Reductions between formal languages

Let EQ = {w : number of a's = number of b's }

Let EQO = { aⁿbⁿ : n ∈  N} (so its Equal and in Order)

Typically we do the following:

Prove EQO is not regular by the pumping lemma.

Then to show EQ is not regular you say: If EQ was regular than EQ INTERSECT a*b*= EQO is regular, hence EQ is not regular (I know you can also show EQ with the Pumping Lemma but thats not important now.)

One can view this as a reduction:

A  ≤  B

If one can take B, apply a finite sequence of closure operations (e.g., intersect with a regular lang,
complement, replace all a with aba, replace all a with e (empty string), ) and get A.

If A is not regular and A≤ B then B is not regular.

Note that

EQO ≤ EQ ≤ EQ

Since EQO is not regular (by pumping ) we get EQ and \overline{EQ} are not regular.

Hence we could view the theory of showing things not-reg like the theory of NP completeness
with reductions and such. However, I have never seen a chain of more than length 2.

BUT- consider the following! Instead of using Pumping Lemma we use Comm. Comp. I have
been able to show (and this was well known) that

EQ is not regular by using Comm. Comp:

EQH = { (x,y) : |x|=|y| and number of a's in xy = number of b's in xy }

Comm Complexity of EQH is known to be log n  + \Theta(1). Important- NOT O(1).

If EQ is regular then Alice and Bob have an O(1) protocol: Alice runs x through the DFA and
transmits to Bob the state, then Bob runs y from that state to the end and transmits 1 if ended up in an accept state, 0 if not.

But I was not able to show EQO is not regular using Comm Complexity. SO imagine a bizzaro world where I taught my students the Comm Comp approach but not the Pumping Lemma. Could they prove that EQO is not regular. For one thing, could they prove

EQO ≤ EQ  ?

Or show that this CANNOT be done.

Anyone know?

One could also study the structure of the degrees induced by the equiv classes.
If this has already been done, let me know in the comments.