Logic: changing quantified variable names and moving quantifiers Let's say I have
$$\exists x (x-2=0) \lor \exists x (x-3=0) \tag{1}$$
in the set of real numbers.
I have recently learned that I can rename one of the two $x$:
$$\exists x (x-2=0) \lor \exists y (y-3=0) \tag{2}$$
But I also know that I can start from $(1)$ and use only one $\exists$:
$$\exists x (x-2=0 \lor x-3=0) \tag{3}$$
And I also know that I can start from $(2)$ and move both $\exists$ at the beginning:
$$\exists x \exists y (x-2=0 \lor y-3=0) \tag{4}$$
Based on what I know, $(1)$, $(2)$, $(3)$ and $(4)$ are alla equivalent each other. And that confuses me. For example $(2)$ is true when

*

*$x=2$ and $y$ is a real number such that $y-3=0$ is false (i.e. $y \ne 3$);

*$y=3$ and $x$ is a real number such that $x-2=0$ is false (i.e. $x \ne 2$);

*$x=2$ and $y=3$.

While $(3)$ is true only in two ways:

*

*with $x=2$ that makes $x-2=0$ true and $x-3=0$ false

*with $x=3$ that makes $x-2=0$ false and $x-3=0$ true

so I miss the case where both $x-2=0$ and $x-3=0$ are true!
Also $(1)$ is true in three cases:

*

*when the first $x$ is equal to $2$ and the second $x$ is not equal to $3$

*when the first $x$ is not equal to $2$ and the second $x$ is equal to $3$

*when the first $x$ is equal to $2$ and the second $x$ is equal to $3$
while I know for sure that $(1)$ and $(3)$ are equivalent because of the "distributive property" of $\exists$ over $\lor$, so how is it possible that $(1)$ is true in three cases and $(3)$ is true only in two cases? How can $(1)$ and $(3)$ be equivalent if they are not true in the same cases?
So it seems clear to me that I'm missing something, that I don't understand something. What am I doing wrong? What formulas are equivalent and what are not equivalent? And why?
Thanks
 A: It is not correct to say that your statements are true in same cases
and in other cases they are false. All cases (1) to (4) are true statements.
There is no alternative.
If you had the open formulas "$x-2=0\vee y-3=0$"
and "$x-2=0\vee x-3=0$", then you would have the cases you speak
about. This is because, in the open formulas, there are
free variable which can take various values. Of course, these are different: in the second, both sides of the disjunction can not be true at the same time, while in the first, this is possible.
But your question is about statements, not open formulas. There are no free variables there,
only the constants 0,2,3 and the bounded variables, and that is why
they are statements and not open formulas. So, it is true that "there
is an $x$ such that $x-2=0$ or there is an $x$ such that $x-3$",
because there are indeed such numbers.
To make it more clear, lets see a different set of examples. Let $c_{1}$
and $c_{2}$ be some constants, from the set of natural numbers. Then
we can prove that the following statements are equivalent:

*

*$\exists x\in\mathbb{N}\left(2-x=c_{1}\right)\vee\exists x\in\mathbb{N}\left(3-x=c_{2}\right)$

*$\exists x\in\mathbb{N}\left(2-x=c_{1}\right)\vee\exists y\in\mathbb{N}\left(3-y=c_{2}\right)$

*$\exists x\in\mathbb{N}\left(2-x=c_{1}\vee3-x=c_{2}\right)$

*$\exists x\in\mathbb{N}\exists y\in\mathbb{N}\left(2-x=c_{1}\vee3-y=c_{2}\right)$
These statements may be true or they may be false, depending on $c_{1}$
and $c_{2}$. Let try some values: If $c_{1}=0$ and $c_{2}=0$, then
all four of them are true. If $c_{1}=8$ and $c_{2}=1$, then all
four of them are true. If $c_{1}=8$ and $c_{2}=9$, then all four
of them are false. There is no case in which some of them are true
and some of them are false. They are either all true or all false.
That is why they are equivalent. In your example, because the statements
are true anyway, the equivalence is trivial.
