The statement is wrong, among other reasons, because the statement
$C(x)$ implies $D(x)$
is true for any $x$ that is not a computer science student, and for any $x$ that takes discrete math. In particular, the statement
There exists $x\in S$ such that $C(x)$ implies $D(x)$. $\qquad\qquad\qquad$ (1)
would be true if nobody is a computer science major. However, I think most people agree that
Some computer science majors take discrete math
should be considered false if there are no computer science majors at all.
Also, the statement (1) would be true if there is at least one person taking discrete math, whether or not that person is a computer science major. So, in a university in which at least one person takes Discrete Math, but no computer science major does, the statement "There exists $x\in S$ such that $C(x)$ implies $D(x)$" would be true, but the statement "Some computer science majors take discrete math" would be false.
What you need to remember is that an implication is true if the antecedent is false, or if the consequent is true. You don't need the antecedent to be true.
What you actually want is:
There exists $x\in S$ such that $C(x)$ and $D(x)$.
I guess they are making the following general point (to tinker with an answer I gave to another question here):
As background, start with the trite remark that mathematicians are very, very often concerned with statements of generality and especially statements of multiple generality – you know the kind of thing, e.g. the definition of continuity that starts for any $\epsilon$ ... there is a $\delta$ ... And the formalized quantifier-variable notation which we all learn serves mathematicians brilliantly to regiment such statements of multiple generality and make them utterly unambiguous and transparent (as contrasted with some of their ordinary-language counterparts).
OK, so now think about restricted quantifiers that talk about only some of a domain (e.g. talk not about all numbers but just about all the even ones). How might we render Goldbach's Conjecture, say? As a first step, we might try
$\forall n$(if $n$ is even and greater than 2, then $n$ is the sum of
two primes)
Here restrict the quantifier using a conditional, and this seems at first sight to give us what we want. But now think about the embedded conditional here. In particular, ask: what if $n$ is odd, so the antecedent of the conditional is false??? If we say this instance of the conditional lacks a truth-value, or may be false, then the quantification would have non-true instances and so would not be true! But of course we, we all agree that we can't refute Goldbach's Conjecture by looking at odd numbers!! So if the quantified conditional is indeed to come out true when Goldbach is right, then we'll have to say that the irrelevant instances of the conditional with a false antecedent have to be treated as non-false by default. And in classical logic, non-falsity implies truth. Which means that the embedded conditional will have to be treated as a material (truth-functional) conditional.
So the moral seems to be: if mathematicians are to deal nicely with expressions of generality using the quantifier-variable notation they will have to get used to using material conditionals too.
Best Answer
One way of approaching a formal proof of this, is by induction, I'll do negation for you, and you can see if you can modify it to do conjunction.
Notice that if you have an expression $\phi$ which is equivalent to $\neg A$, which has any occurrences of any propositional variables other than $A$, then, you can just replace them with $A$. Hence, we only need to consider sentences in $A$ and $\rightarrow$.
Claim: Any sentence in $A$ and $\rightarrow$ evaluates to true, when $A$ is true.
Proof: By induction on the length of the sentence. The base case is the sentence $A$, and clearly this is true with $A$ is true. In the induction step the expression is of the form $\phi \rightarrow \psi$, for some sentences $\phi$ and $\psi$ in only $A$ and $\rightarrow$. Also $\phi$ and $\psi$ are shorter, so, we can apply the induction hypothesis. In particular when $A$ is true $\phi$ and $\psi$ are both true. But then, by the semantics for $\rightarrow$, when $A$ is true $\phi\rightarrow \psi$ is true too.
Q.E.D.
So, any expression we can make preserves truth, and so, can't be equivalent to negation.