Express the statements using quantifiers

Question

The statement :

Every student in this class has taken exactly two mathematics classes at this school.

My idea of answer :

It should probably be using a bi-conditional but I cant understand how to frame it. Probably something like :

$$∀x∀y( T(x,y) <=> ( M1(y) \lor M2(y) ) )$$

where T(x,y) means X is taking class y, M1(x) means x is first math class, M2(x) means x is second maths class. The domain for x is all studint in this class and for y is maths classes in this school.

I'm not at all convinced by the answer. Couldn't find anywhere else either.

Answer 1

Here is another way to write it, using the biconditional that you intuitively 'feel' is involved here (I use $M(x)$ for '$x$ is a math course at this school)

$$\forall x (S(x) \to \exists y \exists z (y \not = z \land \forall w ((w = y \lor w = z) \leftrightarrow (M(w) \land T(x,w))))$$

This really is just a condensed version of Pe's second answer:

Going from left to right: if $w$ is either $y$ or $z$, then that will make $z$ (and thus both $y$ and $z$) a math course taken by the student at this school, and hence the student has taken at least two math courses at this school.

Going from right to left: if $w$ is a a math course taken by the student at this school, then it has to be either $y$ or $z$, and hence the student has taken at most two math courses at this school.

Combined, this gives that the student has taken exactly two math courses at this school.

Answer 2

There are a number of ways that it can be written, but the following expresses it as variable which can be used elsewhere:

• Sx = x is a student in this class
• $q_x$ = quantity of math classes taken by x

$\forall x[Sx \to q_x=2] \\$

In response to your comment, the following expresses the same idea as a simple predicate.

• Mx = x has taken two math classes.

$\forall x[Sx \to Mx] \\$

Here's another way to express it:

• Cx = x is a math class
• Txy = x has taken y

$\forall x[Sx \to \exists yz[Cy \land Cz \land y \neq z \land Txy \land Txz \land \forall w[(Cw \land Txw) \to (w=y \lor w=z)]]] \\$

Notice that the final expression concerning $w$ is a common way to express the idea that there are exactly two and no more. In other words, if there is any such $w$, it must be identical to either $y$ or $z$. Otherwise, there would be more than just two math classes taken by $x$.