Are these formal formulas equivalent? My textbook gave the following
$ \forall x_0 (\exists x_1 \ x_0=(\mathbf{O''} \cdot x_1) \vee \exists x_1 \ x_0=((\mathbf{O''} \cdot x_1)+\mathbf{O'}))  $,
then commented on the syntax and why the brackets are so important here, etc...
However, for this example, I gave 
$ \forall x_0 \ \exists x_1 (\ x_0=(\mathbf{O''} \cdot x_1) \vee x_0=(\mathbf{O''} \cdot x_1)+\mathbf{O'}) $.
To me these are equivalent, but my text didn't mention this form and didn't necessarily suggest it's equivalent to the first. 
So are they equivalent or the way I'm using $ \ \exists x_1 $  gives my formula a different meaning?
Thanks

Edit
In other words, can the existential quantifier be "factored" like I did? 
 A: The claim that either there is some object that is $P$ or there is some object that is $Q$, $\exists x_1 P(x_1) \lor \exists x_1 Q(x_1)$, is logically equivalent to the claim that there is some object that is either $P$ or $Q$, $\exists x_1 (P(x_1) \lor Q(x_1))$:
$\exists x_1 P(x_1) \lor \exists x_1 Q(x_1) \iff \exists x_1 (P(x_1) \lor Q(x_1))$.

Notice the same is not true for existential quantification with conjunction. For example, there exists a cat and there exists a dog, $\exists x \text{Cat}(x) \land \exists x \text{Dog}(x)$, but this does not exactly mean there exists something which is both a cat and a dog, $\exists x (\text{Cat}(x) \land \text{Dog}(x))$.

As a caveat to the preceding image, it would be true to say that all objects are both a cat and a dog, $\forall x (\text{Cat}(x) \land \text{Dog}(x))$, if and only if all objects are a cat and all objects are a dog, $\forall x \text{Cat}(x) \land \forall x \text{Dog}(x).$
A: The existential quantifier distributes over disjunction. So the two sentences are indeed logically equivalent.
