The order of quantifiers I am reviewing for an exam, and I have come across a question that does not contain a solution, so I wanted to verify my answer.
Question 1: If $\exists y \forall x P(x, y)$ is true, then $\forall x \exists y P(x, y)$ is also true.
To me that appears true, because if there exists at least one particular value of $y$ that works for every $x$, then for every $x$ there is at least one $y$ value that satisfies $P(x, y)$.
Question 2: If $\forall x \exists y P(x, y)$ is true, then $\exists y \forall x P(x, y)$ is also true.
I think this one is false. There can exists some value of $y$ that satisfies $P(x, y)$ for every value of $x$, but that does not mean that the value is the same for every $x$. 
Am I correct?
 A: Yes. The two statements read as:
1) There exists a $y$ such that for all $x \dots$
2) For all $x$, there exists a $y \dots$
The first statement says that I can pick a $y$ so that no matter what $x$ you give me, $P(x,y)$ will be true.
The second says that you can pick any $x$, and I can pick a $y$ that makes $P(x,y)$ true.
We might rewrite it to exemplify this:
1) There exists a $y$ such that for all $x^* \dots$
2) For all $x^*$, there exists a $y^* \dots$
So you can see that, in the first statement, $y$ exists independent of $x$ chosen, while in the second, $y^*$ is the value corresponding to $x^*$.
Thus, the implication is that $(1)\Rightarrow (2)$ (simply take $y^*$ to be $y$) but not vice versa.
A: Correct.  To demonstrate that something does not follow, it is often helpful to provide a concrete counterexample, e.g. you could assume that $P(x,y)$ stands for $x$ has $y$ as a parent. So then $\forall x \exists y P(x,y)$ becomes the claim that everyone has a parent (true), but $\exists y \forall x  P(x,y)$ becomes the claim that there is someone who is the parent of everyone (false)
