why are $\forall x(P(x)\rightarrow \exists y(Q(y)∧R(x,y)))\text{ and }\exists y(Q(y)∧\forall x(P(x)\rightarrow (R(x,y)))$ not logically equivalent?

Question

been sitting here for hours and still can't figure this out. is the order of $\forall x$ and $\exists y$ important in this case?

all I can think of now is "all P is R of some Q", but I don't think this is right. Would there be any counter examples?

Answer 1

Is the order of ∀x and ∃y important in this case?

Yes, in general the order of quantifiers is important : $∃y∀x A(x,y)$ is not equivalent to : $∀x∃y A(x,y)$.

Would there be any counter examples?

Here is a simple counter-example with domain $\mathbb N$ :

$∀x∃y (x < y)$ vs $∃y∀x (x < y)$.

Answer 2

In the first one, each $x$ has its own $y$. In the second one, there is one $y$ that works for everyone.

Answer 3

To get a feel for the difference in these statements:

take as domain $\Bbb N$ and as the predicates $P(x)$: "$x$ is even" and $Q(x)$: "$x$ is odd" and $R(x,y)$ : $x < y$.

Then $\forall x:(P(x)\rightarrow \exists y: (Q(y)\land R(x,y)))$ means that for every even number there is some odd number larger than it, which is true: for $2n$ take $2n+1$, e.g.

While $\exists y: (Q(y)\land \forall x(P(x)\rightarrow (R(x,y)))$ means that there is some odd $y$ that is larger than any even number, which is patently false.