# Why is $\exists x\exists y\colon A(x, y)$ an appropriate formalization of the sentence: 'There is an $x$ and an $y$ such that $A(x, y)$ is true'?

Suppose that I state in natural language the following theorem:

There is an $x$ and an $y$ such that $A(x, y)$ is true.

Why is $\exists x\exists y\colon A(x, y)$ an appropriate formalization of this sentence?

I interpret the formula $\exists x\exists y\colon A(x, y)$ as saying "there is an $x$ for which there is an $y$ such that $A(x, y)$". On the other hand, I interpret the above natural language sentence as asserting "there is a pair of objects $(x,y)$ for which $A(x, y)$ is true". Can you convince me that the two are essentially the same?

• Stating "$\exists{x,y}:A(x,y)$" is good enough... – barak manos Sep 20 '16 at 18:03
• @barak: Could please explain your point? – user370728 Sep 20 '16 at 18:03
• Interesting: About ten seconds after I posted this question, I got a dislike. Seems like this person hasn't read my question. – user370728 Sep 20 '16 at 18:04
• I don't think that the down-voter hasn't read the question. More likely, he or she couldn't really spot any question (to be honest, I'm not sure I can either). But in general, try not to get yourself too annoyed with down-votes on this website. I've seen many such unjustified down-votes here, and unfortunately, the rules do not impose that the down-voter must explain the reason. If the community likes your question, then you'll get back your points sooner or later :) – barak manos Sep 20 '16 at 18:08
• @user370728: Long observation suggests that a number of users downvote questions that they consider so trivial that they shouldn’t have needed to be asked, and I suspect that that’s what happened here. (Note: I am not endorsing that judgement: I’m just suggesting a likely reason for the downvote.) – Brian M. Scott Sep 20 '16 at 18:10

Both $\exists x\,\exists y\,A(x,y)$ and $\exists y\,\exists x\,A(x,y)$ say in effect that there are objects $a$ and $b$ in the universe of discourse such that $A(a,b)$ is a true statement about $a$ and $b$; changing the order of the two existential quantifiers has no effect on the meaning of the expression. Perhaps the easiest way to see this is to ask what it means for each of the versions to be false.

$\exists x\,\exists y\,A(x,y)$ is false if no matter what object $a$ you choose for $x$, $\exists y\,A(a,y)$ is false. And $\exists y\,A(a,y)$ is false if no matter what value $b$ you choose for the variable $y$, $A(a,b)$ is false. The net effect is that $\exists x\,\exists y\,A(x,y)$ is false if no matter how you choose values for the variables $x$ and $y$, the statement $A(x,y)$ is false for those values.

$\exists y\,\exists x\,A(x,y)$ is false if no matter what object $b$ you choose for $y$, $\exists x\,A(x,y)$ is false. And $\exists x\,A(x,b)$ is false if no matter what value $a$ you choose for the variable $x$, $A(a,b)$ is false. Once again the net effect is that $\exists x\,\exists y\,A(x,y)$ is false if no matter how you choose values for the variables $x$ and $y$, the statement $A(x,y)$ is false for those values.

This shows that $\exists x\,\exists y\,A(x,y)$ and $\exists y\,\exists x\,A(x,y)$ are false under exactly the same circumstances, so they must also be true under exactly the same circumstances. In other words, they are equivalent statements: both simply say that there is some way to pick values $a$ and $b$ for $x$ and $y$, respectively, so that $A(a,b)$ is a true statement about those values.

People often write simply $\exists x,y\,A(x,y)$; this is partly for brevity, but it’s also partly to try to emphasize that we’re asserting the simultaneous existence of $x$ and $y$ with the desired property.

Is the following convincing enough?

Assume there's an $x$ and a $y$, such that $A(x,y)$ is true. Then take that $x$ and (for that $x$, the) $y$ and thus you have an $x$ for which there's a $y$ such that $A(x,y)$ is true.

Assume there's an $x$ for which there's a $y$ such that $A(x,y)$ is true. Then take those, form a pair, and so there's an $(x,y)$ such that $A(x,y)$ is true.