Frankly, I think the most popular textbooks, Johsonbaugh, Rosen, and Epps are wretched. They make a vibrant subject dull as dishwater. Grimaldi is a better book, though some students find it harder, I gather. Frankly, the best intro to discrete math I've ever seen is what I think was the first textbook on the subject, "Finite Mathematical Structures," by Kemeney, Snell and Thompson. At this moment, there's a cheap copy on amazon. You might give it a try.
Apart from that, look at the resources on the Art of Problem Solving website.
First, a little comment on this:
where the domain for $x$ contain all students in the college, $y$ contains all classes in the college, and $z$ contains all mobile applicatons.
OK, the problem is that you really can't just say that $x$ is used for students, and $y$ for classrooms, etc. Typically, whenever we use a variable, it is assumed to be just one of the objects of the domain, and if that domain includes students, classrooms, mobile apps, and what have you, then $x$ can be any of those, and the same for $y$, $z$, or any other variable.
This is exactly why we use a predicate like $S(x,y)$ for '$x$ is a student in classroom $y$' to make clear that we want $x$ to be a student, and $y$ a classroom
Next: you're off to an excellent start! Yes, let's first symbolize '$y$ is a classroom in which no student has installed the "Firegram" application' ... and your formula:
$\neg \exists x(C(x,y)∧ A(x,"Firegram))$
is exactly correct! Good job!
OK, but can we say that there is exactly one of these?
Well, in general, there are several ways to translate 'there is exactly one $P(x)$'
First, you can take the idea of "there is one $P$, but no second (different!) $P$. This gives you:
$$\exists x (P(x) \land \neg \exists y (y \neq x \land P(y)))$$
Second, you can say: "there is one $P$, and anything that is a $P$ will have to be that first one". This gives you:
$$\exists x (P(x) \land \forall y (P(y) \to y = x))$$
Finally, and most efficiently, you can do:
$$\exists x \forall y (P(y) \leftrightarrow y = x)$$
This one is probably the least immediately understandable, but here's why it works: Given that the formula contains $\forall y$, let's consider all possible valuies of $y$. Well, there are only two interesting cases. First, when you pick $y=x$, then that means that the right side of the biconditional is true, and hence we can infer the left side, meaning that $P(y)$, and thus $P(x)$. So, this statement does imply that there is at least one $P$. Second, for any $y$ that is different from $x$, the statement $y=x$ is false, and hence the left side of the biconditional should be false as well, and so we don't have $P(y)$ for any such $y$. In other words, anything other than this $x$ does not have property $P$. Together, this means that there is exactly one $P$.
So, using the last format, and plugging in your formula for '$y$ is a classroom in which no student has installed the "Firegram" application', we get:
$$\exists x \forall y (\neg \exists z(C(z,y)∧ A(z,"Firegram)) \leftrightarrow y = x)$$
(obviously, I had to change the variable $x$ you used for the students to something else so I changed that to $z$)
Best Answer
The statement is wrong, among other reasons, because the statement
is true for any $x$ that is not a computer science student, and for any $x$ that takes discrete math. In particular, the statement
would be true if nobody is a computer science major. However, I think most people agree that
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: