How to interpret the given Well Formed Formula? How to interpret the first order logic given by :- 
$$∀x∀yP(x,y)→∀x∀yP(y,x)$$

I can see that this is not a logical implication .
LHS can be interpreted as : (Just assuming)
Take Domain : ${1,2,3,4,5,6,7}$ and $P(x,y) = x<y$
So, $P(1,2) = P(3,7)$ is true and this make $RHS$ false.

But, How to read it.
Is this like $$∀x∀y(P(x,y)→∀x∀yP(y,x))$$ OR
$$(∀x∀yP(x,y))→(∀x∀yP(y,x))$$
What is the difference between them ?
 A: The second interpretation is what it should be. Without parentheses the scope of the quantifier is always just the formula right after it.
Also, as a general equivalence principle we have:
Swapping Quantifiers of Same Type
For any formula $\varphi$:
$\forall x \forall y \ \varphi \Leftrightarrow \forall y \forall x \ \varphi$
$\exists x \exists y \ \varphi \Leftrightarrow \exists y \exists x \ \varphi$
(That is: you can always swap two quantifiers of the same type that are right next to each other and retain equivalence)
So, since the LHS and RHS of your conditional are equivalent, your statement is a logical truth.
THe first interpretation (which again, is not how you should read the original conditional) says: As soon as for any $x$ and $y$ we have $P(x,y)$, then $P(x,y)$ holds for all $x$ and $y$ (indeed, it is equivalent to $\exists x \exists y \rightarrow \forall x \forall y P(x,y)$) ... which is clearly not a logical truth. USing your interpretation: just because there are some $x$ and $y$ with $x<y$ does not mean that $x<y$ for all $x$ and $y$.
