Express the statements using quantifiers
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.
$\endgroup$3 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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 $x$.
$\endgroup$ 6 $\begingroup$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
$$ \forall x \exists y \exists z ( y \ne z \land \forall w (M(x,w) \leftrightarrow ( w = y \lor w=z))) $$
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