can ∃x∀y and ∀y∃x sometime be equivalence?

Question

I already know that I can't swap the position of the quantifier and expected it to be equivalence. However, I wonder if it is actually equivalence for some special cases.

Example: Let Q(x,y) denote "x has sent a message to y.", domain for both x and y is all student in your class.

∃x∀yQ(x,y) means: There is a student in your class send a message to all students in your class.

∀y∃xQ(x,y) means: Every student in your class has been sent a message by a student.

Doesn't the translation of these 2 statements seem to carry the same information, and is it just a different way of saying the same thing? Or is there something that I fail to realize?

Answer 1

These do not convey the same information. $\exists x \forall y Q(x, y)$ means "there exists a single student who has sent a message to everyone in the class". $\forall y \exists x Q(x, y)$ means "every student in the class has been sent a message by someone".

In the former, there is one choice of $x$ that satisfies $Q(x, y)$ for all $y$, meaning that there is a single student who has been messaging everyone in class.

In the latter, for each $y$ we can find an $x$ that satisfies $Q(x, y)$, but there's no guarantee that it will be the same $x$ for different choices of $y$, so maybe Sally received a message from Billy but Eve received one from Johnny.

If the former is true then the latter is also true (once we know that there's a student who sent messages to everyone, we know that everyone in the class received a message from someone), but it doesn't work the other way around.

Answer 2

Another way to think of the difference between the two statements is how many choices you are allowed to make to fufill the "there exists" clause in each case.

• In the $∃x∀yQ(x,y)$ case, we are implying that we can choose one student that fufills a particular condition, i.e. that that one student has messaged all students in the class. If you were to imagine a conversation where I wanted you to prove that this statement was true, then you have to pick that one student $x$ first and I would get free choice of trying to find a counterexample $y$ where it wasn't true.
• In the $∀y∃xQ(x,y)$ case, we get to make a choice per student. Analogously, if I were to challenge you to find a student $x$, I have to pick a $y$ first, which means that you can make different choices for which student should be $x$ (i.e. has messaged $y$) depending on which I have chosen.

The former case being true implies the latter is true, but not vice-versa. The latter might be true when the former is false when you can, given any student, name another student who has messaged them, but not always the same student.