Did I analyze the logical form of a statement well?

Question

Analyze the logical forms of the following statements. You may use the symbols ∈, !∈, =, !=, ∧, ∨, →, ↔, ∀, and ∃ in your answers, but not ⊆, ⊆, P , ∩, ∪, \, {, }, or ¬. (Thus, you must write out the definitions of some set theory notation, and you must use equivalences to get rid of any occurrences of ¬.)

Did I analyze the logical form of a statement well?

(d) P(∪i∈I Ai) ⊆ ∪i∈I P(Ai).

∀x(x∈P(∪i∈I Ai) -> x∈∪i∈I P(Ai))

∀x(∀y(y∈x -> y∈∪i∈I Ai) -> ∃i∈I x∈P(Ai))

∀x(∀y(y∈x -> ∃i∈I(y∈Ai)) -> ∃i∈I∀y(y∈x->y∈Ai))

∀x(¬∀y(y∈x -> ∃i∈I(y∈Ai)) ∨ ∃i∈I∀y(y∈x->y∈Ai))

∀x(¬∀y(y!∈x or ∃i∈I(y∈Ai)) ∨ ∃i∈I∀y(y∈x->y∈Ai))

∀x(∃y¬(y!∈x or ∃i∈I(y∈Ai)) ∨ ∃i∈I∀y(y∈x->y∈Ai))

∀x(∃y(y∈x ^ ¬∃i∈I(y∈Ai)) ∨ ∃i∈I∀y(y∈x->y∈Ai))

∀x(∃y(y∈x ^ ∀i∈I(y!∈Ai)) ∨ ∃i∈I∀y(y∈x->y∈Ai))

The textbook's answer is ∃x(∀y(y ∈ x → ∃i∈I (y∈Ai)) ∧ ∀i∈I∃y(y∈x ∧ y!∈Ai))

I seem to have got the negation of textbook answer.

Answer 1

Long comment

Consider the subformula inside the outer quantifier, after your third step you have :

$∀y(y∈x \to ∃i(y∈A_i)) \to ∃i∀y(y∈x \to y∈A_i))$.

Instead of using the equivalence : $p \to q$ and $\lnot p \lor q$, we can use the following one : $p \to q$ and $\lnot (p \land \lnot q)$.

Thus, we get :

$\lnot [∀y(y∈x \to ∃i(y∈A_i)) \land \lnot ∃i∀y(y∈x \to y∈A_i))]$,

i.e.

$\lnot [∀y(y∈x \to ∃i(y∈A_i)) \land ∀i∃y(y∈x \land y \notin A_i))]$.

Now, if we restore the outer quantifier, we have :

$\forall x \lnot [∀y(y∈x \to ∃i(y∈A_i)) \land ∀i∃y(y∈x \land y \notin A_i))]$,

i.e.

$\lnot \exists x [∀y(y∈x \to ∃i(y∈A_i)) \land ∀i∃y(y∈x \land y \notin A_i))]$.

You are right : it seems that there is a typo in the textbook's answer : a missing $\lnot$ ...