Understanding logical form of "Nobody in the calculus class is smarter than everybody in the discrete math class"

Question

I'm self studying How to Prove book and have been working out the following problem in which I have to analyze it to logical form:

Nobody in the calculus class is smarter than everybody in the discrete math class

Now, this is how, I started solving it:

¬(Somebody in the calculus class is smarter than everybody in the discrete math class) ¬(If x is in calculus class then x is smartert than everybody in the discrete maths class)

C(x) = x is in calculus class. D(y) = y is in discrete class. S(x,y) = x is smarter than y

¬∃x(C(x) -> ∀y( D(y) ∧ S(x,y)))

But this is the solution given in the Velleman's book:

¬∃x[C(x) ∧ ∀y(D(y) → S(x, y))]

I cannot understand how that answer is correct. Can someone explain the thing I'm missing there ?

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There is also a related question asked there but that doesn't discuss the Velleman's answer per se.

Answer 1

Your answer asserts that there does not exist anyone $x$, who, iF $x$ is in Calculus, then (all students y are both in Discrete math and x is smarter than them.) This is clearly not what is conveyed in the original statement.

What we need, essentially, is "There does not exist someone $x$ who is enrolled in Calculus AND such that, for all students y, if y is enrolled in Discrete math, then x is smarter than y. $$\lnot \exists x\Big(C(x) \land \forall y(D(x) \rightarrow S(x, y))\Big)$$

Answer 2

Let $C$ denote the set of members of calculus class and let $D$ the set of members of discrete math class.

The following statements are equivalent (explore step by step) and the last one is the Velleman answer:

1) Nobody in the calculus class is smarter than everybody in the discrete math class

2) For every person $x$ in $C$ there is a person $y$ in $D$ such that $\neg S\left(x,y\right)$

3) $\forall x\in C\exists y\in D\left[\neg S\left(x,y\right)\right]$

4) $\forall x\left[x\in C\Rightarrow\exists y\left[y\in D\wedge\neg S\left(x,y\right)\right]\right]$

5) $\neg\exists x\neg\left[x\in C\Rightarrow\exists y\left[y\in D\wedge\neg S\left(x,y\right)\right]\right]$

6) $\neg\exists x\neg\left[x\notin C\vee\exists y\left[y\in D\wedge\neg S\left(x,y\right)\right]\right]$

7) $\neg\exists x\left[x\in C\wedge\neg\exists y\left[y\in D\wedge\neg S\left(x,y\right)\right]\right]$

8) $\neg\exists x\left[x\in C\wedge\forall y\left[y\notin D\vee S\left(x,y\right)\right]\right]$

9) $\neg\exists x\left[x\in C\wedge\forall y\left[y\in D\Rightarrow S\left(x,y\right)\right]\right]$

There is quite some redundancy here, but I hope this give you understanding about the correctness of the answer.