Vtyjkyuk

My textbook gave the following

$ forall x_0 (exists x_1 x_0=(mathbfO'' cdot x_1) vee exists x_1 x_0=((mathbfO'' cdot x_1)+mathbfO')) $,

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=(mathbfO'' cdot x_1) vee x_0=(mathbfO'' cdot x_1)+mathbfO') $.

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?

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, $exists x textCat(x) land exists x textDog(x)$, but this does not exactly mean there exists something which is both a cat and a dog, $exists x (textCat(x) land textDog(x))$.

As a caveat to the preceding image, it would be true to say that all objects are both a cat and a dog, $forall x (textCat(x) land textDog(x))$, if and only if all objects are a cat and all objects are a dog, $forall x textCat(x) land forall x textDog(x).$

The existential quantifier distributes over disjunction. So the two sentences are indeed logically equivalent.