Are these two logical statements involving the unique existential quantifier equivalent? Let $\mathit T(x,y)$ mean "x is a teacher of y." Are the two following logical statements equivalent? 
$$ \exists!x\exists!yT(x,y)  $$
and 
$$ \exists x\exists y\biggl(T(x,y) \land \lnot\Bigl(\exists u\exists v\bigl(T(u,v) \land (u\neq x \lor v \neq y\bigr)\Bigr)\biggr)$$
My class today had a big discussion about this and there is still some deabate on whether or not it they are both equivalent.
 A: Let us examine the situation with the example you gave in a comment:
$$M := \{ x, y, u, v, f \}, \\
T := \{ (x, y), (u, v), (u, f) \}.$$
Now in order to avoid confusion with the members of $M$ let us rename the variables so that the first statement reads $\exists! a \exists! b T(a, b)$.
In order to evaluate the truth value of this statement, let us start with the inner unique existential quantifier, $\exists! b T(a, b)$.  This has a free variable $a$, and we will eventually need its evaluation when $a$ ranges over all values of $M$.  Check that:
$$(\exists!b T(a, b))[a/x] = \exists!b T(x, b) = T,\\
(\exists!b T(a, b))[a/y] = \exists!b T(y, b) = F,\\
(\exists!b T(a, b))[a/u] = \exists!b T(u, b) = F,\\
(\exists!b T(a, b))[a/v] = \exists!b T(v, b) = F,\\
(\exists!b T(a, b))[a/f] = \exists!b T(f, b) = F.$$
From this, you can conclude that $\exists! a \exists! b T(a, b)$ is true.
However, as you observe, $\exists a \exists b (T(a, b) \wedge \lnot(\exists c \exists d (T(c, d) \wedge (a \ne c \vee b \ne d))))$ is false.  To do this in full detail as in the previous part would require over $5^4$ evaluations, yet it should hopefully be clear that since $T$ has more than one element, $\exists c \exists d (T(c, d) \wedge (a \ne c \vee b \ne d))$ is always true no matter what $a$ and $b$ are, and from there you can conclude the overall statement is false.

Intuitively, you might think of the first statement as expressing "there is exactly one teacher who has exactly one student", whereas the second statement expresses "there is exactly one teacher-student pair", and convince yourself why the two ideas are expressing different things.

Some other counterexamples you might consider: if the domain of discourse is $\mathbb{N}$, then $\exists! x \exists! y, x \ge y$ is true, and so is $\exists! x \exists! y, x > y$.  Yet in both cases, there are clearly numerous distinct pairs of $x, y$ such that $x \ge y$, or distinct pairs of $x, y$ such that $x > y$.
In the original example of $M$ and $T$ above, we have $\exists! a \exists! b T(a, b)$ is true but $\exists! b \exists! a T(a, b)$ is false.  Therefore, $\exists! a$ and $\exists! b$ do not "commute".

The second statement above would indeed be a good formalization of the notion that there exists a unique combination of $x$ and $y$ satisfying $T$.  Thus, if that was what you meant in some context, you would want to avoid $\exists! x \exists! y$ because that doesn't mean the same thing.  You might see notations such as $\exists! (x, y), T(x, y)$ for the second.
