Are these two logical statements involving the unique existential quantifier equivalent?

Question

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.

Answer 1

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.

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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.

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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".

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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.