Nested Quantification of exactly one.

Question

Suppose my domain is "All students in the class" and P(x, y):= x has emailed y.

So, how do i define:

1. Every student has emailed exactly one student.
2. Exactly one student has emailed every one.

A proper explanation would be really helpful.

Answer 1

"Exactly one" means "at least one, and at most one".

We express "there is at least one $z$ such that $\phi(z)$" with $$ \exists z\, \phi(z). $$ We express "there is exactly one $z$ such that $\phi(z)$" as follows: $$ \exists z\, (\phi(z) \land \forall u\,(\phi(u)\to u=z)).\tag{*} $$ This says, there is something $z$ such that $\phi(z)$ (at least one), and anything else $u$ such that $\phi(u)$ must equal $z$ (at most one).

Now for your two statements.

Every student has emailed exactly one student.

Using variables, the statement is: Every student $x$ has emailed exactly one student $y$. I take this to mean: For every student $x$, there is exactly one $y$ such that $x$ has emailed $y$. Using the analysis of "exactly one" above, we can render this symbolically as follows, using $\phi(y) := P(x,y)$ and universally quantifying: $$ \forall x \exists y\, (P(x,y) \land \forall u\,(P(x,u)\to u=y)). $$

Exactly one student has emailed everyone.

We want a formula for "$x$ has emailed everyone" (i.e. every student, presumably including self). $\phi(x) := \forall y\,P(x,y)$ says that. Now we only have to say there's exactly one such $x$: $$ \exists x\, (\forall y\,P(x,y) \land \forall u\,(\forall y\,P(u,y)\to u=x)). $$