How to logically analyze the statement: "Nobody in the calculus class is smarter than everybody in the discrete math class." I am self-studying Daniel Velleman's "How to Prove It." 
In the exercises for section 2.1, for question # 1b, I got a different answer than he did (his answer is in the back of the book). 
I think my answer is equivalent to his, and I also think I see yet another equivalent answer.
But since I'm learning, I'm not confident enough that my answers really are equivalent, so I hope someone here can help me.
The question is to "analyze the logical form" of the following statement: "Nobody in the calculus class is smarter than everybody in the discrete math class."
Velleman's answer is:
$$
\lnot \exists x (C(x) \land \forall y (D(y) \to S(x,y))))
$$
I can see that, but it appears the following are fine too. In fact, it looks to me like you have a choice whether or not to use If-Then at all, and if you want to use it, you have 2 separate places where you could put it.
Is this also correct?
$$
\lnot \exists x (C(x) \land (\forall y (D(y) \land S(x,y))))
$$
And is this also correct? 
$$
\forall x (C(x) \to (\lnot \forall y D(y) \land S(x,y))) 
$$
EDIT: Here's a third attempt, added later. Is this equivalent?
$$
\forall x (C(x) \to \exists y (D(y) \land S(y, x)))
$$
where S(y,x) is defined as "y is as smart as or smarter than x"
If those are not correct, please help me understand why not?!?!*
Thanks.
 A: The first of your proposed alternatives says that

There is no person $x$ who is both in the calculus class, and for whom it is true that:
     "every person $y$ is both in the discrete math class and is dumber than $x$".

The second one says that

For any person $x$, if they are in the calculus class, then it is not  the case that
     "every person $y$ is both in the discrete math class and is dumber than $x$".

As long as there exists at least one person not in the discrete math class, then both of these proposed alternatives will necessarily be true, whereas the original statement can still be false; thus, they are not equivalent.
A: As a general rule of thumb, most (negated) universal English language sentences will be translated as a (negated) universal conditional, while (negated) existential sentences will be translated as a (negated) existential conjunction. If you get universal conjunctions or existential conditionals in your translation, then you might want to double check your answer. For instance, the universal fragment of your first answer, "$\forall y (D(y) \wedge S(x,y))$," translates as "Everyone is in the discrete math class and $x$ is smarter than everyone," whereas you wanted to say "$x$ is smarter than anyone who is in the discrete math class," or equivalently, "Everyone who is in the discrete math class is such that $x$ is smarter than them." 
In general, if your English sentence has the form "Everything that is $\phi$ is $\psi$," your translation will be along the lines "$\forall x (\phi \rightarrow \psi)$," whereas "$\forall x (\phi \wedge \psi)$" says, "Everything is both $\phi$ and $\psi$," which is a much stronger claim. Similarly, "Some $\phi$ is $\psi$" gets translated as "$\exists x (\phi \wedge \psi)$," whereas "$\exists x (\phi \rightarrow \psi)$" says, "Something is such that if it's $\phi$, then it's $\psi$," or rather "Something is either not $\phi$ or $\psi$," which is a much weaker claim. "No one that is $\phi$ is $\psi$" is typically translated as "$\forall x (\phi \rightarrow \neg\psi)$". But there are equivalent formulations of each of these sentences as well.
Your edited answer is almost correct. However, "$S(y,x)$" is not logically equivalent to "$\neg S(x,y)$", and you want the effect that your $x$ is only not smarter than your $y$. But if you just replace "$S(y,x)$" with "$\neg S(x,y)$", then your new answer will be correct. Also acceptable translations of this sentence include:


*

*$\forall x (C(x) \rightarrow \neg \forall y(D(y) \rightarrow S(x,y)))$

*$\neg \exists x(C(x) \wedge \neg \exists y (D(y) \wedge \neg S(x,y)))$

A: I'm familiar with this book.  The problems in this section are getting you ready for the manipulations to come in the subsequent sections.  If you haven't gotten that far, it's difficult to give an explanation using the very few tools developed up to this point in the text.  The good news is that even if you don't quite get this problem, as you continue in the book you'll gain insight and eventually it will make sense.
Unfortunately neither of your alternate solutions is correct.  You really need an implication, not $\wedge$.  It turns out that $x\rightarrow y$ is equivalent to $\neg x \vee y$, while you have $x\wedge y$, which is quite different.
As the other solution points out, in certain situations your solution differs from the original.
