I am new to discrete math.
Let $C(x,y)$ be the predicate "$x$ is a student in Class $y$"
$A(x,z)$ be the predicate "$x$ has installed the application $z$ on his/her mobile"
where the domain for $x$ contain all students in the college, $y$ contains all classes in the college, and $z$ contains all mobile applicatons.
The question is:
In the college, there is exactly one class in which no student has installed the application "Firegram".
What I got is:
$¬∃x(C(x,y)∧ A(x,"Firegram))$
But I don't know how to continue with the term "exactly one".
Can someone help and explain it?
Best Answer
First, a little comment on this:
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:
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 $z$)