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Maybe I'm just need to buff up on my logic notation, but I don't fully understand the following:

$\exists y\forall z \left(\exists w(z\in w\wedge w\in x)\implies z\in y\right)$

How should I unravel these statements generally? Starting with the innermost parens? As best I can tell the part starting with $\exists w$ means that there exists some $w$ such that $z$ is an element of $w$ and $w$ is an element $x$ which implies that $z$ is an element of $y$. But I dont understand how to parse $\exists y \forall z$ type statements (i.e. when they're up against each other like that). How do I even read that? "There's some element $y$ for all $z$'s"?

As you can tell, I'm generally confused. Can someone provide some guidance?

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Read left to right. Remember that everything is a set. So $x$ is a set of (what else!) sets. For readability I would write $w\in x\land z\in w$ instead of the (equivalent) other way around.

The first thing to note is $\exists y$. It says there is a set with certain properties. In set theory, most of the time one describes the properties by specifying what the elements (of $y$ in this case) are. So next I see $z\in y$. This has something to do with describing the elements $z$ of $y$, though there will be a small complication.

Think of $x$ as a plastic bag full of plastic bags that contain stuff.

The part $\exists w(w\in x \land z\in w)$ says there is a plastic bag $w$ in $x$ such that $z$ is in that bag. Suppose magically the walls of the plastic bags in $x$ decay. Then all the $z$'s that were contained in any of these bags spill out. The union that the Axiom of Union produces is almost the combined contents of the plastic bags in $x$. But not quite.

The formula $(w\in x \land z\in w)\implies z\in y$actually only says that $y$ includes these contents. That's because by a separate axiom (usually called Separation, or Cut, or a consequence of Replacement) we can cut down $y$ to be precisely these contents.

It might have been better for clarity to say that $z\in y$ iff $\dots$. But it is considered unfashionable by some people to use a stronger-seeming axiom when a weaker one will do.

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    Correct, if we assume $\implies$ is replaced by biconditional. What it actually says is that $y$ is **a** set such that (any $z$ for which there is some $w$ such that $z$ is in $w$) is in $y$.2012-12-05