$.  The problem is that it asserts that everyone is a faculty member (even the janitors!)
To express "for all faculty members $y$,..." you should instead write "for all $y$, if $y$ is a faculty member, then...."  Admittedly, this is awkward English because first it asks us to consider all $y$ in the domain of discourse (even the janitors,) and then it becomes clear that we only care about faculty members $y$; that is, we only care about $y$ if $y$ is a faculty member.  This is where the if comes in.
So to express "$x$ has never been asked a question by a faculty member" we would write
(1) $\forall y\,(F(y) \implies \neg A(y,x))$.
This formula (1) translates to the awkward but correct English phrase "for all $y$, if $y$ is a faculty member, then $y$ has not asked $x$ a question."
Slightly less formally, and abusing notation by using $F$ for the set of faculty members rather than for the predicate "is a faculty member," we could write
(2) $\forall y \in F\; \neg A(y,x)$.
This formula (2) translates to the less awkward English phrase "for every faculty member $y$, $y$ has not asked $x$ a question."  However, in formal logic (2) is just defined to be an abbreviation for (1), so at its root the implication is still there.
A: Your attempted solution is wrong on one point, and equivalent on the other.
The placement of the "All y" part is arbitrary between the two options - as $S(x)$ does not depend on $y$, it can be placed on either side of it (for lack of a better explanation).
For the implication, your implication is inaccurate - it implies that $F(y)$ for all $y$, as it can only be true if ALL of the conditions are true, since you used "and" for all of them.
On the other hand, here's how you'd write the book's answer without explicitly using implication:
$$\exists x \text{ s.t. } (S(x) \land \forall y, (\lnot F(y) \lor \lnot A(y,x)))$$
The english version of this is "There is a person who is a student and who, if you look at all other people, they are either not faculty or have not asked the student a question".
