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.
3 Answers
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.
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 $x$.
Let x be the domain of students in this class, y and z be the domain math classes, and the predicate M(x,y) student x taken math class y
$$ \forall x \exists y \exists z ( y \ne z \land \forall w (M(x,w) \leftrightarrow ( w = y \lor w=z))) $$
