Is $(\forall y\in y)(\exists x \in x)(x=y)$ true in this structure?

Question

Again the sentence is: $(\forall y\in y)(\exists x \in x)(x=y)$

The structure $\mathfrak{A}$ in question is the standard one for the language of set theory $\mathcal{L}_{ST}$ but where the universe $|\mathfrak{A}|$ is the following set:

$|\mathfrak{A}|=\{u,v,w,\{u\},\{u,v\},\{u,v,w\}\}$

Further the notation $(\forall y \in b)( \phi(y))$ abbreviates $(\forall y)[y\in b\rightarrow\phi(y)]$ and $(\exists x \in b)(\phi(x))$ abbreviates $(\exists x)[x\in b \land\phi(x)]$.

Initially, I thought this sentence was saying that each element in $A$ is identical to some element in $A$, which seems true. But then, looking a bit closer at the notation, I think rather that it is saying that each set in $A$ is identical to some set in $A$. The only sets in $A$ are $\{u\},\{u,v\},$ and $\{u,v,w\}$. And clearly none of these sets are identical to one another. So the sentence is false.

My questions are two-fold:

First, am I right in thinking the sentence false?

If so my second question regards the abbreviations: In the abbreviations the quantifiers bind a variable that ranges over elements in a set. But in the sentence, the quantifier is binding a variable for sets. Given this is the sentence even well-formed? And if not, then wouldn't it be neither true nor false?

Thanks for the help. For reference this is Exercise 1.7.5 in Leary and Kristiansan's A Friendly Introduction to Mathematical Logic pg. 33.

Answer 1

Hint

The authors say:

In particular, notice that there is no element $x$ of $A$ such that $x \in x$.

The formula $(∀y∈y)(∃x∈x)(x=y)$ is:

$(∀y)[(y∈y) \to (∃x∈x)(x=y)]$

and we know that $(y∈y)$ is always false in $A$.

Thus, we have a conditional with a false antecedent.

Answer 2

(Posted after previous answer was accepted.)

For what it's worth, for any binary relation $R$ in any "universe," we have:

$~~~~\forall x: [x R x \to \exists y: [y R y ~\land~ x=y]]$

Which can be abbreviated:

$~~~~\forall xRx: \exists yRy:x=y$

Proof

1. $a R~ a~~~$ (Premise)

2. $a=a~~~$ (Reflexivity, 1)

3. $a R a ~\land~ a=a~~~$ (Join, 1, 2)

4. $\exists y: [y R y ~\land~ a=y]~~~$ (E Gen, 3)

5. $\forall x: [x R x \to \exists y: [y R y ~\land~ x=y]]~~~$ (Conclusion, 1, 4)