How to prove that we can switch two $\forall$? 
*

*This is true? 

*See a simple proof (High-school level)
Thanks
e.g:
$$\forall x, \forall y\;P\;\text{is true}. \iff \forall y,\forall x\;\text{P is true}$$
 A: $$\DeclareMathOperator{\and}{~And~} \DeclareMathOperator{\inn}{~In~}$$
Here is an informal proof directed at high school level.  First of all, $\forall$ serves the same purpose as $\and$.  To understand, consider the example:
$$\forall x \inn \{1 \dots 10\} ~:~ P(x)$$
Is the same truth as :
$$P(1) \and P(2) \and P(3) \dots \and P(10)$$
To understand your question, consider another example using natural numbers, which are written as $\mathbb{N}$.  Your expression with naturals is:
$$\forall x \inn \mathbb{N} ~:~ \forall y \inn \mathbb{N} ~:~ P(x,y)$$
So expand the above expression using the idea "$\forall$ means and" to get:
$$\begin{array} {ccccccccc}
\bigg(P(0,0) & \and & P(0,1) & \and & P(0,2) & \and & P(0,3) & \and & \dots\bigg)  \\
 &  &  &  & \and & & & &  \\
\bigg(P(1,0) & \and & P(1,1) & \and & P(1,2) & \and & P(1,3) & \and & \dots\bigg)  \\
 &  &  &  & \and & & & &  \\
\bigg(P(2,0) & \and & P(2,1) & \and & P(2,2) & \and & P(2,3) & \and & \dots\bigg)  \\
 &  &  &  & \vdots & & & &  \\
\end{array}$$
You can regroup the parenthesis to get:
$$
\left(\begin{array} {c} P(0,0) \\ \\ \and \\ \\ P(1,0) \\ \\ \and \\ \\ P(2,0) \\ \\ \vdots \end{array}\right)
\and
\left(\begin{array} {c} P(0,1) \\ \\ \and \\ \\ P(1,1) \\ \\ \and \\ \\ P(2,1) \\ \\ \vdots \end{array}\right)
\and
\left(\begin{array} {c} P(0,2) \\ \\ \and \\ \\ P(1,2) \\ \\ \and \\ \\ P(2,2) \\ \\ \vdots \end{array}\right)
\and
\left(\begin{array} {c} P(0,3) \\ \\ \and \\ \\ P(1,3) \\ \\ \and \\ \\ P(2,3) \\ \\ \vdots \end{array}\right)
\dots
$$
which is the same as
$$\forall y \inn \mathbb{N} ~:~ \forall x \inn \mathbb{N} ~:~ P(x,y)$$
So you are just rearranging the order of a gigantic $\and$ expression.
A: HINT
Suggested first 2 lines of a proof by contradiction:


*

*Suppose $\forall x: \forall y: P(x,y)$

*Suppose to the contrary that $\neg\forall y: \forall x: P(x,y)$, or equivalently $\exists y: \exists x: \neg P(x,y)$
