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Does the quantifier order matter in this problem? I don't think it does.

(∃y in R)(∀x in R)(x+y=x) For some real number y and for all real numbers x, x plus y equals x. I assert this to be true. (I won't bother with the proof unless somebody really wants it)

I feel that, although the quantifiers are switched, the sentence is still true. Is this right?

(∀x in R)(∃y in R)(x+y=x) For all real numbers x there is a real number y such that x plus y = x

I also feel that these sentences are also all true, regardless of their order. Am I missing something here?

(∃y in R)(∀x in R)(x+y=0) and (∀x in R)(∃y in R)(x+y=0).

Plus these two sentences (∃y in R)(∀x in R)(xy=1) and (∀x in R)(∃y in R)(xy=1)

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    Yes, in this **particular** case the order does not matter – 2017-01-31
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    It does not matter in the sense that both are true, but they still mean different things. – 2017-01-31
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    By the way, whenever you have ∀ followed by ∃, you can think of that as a function (assuming the axiom of choice). For instance, if you have ∀x. ∃y. x + y = x, then that means that for each x, you have a value named y -- ie, f(x) such that x + y = x -- ie, x + f(x) = x. In this case, since f(x) = 0 satisfies this, you can use 0 as your y, regardless of what x is. – 2017-01-31
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    @Daniel l also feel that these sentences are also all true, regardless of their order. Am I missing something here? (∃y in R)(∀x in R)(x+y=0) and (∀x in R)(∃y in R)(x+y=0). Plus these two sentences (∃y in R)(∀x in R)(xy=1) and (∀x in R)(∃y in R)(xy=1) – 2017-01-31
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    @frillybob the last two sentences are both false, but of the first pair one is true and the other false. – 2017-01-31

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Just because these two sentences have the same truth-value (true for both) does not mean they are equivalent. Indeed, please do not think that just because the truth-value didn't change for this example, that they will never change in general, as in general, swapping quantifiers will change the meaning of the statement.

Indeed, your follow-up examples are a perfect case in point:

$\exists y \in R \: \forall x \in R \: x+y=0$ is false, (since there is no $y$ which has the property such that if you add it to any $x$, it will add to 0)

But

$\forall x \in R \: \exists y \in R \: x+y=0$ is true! (since for any $x$ we can pick $y =-x$)