I want to be sure that I understand how to show that a structure satisfies a sentence under a variable assignment, and suspect that I'm handling the computation of multiple quantifiers incorrectly. Consider the structure,
(1) $M = \langle \{a\}, \emptyset \rangle$ (i.e., the structure whose domain contains only $a$ and whose "membership" relation is just the empty set)
and the sentence,
(2) $\exists y \forall x (x \not \in y)$ (i.e., the Empty Set Axiom)
Now, (1) is evidently a model of (2). But is the following argument mistaken (where $a$ is any variable assignment, and $a~ (x/b)$ is the variable assignment just like $a$ but which maps $x$ to $b$)?
$$\langle \{a\},~ \emptyset \rangle \models \exists y \forall x (x \not \in y)~[a]$$
If and only if for some $b \in \{a\}$:
$$\langle \{a\},~ \emptyset \rangle \models \forall x (x \not \in y)~[a~(y/b)]$$
If and only if for some $b \in \{a\}$ and for all $c \in \{a\}$:
$$\langle \{a\},~ \emptyset \rangle \models (x \not \in y)~[a~(y/b)~(x/c)]$$
If and only if for some $b \in \{a\}$ and for all $c \in \{a\}$:
$$\lnot\bigg(\langle \{a\},~ \emptyset \rangle \models (x \not \in y)~[a~(y/b)~(x/c)]\bigg)$$
which is true, since all $c \in \{a\}$ are equal to $a$, and $a$ is not in $\emptyset$.