"Commutativity" of quantifiers I'm learning logic from Suppes' Introduction to Logic. I can show within his system of rules, that $\models (\forall x)(\forall y)(Pxy) \rightarrow (\forall y)(\forall x)(Pxy)$ and similarly for the existential quantifier. I can also understand you can "move" the existential quantifier to the right to get a "weaker version" of a premise: $\models(\exists x)(\forall y)(Pxy) \rightarrow (\forall y)(\exists x)(Pxy) $.
I'm wondering how one could go about proving "general commutativity" of a finite number of quantifiers (i.e. interchanging quantifiers is possible, with the exception you can't "move" existential quantifiers to the left). Is there an accessible source to study this from?
It seems that induction might work, but not sure how to create a bijection from $\mathbb{N}$ to the number of quantifiers.
Thanks!
 A: Disclaimer: 
I just had a quick look over the book you've mentioned. It seems that it is not dealing with logic in a pedantic mathematical way, so I'm a bit scared of confusing you even more. If so, you should ignore this answer.
Meta- vs. Object-language
As it happens often to me, I assume that the very core of your question comes from a confusion of meta-language with object-language. Think of (now I take my quick review on Suppes as a foundation -- it might have appeared in the book) of object-language as a certain instance of symbols plus all the logical symbols like $\rightarrow, \wedge, \forall$ etc. For example: 

This apple tastes good.

If we want to formalize this sentence, we need a constant $a = \texttt{this apple}$ and a relation $Gx = \texttt{x tastes good}$. Hence, our object-language is $L = (a, G)$ (I omit the logical symbols since they are part of any language). 
So, a statement within the object-language is a formula using only logical symbols and $a$ or $G$, like $(\forall x) (Gx)$, $(Ga)\wedge(\exists x)(Gx)$, etc.
And a proof within the object-language is a finite sequence of statements within the object-language where each statement of the sequence is either a so called assumption or is the conclusion from applying a rule of inference to (some of) the assumptions. Whatever: You infer only with statements written in the object-language.
On the other side, any statement which states something about languages in general (and therefore not in a certain language) is a statement in meta-language. Consider the statement you've already shown:

For every binary relation(symbol) $P$ it holds that
  \begin{align*}\models (\forall x)(\forall y)(Pxy) \rightarrow (\forall y)(\forall x)(Pxy)\end{align*}

or clearer it states:

For every binary relation(symbol) $P$ it holds that one can infer $(\forall x)(\forall y)(Pxy) \rightarrow (\forall y)(\forall x)(Pxy)$ without any assumptions.

Now, this is clearly not a statement in any object-language. It is a statement about object-languages! For example about $L=(P,a,b)$ or $L=(Q)$ as long as $P,Q$ are binary relations(symbols).
So, actually, the statement is:

For every language $L$ and every relation (symbol) $R$ within $L$, it holds that ...

[Side-kick one could ignore: Of course, one can formalize the meta-language, so it becomes object-language but "one level up". There has to be another meta-language to formalize the former meta-language -- and anyway it is still the meta-language for the former object-language. So: these aren't absolute but relative terms. But for now, consider them as absolute in the above sense.]
Back to the question
Okay, so far, now back to your question: What I would say what has brought up your question is a confusion about meta- and object-language.
In fact, the statement you want to prove within your question is a statement in meta-language as it states something about languages in general. Here: About logical truths. 
The point is that within the meta-language you know about natural numbers and all that stuff. Actually, you know about set-theory. It's like "usual mathematics", it's the language mathematicians use to proof things plus a lot of everyday language (at least in our context here, if you've read the side-kick).
Your statement in question is something like:

For every $n\in\mathbb{N}\setminus \{0\}$, every bijective $\sigma : \{1,\ldots,n\} \rightarrow \{1,\ldots,n\}$, every language $L$ and every $n$-ary relation(symbol) R within the language $L$ it holds that
  \begin{align*}\models (\forall x_1) \ldots (\forall x_n) (Rx_1\ldots x_n)\rightarrow (\forall x_{\sigma(1)})\ldots (\forall x_{\sigma(n)}) (Rx_1\ldots x_n)\end{align*}

So, feel free to use the natural numbers (and therefore induction and so on) to prove this.
