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Let C(x,y) be student x is in class y. Where the domain for x is all the students in my school and domain of y is the set of all classes in my school.

Given ∃x,y ∀z((x≠y)∧(C(x,z)↔C(y,z))) , the answer given to me is there are exactly 2 students that take the same classes z. Why there can't be more than 2?

Given ∃x,y ∀z((x≠y)∧(C(x,z)->C(y,z))), the answer for this is that there are at least 2 students that take the same classes

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    For the second, the given answer is not right either. The sentence says there are (at least) two students, say Xavier and Yolande, such that whenever Xavier is in a class, so is Yolande. But Yolande might be taking **more** classes than Xavier.2012-10-25

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What you were told is incorrect: neither sentence means what you were told. The sentence

$\exists x,y~\forall z\Big((x\ne y)\land\big(C(x,z)\leftrightarrow C(y,z)\big)\Big)$

says that there are (at least) two distinct students who take exactly the same classes: $x$ takes class $z$ if and only if $y$ takes class $z$. There may be any number of other students who also take exactly this set of classes; nothing in the sentence excludes that possibility.

The sentence

$\exists x,y~\forall z\Big((x\ne y)\land\big(C(x,z)\to C(y,z)\big)\Big)$

says that there are students $x$ and $y$ who are not the same student and who are such that if $x$ takes a class, $y$ also takes that class. In other words, $y$ takes every class that $x$ takes, and possibly (but not necessarily) some that $x$ doesn’t take. This means, for instance, that we could not have just three students, $a,b$, and $c$, and three classes, $C_1,C_2$, and $C_3$, such that $a$ takes $C_1$ and $C_2$, $b$ takes $C_2$ and $C_3$, and $c$ takes $C_3$ and $C_1$: in that setup no student takes all of the classes that some other student takes.

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    Just to emphasize what's implicit in your answer and contrary to the interpretations cited in the question: The second sentence can be true even if no two students take exactly the same classes.2016-07-29